Concrete design

Manual for Design and
Detailings of
Reinforced Concrete to
Code of Practice for
Structural Use of Concrete
2004

Housing Department
May 2008
(Version 2.3)

Acknowledgement

We would like to express our greatest gratitude to Professor A.K.H. Kwan
of The University of Hong Kong who has kindly and generously provided
invaluable advice and information during the course of our drafting of
the Manual. His advice is most important for the accuracy and
completeness of contents in the Manual.

Contents
Page
1.0 Introduction 1
2.0 Some highlighted aspects in Basis of Design 3
3.0 Beams 10
4.0 Slabs 49
5.0 Columns 68
6.0 Column Beam Joints 93
7.0 Walls 102
8.0 Corbels 116
9.0 Cantilever Structures 124
10.0 Transfer Structures 132
11.0 Footings 137
12.0 Pile Caps 145
13.0 General R.C. Detailings 156
14.0 Design against Robustness 163
15.0 Shrinkage and Creep 168
16.0 Summary of Aspects having significant Impacts on Current Practices 184
References 194

Appendices

Appendix A – Clause by Clause Comparison between “Code of Practice for
Structural Use of Concrete 2004” and BS8110
Appendix B – Assessment of Building Accelerations
Appendix C – Derivation of Basic Design Formulae of R.C. Beam sections
against Flexure
Appendix D – Underlying Theory and Design Principles for Plate Bending Element
Appendix E – Moment Coefficients for three side supported Slabs
Appendix F – Derivation of Design Formulae for Rectangular Columns to Rigorous
Stress Strain Curve of Concrete
Appendix G – Derivation of Design Formulae for Walls to Rigorous Stress Strain
Curve of Concrete
Appendix H – Estimation of support stiffnesses of vertical support to transfer
structures
Appendix I – Derivation of Formulae for Rigid Cap Analysis
Appendix J – Mathematical Simulation of Curves related to Shrinkage and Creep
Determination

Version 2.3 May 2008

1.0 Introduction

1.1 Promulgation of the Revised Code

A revised concrete code titled “Code of Practice for Structural Use of Concrete
2004” was formally promulgated by the Buildings Department of Hong Kong
in late 2004 which serves to supersede the former concrete code titled “The
Structural Use of Concrete 1987”. The revised Code, referred to as “the Code”
hereafter in this Manual will become mandatory by 15 December 2006, after
expiry of the grace period in which both the revised and old codes can be used.

1.2 Main features of the Code

As in contrast with the former code which is based on “working stress” design
concept, the drafting of the Code is largely based on the British Standard
BS8110 1997 adopting the limit state design approach. Nevertheless, the
following features of the Code in relation to design as different from BS8110
are outlined :

(a) Provisions of concrete strength up to grade 100 are included;
(b) Stress strain relationship of concrete is different from that of BS8110
for various concrete grades as per previous tests on local concrete;
(c) Maximum design shear stresses of concrete ( v max ) are raised;
(d) Provisions of r.c. detailings to enhance ductility are added, together
with the requirements of design in beam-column joints (Sections 9.9
and 6.8 respectively);
(e) Criteria for dynamic analysis for tall building under wind loads are
added (Clause 7.3.2).

As most of our colleagues are familiar with BS8110, a comparison table
highlighting differences between BS8110 and the Code is enclosed in
Appendix A which may be helpful to designers switching from BS8110 to the
Code in the design practice.

1.3 Outline of this Manual

This Practical Design Manual intends to outline practice of detailed design and
detailings of reinforced concrete work to the Code. Detailings of individual

1


types of members are included in the respective sections for the types, though
Section 13 in the Manual includes certain aspects in detailings which are
common to all types of members. Design examples, charts are included, with
derivations of approaches and formulae as necessary. Aspects on analysis are
only discussed selectively in this Manual. In addition, as the Department has
decided to adopt Section 9.9 of the Code which is in relation to provisions for
“ductility” for columns and beams contributing in the lateral load resisting
system in accordance with Cl. 9.1 of the Code, conflicts of this section with
others in the Code are resolved with the more stringent ones highlighted as
requirements in our structural design.

As computer methods have been extensively used nowadays in analysis and
design, the contents as related to the current popular analysis and design
approaches by computer methods are also discussed. The background theory
of the plate bending structure involving twisting moments, shear stresses, and
design approach by the Wood Armer Equations which are extensively used by
computer methods are also included in the Appendices in this Manual for
design of slabs, flexible pile caps and footings.

To make distinctions between the equations quoted from the Code and the
equations derived in this Manual, the former will be prefixed by (Ceqn) and
the latter by (Eqn).

Unless otherwise stated, the general provisions and dimensioning of steel bars

are based on high yield bars with f y = 460 N/mm2.

1.4 Revision as contained in Amendment No. 1 comprising major revisions
including (i) exclusion of members not contributing to lateral load resisting
system from ductility requirements in Cl. 9.9; (ii) rectification of ε0 in the
concrete stress strain curves; (iii) raising the threshold concrete grade for
limiting neutral axis depths to 0.5d from grade 40 to grade 45 for flexural
members; (iv) reducing the x values of the simplified stress block for
concrete above grade 45 are incorporated in this Manual.

2


2.0 Some highlighted aspects in Basis of Design

2.1 Ultimate and Serviceability Limit states

The ultimate and serviceability limit states used in the Code carry the usual
meaning as in BS8110. However, the new Code has incorporated an extra
serviceability requirement in checking human comfort by limiting acceleration
due to wind load on high-rise buildings (in Clause 7.3.2). No method of
analysis has been recommended in the Code though such accelerations can be
estimated by the wind tunnel laboratory if wind tunnel tests are conducted.
Nevertheless, worked examples are enclosed in Appendix B, based on
approximation of the motion of the building as a simple harmonic motion and
empirical approach in accordance with the Australian Wind Code AS/NZS
1170.2:2002 on which the Hong Kong Wind Code has based in deriving
dynamic effects of wind loads. The relevant part of the Australian Code is
Appendix G of the Australian Code.

2.2 Design Loads

The Code has made reference to the “Code of Practice for Dead and Imposed
Loads for Buildings” for determination of characteristic gravity loads for
design. However, this Load Code has not yet been formally promulgated and
the Amendment No. 1 has deleted such reference. At the meantime, the design
loads should be therefore taken from HKB(C)R Clause 17. Nevertheless, the
designer may need to check for the updated loads by fire engine for design of
new buildings, as required by FSD.

The Code has placed emphasize on design loads for robustness which are
similar to the requirements in BS8110 Part 2. The requirements include design
of the structure against a notional horizontal load equal to 1.5% of the
characteristic dead weight at each floor level and vehicular impact loads
(Clause 2.3.1.4). The small notional horizontal load can generally be covered
by wind loads required for design. Identification of key elements and design
for ultimate loads of 34 kPa, together with examination of disproportionate
collapse in accordance with Cl. 2.2.2.3 can be exempted if the buildings are
provided with ties determined by Cl. 6.4.1. The usual reinforcement provisions
as required by the Code for other purposes can generally cover the required
ties provisions.

3


Wind loads for design should be taken from Code of Practice on Wind Effects
in Hong Kong 2004.

It should also be noted that there are differences between Table 2.1 of the
Code that of BS8110 Part 1 in some of the partial load factors γf. The
beneficial partial load factor for earth and water load is 1. However, lower
values should be used if the earth and water loads are known to be
over-estimated.

2.3 Materials – Concrete

Table 3.2 has tabulated a set of Young’s Moduli of concrete up to grade 100.
The values are generally smaller than that in BS8110 by more than 10% and
also slightly different from the former 1987 Code. The stress strain curve of
concrete as given in Figure 3.8 of the Code, whose initial tangent is
determined by these Young’s Moduli values is therefore different from Figure
2.1 of BS8110 Part 1. Furthermore, in order to achieve smooth (tangential)
connection between the parabolic portion and straight portion of the stress
strain curve, the Code, by its Amendment No. 1, has shifted the ε 0 value to

1.34( f cu / γ m ) f cu
instead of staying at 2.4 × 10 − 4 which is the value in
Ec γm

BS8110. The stress strain curves for grade 35 by the Code and BS8110 are
plotted as an illustration in Figure 2.1.

Comparison of stress strain profile between the Code and
BS8110 for Grade 35
The Code BS8110
18
16
14
12
Stress (MPa)

10
8
6
4
2
0
0 0.2 0.4 0.6 0.8 1
Distance ratio from neutral axis
Figure 2.1 - Stress Strain Curves of Grade 35 by the Code and
BS8110

4


From Figure 2.1 it can be seen that stress strain curve by BS8110 envelops that
of the Code, indicating that design based on the Code will be slightly less
economical. Design formulae for beams and columns based on these stress
strain curves by BS8110, strictly speaking, become inapplicable. A full
derivation of design formulae and charts for beams, columns and walls are
given in Sections 3, 5 and 7, together with Appendices C, F and G of this
Manual.

Table 4.2 of the Code tabulated nominal covers to reinforcements under
different exposure conditions. However, reference should also be made to the
“Code of Practice for Fire Resisting Construction 1996”.

To cater for the “rigorous concrete stress strain relation” as indicated in Figure
2.1 for design purpose, a “simplified stress approach” by assuming a
rectangular stress block of length 0.9 times the neutral axis depth has been
widely adopted, as similar to BS8110. However, the Amendment No. 1 of the
Code has restricted the 0.9 factor to concrete grades not exceeding 45. For 45
< fcu ≤ 70 and 70 < fcu, the factors are further reduced to 0.8 and 0.72
respectively as shown in Figure 2.2

0.0035 for fcu ≤ 60
0.0035 – 0.0006(fcu – 60)1/2 for fcu > 60 0.67fcu/γm

0.9x for fcu ≤ 45;
0.8x for 45 < fcu ≤ 70;
0.72x for 70 < fcu

strain stress

Figure 2.2 – Simplified stress block for ultimate reinforced concrete design

2.4 Ductility Requirements (for beams and columns contributing to lateral load
resisting system)

As discussed in para. 1.3, an important feature of the Code is the incorporation
of ductility requirements which directly affects r.c. detailings. By ductility we
refer to the ability of a structure to undergo “plastic deformation”, which is

5


comparatively larger than the “elastic” one prior to failure. Such ability is
desirable in structures as it gives adequate warning to the user for repair or
escape before failure. The underlying principles in r.c. detailings for ductility
requirements are highlighted as follows :

(i) Use of closer and stronger transverse reinforcements to achieve better
concrete confinement which enhances both ductility and strength of
concrete against compression, both in columns and beams;

axial compression
confinement by transverse
re-bars enhances concrete
strength and ductility of the
concrete core within the
transverse re-bars

Figure 2.3 – enhancement of ductility by transverse reinforcements

(ii) Stronger anchorage of transverse reinforcements in concrete by means
of hooks with bent angles ≥ 135o for ensuring better performance of
the transverse reinforcements;

(a) 180o hook (b) 135o hook (c) 90o hook

Anchorage of link in concrete : (a) better than (b); (b) better than (c)

Figure 2.4 – Anchorage of links in concrete by hooks

(In fact Cl. 9.9.1.2(b) of the Code has stated that links must be
adequately anchored by means of 135o or 180o hooks and anchorage by
means of 90o hooks is not permitted for beams. Cl. 9.5.2.2, Cl. 9.5.2.3
and 9.9.2.2(c) states that links for columns should have bent angle at

6


least 135o in anchorage. Nevertheless, for walls, links used to restrain
vertical bars in compression should have an included angle of not more
than 90o by Cl. 9.6.4 which is identical to BS8110 and not a ductility
requirement;
(iii) More stringent requirements in restraining and containing longitudinal
reinforcing bars in compression against buckling by closer and
stronger transverse reinforcements with hooks of bent angles ≥ 135o;
(iv) Longer bond and anchorage length of reinforcing bars in concrete to
ensure failure by yielding prior to bond slippage as the latter failure is
brittle;

Ensure failure by yielding here
instead of bond failure behind

bar in tension
Longer and stronger
anchorage

Figure 2.5 – Longer bond and anchorage length of reinforcing bars

(v) Restraining and/or avoiding radial forces by reinforcing bars on
concrete at where the bars change direction and concrete cover is thin;

Radial force by bar
tending to cause concrete
spalling if concrete is
relatively thin

Radial force by bar
inward on concrete
which is relatively thick

Figure 2.6 – Bars bending inwards to avoid radial forces on thin concrete cover

(vi) Limiting amounts of tension reinforcements in flexural members as
over-provisions of tension reinforcements will lead to increase of

7


neutral axis and thus greater concrete strain and easier concrete failure
which is brittle;

εc εc

x x

Lesser amount of tensile Greater amount of tensile
steel, smaller x, smaller εc steel, greater x, greater εc

Figure 2.7 – Overprovision of tensile steel may lower ductility

(vii) More stringent requirements on design using high strength concrete
such as (a) lowering ultimate concrete strain; (b) restricting percentage
of moment re-distribution; and (c) restricting neutral axis depth ratios
to below 0.5 as higher grade concrete is more brittle.

Often the ductility requirements specified in the Code are applied to locations
where plastic hinges may be formed. The locations can be accurately
determined by a “push over analysis” by which a lateral load with step by step
increments is added to the structure. Among the structural members met at a
joint, the location at which plastic hinge is first formed will be identified as the
critical section of plastic hinge formation. Nevertheless, the determination can
be approximated by judgment without going through such an analysis. In a
column beam frame with relatively strong columns and weak beams, the
critical sections of plastic hinge formation should be in the beams at their
interfaces with the columns. In case of a column connected into a thick pile
cap, footing or transfer plate, the critical section with plastic hinge formation
will be in the columns at their interfaces with the cap, footing or transfer plate
as illustrated in Figure 2.8.

8


Critical section with
plastic hinge formation

Pile cap / footing /
transfer structure

Strong column / weak beam

Figure 2.8 – locations of critical section with plastic hinge formation

2.5 Design for robustness

The requirements for design for robustness are identical to BS8110 and more
detailed discussions are given in Section 14.

2.6 Definitions of structural elements

The Code has included definitions of slab, beam, column and wall in
accordance with their dimensions in Clause 5.2.1.1, 5.4 and 5.5 which are
repeated as follows for ease of reference :

(a) Slab : the minimum panel dimension ≥ 5 times its thickness;
(b) Beam : for span ≥ 2 times the overall depth for simply supported span
and ≥ 2.5 times the overall depth for continuous span, classified as
shallow beam, otherwise : deep beam;
(c) Column : vertical member with section depth not exceeding 4 times its
width;
(d) Wall : vertical member with plan dimensions other than that of column.
(e) Shear Wall : wall contributing to the lateral stability of the structure.
(f) Transfer Structure : horizontal element which redistributes vertical loads
where there is a discontinuity between the vertical structural elements
above and below.

This Manual is based on the above definitions in delineating structural
members for discussion.

9


3.0 Beams

3.1 Analysis (Cl. 5.2.5.1 & 5.2.5.2)

Normally continuous beams are analyzed as sub-frames by assuming no
settlements at supports by walls, columns (or beams) and rotational stiffness
by supports provided by walls or columns as 4 EI / L (far end of column /
wall fixed) or 3EI / L (far end of column / wall pinned).

Figure 3.1 – continuous beam analyzed as sub-frame

In analysis as sub-frame, Cl. 5.2.3.2 of the Code states that the following
loading arrangements will be adequate for seeking for the design moments :

1.4GK+1.6QK 1.4GK+1.6QK 1.4GK+1.6QK 1.4GK+1.6QK 1.4GK+1.6QK 1.4GK+1.6QK

Figure 3.2a – To search for maximum support reactions

1.4GK+1.6QK 1.0GK 1.4GK+1.6QK 1.0GK 1.4GK+1.6QK 1.0GK

Figure 3.2b – To search for maximum sagging moment in spans with
1.4GK+1.6QK

1.0GK 1.0GK 1.4GK+1.6QK 1.4GK+1.6QK 1.0GK 1.0GK

Figure 3.2c – To search for maximum hogging moment at support
adjacent to spans with 1.4GK+1.6QK

10


However, most of the commercial softwares can actually analyze individual
load cases, each of which is having live load on a single span and the effects
on itself and others are analyzed. The design value of shears and moments at
any location will be the summation of the values of the same sign created by
the individual cases. Thus the most critical loads are arrived at easily.

With wind loads, the load cases to be considered will be 1.2(GK+QK+WK) and
1.0GK+1.4WK on all spans.

3.2 Moment Redistribution (Cl. 5.2.9 of the Code)

Moment redistribution is allowed for concrete grade not exceeding 70 under
conditions 1, 2 and 3 as stated in Cl. 5.2.9.1 of the Code. Nevertheless, it
should be noted that there would be further limitation of the neutral axis depth
ratio x / d if moment redistribution is employed as required by (Ceqn 6.4)
and (Ceqn 6.5) of the Code which is identical to the provisions in BS8110. The
rationale is discussed in Concrete Code Handbook 6.1.2.

3.3 Highlighted aspects in Determination of Design Parameters of Shallow Beam

(i) Effective span (Cl. 5.2.1.2(b) and Figure 5.3 of the Code)

For simply supported beam, continuous beam and cantilever, the
effective span can be taken as the clear span plus the lesser of half of the
structural depth and half support width except that on bearing where the
centre of bearing should be used to assess effective span;

(ii) Effective flange width of T- and L-beams (Cl. 5.2.1.2(a))

Effective flange width of T- and L-beams are as illustrated in Figure 5.2.
of the Code as reproduced as Figure 3.3 of this Manual:
beff
beff,1 beff,2

beff,1=0.2×b1+0.1lpi
beff,2=0.2×b2+0.1lpi
b1 b1 bw b2 b2 beff, =bw+beff,1+beff,2

Figure 3.3 – Effective flange Parameters

11


Effective width (beff) = width of beam (bw) + ∑(0.2 times of half the
centre to centre width to the next beam (0.2bi) + 0.1 times the span of
zero moment (0.1lpi), with the sum of the latter not exceeding 0.2 times
the span of zero moment and lpi taken as 0.7 times the effective span of
the beam). An example for illustration as indicated in Figure 3.4 is as
indicated :

Worked Example 3.1

400 2000 400 2000 400 2000 400

Figure 3.4 – Example illustrating effective flange determination

The effective spans are 5 m and they are continuous beams.
The effective width of the T-beam is, by (Ceqn 5.1) of the Code :
l pi = 0.7 × 5000 = 3500 ;
beff ,1 = beff , 2 = 0.2 × 1000 + 0.1× 3500 = 550
As beff ,1 = beff , 2 = 550 < 0.2 × 3500 = 700 , ∴ beff ,1 = beff , 2 = 550 ;
beff = 400 + 550 × 2 = 400 + 1100 = 1500
So the effective width of the T-beam is 1500 mm.

Similarly, the effective width of the L-beam at the end is

bw + beff ,1 = 400 + 550 = 950 .

(iii) Support Moment Reduction (Cl. 5.2.1.2 of the Code)

The Code allows design moment of beam (and slab) monolithic with its
support providing rotational restraint to be that at support face if the
support is rectangular and 0.2Ø if the support is circular with diameter Ø.
But the design moment after reduction should not be less than 65% of
the support moment. A worked example 3.2 as indicated by Figure 3.5
for illustration is given below :

Worked Example 3.2

12


250 kNm at 0.2 Ø into
the support face
350 kNm at
support
200 kNm at
support face

0.2×800
centre line of beam
column elements
idealized as line 800 Bending Moment
elements in analysis Diagram

Figure 3.5 – Reduced moment to Support Face for
support providing rotational restraint

In Figure 3.5, the bending moment at support face is 200 kNm which can
be the design moment of the beam if the support face is rectangular.
However, as it is smaller than 0.65×350 = 227.5 kNm. 227.5 kNm
should be used for design.

If the support is circular and the moment at 0.2Ø into the support and the
bending moment at the section is 250 kNm, then 250 kNm will be the
design moment as it is greater than 0.65×350 = 227.5 kNm.

For beam (or slab) spanning continuously over a support considered not
providing rotational restraint (e.g. wall support), the Code allows
moment reduction by support shear times one eighth of the support width
to the moment obtained by analysis. Figure 3.6 indicates a numerical
Worked Example 3.3.

Worked Example 3.3

By Figure 3.6, the design support moment at the support under
0.8
consideration can be reduced to 250 − 200 × = 230 kNm.
8

13


FEd,sup = 200 kN
250 kNm 230 kNm

800

Figure 3.6 – Reduction of support moment by support shear for support
considered not providing rotational restraint

(iv) Slenderness Limit (Cl. 6.1.2.1 of the Code)

The provision is identical to BS8110 as

1. Simply supported or continuous beam :
Clear distance between restraints ≤ 60bc or 250bc2/d if less; and
2. Cantilever with lateral restraint only at support :
Clear distance from cantilever to support ≤ 25bc or 100bc2/d if less
where bc is the breadth of the compression face of the beam and d is
the effective depth.

Usually the slenderness limits need be checked for inverted beams or
bare beam (without slab).

(v) Span effective depth ratio (Cl. 7.3.4.2 of the Code)

Table 7.3 under Cl. 7.3.4.2 tabulates basic span depth ratios for various
types of beam / slab which are deemed-to-satisfy requirements against
deflection. The table has provisions for “slabs” and “end spans” which
are not specified in BS8110 Table 3.9. Nevertheless, calculation can be
carried out to justify deflection limits not to exceed span / 250. In
addition, the basic span depth ratios can be modified due to provision of
tensile and compressive steels as given in Tables 7.4 and 7.5 of the Code
which are identical to BS8110. Modification of the factor by 10/span for

14


span > 10 m except for cantilever as similar to BS8110 is also included.

Flanged Beam One or two-way
Support condition Rectangular Beam bw/b < 0.3 spanning solid
slab
Cantilever 7 5.5 7

Simply supported 20 16 20

Continuous 26 21 26

End span 23 18.5 23(2)
Note :
1. The values given have been chosen to be generally conservative and calculation may
frequently show shallower sections are possible;
2. The value of 23 is appropriate for two-way spanning slab if it is continuous over one long side;
3. For two-way spanning slabs the check should be carried out on the basis of the shorter span.

Table 3.1 – effective span / depth ratio

(vi) Maximum spacing between bars in tension near surface, by Cl. 9.2.1.4 of
the Code, should be such that the clear spacing between bar is limited by
70000β b
clear spacing ≤ ≤ 300 mm where β b is the ratio of moment
fy
47000
redistribution. Or alternatively, clear spacing ≤ ≤ 300 mm. So the
fs
70000β b 70000 × 1
simplest rule is = = 152 mm when using high yield
fy 460
bars and under no moment redistribution.

(vii) Concrete covers to reinforcements (Cl. 4.2.4 and Cl. 4.3 of the Code)

Cl. 4.2.4 of the Code indicates the nominal cover required in accordance
with Exposure conditions. However, we can, as far as our building
structures are concerned, roughly adopt condition 1 (Mild) for the
structures in the interior of our buildings (except for bathrooms and
kitchens which should be condition 2), and to adopt condition 2 for the
external structures. Nevertheless, the “Code of Practice for Fire Resisting
Construction 1996” should also be checked for different fire resistance
periods (FRP). So, taking into account our current practice of using
concrete not inferior than grade 30 and maximum aggregate sizes not
exceeding 20 mm, we may generally adopt the provision in our DSEG
Manual (DSEDG-104 Table 1) with updating by the Code except for
compartment of 4 hours FRP. The recommended covers are summarized
in the following table :

15


Description Nominal Cover (mm)
Internal 30 (to all rebars)
External 40 (to all rebars)
Simply supported (4 hours FRP) 80 (to main rebars)
Continuous (4 hours FRP) 60 (to main rebars)
Table 3.2 – Nominal Cover of Beams

3.4 Sectional Design for Rectangular Beam against Bending

3.4.1 Design in accordance with the Rigorous Stress Strain curve of Concrete

The stress strain block of concrete as indicated in Figure 3.8 of the Code is
different from Figure 2.1 of BS8110. Furthermore, in order to achieve smooth
connection between the parabolic and the straight line portions, the Concrete
Code Handbook has recommended to shift the ε0 to the right to a value of
1.34 f cu
, which has been adopted in Amendment No. 1. With the values of
γ m Ec
Young’s Moduli of concrete, E c , as indicated in Table 3.2 of the Code, the
stress strain block of concrete for various grades can be determined. The stress
strain curve of grade 35 is drawn as shown in Figure 3.7.

Stress Strain Profile for Grade 35

18
16
14
12
Stress (MPa)

10
8
6
4 0.3769 where
ε0 = 0.001319
2
0
0 0.2 0.4 0.6 0.8 1
Distance Ratio from Neutral axis

Figure 3.7 – Stress strain block of grades 35

Based on this rigorous concrete stress strain block, design formulae for beam

16


can be worked out as per the strain distribution profile of concrete and steel as
indicated in Figure 3.8.

ε ult = 0.0035
d’

x

d neutral axis

Stress Diagram Strain Diagram

Figure 3.8 – Stress Strain diagram for Beam

x
The solution for the neutral axis depth ratio for singly reinforced beam is
d
the positive root of the following quadratic equation where ε ult = 0.0035 for
concrete grades not exceeding 60 (Re Appendix C for detailed derivation) :

0.67 f cu  1 1 ε 0  x  2 0.67 f
2
1 ε   1 ε0 x M
− + 
−  0 
   + cu
1 −
 3ε  − 2 =0
 d bd
γ m  2 3 ε ult 12  ε ult   d  γm  
  ult

(Eqn 3-1)
With neutral axis depth ratio determined, the steel ratio can be determined by

Ast 1 0.67 f cu  1 ε0 x
= 1 −
 3ε 
d (Eqn 3-2)
bd 0.87 f y γ m  ult 
x
As is limited to 0.5 for singly reinforcing sections for grades up to 45
d
under moment redistribution not greater than 10% (Clause 6.1.2.4 of the Code),
M
by (Eqn 3-1), will be limited to K ' values as in
bd 2 f cu
K ' = 0.154 for grade 30; K ' = 0.152 for grade 35;
K ' = 0.151 for grade 40; K ' = 0.150 for grade 45
which are all smaller than 0.156 under the simplified stress block.

However, for grades exceeding 45 and below 70 where neutral axis depth ratio
is limited to 0.4 for singly reinforced sections under moment redistribution not

17


M
greater than 10% (Clause 6.1.2.4 of the Code), again by (Eqn 3-1)
bd 2 f cu
will be limited to
K ' = 0.125 for grade 50; K ' = 0.123 for grade 60;
K ' = 0.121 for grade 70.
which are instead above 0.120 under the simplified stress block as Amendment
No. 1 has reduce the x / d factor to 0.8. Re discussion is in Appendix C.

It should be noted that the x / d ratio will be further limited if moment
redistribution exceeds 10% by (Ceqn 6.4) and (Ceqn 6.5) of the Code (with
revision by Amendment No. 1) as
≤ (β b − 0.4) for f cu ≤ 45 ; and
x
d
≤ (β b − 0.5) for 45 < f cu ≤ 70
x
d
where β b us the ratio of the moment after and before moment redistribution.

M
When exceeds the limited value for single reinforcement,
bd 2 f cu
compression reinforcements at d ' from the surface of the compression side
should be added. The compression reinforcements will take up the difference

between the applied moment and K ' bd 2 f cu and the compression

reinforcement ratio is
 M 
 2 
Asc   bd f − K '  f cu

=
cu
(Eqn 3-3)
bd  d' 
0.87 f y 1 − 
 d
And the same amount of reinforcement will be added to the tensile
reinforcement :
 M 
 2 
 bd f − K '  f cu
Ast 1 0.67 f cu  1 ε 0   
= 1 − 3 ε η +
 cu
 (Eqn 3-4)
bd 0.87 f y γ m  ult   d'
0.87 f y 1 − 
 d
where η is the limit of neutral axis depth ratio which is 0.5 for f cu ≤ 45 , 0.4
for 45 < f cu ≤ 70 and 0.33 for 70 < f cu ≤ 100 where moment redistribution
does not exceed 10%.

It follows that more compressive reinforcements will be required for grade 50
than 45 due to the limitation of neutral axis depth ratio, as illustrated by the
following Chart 3-1 in which compression reinforcement decreases from grade

18


M
30 to 40 for the same , but increases at grade 45 due to the change of the
bd 2
limit of neutral axis depth ratio from 0.5 to 0.4 with moment redistribution not
exceeding 10%. The same phenomenon applies to tensile steel also. With
moment redistribution exceeding 10%, the same trend will also take place.

Reinforcement Ratios for Doubly Reinforced Beams d'/d = 0.1
Grade 30 Ast/bd Grade 30 Asc/bd Grade 35 Ast/bd Grade 35 Asc/bd Grade 40 Ast/bd
Grade 40 Asc/bd Grade 45 Ast/bd Grade 45 Asc/bd Grade 50 Ast/bd Grade 50 Asc/bd

14

12

10

8
2
M/bd

6

4

2

0
0 0.5 1 1.5 2 2.5 3 3.5 4
Reinforcement ratios A/bd (%)

Chart 3-1 – Reinforcement Ratios of Doubly Reinforced Beams for Grade 30
to 50 with Moment Redistribution limited to 10% or below
z
As similar to BS8110, there is an upper limit of “lever arm ratio” which is
d
the depth of the centroid of the compressive force of concrete to the effective
depth of the beam section of not exceeding 0.95. Thus for calculated values of
z x
≥ 0.95 or ≤ 0.111 in accordance with the simplified stress block
d d
A M
approach, st =
bd 0.87 f y (0.95d )bd

Design Charts for grades 30 to 50 comprising tensile steel and compression
Ast A
steel ratios and sc are enclosed at the end of Appendix C.
bd bd

3.4.2 Design in accordance with the Simplified Stress Block

The design will be simpler and sometimes more economical if the simplified

19


rectangular stress block as given by Figure 6.1 of the Code is adopted. The
design formula becomes :

M
For singly reinforced sections where K = ≤ K ' where K ' = 0.156
f cu bd 2
for grades 45 and below and K ' = 0.120 for 45 < f cu ≤ 70; K ' = 0.094 for
70 < f cu ≤ 100.
z K
= 0.5 + 0.25 − ≤ 0.95 ;
d 0 .9
x  z 1  K  1 M
= 1 −  =  0.5 − 0.25 −

 ; Ast = (Eqn 3-5)
d  d  0.45  0.9  0.45
 0.87 f y z

M
For doubly reinforced sections K = > K',
f cu bd 2
z K' x  z 1
= 0.5 + 0.25 − = 1 − 
d 0 .9 d  d  0.45

(K − K ') f cu bd 2 K ' f cu bd 2
Asc = Ast = + Asc (Eqn 3-6)
0.87 f y (d − d ') 0.87 f y z

3.4.3 Ductility Requirement on amounts of compression reinforcement

In accordance with Cl. 9.9.1.1(a) of the Code, at any section of a beam
(participating in lateral load resisting system) within a “critical zone” the
compression reinforcement should not be less than one-half of the tension
reinforcement at the same section. A “critical zone” is understood to be a zone
where a plastic hinge is likely to be formed and thus generally include sections
near supports or at mid-span. The adoption of the clause will likely result in
providing more compression reinforcements in beams (critical zones).

3.4.4 Worked Examples for Determination of steel reinforcements in Rectangular
Beam with Moment Redistribution < 10%

Unless otherwise demonstrated in the following worked examples, the
requirement in Cl. 9.9.1.1(a) of the Code as discussed in para. 3.4.3 by
requiring compression reinforcements be at least one half of the tension
reinforcement is not included in the calculation of required reinforcements.

20


Worked Example 3.4

Section : 500 (h) × 400 (w), f cu = 35 MPa
cover = 40 mm (to main reinforcement)

(i) M 1 = 286 kNm;
d = 500 − 40 − 16 = 444
1.34 f cu 1.34 × 35 ε0
ε0 = = = 0.0013192 = 0.3769
γ m Ec 1.5 × 23700 ε ult
M1 286 × 10 6
= = 0.104 < 0.152 , so singly reinforced
f cu bd 2 35 × 400 × 444 2
x
Solving the neutral axis depth ratio by (Eqn 3-1)
d
0.67 f cu  1 1 ε 1 ε 
2

− + 0
−  0   = −60.38 ;
γ m  2 3 ε ult 12  ε ult  
  

0.67 f cu  1 ε 0  M 286 × 10 6
1 −  = 13.669 ; − 2 = = −3.627
γ m  3 ε ult 
  bd 400 × 444 2
x − 13.699 + 13.699 − 4 × (− 60.38) × (− 3.627 )
2

= = 0.307 ≤ 0.5
d 2 × (− 60.38)
Ast 1 0.67 f cu  1 ε0  x 1
=  3 ε  d = 0.87 × 460 ×13.699 × 0.307 = 0.0105
1 − 
bd 0.87 f y γ m  ult 

⇒ Ast = 1865 mm2 Use 2T32 + 1T25

(ii) M 2 = 486 kNm;
d = 500 − 40 − 20 = 440
1.34 f cu 1.34 × 35 ε0
ε0 = = = 0.0013192 = 0.3769
γ m Ec 1.5 × 23700 ε ult
M2 486 ×10 6
= = 0.179 > 0.152 , so doubly reinforced
f cu bd 2 35 × 400 × 440 2
d ' 50
d ' = 40 + 10 = 50 = = 0.114 (assume T20 bars)
d 440
 M 
 2 
− K  f cu
Asc  bd f cu
  (0.179 − 0.152)× 35 = 0.267 %
By (Eqn 3-3) = =
bd  d' 0.87 × 460 × (1 − 0.114)
0.87 f y 1 − 
 d
Asc = 0.00267 × 400 × 440 = 469 mm2 Use 2T20

21


 M 
 2 − K  f cu
 1 ε0   bd f 
1 0.67 f cu
η +  
Ast
By (Eqn 3-4) = 1 −
 3ε 
cu

bd 0.87 f y γ m  ult   d'
0.87 f y 1 − 
 d
Ast 1
= 13.699 × 0.5 + 0.00267 = 1.978 %
bd 0.87 × 460
Ast = 0.01978 × 400 × 440 = 3481 mm2 Use 3T40

Worked Example 3.5

(i) and (ii) of Worked Example 3.4 are re-done in accordance with Figure 6.1
of the Code (the simplified stress) block by (Eqn 3-5) and (Eqn 3-6)

z K 286 ×10 6
(i) = 0.5 + 0.25 − = 0.5 + 0.25 − = 0.867
d 0.9 35 × 400 × 444 2 × 0.9
Ast M 286 × 10 6
= = = 0.01045
bd bd 2 × 0.87 f y ( z / d ) 400 × 444 2 × 0.87 × 460 × 0.867
⇒ Ast = 1856 mm2 Use 2T32 + 1T25

M 486 × 10 6
(ii) K = = = 0.179 > 0.156 , so doubly reinforcing
f cu bd 2 35 × 400 × 440 2
z
section required, = 1 − 0.5 × 0.9 × 0.5 = 0.775
d
(K − K ') f cu bd 2 (0.179 − 0.156)× 35 × 400 × 440 2
Asc = = = 399 mm2 >
0.87 f y (d − d ') 0.87 × 460 × (440 − 50 )
0.2% in accordance with Table 9.1 of the Code, Use 2T16
K ' f cu bd 2 0.156 × 35 × 400 × 440 2
Ast = + Asc = + 399 = 3498 mm2
0.87 f y z 0.87 × 460 × 0.775 × 440
Use 3T40

(Note : If the beam is contributing in lateral load resisting system and the
section is within “critical zone”, compressive reinforcements has to be at
least half of that of tension reinforcements Asc = 3498 / 2 = 1749 mm2 by
Cl. 9.9.1.1(a) in the Code (D). So use 2T25 + 1T32.)

Results of comparison of results from Worked Examples 3.4 and 3.5 (with the
omission of the requirement in Cl. 9.9.1.1(a) that compressive reinforcements
be at least half of that of tension reinforcements) are summarized in Table 3.3,
indicating differences between the “Rigorous Stress” and “Simplified Stress”
Approach :

22


Singly Doubly
Reinforced Reinforced
Ast (mm2) Asc (mm2) Ast (mm2) Total
(mm2)
Based on Rigorous 1865 469 3481 3950
Stress Approach
Based on Simplified 1856 399 3498 3897
stress Approach
Table 3.3 – Summary of Results for comparison of Rigorous stress and
simplified stress Approaches.

Results by the two approaches are very close. The approach based on the
simplified stress block are slightly more economical.

3.4.5 Worked Example 3.6 for Rectangular Beam with Moment Redistribution >
10%

If the Worked Example 3.4 (ii) has undergone a moment redistribution of 20%
> 10%, i.e. β b = 0.8 , by (Ceqn 6.4) of the Code, the neutral axis depth is

≤ (β b − 0.4) ⇒ ≤ 0.8 − 0.4 = 0.4 ,
x x
limited to
d d
z
and the lever arm ratio becomes = 1 − 0.4 × 0.9 × 0.5 = 0.82 .
d

M K
So the K = 2
value become 0.5 + 0.25 − = 0.82 ⇒ K = 0.132
bd f cu 0 .9

(K − K ') f cu bd 2 (0.176 − 0.132)× 35 × 400 × 440 2
Asc = = = 764 mm2 > 0.2 %
0.87 f y (d − d ') 0.87 × 460 × (440 − 50)

as required by Table 9.1 of the Code;

K ' f cu bd 2 0.132 × 35 × 400 × 440 2
Ast = + Asc = + 764 = 3242 mm2
0.87 f y z 0.87 × 460 × 0.82 × 440

So total amount of reinforcement is greater.

3.5 Sectional Design of Flanged Beam against Bending

3.5.1 Slab structure adjacent to the beam, if in flexural compression, can be used to
act as part of compression zone of the beam, thus effectively widen the
structural width of the beam. The use of flanged beam will be particularly
useful in eliminating the use of compressive reinforcements, as in addition to

23


reducing tensile steel due to increase of lever arm. The principle of sectional
design of flanged beam follows that rectangular beam with an additional

flange width of beff − bw as illustrated in Figure 3.9.

0.67 f cu
beff γm

hf
0 .9 x x
d

bw

Figure 3.9 – Analysis of a T or L beam section

Design formulae based on the simplified stress block are derived in Appendix
C which are summarized as follows :

(i) Singly reinforcing section where η × neutral axis depth is inside
flange depth by checking where η = 0.9 for f cu ≤ 45 ; η = 0.8 for
45 < f cu ≤ 70 ; η = 0.72 for 70 < f cu ≤ 100 .
x K hf M
η = 1− 1− ≤ where K = (Eqn 3-7)
d 0.225 d f cu beff d 2
If so, carry out design as if it is a rectangular beam of width beff .
(ii) Singly reinforcing section where η × neutral axis depth is outside
x hf
flange depth, i.e. η > and
d d

M 0.67 f cu  beff  hf  1 hf  0.67 f cu  x  η x 
= 
 b − 1 d  1 − 2 d
  +
 η 1 − 
bw d 2
γm  w    γ m  d  2 d 

x
be solved by the quadratic equation :
d
0.67 f cu η 2  x  0.67 f cu x M − M f
2

  − η + =0 (Eqn 3-8)
γm 2 d  γm d bw d 2

Mf 0.67 f cu h f  beff  1 h f 
where =  − 11 −  (Eqn 3-9)
bw d 2 γm d  bw


 2 d 


24


Ast 0.67 f cu  beff  hf x
And = 
 − 1 + η 
d (Eqn 3-10)
bw d γ m 0.87 f y  bw  d

(iii) Doubly reinforcing section :
x
By following the procedure in (ii), if obtained by (Eqn 3-8)
d
exceeds ϕ where ϕ = 0.5 for f cu > 45 ; 0.4 for f cu > 70 and 0.33
for f cu > 100 , then double reinforcements will be required with
required Asc and Ast as
Asc 1  M 0.67 f cu  beff  hf  1 hf   1  
= 
bw d 0.87 f y (1 − d ' / d )  bw d 2
−
γm

 b − 1 d
  2 d  + ηϕ 1 − 2 ϕ  
1 −    
  w    
(Eqn 3-11)
Ast 0.67 f cu  beff  hf  Asc
=
bw d γ m 0.87 f y  b − 1 d + ηϕ  + b d
  (Eqn 3-12)
 w   w

3.5.2 Worked Examples for Flanged Beam, grade 35 (η = 0.9 )
x hf
(i) Worked Example 3.7 : Singly reinforced section where 0.9 ≤
d d
Consider the previous example done for a rectangular beam 500 (h) ×
400 (w), f cu = 35 MPa, under a moment 486 kNm, with a flanged
section of width = 1200 mm and depth = 150 mm :

bw = 400 , d = 500 − 40 − 20 = 440 , beff = 1200 h f = 150

x hf
First check if 0.9 ≤ based on beam width of 1200,
d d
M 486 × 10 6
K= = = 0.0598
f cu beff d 2 35 × 1200 × 440 2
x  K  1
By (Eqn 3-5), =  0.5 − 0.25 −  = 0.159 ;
d   0.9  0.45

x h f 150 z x
∴ 0.9 = 0.143 < = = 0.341 . = 1 − 0.45 = 0.928 ; Thus
d d 440 d d
Ast M 486 × 10 6
= = = 0.00563
beff d beff d × 0.87 f y ( z / d ) 1200 × 440 × 0.87 × 460 × 0.928
2 2

bw 400
> 0.18% (minimum for = = 0.33 < 0.4 in accordance with
beff 1200
Table 9.1 of the Code)
∴ Ast = 2974 mm2. Use 2T40 + 1T25
As in comparison with the previous example based on rectangular

25


section, it can be seen that there is saving in tensile steel (2974 mm2 vs
3498 mm2) and the compression reinforcements are eliminated.

x hf
(ii) Worked Example 3.8 – Singly reinforced section where η > , and
d d
η = 0.9 for grade 35.
Beam Section : 1000 (h) × 600 (w), flange width = 2000 mm,
flange depth = 150 mm f cu = 35 MPa under a moment 4000 kNm
bw = 600 , d = 1000 − 50 − 60 = 890 , beff = 2000 h f = 150
hf 150 beff 2000
= = 0.169 ; = = 3.333
d 890 bw 600
x hf
First check if 0.9 ≤ based on beam width of bw = beff = 2000
d d
M 4000 × 10 6
K= = = 0.0721
f cu beff d 2 35 × 2000 × 890 2
By (Eqn 3-7)
x  K  h 150
0.9 = 2 0.5 − 0.25 −

 = 0.176 > f = = 0.169
d  0.9  d 890
So 0.9 × neutral axis depth extends below flange.
Mf 0.67 f cu h f  beff  1 h f 
=  − 11 − 
bw d 2 γm d  bw  2 d  ⇒ M f = 2675.65 kNm
  
x
Solve by (Eqn 3-8) with η = 0.9 .
d
x M −M f
2
x
0.1809 f cu   − 0.402 f cu + =0
d  d bw d 2
x (4000 − 2675.65) × 10 6
2
x
⇒ 0.1809 × 35  − 0.402 × 35 + = 0;
d  d 600 × 890 2
x
⇒ = 0.2198 ;
d
By (Eqn 3-10)
Ast 1 0.67 f cu  beff  hf 
=  b − 1 d + 0.9 × 0.2198 = 0.02309
 
bw d 0.87 f y γ m  w  
2
Ast = 12330 mm , Use 10-T40 in 2 layers

(iii) Worked Example 3.9 – Doubly reinforced section

Beam Section : 1000 (h) × 600 (w), flange width = 1250 mm,
flange depth = 150 mm f cu = 35 MPa under a moment 4000 kNm
bw = 600 , d = 1000 − 50 − 60 = 890 , beff = 1250 h f = 150

26


hf 150 beff 1250
= = 0.169 ; = = 2.083 ; η = 0.9
d 890 bw 600
x hf
First check if η ≤ based on beam width of beff = 1250
d d
M 4000 ×10 6
K= = = 0.115
f cu beff d 2 35 ×1250 × 890 2
By (Eqn 3-7)
x 0.115 h f 150
0 .9 = 1 − 1 − = 0.302 > = = 0.169
d 0.225 d 890
So 0.9 × neutral axis depth extends below flange.
Mf 0.67 f cu h f  beff  1 h f 
=  − 11 − 
bw d 2 γm d  bw  2 d  ⇒ M f = 1242.26 kNm
  
x
Solve by (Eqn 3-8) with η = 0.9
d
x M −M f
2
x
0.1809 f cu   − 0.402 f cu + =0
d  d bw d 2
x (4000 − 1242.26)× 10 6
2
x
⇒ 0.1809 f cu   − 0.402 × 35 + =0
d  d 600 × 890 2
x
= 0.547 > 0.5 . Double reinforcement required. d ' = 50 + 20 = 70
d
By (Eqn 3-11)
 M 0.67 f cu  beff  hf  1 hf  1 
 + ηϕ 1 − ϕ  
Asc 1
=  −  − 1 1 −
d  2 d   
bw d 0.87 f y (1 − d ' / d )  bw d

2
γ m  bw     2   
= 0.001427 = 0.143 %
Asc = 763 mm2 > 0.4% on flange as per Table 9.1 of the Code which is
0.004 × 1250 × 150 = 750 mm2. Use 6T20
By (Eqn 3-12)

Ast 0.67 f cu  beff  hf  A
= 
 − 1 + ηϕ  + sc = 0.02614
 d
bw d γ m 0.87 f y  bw   bw d
⇒ Ast = 13958 mm2 , Use 10-T40 + 2-T32 in 2 layers (2.65%)

3.6 Detailings of longitudinal steel for bending
The followings should be observed in placing of longitudinal steel bars for
bending. Re Cl. 9.2.1 and 9.9.1 of the Code. The requirements arising from
“ductility” requirements are marked with “D” for beams contributing in lateral
load resisting system:

(i) Minimum tensile steel percentage : For rectangular beam, 0.13% in
accordance with Table 9.1 of the Code and 0.3% in accordance with Cl.
27


9.9.1 of the Code (D); except for beams subject to pure tension which
requires 0.45% as in Table 9.1 of the Code;
(ii) Maximum tension steel percentage : 2.5% (Cl. 9.9.1.1(a)) for beams
contributing in lateral load resisting system(D); and 4% for others (Cl.
9.2.1.3 of the Code);
(iii) Minimum compressive steel percentage : When compressive steel is
required for ultimate design, Table 9.1 of the Code should be followed
by providing 0.2% for rectangular beam and different percentages for
others. In addition, at any section of a beam within a critical zone (e.g.
a potential plastic hinge zone as discussed in Section 2.4) the
compression reinforcement ≥ one-half of the tension reinforcement in
the same region (Cl. 9.9.1.1(a) of the Code) (D);
(iv) For flanged beam, Figure 3.10 is used to illustrate the minimum
percentages of tension and compression steel required in various parts
of flanged beams (Table 9.1 of the Code), but not less than 0.3% in
accordance with Cl. 9.9.1.1(a) of the Code (D);
Longitudinal bars in flange
beff Ast ≥ 0.0026bw h (T-beam)
Ast ≥ 0.002bw h (L-beam)
Asc ≥ 0.004beff h f
hf
Transverse bars in flange
As ≥ 0.0015h f × 1000 per
h
unit metre of flange length

Longitudinal bars in web:
Ast ≥ 0.0018bw h if bw / beff < 0.4
Ast ≥ 0.0013bw h if bw / beff ≥ 0.4
bw
Asc ≥ 0.002bw h

Figure 3.10 – Minimum steel percentages in various parts of flanged beams

(v) For beams contributing in lateral load resisting system, calculation of
anchorage lengths of longitudinal bars anchored into exterior columns,
bars must be assumed to be fully stressed as a ductility requirement
according to Cl 9.9.1.1(c) of the Code. That is, stresses in the steel

should be f y instead of 0.87 f y in the assessment of anchorage

length. As such, the anchorage and lap lengths as indicated in Tables
8.4 and 8.5 of the Code should be increased by 15% as per (Ceqn 8.4)

28


f yφ
of the Code in which l b ≥ which is a modification (by changing
4 f bu

0.87 f y to f y ) where f bu = β f cu and β is 0.5 for tension

anchorage and 0.63 for compression anchorage for high yield bars in
accordance with Table 8.3 of the Code. Lap lengths can be taken as
identical to anchorage lengths (D);
(vi) Full strength welded splices may be used in any location according to
Cl. 9.9.1.1(d) of the Code;
(vii) For beams contributing in lateral load resisting system, no portion of
the splice (laps and mechanical couplers) shall be located within the
beam/column joint region or within one effective depth of the member
from the critical section of a potential plastic hinge (discussed in
Section 2.4) in a beam where stress reversals in lapped bars could
occur (Cl. 9.9.1.1(d) of the Code). However, effects due to wind load
need not be considered as creating stress reversal (D);
no lap /
potential mechanical stress reversal could
plastic hinge coupler zone occur
section

d

d

Figure 3.11 – Location of no lap / mechanical coupler zone in beam
contributing to load resisting system

(viii) For beams contributing in lateral load resisting system, longitudinal
bars shall not be lapped in a region where reversing stresses at the

ultimate state may exceed 0.6 f y in tension or compression unless

each lapped bar is confined by links or ties in accordance with (Ceqn
9.6) reproduced as follows (D) :

Atr φ ⋅ f y
≥ (Eqn 3-13)
s 48 f yt

According to the definitions in the Code, φ is the diameter of the
longitudinal bar; Atr is the smaller of area of transverse
reinforcement within the spacing s crossing a plane of splitting

29


normal to the concrete surface containing extreme tension fibres, or
total area of transverse reinforcements normal to the layer of bars
within a spacing s , divided by n (no. of longitudinal bars) in mm2;
s is the maximum spacing of transverse reinforcement within the lap

length, f yt is the characteristic yield strength of the transverse

reinforcement.
As the “just adequate” design is by providing steel so that the

reinforcing bars are at stress of 0.87 f y , overprovision of the section

by 0.87/0.6 – 1 = 45% of reinforcing bars at the laps should fulfill the
requirement for lapping in regions with reversing stress. Or else,
transverse reinforcement by (Ceqn 9.6) will be required. Figure 3.12
shows the occurrence of the plane of splitting at lapping.

(a)
Potential split faces by the bar force
transmitting lapping force by shear
friction

(b)
Figure 3.12 – splitting of concrete by shear friction in lapping of bars

Consider the example (a) illustrated in Figure 3.12, transverse

Atr φ ⋅ f y φ
reinforcement required will simply be ≥ = if high yield
s 48 f yt 48

bars are used for both longitudinal and transverse reinforcements. If
Atr 40
φ = 40 (i.e. the longitudinal bars are T40), ≥ = 0.833 . The
s 48

total area of transverse reinforcement is ∑A tr = 4 Atr as there are 4

30


no. of bars. So
∑A tr
≥ 0.833 × 4 = 3.333 . Using T12 – 4 legs – 125 is
s

adequate as
∑A tr
provided is 3.619. It should be noted that case (b)
s
is generally not the controlling case.
(ix) At laps in general, the sum of reinforcement sizes in a particular layer
should not exceed 40% of the beam width as illustrated by a numerical
example in Figure 3.13 (Cl. 9.2.1.3 of the Code);

bar diameter bar diameter
d = 40 d = 40 Sum of reinforcement
sizes = 40 × 8 = 320
< 0.4 × 900 = 360.
So O.K.

beam width b = 900

Figure 3.13 – Illustration of sum of reinforcement sizes at laps ≤ 0.4 of
beam width

(x) Minimum clear spacing of bars should be the greatest of bar diameter,
20 mm and aggregate size + 5 mm (Cl. 8.2 of the Code);
(xi) Maximum clear spacing between adjacent bars near tension face of a
beam ≤ 70000βb/fy ≤ 300 mm where βb is the ratio of moment
redistribution (ratio of moment after redistribution to moment before
redistribution) or alternatively ≤ 47000/fs ≤ 300 mm where

2 f y As ,req 1
fs = × . Based on the former with βb = 1 (no
3 As , prov βb

redistribution), the maximum clear spacing is 152 mm (Cl. 9.2.1.4 of
the Code);
(xii) Requirements for containment of compression steel bars is identical to
that of columns (Cl. 9.5.2.2 of the Code) : Every corner bar and each
alternate bar (and bundle) in an outer layer should be supported by a
link passing around the bar and having an included angle ≤ 135o. Links
should be adequately anchored by means of hook through a bent angle
≥ 135o. No bar within a compression zone be more than 150 mm from
a restrained bar (anchored by links of included angle ≥ 135o) as
illustrated in Figure 3.14;

31


(xiii) No tension bars should be more than 150 mm from a vertical leg which
is also illustrated in Figure 3.14 (Cl. 9.2.2 of the Code);
Links bent through Spacing of tension Alternate bar in an outer layer
angle ≥ 135o for bar ≤150 from a restrained by link of included
anchorage in concrete vertical leg angle ≤135o

compression zone
≤135o

≤ 150 ≤ 150
≤ 150 restrained longitudinal
≤ 150
compression bars be
anchored by links of
included angle ≤ 135o

bar in compression ≤ 150
from a restrained bar
≤ 150 ≤ 150

Figure 3.14 – Anchorage of longitudinal bar in beam section

(xiv) At an intermediate support of a continuous member, at least 30% of the
calculated mid-span bottom reinforcement should be continuous over
the support as illustrated in Figure 3.15 (Cl. 9.2.1.8 of the Code);

≥ 0.3 As1 and
Calculated ≥ 0.3 As 2 Calculated
mid-span steel mid-span steel
area As 2 area As1

Figure 3.15 – At least 30% of the calculated mid-span bottom bars be
continuous over intermediate support

(xv) In monolithic construction, simple supports should be designed for
15% of the maximum moment in span as illustrated in Figure 3.16 (Cl.
9.2.1.5 of the Code);

32


section designed
for 0.15 Mmax

maximum bending
moment Mmax

Bending moment diagram
Simple support by
beam or wall

Figure 3.16 – Simple support be designed for 15% of the maximum span
moment

(xvi) For flanged beam over intermediate supports, the total tension
reinforcements may be spread over the effective width of the flange
with at least 50% inside the web as shown in Figure 3.17 reproduced
from Figure 9.1 of the Code;

beff

at most 50% of at least 50% of
reinforcements reinforcements
outside the web inside the web

b

Figure 3.17 – distribution of tension rebars of flanged beam over support

(xvii) For beam with depths > 750 mm, provision of sides bars of size (in

mm) ≥ sb b / f y where sb is the side bar spacing (in mm) and b

is the lesser of the beam breadth (in mm) under consideration and 500

mm. f y is in N/mm2. In addition, it is required that sb ≤ 250 mm and

side bars be distributed over two-thirds of the beam’s overall depth
measured from its tension face. Figure 3.18 illustrate a numerical
example (Cl. 9.2.1.2 of the Code);

33


b is the lesser of 600 and 500, so
b = 500
s b chosen to be 200 mm ≤ 250mm,
So size of side bar is
s b b / f y = 200 × 500 / 460
1500
= 14.74
1000 Use T16.
T16 The side bars be distributed over
2
× 1500 = 1000 from bottom
3
which is the tension side.
tension side
600

Figure 3.18 – Example of determination of side bars

(xviii) When longitudinal bars of beams contributing to lateral load resisting
system are anchored in cores of exterior columns or beam studs, the
anchorage for tension shall be deemed to commence at the lesser of 1/2
of the relevant depth of the column or 8 times the bar diameter as
indicated in Figure 3.19. In addition, notwithstanding the adequacy of
the anchorage of a beam bar in a column core or a beam stud, no bar
shall be terminated without a vertical 90o standard book or equivalent
anchorage device as near as practically possible to the far side of the
column core, or the end of the beam stud where appropriate, and not
closer than 3/4 of the relevant depth of the column to the face of entry.
Top beam bars shall be bent down and bottom bars must be bent up as
indicated in Figure 3.19. (Cl. 9.9.1.1(c) of the Code) (D);

anchorage
D commences at this
section generally.

Not
permitted
≥ 0.5D ≥ 500mm or h X h
or 8Ø

≥ 0.75 D anchorage can commence at this section
if the plastic hinge (discussed in Section
2.4) of the beam is beyond X

Figure 3.19 – Anchorage of reinforcing bars at support for beams contributing to
lateral load resisting system

34


(xix) Beam should have a minimum support width by beam, wall, column as
shown in Figure 3.20 as per Cl. 8.4.8 of the Code;

≥2(4Ø+c) if Ø ≤ 20
≥2(5Ø+c) if Ø > 20 centre line of
support
c

bar of
3Ø if Ø ≤ 20; diameter Ø
4Ø if Ø > 20

≥0

Figure 3.20 – Support width requirement

(xx) Curtailment of tension reinforcements except at end supports should be
in accordance with Figure 3.21 (Cl. 9.2.1.6 of the Code).

Section beyond
Bar of diameter Ø which the bar is no
d longer required

≥12Ø and d at least;

if the bar is inside tension
zone, ≥ 1.0 bond length

Figure 3.21 – curtailment of reinforcement bars

Worked Example 3.10

Worked example 3.10 is used to illustrate the arrangement of longitudinal bars
and the anchorages on thin support for the corridor slab beam of a typical
housing block which functions as coupling beam between the shear walls on
both sides. Plan, section and dimensions are shown in Figure 3.22(a). Concrete
grade is 35.

35


300

Slab beam

1400 Plan 200
Section

Figure 3.22(a) – Layout of the slab beam in Worked Example 3.10

The designed moment is mainly due to wind loads which is 352 kNm,
resulting in required longitudinal steel area of 3910 mm2 (each at top and
bottom). The 200 mm thick wall can accommodate at most T16 bars as
2(4 × 16 + 25) = 178 < 200 as per 3.6(xix). So use 20T16 ( Ast provided is
4020 mm2. Centre to centre bar spacing is (1400 − 25 × 2 − 16 ) / 19 = 70 mm.

For anchorage on support, lap length should be 34 × 16 = 544 mm. The factor
34 is taken from Table 8.4 of the Code which is used in assessing anchorage
length. Anchorage details of the longitudinal bars at support are shown in
Figure 3.22(b);

20T16

T16 544
cross bar
T10 – 10 legs – 200 c/c

25 64 11
Anchorage commences at
centre line of wall as 20T16
200/2=100<16×8=128
200

Figure 3.22(b) – Anchorage Details at Support for Worked Example 3.10

3.7 Design against Shear

36


3.7.1 Checking of Shear Stress and provision of shear reinforcements

Checking of shear in beam is based on the averaged shear stress calculated
from (Ceqn 6.19)
V
v=
bv d
where v is the average shear stress, V is the ultimate shear, d is the
effective depth of the beam and bv is beam width. bv should be taken as the
averaged width of the beam below flange in case of flanged beam)

If v is greater than the values of v c , termed “design concrete shear stress” in
Table 6.3 of the Code which is determined by the formula
1 1 1
 f  3  100 As  3  400  4 1
v c = 0.79 cu    
  d  γ listed in Table 6.3 of the Code with
 25   bv d   m
the following limitations :
(i) γ m = 1.25 ;
100 As
(ii) should not be taken as greater than 3;
bv d
1
 400  4
(iii)   should not be taken as less than 0.67 for member without shear
 d 
reinforcements and should not be taken as less than 1 for members with
minimum links. For others, calculate according to the expression;
A b (v − v c ) bv
Then shear links of sv = v ≥ v r should be provided (Table
sv 0.87 f yv 0.87 f yv
6.2 of the Code) where v r = 0.4 for f cu ≤ 40 MPa and 0.4(fcu/40)2/3 for
f cu > 40 , or alternatively, less than half of the shear resistance can be taken up
d − d'
by bent up bars by 0.5V ≥ Vb = Asb (0.87 f yv )(cos α + sin α cot β ) as per
sb
(Ceqn 6.20) and Cl. 6.1.2.5(e) of the Code and the rest by vertical links.

Maximum shear stress not to exceed v tu = 0.8 f cu or 7 MPa, whichever is

the lesser by Cl. 6.1.2.5(a).

3.7.2 Minimum shear reinforcements (Table 6.2 of the Code)

If v < 0.5v c , no shear reinforcement is required;

If 0.5v c < v < (v c + v r ) , minimum shear links of
Asv bv
= v r along the
sv 0.87 f yv

37


whole length of the beam be provided where v r = 0.4 for f cu ≤ 40 and

0.4( f cu / 40 )
2/3
for f cu > 40 , but not greater than 80;

3.7.3 Enhanced shear strength close to support (Cl. 6.1.2.5(g))

At sections of a beam at distance a v ≤ 2d from a support, the shear strength
2d
can be increased by a factor , bounded by the absolute maximum of
av
v tu = 0.8 f cu or 7 MPa as illustrated by Figure 3.23.

d

section under consideration, shear strength
av
2d
enhanced to vc
av

Figure 3.23 – Shear enhancement near support

3.7.4 Where load is applied to the bottom of a section, sufficient vertical
reinforcement to carry the load should be provided in addition to any
reinforcements required to carry shear as per Cl. 6.1.2.5(j);

Vertical rebars to
resist beam
bottom loads
which may
increase required
provisions of
links

Hanging load at beam bottom

Figure 3.24 – Vertical rebars to resist hanging load at beam bottom
(e.g. inverted beam)

3.7.5 Worked Examples for Shears

38


(i) Worked Example 3.11 – shear design without shear enhancement in
concrete

Section : b = 400 mm;
100 Ast
d = 700 − 40 − 16 = 644 mm; = 1.5 ; f cu = 35 MPa;
bd
V = 700 kN;
1 1 1
 f  3  100 As  3  400  4 1
vc = 0.79 cu    
  = 0.81 MPa;
 25   bv d   d  γm
1
 400  4
where   be kept as unity for d > 400 .
 d 
700 × 10 3
v= = 2.72 MPa,
400 × 644
Asv b(v − vc ) 400(2.72 − 0.81)
= = = 1.91 ; Use T12 – 200 c/c d.s.
sv 0.87 f yv 0.87 × 460

(ii) Worked Example 3.12 – shear design with shear enhancement in
concrete.

Re Figure 3.25 in which a section of 0.75 m from a support, as a heavy
point load is acting so that from the support face to the point load along
the beam, the shear is more or less constant at 700 kN.

av = 750 700 kN

d = 644
T32

Figure 3.25 – Worked Example 3.11

Section : b = 400 mm; cover to main reinforcement = 40 mm;
100 Ast
d = 700 − 40 − 16 = 644 mm; = 1.5 ; f cu = 35 MPa; V = 700 kN;
bd
1 1 1
 f  3  100 As  3  400  4 1
vc = 0.79 cu    
  = 0.81 MPa as in Worked Example
 25   bv d   d  γm
3.11.

39


2d 2 × 644
Concrete shear strength enhanced to vc = × 0.81 = 1.39 MPa <
av 750
7 MPa and 0.8 f cu = 0.8 35 = 4.7 MPa
700 × 10 3
v= = 2.72 MPa,
400 × 644
Asv b(v − vc ) 400(2.72 − 1.39 )
= = = 1.33 ; Use T12 – 150 c/c s.s
sv 0.87 f yv 0.87 × 460

(iii) Worked Example 3.13 – inclusion of bent-up bars (Cl. 6.1.25(e) of the
Code)

If half of the shear resisted by steel in Worked Example 3.11 is taken up
by bent-up bars, i.e. 0.5 × (2.72 − 0.81)× 400 × 644 ×10 −3 = 246 kN to be
taken up by bent-up bars as shown in Figure 3.26.

sb = 441 sb = 441

d – d’= 644 – 40 – 16 = 588
β = 60 o
α = 45 o

St = 882

Figure 3.26 – Worked Example 3.12

By (Ceqn 6.20) of the Code,
d − d'
Vb = Asb (0.87 f yv )(cos α + sin α cot β )
sb
246000 × 441
⇒ Asb = = 413 mm2
0.87 × 460(cos 45 + sin 45 cot 60 )× 588
0 0 0

Use 6 nos. of T10 at spacing of sb = 441 mm as shown.

3.8 Placing of Shear reinforcements

The followings should be observed for the placing of shear reinforcements :
(i) Bar size ≥ the greater of 1/4 of the bar size of the maximum
longitudinal bar and 6 mm (BS8110 Cl. 3.12.7.1);
(ii) The minimum provision of shear reinforcements (links or bent up bars)
vbs
in beams should be given by Asv ≥ r v v where v r = 0.4 for
0.87 f yv
f cu ≤ 40 and v r = 0.4( f cu / 40)
2/3
for 80 ≥ f cu > 40 (Cl. 6.1.2.5(b)
of the Code);

40


(iii) At least 50% of the necessary shear reinforcements be in form of links
(Cl. 6.1.2.5(e) of the Code);
(iv) The maximum spacing of links in the direction of span of the beam
should be the least of the followings as illustrated by a numerical
example in Figure 3.27 (Cl. 6.1.2.5(d), 9.2.1.10, 9.5.2.1 (BS8110 Cl.
3.12.7.1), 9.5.2.2, 9.9.1.2(a) of the Code) :
(a) 0.75d;
(b) the least lateral dimension of the beam (D);
(c) 16 times the longitudinal bar diameter (D);
(d) 12 times the smallest longitudinal bar diameter for containment
of compression reinforcements.
For beam contributing to
width of web of beam lateral load resisting
= 400 system, maximum link
3T32 (tension bar) spacing be the least of
(a) 0.75d = 483;
(b) b = 400;
(c) 16 × 32 = 512;
d = 644 (d) 12 × 25 = 300
So maximum link spacing
should be 300 mm.

For beam not contributing
3T25 (compression bar) to lateral load resisting
system, (b) & (c) not count

Figure 3.27 – Maximum spacing of shear links in the span direction of beam

(v) At right angle of the span, the horizontal spacing of links should be
such that no longitudinal tension bar should be more than 150 mm
from a vertical leg and ≤ d as per Cl. 6.1.2.5(d) of the Code and shown
in Figure 3.28;

≤d ≤d

≤150 ≤150

Figure 3.28 – Maximum spacing of shear links at right angle to the span
direction of beam

41


(vi) Links or ties shall be arranged so that every corner and alternate
longitudinal bar that is required to function as compression
reinforcement shall be restrained by a leg as illustrated Figure 3.14;
(vii) By Cl. 9.9.1.2(b) of the Code, links in beams contributing to lateral
load resisting system should be adequately anchored by means of 135o
or 180o hooks in accordance with Cl. 8.5 of the Code as shown in
Figure 3.29 (D);
(viii) Anchorage by means of 90o hook is only permitted for tensile steel in
beams not contributing to lateral load resisting system;
(ix) Links for containment of compression longitudinal bars in general
must be anchored by hooks with bent angle ≥ 135o in accordance with
Cl. 9.2.1.10 and 9.5.2.1 of the Code. Links with different angles of
hooks are shown in Figure 3.29. (Reference to Cl. 9.5.2.1 should be
included in Cl. 9.2.1.10 as per Cl. 3.12.7.1 of BS8110)

Link with 180o Link with 135o Link with 90o hooks (not
hooks hooks permitted for containment
of compression bars of
beam in general and all
links in beams contributing
to lateral load resisting
system)

Figure 3.29 – Links with hooks with different bent angles

3.9 Design against Torsion

3.9.1 By Cl. 6.3.1 of the Code, in normal slab-and-beam and framed construction,
checking against torsion is usually not necessary. However, checking needs be
carried out if the design relies entirely on the torsional resistance of a member
such as that indicated in Figure 3.30.

42


Beam carrying
unbalanced
torsion induced
by slab needs be
checked

Figure 3.30 – Illustration for necessity of checking against torsion

3.9.2 Calculation of torsional rigidity of a rectangular section for analysis (in
grillage system) is by (Ceqn 6.64) of the Code

1
C= βhmin 3 hmax where β is to be read from Table 6.16 of the Code
2
reproduced as Table 3.4 of this Manual.

hmax/hmin 1 1.5 2 3 5 >5
β 0.14 0.20 0.23 0.26 0.29 0.33
Table 3.4 – Values of coefficient β

3.9.3 Calculation of torsional shear stress

Upon arrival of the torsion on the rectangular section, the torsional shear stress
is calculated by (Ceqn 6.65) of the Code
2T
vt =
2 h 
hmin  hmax − min 
 3 
and in case of a section such as T or L sections made of rectangles, the section

∑h
3
should be broken up into two or more rectangles such that the min hmax is

maximized and the total Torsional moment T be apportioned to each
 hmin 3 hmax 
rectangle in accordance with (Ceqn 6.66) of the Code as T ×  .
 ∑ h 3h 
 min max 

43


If the torsional shear stress exceeds 0.067 f cu (but not more than 0.6MPa),

torsional reinforcements will be required (Table 6.17 of the Code).

Furthermore, the torsional shear stress should be added to the shear stress

induced by shear force to ensure that the absolute maximum v tu = 0.8 f cu

or 7MPa is not exceeded, though for small section where y1 (the larger
centre-to-centre dimension of a rectangular link) < 550mm, v tu will be
decreased by a factor y1 / 550 . Revision of section is required if the absolute
maximum is exceeded (Table 6.17 of the Code).

3.9.4 Calculation of torsional reinforcements

Torsional reinforcement in forms of close rectangular links and longitudinal
bars are to be calculated by (Ceqn 6.67) and (Ceqn 6.68) of the Code as
Asv T
= (Ceqn 6.67)
sv 0.8 x1 y1 (0.87 f yv )
( Asv is the area of the 2 legs of the link)

Asv f yv ( x1 + y1 )
As = (Ceqn 6.68)
sv f y

It should be noted that there is no reduction by shear strength ( v c ) of concrete.
The derivation of the design formula (Ceqn 6.67) of the Code for close
rectangular links is under the assumption of a shear rupture length of stirrup
width + stirrup depth x1 + y1 as shown in Figure 3.31. A spiral torsional
failure face is along the heavy dotted line in the figure. It is also shown in the
figure that the torsional moment of resistance by the stirrups within the
Regions X and Y are identical and is the total resistance is therefore

Asv 0.87 f y x1 y1 Asv 0.87 f y x1 y1 Asv T
. So T = ⇒ = . An additional
sv sv sv 0.87 f y x1 y1

factor of 0.8 is added and the equation becomes (Ceqn 6.67) by which
Asv T
= . The derivation of the longitudinal bars is based on the
sv 0.8 x1 y1 (0.87 f y )
use of same quantity of longitudinal bars as that of stirrups with even
distribution along the inside of the stirrups. Nevertheless, the Code allows
merging of the flexural steel with these longitudinal bars by using larger

44


diameter of bars as will be illustrated in the Worked Example 3.14.

Moment provided by this
0.5 Asv 0.87 f y stirrup in region X is
0.5 Asv 0.87 f y y1 . Total
y1 nos. of stirrup within X is
shear rupture x1 / sv . So total moment
spiral face 0.5 Asv 0.87 f y is 0.5 Asv 0.87 f y y1x1 / sv

45o

x1
0.5 Asv 0.87 f y 0.5 Asv 0.87 f y
Moment provided by this
stirrup in region Y is
y1 x1 0.5 Asv 0.87 f y x1 . Total
Region X
nos. of stirrup within Y is
T y1 / sv . So total moment
45o x1
y1 is 0.5 Asv 0.87 f y x1 y1 / sv
Region Y

Figure 3.31 – Derivation of Formulae for torsional reinforcements

3.9.5 Worked Example 3.14 – Design for T-beam against torsion

A total torsion of T = 200 kNm on a T-section as shown in Figure 3.32 with
an average vertical shear stress on the web of 0.82 N/mm2. The section is also
under bending requiring flexural steel area of 2865 mm2 at bottom. Concrete
grade is 35.

1500

400

1000

450 Option A Option B

Figure 3.32 – Section of a T section resisting torsion for Worked Example 3.14

100 As 2865 × 100
For vertical shear, taking = = 0.477
bv d 450 ×1334

45


1 1 1 1
 f  3  100 As  3  400  4 1  400  4
vc = 0.79 cu    
  d  γ = 0.55 , again taking  d  as unity.
 25   bv d   m  
Asv b(v − vc ) 450(0.82 − 0.55)
= = = 0 .3
sv 0.87 f yv 0.87 × 460

For torsion, Option A is made up of two rectangles of 525× 400 and one
rectangle of 450 × 1400 .

∴ (∑ h min
3
hmax )
optionA
= 2 × 525 × 400 3 + 1400 × 450 3 = 1.94775 × 1011 mm4

Option B is made up of one rectangle of 1500 × 400 and one rectangle of
450 × 1000 .

∴ (∑ h min
3
hmax )
optionB
= 1500 × 400 3 + 1000 × 450 3 = 1.87125 × 1011 mm4

As Option A has a larger torsional stiffness, it is adopted for design.

The torsional moment is apportioned to the three rectangles of Option A as :
525 × 400 3
For the two 525× 400 rectangles T1 = 200 × = 34.50 kNm;
1.94775 × 1011
Torsional shear stress is
2T1 2 × 34.5 × 10 6
vt1 = = = 1.101 N/mm2
2 h   400 
hmin  hmax − min  400 2  525 − 
 3   3 

> 0.067 f cu = 0.396 N/mm2 (< 0.6 N/mm2)

So torsional shear reinforcement is required
x1 = 400 − 40 × 2 − 6 × 2 = 308 ; y1 = 525 − 40 × 2 − 6 × 2 = 433

Asv T1 34.5 ×10 6
= = = 0.808
sv 0.8 x1 y1 (0.87 f yv ) 0.8 × 308 × 433 × 0.87 × 460

Use T12 – 200 C.L. x1 = 308 , y1 / 2 = 525 / 2 = 262.5 ; use sv = 200 ≤ 200 ;
≤ x1 and ≤ y1 / 2 as per Cl. 6.3.7 of the Code.

Asv f yv ( x1 + y1 ) 0.808 × 460 × (308 + 525)
As = = = 673 mm2
sv f y 460

Use 4T16

46


1400 × 450 3
For the 1400 × 450 rectangle T2 = 200 × = 131 kNm
1.94775 ×1011
2T2 2 × 131× 10 6
vt1 = = = 1.035 N/mm2
2 h   450 
hmin  hmax − min  450 2 1400 − 
 3   3 
The total shear stress is 1.035 + 0.82 = 1.855 N/mm2 < vtu = 4.73 MPa

As 1.035 > 0.067 f cu = 0.396 N/mm2, torsional shear reinforcement is

required.
x1 = 450 − 40 × 2 − 6 × 2 = 358 mm; y1 = 1400 − 40 × 2 − 6 × 2 = 1308 mm

Asv T2 131× 10 6
= = = 0.87 mm
sv 0.8 x1 y1 (0.87 f yv ) 0.8 × 358 × 1308 × 0.87 × 460

Asv
Adding that for vertical shear, total = 0.87 + 0.3 = 1.17
sv
Use T12 – 175 C.L. x1 = 358 , y1 / 2 = 1308 / 2 = 654 ; use sv = 175 ≤ 200 ;
≤ x1 and ≤ y1 / 2 as per Cl. 6.3.7 of the Code.
It should be noted that the torsional shear link should be closed links of shape
as indicated in Figure 9.3 of the Code.

Asv f yv ( x1 + y1 ) 0.87 × 460 × (358 + 1308)
As = = = 1449 mm2. Use 13T12
sv f y 460

Incorporating the bottom 3T12 into the required flexural steel, the bottom steel
area required is 2865 + 113.1× 3 = 3205 mm2. So use 4T32 at bottom and
10T12 at sides.

The sectional details is shown in Figure 3.33.

1500

T12
400
T16

T12 – 200 C.L.
1000 T32

4T32

450

Figure 3.33 – Arrangement of torsional reinforcements

47


It should be borne in mind that these torsional reinforcements are in addition
to others required for flexure and shear etc.

3.10 Placing of Torsional reinforcements

The followings (in Cl. 6.3.7, Cl. 6.3.8 and Cl. 9.2.3 of the Code) should be
observed for the placing of shear reinforcements :

(i) The torsional shear link should form the closed shape as in Figure 9.1
of the Concrete Code Handbook reproduced as Figure 3.34. It should
be noted that the second configuration in the Figure is not included in
Figure 9.3 of the Code though it should also be acceptable;

Full lap length Full lap length

Full
anchorage
length

Figure 3.34 – Shape of Torsional shear links

(ii) The value s v for the closed link should not exceed the least of x1 ,
y1 / 2 or 200 mm as per Cl. 6.3.7 of the Code;
(iii) In accordance with Cl. 9.2.3 of the Code, provision of the longitudinal
torsion reinforcement should comply the followings :
(a) The bars distributed should be evenly round the inside perimeter
of the links as illustrated in Figure 3.33;
(b) Clear distance of the bars not to exceed 300 mm;
(c) Additional longitudinal bars required at the level of the tension or
compression reinforcements may be provided by using larger
bars than those required for bending alone, as illustrated in
Worked Example 3.14 and Figure 3.33;
(d) The longitudinal bars should extend a distance at least equal to
the largest dimension of the section beyond where it theoretically
ceases to be required.

48


4.0 Slabs

4.1 Types of Slabs

Slabs can be classified as “one way slab”, “two way slab”, “flat slab”, “ribbed
slab” with definition in Cl. 5.2.1.1 of the Code.

4.1.1 One way slab is defined by the Code as one subjected predominantly to u.d.l.
either
(i) it possesses two free and parallel edge; or
(ii) it is the central part of a rectangular slab supported on four edges with
a ratio of the longer to the shorter span greater than 2.

4.1.2 Two way slab is a rectangular one supported on four sides with length to
breadth ratio smaller than 2.

4.1.3 Flat slab is a slab supported on columns without beams.

4.1.4 Ribbed or Waffled Slab is a slab with topping or flange supported by closely
spaced ribs. The Code allows idealization of the ribbed slab or waffled slab as
a single slab without treatment as discretized ribs and flanges in analysis in Cl.
5.2.1.1(d) of the Code. If the stiffness of the ribbed or waffled slab is required
for input, the bending stiffness in the X and Y directions can be easily found
by summing the total bending stiffness of the composite ribs and flange
structure per unit width as illustrated in Figure 4.1. The twisting stiffness is
more difficult to assess. However, it should be acceptable to set the twisting
stiffness to zero which will end up with pure bending in the X and Y directions
as the slab, with its ribs running in the X and Y directions are clearly
predominantly strong in bending in the two directions.

Figure 4.1 illustrates the computation of “I” value of a waffle slab about the
X-direction which is the total stiffnesses of the nos. of “flanged ribs” within
one metre. “I” value in the Y-directions can be worked out similarly.

49


Centroid of a flanged rib is at
750 0.55 × 0.12 / 2 + 0.2 × 0.6 2 / 2
0.55 × 0.1 + 0.2 × 0.6
750
=0.2214m from top
100 I of a rib is
1
12
(0.55 × 0.13 + 0.2 × 0.63 )
+ 0.55 × 0.1× (0.2214 − 0.05)2
+ 0.2 × 0.6 × (0.3 − 0.2214)2
200 750 =0.006m4
600
Y 200
Within one metre, there will be
200 1000/750 = 1.333 nos. of ribs.
600
Dimensions of So the I per metre width is
a flanged rib 0.006 × 1.333 = 0.008 m4/m
X
200

Figure 4.1 – Illustration of calculation of I value about X-direction of a waffle slab

4.2 Analysis of Slabs without the use of computer method

4.2.1 One way slab can be analyzed as if it is a beam, either continuous or single
span. As we aim at simple analysis for the slab, we tend to treat it as a single
element without the necessity to consider the many loading cases for
continuous spans, Cl. 6.1.3.2(c) of the Code allows the design against moment
and shear arising from the single-load case of maximum design load on all
spans provided that :
(i) the area of each bay (defined in Figure 6.5 of the Code and reproduced
in Figure 4.2) > 30 m2;
(ii) the ratio of the characteristic imposed load to characteristic dead load ≤
1.25; and
(iii) the characteristic imposed load ≤ 5 kN/m2 excluding partitions.

bay
panel

Figure 4.2 – Definition of panels and bays

50


4.2.2 Two way rectangular slab is usually analyzed by treating it as if it is a single
slab in the absence of computer method. Bending moment coefficients for
calculation of bending moments are presented in Table 6.6 of the Code for
different support restraint conditions. Nevertheless, simplified formulae for the
bending coefficients in case of rectangular simply supported two way slab are
available in the Code (Ceqn 6.26 and 6.27) and reproduced as follows :
m x = α sx nl x and m y = α sy nl x
2 2
where n is the u.d.l. l x and l y are
respectively the shorter and longer spans and
(l y / l x )4 (l y / l x )2
α sx = ; α sy =
[
8 1 + (l y / l x )
4
] [
8 1 + (l y / l x )]
4
.

4.2.3 Flat slabs, if of regular arrangement, can be analyzed as frames in the
transverse and longitudinal directions by such methods as moment distribution
method as if they are separate frames. Analyzed moments and shears should
be apportioned to the “column strip” and “Middle strip” as per Table 6.10 of
the Code. In addition, the bending moment and shear force coefficients for the
one way slab can also be used as a simplified approach.

4.2.4 More bending moment and shear force coefficients of rectangular slabs with
various different support and loading conditions can be found from other
published handbooks, the most famous one being “Tables for the Analysis of
Plates, Slabs and Diaphragms based on the Elastic Theory”.

4.3 Analysis of Slabs with the use of the computer method

Analysis of slabs with the use of the computer method is mainly by the finite
element method in which the slab is idealized as an assembly of discrete “plate
bending elements” joined at nodes. The support stiffnesses by the supporting
walls and columns are derived as similar to that for beams as “sub-frames”. A
complete set of results including bending moments, twisting moment, shear
force per unit width (known as “stress” in finite element terminology) can be
obtained after analysis for design purpose. The design against flexure is most
commonly done by the Wood Armer Equations which calculate design
moments in two directions (conveniently in two perpendicular directions) and
they are adequate to cater for the complete set of bending and twisting
moments. The design based on node forces / moments should be avoided due
to its inadequacy to cater for twisting effects which will result in under-design.
A discussion of the plate bending theory and the design approach by the Wood

51


Armer Equations is enclosed in Appendix D, together with the “stress
approach” for checking and designing against shear in the same appendix. An
example of the mathematical modeling of a floor slab by the software SAFE
and results of subsequent analysis is illustrated in Figure 4.3.

Figure 4.3 – Modeling of an irregular floor slab as 2-D mathematical model,
subsequent analytical results of bending moments and twisting moment, design of
reinforcements by the Wood Armer Equations.

The finite element mesh of the mathematical model is often very fine. So it is
a practice of “lumping” the design reinforcements of a number of nodes over
certain widths and evenly distributing the total reinforcements over the widths,

52


as is done by the popular software “SAFE”. However, care must be taken in
not taking widths too wide for “lumping” as local effects may not be well
captured.

4.4 Detailing for Solid Slabs

Generally considerations in relation to determination of “effective span”,
“effective span depth ratio”, “moment redistribution”, “reduced design
moment to support”, “maximum and minimum steel percentages”, “concrete
covers” as discussed in Section 3.3 for design of beam are applicable to design
of slab. Nevertheless, the detailing considerations for slabs are listed as
follows with h as the structural depth of the slab (Re 9.3.1.1 of the Code) :

(i) Minimum steel percentage (Cl. 9.3.1.1(a) of the Code):
Main Reinforcing bars:

0.24% for f y = 250 MPa and 0.13% for f y = 460 MPa;

Distribution bars in one way slab ≥ 20% of the main reinforcements
(ii) Maximum reinforcements spacing (Cl. 9.3.1.1(b) of the Code):
(a) In general areas without concentrated loads :
the principal reinforcement, max. spacing ≤ 3h ≤ 400 mm; and
the secondary reinforcement, max. spacing ≤ 3.5h ≤ 450 mm.
(b) In areas with concentrated loads or areas of maximum moment:
the principal reinforcement, max. spacing ≤ 2h ≤ 250 mm; and
for the secondary reinforcement, max. spacing ≤ 3h ≤ 400 mm.
(iii) In addition to (ii), if either :
(a) h ≤ 250 mm (grade 250 steel);
(b) h ≤ 200 mm (grade 460 steel); or
(c) the percentage of required tension reinforcement is less than
0.3%.
no direct crack widths check by calculation is required. If none of
conditions in (a), (b) & (c) is satisfied, bar spacing to comply with Cl.
9.2.1.4 of the Code as discussed in 3.3(vi) of this Manual if steel
percentage > 1%. Otherwise, increase the spacing by 1/percentage;
(iv) Requirements pertaining to curtailment and anchoring of tension
reinforcements should be similar to that of beams;
(v) Reinforcements at end supports (Cl. 9.3.1.3 of the Code)
(a) At least 50% of the span reinforcements should be provided and

53


well anchored on supports of simply supported slabs and end
supports of continuous slabs as illustrated in Figure 4.4;
(b) If support shear stress v < 0.5vc , the arrangement in Figure 4.4
can be considered as effective anchorage.
(vi) Minimum bottom reinforcements at internal supports : 40% of the
calculated mid-span bottom reinforcements as illustrated in Figure 4.4.
(Cl. 9.3.1.4 of the Code)

at least 0.5 As
b provided at top of
end support, dia. Ø

As at span
= the greater of 1/3b and
30 mm if v < 0.5v c , at least 0.5 As
at least 0.4 As
otherwise 12Ø anchored into end
span support continuous through
internal support

Figure 4.4 – Anchorage of bottom reinforcements into supports

(vii) Reinforcements at free edge should be as shown in Figure 4.5 (Cl.
9.3.1.6 of the Code)

h

≥2h

Figure 4.5 – Free edge reinforcements for Slabs

(viii) Shear reinforcements not to be used in slabs < 200 mm. (Cl. 9.3.2 of
the Code)

4.5 Structural Design of Slabs

The structural design of slab against flexure is similar to that of beam. The
determination of reinforcements should be in accordance with Section 3.4 of

54


this Manual listing the options of either following the rigorous or simplified
“stress strain” relationship of concrete. Design against shear for slabs under
line supports (e.g. one-way or two-way) is also similar to that of beam.
However for a flat slab, the checking should be based on punching shear in
accordance with the empirical method of the Code or based on shear stresses
revealed by the finite element method. They are demonstrated in the Worked
Examples in the following sub-Section 4.6 :

4.6 Worked Examples

Worked Example 4.1 – One Way Slab

A one-way continuous slab with the following design data :
(i) Live Load = 4.0 kN/m2;
(ii) Finishes Load = 1 kN/m2;
(iii) Concrete grade : 35 with cover 25 mm;
(iv) Slab thickness : 200 mm;
(v) Fire rating : 1 hour, mild exposure;
(vi) Span : 4 m

4m 4m 4m 4m

Figure 4.6 – Slab in Worked Example 4.1

Sizing : Limiting Span depth ratio = 23 × 1.1 = 25.3 (by Table 7.3 and Table
7.4 of the Code, assuming modification by tensile reinforcement to be 1.1 as
the slab should be lightly reinforced). Assuming 10mm dia. bars under 25mm
concrete cover, effective depth is d = 200 − 25 − 5 = 170 . Span effective
depth ratio is 4000 / 170 = 23.5 < 25.3 . So OK.

Loading : D.L. O.W. 0.2 × 24 = 4.8kN/m2
Fin. 1.0 kN/m2
Total 5.8 kN/m2
L.L. 4.0 kN/m2
The factored load on a span is F = (1.4 × 5.8 + 1.6 × 4.0 ) × 4 = 58.08 kN/m.

55


Based on coefficients of shear and bending moment in accordance with Table
6.4 of the Code listed as follows :

B.M. -0.04Fl 0.075 Fl -0.086Fl 0.063Fl -0.063Fl 0.063Fl -0.086Fl 0.086 Fl 0Fl

S.F. 0.46F 0.6F 0.6F 0.5F 0.5F 0.6F 0.6F 0.4F

Continuous
(e.g. over
wall) simply
supported

Figure 4.7 – Bending Moment and Shear Force coefficients for Continuous Slab

(a) End span support moment (continuous) = 0.04 × 58.08 × 4 = 9.29 kNm/m
M 9.29 × 10 6
K= = = 0.0092
f cu bd 2 35 × 1000 × 170 2
z K
= 0.5 + 0.25 − = 0.989 > 0.95
d 0 .9
Ast M 9.29 × 10 6
= = = 0.08% < 0.13%
bd bd 2 × 0.87 f y z / d 1000 × 170 2 × 0.87 × 460 × 0.95
Ast = 0.13 ÷ 100 × 1000 × 170 = 221 mm2 Use T10 – 300
( Ast provided = 262mm2)
(b) End span span moment = 0.086 × 58.08 × 4 = 19.98 kNm/m
M 19.98 × 10 6
K= = = 0.0198
f cu bd 2 35 × 1000 × 170 2
z K
= 0.5 + 0.25 − = 0.978 > 0.95
d 0 .9
Ast M 19.98 × 10 6
= 2 = = 0.18% > 0.13%
bd bd × 0.87 f y z / d 1000 × 170 2 × 0.87 × 460 × 0.95

Ast = 0.18 ÷ 100 × 1000 × 170 = 309 mm2 Use T10 – 250
(c) First interior support moment = 0.086 × 58.08 × 4 = 19.98 kNm/m, same
reinforcement as that of end span reinforcement.
(d) Interior span or support moment = 0.063 × 58.08 × 4 = 14.64 kNm/m;
Ast M 14.64 × 10 6
= = = 0.133% > 0.13%
bd bd 2 × 0.87 f y z / d 1000 × 170 2 × 0.87 × 460 × 0.95
Ast = 0.133 ÷ 100 × 1000 × 170 = 227 mm2 Use T10 – 300
(e) End span span moment to continuous support

56


= 0.075 × 58.08 × 4 = 17.42 kNm/m
Ast M 17.42 × 10 6
= = = 0.159% > 0.13%
bd bd 2 × 0.87 f y z / d 1000 × 170 2 × 0.87 × 460 × 0.95
Ast = 0.159 ÷ 100 × 1000 × 170 = 270 mm2. Use T10 – 250
(f) Check Shear
Maximum shear = 0.6 × 58.08 = 34.85 kN/m.
1 1 1
 100 As  3  400  4 1  f cu  3
By Table 6.3 of the Code vc = 0.79     
 bd   d  γ m  25 

1 1
 400  4 1  35  3
1
vc = 0.79(0.13) 
3 
2
  = 0.44 N/mm , based on minimum
 170  1.25  25 
34850
steel 0.13%; v= = 0.205 N/mm2 < vc = 0.44 N/mm2.
1000 × 170
No shear reinforcement required.

Worked Example 4.2 – Two Ways Slab (4 sides simply supported)

A two-way continuous slab with the following design data :

(iii) Concrete grade : 35;
(iv) Slab thickness : 200 mm
(v) Fire rating : 1 hour, mild exposure, cover = 25mm;
(vi) Span : Long way : 4 m, Short way, 3 m

Sizing : Limiting Span depth ratio = 20 (by Table 7.3). So effective depth
taken as d = 200 − 25 − 5 = 170 as 3000/170 = 17.65 < 20.

Loading : D.L. O.W. 0.2 × 24 = 4.8kN/m2
Fin. 1.0 kN/m2
Total 5.8 kN/m2
L.L. 4.0 kN/m2
The factored load is F = (1.4 × 5.8 + 1.6 × 4.0 ) = 14.52 kN/m2

(Ceqn 6.26) and (Ceqn 6.27) of the Code are used to calculate the bending

57


moment coefficients along the short and long spans :

(l /lx)
4
(l /lx)
2

α sx = = 0.095 ; α sy =
] = 0.053
y y

[
8 1 + (l y / l x )
4
] [
8 1 + (l y / l x )
4

So the bending moment along the short span is

M x = 0.095 × 14.52 × 3 2 = 12.41 kNm/m

M 12.41 × 10 6
K= = = 0.0123
f cu bd 2 35 × 1000 × 170 2
z K
= 0.5 + 0.25 − = 0.986 > 0.95
d 0 .9
Ast M 12.41 × 10 6
= = = 0.113% < 0.13%
bd bd 2 × 0.87 f y z / d 1000 × 170 2 × 0.87 × 460 × 0.95
Ast = 0.13 ÷ 100 × 1000 × 170 = 221 mm2 Use T10 – 300
M y < M x , so same provision, despite the slight reduction of effective depth.

Worked Example 4.3 – Two Ways Slab (3 sides supported)

A two-way slab with the following design data :
(iii) Concrete grade : 35;
(iv) Slab thickness : 200 mm
(v) Span : Long way : 5 m, Short way, 4 m
(iv) Fire rating : 1 hour, mild exposure, cover = 25mm;

5m

free edge

4m

simply supported edge
continuous
edge

Figure 4.8 – Plan of 3-sides supported slab for Worked Example 4.3

58


Loading : D.L. O.W. 0.2 × 24 = 4.8kN/m2
Fin. 1.0 kN/m2
Sum 5.8 kN/m2
L.L. 4.0 kN/m2

From Table 1.38 of “Tables for the Analysis of Plates, Slabs and Diaphragms
based on Elastic Theory” where γ = 4 / 5 = 0.8 , the sagging bending moment
coefficient for short way span is maximum at mid-span of the free edge which
0.1104 (linear interpolation between γ = 0.75 and γ = 1.0 ). The coefficients
relevant to this example are interpolated and listed in Appendix E.

M x = 0.1104 × 14.52 × 4 2 = 25.65 kNm/m

M 25.65 × 10 6
K= = = 0.0254
f cu bd 2 35 × 1000 × 170 2
z K
= 0.5 + 0.25 − = 0.971 > 0.95
d 0 .9
Ast M 25.65 × 10 6
= = = 0.233% > 0.13%
bd bd 2 × 0.87 f y z / d 1000 × 170 2 × 0.87 × 460 × 0.95
Ast = 0.233 ÷ 100 × 1000 × 170 = 397 mm2 Use T10 – 175

At 2 m and 4 m from the free edge, the sagging moment reduces to
0.0844 × 14.52 × 4 2 = 19.608 kNm/m and 0.0415 × 14.52 × 4 2 = 9.64 kNm/m
and Ast required are reduces to 303 mm2 and 149 mm2.
Use T10 – 250 and T10 – 300 respectively.

The maximum hogging moment (bending along long-way of the slab) is at
mid-way along the supported edge of the short-way span
M y = 0.0729 × 14.52 × 5 2 = 26.46 kNm/m

M 26.46 × 10 6
K= = = 0.0262
f cu bd 2 35 × 1000 × 170 2
z K
= 0.5 + 0.25 − = 0.97 > 0.95
d 0 .9
Ast M 26.46 × 10 6
= = = 0.241% > 0.13%
bd bd 2 × 0.87 f y z / d 1000 × 170 2 × 0.87 × 460 × 0.95
2
Ast = 0.241 ÷ 100 × 1000 × 170 = 409 mm Use T10 – 175

59


The maximum sagging moment along the long-way direction is at 2 m from
the free edge which is
M y = 0.0188 ×14.52 × 5 2 = 6.82 kNm/m. The moment is small.
Use T10 – 300

Back-check compliance of effective span ratio (Re Tables 7.3 and 7.4 of the
Code) by considering only the short span which is simply supported,

2 f y Ast ,req 1 2 × 460 × 397 1
fs = × = × = 271 N/mm2;
3 Ast , prov βb 3 × 449 1

The modification factor (Table 7.4) is

0.55 +
(477 − f s ) = 0.55 + (477 − 271) = 1.51
 M  120(0.9 + 0.0254 × 35)
120 0.9 + 2 
 bd 
4000
Allowable effective span depth ratio is 1.51× 20 = 30.2 > = 23.5 . O.K.
170

Finally the reinforcement arrangement on the slab is (Detailed curtailment, top
support reinforcements at simple supports (0.5As) omitted for clarity.)
T10 – 300 T1

1200

800
T10 – 300B2
T10 – 175T1

800

T10 – 300 B1 T10 – 250 B1 T10 – 175B1 1200

1000 2000 2000

Figure 4.9 – Reinforcement Details for Worked Example 4.3

Worked Example 4.4 – Flat Slab by Simplified Method (Cl. 6.1.5.2(g))

60


Flat slab arrangement on rectangular column grid of 7.5 m and 6 m as shown
in Figure 4.10 with the following design data :
(i) Finish Load = 1.5 kPa
(ii) Live Load = 5.0 kPa.
(iii) Column size = 550 × 550
(iv) Column Drop size = 3000 × 3000 with dh = 200 mm
(v) Fire rating : 1 hour, mild exposure, cover = 25 mm
(vi) Concrete grade 35;

As the number of panel is more than 3 and of equal span, the simplified
method for determining moments in accordance with Cl. 6.1.5.2(g) of the
Code is applicable and is adopted in the following analysis.

Effective dimension of column head l h max = l c + 2(d h − 40 ) (Ceqn 6.37)
= 550 + 2(200 − 40 ) = 870 mm
Effective diameter of column head (Cl. 6.1.5.1(c))
870 2 × 4 1
hc = = 982 mm < × 6000 = 1500 mm
π 4

lho=3000
7500

Lhmax=870

250 dh=200

1500
7500
1500
lc=550 1500

Column head details 1500

6000 6000 6000

Figure 4.10 – Flat Slab Plan Layout for Worked Example 4.4

In the simplified method, the flat slab is effectively divided into (i) “column
strips” containing the columns and the strips of the linking slabs and of “strip
widths” equal to the widths of the column drops; and (ii) the “middle strips”

61


between the “column strips”. (Re Figure 6.9 of the Code.) They are designed
as beams in flexural design with assumed apportionment of moments among
the strips. However, for shear checking, punching shears along successive
“critical” perimeters of column are carried out instead.

Sizing : Based on the same limiting span depth ratio for one way and two way
slab which is 26 × 1.15 = 30 (by Table 7.3 and Table 7.4 of the Code,
6000
assuming modification by tensile reinforcement to be 1.15), d = = 200 .
30
As cover = 25 mm, assuming T12 bars, structural depth should at least be
200 + 25 + 12 ÷ 2 = 231 mm, so use structural depth of slab of 250 mm.

Loading : D.L. O.W. 0.25 × 24 = 6.0kN/m2
Fin. 1.5 kN/m2
Total 7.5 kN/m2
L.L. 5.0 kN/m2

Design against Flexure (Long Way)

The bending moment and shear force coefficients in Table 6.4 will be used as
per Cl. 6.1.5.2(g) of the Code. Total design ultimate load on the full width of
panel between adjacent bay centre lines is F = 18.5 × 7.5 × 6 = 832.5 kN. Thus
the reduction to support moment for design, as allowed by Cl. 6.1.5.2(g) of the
Code, is 0.15Fhc = 0.15 × 832.5 × 0.982 = 122.63 kNm for internal support and
0.15Fhc = 0.15 × 832.5 × 0.982 / 2 = 61.32 kNm for outer support.

The design moment at supports are :

Total moment at outer support is 0.04 × 832.5 × 7.5 = 249.75 kNm, which can
be reduced to 249.75 − 61.32 = 188.43 kNm;

Total moment at first interior support is 0.086 × 832.5 × 7.5 = 536.96 kNm,
which can be reduced to 536.96 − 122.63 = 414.33 kNm

Total moment at interior support is 0.063 × 832.5 × 7.5 = 393.36 kNm, which
can be reduced to 393.36 − 122.63 = 270.73 kNm

62


The flat slab is divided into column and mid strips in accordance with Figure
6.9 of the Code which is reproduced as Figure 4.11 in this Manual.

middle strip
column column
ly – lx/2
strip strip

lx/4
column
strip
lx/4

middle
strip lx

lx/4
column
strip
lx/4

lx/4 lx/4 ly lx/4 lx/4

lx is the shorter Flat slab without drop
span whilst ly is
the longer span
Column strip = drop size
Ignore drop if dimension < lx/3

Drop
middle strip
– drop size
lx
Column strip
= drop size
Ignore drop if
dimension <
lx/3

Drop

Flat slab with drop

Figure 4.11 – Division of panels

The total support moments as arrived for the whole panel are to be
apportioned to the middle and column strips with the percentages of 75% and

63


25% respectively as per Table 6.10 of the Code,
Column Strip (75%) Mid Strip (25%)
Total Mt Mt/width Total Mt Mt/width
Outer Support 141.32 47.11 47.11 15.70
st
1 interior support 310.75 103.58 103.58 34.53
Middle interior support 203.05 67.68 67.68 22.56

The reinforcements – top steel are worked out as follows, (minimum of 0.13%
in brackets) ( d = 450 − 25 − 6 = 419 over column support and
d = 250 − 25 − 6 = 219 in other locations)
2
Area (mm )/m Steel Area (mm2)/m Steel
Outer Support 281 (548) T12 – 200 178 (548) T12 – 200
st
1 interior support 618 T12 – 150 392 (548) T12 – 200
Middle interior support 404 (548) T12 – 200 256 (548) T12 – 200

Sagging Moment :
Total moment near middle of end span is 0.075 × 832.5 × 7.5 = 468.28 kNm
Total moment near middle of interior span 0.063 × 832.5 × 7.5 = 393.36 kNm
These moments are to be apportioned in the column and mid strips in
accordance with the percentages of 55% and 45% respectively as per Table
6.10, i.e.
Total Mt Mt/width Total Mt Mt/width
Middle of end span 257.55 85.85 210.73 70.24
Middle of interior span 216.35 72.12 177.01 59.00
The reinforcements – bottom steel are worked out as follows :
Area (mm2)/m Steel Area (mm2)/m Steel
Middle of end span 980 T12 – 100 801 T12 – 125
Middle of interior span 823 T12 – 125 673 T12 – 150
Design in the short way direction can be carried out similarly.

Design against Shear

Design of shear should be in accordance with Cl. 6.1.5.6 of the Code which is
against punching shear by column. For the internal column support, in the

64


absence of frame analysis, the shear for design will be Veff = 1.15Vt where Vt

is the design shear transferred to column calculated on the assumption of all
adjacent panels being fully loaded by Cl. 6.1.5.6(b) of the Code.

Vt = 7.5 × 6 × 18.5 = 832.5 kN; Veff = 1.15Vt = 957.38 kN

Check on column perimeter as per Cl. 6.1.5.6(d) of the Code :
Veff 957.38 × 10 3
≤ 0.8 f cu or 7 = 1.04 ≤ 0.8 f cu = 4.73 MPa; O.K.
ud (4 × 550) × 419
Check on 1st critical perimeter – 1.5d from column face, i.e.
1.5 × 0.419 = 0.6285 . So side length of the perimeter is
(550 + 628.5 × 2) = 1807 mm
Length of perimeter is 4 ×1807 = 7228 mm
Shear force to be checked can be the maximum shear 957.38 kN after
deduction of the loads within the critical perimeter which is
957.38 − 18.5 ×1.807 2 = 896.97 kN
896.97 × 10 3
Shear stress = = 0.296 N/mm2. < vc = 0.48 N/mm2 in accordance
7228 × 419

with Table 6.3 ( 0.43 × (35 / 25)
1/ 3
= 0.48 ). No shear reinforcement is required.

No checking on further perimeter is required.

Worked Example 4.5 – Design for shear reinforcement
(Ceqn 6.44) and (Ceqn 6.45) of the Code gives formulae for reinforcement
design for different ranges of values of v .
(v − v )ud
For v ≤ 1.6vc , ∑ Asv sin α ≥ 0.87c f
y

5(0.7v − vc )ud
For 1.6vc < v ≤ 2.0vc , ∑A sv sin α ≥
0.87 f y
As a demonstration, if v = 0.7 N/mm2 in the first critical perimeter which is <
1.6vc = 0.77 N/mm2 but > vc = 0.48 N/mm2 in the Example 4.4. By Table 6.8
of the Code, as v − vc < 0.4 , vr = 0.4
vr ud 0.4 × 7228 × 419
∑ Asv sin α ≥ 0.87 f = 0.87 × 460 = 3027 mm2. If vertical links is chosen
y

as shear reinforcement, α = 90 0 ⇒ sin α = 1 . So the 3027 mm2 should be
distributed within the critical perimeter as shown in Figure 4.12.

In distributing the shear links within the critical perimeter, there are
recommendations in Cl. 6.1.5.7(f) of the Code that

65


(i) at least two rows of links should be used;
(ii) the first perimeter should be located at approximately 0.5d from the
face of the loaded area (i.e. the column in this case) and should contain
not less than 40% of the calculated area of reinforcements.

So the first row be determined at 200 mm from the column face with total row
length 950 × 4 = 3800 . Using T10 – 225 spacing along the row, the total steel
( )
area will be 10 2 / 4 π × 3800 / 225 = 1326 mm2 > 40% of 3027 mm2.
The second row be at further 300 mm (≤ 0.75d = 314) away where row length
is 1550 × 4 = 6200 . Again using T10 – 225 spacing along the row, the total
( )
steel area will be 10 2 / 4 π × 6200 / 225 = 2164 mm2 > 60% of 3027 mm2.

Total steel area is 1326 + 2164 = 3490 mm2 for shear. The arrangement is
illustrated in Figure 4.12.

1st row , using
T10 – 225 (or 17 nos. of T10)
(area = 1335mm2)
628.5 >40% of 3027 mm2

2nd row , using
T10 – 225 (or 27 nos. of T10)
(area = 2120mm2)
550
Note : It should be noted
that the link spacings as
arrived in this Example
are for demonstration
628.5 purpose. In actual
practice, they should
match with the
longitudinal bar spacing.
275 200 300
200 ≈ 0.5d
1st critical
300 ≤ 0.75d perimeter

T10 – 225 link

1.5d = 628.5

Figure 4.12 – Shear links arrangement in Flat Slab for Worked Example 4.5

66


Design for Shear when ultimate shear stress exceeds 1.6vc

It is stated in (Ceqn 6.43) in Cl. 6.1.5.7(e) that if 1.6 c < v ≤ 2.0vc ,
5(0.7v − v )ud
∑ Asv sin α ≥ 0.87 f c which effectively reduces the full inclusion of
y

vc for reduction to find the “residual shear to be taken up by steel” at
v = 1.6vc to zero inclusion at v = 2.0vc .

Worked Example 4.6 – when 1.6vc < v ≤ 2.0vc

In the previous Example 4.5, if the shear stress v = 0.85 N/mm2 which lies
between 1.6vc = 1.6 × 0.48 = 0.77 N/mm2 and 2.0vc = 2 × 0.48 = 0.96 N/mm2
5(0.7v − v )ud 5(0.7 × 0.85 − 0.48)× 7228 × 419
∑ Asv sin α ≥ 0.87 f c = 0.87 × 460
= 4351 mm2. If
y

arranged in two rows as in Figure 4.12, use T12 – 225 for both rows : the inner
( )
row gives 12 2 / 4 π × 3800 / 225 = 1908 mm2 > 40% of 4351 mm2; the outer
( )
row gives 12 2 / 4 π × 6200 / 225 = 3114 mm2 . The total area is 1908 + 3114 =
5022mm2 > 4351 mm2.

Cl. 6.1.5.7(e) of the Code says, “When v > 2vc and a reinforcing system is
provided to increase the shear resistance, justification should be provided to
demonstrate the validity of design.” If no sound justification, the structural
sizes need be revised.

67


5.0 Columns

5.1 Slenderness of Columns

Columns are classified as short and slender columns in accordance with their
“slenderness”. Short columns are those with ratios l ex / h and l ey / b < 15
(braced) and 10 (unbraced) in accordance with Cl. 6.2.1.1(b) of the Code
where l ex and l ey are the “effective lengths” of the column about the major
and minor axes, b and h are the width and depth of the column.

As defined in Cl. 6.2.1.1 of the Code, a column may be considered braced in a
given plane if lateral stability to the structure as a whole is provided by walls
or bracing or buttressing designed to resist all lateral forces in that plane. It
would otherwise be considered as unbraced.

The effective length is given by (Ceqn 6.46) of the Code as
l e = β ⋅ l 0 where l 0 is the clear height of the column between restraints and
the value β is given by Tables 6.11 and 6.12 of the Code which measures the
restraints against rotation and lateral movements at the ends of the column.

Generally slenderness limits for column : l 0 / b ≤ 60 as per Cl. 6.2.1.1(f) of
100b 2
the Code. In addition, for cantilever column l 0 = ≤ 60b .
h

Worked Example 5.1 : a braced column of clear height l 0 = 8 m and sectional
dimensions b = 400 mm, h = 550 mm with its lower end connected
monolithically to a thick cap and the upper end connected monolithically to
deep transfer beams in the plane perpendicular to the major direction but beam
of size 300(W) by 400(D) in the other direction.

By Tables 6.11 and 6.12 of the Code
Lower end condition in both directions : 1
Upper end condition about the major axis : 1
Upper end condition about the minor axis : 2

For bending about the major axis : end conditions 1 – 1, β x = 0.75 ,
l ex = 0.75 × 8 = 6
l ex / 550 = 10.91 < 15 . ∴ a short column.
For bending about the minor axis : end conditions 1 – 2, β y = 0.8 ,

68


l ey = 0.8 × 8 = 6.4
l ey / 400 = 16 > 15 ∴ a slender column. l ey / 400 = 16 < 60 , O.K.

For a slender column, an additional “deflection induced moment” M add will
be required to be incorporated in design, as in addition to the working
moment.

5.2 Design Moments and Axial Loads on Columns

5.2.1 Determination of Design moments and Axial Loads by sub-frame Analysis

Generally design moments, axial loads and shear forces on columns are that
obtained from structural analysis. In the absence of rigorous analysis, (i)
design axial load may be obtained by the simple tributary area method with
beams considered to be simply supported on the column; and (ii) moment may
be obtained by simplified sub-frame analysis as illustrated in Figure 5.1 :

Ku Ku

1.4Gk+1.6Q 1.0Gk 1.4Gk+1.6Q

Kb K b2 K b1

KL
M e Ku KL M es K u
Mu = Mu =
K L + K u + 0.5K b K L + K u + 0.5 K b1 + 0.5 K b 2
MeKL M es K L
ML = ML =
K L + K u + 0.5 K b K L + K u + 0.5 K b1 + 0.5 K b 2

Symbols:
M e : Beam Fixed End Moment. K u : Upper Column Stiffness
M es : Total out of balance Beam Fixed End Moment. K L : Upper Column Stiffness
M u : Upper Column Design Moment K b1 : Beam 1 Stiffness
M L : Upper Column Design Moment K b 2 : Beam 2 Stiffness

Figure 5.1 – Diagrammatic illustration of determination of column design
moments by Simplified Sub-frame Analysis

Worked Example 5.2 (Re Column C1 in Plan shown in Figure 5.2)

Design Data :
Slab thickness : 150 mm Finish Load : 1.5 kN/m2

69


Live Load : 5 kN/m2 Beam size : 550(D) × 400(W)
Upper Column height : 3 m Lower Column Height : 4 m
Column size : 400(W) × 600(L)
Column Load from floors above D.L. 443 kN L.L. 129 kN

B3
5m
B4

C1

B1 B2

4m 3m 3m

Figure 5.2 – Plan for illustration for determination of design axial load and
moment on column by the Simplified Sub-frame Method

Design for Column C1 beneath the floor

Check for slenderness : As per Cl. 6.2.1.1(e) of the Code, the end conditions of
the column about the major and minor axes are respectively 2 and 1 at the
upper end and 1 at the lower end for both axes (fixed on pile cap). The clear
height between restraints is 4000 − 550 = 3450 . The effective heights of the
column about the major and minor axes are respectively 0.8 × 3.45 = 2.76 m
and 0.75 × 3.45 = 2.59 m. So the slenderness ratios about the major and minor
2760 2590
axes are = 4.6 < 15 and = 6.475 < 15 . Thus the column is not
600 400
slender in both directions.

Loads :
Slab: D.L. O.W. 0.15 × 24 = 3.6 kN/m2
Fin. 1.5 kN/m2
5.1 kN/m2
L.L. 5.0 kN/m2

Beam B1 D.L. O.W. 0.4 × 0.55 × 24 × 4 = 21.12 kN

70


End shear of B1 on C1 is D.L. 21.12 ÷ 2 = 10.56 kN

Beam B3 D.L. O.W. 0.4 × (0.55 − 0.15) × 24 = 3.84 kN/m
Slab 5.1 × 3.5 = 17.85 kN/m
21.69 kN/m
L.L. Slab 5.0 × 3.5 = 17.5 kN/m
End shear of B3 on C1 D.L. 21.69 × 5 ÷ 2 = 54.23 kN
L.L. 17.50 × 5 ÷ 2 = 43.75 kN

Beam B4 D.L. O.W. 0.4 × (0.55 − 0.15) × 24 = 3.84 kN/m
Slab 5.1 × 3 = 15.3 kN/m
19.14 kN/m
L.L. Slab 5.0 × 3 = 15.0 kN/m
End shear of B4 on B2 D.L. 19.14 × 5 ÷ 2 = 47.85 kN
L.L. 15.00 × 5 ÷ 2 = 37.50 kN

Beam B2 D.L. O.W. 0.4 × 0.55 × 24 × 6 = 31.68 kN
B4 47.85 kN
79.53 kN
L.L. B4 37.50 kN
End shear of B2 on C1 , D.L. 79.53 ÷ 2 = 39.77 kN
L.L. 37.5 ÷ 2 = 18.75 kN

Total D.L. on C1 O.W. 0.4 × 0.6 × 24 × 4 = 23.04 kN
B1 + B2 + B3 10.56 + 39.77 + 54.23 = 104.56 kN
Floor above 443.00 kN
Sum 570.60 kN

Total L.L. on C1 B1 + B2 +B3 0 + 18.75 + 43.75 = 62.5 kN
Floor above 129.00 kN
Sum 191.50 kN

So the factored axial load on the lower column is
1.4 × 570.6 + 1.6 × 191.5 = 1105.24 kN

Factored fixed end moment bending about the major axis (by Beam B3 alone):
1
Me = × (1.4 × 21.69 + 1.6 × 17.5) × 5 2 = 121.6 kNm
12

71


Factored fixed end moment bending about the minor axis by Beam B2:
1 1  1 
M eb 2 = 1.4 ×  × 31.68 + × 47.85  × 6 + 1.6 ×  × 37.5  × 6 = 95.24 kNm
 12 8  8 
Factored fixed end moment bending about the minor axis by Beam B1:
1 
M eb 2 = 1.0 ×  × 21.12  × 4 = 7.04 kNm
 12 
So the unbalanced fixed moment bending about the minor axis is
95.24 − 7.04 = 88.2 kNm

The moment of inertia of the column section about the major and minor axes
0.4 × 0.6 3 0.6 × 0.4 3
are I cx = = 0.0072 m4, I cy = = 0.0032 m4
12 12
The stiffnesses of the upper and lower columns about the major axis are :
4 EI cx 4 E × 0.0072
K ux = = = 0.0096 E
Lu 3
4 EI cx 4 E × 0.0072
K Lx = = = 0.0072 E
LL 4
The stiffnesses of the upper and lower columns about the minor axis are :
4 EI cy 4 E × 0.0032
K uy = = = 0.004267 E
Lu 3
4 EI cy 4 E × 0.0032
K Ly = = = 0.0032 E
LL 4
The moment of inertia of the beams B1, B2 and B3 are
0.4 × 0.55 3
= 0.005546 m4
12
The stiffness of the beams B1, B2 and B3 are respectively
4 E × 0.05546 4 E × 0.05546
= 0.005546 E ; = 0.003697 E ; and
4 6
4 E × 0.05546
= 0.004437 E
5

Distributed moment on the lower column about the major axis is
M ex K Lx 121.6 × 0.0072 E
M Lx = =
K ux + K Lx + 0.5 K b 3 0.0096 E + 0.0072 E + 0.5 × 0.004437 E
= 46.03 kNm
Distributed moment on the lower column about the minor axis is

M ey K Ly
M Ly =
K uy + K Ly + 0.5(K b1 + K b 2 )

72


88.2 × 0.0032 E
= = 23.35 kNm
0.004267 E + 0.0032 E + 0.5 × (0.005546 + 0.003697 )E

So the lower column should be checked for the factored axial load of
1105.24kN, factored moment of 46.03 kNm about the major axis and factored
moment of 23.35 kNm about the minor axis.

5.2.2 Minimum Eccentricity

A column section should be designed for the minimum eccentricity equal to
the lesser of 20 mm and 0.05 times the overall dimension of the column in the
plane of bending under consideration. Consider Worked Example in 5.2, the
minimum eccentricity about the major axis is 20 mm as
0.05 × 600 = 30 > 20 mm and that of the minor axis is 0.05 × 400 = 20 mm. So
the minimum eccentric moments to be designed for about the major and minor
axes are both 1105.24 × 0.02 = 22.1 kNm. As they are both less than the
design moment of 46.03 kNm and 23.35 kNm, they can be ignored.

5.2.3 Check for Slenderness

In addition to the factored load and moment as discussed in 5.2.1, it is required
by Cl. 6.2.1.3 of the Code to design for an additional moment M add if the
column is found to be slender by Cl. 6.2.1.1. The arrival of M add is an
eccentric moment created by the ultimate axial load N multiplied by a
pre-determined lateral deflection au in the column as indicated by the
following equations of the Code.

M add = Nau (Ceqn 6.52)
au = β a Kh (Ceqn 6.48)
2
1  le 
βa =   (Ceqn 6.51)
2000  b 
N uz − N
K= ≤ 1 (conservatively taken as 1) (Ceqn 6.49)
N uz − N bal
or by N uz = 0.45 f cu Anc + 0.87 f y Asc (Ceqn 6.50)
N bal = 0.25 f cu bd

Final design moment M t will therefore be the greatest of
(1) M 2 , the greater initial end moment due to design ultimate load;

73


(2) M i + M add where M i = 0.4M 1 + 0.6M 2 ≥ 0.4M 2 (with M2 as
positive and M 1 negative.)
(3) M 1 + M add / 2
in which M 1 is the smaller initial end moment due to design ultimate
load.
(4) N × emin (discussed in Section 5.2.2 of this Manual)
where the relationship between M 1 , M 2 , M add and the arrival of the
critical combination of design moments due to M add are illustrated in Figure
5.3 reproduced from Figure 6.16 of the Code.

End Initial moment Additional Design moment
conditions of (from analysis) moment envelope
column

+ M add =

M2

+ =
Mi M add M i + M add

Larger
moment M 2

Mi
M add M i + M add

Smaller
moment M1 0.5M add M1 + 0.5M add

Figure 5.3 – Braced slender columns

In addition to the above, the followings should be observed in the
determination of M t as the enveloping moment of the 4 cases described in
the previous paragraph (Re Cl. 6.2.1.3 of the Code) :

(i) In case of biaxial bending (moment significant in two directions), M t

74


should be applied separately for the major and minor directions with
b in Table 6.13 of the Code be taken as h , the dimension of the
column in the plane considered for bending. Re Ceqn 6.48;
(ii) In case of uniaxial bending about the major axis where l e / h ≤ 20 and
h < 3b , M t should be applied only in the major axis;
(iii) In case of uniaxial bending about the major axis only where either
l e / h ≤ 20 or h < 3b is not satisfied, the column should be designed
as biaxially bent, with zero M i in the minor axis;
(iv) In case of uniaxial bending about the minor axis, M add obviously be
applied only in the minor axis only.

Worked Example 5.3 :

A slender braced column of grade 35, cross sections
b = 400 , h = 500
l ex = l ey = 8 m, N = 1500 kN

(i) Moment due to ultimate load about the major axis only, the greater and
smaller bending moments due to ultimate load are respectively
M 2 x = 153 kNm and M 1x = 96 kNm

As l ex / h = 16 ≤ 20 ; h = 500 < 3b = 1200
So needs to check for additional bending in the major axis but with
M add based on the minor axis.
Take K = 1
2 2
1  le  1  8000 
βa =   =   = 0 .2
2000  b  2000  400 

au = β a Kh = 0.2 × 1 × 0.5 = 0.1
M addx = Na u = 1500 × 0.1 = 150
M ix = 0.4M 1 + 0.6M 2 = 0.4(− 96 ) + 0.6 × 153 = 53.4 < 0.4M 2 = 0.4 × 153 = 61.2

The design moment about the major axis will be the greatest of :
(1) M 2 x = 153
(2) M ix + M addx = 61.2 + 150 = 211.2
(3) M 1x + M addx / 2 = 96 + 150 / 2 = 171
(4) N × emin = 1500 × 0.02 = 30 as emin = 20 < 0.05 × 500 = 25
So the greatest design moment is case (2) M ix + M addx = 211.2
Thus the section needs only be checked for uniaxial bending with
N = 1500 kN and M x = 211.2 kNm bending about the major axis.

75


(ii) Moments due to ultimate loads about the minor axis only, the greater and
smaller moments are identical in magnitudes to that in (i), but about the
minor axis, repeating the procedure :

M 2 y = 143 kNm and M 1 y = 79 kNm

As l ex / h = 16 ≤ 20 ; h = 500 < 3b = 1200
So needs to checked for additional bending in the major axis.
Take K = 1
2 2
1  le  1  8000 
βa =   =   = 0.2
2000  b  2000  400 
a u = β a Kh = 0.2 × 1× 0.4 = 0.08
M addy = Na u = 1500 × 0.08 = 120
M iy = 0.4M 1 y + 0.6M 2 y = 0.4 × (− 79) + 0.6 × 143 = 54.2 < 0.4M 2 y = 0.4 × 143 = 57.2

The design moment will be the greatest of :
(1) M 2 y = 143
(2) M iy + M addy = 57.2 + 120 = 177.2
(3) M 1 y + M addy / 2 = 79 + 120 / 2 = 139
(4) N × emin = 1500 × 0.02 = 30 as e min = 20 ≤ 0.05 × 400 = 20
So the greatest design moment is case (2) M iy + M addy = 177.2
Thus the section need only be checked for uniaxial bending with
N = 1500 kN and M y = 177.2 kNm bending about the minor axis.

(iii) Biaxial Bending, there are also moments of M 2 x = 153 kNm and
M 1x = 96 kNm; M 2 y = 143 kNm and M 1 y = 79 kNm. By Cl. 6.1.2.3(f),
M add about the major axis will be revised as follows :

Bending about the major axis :
2 2
1  le  1  8000 
βa =   =   = 0.128
2000  h  2000  500 

au = β a Kh = 0.128 × 1 × 0.5 = 0.064
M addx = Na u = 1500 × 0.064 = 96 kNm.
Thus items (2) and (3) in (i) are revised as
(2) M ix + M addx = 61.2 + 96 = 157.2
(3) M 1x + M addx / 2 = 96 + 96 / 2 = 144
So the moment about major axis for design is 157.2 kNm

76


Bending about the minor axis :

M 2 y = 143 kNm and M 1 y = 79 kNm; same as (ii);

Thus the ultimate design moment about the major axis is 157.2 kNm and
that about the minor axis is 177.2 kNm.


A slender braced column of grade 35, cross section
b = 400 , h = 1200
l ex = l ey = 8 m, N = 1500 kN, M 2 x = 153 kNm and M 1x = 96 kNm

As 3b = h , Cl. 6.2.1.3(e) should be used.
Take K = 1
2 2
1  le  1  8000 
βa =   =   = 0 .2
2000  b  2000  400 

a u = β a Kb = 0.2 ×1× 0.4 = 0.08 m > 20 mm

M addy = Na u = 1500 × 0.08 = 120

So the minor axis moment is 120 kNm
l e 8000
As = = 6.67 , the column is not slender about the major axis.
h 1200
So the major axis moment is simply 153 kNm.

5.3 Sectional Design

Generally the sectional design of column utilizes both the strengths of
concrete and steel in the column section in accordance with stress strain
relationship of concrete and steel in Figures 3.8 and 3.9 of the Code
respectively. Alternatively, the simplified stress block for concrete in Figure
6.1 of the Code can also be used.

5.3.1 Design for Axial Load only

(Ceqn 6.55) of the Code can be used which is

77


N = 0.4 f cu Ac + 0.75 Asc f y . The equation is particularly useful for a column

which cannot be subject to significant moments in such case as the column
supporting a rigid structure or very deep beams. There is a reduction of
approximately 10% in the axial load carrying capacity as compared with the

normal value of 0.45 f cu Ac + 0.87 Asc f y accounting for the eccentricity of

0.05h .

Furthermore, (Ceqn 6.56) reading N = 0.35 f cu Ac + 0.67 Asc f y which is

applicable to columns supporting an approximately symmetrical arrangement
of beams where (i) beams are designed for u.d.l.; and (ii) the beam spans do
not differ by more than 15% of the longer. The further reduction is to account
for extra moment arising from asymmetrical loading.

5.3.2 Design for Axial Load and Biaxial Bending :

The general section design of a column is accounted for the axial loads and
biaxial bending moments acting on the section. Nevertheless, the Code has
reduced biaxial bending into uniaxial bending in design. The procedure for

determination of the design moment, either M x ' or M y ' bending about the

major or minor axes is as follows :
b
Determine b' and h' as defined by the
diagram. In case there are more than one Mx
row of bars, b' and h' can be measured to
the centre of the group of bars. h' h

Mx My
(i) Compare and .
h' b'
Mx M y h'
If ≥ , use M x ' = M x + β M y My
h' b' b' b'
Mx M y b'
If < , use M y ' = M y + β M x
h' b' h'
where β is to be determined from Table 5.1 which is reproduced
N
from Table 6.14 of the Code under the pre-determined ;
bhf cu

78


N / (bhf cu ) 0 0.1 0.2 0.3 0.4 0.5 ≥ 0 .6
β 1.00 0.88 0.77 0.65 0.53 0.42 0.30
Table 5.1 – Values of the coefficients β

(ii) The M x ' or M y ' will be used for design by treating the section as
either (a) resisting axial load N and moment M x ' bending about
major axis; or (b) resisting axial load N and moment M y ' bending
about minor axis as appropriate.h

5.3.3 Concrete Stress Strain Curve and Design Charts

The stress strain curve for column section design is in accordance with Figure
3.8 of the Code. It should be noted that Amendment No. 1 has revised the
1.34( f cu / γ m )
Figure by shifting ε 0 to . With this revision, the detailed
Ec
design formulae and design charts have been formulated and enclosed in
Appendix F. Apart from the derivation for the normal 4-bar column, the
derivation in Appendix F has also included steel reinforcements uniformly
distributed along the side of the column idealized as continuum of
reinforcements with symbol Ash . The new inclusion has allowed more
accurate determination of load carrying capacity of column with many bars
along the side as illustrated in Figure 5.4 which is particularly useful for
columns of large cross sections. The user can still choose to lump the side
reinforcements into the 4 corner bars, with correction to the effective depth as
in conventional design by setting Ash = 0 in the derived formulae.

idealized as

continuum of steel
“strip” with area
equivalent to the row
of bars

Figure 5.4 – Idealization of steel reinforcements in large column

Figure 5.5 shows the difference between the 2 idealization. It can be seen that
the continuum idealization is more economical generally except at the peak
moment portion where the 4 bar column idealization shows slight over-design.

79


Comparison of Load Carrying Capacities of Rectangular Shear Walls with Uniform Vertical
Reinforcements Idealized as 4 bar column with d/h = 0.75 and Continuum to the Structural Use of
Concrete 2004 Concrete Grade 35
50
0.4% steel as 4 bar column
0.4% steel as continuum
45
1% steel as 4 bar column
1% steel as continuum
40
35
30
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5
2 2
M/bh N/mm

Figure 5.5 – Comparison between Continuum and 4-bar column Idealization


Consider a column of sectional size b = 400 mm, h = 600 mm, concrete
grade 35 and under an axial load and moments of

N = 4000 kN, M x = 250 kNm, M y = 150 kNm,

cover to longitudinal reinforcements = 40 mm
Assume a 4-bar column and T40 bars, h' = 600 − 40 − 20 = 540 mm;
b' = 400 − 40 − 20 = 340 mm;
N 4000000
= = 0.476 ;
f cu bh 35 × 400 × 600
β = 0.446 from Table 5.1 or Table 6.14 of the Code;
Mx My
= 0.463 > = 0.441 ;
h' b'
h' 540
∴M x '= M x + β M y = 250 + 0.446 × × 150 = 356 kNm
b' 340
N M 356 × 10 6 d 540
= 16.67 ; = = 2.47 ; = = 0.9
bh bh 2 400 × 600 2 h 600

Use Chart F-9 in Appendix F as extracted in Figure 5.6, 1.8% steel is
approximated which amounts to 400 × 600 × 0.018 = 4320 mm2, or 6-T32
(Steel provided is 4826mm2) The section design is also shown in Figure 5.6,

80


with two additional T20 bars to avoid wide bar spacing.

Design Chart of Rectangular Column to Code of Practice for Structural Use of Concrete 2004
Concrete Grade 35, 4-bar column, d/h = 0.9
50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15
2 2
M/bh N/mm

T32

Extra T20

Figure 5.6 – Design Chart and Worked Re-bar details for Worked Example 5.5


Consider a column of sectional size b = 800 mm, h = 1000 mm, concrete
grade 40 and under an axial load and moments

N = 14400 kN, M x = 2000 kNm, M y = 1500 kNm,

concrete cover to longitudinal reinforcement = 40 mm;
Approximation as a 4-bar column and assume d / h = 0.8
b' = 0.8 × 800 = 640 mm; h' = 0.8 × 1000 = 800 mm;
N 14400000
= = 0.45 ; β = 0.475 from Table 5.1 or Table 6.14 of
f cu bh 40 × 800 × 1000
the Code;

81


M x 2000 M y 1500
= = 2.5 > = = 2.34 ;
h' 800 b' 640
h' 800
∴M x '= M x + β M y = 2000 + 0.475 × × 1500 = 2890.6 kNm
b' 640
N M 2890.6 × 10 6
= 18 ; = = 3.61 ;
bh bh 2 800 × 1000 2
Use Chart F-12 in Appendix F as extracted in Figure 5.7, 3.0% steel is
approximated which amounts to 0.031× 800 ×1000 = 24,800 mm2, or 32-T32
(Steel provided is 25,736mm2) The arrangement of steel bars is also shown in
Figure 5.7. It should be noted that alternate lapping may be required if the
column is contributing in lateral load resisting system as the steel percentage
exceeds 2.6% as per discussion in 5.4(ii) of this Manual.

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
2 2
M/bh N/mm

Centre of mass of the reinforcements in one half
of the section below the centre line is
5 × 434 + 5 × 370 + 4 × 222 + 2 × 74
= 316
16

74 So h' = 316 + 500 = 816
d
148 = 0.816 > 0.8
h
148 d
So the original use of = 0.8 is OK
64 h
56

Figure 5.7 – Chart and Column Section for Worked Example 5.6

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d
The back-calculation in Figure 5.7 has shown that the ratio is the steel bar
h
arrangement is 0.816 which is greater than the original assumed value of 0.8.
So the use of the chart is conservative.

5.3.4 Alternatively, the design of reinforcements can be based on formulae derived
in Appendix F. However, as the algebraic manipulations are very complicated
(may involve solution of 4th polynomial equations) and cases are many, the
approach is practical only by computer methods. Nevertheless, spread sheets
have been prepared and 2 samples are enclosed at the end of Appendix F.

5.3.5 The approach by the previous British Code CP110 is based on interaction
formula by which the moments of resistance in both directions under the axial
loads are determined with the pre-determined reinforcements and the
“interaction formula” is checked. The approach is illustrated in Figure 5.8.

α α
P / bd P  Mx  M 
+
  + y

 ≤1
bd  M ux M 
  uy 

M x / bd 2

M y / b2d

Figure 5.8 – Interaction formula for design of biaxial bending

5.3.6 Direct sectional analysis to Biaxial Bending without the necessity of
converting the biaxial bending problem into a uniaxial bending problem :

Though the Code has provisions for converting the biaxial bending problem
into a uniaxial bending problem by
(i) searching for the controlling bending axis; and
(ii) aggravate the moments about the controlling bending axis as appropriate
to account for the effects of bending in the non-controlling axis;
a designer can actually solve the biaxial bending problem by locating the
orientation and the neutral axis depth (which generally does not align with the
resultant moment except for circular section) of the column section by
balancing axial load and the bending in two directions. Theoretically, by

83


balancing axial load and the 2 bending moments, 3 equations can be obtained
for solution of the neutral axis orientation, neutral axis depth and the required
reinforcement. However the solution process, which is often based on trial and
error approach, will be very tedious and not possible for irregular section
without computer methods. Reinforcements generally need be pre-determined.
Figure 5.9 illustrates the method of solution.
The total sectional resistance of the section
under the stress strain profile resists the
applied axial loads and moments
Neutral axis

Strain profile on section

strain profile
across section

concrete
stress
profile

Figure 5.9 – General Biaxial Bending on irregular section

5.4 Detailing requirements for longitudinal bars in columns (generally by Cl. 9.5
and Cl. 9.9.2.1(a) of the Code, the ductility requirements applicable to columns
contributing in lateral load resisting system are marked with “D”)

(i) Minimum steel percentage based on gross area of a column is 0.8% (Cl.
9.5.1 of the Code);
(ii) Maximum steel based on gross area of a column is (a) 4% except at lap
which can be increased to 5.2% (D) for columns contributing to lateral
load resisting system (Cl. 9.9.2.1(a) of the Code); and (b) 6% without
laps and 10% at laps for other columns (Cl. 9.5.1 of the Code);
(iii) Bar diameter ≥ 12 mm (Cl. 9.5.1 of the Code);
(iv) The minimum number of bars should be 4 in rectangular columns and 6
in circular columns. In columns having a polygonal cross-section, at
least one bar be placed at each corner (Cl. 9.5.1 of the Code);
(v) In any row of longitudinal bars in columns contributing to lateral load

84


resisting system, the smallest bar diameter used shall not be less than 2/3
of the largest bar diameter used (Cl. 9.9.2.1(a) of the Code). For example,
T40 should not be used with T25 and below (D);
(vi) At laps, the sum of reinforcement sizes in a particular layer should not
exceed 40% of the breadth at that section (Cl. 9.5.1 of the Code). The
requirement is identical to that of beam as illustrated by Figure 3.13;
(vii) For columns contributing to lateral load resisting system, where the
longitudinal bars pass through the beams at column beam joints, column

bars shall satisfy φ ≤ 3.2h 0.8 f cu / f y as per Ceqn 9.7 where h is the

beam depth. For grade 35 concrete and based on high yield bar, the
limiting bar diameter is simply φ ≤ 0.0368h , i.e. if beam depth is 600
mm, φ ≤ 22.1 implying maximum bar size is 20 mm. If the column is
not intended to form a plastic hinge, the bar diameter can be increased by
25% (Cl. 9.9.2.1(a) of the Code) (D);
(viii) For columns contributing to lateral load resisting system, where the
longitudinal bars terminate in a joint between columns and foundation
members with possible formation of a plastic hinge in the column, the
anchorage of the column bars into the joint region should commence at
1/2 of the depth of the foundation member or 8 times the bar diameter
from the face at which the bars enter the foundation member. Where a
plastic hinge adjacent to the foundation face cannot be formed,
anchorage can commence at the interface with the foundation (Cl.
9.9.2.1(c) of the Code) as illustrated in Figure 5.10 (D);

Anchorage can
commence at this point if
plastic hinge cannot Anchorage should
generally commence at
occur at the column face
this point

≥0.5D
or 8ø
foundation
D element
L

The bends can be eliminated if
L ≥ anchorage length

Figure 5.10 – Longitudinal Bar anchorage in foundation for columns contributing
to lateral load resisting system

85


(ix) For columns contributing to lateral load resisting system, where the
longitudinal bars anchor into beam (transfer beam or roof beam), in
addition to the requirement in (viii), the bars should not be terminated in
a joint area without a horizontal 90o standard hook or an equivalent
device as near as practically possible to the far side of the beam and not
closer than 3/4 of the depth of the beam to the face of entry. Unless the
column is designed to resist only axial load, the direction of bend must
always be towards the far face of the column (Cl. 9.9.2.1(c) of the Code)
as illustrated in Figures 5.11 and 5.12 (D);

Anchorage should
generally commence at
Anchorage can commence this point
at this point if plastic hinge
cannot occur adjacent to Bar can bend outwards if
beam face column designed for
axial load only

≥0.5D
≥0.75D if bar or 8ø
anchored in Beam
D beam element

Figure 5.11 – Longitudinal Bar anchorage in Beam (Transfer Beam) for columns
contributing to lateral load resisting system

Bar can bend outwards if
column designed for
axial load only
Roof Beam
or transfer
≥0.75D if bar beam
anchored in D
beam ≥0.5D
or 8ø

Anchorage should
Anchorage can commence generally commence at
at this point if plastic hinge this point
cannot occur adjacent to
beam face

Figure 5.12 – Longitudinal Bar anchorage in Beam (Transfer Beam / Roof Beam)
for columns contributing to lateral load resisting system

86


(x) For laps and mechanical couplers in a column contributing to lateral load
resisting system, the centre of the splice must be within the middle
quarter of the storey height of the column unless it can be shown that
plastic hinges cannot develop in the column adjacent to the beam faces.
As per discussion in Section 2.4, such lapping arrangement should be
followed in locations such as column joining at pile caps or thick
structures. Normal lapping at other floors can usually be followed unless
there are very stiff beams, e.g. transfer beams (Cl. 9.9.2.1(d) of the
Code). Examples are illustrated in Figure 5.13 (D);

Consider a column of
storey height 3m,
grade 40 and with
T20, T32 and T40
bars. Half lap lengths
(tension) are
T20 T40 respectively
1125
1.4×32×20÷2=448;
T32 1.4×32×32÷2=716.8;
1.4×32×40÷2=896;

The middle quarter of
3000 the column is at a level
750 of 1125 from the lower
floor as shown. If the
centre of lap is within
the middle quarter, the
≤448 bars need be lapped at
≤716.8 level higher than the
≤896 floor level.
1125
Section where plastic
hinge may develop

Figure 5.13 – Centre of lapping be within middle quarter of floor height in Column

(xi) Full strength welded splices may be used in any location (Cl. 9.9.2.1(d)
of the Code);
(xii) As similar to limitation of lapping of bars in beams as described in
Section 3.6(vii), longitudinal bars in columns contributing to lateral load
resisting system shall not be lapped in a region where reversing stresses

at the ultimate limit state may exceed 0.6 f y in tension or compression

87


unless each lapped bar is confined by adequate links or ties satisfying
(Ceqn 9.6), as explained in Section 3.6(vii) and illustrated by Figure 3.11.
Summing up, lapping should be avoided from region with potential
plastic hinge and with reversing stresses (Cl. 9.9.2.1(a) of the Code) (D);
(xiii) Minimum clear spacing of bars should be the greatest of (1)bar diameter;
(2) 20 mm; and (3) aggregate size + 5 mm (Cl. 8.2 of the Code).

5.5 Detailing Requirements for transverse reinforcements in columns include the
general requirements by Cl. 9.5.2 and the ductility requirements in Cl. 9.9.2.2 of
the Code (marked with “D”) for columns contributing to lateral load resisting
system. Items (i) to (iv) below are requirements for columns not within “critical
regions”. “Critical region” is defined in item (v) and Figure 5.15 for columns
contributing to lateral load resisting system:

(i) Diameter of transverse reinforcements ≥ the greater of 6 mm and 1/4 of
longitudinal bar diameter (Cl. 9.5.2.1 of the Code);
(ii) The spacing of transverse reinforcement shall not exceed 12 times the
diameter of the smallest longitudinal bar (Cl. 9.5.2.1 of the Code);
(iii) For rectangular or polygonal columns, every corner bar and each
alternate bar (or bundle) shall be laterally supported by a link passing
around the bar and having an included angle ≤ 135o. No bar within a
compression zone shall be further than 150 mm from a restrained bar.
Links shall be adequately anchored by hooks through angles ≥ 135o. See
Figure 5.14 which is reproduced from Figure 9.5 of the Code (Cl. 9.5.2.2
of the Code);
(iv) For circular columns, loops or spiral reinforcement satisfying (i) to (ii)
should be provided. Loops (circular links) should be anchored with a
mechanical connection or a welded lap by terminating each end with a
135o hook bent around a longitudinal bar after overlapping the other end
of the loop. Spiral should be anchored either by welding to the previous
turn or by terminating each end with a 135o hook bent around a
longitudinal bar and at not more than 25 mm from the previous turn.
Loops and spirals should not be anchored by straight lapping, which
causes spalling of the concrete cover (Cl. 9.5.2.2 of the Code). The
details are also illustrated in Figure 5.14;

88


≤135o, longitudinal
bar considered to be
restrained

restraining
≥135o bar

≥135o

>135o
≤150
longitudinal bar restraining
not considered link
bar
restrained since anchorage
required
enclosing angle
>135o

Figure 5.14 – Column transverse reinforcements outside “Critical Regions”

(v) Transverse reinforcements in “critical regions” within columns of limited
ductile high strength concrete (contributing to lateral load resisting system)
as defined in Figure 5.15 (Re Cl. 9.9.2.2 of the Code) shall have additional
requirements as :
(a) For rectangular or polygonal columns, each (not only alternate)
longitudinal bar or bundle of bars shall be laterally supported by a
link passing around the bar having an included angle of not more
than 135o. As such, Figure 5.16 shows the longitudinal bar
anchorage requirements in “critical region” (Cl. 9.9.2.2(b) of the
Code) (D);
(b) Spacing ≤ 1/4 of the least lateral column dimension in case of
rectangular or polygonal column and 1/4 of the diameter in case of
a circular column and 6 times the diameter of the longitudinal bar
to be restrained (Cl. 9.9.2.2(b) of the Code) (D);

89


Height of “critical
H, Critical regions
with enhanced
region” , H, depends
transverse on N/Agfcu ratio :
reinforcements
(a) 0<N/Agfcu ≤ 0.1,
x = 0.85
H ≥ hm and
H ≥ h or D
Normal
transverse (b) 0.1 <N/Agfcu ≤ 0.3,
reinforcement x = 0.75
H ≥ hm and
xM max H ≥ 1.5h or 1.5D
(c) 0.3 <N/Agfcu ≤ 0.6,
x = 0.65
H, Critical regions
with enhanced H ≥ hm and
hm
transverse H ≥ 2h or 2D
reinforcements

M max

h b
if rectangular

if circular
D

Figure 5.15 – “Critical Regions (Potential Plastic Hinge Regions)” in Columns

≤135o, longitudinal
bar considered to be
restrained

restraining
≥135o bar

≥135o

>135o

longitudinal bar not
considered restrained link
since enclosing angle anchorage
>135o

Figure 5.16 – Enhanced transverse reinforcements inside “Critical Regions” in
columns contributing to lateral load resisting system

90


Worked Example 5.7 – for determination of “critical regions” within columns of
limited ductile and high strength concrete

Consider a rectangular column of the following details :
Cross section 500× 600 mm; height 3 m; grade 65; re-bars : T32

Loads and moments are as follows :
Axial Load 4875 kN
M x = 800 kNm (at top), M x = 500 kNm (at bottom)

M y = 450 kNm (at top) M y = 300 kNm (at bottom)

N 4875 × 10 3
= = 0.25 ∴ x = 0.75 for determination of critical regions
Ag f cu 500 × 600 × 40

hm for bending about X and Y directions are determined as per Figure 5.17.

800kNm 450kNm
hm=462 hm=450

1846 1800
800×0.75=600kNm 450×0.75=338kNm

500×0.75=375kNm 300×0.75=225kNm
1154 1200
hm=289 hm=300
500kNm 300kNm

Bending about X-X Bending about Y-Y

Figure 5.17 – Determination of critical heights in Worked Example 5.7

As the hm are all less than 1.5h = 1.5×600 = 900, so the critical regions should
then both be 1200 mm from top and bottom and the design of transverse
reinforcements is as indicated in Figure 5.18 :

91


Transverse re-bars

(i) Within critical region (for columns of
T10 @ 125 limited high strength concrete and
900 contributing to lateral load resisting
system only) :
Bar size 0.25×32 = 8 mm > 6 mm
Spacing : the lesser of
0.25×500 = 125 mm
T10 @ 350 6×32 = 192 mm
1200 So spacing is 125 mm

(ii) Within normal region (regardless of
whether the column is contributing to
lateral load resisting system) :
T10 @ 125 Bar size 0.25×32 = 8 mm > 6 mm
900 Spacing 12×32 = 384 mm

Figure 5.18 –Transverse Reinforcement arrangement to Worked Example 5.7

92


6.0 Column-Beam Joints

6.1 General

The design criteria of a column-beam joint comprise (i) performance not inferior
to the adjoining members at serviceability limit state; and (ii) sufficient strength
to resist the worst load combination at ultimate limit state. To be specific, the
aim of design comprise (a) minimization of the risk of concrete cracking and
spalling near the beam-column interface; and (b) checking provisions against
diagonal crushing or splitting of the joint and where necessary, providing
vertical and horizontal shear links within the joint and confinement to the
longitudinal reinforcements of the columns adjacent to the joint.

6.2 The phenomenon of “diagonal splitting” of joint

Diagonal crushing or splitting of column-beam joints is resulted from “shears”
and unbalancing moment acting on the joints as illustrated in Figure 6.1(a) and
6.1(b) which indicate typical loadings acting on the joint. Figure 6.1(a) shows a
joint with hogging moment on the right and sagging moment on the left, which
may be due to a large applied horizontal shear from the right. In contrast, Figure
6.1(b) shows a joint with hogging moment on both sides which is the normal
behaviour of a column beam joint under dominant gravity loads. However, it
should be noted that the hogging moments on both sides may not balance.

Column Potential failure
surface (tension)
Shear Vc1

C BL TBR
sagging moment
in beam Vb2 Vb1 hogging moment
in beam
TBL C BR

Column
Shear Vc2

Figure 6.1(a) – Phenomenon of Diagonal Joint Splitting by moments of opposite
signs on both sides of joint

93


Potential failure
Column surface (tension)
Shear Vc1
TBR > TBL
T BL
TBR
hogging moment
in beam Vb2 Vb1 hogging moment
in beam
C BL C BR

Column
Shear Vc2

Figure 6.1(b) – Phenomenon of Diagonal Joint Splitting by moments of same sign
on both sides of joint

In both cases, the unbalanced forces due to unbalanced flexural stresses by the
adjoining beams on both sides of the joint tend to “tear” the joint off with a
potential tension failure surface, producing “diagonal splitting”. In co-existence
with the bending moments, there are shears in the columns which usually tend to
act oppositely. The effects by such shears can help to reduce the effects of shears
on the column joints created by bending. Reinforcements in form of links may
therefore be necessary if the concrete alone is considered inadequate to resist the
diagonal splitting.

6.3 Design procedures :

(i) Work out the total nominal horizontal shear force across the joint V jh in

X and Y directions generally. V jh should be worked out by considering

forces acting on the upper half of the joint as illustrated in Figures 6.2(a)
and 6.2(b). Figure 6.2(a) follows the case of Figure 6.1(a) in which the
moments in the beams on both sides of the joint are of different signs (i.e.
one hogging and one sagging). There is thus a net “shear” of

V jh = TBL + TBR − Vc acting on the joint where T BR = f y A sR and

C BL = TBL = f y AsL are the pull and push forces by the beams in which AsR

and AsL are the steel areas of the beams. This approach which originates

94


from the New Zealand Code NSZ 3103 requires TBR and TBL be
increased by 25% under the load capacity concept in which the reinforcing
bars in the beam will be assumed to have steel stress equal to 125% yield
strength of steel if such assumption will lead to the most adverse
conditions. Thus the following equation can be listed :

V jh = TBL + TBR − Vc = 1.25 f y ( AsL + AsR ) − Vc (Eqn 6.1)

Column
shear Vc

C BL = T BL T BR = 1.25 f y AsR
or f y AsR

TBL = 1.25 f y AsL
C BR
or f y AsR
sagging moment hogging moment
in beam Column in beam
shear Vc’

Figure 6.2(a) – Calculation of V jh , opposite sign beam moments on both sides

However, there is a comment that New Zealand is a country of severe
seismic activity whilst in Hong Kong the dominant lateral load is wind
load. The 25% increase may therefore be dropped and (Eqn 6.1) can be
re-written as

V jh = TBL + TBR − Vc = f y ( AsL + AsR ) − Vc (Eqn 6.2)

Furthermore, as Vc counteracts the effects of TBR and TBL and Vc is
generally small, Vc can be ignored in design. Nevertheless, the inclusion
of Vc can help to reduce steel congestion in case of high shear.

Similarly Figure 6.2(b) follows the case of Figure 6.1(b) which may be the
case of unbalancing moments due to gravity load without lateral loads or
even with the lateral loads, such loads are not high enough to reverse any
of the beam moments from hogging to sagging. By similar argument and
formulation as for that of Figure 6.2(a), (Eqn 6.3) and (Eqn 6.4) can be
formulated for Figure 6.2(b)
M
V jh = TBR − TBL − Vc = 1.25 f y AsR − L − Vc (Eqn 6.3)
zL

95


ML
V jh = TBR − TBL − Vc = f y AsR − − Vc (Eqn 6.4)
zL

Column
shear Vc
TBR > TBL
T BL
T BR = 1.25 f y AsR
M
T BL = L zL zR or f y AsR
zL
C BR

hogging moment CBL hogging moment
in beam ML in beam

Figure 6.2(b) – Calculation of V jh , same sign beam moments on both sides

Equations (Eqn 6.2) and (Eqn 6.4) will be used in this Manual.

(ii) With the V jh determined, the nominal shear stress is determined by
V jh
(Ceqn 6.71) in the Code. v jh =
b j hc
where hc is the overall depth of the column in the direction of shear
b j = bc or b j = bw + 0.5hc whichever is the smaller when bc ≥ bw ;
b j = bw or b j = bc + 0.5hc whichever is the smaller when bc < bw ;
where bc is the width of column and bw is the width of the beam.

Cl. 6.8.1.2 of the Code specifies that “At column of two-way frames,
where beams frame into joints from two directions, these forces need be

considered in each direction independently.” So v jh should be calculated

independently for both directions even if they exist simultaneously and
both be checked that they do not exceed 0.25 f cu .

(iii) Horizontal reinforcements based on Ceqn 6.72 reading
*
V jh  C N* 
A jh =  0.5 − j  should be worked out in both directions and
0.87 f yh 
 Ag f cu 


96


*
be provided in the joint as horizontal links. In Ceqn 6.72, V jh should be

the joint shear in the direction (X or Y) under consideration and N * be
the minimum column axial load. If the numerical values arrived at is

positive, shear reinforcements of cross sectional areas A jh should be

provided. It may be more convenient to use close links which can serve as
confinements to concrete and horizontal shear reinforcements in both
directions. If the numerical values arrived by (Ceqn 6.72) becomes
negative, no horizontal shear reinforcements will be required;

(iv) Similarly vertical reinforcements based on (Ceqn 6.73) reading

0.4(hb / hc )V jh − C j N *
*

A jv = should be worked out in both directions and
0.87 f yv

be provided in the joint as vertical links or column intermediate bars (not
corner bars). Again if the numerical values arrived by (Ceqn 6.73) is
negative, no vertical shear reinforcements will be required;

(v) Notwithstanding the provisions arrived at in (iii) for the horizontal
reinforcements, confinements in form of closed links within the joint
should be provided as per Cl. 6.8.1.7 of the Code as :
(a) Not less than that in the column shaft as required by Cl. 9.5.2 of the
Code, i.e. Section 5.5 (i) to (iv) of this Manual if the joint has a free
face in one of its four faces;
(b) Reduced by half to that provisions required in (a) if the joint is
connected to beams in all its 4 faces;
(c) Link spacing ≤ 10Ø (diameter of smallest column bar) and 200 mm.

Longitudinal
bar dia. Ø

Transverse reinforcements
ø ≥1/4(maxØ) and 6mm;
with spacing ≤ 10(minØ)
and 200mm

Figure 6.3 – Minimum transverse reinforcements in Column Beam Joint

97


6.4 Worked Example 6.1:

Consider the column beam joints with columns and beams adjoining as
indicated in Figure 6.4 in the X-direction and Y-directions. Concrete grade is 40.
All loads, shears and moments are all ultimate values. The design is as follows :

X-direction N * = 6000 kN

Beam moment 550 kNm
(hogging)
Vcx = 300 kN
Beam size 700 × 500
(effective depth 630)
Beam moment 300 kNm
(sagging) Column size 900 × 800

800 900

Column size 900 × 800

Y-direction N * = 6000 kN

Beam moment 550 kNm
(hogging)
Beam moment 300 kNm
(hogging)
Beam size 700 × 500 Vcy = 0 kN (effective depth 630)

Column size 900 × 800

900 800

Figure 6.4 – Design Example for Column Beam Joint

(i) Check nominal shear stress :

X-direction

98


The moments on the left and right beams are of opposite signs. So Figure
6.2(a) is applicable. The top steel provided on the right beam is 3T32, as
designed against the ultimate hogging moment of 550kNm with
AsR = 2413 mm2 whilst the bottom steel provided on the left beam is
4T20 with AsL = 1257 mm2, again as designed against the ultimate
sagging moment of 300kNm.

C BL TBR

TBL C BR

TBR = 460 × 2413 × 10 −3 = 1109.98 kN; C BR = TBR
TBL = 460 × 1257 ×10 −3 = 578.22 kN; C BL = TBL
So the total shear is

V jx = TBL + TBR − Vcx = 1109.98 + 578.22 − 300 = 1388.2 kN

In the X-direction hc = 900
As bc = 800 > bw = 500 ,the effective joint width is the smaller of

bc = 800 and bw + 0.5hc = 500 + 0.5 × 900 = 950 , so b j = 800

So, checking against Cl. 6.8.1.3 of the Code,
V jx 1388.2 ×10 3
v jx = = = 1.93 MPa < 0.25 f cu = 10 MPa
b j hc 800 × 900

Y-direction

The moments on the left and right beams are of equal sign, both hogging.
So Figure 6.2(b) is applicable. As the moment on the right beam is
higher, the potential plastic hinge will be formed on the right beam.
Again the top steel provided in the right beam is 3T32 as designed
against the ultimate hogging moment of 550 kNm.

99


T BL TBR

zL TBR > TBL

CBR

CBL

TBR = 460 × 2413 × 10 −3 = 1109.98 kN; C BR = TBR
TBL is to be determined by conventional beam design method for the
ultimate hogging moment of 300 kNm
M z
2
= 1.512 MPa, = 0.948 , AsL = 1254.49 mm2;
bd d
TBL = 0.87 f y AsL = 502.05 kN
As the column shear is zero, by (Eqn 6.3)
V jy = 1109.98 − 502.05 = 607.93 kN
In the Y-direction hc = 800 , bc = 900
and bw + 0.5hc = 500 + 0.5 × 800 = 900 , so b j = 900
So, checking against Cl. 6.8.1.3 of the Code,
V jy 607.93 × 10 3
v jy = = = 0.84 MPa < 0.25 f cu = 10 MPa
b j hc 900 × 800

(ii) To calculate the horizontal joint reinforcement by Ceqn 6.72, reading
V jh 
*
C N* 
A jh =  0.5 − j 
0.87 f yh 
 Ag f cu 

V jh
where C j =
V jx + V jy

X-direction
V jx 1388.2
C jx = = = 0.695
V jx + V jy 1388.2 + 607.93
*
 C N* 
 0.5 − jx  = 1388.2 × 10  0.5 − 0.695 × 6000000 
3
V jhx
A jhx =  
0.87 f yh 
 Ag f cu  0.87 × 460 
 900 × 800 × 40 
= 1232 mm2

Y-direction
V jy 607.93
C jy = = = 0.305
V jx + V jy 1388.2 + 607.93

100


*

V jhy C N *  607.93 × 10 3  0.305 × 6000000 
A jhy =  0.5 − jy =  0.5 − 
0.87 f yh 
 Ag f cu  0.87 × 460 
 900 × 800 × 40 
= 663 mm2

Use 6T12 close stirrups (Area provided = 1357 mm2) which can
adequately cover shear reinforcements in both directions

(iii) To calculate the vertical joint reinforcement by (Ceqn 6.73 of the Code),
0.4(hb / hc )V jh − C j N *
*

reading A jv =
0.87 f yh

X-direction

0.4(hb / hc )V jh − C j N *
*
0.4(700 / 900 ) × 1388200 − 0.695 × 6000000
A jvx = =
0.87 f yh 0.87 × 460

= −9341 . So no vertical shear reinforcement is required.

Y-direction

0.4(hb / hc )V jh − C j N *
*
0.4(700 / 800 ) × 607930 − 0.305 × 6000000
A jvy = =
0.87 f yh 0.87 × 460
= −4041 . Again no vertical shear reinforcement is required.

(iv) The provision of outermost closed stirrups in the column shaft is T12 at
approximately 120mm which is in excess of the required confinement as
listed in 6.3(v). So no additional confinement is requirement.

Closed links 6T12
(spacing = 120 < 200
and 10Ø = 320)

T32

Figure 6.5 – Details of Column Beam Joint Detail for Column Beam Joint
(Plan) Design – Other details omitted for clarity

101


7.0 Walls

7.1 Design Generally

7.1.1 Similar to column by design to resist axial loads and moments.

7.1.2 The design ultimate axial force may be calculated on the assumption that the
beams and slabs transmitting force to it are simply supported. (Re Cl. 6.2.2.2(a)
and Cl. 6.2.2.3(a) of the Code).

7.1.3 Minimum eccentricity for transverse moment design is the lesser of 20 mm or
h / 20 , as similar to columns.

7.2 Categorization of Walls

Walls can be categorized into (i) slender walls; (ii) stocky walls; (iii)
reinforced concrete walls; and (iv) plain walls.

7.3 Slender Wall Section Design

7.3.1 Determination of effective height l e (of minor axis generally which
controls) –
(i) in case of monolithic construction, same as that for column; and
(ii) in case of simply supported construction, same as that for plain wall.

7.3.2 Limits of slender ratio (Re Table 6.15 of the Code) –
(i) 40 for braced wall with reinforcements < 1%;
(ii) 45 for braced wall with reinforcements ≥ 1%;
(iii) 30 for unbraced wall.

7.3.3 Other than 7.3.1 and 7.3.2, reinforced concrete design is similar to that of
columns.

7.4 Stocky Wall

7.4.1 As similar to column, stocky walls are walls with slenderness ratio < 15 for
braced walls and slenderness ratio < 10 for unbraced walls;

102


7.4.2 Stocky reinforced wall may be designed for axial load n w only by (Ceqn
6.59) of the Code provided that the walls support approximately symmetrical
arrangement of slabs with uniformly distributed loads and the spans on either
side do not differ by more than 15%;

n w ≤ 0.35 f cu Ac + 0.67 f y Asc

7.4.3 Other than 7.4.2 and the design for deflection induced moment M add , design
of stocky wall is similar to slender walls.

7.5 Reinforced Concrete Walls design is similar to that of columns with
categorization into slender walls and stocky walls.

7.6 Plain Wall – Plain wall are walls the design of which is without consideration
of the presence of the reinforcements.

7.6.1 Effective height of unbraced plain wall, where l 0 is the clear height of the
wall between support, is determined by :
(a) l e = 1.5l 0 when it is supporting a floor slab spanning at right angles to it;
(b) l e = 2.0l 0 for other cases.

Effective height ratio for braced plain wall is determined by
(a) l e = 0.75l 0 when the two end supports restraint movements and rotations;
(b) l e = 2.0l 0 when one end support restraint movements and rotations and
the other is free;
(c) le = l0 ' when the two end supports restraint movements only;
(b) le = 2.5l0 ' when one end support restraint movements only and the other
is free; where l0 ' in (c) and (d) are heights between centres of supports.

7.6.2 For detailed design criteria including check for concentrated load, shear, load
carrying capacities etc, refer to Cl. 6.2.2.3 of the Code.

7.7 Sectional Design

The sectional design of wall section is similar to that of column by utilizing
stress strain relationship of concrete and steel as indicated in Figure 3.8 and
3.9 of the Code. Alternatively, the simplified stress block of concrete as
indicated in Figure 6.1 can also be used. Nevertheless, the Code has additional

103


requirements in case both in-plane and transverse moments are “significant”
and such requirements are not identical for stocky wall and slender wall.

7.7.1 Wall with axial load and in-plane moment

Conventionally, walls with uniformly distributed reinforcements along its
length can be treated as if the steel bars on each side of the centroidal axis are
lumped into two bars each carrying half of the steel areas as shown in Figure
7.1 and design is carried out as if it is a 4 bar column. Nevertheless, it is
suggested in this Manual that the reinforcements can be idealized as a
continuum (also as shown in Figure 7.1) which is considered as a more
realistic idealization. Derivation of the formulae for the design with
reinforcements idealized as continuum is contained in Appendix G, together
with design charts also enclosed in the same Appendix.

d = 0.75h

Shear Wall Section Current idealization Proposed idealization
based on 4-bar with reinforcing bars as
column design chart continuum with areas
equal to the bars

Figure 7.1 – Idealization of Reinforcing bars in shear wall

Worked Example 7.1

Consider a wall of thickness 300 mm, plan length 3000 mm and under an axial
load P = 27000 kN and in-plane moment M x = 4500 kNm. Concrete grade is
45. The problem is an uniaxial bending problem. Then

104


P 27000 × 10 3 M 4500 × 10 6
= = 30 and = = 1.67 .
bh 300 × 3000 bh 2 300 × 3000 2
If based on the 4-bar column chart with d / h = 0.75 , p = 3.8 %, requiring
T32 – 140 (B.F.)

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
2 2
M/bh N/mm

If use chart based on continuum of bars, the reinforcement ratio can be slightly
reduced to 3.7%.

Design Chart of Rectangular Shear Wall with Uniform Vertical Reinforcements to Code of Practice
for Structural Use of Concrete 2004, Concrete Grade 45
55
0.4% steel
1% steel
50
2% steel
45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
2 2
M/bh N/mm

By superimposing the two design charts as in Figure 7.2, it can be seen that the
idealization of steel re-bars as continuum is generally more conservative.

105


Comparison of Idealization as 4-bar columns and Continuum of Steel to Code of Practice for
Structural Use of Concrete 2004, Concrete Grade 45
55
0.4% steel - 4 bar column
50
0.4% steel - wall
2% steel - 4 bar column
45
2% steel - wall
40 5% steel - 4 bar column
5% steel - wall
35
8% steel - 4 bar column
2
N/bh N/mm

30 8% steel - wall

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10

2 2
M/bh N/mm

Figure 7.2 – Comparison of design curve between idealization of steel bars as 4 bar
column and continuum

7.7.2 Wall with axial load and transverse moment

The design will also be similar to that of column with the two layers of
longitudinal bars represented by the bars in the 4-bar column charts as shown
in Figure 7.3

idealized as
Bars carry total steel total
area of the row of steel

Figure 7.3 – Sectional design for column with axial load and transverse moment

7.7.3 Wall with significant in-plane and transverse moments

106


The Code has not defined the extent of being “significant”. Nevertheless, if
significant in-plane and transverse moments exist, the Code effectively
requires the wall section be examined at various points (for stocky wall) and
unit lengths (for slender wall) along the length of the wall at the splitting up of
the axial load and in-plane moment as demonstrated in Figure 7.4.

P
M

wall
By elastic analysis
P Mx
± 3
L L / 12

P 6M
P 6M + 2
− ≥0 L L
L L2

x
L
OR

2P
P 6M L M 
− <0 3 − 
L L2 2 P 

L M 
3 − 
2 P 

L

Figure 7.4 – conversion of axial load (kN) and in-plane moment (kNm) into linear
va rying load (kN/m) along wall section

Worked Example 7.2

Consider a grade 45 wall of thickness 300 mm, plan length 3000 mm and

107


under an axial load P = 27000 kN and in-plane moment M x = 4500 kNm and

transverse moment M y = 300 kNm as shown in Figure 7.5. By elastic analysis,

the load intensities at the 4 points as resolution of P and M x are :
27000 6 × 4500
A: + = 12000 kN/m;
3 32
27000 4500 × 0.5
B: + = 10000 kN/m
3 33 / 12
27000 4500 × 0.5
C: − = 8000 kN/m;
3 33 / 12
27000 6 × 4500
D: − = 6000 kN/m
3 32
The varying load intensities are as indicated in Figure 7.5.

P = 27000 kN

M x = 4500 kNm

wall
By elastic analysis

10000kN/m
8000kN/m

12000kN/m
6000kNm

A
D
C B

1000 1000 1000

Figure 7.5 – Conversion of axial load (kN) and in-plane moment (kNm) into linear
varying load (kN/m) along wall section for Worked Example 7.2

(i) If the wall is considered stocky, each of the points with load
intensities as determined shall be designed for the load intensities as
derived from the elastic analysis and a transverse moment of
300 ÷ 3 = 100 kNm/m by Clause 6.2.2.2(f)(iv) of the Code. Consider
one metre length for each point, the 4 points shall be designed for the
following loads with section 1000 mm by 300 mm as tabulated in

108


Table 7.1, i.e. all the points are undergoing uniaxial bending and the
sectional design are done in the same Table in accordance with the
chart extracted from Appendix F:

Point A B C D
Axial Load 12000 10000 8000 6000
In-plane Mt 0 0 0 0
Transverse Mt 100 100 100 100
N / bh 40 33.33 26.67 20
M / bh 2 1.11 1.11 1.11 1.11
p (%) 5.9 4.1 2.4 0.6
Re-bars (BF) T40 – 140 T40 – 200 T32 – 225 T20 – 300
Table 7.1 – Summary of Design for Worked Example 7.2 as a stocky Wall

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 A 5% steel
6% steel
35 7% steel
B 8% steel
2
N/bh N/mm

30

25
C
20
D
15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5
2 2
M/bh N/mm

The Code is not clear in the assignment of reinforcements at various
segments of the section based on reinforcements worked out at
various points. The assignment can be based on the tributary length
principle, i.e. the reinforcement derived from A shall be extended
from A to mid-way between A and B; the reinforcement derived from
B be extended from mid-way between A and B to mid-way between
B and C etc. As such, the average reinforcement ratio is 3.25%.
Nevertheless, as a more conservative approach, the assignment of
reinforcement design between A and B should be based on A and that

109


of B and C be based on B etc. As such the reinforcement ratio of the
whole section will be increased to 4.13% and the reinforcement ratio
at D is not used.

(ii) If the wall is slender, by Cl. 6.2.2.2(g)(i) of the Code, the wall should
be divided into “unit lengths” with summing up of loads. Consider
the three units AB, BC and CD. The loads and in-plane moments
summed from the trapezoidal distribution of loads are as follows,
with the assumption that the transverse moment of 300 kNm has
incorporated effects due to slenderness :

For Unit Length AB :
12000 + 10000
Summed axial load = × 1 = 11000 kN
2
12000 − 10000 2 1
Summed in-plane moment × 1×  −  × 1 = 167 kNm.
2 3 2
The summed axial loads and moments on the unit lengths BC and
CD are similarly determined and design is summarized in Table 7.2,
with reference to the design chart extracted from Appendix F. In the

computation of M x / h' and M y / b' , h' and b' are taken as 750

and 225 respectively.

Unit Length AB BC CD
Axial Load 11000 9000 7000
In-plane Mt ( M x ) 167 167 167
Transverse Mt ( M y ) 100 100 100
M x / h' 0.227 0.227 0.227
M y / b' 0.444 0.444 0.444
N / f cu bh 0.272 0.222 0.172
β 0.684 0.744 0.801
M y ' = M y + β (b' / h')M x 134.2 137.2 140.1
N / bh 36.67 30 23.33
M y ' / hb 2 1.49 1.524 1.556
p (%) 5.4 3.7 1.9
Re-bars (BF) T40 – 155 T32 – 145 T25 – 175
Table 7.2 – Design of Wall for Worked Example 7.2 as a slender wall

110


55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
AB 6% steel
35 7% steel
8% steel
2
N/bh N/mm

30
BC
25
CD
20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5
2 2
M/bh N/mm

The average steel percentage is 3.67%.
So the reinforcement worked out by Clause 6.2.2.2(g)(i) of the Code
for a slender wall is between the results of the two methods of
reinforcement ratios assignments as described in sub-section (i)
based on Clause 6.2.2.2(f)(iv) of the Code.

(iii) Summary of the reinforcements design of the three approaches

stocky wall –
tributary
length T20 –30 BF T32 – 225 BF T40 –200 BF T40 – 140 BF

stocky wall –
conservative
approach
T32 – 225 BF T40 –200 BF T40 – 140 BF

slender
wall
T25 – 175 BF T32 – 145 BF T40 – 155 BF

Figure 7.6 – Summary of reinforcement details of Worked Example 7.2

(iv) The approach recommended in the Code appears to be reasonable
and probably economical as higher reinforcement ratios will be in
region of high stresses. However, it should be noted that if moment
arises from wind loads where the direction can reverse, design for the

111


reversed direction may result in almost same provisions of
reinforcements at the other end.

As the division of segments or points as recommended by the Code
for design of wall with significant transverse and in-plane moments
is due to the inaccurate account by the biaxial bending formula used
for design of column, more accurate analysis can be done by true
biaxial bending analysis as discussed in Section 5.3.5 and Figure 5.8
of this Manual, so long the “plane remain plane” assumption is valid,
though the design can only be conveniently done by computer
methods. The sections with reinforcement ratios arrived at in (i) and
(ii) have been checked against by the software ADSEC, the section in
(ii) has yielded an applied moment / moment capacity ratio of 0.8
showing there is room for slight economy. Nevertheless, the first
reinforcement ratio in (i) is inadequate as checked by ADSEC whilst
the second one yielded an over design with applied moment /
moment capacity ratio up to 0.68.

7.8 The following Worked Example 7.3 serves to demonstrate the determination
of design moment for a slender wall section, taking into account of additional
moment due to slenderness.

Worked Example 7.3

Wall Section : thickness : 200 mm, plan length : 2000 mm;
Wall Height : 3.6 m,
Concrete grade : 35
Connection conditions at both ends of the wall : connected monolithically with
floor structures shallower than the wall thickness.

Check for slenderness

Generally only necessary about the minor axis.
End conditions are 2 for both ends, β = 0.85 (by Table 6.11 of the Code);
l e = 0.85 × 3.6 = 3.06 m

Axial Load : N = 7200 kN, M x = 1800 kNm at top and 1200 kNm at

bottom , M y = 25 kNm at top and 24 kNm at bottom.

112


Determination of final design moment M t about the major and minor axes is
similar to (i)

For bending about the major axis, le / h = 3060 / 2000 = 1.53 < 15 , so
M add = 0 , M x will be the greatest of
(1) M 2 = 1800 ;
(2) M i + M add = 0.4 × (− 1200 ) + 0.6 ×1800 + 0 = 600 ;
< 0.4 × 1800 = 720
(3) M 1 + M add / 2 = 1200 + 0 = 1200 ; and
(4) N × emin = 7200 × 0.02 = 144 .
So M x = 1800 kNm for design.

For bending about the minor axis, le / b = 3060 / 200 = 15.3 > 15 ,
2 2
1  le  1  3060 
βa =   =   = 0.117
2000  b  2000  200 

au = β a Kh = 0.117 × 1 × 0.2 = 0.0234

M add = Nau = 7200 × 0.0234 = 168.48 kNm, M y will be the greatest of

(1) M 2 = 25 ;
(2) M i + M add = 0.4 × 25 + 168.48 = 178.48 ;
as 0.4 × (− 24 ) + 0.6 × 25 = 5.4 < 0.4 × 25 = 10
(3) M 1 + M add / 2 = 24 + 168.48 / 2 = 108.3 ; and

(4) N × emin = 7200 × 0.02 = 144 . So M y = 178.48 kNm for design.

So the factored axial load and moments for design are

N = 7200 kN; M x = 1800 kNm; M y = 178.48 kNm

Design can be performed in accordance with Cl. 6.2.2.2(g) of the Code as
demonstrated in Worked Examples 7.2 and by calculations with the formulae
derived in Appendices F and G. However, the calculations are too tedious and
cases to try are too many without the use of computer methods. Spread sheets
have been devised to solve the problem with a sample enclosed in Appendix G.

7.9 Detailing Requirements

There are no ductility requirements in the Code for walls. The detailing
requirements are summarized from Cl. 9.6 of the Code :

113


Vertical reinforcements for reinforced concrete walls :

(i) Minimum steel percentage : 0.4%. When this reinforcement controls
the design, half of the steel area be on each side;
(ii) Maximum steel percentage : 4%;
(iii) All vertical compression reinforcements should be enclosed by a link
as shown in Figure 7.7;
(iv) Maximum distance between bars : the lesser of 3 times the wall
thickness and 400 mm as shown in Figure 7.7.

≤ 3h and 400 mm

h

Figure 7.7 – Vertical reinforcements for walls

Horizontal and transverse reinforcements for reinforced concrete walls

(i) If the required vertical reinforcement does not exceed 2%, horizontal
reinforcements be provided as follows and in accordance with Figure
7.8 :
(a) Minimum percentage is 0.25% for f y = 460 MPa and 0.3% for
f y = 250 MPa;
(b) bar diameter ≥ 6 mm and 1/4 of vertical bar size;
(c) spacing ≤ 400 mm.

114


h

(a) 0.25% for fy = 460 MPa and 0.3% for fy = 250 MPa;
(b) bar diameter ≥ 6 mm and 1/4 of vertical bar size;
(c) spacing in the vertical direction ≤ 400 mm

Figure 7.8 – Horizontal reinforcements for walls with vertical
reinforcement ≤ 2%
(ii) If the required vertical reinforcement > 2%, links be provided as
follows as shown in Figure 7.9 :
(a) to enclose every vertical compression longitudinal bar;
(b) no bar be at a distance further than 200 mm from a restrained bar
at which a link passes round at included angle ≤ 90o;
(c) minimum diameter : the greater of 6 mm and 1/4 of the largest
compression bar;
(d) maximum spacing : twice the wall thickness in both the
horizontal and vertical directions. In addition, maximum spacing
not to exceed 16 times the vertical bar diameter in the vertical
direction.
(a) Spacing in vertical direction ≤ 2h
and 16 Ø;
Links of included restrained
(b) bar diameter ≥ 6 mm or 1/4 Ø
angle ≤ 90o to vertical
restrain vertical bars
≤ 200 ≤ 200
bars

h

≤ 200 ≤ 200
≤ 2h

Figure 7.9 – Anchorage by links on vertical reinforcements of more than 2%

Plain walls

If provided, minimum reinforcements : 0.25% for f y = 460 MPa and 0.3%
for f y = 250 MPa in both directions generally.

115


8.0 Corbels

8.1 General – A corbel is a short cantilever projection supporting a load-bearing
member with dimensions as shown :

av < d Applied
Load

≥ 0 .5 h
h
d

Top steel bar

Figure 8.1 – Dimension requirement for a Corbel

8.2 Basis of Design (Cl. 6.5.2 of the Code)

8.2.1 According to Cl. 6.5.2.1 of the Code, the basis of design method of a corbel is
that it behaves as a “Strut-and-Tie” model as illustrated in Figure 8.2. The strut
action (compressive) is carried out by concrete and the tensile force at top is
carried by the top steel.

av Applied
Steel strain to be
determined by linear Load Vu
extrapolation Tie action by
reinforcing bar
T

d Strut action by
Vu
neutral axis concrete
Fc

x β
0 .9 x Balancing force polygon

Concrete ultimate Concrete stress block at corbel
strain ε ult = 0.0035 support

Figure 8.2 – Strut-and-Tie Action of a Corbel

116


8.2.2 Magnitude of resistance provided to the horizontal force should be not less
than one half of the design vertical load, thus limiting the value of the angle
β in Figure 8.2 or in turn, that the value of a v cannot be too small.

8.2.3 Strain compatibility be ensured.

8.2.4 In addition to the strut-and tie model for the determination of the top steel bars,
shear reinforcements should be provided in form of horizontal links in the
upper two thirds of the effective depth of the corbel. The horizontal links
should not be less than one half of the steel area of the top steel.

8.2.5 Bearing pressure from the bearing pad on the corbel should be checked and
properly designed in accordance with “Code of Practice for Precast Concrete
Construction 2003” Cl. 2.7.9. In short, the design ultimate bearing pressure to
ultimate loads should not exceed
(i) 0.4 f cu for dry bearing;
(ii) 0.6 f cu for bedded bearing on concrete;
(iii) 0.8 f cu for contact face of a steel bearing plate cast on the corbel with
each of the bearing width and length not exceeding 40% of the width and
length of the corbel.

The net bearing width is obtained by
ultimate load
effective bearing length × ultimate bearing stress

The Precast Concrete Code 2003 (in Cl. 2.7.9.3 of the Precast Concrete Code)
has specified that the effective bearing length of a bearing be the least of :
(i) physical bearing length;
(ii) one half of the physical bearing length plus 100 mm;
(iii) 600 mm.

8.3 Design Formulae for the upper steel tie

The capacity of concrete in providing lateral force as per Figure 8.2 is
0.45 f cu × b × 0.9 x = 0.405 f cu bx where b is the length of the corbel.
The force in the compressive strut is therefore Fc = 0.405 f cu bx cos β .
By the force polygon, Fc sin β = Vu ⇒ 0.405 f cu bx sin β cos β = Vu

117


d − 0.45 x av
As tan β = ; cos β =
a v + (d − 0.45 x )
av 2 2

sin β =
(d − 0.45 x )
a v + (d − 0.45 x )
2 2

a v (d − 0.45 x ) 0.405 f cu bxa v (d − 0.45 x )
So 0.405 f cu bx = Vu ⇒ Vu =
a v + (d − 0.45 x ) a v + (d − 0.45 x )
2 2 2 2

Expanding and re-arranging

(0.2025Vu + 0.18225 f cu bav )x 2 − 0.9d (Vu + 0.45 f cu bav )x + Vu (av 2 + d 2 ) = 0
Putting A = 0.2025Vu + 0.18225 f cu bav ; B = −0.9d (Vu + 0.45 f cu bav )

( 2
C = Vu av + d 2 )
− B − B 2 − 4 AC
x= (Eqn 8-1)
2A
Vu av
By the equilibrium of force, the top steel force is T = Vu cot β =
d − 0.45 x
(Eqn 8-2)
The strain at the steel level is, by extrapolation of the strain diagram in Figure
d−x d−x
8.2 is ε s = ε ult = × 0.0035 (Eqn 8-3)
x x

8.4 Design Procedure :

(i) Based on the design ultimate load and av , estimate the size of the corbel
and check that the estimated dimensions comply with Figure 8.1;
(ii) Check bearing pressures;
(iii) Solve the neutral axis depth x by the equation (Eqn 8-1).
(iv) By the assumption plane remains plane and that the linear strain at the
base of the corbel is the ultimate strain of concrete ε ult = 0.0035 , work
out the strain at the top steel level as ε s ;
(v) Obtain the steel stress as σ s = E s ε s where E s = 200 × 10 6 kPa.
However, the stress should be limited to 0.87 f y even ε s ≥ 0.002 ;
(vi) Obtain the force in the top steel bar T by (Eqn 8-2)
(vii) Check that T ≥ 0.5Vu ;
T
(viii) Obtain the required steel area of the top steel bars Ast by Ast =
σs

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Vu
(ix) Check the shear stress by v = . If v > v c (after enhancement as
bd
A b(v − v c )
applicable), provide shear reinforcements by sv = over the
sv 0.87 f y
2
upper d where Asv is the cross sectional area of each link and s v
3
is the link spacing.
A
(x) Check that the total shear area provided which is sv d is not less than
sv
d 1
half of the top steel area, i.e. Asv × ≥ Ast even if v < vr .
sv 2

8.5 Detailing Requirements

(i) By Cl. 6.5.2.2 of the Code, anchorage of the top reinforcing bar should
either
(a) be welded to a transverse bar of equivalent strength or diameter.
The bearing area of the load should stop short of the transverse bar
by a distance equal to the cover of the tie reinforcement as shown
in Figure 8.3(a); or
(b) bent back to form a closed loop. The bearing area of the load
should not project beyond the straight portion of the bars forming
the tension reinforcements as shown in Figure 8.3(b).

(ii) By Cl. 6.5.2.3 of the Code, shear reinforcements be provided in the
upper two thirds of the effective depth and total area not less than half of
the top bars as shown in Figure 8.3(a) and 8.3(b).

119


Vu
>c
Top main bar av c, cover to
transverse bar

c
transverse bar welded to
2 the main tension bar of
d equal diameter or
3 strength
d
Additional bar for shear
link anchorage

Shear reinforcements

Figure 8.3(a) – Typical Detailing of a Corbel

Vu
av >0
Top main bar

c

2
d
3
d

Shear reinforcements

Figure 8.3(b) – Typical Detailing of a Corbel

8.6 Worked Example 8.1

Design a corbel to support an ultimate load of 600 kN at a distance 200 mm
from a wall support, i.e. Vu = 600 kN, av = 200 mm. The load is transmitted
from a bearing pad of length 300 mm. Concrete grade is 40.

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av = 200 < d = 450
Vu = 600

250 ≥ 0.5h
Net bearing
h = 500 width
d = 450

Figure 8.4 – Worked Example. 8.1

1. The dimensions of the corbel are detailed as shown which comply with
the requirement of Cl. 6.5.1 of the Code with length of the corbel
b = 300 mm;
2. Check bearing stress :
Design ultimate bearing stress is 0.8 f cu = 0.8 × 40 = 32 MPa
600 ×10 3
Net bearing width is = 62.5 mm.
300 × 32
So use net bearing width of bearing pad 70 mm.
3. With the following parameters :
Vu = 600 kN; f cu = 40 MPa; b = 300 mm;
av = 200 mm; d = 450 mm
substituted into (Eqn 8-1)

(0.2025Vu + 0.18225 f cu bav )x 2 − 0.9d (Vu + 0.45 f cu bav )x + Vu (av 2 + d 2 ) = 0
Solving x = 276.77 mm.
4. The strain at steel level,
d−x 450 − 276.77
εs = ε ult = × 0.0035 = 0.00219 > 0.002
x 276.77

5. The stress in the top steel is 0.87 f y as ε s > 0.002 ;

(if ε s ≤ 0.002 , f s = E s × ε s where E s = 200 GPa)
6. The force in the top steel is

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Vu av 600 × 200
T= = = 368.71 kN > 0.5 × 600 = 300 kN;
d − 0.45 x 450 − 0.45 × 276.77
368710
7. Steel area required is = 921.32 mm2, provide 3T20 (0.7%);
0.87 × 460

vc = 0.556 × (40 / 25)
1/ 3
8. = 0.65 MPa without enhancement. With

2d
enhancement, it becomes × 0.65 = 2.925 MPa.
av
9. Check shear stress
600000
= 4.444 MPa > vc = 2.925 MPa.
450 × 300
A b(v − vc ) 300(4.444 − 2.925)
So shear reinforcement sv = = = 1.14 mm;
sv 0.87 f y 0.87 × 460
Asv = 1.14 × 450 = 513 mm2. So use 3T12 closed links over the top 300
mm.
10. Area of 3T12 closed link is 678 mm2 > half of area of tensile top steel =
0.5×3×314 = 471 mm2.

The details of the Corbel is finally as shown in Figure 8.5.

Vu
av
3T20

300
450 T20 anchor bar

T12 closed links

Figure 8.5 – Detailing of Worked Example 8.1

8.7 Resistance to horizontal forces

Cl. 9.8.4 requires additional reinforcement connected to the supported member

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to transmit external horizontal force exerted to the corbel in its entirety.
However, it should be on the conservative side if strain compatibility is also
considered in designing the corbel to resist also this horizontal force N c as in
addition to the vertical load Vu . This is in consistency with the Code
requirement. The force polygon as modified from Figure 8.2 will becomes

Applied Vu
Steel strain to be
av
determined by linear
extrapolation Tie action by
reinforcing bar
Applied N c
T

R
d Strut action by
concrete
Vu
neutral axis Fc
Nc
x β
0 .9 x
Balancing force polygon
Concrete ultimate Concrete stress block at corbel
strain ε ult = 0.0035 support

Figure 8.6 – Strut-and-Tie Action of a Corbel with inclusion of horizontal force

From Figure 8.6 and formulae derivation in Section 8.3 of this Manual, it can
be seen that the determination of the neutral axis depth x and subsequently
the strain profile of the root of the corbel is independent of N c . Thus the steps
(i) to (v) in Section 8.4 of this Manual can be followed in calculation of x ,
ε s and σ s as if N c does not exist.
Vu a v
However, the tension in the top bar will be T = N c + (Eqn 8-4)
d − 0.45 x
T
And the steel area of the top bar can be worked out as Ast =
σs


If an additional horizontal force of 200kN is exerted on the corbel in Example
8.1, tending to pull away from the root of the corbel, the total tensile force to
be resisted by the top bars will be T = 368.71 + 200 = 568.71 kN and the top
568.71×10 3
bar area required is = 1421.06 mm2, as the strain at the steel level
0.87 × 460
has exceed 0.002. The top bar has to be increased from 3T25.

123


9.0 Cantilever Structures

9.1 Cl. 1.4 of the Code defines “Cantilever Projection” as “a structural element that
cantilevers from the main structure, for example, canopies, balconies, bay
windows, air conditioning platforms.” In addition, PNAP 173 which refers to
cantilevered reinforced concrete structures in general indicates more clearly
design and construction criteria to be complied with.

9.2 Design Considerations

Design considerations for a cantilevered structure from the Code (Table 7.3, Cl.
9.4 etc. of the Code) and PNAP 173 are summarized as follows :

Slabs and Beams in General

(i) The span to overall depth of cantilever beams or slabs should not be
greater than 7;
(ii) For cantilever span more than 1000 mm, a beam-and-slab type of
arrangement should be used instead of pure slab cantilever where
practicable (PNAP173 App. A 1(a));
(iii) The minimum percentage of top tension longitudinal reinforcement based
on the gross cross-sectional concrete area should be 0.25% for all
reinforcement grades generally (PNAP173 App. A 6(c)). However, if the
cantilever structure is a flanged beam where the flange is in tension, the
minimum steel percentage is 0.26% for T-section and 0.2% for L-section
but based on the gross area of the rectangular portion of width of the web
times the structural depth as per Table 9.1 of the Code. The more stringent
requirement shall prevail;
(iv) Diameter of the longitudinal reinforcement ≥ 10 mm as illustrated in
Figure 9.1 (PNAP173 App. A 6(c));
(v) The centre-to-centre spacing of the top tension longitudinal bars ≤ 150mm
as illustrated in Figure 9.1 (PNAP173 App. A 6(c));
(vi) For cantilevered structure exposed to weathering, cover to all
reinforcement ≥ 40 mm (PNAP173 App. A 8(a));
(vii) Anchorage of tension reinforcement shall be based on steel stress of

0.87 f y and (a) full anchorage length should be provided with location of

commencement in accordance with Cl. 9.4.3 of the Code as illustrated in

124


Figures 9.1 and 9.2; and (b) minimum anchorage length of 45 times the
longitudinal bar diameter in accordance with PNAP 173 App. A 6(d). The
different commencement points of anchorage lengths as indicated by
PNAP 173 Appendices B and C are not adopted in this Manual. However,
requirements for the lengths of curtailment of tension reinforcement bars
PNAP173 and Cl. 9.2.1.6 of the Code in relation to curtailment of tension
reinforcements are amalgamated. They are shown in Figures 9.1 and 9.2.

bar dia. Ø ≥ 10 mm
cover to all reinforcements
≥ 40 if the beam is subject ≤150
T.L. to weathering
≥0.5d
≤150
or 0.5L

d

(slab similar)
Support providing K
rotational restraint
L

bar dia. Ø ≥ 10 mm
cover to all reinforcements
≥ 40 if the beam is subject
T.L. to weathering

T.L. should be the greatest of
(i) To point of zero moment +
d the greater of anchorage length
and d; (ii) 1.5K; (iii) 45Ø; (iv)
0.3×next span length (for slab
Support not providing only)
rotational restraint K

L

Figure 9.1 – Anchorage and maximum longitudinal bar spacing in Cantilevers as
required by the Code and PNAP 173

Beam in particular

(viii) The overall depth at support should be at least 300 mm as shown in Figure
9.2;
(ix) For cantilever beam connected with continuous beams, requirements for
curtailment of longitudinal bars into the next continuous span are similar
to slab except that half of the bars can be curtailed at 0.75K + L/2 as

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shown in Figure 9.2;

Half of bars be cover to all reinforcements ≥ 40 if
curtailed at 0.75K the beam is subject to weathering

≥300
d

L K

Figure 9.2 – Particular requirements for cantilever beams as required by the Code
and PNAP 173

Slab in particular

(x) Minimum overall slab thickness (PNAP173 App. A 6(a)):
(a) 100 mm for span ≤ 500mm;
(b) 125 mm for 500 mm < span ≤ 750mm;
(c) 150 mm for span > 750 mm;
(xi) Reinforcements be high yield bars in both faces and in both directions
(PNAP173 App.A 6(c));
(xii) Particular attentions to loads as shown in Figure 9.3 should be given :

Care be taken not
to ignore loads
from this parapet

Care be taken (1) not to
ignore loads from this
area; (2) change of
direction of main bars

Figure 9.3 – Loads on cantilever slab (PNAP173 App.A 6(e))

(xiii) For a cantilever slab with a drop at the supporting end, top reinforcement
bars ≤ 16 mm in diameter should be used in order that an effective and
proper anchorage into the supporting beam and internal slab can be

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developed as illustrated in Figure 9.4. (PNAP173 App. A 6(d))

bar dia. ≤ 16 mm

Figure 9.4 – Cantilever slab with drop at supporting end

(xiv) Cantilevered slabs exposed to weathering should satisfy :
(1) maximum crack width at the tension face ≤ 0.1 mm under
serviceability check OR stress of deformed high yield steel bar ≤ 100
MPa when checking the flexural tension under working load
condition (PNAP173 App. A 8(a));
(2) Cover to all reinforcement at the exposed surface ≥ 40 mm.
(PNAP173 App. A 8(a)).


R.C. design of a cantilevered slab as shown in Figure 9.5 is subject to
weathering. Concrete grade is 35.

100
1000
(150) 900

200 wall support

Plan Elevation

Figure 9.5 – Cantilever slab in Worked Example 9.1

Loading D.L. O.W. 0.15 × 24 = 3.6 kN/m2
Fin 2.0 kN/m2
5.6 kN/m2
Para. 0.1× 1.0 × 24 = 2.4 kN/m

127


L.L. 1.5 kN/m2
Effective span is taken to be 900 + 0.5 ×150 = 975
Moment = (1.4 × 5.6 + 1.6 × 1.5) × 0.975 × 0.975 / 2 + 1.4 × 2.4 × 0.925
= 7.975 kNm/m
Design for ultimate state,
d = 150 − 40 − 5 = 105
M 7.975 × 10 6
= = 0.723
bd 2 1000 × 105 2
7.975 × 10 6
Ast = = 200 mm2/m.
0.87 × 460 × 0.95 × 105
Use T10 – 150 (Area provided is 523 mm2/m)
If the slab is subject to weathering, check the service stress by equation
in item (2) in Table 7.4 of the Code reading

2 f y Ast ,req 1
fs = ×
3 Ast , prov βb

Note : β b = 1 as no moment redistribution in cantilever.

If f s is to be limited to 100 N/mm2, Ast ,req = 200 mm2/m

2 f y Ast ,req 1 2 × 460 × 200 1
Ast , prov = × = × = 613 mm2
3 fs βb 3 ×100 1
Use T10 – 100 (area provided is 785 mm2/m or 0.52%)

Alternatively, crack width is checked by (Ceqn 7.1) and (Ceqn 7.2)
To calculate crack width, it is first necessary to assess the neutral axis
depth x by the elastic theory in accordance with the cracked section
of Figure 7.1 of the Code on the basis of a cracked section.

ε c = f c / Ec fc

x
h d

f s / Es fs
strain stress

Figure 9.6 – Stress/strain relation of a cracked R.C. section

128


E c is the long term value which, by Cl. 7.2.3 of the Code is taken as
half of the instantaneous value which is 23.7 ÷ 2 = 11.85 GPa

E s = 200 kN/mm2
Consider equilibrium of the section in Figure 9.6.
1 1 ε (d − x )
f c bx = f s Ast ⇒ E c ε c bx = E s c Ast
2 2 x
1
⇒ E c bx 2 + E s Ast x − E s dAst = 0 (Eqn 9.1)
2
Consider 1 m width of the section in Worked Example 9.1, b = 1000
(Eqn 9.1) becomes
1
× 11.85 × 1000 x 2 + 200 × 785 x − 200 × 105 × 785 = 0
2
Solving x = 41.14 mm
Taking moment about the centroid of the triangular concrete stress
block (the moment should be the unfactored moment which is
5.817kNm/m as it is a checking on serviceability limit state), the steel
tensile stress can be worked out as
 x M 5.817 ×10 6
M = f s Ast  d −  ⇒ f s = = (Eqn 9.2)
 3  x  41.14 
Ast  d −  785105 − 
 3  3 
= 81.17 N/mm2

So the strain of the steel is
81.17 0.8 f y
εs = = 0.000406 < = 0.0184 .
200 × 10 3 Es
So checking of crack width by (Ceqn 7.1) is applicable.
At the extreme fibre of the concrete at the tension side, the strain is

ε1 = ε s
(h − x ) = 0.000406 × 150 − 41.14 = 0.000692
(d − x ) 105 − 41.14
By (CEqn 7.2), to include the stiffening effect of cracked concrete,
b (h − x )(a'− x ) 1000(150 − 41.14 )(150 − 41.14)
ε m = ε1 − t = 0.000692 −
3E s As (d − x ) 3 × 200 × 10 3 × 785 × (105 − 41.14)
= 0.000298

The expected shrinkage strain, in accordance with Cl. 3.1.8 of the

Code is ε cs = c s K L K c K e K j K s where

129


c s = 3.0 ;
K L = 275 × 10 −6 for normal air according to Figure 3.6 of the Code;
K c = 1.17 according to Figure 3.3 of the Code for cement content 434
kg/m3 and water cement ratio 0.47 for grade 35;
K e = 0.91 according to Figure 3.7 of the Code for he = 150 ;

K j = 1 according to Figure 3.5 at time at infinity.

1 1
Ks = = = 0.919 according to (Ceqn 3.3)
1 + ρα e 200
1 + 0.0052 ×
11.85
So the expected shrinkage strain is

ε cs = 3.0 × 275 × 10 −6 × 1.17 × 0.91× 1.0 × 0.919 = 0.000807 > 0.0006 .

Thus it is subjected to “abnormally high shrinkage” according to the
Code and half of the expected strain be added to ε m .
∴ ε m = 0.000298 + 0.000807 × 0.5 = 0.000702

The cracked width should be the greatest at the concrete surface
mid-way between steel bars as illustrated in Figure 9.7;

acr = 50 2 + 40 2 = 64

40

100

Figure 9.7 – Illustration of a cr in Worked Example 9.1

By (CEqn 7.1) the cracked width is
3acr ε m 3 × 64 × 0.000702
ω= = = 0.0935 mm ≤ 0.1 mm as
 acr − cmin   64 − 40 
1 + 2  1 + 2 
 h−x   150 − 41.14 
required by PNAP 173. So O.K.

As PNAP 173 requires either checking of working stress below 100
MPa or crack width ≤ 0.1 mm, it should be adequate if any one of the
conditions is satisfied. Apparently it would be simpler to check only

130


the former.

Summing up, reinforcement details is as shown :

T10 – 100

Adequate anchorage
length as determined by
Figure 9.1 and 9.2

T10 – 300

Figure 9.8 – Reinforcement Details for Worked Example 9.1

9.4 R.C. Detailing

Apart from the requirements stipulated in the preceding sections, reference can
also be made to the drawings attached at the Appendices B and C of PNAP 173,
especially for the locations of anchorage length commencement. However, it
should be noted that not all sketches in PNAP 173 indicate locations of
anchorage length commence from mid-support widths.

131


10.0 Transfer Structures

10.1 According to Cl. 5.5 of the Code, transfer structures are horizontal elements
which redistribute vertical loads where there is a discontinuity between the
vertical structural elements above and below.

10.2 In the analysis of transfer structures, consideration should be given to the
followings as per Cl. 5.5 of the Code :
(i) Construction and pouring sequence – the effects of construction
sequence can be important in design of transfer structures due to the
comparatively large stiffness of the transfer structure and sequential built
up of stiffness of structures above the transfer structure as illustrated in
Figure 10.1;
(ii) Temporary and permanent loading conditions – especially important
when it is planned to cast the transfer structures in two shifts and use the
lower shift to support the upper shift as temporary conditions, thus
creating locked-in stresses;
(iii) Varying axial shortening of elements supporting the transfer structures –
which leads to redistribution of loads. The phenomenon is more serious
as the transfer structure usually possesses large flexural stiffness in
comparison with the supporting structural members, behaving somewhat
between (a) flexible floor structures on hard columns; and (b) rigid
structures (like rigid cap) on flexible columns;
(iv) Local effects of shear walls on transfer structures – shear walls will
stiffen up transfer structures considerably and the effects should be taken
into account in more accurate assessment of structural behaviour;
(v) Deflection of the transfer structures – will lead to redistribution of loads
of the superstructure. Care should be taken if the structural model above
the transfer structure is analyzed separately with the assumption that the
supports offered by the transfer structures are rigid. Re-examination of
the load redistribution should be carried out if the deflections of the
transfer structures are found to be significant;
(vi) Lateral shear forces on the transfer structures – though the shear is lateral,
it will nevertheless create out-of-plane loads in the transfer structures
which needs be taken into account;
(vii) Sidesway of the transfer structures under lateral loads and unbalanced
gravity loads should also be taken into account. The effects should be
considered if the transfer structure is analyzed as a 2-D model.

132


Stage (1) : Stage (2) : Stage (3) :
Transfer Structure (T.S.) Wet concrete of 1/F just 1/F hardened and 2/F wet
just hardened poured concrete just poured

2/F
1/F
1/F
G/F
G/F G/F
Stress/force in T.S. : {FT} +
{F1}, {F1} being force
induced in transfer Stress/force in T.S. being due
Stress/force in T.S. : {FT} due to {FT} + {F1} + {F2}, {F2}
to own weight of T.S. structure due to weight of
1/F structure. being force induced in
Stiffness : the T.S only transfer structure due to
Stiffness : the T.S. only.
weight of 2/F structure.
Stiffness : the T.S. + 1/F.

Stage (4) : Stage (5) : Stage (6) and onwards
2/F hardened and 3/F wet 3/F hardened and 4/F wet Structure above transfer
concrete just poured concrete just poured structure continues to be
built. Final force induced on
T.S. becomes {Fn} + {Fn-1}
4/F + {Fn-2} + ........... + {F2 } +
{F1} + {FT}.
3/F
3/F
2/F 2/F

1/F 1/F

G/F G/F

Stress/force in T.S. : {FT} + Stress/force in T.S. : {FT}
{F1} + {F2} + {F3}, {F3} + {F1} + {F2} + {F3} +
being force induced in T.S. {F4}, {F4} being force
due to weight of 3/F induced in T.S. due to
structure. weight of 4/F structure.
Stiffness : T.S. + 1/F + 2/F Stiffness : T.S. + 1/F + 2/F
+ 3/F

Figure 10.1 – Diagrammatic illustration of the Effects of Construction Sequence of
loads induced on transfer structure

133


10.3 Mathematical modeling of transfer structures as 2-D model (by SAFE) :

The general comments in mathematical modeling of transfer structures as 2-D
model to be analyzed by computer methods are listed :
(i) The 2-D model can only be analyzed against out-of-plane loads, i.e.
vertical loads and out-of-plane moments. Lateral loads have to be
analyzed separately;
(ii) It is a basic requirement that the transfer structure must be adequately
stiff so that detrimental effects due to settlements of the columns and
walls being supported on the transfer structure are tolerable. In view of
the relatively large spans by comparing with pile cap, such settlements
should be checked. Effects of construction sequence may be taken into
account in checking;
(iii) The vertical settlement support stiffness should take the length of the
column/wall support down to a level of adequate restraint against further
settlement such as pile cap level. Reference can be made to Appendix H
discussing the method of “Compounding” of vertical stiffness and the
underlying assumption;
(iv) Care should be taken in assigning support stiffness to the transfer
structures. It should be noted that the conventional use of either 4 EI / L
or 3EI / L have taken the basic assumption of no lateral movements at
the transfer structure level. Correction to allow for sidesway effects is
necessary, especially under unbalanced applied moments such as wind
moment. Fuller discussion and means to assess such effects are discussed
in Appendix H;
(v) Walls which are constructed monolithically with the supporting transfer
structures may help to stiffen up the transfer structures considerably.
However, care should be taken to incorporate such stiffening effect in the
mathematical modeling of the transfer structures which is usually done
by adding a stiff beam in the mathematical model. It is not advisable to
take the full height of the wall in the estimation of the stiffening effect if
it is of many storeys as the stiffness can only be gradually built up in the
storey by storey construction so that the full stiffness can only be
effected in supporting the upper floors. Four or five storeys of walls may
be used for multi-storey buildings. Furthermore, loads induced in these
stiffening structures (the stiff beams) have to be properly catered for
which should be resisted by the wall forming the stiff beams;

134


10.4 Modeling of the transfer structure as a 3-dimensional mathematical model can
eliminate most of the shortcomings of 2-dimensional analysis discussed in
section 10.3, including the effects of construction sequence if the software has
provisions for such effects. However, as most of these softwares may not have
the sub-routines for detailed design, the designer may need to “transport” the
3-D model into the 2-D model for detailed design. For such “transportation”,
two approaches can be adopted :

(i) Transport the structure with the calculated displacements by the 3-D software
(after omission of the in-plane displacements) into the 2-D software for
re-analysis and design. Only the displacements of the nodes with external loads
(applied loads and reactions) should be transported. A 2-D structure will be
re-formulated in the 2-D software for re-analysis by which the structure is
re-analyzed by forced displacements (the transported displacements) with
recovery of the external loads (out-of-plane components only) and subsequently
recovery of the internal forces in the structure. Theoretically results of the two
models should be identical if the finite element meshing and the shape functions
adopted in the 2 models are identical. However, as the finite element meshing of
the 2-D model is usually finer than that of the 3-D one, there are differences
incurred between the 2 models, as indicated by the differences in recovery of
nodal forces in the 2-D model. The designer should check consistencies in
reactions acting on the 2 models. If large differences occur, especially when
lesser loads are revealed in the 2-D model, the designer should review his
approach;

External nodal
External nodal force is {F2D} ≠
force is {F3D} {F3D} after
re-analysis

3-D model (usually coarser meshing) 2-D model (usually finer meshing) with
with displacements at nodes with nodal forces recovered by forced
external loads marked with displacement analysis at nodes marked
with

Figure 10.2 – 3-D model to 2-D with transportation of nodal displacements

135


(ii) Transport the out-of-plane components of the external loads (applied loads and
reactions) acting on the 3-D model to the 2-D model for further analysis. This
type of transportation is simpler and more reliable as full recovery of loads
acting on the structure is ensured. However, in the re-analysis of the 2-D
structure, a fixed support has to be added on any point of the structure for
analysis as without which the structure will be unstable. Nevertheless, no effects
due to this support will be incurred by this support because the support reactions
should be zero as the transported loads from the 3-D model are in equilibrium.

The out-of-plane
components of all loads
acting on the structure
including reactions be
transported

3-D model with external loads obtained 2-D model with out-of-plane components of
by analysis external forces transported from 3-D model and
re-analyzed with a fixed support at any point

Figure 10.3 – 3-D model to 2-D with transportation of nodal forces

10.5 Structural Sectional Design and r.c. detailing

The structural sectional design and r.c. detailing of a transfer structure member
should be in accordance with the structural element it simulates, i.e. it should be
designed and detailed as a beam if simulated as a beam and be designed and
detailed as a plate structure if simulated as a plate structure. Though not so
common in Hong Kong, if simulation as a “strut-and-tie” model is employed,
the sectional design and r.c. detailing should accordingly be based on the tie and
strut forces so analyzed.

The commonest structural simulation of a transfer plate structure is as an
assembly of plate bending elements analyzed by the finite element method. As
such, the analytical results comprising bending, twisting moments and
out-of-plane shears should be designed for. Reference to Appendix D can be
made for the principles and design approach of the plate bending elements.

136


11.0 Footings

11.1 Analysis and Design of Footing based on the assumption of rigid footing

Cl. 6.7.1 of the Code allows a footing be analyzed as a “rigid footing”
provided it is of sufficient rigidity with uniform or linearly varying pressures
beneath. As suggested by the Code, the critical section for design is at column
or wall face as marked in Figure 11.1, though in case of circular columns, the
critical section may need be shifted into 0.2 times the diameter of the column,
as in consistency with Cl. 5.2.1.2(b) of the Code.

critical sections for design

Footing under pure axial load Footing under eccentric load creating
creating uniform pressure beneath linearly varying pressure beneath

Figure 11.1 – Assumed Reaction Pressure on Rigid Footing
As it is a usual practice of treating the rigid footing as a beam in the analysis
of its internal forces, Cl. 6.7.2.2 of the Code requires concentration of steel
bars in areas with high stress concentrations as illustrated in Figure 11.2.
area with 2/3 of the area with 2/3 of the
required required
reinforcements reinforcements

lc is the greater
of lc1 and lc2
1.5d c 1.5d 1.5d c 1.5d
d is the
effective depth

lc2 2lc1 Plan

Figure 11.2 – Distribution of reinforcing bars when lc > (3c/4 + 9d/4)

137


Cl. 6.7.2.4 of the Code requires checking of shear be based on (i) section
through the whole width of the footing (as a beam); and (ii) local punching
shear check as if it is a flat slab. (Re Worked Example 4.5 in Section 4).


Consider a raft footing under two column loads as shown in Figure 11.3.
Design data are as follows :
Column Loads (for each): Axial Load: D.L. 800 kN L.L. 200 kN
Moment D.L. 100kNm L.L. 20 kNm
Overburden soil : 1.5 m deep
Footing dimensions : plan dimensions as shown, structural depth 400 mm,
cover = 75 mm; Concrete grade of footing : grade 35

400 400
1000

D.L. 100 kNm
L.L. 20kNm for 400
each column

1000
1250 2500 1250
Plan

Figure 11.3 – Footing layout for Worked Example 11.1

(i) Loading Summary :
D.L. Column: 2 ×800 = 1600 kN;
O.W. 5.0 × 2.0 × 0.4 × 24 = 96kN
Overburden Soil 5.0 × 2 × 1.5 × 20 = 300 kN
Total 1996 kN
Moment (bending upwards as shown in Figure 11.3)
2 × 100 = 200 kNm
L.L. Column 2 × 200 = 400 kN.
Moment (bending upwards as shown in Figure 11.3)
2 × 20 = 40 kNm
Factored load : Axial load 1.4 ×1996 + 1.6 × 400 = 3434.4 kN
Moment 1.4 × 200 + 1.6 × 40 = 344 kNm

138


(ii) The pressure beneath the footing is first worked out as :
3434.4 6 × 344
At the upper end : + = 343.44 + 103.2 = 446.64 kN/m2
5× 2 5× 2 2

3434.4 6 × 344
At the lower end : − = 343.44 − 103.2 = 240.24 kN/m2
5× 2 5× 2 2

3434.4 344 × 0.2
Critical section + = 343.44 + 20.64 = 364.08 kN/m2
5× 2 5 × 2 / 12
3

The pressures are indicated in Figure 11.3(a)

446.64 kN/m2
400 Section for
critical design 400

364.08
400 kN/m2

1250 2500 1250

240.24 kN/m2

Figure 11.3(a) – Bearing Pressure for Worked Example 10.1

(iv) At the critical section for design as marked in Figure 11.3(a), the total
shear is due to the upward ground pressure minus the weight of the
footing and overburden soil ( 1.4(0.4 × 24 + 1.5 × 20 ) = 55.44 kN/m2)
 446.64 + 364.08 
which is   × 0.8 × 5 − 55.44 × 0.8 × 5 = 1399.68 kN
 2 
The total bending moment is
 446.64 − 364.08 
2
(364.08 − 55.44)× 0.8 ×5 + 
2
 × 0 .8 × × 5
2

2  2  3
= 581.89 kNm

(v) Design for bending : Moment per m width is :
581.89
= 116.38 kNm/m;
5
d = 400 − 75 − 8 = 317 mm, assume T16 bars
M 116.38 × 10 6
K= 2 = = 1.158 ,
bd 1000 × 317 2
By the formulae in Section 3 for Rigorous Stress Approach,
p 0 = 0.306 %; Ast = 969 mm2/m
As l c = 1250 > 3c / 4 + 9d / 4 = 3 × 400 / 4 + 9 × 317 / 4 = 1013 , two thirds

139


of the reinforcements have to be distributed within a zone of c + 1.5 × 2d
from the centre and on both sides of the column, i.e. a total width of
400 + 1.5 × 317 × 2 = 1.351 m about the centre line of the columns.
Total flexural reinforcements over the entire width is
969 × 5 = 4845 mm2, 2/3 of which in 1.351× 2 = 2.702 m.
So 4845 × 2 / 3 / 2.702 = 1195 mm2/m within the critical zone.
So provide T16 – 150.
Other than the critical zone, reinforcements per metre width is
4845 / 3 / (5 − 2.702 ) = 703 mm2/m. Provide T16 – 275.

Design for Strip Shear : Total shear along the critical section is
1399.86 kN, thus shear stress is
1399.68 × 10 3
v= = 0.883 N/mm2
5000 × 317
1 1
 400  4 1  35  3
1
> v c = 0.79 × 0.306  3
 ×   = 0.505 N/mm2 as per
 317  1.25  25 
Table 6.3 of the Code.
So shear reinforcement required is
Asv b(v − v c ) 5000(0.883 − 0.505) 5000 × 0.4
= = < = 4.998 mm2/mm
sv 0.87 f yv 0.87 × 460 0.87 × 460
Within the two-thirds (of total width 2.675 m) with heavier shear
reinforcement :
2
4.998 × ÷ 2.702 = 1.233 mm2/m. Use T10 – 175 s.w. and – 300 l.w.
3
In the rest of the footing,
1
4.998 × ÷ 2.298 = 0.724 mm/m. Use T10 – 300 BWs.
3

(vi) Check punching shear along perimeter of column
Factored load by a column is 1.4 × 800 + 1.6 × 200 = 1440 kN. By Cl.
6.1.5.6(d), along the column perimeter,
Veff 1440 × 10 3
= = 2.84 < 0.8 f cu = 4.7 MPa. O.K.
ud 4 × 400 × 317
Locate the next critical perimeter for punching shear checking as shown
in Figure 11.3(b) which is at 1.5d from the column face.
Weight of overburden soil and weight of footing is
1.3512 × 55.44 − 1.4 × 0.4 2 × 1.5 × 20 = 94.47 kN
3434.4
Upthrust by ground pressure is × 1.3512 = 627.03 kN
5× 2

140


Net load along the critical perimeter is
1440 + 94.47 − 627.03 = 907.44 kN
Critical perimeter for
punching shear checking

400
400 1000

400+1.5d×2=1351 400

1000
1250
400+1.5d×2=1351
Plan

Figure 11.3(b) – checking punching shear for Worked Example 10.1

By (Ceqn 6.40)

 1.5M t 
Veff = Vt 1 +  = 907.441 + 1.5 × 172  = 1098.41 kN
 
 Vt x sp   907.44 × 1.351 
 

1098.41×10 3
Punching shear stress is v = = 0.641 N/mm2
1351× 4 × 317
As v < 1.6vc = 0.808 , use (Ceqn 6.44) in determining punching shear
reinforcement,
(v − vc )ud = (0.641 − 0.489)×1351× 4 × 317 < 0.4 ×1351× 4 × 317
0.87 f yv 0.87 × 460 0.87 × 460
=1712m2. The reinforcement should be distributed in the manner as that
of flat slab, i.e. with 40%, 685mm2 (i.e. 9 nos. of T10) at
0.5d (158.5mm) and others 1027 mm2 (i.e. 13 nos. of T10)at
1.25d (396.25mm) away from the surface of the column as per the
advice in Figure 6.13 of the Code.
The nos. of links
arrived at is for
illustration purpose.
The actual
1351 717 1192.5 arrangement of links
should take the
9-T10 links spacing of
0.5d=158.5 longitudinal bars
into account.
400 d=317 13-T10 links

Figure 11.3(c) – Area for punching shear reinforcement

141


So the provision by the strip shear obtained in (v) which is greater is
adopted as per Cl. 6.7.2.4 of the Code which requires the more “severe”
provision for checking of strip and punching shears.

(vii) Checking of bending and shear in the direction parallel to the line joining
the columns can be carried out similarly. However, it should be noted
that there is a net “torsion” acting on any section perpendicular to the
line joining the two columns due to linearly varying ground pressure. To
be on the conservative side, shear arising due to this torsion should be
checked and designed accordingly as a beam as necessary. Nevertheless,
one can raise a comment that the design has to some extent be duplicated
as checking of bending has been carried out in the perpendicular
direction. Furthermore, for full torsion to be developed for design in
accordance with (Ceqn 6.65) to (Ceqn 6.68) of the Code, the “beam”
should have a free length of beam stirrup width + depth to develop the
torsion (as illustrated in Figure 3.31 in Section 3) which is generally not
possible for footing of considerable width. As unlike vertical shear where
enhancement can be adopted with “shear span” less than 2d or 1.5d ,
no similar strength enhancement is allowed in Code, though by the same
phenomenon there should be some shear strength enhancement. So full
design for bending in both ways together with torsion will likely result in
over-design.

(viii) The flexural and shear reinforcements provisions for the direction
perpendicular to the line joining the columns is

1345 1345

400
400 T16 – 275(B1)

400

T16 – 150(B1)

Plan
Shear links T10 – Shear links T10 –
Shear links T10 –
175 s.w. 300 l.w. 300 BWs. in
175 s.w. 300 l.w.
other areas

Figure 11.3(c) – Reinforcement Details for Worked Example 11.1 (in the
direction perpendicular to the line joining the two columns only)

142


11.3 Flexible Footing Analysis and Design

As contrast to the footing analyzed under the rigid footing assumption, the
analysis of footing under the assumption of its being a flexible structure will
take the stiffness of the structure and the supporting ground into account by
which the deformations of the structure itself will be analyzed. The
deformations will affect the distribution of the internal forces of the structure
and the reactions which are generally significantly different from that by rigid
footing analysis. Though it is comparatively easy to model the cap structure, it
is difficult to model the surface supports provided by the ground because :
(i) the stiffness of the ground with respect to the hardness of the subgrade
and geometry of the footing are difficult to assess;
(ii) the supports are interacting with one another instead of being
independent “Winkler springs” supports. However, we are currently
lacking computer softwares to solve the problem. Use of constant
“Winkler springs” thus becomes a common approach.

As the out-of-plane deformations and forces are most important in footing
analysis and design, flexible footings are often modeled as plate bending
elements analyzed by the finite element method as will be discussed in 11.4 in
more details.

11.4 Analysis and Design by Computer Method

The followings are highlighted for design of footing modeled as 2-D model
(idealized as assembly of plate bending elements) on surface supports:

(i) The analytical results comprise bending, twisting moments and
out-of-plane shears for consideration in design;
(ii) As local “stresses” within the footing are revealed in details, the rules
governing distribution of reinforcements in footing analyzed as a beam
need not be applied. The design at any location in the footing can be
based on the calculated stresses directly. However, if “peak stresses”
(high stresses dropping off rapidly within short distance) occur at certain
locations as illustrated in Figure 11.4 which are often results of finite
element analysis at points with heavy loads or point supports, it would be
reasonable to “spread” the stresses over certain width for design.

143


Nevertheless, care must be taken not to adopt widths too wide for
“spreading” as local effects may not be well captured.

peak stress

width over
which the
peak stress is
designed for

Figure 11.4 – Spreading of peak stress over certain width for design

(iii) The design against flexure should be done by the “Wood Armer
Equations” listed in Appendix D, together with discussion of its
underlying principles. As the finite element mesh of the mathematical
model is often very fine, it is a practice of “lumping” the design
reinforcements of a number of nodes over certain widths and evenly
distributing the total reinforcements over the widths, as is done by the
popular software “SAFE”. Again, care must be taken in not taking
widths too wide for “lumping” as local effects may not be well captured.
The design of reinforcements by SAFE is illustrated on the right portion
of Figure 11.4;
(iv) The principle together with a worked example for design against shear is
included in Appendix D, as illustrated in Figure D5a to D5c. It should be
noted that as the finite element analysis give detailed distribution of
shear stresses on the structure, it is not necessary to carry out shear
distribution into column and mid-strips as is done for flat slab under
empirical analysis in accordance with the Code. The checking of shear
and design of shear reinforcements can be based directly on the shear
stresses revealed by the finite element analysis.

144


12.0 Pile Caps

12.1 Rigid Cap analysis

Cl. 6.7.3 of the Code allows a pile cap be analyzed and designed as a “rigid
cap” by which the cap is considered as a perfectly rigid structure so that the
supporting piles deform in a co-planar manner at their junctions with the cap.
As the deformations of the piles are governed, the reactions by the piles can
be found with their assigned (or assumed) stiffnesses. If it is assumed that the
piles are identical (in stiffnesses), the reactions of the piles follow a linearly
varying pattern. Appendix I contains derivation of formulae for solution of
pile loads under rigid cap assumption.

pile loads,
magnitude follows
linear profile
under assumption
of equal pile
stiffness

Figure 12.1 – Pile load profile under rigid cap assumption

Upon solution of the pile loads, the internal forces of the pile cap structure
can be obtained with the applied loads and reactions acting on it as a free
body. The conventional assumption is to consider the cap as a beam structure
spanning in two directions and perform analysis and design separately. It is
also a requirement under certain circumstances that some net torsions acting
on the cap structure (being idealized as a beam) need be checked. As the
designer can only obtain a total moment and shear force in any section of full
cap width, there may be under-design against heavy local effects in areas
having heavy point loads or pile reactions. The Code (Cl. 6.7.3.3) therefore
imposes a condition that shear enhancement of concrete due to closeness of
applied load and support cannot be applied.

Cl. 6.7.3.5 of the Code requires checking of torsion based on rigid body

145


theory which is similar to discussion in Section 11.2 (vii).

12.2 Worked Example 12.1 (Rigid Cap Design)

The small cap as shown in Figure 12.2 is analyzed by the rigid cap
assumption and will then undergo conventional design as a beam spanning in
two directions.
Design data : Pile cap plan dimensions : as shown
Pile cap structural depth : 2 m
Pile diameter : 2 m
Concrete grade of Cap : 35
Cover to main reinforcements : 75 mm
Column dimension : 2 m square
Factored Load from the central column :
P = 50000 kN
M x = 2000 kNm (along X-axis)
M y = 1000 kNm (along Y-axis)

1500
P1 P2 P3

400 Y
3000

3000
X
P4 P5 P6
1500 400
critical sections
for shear
checking
1500 4000 4000 1500

Figure 12.2 – Pile cap layout of Worked Example 12.1

(i) Factored Loads from the Column :
P = 50000 kN
M x = 2000 kNm (along positive X-axis)
M y = 1000 kNm (along positive Y-axis)
O.W. of Cap 11 × 9 × 2 × 24 = 4752 kN
Weight of overburden soil 11× 9 × 1.5 × 20 = 2970 kN
Factored load due to O.W. of Cap and soil is

146


1.4 × (4752 + 2970) = 10811 kN
So total axial load is 50000 + 10811 = 60811 kN

(ii) Analysis of pile loads – assume all piles are identical
(Reference to Appendix I for general analysis formulae)
I x of pile group = 6 × 3 2 = 54
I y of pile group = 4 × 4 2 + 2 × 0 = 64
60811 2000 × 4 1000 × 3
Pile Loads on P1 : − + = 10065.72 kN
6 64 54
60811 2000 × 0 1000 × 3
P2: − + = 10190.72 kN
6 64 54
60811 2000 × 4 1000 × 3
P3: + + = 10315.72 kN
6 64 54
60811 2000 × 4 1000 × 3
P4: − − = 9954.61 kN
6 64 54
60811 2000 × 0 1000 × 3
P5: − − = 10079.61 kN
6 64 54
60811 2000 × 4 1000 × 3
P6: + − = 10204.61 kN
6 64 54

(iii) Design for bending along the X-direction
The most critical section is at the centre line of the cap
Moment created by Piles P3 and P6 is
(10315.72 + 10204.61)× 4 = 82081.32 kNm
Counter moment by O.W. of cap and soil is
10811 ÷ 2 × 2.75 = 14865.13 kNm
The net moment acting on the section is
82081.32 − 14865.13 = 67216.19 kNm
d = 2000 − 75 − 60 = 1865 (assume 2 layers of T40); b = 9000
M 67216.19 ×10 6 z
= = 2.147 ; = 0.926 p = 0.58 %
bd 2
9000 ×1865 2
d
Ast = 97210 mm2, provide T40 – 200 (2 layers, B1 and B3)

(iv) Design for shear in the X-direction
By Cl. 6.7.3.2 of the Code, the critical section for shear checking is at
20% of the diameter of the pile inside the face of the pile as shown in
Figure 12.2.
Total shear at the critical section is :

147


Upward shear by P3 and P6 is 10315.72 + 10204.61 = 20520.33 kN
Downward shear by cap’s O.W. and soil is
2.1
10811× = 2063.92 kN
11
Net shear on the critical section is 20520.33 − 2063.92 = 18456.41 kN
18456.41× 10 3
v= = 1.10 N/mm2 > vc = 0.58 N/mm2 by Table 6.3 of
9000 × 1865
the Code.
No shear enhancement in concrete strength can be effected as per Cl.
6.7.3.3 of the Code because no shear distribution across section can be
considered.

Shear reinforcements in form of links per metre width is
Asv b(v − vc ) 1000(1.10 − 0.58)
= = = 1.299
sv 0.87 f yv 0.87 × 460
Use T12 links – 200 in X-direction and 400 in Y-direction by which
Asv
provided is 1.41.
sv

(v) Design for bending along the Y-direction
The most critical section is at the centre line of the cap
Moment created by Piles P1, P2 and P3
(10065.72 + 10190.72 + 10315.72)× 3 = 30572.16 × 3 = 91716.48 kNm
Counter moment by O.W. of cap and soil is
10811 ÷ 2 × 2.25 = 12162.38 kNm
The net moment acting on the section is
91716.48 − 12162.38 = 79554.11 kNm
d = 2000 − 75 − 60 − 40 = 1825 ; (assume 2 layers of T40) b = 11000
M 76554.11× 10 6 z
= = 2.09 ; = 0.929 p = 0.55%
bd 2
11000 × 1825 2
d
Ast = 110459 mm2, provide T40 – 200 (2 layers, B2 and B4)

(vi) Checking for shear in the Y-direction
By Cl. 6.7.3.2 of the Code, the critical section for shear checking is at
20% of the diameter of the pile inside the face of the pile as shown in
Figure 12.2
Total shear at the critical section is :
Upward shear by P1, P2 and P3 is 30572.16 kN

148


Downward shear by cap’s O.W. and soil is
2.1
10811× = 2522.57 kN
9
Net shear on the critical section is 30572.16 − 2522.57 = 28049.59 kN
28049.59 × 103
v= = 1.397 N/mm2 > vc = 0.579 N/mm2 by Table 6.3
11000 × 1825
of the Code.

Similar to checking of shear checking in X-direction, no shear
enhancement of concrete strength can be effected.

Shear reinforcements in form of links per metre width is
Asv b(v − vc ) 1000(1.397 − 0.579 )
= = = 2.044
sv 0.87 f yv 0.87 × 460
Asv
As in Y-direction is greater than that in X-direction, so adopt this
sv
for shear reinforcement provision.
Asv
Use T12 links – 200 BWs by which provided is 2.82.
sv

(vii) Punching shear :
Punching shear check for the column and the heaviest loaded piles at
their perimeters in accordance with Cl. 6.1.5.6 of the Code :
1.25 × 50000 × 10 3
Column : = 4.28 MPa < 0.8 f cu = 4.73 MPa.
4 × 2000 × 1825
1.25 ×10315 × 10 3
Pile P3 : = 1.12 MPa < 0.8 f cu = 4.73 MPa.
2000π ×1825
Not necessary to check punching shear at the next critical perimeters
as the piles and column overlap with each other to very appreciable
extents;

(viii) Checking for torsion : There are unbalanced torsions in any full width
sections at X-Y directions due to differences in the pile reactions.
However, as discussed in sub-section 11.2(vii) of this Manual for
footing, it may not be necessary to design the torsion as for that for
beams. Anyhow, the net torsion is this example is small, being
361.11× 3 = 1083.33 kNm (361.11kN is the difference in pile loads
between P3 and P4), creating torsional shear stress in the order of

149


2T 2 ×1083.33 ×10 6
vt = = = 0.065 N/mm2. So the
2 h   2000 
hmin  hmax − min  2000 2  9000 − 
 3   3 
torsional shear effects should be negligible;

(ix) Finally reinforcement details are as shown in Figure 12.3,

1500
P1 P2 P3

Y

T40-200 (T1,B1,B3)
3000

T40-200 (T2, B2,B4)
3000
X
P4 P5 P6
1500 Shear links
T12 – 200 BWs on
the whole cap

1500 4000 4000 1500

Figure 12.3 – Reinforcement Design of Worked Example 12.1

12.3 Strut-and-Tie Model

Cl. 6.7.3.1 of the Code allows pile cap be designed by the truss analogy, or
more commonly known as “Strut-and-Tie Model” (S&T Model) in which a
concrete structure is divided into a series of struts and ties which are
beam-like members along which the stress are anticipated to follow. In a
S&T model, a strut is a compression member whose strength is provided by
concrete compression and a tie is a tension member whose strength is
provided by added reinforcements. In the analysis of a S&T model, the
following basic requirements must be met (Re ACI Code 2002):
(i) Equilibrium must be achieved;
(ii) The strength of a strut or a tie member must exceed the stress induced
on it;
(iii) Strut members cannot cross each other while a tie member can cross
another tie member;
(iv) The smallest angle between a tie and a strut joined at a node should
exceed 25o.

150


The Code has specified the following requirements in Cl. 6.7.3.1 :
(i) Truss be of triangular shape;
(ii) Nodes be at centre of loads and reinforcements;
(iii) For widely spaced piles (pile spacing exceeding 3 times the pile
diameter), only the reinforcements within 1.5 times the pile diameter
from the centre of pile can be considered to constitute a tension
member of the truss.

12.4 Worked Example 12.2 (Strut-and-Tie Model)

Consider the pile cap supporting a column factored load of 6000kN
supported by two piles with a column of size 1m by 1 m. The dimension of
the cap is as shown in Figure 12.4, with the width of cap equal to 1.5 m.

6000kN

2500

Elevation

3000 3000
1000 1000
dia. dia.

1500

Plan

Figure 12.4 – Pile Cap Layout of Worked Example 12.2

(i) Determine the dimension of the strut-and-tie model
Assume two layers of steel at the bottom of the cap, the centroid of the
two layers is at 75 + 40 + 20 = 135 mm from the base of the cap. So the

151


effective width of the tension tie is 135 × 2 = 270 mm. The dimensions
and arrangement of the ties and struts are drawn in Figure 12.5.

(ii) A simple force polygon is drawn and the compression in the strut can be
simply worked out as ( C is the compression of the strut) :
2C sin 38.250 = 6000 ⇒ C = 4845.8 kN;
And the tension in the bottom tie is T = C cos 38.250 = 3805.49 kN.

1000

top strut 6000kN concrete
width = strut
619.09mm

2230 2500

270

bottom tie, strength
Bottom strut be provided by steel
width =
831.13mm
3000 3000
1000 1000
dia. dia.
Elevation

6000 kN

C = 4845.8kN C = 4845.8kN 2230+
270÷2
=2365

38.25o 38.25o
T = 3805.49kN

3000 kN 3000 kN

3000 3000

Figure 12.5 – Analysis of strut and tie forces in Worked Example 12.2

(iii) To provide the bottom tension of 3805.49 kN, the reinforcement steel

3805.49 × 103 3805.49 × 10 3
required is = = 9509 mm2. Use 8–T40;
0.87 f y 0.87 × 460

(iv) Check stresses in the struts :

152


Bottom section of the strut, the strut width at bottom is
1000 sin 38.250 + 270 cos 38.250 = 831.13 mm
As the bottom part is in tension, there is a reduction of compressive

strength of concrete to 1.8 f cu = 1.8 35 = 10.08 MPa as suggested by

OAP, which is an implied value of the ultimate concrete shear strength of

0.8 f cu as stated in the Code and BS8110.

As a conservative approach, assuming a circular section at the base of
the strut since the pile is circular, the stress at the base of the strut is
4845.8 ×10 3
= 8.93 MPa < 10.08MPa
8312 π / 4
For the top section of the strut, the sectional width is
2 × 500 sin 38.25 = 619.09 mm
0

As the sectional length of the column is 1 m, it is conservative to assume
a sectional area of 1000mm × 619.09 mm.
The compressive stress of the strut at top section is
4845.8 × 103
= 7.83 MPa < 0.45 f cu = 15.75 MPa
1000 × 619.09

(v) The reinforcement details are indicated in Figure 12.6. Side bars are
omitted for clarity.

1000

4T25 T1

T16 s.s. – 200
2500

4T40 B1 & 4T40 B2

3000 3000
1000 1000
dia. dia.
Elevation

Figure 12.6 – Reinforcement Details of Worked Example 12.2

153


12.5 Flexible Cap Analysis

A pile cap can be analyzed by treating it as a flexible structure, i.e., as in
contrast to the rigid cap assumption in which the cap is a perfectly rigid body
undergoing rigid body movement only with no deformation upon the
application of loads, the flexible pile cap structure will deform and the
deformations will affect the distribution of internal forces of the structure and
the reactions. Analysis of the flexible cap structure will require input of the
stiffness of the structure which is comparatively easy. However, as similar to
that of footing, the support stiffness of the pile cap which is mainly offered
by the supporting pile is often difficult, especially for the friction pile which
will interact significantly with each other through the embedding soil. Effects
by soil restraints on the piles can be considered as less significant in
end-bearing piles such large diameter bored piles.

Similar to the flexible footing, as the out-of-plane loads and deformation are
most important in pile cap structures, most of the flexible cap structures are
modeled as plate structures and analyzed by the finite element method.

12.6 Analysis and Design by Computer Method

Analysis and design by computer method for pile cap are similar to Section
11.3 for footing. Nevertheless, as analysis by computer methods can often
account for load distribution within the pile cap structure, Cl. 6.7.3.3 of the
Code has specified the followings which are particularly applicable for pile
cap design :

(i) shear strength enhancement of concrete may be applied to a width of
3φ for circular pile, or pile width plus 2 × least dimension of pile as
shown in Figure 12.7 as shear distribution across section has generally
been considered in flexible cap analysis;

154


Area where shear
enhancement may
φ
apply B

φ B

φ B

av av av av

Figure 12.7 – Effective width for shear enhancement in pile cap around a pile

(ii) averaging of shear force shall not be based on a width > the effective
depth on either side of the centre of a pile, or as limited by the actual
dimension of the cap.

d : effective
depth of cap

Width over which
d
X φ the shear force can
be averaged in the
X
cap for design
≤d

peak shear at Shear force diagram along X-X
pile centre

Figure 12.8 – Width in cap over which shear force at pile can be
averaged for Design

Illustration in Figure 12.8 can be a guideline for determination of
“effective widths” adopted in averaging “peak stresses” as will often
be encountered in finite element analysis for pile cap structure modeled
as an assembly of plate bending elements under point loads and point
supports, as in the same manner as that for footing discussed in 11.4(ii)
of this Manual.

155


13.0 General Detailings

13.1 In this section, the provisions of detailing requirements are general ones
applicable to all types of structural members. They are mainly taken from
Section 8 of the Code. Requirements marked with (D) are ductility ones for
beams and columns contributing in lateral load resisting system.

13.2 Minimum spacing of reinforcements (Cl. 8.2 of the Code) – clear distance
(horizontal and vertical) is the greatest of
(i) maximum bar diameter;
(ii) maximum aggregate size (hagg) + 5 mm;
(iii) 20 mm.

13.3 Permissible bent radii of bars. The purpose of requiring minimum bend radii
for bars are
(i) avoid damage of bar;
(ii) avoid overstress by bearing on concrete in the bend.
Table 8.2 of the Code requires the minimum bend radii to be 3φ for
φ ≤ 20 mm and 4φ for φ > 20 mm (for both mild steel and high yield bar)
and can be adopted without causing concrete failures if any of the conditions
shown in Figure 13.1 is satisfied as per Cl. 8.3 of the Code.

Bar of Bar of
≥8∅ or D/2 diameter ∅
diameter ∅
TL2

Point beyond which bar
TL1 assumed not be stressed
4∅ 4∅ at ultimate limit state

condition (a) condition (b)

TL1 ≥ required
anchorage length for Bar of
beam contributing to diameter ∅
lateral load resisting
system;
TL2 ≥ required cross bar of diameter
anchorage length for ≥ ∅ inside the bend
beam not
contributing to
lateral load resisting condition (c)

Figure 13.1 – Conditions by which concrete failure be avoided by bend of bars

156


If the none of the conditions in Figure 13.1 is fulfilled, (Ceqn 8.1) of the Code,
reproduced as (Eqn 13.1) in this Manual should be checked to ensure that
bearing pressure inside the bend is not excessive.

Fbt 2 f cu
bearing stress = ≤ (Eqn 13.1)
rφ  φ 
1 + 2
 

 ab 
In (Eqn 13.1), Fbt is the tensile force in the bar at the start of the bend; r
the internal bend radius of the bar; φ is the bar diameter, a b is centre to
centre distance between bars perpendicular to the plane of the bend and in case
the bars are adjacent to the face of the member, a b = φ + cover.
Take an example of a layer of T40 bars of centre to centre separation of 100
mm and internal bend radii of 160mm in grade 35 concrete.
Fbt = 0.87 × 460 × 1257 = 503051 N

Fbt 503051 2 f cu 2 × 35
= = 78.6 > = = 38.89
rφ 160 × 40  φ   40 
1 + 2
 
 1 + 2 × 
 ab   100 

So (Ceqn 8.1) is not fulfilled. Practically a cross bar should be added as in
Figure 13.1(c) as conditions in Figure 13.1(a) and 13.1(b) can unlikely be
satisfied.

13.4 Anchorage of longitudinal reinforcements

(i) Anchorage is derived from ultimate anchorage bond stress with concrete
assessed by the (Ceqn 8.3) of the Code.
f bu = β f cu where for high yield bars β = 0.5 for tension and
β = 0.65 for compression. For example, f bu = 0.5 35 = 2.96 MPa for
grade 35. For a bar of diameter φ , the total force up to 0.87 f y
 φ 2π 
is 0.87 f y 
 4  . The required bond length L will then be related by

 
φ π 
2 0.87 f y φ

0.87 f y   = β f cu πφL ⇒ L = = 33.8φ ≈ 34φ which agrees
 4  4β f cu
with Table 8.5 of the Code;

(ii) Notwithstanding provision in (i), it has been stated in 9.9.1.1(c) of the
Code which contains ductility requirements for longitudinal bars of
beams (contributing in lateral load resisting system) anchoring into

157


exterior column requiring anchorage length to be increased by 15% as
discussed in Section 3.6 (v); (D)

(iii) With the minimum support width requirements as stated in Cl. 8.4.8 of
the Code, bends of bars in end supports of slabs or beams will start
beyond the centre line of supports offered by beams, columns and walls.
By the same clause the requirement can be considered as not confining
to simply supported beam as stated in Cl. 9.2.1.7 of the Code as
illustrated in Figure 13.2.

≥2(4Ø+c) if Ø ≤ 20
≥2(5Ø+c) if Ø > 20

c

3Ø if Ø ≤ 20;
4Ø if Ø > 20

≥0 centre line of support

Figure 13.2 – Support width requirement

13.5 Anchorage of links – Figure 8.2 of the Code displays bend of links of bend
angles from 90o to 180o. However, it should be noted that the Code requires
anchorage links in beams and columns contributing in lateral load resisting
system to have bent angles not less than 135o as ductility requirements (D);

13.6 Laps arrangement – Cl. 8.7.2 of the Code requires laps be “normally”
staggered with the followings requirement for 100% lapping in one single
layer:

(i) Sum of reinforcement sizes in a particular layer must not exceed 40% of
the breadth of the section at that level, otherwise the laps must be
staggered;
(ii) Laps be arranged symmetrically;
(iii) Details of requirements in bar lapping are indicated in Figure 8.4 of the
Code reproduced in Figure 13.3 for ease of reference;

158


≥0.3l0 l0 If clear distance
≤50mm between 2 lapping
≤4ø bars > 4ø or 50
mm by x, l0
should be
a : distance between increased by x
adjacent laps

≥20mm
≥2ø

Figure 13.3 – Lapping arrangement for tension laps

(iv) When Figure 13.3 is complied with, the permissible percentage of lapped
bars in tension may be 100% (but still required to be staggered, i.e. not
in the same section)where the bars are all in one layer. When the bars are
in several layers, the percentage should be reduced to 50%;
(v) Compression and secondary reinforcements can be lapped in one section.

The Clause effectively requires tension laps to be staggered with arrangement
as shown in Figure 13.3 which is applicable in to the flexural steel bars in
beams, slabs, footings, pile caps etc. Fortunately, the Code allows compression
and secondary bars be lapped in one section, i.e. without the necessity of
staggered laps. As such staggered laps can be eliminated in most of the
locations in columns and walls.

13.7 Lap Lengths (Cl. 8.7.3 of the Code)

The followings should be noted for tension lap lengths:

(i) Absolute minimum lap length is the greater of 15φ and 300 mm;
(ii) Tension lap length should be at least equal to the design tension
anchorage length and be based on the diameter of the smaller bar;
(iii) Lap length be increased by a factor of 1.4 or 2.0 as indicated in Figure
13.4 which is reproduced from Figure 8.5 of the Code.

159


Top bars Note
< 2φ ≥ 2φ Condition 1:
Lap at top as cast
and cover < 2φ ;
< 2φ < 75 ≥ 75
and 6φ ≥ 75 ≥ 75 < 2φ Condition 2 :
or 6φ
and 6φ and 6φ Lap at corner and
cover < 2φ
Factor 2 2 1.4 1.0 1.4
Condition 3 :
Clear distance
Bottom bars between adjacent
laps < 75 or 6φ
< 75 ≥ 75 ≥ 75
< 2φ > 2φ Any one of the 3
or 6φ and 6φ and 6φ
conditions, factor
is 1.4.

Condition 1 + 2 or
< 2φ
conditions 1 + 3 :
factor is 2.0
Factor 1.4 1.4 1.0 1.4

Figure 13.4 – Factors for tension lapping bars

The compression lap length should be at least 25% greater than the design
compression anchorage length as listed in Table 8.4 of the Code.

13.8 Transverse reinforcement in the tension lap zone (Cl. 8.7.4 of the Code)

For lapped longitudinal bars in tension, the transverse reinforcement is used to
resist transverse tension forces. 3 cases be considered as :

(i) No additional transverse reinforcement is required (existing transverse
reinforcement for other purpose can be regarded as sufficient to resist
the transverse tension forces) when the longitudinal bar diameter
φ < 20 mm or percentage of lapping in any section < 25%;
(ii) When φ ≥ 20 mm, the transverse reinforcement should have area

∑A st ≥ As where As is the area of one spiced bar and be placed

between the longitudinal bar and the concrete surface as shown in
Figure 13.5;

160


For T40 bars in tension lap in concrete
grade 35 with spacing 200 mm (≤ 10ø
= 400mm) and lap length
l0 = 1.4×standard lap = 1920 mm.
≤ 4φ
≤ 4φ Transverse reinforcement area
required is∑ Ast = 1257 mm2.

l0 Use 12T12, spacing is 175 mm along
the lapped length.
One bar be outside lap if
the lap is in compression

Transverse bars between
longitudinal bars and
concrete surface

Figure 13.5 – Transverse reinforcement for lapped splices –not greater than 50% of
reinforcement is lapped at one section and φ ≥ 20 mm

(iii) If more than 50% of the reinforcement is lapped at one point and the
distance between adjacent laps ≤ 10φ , the transverse reinforcement
should be formed by links or U bars anchored into the body of the
section. The transverse reinforcement should be positioned at the outer
sections of the lap as shown in Figure 13.6;

It should be noted that effectively condition (ii) requires area of transverse
reinforcement identical to that of (iii), except that the bars need not be
concentrated at the ends of the laps and the transverse reinforcements be in
form of links or U bars.

13.9 Transverse reinforcement in the permanent compression lap zone

The requirement will be identical to that of tension lap except for an additional
requirement that one bar of the transverse reinforcement should be placed
outside each of the lap length and within 4φ of the ends of the lap length also
shown in Figure 13.5 and 13.6.

161


For T40 bars in tension lap in
∑ Ast / 2 ∑ Ast / 2 concrete grade 35 with spacing 200
mm with lap length
l0/3 l0/3 l0 = 1.4×standard lap = 1920 mm.
≤150mm
l0/3=640mm
≤ 4φ
Transverse reinforcement area
∑
required is Ast = 1257 mm2.
≤ 4φ
For tension lap, on each
l0/3=640mm, 1257/2 = 629mm2 is
required. So use 6T12, area
provided is 678mm2 over
l0 1950/3=640 mm zone, i.e. spacing
is 128mm < 150mm.

One bar be outside lap if the For compression lap, also use 7T10,
lap is in compression (628mm2=1257/2) with 6T12
Transverse bars in
within 640mm (equal spacing =
form of U bars or
128mm) and the 7th T10 at 160mm
links
(=4∅) from the end of lap.

Lapping longitudinal bars

Figure 13.6 – Transverse reinforcement for lapped splices – more than 50% is
lapped at one section and clear distance between adjacent laps ≤ 10∅

162


14.0 Design against Robustness

14.1 The Code defines the requirement for robustness in Clause 2.1.4 as “a
structure should be designed and constructed so that it is inherently robust and
not unreasonably susceptible to the effects of accidents or misuse, and
disproportionate collapse.” By disproportionate collapse, we refer to the
situation in which damage to small areas of a structure or failure of single
elements may lead to collapse of large parts of the structure.

14.2 Design requirements comprise :

(i) building layouts checked to avoid inherent weakness;
(ii) capable to resist notional loads simultaneously at floor levels and roof as
shown in Figure 14.1. (Re Cl. 2.3.1.4(a) of the Code which also requires
that applied ultimate wind loads should be greater than these notional
values);
WR, characteristic
dead weight between
0.015WR
roof and next mid
} floor heights

0.015WN+1
} WN+1, characteristic
dead weight between
mid floor heights

0.015WN
} WN, characteristic dead
weight between mid
floor heights

}
0.015W1 W1, characteristic dead
weight between mid
floor heights

Figure 14.1 – Illustration of notional loads for robustness design

(iii) provides effective horizontal ties (in form of reinforcements embedded
in concrete) (a) around the periphery; (b) internally; (c) to external
columns and walls; and (d) vertical ties as per Cl. 6.4.1 of the Code, the
failure of which will lead to requirement of checking key elements in
accordance with Cl. 2.2.2.3 of the Code.

14.3 Principles in Design of ties (Cl. 6.4.1.2 and Cl 6.4.1.3 of the Code)

163


(i) The reinforcements are assumed to be acting at f y instead of 0.87 f y ;

(ii) To resist only the tying forces specified, not any others;
(iii) Reinforcements provided for other purpose can also act as ties;
(iv) Laps and anchorage of bars as ties similar to other reinforcements;
(v) Independent sections of a building divided by expansion joints have
appropriate tying system.

14.4 Design of ties

(i) Internal Ties be provided evenly distributed in two directions in slabs –
design force is illustrated in Figure 14.2 with example. The tie
reinforcements can be grouped and provided in beam or wall.

40 storeys building with average
floor dead load as 9 kPa and average
floor live load as 3 kPa
Internal ties in Y-direction
n0 = 40
lry1
lrx1 lrx2 Gk = 9 kPa; Qk = 3 kPa
l ry = 6 m
Ft ≤ 60 and 20 + 4n0 = 180
∴ Ft = 60
G k + Qk l r
Ft = 115.2
7.5 5
lry2 1.0 Ft = 60
B1 Design tensile force in ties is
115.2 kN/m
Required Ast for internal ties is
T.A.L.
F 115.2 × 10 3
= = 250 mm2/m
fy 460
Use T10 – 300
lrx1 =6m, lrx2 =4m, (can likely be met by DB or rebars
lrx ≥ lrx1 and lrx2; Design force (kN/m)
provided for strength purpose)
lrx=6m; G + Qk l ry
≥ k Ft and 1.0 Ft
lry1 =lrx1 =6m 7.5 5 If tying bars grouped in beam, total
lry ≥ lry1 and lry2 where Ft ≤ 60 and 20 + 4n0 rebars in the middle beam B1 may
lry=6m be 250 × 5 = 1250 mm2. So likely
and n0 is the no. of storeys
can be met by the longitudinal bars.

Figure 14.2 – Derivation of internal tie reinforcement bars in slabs (evenly distributed)

164


(ii) Peripheral ties – Continuous tie capable resisting 1.0Ft, located within
1.2 m of the edge of the building or within perimeter wall;

Peripheral area of 1.2 m wide
with ties to resist 1.0Ft, i.e.
the lesser of 60kN/m and
(20 + 4n0 ) kN/m; or within
perimeter wall perimeter wall
1.2m

Figure 14.3 – Location and determination of Peripheral ties

(iii) External columns and wall to have ties capable of developing forces as
indicated in Figure 14.4;

Corner column in 40 storey
building, tie design force >
(i) 2.0Ft=2×60=120kN;
Ties to take up (ls/2.5)Ft
tensile force being =(3/2.5)×60=72kN;
the greater of Smaller is 72kN
(i) 2.0Ft or (ls/2.5)Ft (ii) 400×600, grade 40 with
if less; and (ii) 3% 8T32 steel, design
of design ultimate ultimate load is
load of column or (400×600×0.45×40+
wall where ls is floor 0.87×460×6434)×10-3
to ceiling height =6895kN; 3% is 207kN
So design tie force is 207kN.
Provide F/fy =
207×103/460 = 517mm2; can
likely be provided by beam
steel anchoring into the
column.

Figure 14.4 – Ties to external column and wall with example

(iv) Vertical ties provided to wall and column should be continuous and be
capable of carrying exceptional load. Use γf ×[dead load + 1/3 imposed
load + 1/3 wind load] of one floor to determine the design load for the
vertical ties where γf = 1.05.

165


Design load for the
vertical tie of the interior
column C1 is
(i) Dead load from one
7m storey 800kN;
(ii) One third of
imposed load from
C1 one storey 1/3×240
= 80 kN;
(iii) One third of wind
load = 1/3×90=30kN
Designed tensile load is
1.05(800+80+30)
8m = 956kN.
8m 6m
Requiring 956000/460
= 2078mm2

Figure 14.5 – Design of Vertical Ties in Columns and Walls

14.5 Design of “Key Elements”

By Cl. 2.2.2.3 of the Code, when for some reasons it is not possible to
introduce ties, key elements (usually columns or walls), the failure of which
will cause disproportionate collapse should be identified. If layout cannot be
revised to avoid them, design these elements and the supporting building
components to an ultimate load of 34 kN/m2, from any direction, to which no
partial safety factor shall be applied. The Code has not defined the extent of
“disproportionate collapse” for the element to be qualified as a “key element”.
However, reference can be made to the “Code of Practice for the Structural
Use of Steel 2005” Cl. 2.3.4.3 by which an element will be considered a key
element if the removal of it will cause collapse of 15% of the floor area or
70m2, whichever is the greater. The design is illustrated in Figure 14.6.

166


Examples

C1 is identified as the key
element as the collapse of
7m which will lead to
disproportionate collapse
of area around it (more
than 15% collapse of the
C1
floor).
The tributary area is
B1 7.5×7=52.5m2;
The design load is
34×52.5=1785kN.
8m
8m 6m 2m
Similarly, beam B1 is also
required as the removal of
which will cause more than
15% collapse of the floor.
So B1 needs be designed
for a u.d.l. of 34kPa on the
linking slabs.
Figure 14.5 – Design of Key Elements

14.6 Nevertheless, it should be noted that requirements in the Code for robustness
design often poses no additional requirements in monolithic reinforced
concrete design in comparison with the criteria listed in 14.2 :
(i) normally no inherent weakness in the structure for a reasonable
structural layout;
(ii) ultimate wind loads normally applied to the structure according to the
local Wind Code can usually cover the notional loads (1.5%
characteristic dead weight) specified in 14.2(ii);
(iii) requirements for various types of ties can normally be met by the
reinforcements provided for other purposes. Nevertheless, continuity of
the ties should be checked.

167


15.0 Shrinkage and Creep

15.1 Shrinkage

Shrinkage is the shortening movement of concrete as it dries after hardening.
If the movement is restrained, stress and/or cracking will be created.

(Ceqn 3.5) gives estimate of drying shrinkage of plain concrete under
un-restrained conditions. Together with the incorporation of the
“reinforcement coefficient” K s , the equation can be written as

ε s = cs K L K c K e K j K s (Eqn 15-1)

where c s = 3.0 and other coefficients can be found by Figures 3.3, 3.5, 3.6
and 3.7 and (Ceqn 3.4) of the Code, depending on atmospheric humidity,
dimensions, compositions of the concrete, time and reinforcement content. It

should be noted that K j is a time dependent coefficient.

The equation and the figures giving values of the various coefficients are
adopted from BS5400 which in turn are quoted from CEB-FIP International
Recommendation for the Design and Construction, 1970 (CEB 1970) (MC-70).
It should, however, be noted that the coefficient c s is extra to CEB-FIP. The
value accounts for the comparatively higher shrinkage value (3 times as high)
found in Hong Kong.

Shrinkage is always in contraction.

15.2 Creep

Creep is the prolonged deformation of the structure under sustained stress.
(Ceqn 3.2) and (Ceqn 3.3) give estimate of the creep strain :
stress
Creep strain = × φc (Eqn 15-2)
E28

where φc = K L K m K c K e K j K s (Eqn 15-3)

Again (Eqn 15-3) has incorporated the reinforcement coefficient K s .

Thus creep strain depends on the stress in the concrete and various coefficients
related to parameters similar to that of shrinkage (which can be read from

168


Figures 3.1 to 3.5 of the Code) with K m and K j dependent on time. As

stress and strain are inter-dependent, it will be shown that assessment of strain
will require successive time staging in some cases.

Creep creates deformation in the direction of the stress. In case shrinkage
which results in tensile stress under restrained condition such as a floor
structure under lateral restraints, the creep strain will serve to relax the stress
due to shrinkage. Both stress and strain due to shrinkage and creep vary with
time, as can be shown in the analyses that follow.

15.3 The determination of the time dependent coefficients K m and K j as listed

in (Ceqn 3.3) and (Ceqn 3.5) will be tedious in calculation of stress and strain
of a structure in a specified time step which may involve reading the figures
many times. Curves in Figures 3.2 and 3.5 are therefore simulated by
polynomial equations as shown in Appendix J to facilitate determination of the
coefficients by spreadsheets.


A grade 35 square column of size 800 × 800 in a 4 storey building with
reinforcement ratio 2% is under an axial stress from the floors as follows :

Floor Height (m) Time of stress creation from floor (days) Stress (MPa)
G 4 28 3.5
st
1 3 56 2.1
nd
2 3 84 2.1
rd
3 3 120 3.5

Strain and shortening of the G/F column due to shrinkage and creep at 360
days are determined as follows :

Shrinkage
The coefficients for determination of the free shrinkage strain are as follows :
K L = 0.000275 for normal air from Figure 3.6;
Based on empirical formulae, for grade 35:
Water / Cement ratio = − 0.0054 f cu + 0.662 = −0.0054 × 35 + 0.662 = 0.473

169


Cement content = 3.6 f cu + 308 = 3.6 × 35 + 308 = 434 kg/m3
From Figure 3.3 K c = 1.17 ;

For the 800 × 800 column, the effective thickness he , defined as the ratio of

the area of the section A, to the semi-perimeter, u/2 (defined in Cl. 3.1.7 of the
800 × 800
Code) is = 400 mm. So from Figure 3.7, K e = 0.55 ;
800 × 4 / 2

From Figure 3.5, time at 360 days K j = 0.51 ;

1 200
Ks = = 0.856 ; where ρ = 0.02 (2% steel) and α e = = 8.44
1 + ρα e 23.7
So the shrinkage strain under perfectly free condition is :

ε s = cs K L K c K e K j K s = 3.0 × 0.000275 × 1.17 × 0.55 × 0.51× 0.856 = 231.76 ×10 −6

Creep
stress
For estimation of creep strain, Creep strain εc = × φc
E28

where φc = K L K m K c K e K j K s

E 28 = 23.7 GPa for grade 35 concrete.
All coefficients are same as that for shrinkage except K L = 2.3 (Figure 3.1),
K m = 1.0 (Figure 3.2 – loaded at 28 days) and K e = 0.72 (Figure 3.4)

Concrete
Load Time since Stress by
age at time εc
from Km Loading Kj φc Floor
Floor
of load
(Day) (MPa) (×10-6)
(Day)
1/F 28 1 332 0.489 0.811 3.5 119.73
2/F 56 0.85 304 0.467 0.658 2.1 58.30
3/F 84 0.761 276 0.443 0.560 2.1 49.61
Roof 120 0.706 240 0.412 0.482 3.5 71.15
∑ εc = 298.80

So the creep strain at 360 days is ε c = 298.80 × 10 −6

Elastic strain
σ 3.5 + 2.1 + 2.1 + 3.5
The elastic strain is simply ε e = = = 472.57 × 10 −6
E 23700
So the total strain is

170


ε = ε s + ε c + ε e = 231.76 ×10 −6 + 298.80 ×10 −6 + 472.57 ×10 −6 = 1003.13 ×10 −6 .

Total shortening of the column at G/F is at 360 days is
ε × H = 1003.13 ×10 −6 × 4000 = 4.01 mm.

15.5 Estimation of shrinkage and creep effect on restrained floor structure

It is well known that shrinkage and creep effects of long concrete floor
structures can be significant. The following derivations aim at providing a
design approach to account for such effects based on recommendations by the
Code.

Consider a floor structure spanning on vertical members of lateral support

stiffness K sup1 and K sup 2 as shown in Figure 15.1.

Let the lateral deflections at supports 1 and 2 be δ 1 and δ 2 . At any time
when the floor structure has an internal stress σ , a free shrinkage strain ε s ,
creep strain ε c , elastic strain ε e , internal force, by displacement
compatibility, the followings can be formulated :
σ σ σA σA
εc = φc , ε e = , K sup1δ 1 = K sup 2δ 2 = F = σA ⇒ δ 1 = ; δ2 =
E E K sup1 K sup 2
(ε s − ε c − ε e )L = δ 1 + δ 2
 σ σ σA σA Eε s
⇒  ε s − φc −  L = δ 1 + δ 2 = + ⇒σ =
 E E K sup1 K sup 2 AE  1
 1  
1 + φc + +
L  K sup1 K sup 2 
 

(Eqn 15-4)

If the floor structure undergoes no net lateral deflection at a point P at Le
from support 2, it can be visualized as if the floor structure is divided into 2
floor structures both fixed at P and undergoes deflection δ 1 at the left portion
and δ 2 at the right portion. By constant strain (implying linearly varying
displacement) in the floor structure :

δ2 L  1 1 
Le = L=  +  (Eqn 15-5)
δ1 + δ 2 K sup 2 K K sup 2 
 sup1 
Substituting (Eqn 15-5) into (Eqn 15-4)

171


Eε s Eε s
σ= = (Eqn 15-6)
AE K
1 + φc + 1 + φc + b
Le K sup 2 K sup 2
AE
where K b = , the equivalent axial stiffness of the floor.
Le
So, as an alternative to using (Eqn 15-4), we may use (Eqn 15-5) to find out
Le and (Eqn 15-6) to calculate internal stress of the floor structure.

Floor structure of cross sectional area A and axial stiffness Kb

P
H
Supporting members Supporting member
providing lateral providing lateral
restraints of stiffness restraints of stiffness
Ksup1 Ksup2
L

idealized as

Kb

Ksup1 P Ksup2
Le

∆δ

L  1 1 
Le =  + 
K sup 2 K K sup 2 
 sup 1  Kb
Ksup

∆ε s

Le

Figure 15.1 – Idealization of floor structure for shrinkage and creep estimation

In the determination of stress due to shrinkage and creep, the main difficulty
lies in the determination of φc which is time dependent. Stress in concrete has
therefore to be determined in successive time steps and with numerical method
as demonstrated in Figure 15.2 for calculation of the creep strains. Instead of

172


being treated as continuously increasing, the stress is split up into various
discrete values, each of which commences at pre-determined station of times.
Fine divisions of time steps will create good simulation of the actual
performance.

∆σ 2

∆σ1 ∆σ1

t1 t1 t1 + t2
t1 t2
2 2 2

(a) at t = t1 – constant stress at ∆σ 1 (b) at t = t 2 – constant stress at ∆σ 1
from t1/2 to t1. from t1/2 to t2 + constant stress ∆σ 2
from (t1+ t2)/2 to t2

∆σ n

∆σ 3 ∆σ 3

∆σ 2 ∆σ 2

∆σ1 ∆σ1

t1 t1 + t2 t2 + t3
t3 t1 t1 + t2 t2 + t3 tn −1 + tn
2 2 2 tn
2 2 2 2

(c) at t = t 3 – constant stress at ∆σ 1 (d) at t = t n – similarly adding up effects
from t1/2 to t3 + ∆σ 2 from (t1+ t2)/2 to t3 of all stress increments
+ ∆σ 3 from (t2+ t3)/2 to t3

Figure 15.2 – Estimation of Creep Strains by Numerical Method

Consider the floor structure shrinks for ε s1 at the time interval from time
t = 0 to t = t1 , φc = K L K m K c K e K j K s should be determined at concrete age
t1 t
(for determination of K m ) and with the time since loading 1 to t1 (for
2 2
t1
determination of K j ) which is the φc value for t1 − and denoted by
2
 t1 
φc  t1 −  . The timing for determination of φc is illustrated in Figure
 2
15-2(a). So by (Eqn 15-6)

173


Eε s1
∆σ 1 = (Eqn 15-7)
K  t 
1 + b + φc  t1 − 1 
K sup 2  2
At time t 2 after shrinkage commencement when the shrinkage strain is ε s 2 ,
the creep strain ε c 2 can be regarded as made up of two time steps with
stresses ∆σ 1 and ∆σ 2 (increment of concrete stress between time t1 and
∆σ 1  t  ∆σ 2  t +t 
t 2 ) as ε c 2 = φc  t 2 − 1  + φc  t 2 − 1 2  as illustrated in Figure
E  2 E  2 
15-2(b). So, similar to the above, we can list

 ∆σ 1  t  ∆σ 2  t + t  ∆σ + ∆σ 2 (∆σ 1 + ∆σ 2 )A
ε s2 −  φc  t 2 − 1  + φc  t 2 − 1 2   − 1 =
 E  2 E  2  E K sup L

  t  EA    t1 + t 2  EA 
⇒ ∆σ 1 φc  t 2 − 1  + 1 +  + ∆σ 2 φc  t 2 −  +1+  = Eε s 2
 
 2 K sup 2 Le 
  
 2  K sup 2 Le 

(Eqn 15-8)
  t  K 
Eε s 2 − ∆σ 1 φc  t 2 − 1  + 1 + b 
 
 2 K sup 2 

⇒ ∆σ 2 = (Eqn 15-9)
  t1 + t 2  K 
φc  t 2 −  +1+ b 
 
 2  K sup 2 

∆σ 2 can be determined with pre-determination of ∆σ 1 by (Eqn 15-7)
Similarly for time t 3 with 3 time steps where
∆σ 1  t  ∆σ 2  t + t  ∆σ 3  t +t 
ε c3 = φc  t 3 − 1  + φc  t 3 − 1 2  + φc  t 3 − 2 3 
E  2 E  2  E  2 

  t  K    t +t  K 
∴ ∆σ 1 φc  t 3 − 1  + 1 + b  + ∆σ 2 φc  t 3 − 1 2  + 1 + b 
 
 2 K sup 2 
  
 2  K sup 2 


  t +t  K 
+ ∆σ 3 φc  t 3 − 2 3  + 1 + b  = Eε s 3 (Eqn 15-10)
 
 2  K sup 2 


  t  K    t +t  K 
Eε s 3 − ∆σ 1 φc  t 3 − 1  + 1 + b  − ∆σ 2 φc  t 3 − 1 2  + 1 + b 
 
 2 K sup 2 
  
 2  K sup 2 

∆σ 3 =
  t 2 + t3  K 
φc  t 3 −  +1+ b 
 
 2  K sup 2 

(Eqn 15-11)
So for any time t n after shrinkage commencement

174


  t  K    t +t  K 
∆σ 1 φc  t n − 1  + 1 + b  + ∆σ 2 φc  t n − 1 2  + 1 + b 
 
 2 K sup 2 
  
 2  K sup 2 


  t +t  K    t +t  K 
+ ∆σ 3 φc  t n − 2 3  + 1 + b  + ..... + ∆σ n−1 φc  t n − n −2 n−1  + 1 + b 
 
 2  K sup 2 
  
 2  K sup 2 


  t +t  K 
+ ∆σ n φc  t n − n−1 n  + 1 + b  = Eε sn
 
 2  K sup 2 

(Eqn 15-12)
  t  K    t +t  K 
Eε sn − ∆σ 1 φc  t n − 1  + 1 + b  − ..... − ∆σ n −1 φc  t n − n −1 n − 2  + 1 + b 
 
 2 K sup 2 
  
 2  K sup 2 

∆σ n =
  t n −1 + t n  K 
φc  t n −  +1+ b 
 
 2  K sup 2 

(Eqn 15-13)

Thus the solution for ∆σ n can be obtained by successive solution of (Eqn
15-7), (Eqn 15-9), (Eqn 15-11) and (Eqn 15-13) or alternatively, in a more
compact form by a system of linear simultaneous equations of (Eqn 15-7),
(Eqn 15-8), (Eqn 15-10) and (Eqn 15-12). The final stress up to t n is
∆σ 1 + ∆σ 2 + ∆σ 3 + ...... + ∆σ n .


A wide 200 mm slab of grade 35 is supported by 350×600 beams at spacing of
3000mm under restraints at both ends as shown in Figure 15.3. The span of the
slab beam structure between restraints is 10 m. The longitudinal steel ratio is
0.5%. The free shrinkage strain and the stress developed due to shrinkage at
360 days after casting are to be assessed.
3EI 3 × 23.7 × 10 6 × 0.08
K sup1 = = = 210667 kN/m
H3 33
3EI 3 × 23.7 × 10 6 × 0.12
K sup 2 = = = 316000 kN/m
H3 33

L  1 1 
Le =  +  = 4m
K sup 2 K K sup 2 
 sup1 
Area of a portion between centre line of two adjacent beams is

175


A = 350 × 400 + 3000 × 200 = 740000 mm2 for 3 m width;
Half Perimeter of the portion in contact with the atmosphere is
(3000 × 2 + 400 × 2) ÷ 2 = 3400
740000
So the effective thickness is he = = 218 mm.
3400

200

3m 3m 3m

All beams are 600(d)×350(w) Cross section

Floor structure

Supporting members Supporting members 3m
providing lateral providing lateral
restraints of stiffness restraints of stiffness
I value = 0.08m4/m I value = 0.12m4/m
width width

L = 10m

Figure 15.3 – Floor structure of Worked Example 15.2

Determination of the coefficients for free shrinkage strain

ε s = cs K L K c K e K j K s ,

K L = 0.000275 for normal air from Figure 3.6 of the Code
Based on empirical formulae :
Water / Cement ratio = − 0.0054 f cu + 0.662 = −0.0054 × 35 + 0.662 = 0.473
Cement content = 3.6 f cu + 308 = 3.6 × 35 + 308 = 434 kg/m3
K c = 1.17 from Figure 3.3 of the Code;
For he = 218 mm thick slab, from Figure 3.7 of the Code, K e = 0.768 ;

K j is time dependent and is to be read from Figure 3.5 of the Code;

176


1
Ks = = 0.96 from (Ceqn 3.4).
1 + ρα e

Determination of the coefficients for creep strain
stress
Creep strain = × φc where φc = K L K m K c K e K j K s ,
E28
K L = 2.3 for normal air from Figure 3.1 of the Code;
K m is time dependent and is to be read from Figure 3.2 of the Code;
K c = 1.17 from Figure 3.3 of the Code, same as Shrinkage
For he = 218 mm thick slab, from Figure 3.4 of the Code, K e = 0.831

K j is time dependent and is to be read from Figure 3.5 of the Code.

K s = 0.96 , same as Shrinkage.

Stiffness per metre width:
EA 23.7 × 10 6 × 0.2467
Kb = = = 1461500 kN/m
Le 4
3EI 3 × 23.7 × 10 6 × 0.12
K sup 2 = 3 = = 316000 kN/m;
H 33
Kb
∴ = 4.625
K sup 2

The time history to 360 days can be divided into various time points, i.e.
t1 = 3 days, t 2 = 7 days, t 3 = 14 days ….. up to t n = 360 days in accordance
with Figure 15.2. Equations in accordance with the above can be formulated
numerically and stress at 360 days is calculated to be 1726.34 kN/m2. As a
demonstration, the stress increment in the first two intervals are presented :

At t1 = 3 , for he = 218 , for shrinkage K j = 0.0637 (Figure 3.5)
ε s1 = c s K L K c K e K j K s = 3.0 × 0.000275 × 1.17 × 0.768 × 0.0637 × 0.96 = 45.22 × 10 −6
K m = 1.743 at t1 = 3 and K j = 0.0546 for time interval from 1.5 to 3 days.
 t  t   t 
φc  t1 − 1  = K L K m  1  K c K e K j  t1 − 1  K s
 2 2  2
= 2.3 × 1.743 × 1.17 × 0.831× 0.0546 × 0.96 = 0.2046
By (Eqn 15-6)
Eε s1 23.7 × 10 6 × 45.22 × 10 −6
∆σ 1 = = = 183.84 kN/m2;
EA  t  1 + 4.625 + 0.2046
1+ + φc  t1 − 1 
Le K sup 2  2

177


At t 2 = 7 , for he = 218 , for shrinkage K j = 0.0896 (Figure 3.5)
ε s 2 = c s K L K c K e K j K s = 3.0 × 0.000275 × 1.17 × 0.768 × 0.0896 × 0.96 = 63.85 × 10 −6

For time step 1,
K m = 1.743 at t1 = 3 and K j = 0.0818 for time interval from 1.5 to 7 days.
 t  t   t 
φc  t 2 − 1  = K L K m  1  K c K e K j  t 2 − 1  K s
 2 2  2
= 2.3 × 1.743 × 1.17 × 0.831× 0.0818 × 0.96 = 0.306 ;

For time step 2,
t1 + t 2
K m = 1.4667 at t = = 5 and K j = 0.0572 from
2
t1 + t 2 3 + 7
t= = = 5 to t = 7
2 2
 t +t  t +t   t +t 
∴φ c  t 2 − 1 2  = K L K m  1 2  K c K e K j  t 2 − 1 2  K s
 2   2   2 
= 2.3 × 1.4667 × 1.17 × 0.831 × 0.0572 × 0.96 = 0.1800
By (Eqn 15-9)
  t  EA 
Eε s 2 − ∆σ 1 φc  t 2 − 1  + 1 + 
 
 2 Le K sup 2 

∆σ 2 =
  t1 + t 2  EA 
φc  t 2 −  +1+ 
 
 2  Le K sup 2 

23.7 × 10 6 × 63.85 × 10 −6 − 183.84 × (0.306 + 1 + 4.625)
= = 72.85 kN/m2;
(0.1800 + 1 + 4.625)

So the total stress at t 2 = 7 is
∆σ 1 + ∆σ 2 = 183.84 + 72.85 = 256.69 kN/m2

The process can be similarly repeated to calculate stress increments at later
times. As the shrinkage rises rapidly in the beginning and slows down at later
times, the time stations should be more frequent when t is small and be less
frequent when t is high.

The stress finally arrived at 360 days is 1726.34 kN/m2. The exercise stops at
360 days because Figure 3.2 of the Code indicates the values of the coefficient
K m up to 360 days only. The stress induced in the structure is plotted in
Figure 15.4.

178


Stress due to shrinkage and creep on the slab structure under
Elastic Restraint of Worked Example 15.2

2000
1800
1600
1400
Stress (kN/m )
2

1200
1000
800
600
400
200
0
0 50 100 150 200 250 300 350

Days after Cast

Figure 15.4 – Increase of internal stress in concrete due to shrinkage and creep of
Worked Example 15.2

By assuming K m = 0.5 beyond 360 days, the exercise is repeated for various
span lengths up to 80 m and finally at perfect restraint where the span is set at
infinity. The stress curves are plotted as indicated.

Stress due to shrinkage and creep on the slab structure under Elastic
Restraint of various spans

L=10m L=20m L=40m L=60m L=80m L=infinity

6000

5000
Stress (kN/m )

4000
2

3000

2000

1000

0
0 100 200 300 400 500 600 700 800 900 1000

Days after Cast

Figure 15.5 – Increase of internal stress in concrete due to shrinkage and creep of
Worked Example 15.2 for various span lengths

179


15.7 The followings are discussed as revealed from Figure 15.5 :
(i) The stress induced increases with time and the increase becomes less
significant as time goes by;

(ii) The magnitude of the stress increases with decrease of the K b / K sup

ratio. As K b decreases with increases of the floor length, longer floor
length will lead to higher stress. So particular attention in relation to
shrinkage and creep should be paid to long floor structures;

(iii) The particular case of perfect restraint is when K b / K sup = 0 , i.e. K sup

becomes infinity where the floor structure stress becomes maximum;
(iv) Thicker floor structures are less prone to shrinkage and creep as the

coefficients K j and K e decrease with increase of thickness;

(v) Stronger lateral restraints will also induce higher shrinkage and creep
stresses. The strong lateral restraints are often in form of core walls or
shear walls whilst the columns are comparatively weak in lateral
restraints. For rough analysis, the columns can be ignored. Figure 15.6
demonstrates the determination of floor span length for the assessment
of shrinkage and creep effects.

Columns be Stiff
ignored due to corewall
small lateral Span length for shrinkage and creep
restraints

Figure 15.6 – Determination of floor span length for shrinkage and creep

15.8 For single span floor structures, if only the stress at age near to the final one
such as the 360 days age is to be estimated, the design parameters for a
particular concrete grade (ignoring reinforcements) can be reduced to

180


comprising only (i) effective thickness of the floor structure; and (ii) relative
stiffness of the axial stiffness of the floor structure to the lateral stiffness of

“support 2”, i.e. K b K sup 2 . Charts as contained in Figure 15.7 for grades 30,

35, 40 and 45 concrete are produced which can be for general use.

Variation of stress of grade 30 concrete floor due to Shrinkage & Creep with
effective thickness and floor / end restraint ratios at 360 days
Effective thickness = 100mm Effective thickness = 200mm
5500
5000
Concrete Stress at 360 days (kN/m )

4500
2

4000
3500
3000
2500
2000
1500
1000
500
0
0 1 2 3 4 5 6 7 8 9 10

Floor / End Restraint Ratio K b / K sup2

5500
5000

4500
2

4000
3500
3000
2500
2000
1500
1000
500
0
0 1 2 3 4 5 6 7 8 9 10


181


5500
5000

4500
2

4000
3500
3000
2500
2000
1500
1000
500
0
0 1 2 3 4 5 6 7 8 9 10


5500
5000

4500
2

4000
3500
3000
2500
2000
1500
1000
500
0
0 1 2 3 4 5 6 7 8 9 10


Figure 15.7 – Variation of 360 days stress due to shrinkage and creep of structural
floor with effective thickness and span / support stiffness ratios

15.9 The induced stress in the concrete structure can be resisted by the tensile

182


strength of concrete under no cracking conditions. Or if the tensile stress is
excessive, it should be resisted by reinforcements with cracks limited to
various widths according to exposure conditions.

Worked Example 15.3

Consider a grade 35 floor structure of unit width under restraints at ends of the
following design parameters :
Stress induced is 3MPa;
Thickness h = 160 mm;
Longitudinal reinforcement content : T10@100 (B.F.) ρ = 0.982 %;
The floor structure is now checked for pure tension created due to shrinkage
and creep alone :
Crack width is checked in accordance with Cl. 3.2.2 and Appendix B of
BS8007:1987 with limiting crack width of 0.2mm;
Strain for coaxial tension :
σ 2bt h 1  2.5 2 × 1000 × 160 
ε m = ε1 − ε 2 = − =  −  = 0.000934 ,
E s ρ 3E s As 200000  0.00982 3 × 1571 
0.8 f y
< = 0.00184 ;
Es
( ε 1 is the strain due to steel only without consideration of the tensile strength
of the concrete and ε 2 represents the stiffening effect by the cracked
concrete.)

Cover to reinforcement is cmin = 25 mm;
So the greatest value acr (distance from the point under consideration to the
nearest reinforcement) that will lead to greatest crack width is

50 2 + 25 2 = 55.9 mm;

By equation 4 of Appendix B of BS8007, the crack width is
ω = 3acr ε m = 3 × 55.9 × 0.000934 = 0.157 mm < 0.2mm;
The crack width is acceptable for all exposure conditions as required by Table
7.1 of the Code.

183


16.0 Summary of Aspects having significant impacts on current Practice

16.1 General

Though some of the new practices in the Code as different from BS8110 have
significant impacts on our current design, detailing and construction practices,
these practices are however generally good ones resulting in better design and
workmanship. The improvement in design lies mainly in enhancing ductility
of the structure which should be regarded as another “limit state” equally as
important as the “ultimate” and “serviceability” limit states. This section tends
to summarize all these new practices and discuss the various impacts so as to
alert the practitioners in switching from BS8110 to the Code.

The aspects with the most significant impacts by the Code on our current
design are obviously the incorporation of the ductility requirements in Cl. 9.9
of the Code for beams and columns contributing in lateral load resisting
system, and the design of beam column beam joints in Cl. 6.8. Others include
checking building accelerations in Cl. 7.3.2. Nevertheless, minor ones such
more stringent requirements in locations and provisions of transverse
reinforcements in lapping of longitudinal bars should also be noted. In
addition, there are relaxations in design requirements such as raising the
absolute ultimate design shear stress (vtu) to 7N/mm2 and giving clear
guidelines in choosing design moments at or near column faces in Cl. 5.2.1.2.
These aspects are highlighted and briefly discussed in this Section. The effects
of different concrete stress strain curve as indicated in Figure 3.8 of the Code
from that of BS8110 are, however, found to be insignificant on the calculation
of longitudinal bars required in beams and columns.

16.2 Ductility Requirements

The followings are highlighted :

(i) Bending and lapping of reinforcement bars

(a) Though the Code includes BS8666 : 2000 in its list of acceptable
standards for the specifications of bending and dimensioning of
reinforcing bars, Table 8.2 of the Code has, however, indicated simple
rules for the minimum internal bend radii of bar diameter as 3Ø for Ø ≤

184


20 mm and 4Ø for Ø > 20 mm where Ø is the diameter of the reinforcing
bar. The minimum internal bend radii are all greater than that required by
BS8666 : 2000 which ranges from 2Ø to 3.5Ø. Furthermore, as unlike
the British Standards (including the former BS4466), no distinction is
made for mild steel and high tensile steel bar in the Code. The more
stringent requirement in minimum bend radii creates greater difficulties
in r.c. detailings;

(b) Cl. 9.9.1.2 and Cl. 9.9.2.2(c) of the Code under the heading of ductility
requirements for “Beams” and “Columns” (contributing in lateral load
resisting system) state that “Links should be adequately anchored by
means of 135o or 180o hooks in accordance with Cl. 8.5. (Presumably
180o bent hooks can be accepted as better anchorage is achieved.) So the
all links in such beams and columns contributing in lateral load resisting
system should be anchored by links with bent angle ≥ 135o as indicated
in Figure 16.1.

Link with 180o bent hooks Link with 135o bent hooks
Figure 16.1 – Links with hooks for beams and columns contributing to lateral load
resisting system and for containment of beam compression reinforcements

For structural elements other than beams and columns contributing in
lateral load resisting system, the 90o anchorage hooks can still be used
except for containment of compression reinforcements in beams which
should follow Figure 16.1 (Re Cl. 9.2.1.10 and Cl. 9.5.2.2 of the Code.)

Figure 16.2 – 90o bent links : used in structural elements other than beams / columns
contributing in lateral load resisting system and except compression bar containment

185


(c) Cl. 9.9.1.1(c) of the Code under the heading of ductility requirement for
anchorage of longitudinal bars in beams (contributing in lateral load
resisting system) into exterior column states that “For the calculation of
anchorage length the bars must be assumed to be fully stressed”. The
calculation of anchorage length of bars should therefore be based on

f y instead of 0.87 f y as discussed in Sections 8.4.4 and 8.4.5 of this

Manual, resulting in some 15% longer in anchorage and lap lengths as
compared with Table 8.4 of the Code. Thus longer anchorage length is
required for longitudinal bars in beams contributing in lateral load
resisting system anchoring into exterior column;

(d) Cl. 8.7.2 and Figure 8.4 of the Code have effectively required all tension
laps to be staggered which are generally applicable in the flexural steel
bars in beams, slabs, pile caps, footings etc. as per discussion in Section
13.6 of this Manual. The practice is not as convenient as the practice
currently adopted by generally lapping in one single section.
Nevertheless, if staggered lapping is not adopted, lapping will likely be
greater than 50% and clear distance between adjacent laps will likely be
≤ 10φ, transverse reinforcement by links or U bars will be required by Cl.
8.7.4.1 of the Code which may even be more difficult to satisfy.
Fortunately, the requirements for staggered lapping (in Cl. 8.7.2) do not
cover distribution bars and compression bars. So most of the
longitudinal bars in columns and walls can be exempted;

(e) Cl. 8.7.4 of the Code requires additional transverse reinforcements
generally in lap zones of longitudinal bars which is not required by
BS8110. Arrangement and form of transverse reinforcements (straight
bars or U-bars or links) required are in accordance with the longitudinal
bar diameter φ , spacing of adjacent laps and percentage of lapping at
one point. Take an example : when T40 bars of transverse spacing ≤ 400
mm ( 10φ ) are lapped at one section, total area of transverse
reinforcements equal to 1 longitudinal bar which is 1257 mm2 should be
spaced along the lapped length of some 2000 mm. The transverse
reinforcement is therefore T12 – 125 mm spacing (providing 2261mm2)
in form of U-bars or links at ends of the laps as demonstrated in Figure
13.6. Apparently these transverse reinforcements should be in addition to
the transverse reinforcements already provided for other purposes unless

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φ < 20 mm or percentage of lapping at a section < 25%. As it is difficult
to perform lapping with percentage < 25% in any one section, such extra
transverse reinforcement will normally be required for φ ≥ 20 mm.
Nevertheless, with lapping ≤ 50% at one section, at least U-bars or links
can be eliminated.

(ii) Beam

(a) Limitation of neutral axis depths

Neutral axis depths have been reduced from 0.5 to 0.4 for concrete
grades 45N/mm2 < fcu ≤ 70 N/mm2 and further reduced to 0.33 if fcu > 70
N/mm2 as per Cl. 6.1.2.4(b) of the Code under Amendment No. 1. The
effects should be insignificant as it is uncommon to design flexural
members with grade higher than 45.

(b) Reduction of moment arm factors for high grades concrete in sectional
design of beam by the Simplified Stress Block from 0.9 to 0.8 and 0.72.

(c) Steel Percentages

The maximum and minimum tension steel percentages are respectively
2.5% and 0.3% in Cl. 9.9.1.1(a) of the Code for beams contributing to
lateral load resisting system. The lower maximum tension steel
percentages may force the designer to use larger structural sections for
the beams contributing in lateral load resisting system.

In addition, Cl. 9.9.1.1(a) also imposes that “At any section of a beam
within a critical zone (the Handbook gives an example of that “plastic
hinge zone” is a critical zone), the compression reinforcement should not
be less than one-half of the tension reinforcement at the same section.”
The “critical zone” should likely include mid-spans and/or internal
supports in continuous beam. As plastic hinges will likely be extensively
in existence in normal floor beams as per the discussion in Section 2.4,
the requirement is expected to be applicable in many locations in beams
contributing in lateral load resisting system. The adoption of this clause
will obviously increase amounts of longitudinal bars significantly for
these beams.

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(d) End Support Anchorage

The Code has clarified support anchorage requirements of reinforcement
bars of beams as summarized in the following Figure 16.3 which
amalgamates contents in Figures 3.19, 3.20 and 13.1 of this Manual :

D ≥ 2(4Ø+c) if Ø ≤ 20 anchorage
≥2(5Ø+c) if Ø > 20 commences at this Longitudinal
section generally. bar of dia. Ø
c ≥0

support
centre
line ≥0.5D ≥ 500mm or h X h
or 8Ø
≥0
cross bar of
dia. ≥ Ø ≥0
anchorage can commence at
this section if the plastic hinge
≥0.75D of the beam is beyond X

Beam contributing to lateral load resisting system

anchorage
D ≥ 2(4Ø+c) if Ø ≤ 20 commences at this
≥2(5Ø+c) if Ø > 20 Longitudinal
section generally.
bar of dia. Ø
c ≥0

support
centre
line
h

cross bar of
dia. ≥ Ø ≥0

Beam not contributing to lateral load resisting system

Figure 16.3 – Summary of longitudinal bar anchorage details at end support

(1) Cl 8.4.8 clarifies the support widths to beams in form of beams,

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columns and walls as in excess of 2(4Ø+c) if Ø ≤ 20 and 2(5Ø+c) if
Ø > 20 where Ø is the diameter of the longitudinal bar and c is the
concrete cover to the bar. The clause has effectively required bend
of bars to commence beyond the centre-line of support which is an
existing requirement in BS8110 stated for simply supported end (Cl.
3.12.9.4 of the BS). The clause has extended the requirement to
cover beam at supports restrained against rotation. Support widths
may then require to be increased or bar size be reduced to satisfy
the requirement, giving constraints in design;
(2) Cl. 9.9.1.1(c) of the Code requires anchorage of longitudinal bar of
beam contributing to lateral load resisting system to commence at
the centre line of support or 8 times the longitudinal bar diameter
whichever is the smaller unless the plastic hinge at the beam is at
the lesser of 500mm or a beam structural depth from the support
face of the beam. Effectively the requirement covers all such
beams designed to be having rotational restraints at the exterior
columns or walls unless it can be shown that critical section of
plastic hinge is beyond X as shown in Figure 16.3. By the
requirement, anchorage length needs be increased by the lesser of
the half of the support width and or 8 times the longitudinal bar
diameter for most of the end span beams contributing to lateral
load resisting system and anchored into exterior column which is
often relatively stiffer than the beam as per discussion in Section
2.4;
(3) A method of adding a cross bar so as to avoid checking of internal
stress on concrete created by the bend of the longitudinal bar (by
(Ceqn 8.1) of the Code) has been added in Cl. 8.3 of the Code,
even if checks on the bar indicates that anchorage of the
longitudinal bar is still required at 4Ø beyond the bend. The method
is not found in BS8110. The method is quite helpful as the designer
can avoid using large bends of bars to reduce bearing stress in
concrete which may otherwise result in non-compliance with Cl.
8.4.8 of the Code;
(4) Cl. 9.9.1.1 (c) states clearly that top beam bars be bent downwards
and bottom beam bars bent upwards, again applicable to beams
contributing in lateral load resisting system. The requirement may
create difficulties to the conventional construction work as, apart
from aggravating steel bar congestion problems in the column

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beam joint, the top bars may be required to be fixed prior to
column concreting if they have to be bent into the column shaft to
achieve adequate anchorage. The practice is obviously not
convenient in the current construction sequence for buildings.

A Solution to anchorage problem may be adding an “elongation” of the
structural beam, if possible, beyond the end column as shown in Figure 16.4.

Longitudinal
bar of dia. Ø

h

elongation

Figure 16.4 – “Elongation” for anchorage of longitudinal bars beyond end supports

(iii) Column

(a) Steel Percentages

Cl. 9.9.2.1(a) of the Code has required the maximum longitudinal
reinforcements to be 4% of the gross sectional area for columns
contributing to lateral load resisting system which are more stringent
than columns not contributing to lateral load resisting system (6% to
10% in accordance with 9.5.1 of the Code). In addition, the clause also
clarifies that the maximum longitudinal bar percentage at laps is 5.2%
which effectively reduces the maximum steel percentage to 2.6% if the
conventional lapping at single level (not staggered lap) is adopted in
construction for columns contributing to lateral load resisting system.

(b) Anchorage and lapping of longitudinal bars in supporting beams or
foundations

190


Figures 5.9, 5.10 and 5.11 of this Manual illustrate anchorage of
longitudinal bars of columns in supporting beam or foundations as
required by Cl. 9.9.2.1(c) of the Code where the columns contribute in
lateral load resisting system and plastic hinges will occur in the column.
Generally anchorage lengths will be increased as anchorage
commences inside the beam and foundation element instead of column
foundation interfaces for such columns. Furthermore, bars in columns
anchored into intersecting beams must be terminated with 90o standard
hooks (or equivalent anchorage device) and have to be bent inwards
unless the column is designed only for axial loads. All these lead to
longer anchorage lengths and stability problem in reinforcing bars
erection.

(c) Splicing of longitudinal bars

To reduce weakening of the column in reinforcement splicing (lapping
and mechanical coupling) in “critical zones” (potential plastic hinge
formation zones), Cl. 9.9.2.1(d) of the Code requires the longitudinal
splicing locations of columns contributing in lateral load resisting
system should, as far as possible, be away from these “critical zones”
which are near the mid-storey heights as illustrated in Figure 5.9. For
such columns, the current practice of lapping at floor levels in building
construction requires review.

(d) Transverse Reinforcements

Cl. 9.9.2.2 of the Code which is applicable to column contributing to
lateral load resisting system defines “critical regions” along a column
shaft which are near the ends of the column resisting high bending
moments and specifies more stringent transverse reinforcement
requirements in the same clause than the normal region near
mid-heights of the column. In this clause, the definition of “critical
regions” relies on axial stress in the column and has made no reference
to any potential plastic hinge formation zone. As it is not our usual
practice of specifying different transverse reinforcements along the
column shaft and the lengths of the “critical regions” are often more
than half of the column shaft (dictated also by the requirement of one
to two times the greater lateral dimension of the column), it seems

191


sensible to adopt the more stringent requirements along the whole
column shaft. The more stringent requirements of transverse
reinforcements comprise closer spacing and that every longitudinal bar
(instead of alternate bar) must be anchored by a link. Whilst the
maximum spacing in the normal region is 12Ø where Ø is the
longitudinal bar diameter and that in the critical region is the smaller of
6Ø and 1/4 of the least lateral dimension in case of rectangular or
polygonal column and 1/4 of the diameter in case of circular column,
the quantities of transverse reinforcements can be doubled.

(iv) Column Beam Joints

The requirement of providing checking and design in column beam
joints as discussed in Section 6 constitutes a significant impact on the
current design and construction as, apart from increase of construction
cost due to increase of steel contents, the requirement aggravate the
problem of steel congestions in these joints. Enlargement of the
column head as indicated in Figure 16.6 may be required in case the
shear stress computed by Ceqn 6.71 is excessive or the required
reinforcements are too congested. In addition, it should also be noted
that even no shear reinforcement is required as per checking of shear in
the joints in accordance with Cl. 6.8 of the Code, transverse
reinforcements in accordance with Cl. 9.5.2 which are installed in the
column shaft outside “critical regions” shall also be installed within the
column beam joints as shown in Figure 6.3 of this Manual.

Enlarged column
head

Figure 16.6 – Column head enlargement for column beam joint

192


16.3 Building Accelerations

Cl. 7.3.2 of the Code specifies that “where a dynamic analysis is undertaken,
the maximum peak acceleration should be assessed for wind speeds based on a
1-in-10 year return period of 10 minutes duration with the limits of 0.15m/sec2
for residential buildings and 0.25m/sec2 for office or hotel. The term “dynamic
analysis” is not defined in the Code. However, if it is agreed that computation
of wind loads in accordance with Wind Code 2004 Appendix F (titled
“Dynamic Analysis”) is a dynamic analysis, the requirement will be applicable
to all buildings defined as ones with “significant resonant dynamic response”
in Clause 3.3 of the Wind Code, i.e. (i) taller than 100 m; and (ii) aspect ratio >
5 unless it can be demonstrated that the fundamental natural frequency > 1 Hz.
Thus most of the high-rise buildings are included.

Empirical approaches for assessment of building accelerations are described in
Appendix B. The second approach which is taken from the Australian Code
should be compatible to the Hong Kong Wind Code as it is based on the
Australian Code that the Hong Kong Wind Code determines approaches of
dynamic analysis in its Appendix F. Furthermore, it can be shown in the chart
attached in the Appendix that building acceleration generally increases with
building heights and thus pose another compliance criterion. Fortunately, the
accelerations approximated are not approaching the limiting criterion as per
the exercise on a square plan shaped building. However, the effects should be
more significant for buildings with large plan length to breadth ratios.

193


References

This Manual has made reference to the following documents :

1. Code of Practice for the Structural Use of Concrete 2004
2. Concrete Code Handbook by HKIE
3. Hong Kong Building (Construction) Regulations
4. The Structural Use of Concrete 1987
5. Code of Practice on Wind Effects in Hong Kong 2004
6. Code of Practice for the Structural Use of Steel 2005
7. The Code of Practice for Dead and Imposed Loads for Buildings (Draft)
8. BS8110 Parts 1, 2 and 3
9. BS5400 Part 4
10. Eurocode 2
11. Code of Practice for Precast Concrete Construction 2003
12. New Zealand Standard NZS 3101:Part 2:1995
13. ACI Code ACI 318-05
14. Code of Practice for Fire Resisting Construction 1996
15. BS8666 : 2000
16. BS4466 : 1989
17. PNAP 173
18. Practical design of reinforced and prestressed concrete structures – based
on CEP-FIP model code MC78
19. CEB-FIP Model Code 1990
20. Standard Method of Detailing Structural Concrete – A Manual for best
practice – The Institution of Structural Engineers
21. The “Structural Design Manual” – Highways Department of HKSAR
22. Reinforced and Prestressed Concrete 3rd edition – Kong & Evans
23. Reinforced Concrete Design to BS110 – Simply Explained A.H. Allen
24. Design of Structural Concrete to BS8110 – J.H. Bungey
25. Handbook to British Standard BS8110:1985 Structural Use of
Concrete – R.E. Rowe and others – Palladian Publication Ltd.
26. Ove Arup and Partners. Notes on Structures: 17 April 1989
27. Australian/New Zealand Standard – Structural Design Actions Part 2 :
Wind Actions AS/NZS 1170.2.2002
28. Examples for the Design of Structural Concrete with Strut-and-Tie
Models – American Concrete Institute
29. Tables for the Analysis of Plates, Slabs and Diaphragms based on the

194


Elastic Theory – Macdonald and Evans Ltd.
30. Some Problems in Analysis and Design of Thin/Thick Plate, HKIE
Transactions – Cheng & Law, Vol. 12 No. 1 2004

195

Appendix A

Clause by Clause Comparison
between “Code of Practice for
Structural Use of Concrete 2004”
and BS8110

Comparison between Code of Practice for Structural Use of Concrete 2004 and BS8110:1997 (and 1985) Appendix A
HK CoP Structural Use of Concrete 2004 BS8110:1997 (and 1985) Remark
Clause No. Contents Clause No. Contents
1.1 – Scope The clause has explicitly stated that the Code applies Pt. 1 1.1 – The clause only explicitly excludes bridge structures The exclusion of CoPConc2004
only to normal weight concrete, with the exclusion of Scope and structural concrete of high alumina cement. should also be applied to
(i) no fines, aerated, lightweight aggregate concrete BS8110. In addition, BS8110
etc; (ii) bridge and associated structures, precast does not apply to high strength
concrete (under the separate code for precast concrete.
concrete); and (iii) particular aspects of special types
of structures such as membranes, shells.

2.1.5 – The clause states that the Code assumes a design – Nil No similar statement in
Design working life of 50 years. Where design working life BS8110.
working life differs from 50 years, the recommendations should be
modified.

2.2.3.3 – The clause refers to clause 7.3.2 for the usual limits Pt. 1 2.2.3.3 – Reference to specialist literature is required. In CoPConc2004 is more specific.
Response to of H/500 to lateral deflection at top of the building Response to addition Pt. 2 3.2.2.2 stipulates a limit on inter-storey However, method for
wind loads and accelerations of 1-in-10 year return period of 10 wind loads drift of Storey height/500 for avoidance of damage to determination of the
minutes duration of 0.15m/sec2 for residential and non-structural elements. acceleration is not given in the
0.25m/sec2 for office. However, there is no Code and in the HKWC-2004.
requirement on the inter-storey drift, though the draft
steel code has a requirement of storey height/400.

2.3.2.1 – Table 2.1 is generally reproduced from Table 2.1 of Pt. 1 It is stated in the clause that when applying the load 1.0 in CoPConc2004 may not
Loads for BS8110 except that the partial factor for load due to 2.4.3.1.2 – factor, no distinction should be made between be adequately conservative as
ultimate limit earth and water is 1.0 for the beneficial case. Partial factors adverse and beneficial loads. there may even be
state for earth over-estimation in the
pressures determination of the unfactored
soil load. It is even a practice to
set the load to zero in beneficial
case. ICU has raised this
comment during the comment
stage when the draft Code has
exactly the content of BS8110.

2.3.2.3 & The clauses explicitly state that these effects need – No similar clauses in BS8110 The clause in CoPConcrete2004

A-1

2.3.2.4 – only be considered when they are significant for affirms engineers to ignore
Differential ULS. In most other cases they need not be considered consideration of these effects in
settlement of provided ductility and rotational capacity of the normal cases which are the
foundations, structure sufficient. usual practices.
creep,
shrinkage,
temperature
effects

2.4.3.2 – Table 2.2 gives γm for ULS for concrete and re-bars. Pt. 1 2.4.4.1 – Table 2.2 gives γm for ULS for concrete and re-bars. BD has been insisting on the
Values of γm γm for re-bars is 1.15, implying strength of re-bars for Values of γm γm for re-bars is 1.05, implying strength of re-bars for use of 0.87fy even if BS8110
for ULS design remain as 0.87fy. for ULS design remain as 0.95fy. was used before the
promulgation of the new
concrete code

3.1.3 – Table 3.1 states concrete strength grades from 20 – BS8110 has not explicitly stated the concrete grades The coverage of CoPConc2004
Strength MPa up to 100 MPa which is the range covered by covered by the BS, However, concrete grades is wider.
grades the Code. covered by the design charts in Part 3 of the Code
range from grade 25 to 50 whilst other provisions
such as vc (Pt. 1 Table 3.8), lap lengths (Pt. 1 Table
3.27) are up to grade 40.

3.1.4 – It is stated in the 1st paragraph of the clause that for Pt. 1 2.4.3.3 – The clause states that “For the ULS, these effects will BS 8110 has included
Deformat- ULS, creep and shrinkage are minor, and no specific creep, usually be minor and no specific calculations will be temperature effects be a minor
ion of calculation are required. shrinkage and necessary. one that can be ignored in
Concrete temperature calculation.
effects

3.1.5 – Elastic Table 3.2 stipulates short term static Young’s Pt. 1 Figure The determination of short term static Young’s Values in the CoPConc2004
deformat- Modulus of concrete of various grades based on the 2.1 Modulus of concrete is given by the slope gradient in should be used as it is based on
ion formula 3.46√fcu+3.21 in MPa derived from local the figure which is 5.5√(fcu/γm). local research and concrete E
research. values are affected by
constituents which are of local
materials. Nevertheless, it

A-2

should be noted that E values in
the new Code are slightly
higher than the previous ones in
“The Structural Use of
Concrete – 1987” (Table 2.1).

3.1.7 & Though it is stated in 3.1.4 that for ULS creep and – No account has been given for creep and shrinkage, As stated in the CoPConc2004
3.1.8 – Creep shrinkage are minor and require no specific as stated in 2.4.3.3. Cl. 2.3.2.4, effects need only be
and shrinkage calculations, these clauses contain detailed formulae considered if they are
and charts for prediction of creep and shrinkage significant.
strain. The approach is identical to the “Structural
Design Manual” issued by Highways Department and
the charts are extracted from BS5400:Pt 4:1990. As
identical to the previous version of “Structural
Design Manual” (SDM), the cs value is 4.0 to suit
local crushed granite. However, it is noted that
recently the SDM has reduced the factor to 3.0 and
the Code has adopted the change in Amendment No.
1.

3.1.9 – The linear coefficient of thermal expansion given in – No account has been given for temperature, as stated The linear coefficient of
Thermal the Code for normal weight concrete is 10×10-6/oC in 2.4.3.3. thermal expansion given by
Expansion whilst that stated in the “Structural Design Manual” both CoPConc2004 is slightly
issued by Highways Department is 9×10-6/oC in higher than SDM and both are
Clause 2.4.4. independent of concrete grades.

3.1.10 – The short term design stress-strain curve of concrete Pt. 1 2.5.3 – Pt. 1 Fig. 2.1 shows stress-strain relation of normal (i) The Young’s Moduli of
Stress (Fig. 3.8) follows closely the traditional one in Analysis of weight concrete. concrete stipulated in
-strain BS8110, though there are differences in the values of section for CoPConc2004 should be
relationship the Young’s Moduli. Furthermore, the ultimate strain ULS adopted as they are based
for design is limited to below 0.0035 for concrete grade on local data and cover up
exceeding C60. Nevertheless, the “plastic strain”, to grade 100 concrete,
strain beyond which stress is constant remains together with the decrease
identical as BS8110; of ultimate strain for grade

A-3

above C60, to account for
the brittleness of high
strength concrete;
(ii) “Smooth” connection
between the parabolic
curve and the straight line
portion of the stress-strain
curve cannot be effected if
ε0 = 2.4×10-4√fcu/γm is kept
and the Young’s moduli in
Table 3.2 of CoPConc2004
are used. For smooth
connection, ε0 should be
revised 2σult/Ec or
1.34fcu/γmEc where
σult=0.67fcu/γm.
Nevertheless, ε0 has been
rectified to 1.34fcu/γmEc in
Amendment No. 1.

3.2.7 – The clause states that re-bars can be welded provided Pt. 1 3.12.8.16 There are provisions for lapping re-bars by welding Steel complying CS2 is likely
Weldability the types of steel have the required welding 7.6 in Pt. 1 3.12.8.16, and general welding requirements weldable as CS2 does not differ
(of re-bars) properties given in acceptable standards and under in Pt. 1 7.6. Nevertheless, another BS7123 titled significantly from BS4449.
Approval and inspection by competent person. “Metal arc welding of steel for concrete
Further provisions for welding are given in 10.4.6. reinforcement” requires the re-bars be in compliance
with BS4449 or BS4482.

Section 4 – The requirements are general. The followings are Pt. 1 2.2.4 – The requirements are general. However, the The approaches of the two
Durability highlighted :- Durability; followings are highlighted : codes are quite different.
and fire (i) In 4.1.1, it is stated that requirements are based Pt. 1 2.2.6 – (i) 3.1.5.2 states that when cement content > 400 However, CoPConc2004 is
resistance on design working life of 50 years; Fire kg/m3 and section thicker than 600 mm, more related to local practice.
(ii) In Table 4.1 under 4.2.3.2, exposure conditions resistance; Pt. measures for temperature control should be
1 to 5 are classified with headings similar to that 1 3.1.5.2 – implemented;

A-4

in Pt. 1 3.3.4 of BS8110. However, detailed Design for (ii) A full description for fire resistance control is
descriptions are different except the last one Durability given in Pt. 2 Section 4, outlining methods of
“abrasive”; Pt. 1 3.3 – determining fire resistance of structural
(iii) In 4.2.7.3, control of AAR has been Concrete elements and with reference to BS476 Pt. 8;
incorporated from the previous PNAP 180; cover to
(iv) Concrete covers are given in Table 4.2 for rebars
various concrete grades and exposure Pt. 2 Section 4
conditions;
(v) By 4.3, the user has to refer to the current fire
code for additional requirements against fire
resistance.

5.1.3.2 – 3 simplified load cases for Dead + Live loads are Pt. 1 3.2.1.2.2 2 simplified load cases are considered sufficient for CoPConc2004 more reasonable,
Load cases recommended (with DL always present): (i) all spans design of spans and beams (with DL always present): though not truly adequate.
and loaded with LL; (ii) alternate spans loaded with LL; (i) all spans loaded with LL; (ii) alternate spans Nevertheless, the current
combinations (iii) adjacent spans loaded with LL. The 2nd case is to loaded with LL; softwares mostly can account
for beams and seek for max. sagging moment and the 3rd case is for the load case to search for
slabs likely to seek for max. hogging moment. But true max. hogging moment at
max hogging moment should also include alternate support.
spans from the support being loaded.

5.2 – Analysis The clause contains definition of beam, slab 3.2 – Analysis The followings are highlighted : The following remarks are
of Structure (included ribbed, waffle, one-way or two-ways), of Structures (i) Generally provisions are applicable to normal made :
column in accordance with their geometries. The strength concrete; (i) CoPConc2004 is more
following differences with BS8110 are highlighted : 3.4 – Beams (ii) Consideration for high-rise buildings (second suitable for use in Hong
(i) Conditions in relations to rib spacings and order effects) and structures such as shear walls, Kong as it cover high
flange depths for analysis of a ribbed or waffle transfer structures are not given; strengths concrete, high
slab as an integral structural unit are given in (iii) The effective span of a simply supported beam rise buildings, shears,
5.2.1.1(d); is the clear span plus the lesser of half effective transfer structures.
(ii) By definition, the effective flange widths in T depth and half of support width whilst that of (ii) Extended use of effective
and L-beams included in 5.2.1.2 (a) are slightly continuous beam and cantilever are span flange in beam in
greater than that in BS8110 by 0.1×clear rib between supports (presumably mid-support structural analysis and
spacing unless for narrow flange width width) except at end span in case of continuous moment reduced to
controlled by rib spacing. It is also stated that beam and a cantilever forming the end of a support shear are explicitly

A-5

the effective flange can be adopted in structural continuous beam, the effective span is the clear stated in CoPConc2004.
analysis; span plus mid-support width should be used. (iii) Method of analysis in both
(iii) Clearer definition of effective spans of beams is codes are old-fashioned
also included in 5.2.1.2(a) with illustration by ones that can be performed
diagrams, together with reduction of span by hand calculations. Use
moments due to support width in 5.2.1.2(b). of computer methods
In principle, the effective span of a simply (extensively adopted
supported beam, continuous beam or cantilever currently in Hong Kong)
is the clear span plus the minimum of half are not mentioned.
effective depth and half support width except
that on bearing where the span should be up to
the centre of the bearing.
(iv) Furthermore reduction in support moments due
to support width is allowed which is not
mentioned in BS8110;
(v) The definition of effective flange for T- and
L-beams are slightly different from BS8110
though the upper bound values are identical to
that of BS8110. So more stringent.
(vi) Moment redistribution is restricted to grade
below C70 as per 5.2.9.1;
(vii) Condition 2 in 5.2.9.1 relation to checking of
neutral axis depths of beam in adopting moment
re-distribution is different from Cl. 3.2.2.1 b) of
BS8110. More stringent limit on neutral axis is
added for concrete grade higher than C40 as per
6.1.2.4(b);
(viii) Provision for second order effects with axial
loads in 5.3 is added. The provisions are quite
general except the statement “second order
effects can be ignored if they are less than
10%”;
(ix) Provisions for shears walls and transfer
structures are added in 5.4 and 5.5 though the

A-6

provisions are too general.

6.1 – The followings differences with BS8110 are Pt. 1 3.4 – In the design aspects, the Code is limited to grade 40. The followings are highlighted :
Members in identified : Beams (i) Provisions are made in
Flexure (i) Ultimate strains are reduced below 0.0035 for CoPConc2004 for concrete
concrete grade exceeding C60 as per Fig. 6.1; grades higher than 40
(ii) In Figure 6.1, the factor 0.9, concrete stress including limitation of
block depth ratio is revised to 0.8 for 45<fcu≤70 neutral axis depths etc.
and 0.72 for 70<fcu in Amendment No. 1; However, it is also noted
(iii) Neutral axis depths limited to less than 0.5d for that the ultimate stress in
concrete grade higher than C45 as per 6.1.2.4(b) Fig. 6.1 remains
(Originally the Code limits demarcate neutral unchanged for high
axis depths from 0.5d to 0.4d at C40. But the concrete grade though the
demarcation is revised to C45 at Amendment ultimate strain is reduced;
No. 1); (ii) Ultimate concrete shear
(iv) The limitations on neutral axis depth ratio under strengths increased in
redistribution of moments are further restricted CoPConc2004;
for grade > 45 (revised as per Amendment No. (iii) Due to the more stringent
1), as different from BS8110 which makes no limitation on neutral axis
differences among concrete grades in Cl 3.4.4.4 depth for high grade
of BS8110 which if redistribution of moment concrete in CoPConc2004,
exceed 10%; different formulae have
(v) Different design formulae are used for the been devised for x > 0.5d
higher grades (45≤fcu≤70; 70≤fcu<100, as for rectangular beams.
revised by Amendment No. 1) concrete based However, similar formulae
on simplified stress blocks as per 6.1.2.4(c). are not given for flanged
Design charts similar to that BS8110 based on beams;
stress-strain relationship in figure 3.8 are not (iv) 6.1.4.2 of CoPConc2004 is
available; similar to BS8110 Pt. 1
(vi) Table 6.2 in relation to vc of concrete as related 3.6.2. However, it is ICU’s
to tensile reinforcements is identical to Pt. 1 comment to qualify that if
Table 3.8 of BS8110 except that applicability is ribbed slabs are to be
extended to C80 whereas BS8110 limits to designed as two-ways
grade 40. The minimum amount of shear spanning as similar to a

A-7

reinforcements required is reduced to that can flat slab, it should be
provide shear resistance of 0.4(fcu/40)2/3 MPa qualified that the slab has
for grade over 40 whilst that for concrete at and equal structural properties
below grade 40 remains to that can provide 0.4 in two mutually
MPa; perpendicular directions.
(vii) Partial strength factor for steel remains as 0.87fy
in accordance with BS8110:1985 instead of
0.95fy as in accordance with BS8110:1997, for
both flexure and shear;
(viii) Ultimate shear strength of concrete increased to
7.0 MPa as compared 5.0 MPa in BS8110. The
other limitation of 0.8√fcu is identical to
BS8110;
(ix) Table 6.3 is identical to Table 3.8 of BS8110
except, the last note under Notes 2 for the effect
of effective depth (d) on vc where shear
reinforcement is required. Whilst BS8110
effectively gives 1 for the factor (400/d)1/4 for d
≥ 400mm, Table 6.3 requires the factor to
decrease beyond d=400mm. However as
subsequent to discussion with experts, it is
advisable to keep the factor to 1 for d ≥ 400mm
as there are no tests to justify the decrease;
(x) By 6.1.3.5, the minimum shear reinforcement to
cater for shear strength of 0.4 MPa is for
concrete grade below C40 (requirement by
BS8110). Above grade C40, the required shear
strength is increased by factor (fcu/40)2/3 as per
Table 6.2;
(xi) By 6.1.4.2, ribbed slabs (6.1.4.2) can be
designed as two-ways spanning as similar to flat
slab if they have equal structural properties in
two mutually perpendicular directions. BS8110
does not explicitly require equal structural

A-8

properties in mutually perpendicular directions
in adopting design method as flat slab (BS8110,
3.6.2).

6.2 – The followings are highlighted: Pt. 1 3.8 – The followings are highlighted : The followings are highlighted :
Members (i) The CoP contains no design charts for column Columns (i) In the design aspects, the provisions are limited (i) The limitation of neutral
axially loaded design with moments. Strictly speaking, the Pt 2 2.5 – to grade 40; axis depth as for beam is
with or design charts in BS8110 are not applicable (a) Effective (ii) A more tedious approach for determination of clearly not applicable to
without the Young’s Moduli of concrete are different; column height effective height of column by consideration columns. However, it
flexure and (b) the ultimate strain for concrete grade > stiffness of the connecting beams and columns should be noted that
C60 are reduced; is outlined in Pt. 2 2.5. members with axial loads
(ii) Due to the use of lower partial strength factor of creating axial stress < 0.1fcu
steel (γm = 1.15), the factors on steel in may be regarded as beam in
equations for strengths of column sections design. Re 6.1.2.4 of
(6.55, 6.56) are lower than BS8110 (Eqn 28, CoPConc2004.
29);
(iii) The provisions for more accurate assessment of
effective column height in accordance with
BS8110 Pt. 2 2.5 are not incorporated in the
Code.

6.3 – Torsion The followings are highlighted : Pt. 2 2.4 The followings are highlighted CoPConc2004 contains more
and Table 6.17 in relation to limitation of torsional shear (i) Table 2.3 contains specific values for vtmin and specific values for ultimate
Combined stresses contains specific values for grades 25 to 80. vt up to grade 40. The values remain constant concrete shear stress.
Effects For values below grade 40, they are identical to from grade 40 thereon;
BS8110. Above grade 40, the Code provides values (ii)
for vtmin and vt for different grades up to 80, beyond
which the values remain constant whilst BS8110 set
the values to that of grade 40 for grades above 40;

6.4 – Design The followings are highlighted : Pt 2 2.6.3 The followings are highlighted : CoPConcrete 2004 is more
for robustness (i) In 6.4.2 in relation to design of “bridging (i) Pt.2 2.6.3 in relation to design of “bridging stringent.
elements” which is identical to BS8110 Pt.2 elements” applies to buildings of five or more
2.6.3, the words “where required in buildings of storeys;

A-9

five or more storeys” have been deleted. So the
Code is more stringent as consideration to loss
of elements is required for all buildings;

6.6 – The followings are highlighted : Pt 1 3.10.1 The followings are highlighted : It is more reasonable to assume
Staircase (i) 6.6.1 is in relation to design of staircase. There (i) A note in 3.10.1 in relation to design of staircase should include
is no stipulation that the staircase may include staircase has stipulated that a staircase also landing, as in CoPConc2004.
landing; include a section of landing spanning in the
same direction and continuous with flight;

6.7 – The following differences with BS8110 are Pt 1 3.11 The followings are highlighted : The followings are highlighted :
Foundations highlighted : (i) In Pt 1 3.11.2.1, the assumption of uniform or (i) CoPConc2004 is generally
(i) In 6.7.1.1, the assumptions of uniform reaction linearly varying assumption is without the more reasonable as it
or linearly varying reaction of footing and pile pre-requisite that the footing or pile cap is makes provision for the
cap are based on use of rigid footings or pile sufficiently rigid. This is not good enough as modern analysis by treating
caps. So a pre-requisite for the use of these significant errors may arise if the footing or the pile cap as a flexible
assumptions is stated at the end of the clause pile cap is flexible; structure by computer
which reads “if a base or pile cap is considered (ii) In Pt 1 3.11.4.1, there is no mention of pile cap methods;
be of sufficient rigidity.”; rigidity which affects design; (ii) Apparently CoPConc2004
(ii) In 6.7.3.1, a statement has been inserted that a (iii) In Pt 1 3.11.4.3, there is no mention that shear forces the designer to check
pile cap may be designed as rigid or flexible, enhancement shall not be applied to where torsion as if the cap or
depending on the structural configuration. No shear distribution across section has not been footing is a beam under the
similar provision is found in BS8110; considered; rigid cap (footing). This is
(iii) In 6.7.3.3, 2nd dot, it is stated that “where the (iv) In Pt 1 3.11.4.4 b), there is no mention that not too sound as the
shear distribution across section has not been shear enhancement in pile cap shall be applied formulae for beam are
considered, shear enhancement shall not be to under the condition that shear distribution under the assumption that
applied.” No similar provision is found in across section has not been duly considered; torsional cracks can be
BS8110; (v) There is no explicit stipulation on the limit of fully developed for a beam
(iv) In 6.7.3.3 3rd dot, shear enhancement in pile cap effective width for checking shear; length of b+d. If such
can only be applied where due consideration (vi) No explicit statement that shear reinforcement length is not available
has been given to shear distribution across is required if v<vc though it is a normal practice which is very common for
section. No similar provision is found in of not providing shear stirrups in pile caps; cap or footing structures
BS8110; (vii) No explicit statement that torsion is required to which are usually wide
(v) In 6.7.3.3 4th dot, determination of effective be checked, if any. structures. There should be

A-10

width of section for resisting shear is included. shear enhancement if the
No similar provision is found in BS8110; full length cannot
(vi) In 6.7.3.3 5th dot, it is explicitly stated that no mobilized, as similar to
shear reinforcement is required if v<vc. No direct stress. However, no
similar stipulation is found in BS8110; study data is available for
(vii) In 6.7.3.4, the ultimate shear stress is increased torsional shear
to 7 MPa, as compared with BS8110; enhancement.
(viii) In 6.7.3.5, it is stated that torsion for a rigid pile
cap should be checked based on rigid body
theory and where required, torsional
reinforcements be provided.

6.8 – Beam The Code contains detailed provisions for design of – No similar provision. Design checking on
Column beam column joints. No similar provision found in beam-column shall be done if
Joints BS8110. CoPConc2004 is used.

Section 7 – The followings are highlighted : Pt. 2 Section 3 The followings are highlighted : The following remarks are
Serviceability (i) This section contains provisions in BS8110 Pt. (i) General provisions for determination of made :
Limit States 2 Section 3 and the deem-to-satisfy deflections, cracks (including thermal cracking) (i) The omission of the last
requirements (for deflections) in BS8110 Pt. 1 are given; paragraph and Table 3.2
Section 3; (ii) Some of the provisions may be applicable to of BS8110 Pt. 2 3.8.4.1 is
(ii) Limits of calculated crack widths are given in UK only; likely because of the
7.2.1 for member types and exposure (iii) Limit of calculated crack widths is given as different climate and
conditions. The limited values are mostly 0.3mm in 3.2.4.1 as a guide. material properties in
0.3mm as compared with BS8110 except water Hong Kong;
retaining structures and pre-stressed concrete (ii) Equation 7.3 in 7.2.4.2
(0.2 mm); appears to be a refined
(iii) 7.2.4.1 is identical to BS8110 Pt. 2 3.8.4.1 version of Equation 14 of
except that the last paragraph and Table 3.2 of BS8110 Pt. 2 3.8.4.2. The
BS8110 Pt. 2 3.8.4.1 have been omitted. The factor 0.8 accounts for
omitted portion is in relation to estimated discount of strain due to
limiting temperatures changes to avoid the long term strain.
cracking;
(iv) Equation 7.3 in 7.2.4.2 is different from

A-11

Equation 14 of BS8110 Pt. 2 3.8.4.2 where the
difference in temperatures is divided into 2
parts and a factor of 0.8 is employed in the
estimation of thermal strain. Also it is allowed
in the clause to take reinforcements into
consideration;
(v) In 7.3, pre-camber is limited to span/250 to
compensate excessive deflection. The limit is
not given in BS8110 Pt. 2 3.2.1. Also,
deflection limit after construction for avoidance
of damage to structure is limited to span/500,
whilst BS8110 Pt. 2 3.2.1.2 specifies span/500
or span/350 in accordance with brittleness of
finishes for avoidance of damage to
non-structural elements. In addition, 20 mm as
an absolute value is also imposed in BS8110;
(vi) In 7.3.2, limits (0.15 and 0.25 m/s2) on
accelerations (10 years return period on 10 min.
duration) are given as avoidance of “excessive
response” to wind loads whilst no numerical
values are given in BS8110 Pt. 2 3.2.2.1.
Furthermore, deflection limit due to wind load
is given as H/500 whilst BS8110 Pt. 2 3.2.2.2
indicates limit of h/500 as inter-storey drift for
avoidance of damage to non-structural
members;
(vii) In 7.3.3, excessive vibration should be avoided
as similar in BS8110 Pt. 2 3.2.3. However,
there is extra requirement in 7.3.3 that dynamic
analysis be carried out in case structural
frequency less than 6 Hz for structural floor and
footbridge less than 5 Hz;
(viii) Table 7.3 under 7.3.4.2 in relation to
deem-to-satisfy requirement for basic

A-12

span/depth ratios of beam and slab contains,
requirements for “two-way slabs” and “end
spans” are included, as in comparison with
Table 3.9 of BS8110 Pt 1;
(ix) Table 7.4 in relation to modification factor to
effective span depth ratio by tensile
reinforcement is identical to Table 3.10 of
BS8110 except that the row with service stress
307 (fy = 460) has replaced that of 333 (fy =
500);
(x) 7.3.4.6 is identical to BS8110 Pt. 1 3.4.6.7
except that the last sentence in BS8110 is
deleted;
(xi) The provision of deflection calculation in 7.3.5
is identical to BS8110 Pt 2 Cl. 3.7;
(xii) Equation 7.7 in 7.3.6 is not identical to
equation 9 in BS8110 Pt. 2 in the derivation of
shrinkage curvature;

Section 8 – The followings are highlighted : Pt. 1 Section 3 The followings are highlighted : The followings are highlighted :
Reinf’t (i) In 8.1.1, it is declared that the rules given in the Pt. 1 4.10 in (i) The provisions are general; (i) CoPConc2004 requires bar
requirements Section for re-bars detailings do not apply to relation to (ii) Consideration for ductility is not adequate. scheduling to acceptable
seismic, machine vibration, fatigue etc and anchorage of standards. However,
epoxy, zinc coated bars etc. No similar tendons in provisions in 8.3, 9.2.3 etc
exclusion is found in BS8110; prestressed have requirements for bend
(ii) In 8.1.2, it is stated that bar scheduling should concrete of bars;
be in accordance with acceptable standards (ii) ICU has commented that
whilst BS8110 Pt. 1 3.12.4.2 requires standard requirement in BS8110 Pt.
be in accordance with BS4466; 1 3.12.9.4(a) should extend
(iii) In 8.2, the minimum spacing of bars should be to all support conditions.
the lesser of bar diameter, hagg+5 mm and 20 8.4.8 of the Code seems to
mm. BS8110 Pt. 1 3.12.11.1 does not include incorporate the comment;
20 mm and bar diameter is only required when (iii) ICU has also suggested the
bar size > hagg+5 mm; use of torsional links

A-13

(iv) In 8.2, it is stated that the distance between similar to that in ACI code
horizontal layer of bars should be sufficient (135o hood) which is of
without quantification whilst BS8110 Pt. 1 shape other than shape
3.12.11.1 requires minimum be 2hagg/3; code 77 of BS4466.
(v) 8.3 is essentially identical to BS8110 Pt. 1
3.12.8.22, 24, 25, except that
(a) an additional provision of installing a cross
bar inside a bend can eliminate checking of
bearing stresses of the bend in concrete;
(b) a single Table 8.2 (in relation to minimum
bend radii of re-bars) to replace Table 3 of
BS4466 to which BS8110 is making
reference. As such, no distinction is made
between mild steel bars and HY bars –
both adopting minimum radii of HY bars.
Provisions to the newer BS – BS8666
where the minimum bend radii are
generally smaller is not adopted;
(vi) 8.4 is essentially identical to BS8110 Pt. 1
3.12.8 except
(a) It is mentioned in 8.4.1 that when
mechanical device is used, their
effectiveness has to be proven. No similar
provision is found in BS8110;
(b) Type 1 bars are not included in the Code;
(c) 8.4.6 is added with Figure 8.1 for
illustration of bend anchorage;
(d) 8.4.8 in relation to minimum support
widths requires any bend inside support be
beyond the centre line of the support. This
effectively extend the requirement of
BS8110 Pt. 1 3.12.9.4(a) to support
conditions other than simply support;
(vii) 8.5 in relation to anchorage of links and shear

A-14

reinforcements contains more stringent
requirement for the length of the link beyond
bends of bars than BS8110 Pt. 1 3.12.8.6. – (1)
the greater of 4φ or 50 mm for 180o bend in the
Code but only 4φ in BS8110; (2) the greater of
10φ or 70 mm for 90o bend in the Code but
only 8φ in BS8110. Provisions for 135o bend
and welded bars are also added;
(viii) 8.6 contains requirements for anchorage by
welded bars which is not found in BS8110;
(ix) Except for the requirement that the sum of
re-bar sizes at lapping < 40% of the breadth of
the section, 8.7 contains more requirements in
terms of “staggering laps”, minimum
longitudinal and transverse distances between
adjacent lapping of bars which are not found in
BS8110. The requirements are also
schematically indicated in Fig. 8.4. Effectively
the clause requires tension laps be always
staggered. Fortunately compression laps and
secondary rebar lapping can be in one section;
(x) 8.7.4.1 contains different and more detailed
requirements for transverse reinforcements in
lapped zone than BS8110 Pt. 2 3.12.8.12;
(xi) 8.8 and 8.9 in relation to large diameter bars
(>40φ) and bundle bars which are not found in
BS8110;
(xii) The provision in BS8110 Pt. 1 3.12.8.16 for
butt joints of re-bars is not found in the Code.

Section 9 – The followings are highlighted : Pt. 1 Section 3 The followings are highlighted : The followings are highlighted :
Detailing of (i) Table 9.1 under Cl. 9.2 tabulates minimum steel and 5.2.7 (i) The analysis procedures are largely (i) The stress reduction in
Members and percentage equal to Table 3.25 of BS8110 for old-fashioned relying on old theories of design of cantilevered

A-15

particular beams; Johansen, Hillerborg. Detailings to cater for projecting structures in
rules (ii) In 9.2.1.4 in relation to maximum distance of behaviours not well understood or quantified PNAP173 is not
bars in tension as similar to BS8110 Pt. 1 are thus provided, though the determination of incorporated is likely
3.12.11.2, the stipulation in BS8110 that which are largely empirical or from past because the PNAP is based
demonstration of crack width < 0.3 mm can be experiences; on working stress design
accepted is omitted; (ii) Though ductility is not a design aid explicitly method. So there should be
(iii) In 9.2.1.5, a requirement of 15% span moment stated in the BS, the BS does requires 135o some other approaches and
be used to design beam support even under bend of links in anchoring compression bars in this is not mentioned in the
simply supported assumption is not found in columns and beams (Pt. 1 3.12.7.2). CoPConc2004;
BS8110. Furthermore, it is also stated in the (ii) Ductility is more
clause that total tension re-bars of a flanged emphasized in
beam over intermediate supports can be spread CoPConc2004 9.9 which
over the effective width of the flange provided largely stem from seismic
that half of the steel within the web width. design.
There is also no such provision in BS8110;
(iv) 9.2.1.8 requiring 30% of the calculated
mid-span re-bars be continuous over support
appears to be adopted from Fig. 3.24 a) of
BS8110. However, the circumstances by which
the Figure is applicable as listed in 3.12.10.2 of
the BS is not quoted;
(v) 9.2.1.9 requires top steel of cantilever to extend
beyond the point of contraflexure of the
supporting span whilst Fig. 3.24 c) requires at
least half of the top steel to extend beyond half
span of the cantilever or 45φ;
(vi) In 9.2.2, maximum spacing of bent-up bars is
stipulated whilst no such requirement is found
in BS8110;
(vii) Torsional links has to be closed links (shape
code 77 of BS4466) as required by BS8110 Pt.
2 2.4.8. However, 9.2.3 of the Code provides an
alternative of using closed links of 135o bend;
(viii) In 9.3.1.1 (b) in relation to maximum spacing

A-16

of re-bars in slab is more detailed than BS8110
Pt. 1 3.12.11.2.7 and appears more reasonable.
The provisions are, in fact, more stringent :
(a) for principal re-bars, 3h ≤ 400 mm whilst
BS8110 is 3h ≤ 750 mm;
(b) for secondary re-bars 3.5h ≤ 450 mm
whilst no provision in BS8110;
(c) more stringent requirements are added for
slabs with concentrated loads or areas of
maximum moments whilst no similar
requirements are found in BS8110;
(ix) The first para. in 9.3.1.3 requires half of the
area of the calculated span re-bars be provided
at the simply supported slabs and end support
of continuous slabs. The requirement is
identical to BS8110 Pt. 1 3.12.10.3.2. However,
the provision in BS8110 is under the condition
listed in 3.12.10.3.1 that the slabs are designed
predominantly to carry u.d.l. and in case of
continuous slabs, approximately equal span.
These conditions are not mentioned in the
Code;
(x) In 9.3.1.3, there is also a provision that if the
ultimate shear stress < 0.5vc at support, straight
length of bar beyond effective anchorage for
1/3 of support width or 30 mm (whichever is
the greater) is considered effective anchorage.
No similar provision is found in BS8110;
(xi) 9.3.1.6 requiring closed loops of longitudinal
rebars at free edge of slab is not found in
BS8110;
(xii) 9.3.2 is in relation to shear in slabs which
should be identical to that for beams. However
it is stated that shears should be avoided in

A-17

slabs < 200 mm thick;
(xiii) 9.4 in relation to cantilevered projecting
structures has incorporated requirements for
minimum thickness, minimum steel areas,
maximum bar spacing, anchorage length from
PNAP173. However, other requirements such
as design with reduced stresses are not found;
(xiv) 9.5.2.3 contains more stringent requirements of
links in circular columns than that in BS8110
Pt. 1 3.12.7.3 as the termination of links should
be at 135o hook;
(xv) There is an extra requirement in the maximum
spacing of traverse reinforcement in wall in
9.6.3 which is 400 mm, as in comparison with
BS8110 Pt. 1 3.12.7.4;
(xvi) 9.6.5 in relation to reinforcement provisions to
plain walls include BS8110 Pt. 1 3.9.4.19 to 22.
However, 3.9.4.23 in relation to plain walls
with more than 1/10 in tension to resist flexure
is not included. Anyhow, this is not important
as the wall should not be regarded as plain
wall;
(xvii) In 9.7.1 and 9.7.2 in relation to pile caps and
footings, it is stipulated that the minimum steel
percentage should refer to Table 9.1 though
Table 9.1 is under the heading “Beam”. So the
minimum steel percentage of 0.13% for HY
bars should be observed. There is no explicit
provision in BS8110 for minimum steel in pile
caps and footings;
(xviii) 9.7.3 in relation to tie beams has included
a requirement that the tie beam should be
designed for a load of 10 kN/m if the action of
compaction machinery can cause effects to the

A-18

tie beams. This is not found in BS8110;
(xix) 9.8 in relation to design of corbels carries the
same content as 6.5.2.2 to 6.5.2.4 and is
identical to BS8110 Pt. 1 5.2.7;
(xx) 9.9 contains requirements more stringent than
BS8110 in detailing with the aim to enhance
ductility. The followings are highlighted :
(a) 9.9.1.1(a) requires steel percentage in
beam > 0.3% and percentage of tensile
steel < 2.5%;
(b) 9.9.1.1(c) requires anchorage of beam bar
into exterior column to commence beyond
centre line of column or 8φ instead of
column face unless the moment plastic
hinge can be formed at 500 mm or half
beam depth from column face;
(c) 9.9.1.1(d) imposes restriction in locations
of laps and mechanical couplers – (i) not
within column/beam joints, (ii) not within
1 m from potential plastic hinge locations;
(iii) reversing stresses exceeding 0.6fy
unless with specified confinement by links.
In addition, bars be terminated by a 90o
bend or equivalent to the far face of
column;
(d) 9.9.1.2(e) requires distribution and
curtailment of flexural re-bars be attained
in critical sections (potential plastic hinge
regions);
(e) 9.9.1.2(a) states the link spacing in beam <
the lesser of 16φ and beam width or depth
and corner and alternate compression
re-bars be anchored by links;
(f) 9.9.1.2(b) states that links be adequately

A-19

anchored by means of 135o or 180o hooks.
Anchorage by 90o hooks or welded cross
bars not permitted;
(g) 9.9.2.1(a) states min. (0.8%) and max.
steel% (4% with increase to 5.2% at lap)in
column;
(h) 9.9.2.1(a) requires the smallest dia. of any
bars in a row > 2/3 of the largest bar;
(i) 9.9.2.1(a) limits max. dia. of column re-bar
through beam by (eqn 9.7) dependent on
beam depth, with increase by 25% if not
forming plastic hinge;
(j) 9.9.2.1(b) requires spacing of links to
longitudinal bars not be spaced further than
1/4 of the adjacent column dimension or
200 mm;
(k) 9.9.2.1(c) requires anchorage of column
bar into exterior beam or foundation to
commence beyond centre line of beam or
foundation or 8φ instead of interface unless
the moment plastic hinge can be formed at
500 mm or half beam depth from column
face;
(l) 9.9.2.1(d) states restrictions in locations of
laps;
(m) 9.9.2.2 describes the establishment of
“critical regions” in columns where there
are extra requirements on links – (i) link
spacing in column < the lesser of 6φ and
least 1/4 of column lateral dimension; (ii)
each longitudinal bar be laterally supported
by a link passing around the bar and
having an included angle < 135o. (Regions
other than “critical regions” fallow 9.5.2)

A-20


Section 10 The followings are highlighted : Pt. 1 2.3, The followings are highlighted : The followings are highlighted :
(i) 10.2 lists figures for construction tolerances Section 6, (i) Pt. 1 2.3 lists general requirements for (i) The first part of Section 10
whilst BS8110 refers most of the requirements Section 7, inspection of construction; of CoPConc2004 mainly
to other BS; Section 8 (ii) References to other BS are often stated in stems from HKB(C)R,
(ii) 10.3.4 in relation to sampling, testing and Section 6 and 7; CS1, CS2 whilst the second
compliance criteria of concrete. They are (iii) Provisions of works in extreme temperatures part incorporates
extracted from HKB(C)R but with are given which are deleted in CoPConc2004. workmanship requirements
incorporation of 100 mm test cubes. Such listed in BS8110 Pt. 1
provision is not found in BS8110; Section 6;
(iii) The sub-clause on “Concreting in cold (ii)
weather” in BS8110 is not incorporated. 10.3.7 (iii)
on “Concreting in hot weather” is modified
from BS8110 Pt. 1 6.2.5 (reference to BS
deleted);
(iv) Table 10.4 is similar to BS8110 Pt. 1 Table 6.1.
However the parameter t (temperature) is
deleted and the categorization of cement is
OPC and “others” instead of the few types of
cement in BS8110;
(v) 10.3.8.1 contains general requirements for
“Formwork and falsework” similar (but not
identical) to BS8110 Pt. 1 6.2.6.1;
(vi) 10.3.8.2 lists criteria for striking of formwork
identical to that in BS8110 Pt. 1 6.2.6.3.1. In
addition, provisions for using longer or shorter
striking time for PFA concrete and climbing
formwork are included;
(vii) Minimum striking time in 10.3.8.2 are in
accordance with HKB(C)R Table 10 (with the
addition of props to cantilever requiring 10
days striking time) instead of BS8110 Pt. 1
Table 6.2. Furthermore, BS8110 Pt. 1 Table 6.2
gives temperature dependent striking time

A-21

whilst the striking times in CoPConc2004 are
not temperature dependent;
(viii) The contents of 10.3.9 in relation to surface
finish are extracted from BS8110 Pt. 1 6.2.7.
However, the general requirements are
differently written and the “classes of finish”
have been deleted;
(ix) 10.3.10 and 10.3.11 in relation to joints are
identical to BS8110 Pt. 1 6.2.9 and 6.2.10
though the wordings are different, except the
last sentence of 6.2.9 last para. in relation to
treating vertical joint as movement joint;
(x) 10.4.1 contains general requirements on re-bars
to standards CS2 and other acceptable
standards whilst BS8110 Pt. 1 7.1 requires
conformance to other BS;
(xi) 10.4.2 in relation to cutting and bending of
re-bars is identical to BS8110 Pt. 1 7.2 except
(a) conformance is not restricted to BS but to
acceptable standards; and (b) the requirement
of pre-heating re-bars at temperatures below
5oC is deleted;
(xii) 10.4.3 is effectively identical to BS8110 Pt. 1
7.3 except that the requirement for spacer
blocks be of concrete of small aggregates of
equal strength to the parental concrete is
replaced by spacer blocks to acceptable
standards;
(xiii) 10.4.6 is effectively identical to BS8110 Pt. 1
7.6 except (a) conformance to BS changed to
acceptable standards; (b) detailed descriptions
of the types of welding omitted; and (c)
requirement to avoid welding in re-bar bends
omitted;

A-22

(xiv) 10.5.1 is identical to BS8110 Pt. 1 8.1 except
conformance to BS is changed to acceptable
standards;
(xv) 10.5.5.3 in relation to tensioning apparatus of
prestressing tendons is effectively identical to
BS8110 Pt. 1 8.7.3 except that CoPConc2004
has an additional requirements that apparatus
be calibrated within 6 months;
(xvi) 10.5.5.4 in relation to pre-tensioning of
deflected tendons, compressive and tensile
stresses should be ensured not to exceed
permissible limits during transfer of
prestressing force to the concrete with the
release of holding-up and down forces. BS8110
Pt. 1 8.7.4.3 has omitted the compressive
forces;
(xvii) 10.5.5.5(b) requires anchorage of post
tensioning wires to conform to acceptable
standards whilst BS8110 Pt. 1 8.7.5.2 requires
compliance with BS4447
(xviii) 10.5.5.5(d) in relation to tensioning
procedures which is identical to BS8110 Pt. 1
8.7.5.4, the requirement of not carrying out
tensioning below 0oC is omitted. Further, the
paragraph in BS8110 stipulating that when full
force cannot be developed in an element due to
breakage, slip during the case when a large no.
of tendons is being stressed is omitted in
CoPConc2004;
(xix) 10.5.7 contains detailed provisions for grouting
of prestressed tendons whilst BS8110 Pt. 1 8.9
requires compliance to BS EN 445, 446.

Section 11 – This section outlines measures and procedures in – No similar provisions in BS8110. The control in CoPConc2004

A-23

Quality general quality assurance and control, with reference are summaries of local good
Assurance to local practice. The followings are highlighted : practice.
and Control (i) Control are on design, construction and
completed products;
(ii) Control can be by independent organization;
(iii) Concrete must be from supplier certified under
the Quality Scheme for the Production and
Supply of Concrete (QSPSC);
(iv) Control on construction includes surveillance
measures.

Section 12 – This section is basically identical to Section 4 of Pt. 1 Section 4 The provisions are general. CoPConc2004 follows quite
Prestressed BS8110 Pt. 1. The followings are highlighted : closely the provisions in
Concrete (i) 12.1.5 in relation to durability and fire BS8110 except for minor
resistance makes reference to previous changes.
recommendations in Sections 4 and 10 whilst
BS8110 makes reference also to Part 2 of
BS8110;
(ii) 12.2.3.1 in relation to redistribution of moments
is restricted to concrete grade C70, as in
consistency with reinforced concrete. BS8110
Pt. 1 4.2.3.1 does not have this limitation. But
the BS covers grades up to 40;
(iii) The first loading arrangement in 12.3.3 for
continuous beam is not found BS8110 Pt. 1
4.3.3. The loading arrangement is in consistency
with 5.1.3.2 for reinforced concrete beams.
Though not truly adequate (per similar
argument as above), CoPConc2004 is more
conclusive than BS8110;
(iv) 12.3.8.2 gives ultimate concrete stress 7.0 MPa,
as similar to r.c. works;
(v) 12.8.2.2 in relation to 1000 h relaxation value
which is identical to BS8110 Pt. 1 4.8.2.2, “UK”

A-24

has been deleted in description of
manufacturer’s appropriate certificate;
(vi) 12.8.4 and 12.8.5 in relation to shrinkage and
creep of concrete make reference to 3.1.8 and
3.1.7 whilst BS8110 Pt 1. 4.8.4 and 4.8.5 list
UK data;
(vii) 12.10 makes reference to 8.10.2.2 for
transmission lengths in pre-stressed members
which is titled “transfer of prestress” which is
identical to BS8110 Pt. 1 4.10.1 except that the
2nd paragraph of the BS in relation to the
difficulty of determination of transmission
length has been deleted;
(viii) 12.12.3.1(a) is identical to BS8110 Pt. 1
4.12.3.1.1 except that not only protection
against corrosion is added. In 12.12.3.1(c),
reference for protection against fire is not
identical to BS8110;

Section 13 – This section contains testing of structures during – No similar provisions in BS8110.
Load Tests of construction stage under circumstances such as
Structures or sub-standard works are suspected and visible defects
parts of are identified.
structures

A-25

Appendix B

Assessment of Building
Accelerations

Appendix B

Assessment of “along wind” acceleration of Buildings (at top residential floor)

Underlying principles :

Two Approaches are outlined in this Appendix :

(i) The first one is based on the assumption that the building will undergo simple
harmonic motion under wind loads. Thus the equation of governing simple
harmonic motion which is && = −ω 2 x where && is the acceleration, x is the
x x
displacement of the motion, ω is the circular frequency of the building equal
to 2πf ( f is the natural frequency of the building) can be used. However,
generally only the “dynamic resonant component” of the motion is needed for
calculating the acceleration. So if the G factor which is equal to
2
g f SE
1 + 2I h g v B +
2
in Appendix F of the Wind Code 2004 is used to
ζ

arrive at a total displacement which can be considered to be made of up of
three components : (a) the static part which is 1 in the equation; (b) the
2
dynamic background component which is 2 I h g v B ; and (c) the dynamic

2
g f SE
resonant component 2 I h , it is the last component that should be
ζ

multiplied to ω 2 to arrive at the acceleration causing discomfort. So it is
only necessary to calculate the displacement due to the dynamic resonant
component by multiplying the total displacement by the factor
2  2 
2I h
g f SE 1 + 2 I g 2 B + g f SE  . Alternatively, the same result can be
ζ  h v
ζ 
 
2
g f SE
obtained by multiplying the factor 2 I h to the static wind pressure,
ζ
i.e. Table 2 of the Hong Kong Wind Code 2004. The circular frequency, ω of
the building can either be obtained by detailed dynamic analysis or by some
empirical formula such as 460/h.

(ii) The second approach is that listed in Australian Wind Code AS/NSZ

B-1

Appendix B

3 ˆ
1170.2:2002 Appendix G2. It is based on the simple formula a = Mb
m0 h 2
where m0 is the average mass per unit height of the building, h is the

ˆ
average roof height of the building above ground M b is the resonant

component of peak base bending moment. By the “resonant component”, the
approach is also based on the same principle by using only the dynamic

SE
resonant component in deriving acceleration as the factor g R is
ζ

multiplied to the overturning moment for assessment of acceleration. The
parameters comprising m 0 and h are used for assessment of the dynamic
properties of the building. In addition, there is a denominator of 1+ 2 g v I h in

ˆ
the expression for M b in the Australian Code as different from Hong Kong

Wind Code, the reason being that the Australian Code is based on V des ,θ

which is 3 second gust whilst Hong Kong Code is based on hourly mean wind
speed. So this factor should be ignored when using Hong Kong Code which is
based on hourly mean speed.

Furthermore, two aspects should also be noted :

(i) The Concrete Code requires the wind load for assessment of acceleration to be
1-in-10 year return period of 10 minutes duration whilst the wind load arrived
for structural design in the Hong Kong Wind Code is based on 1-in-50 year
return period of hourly duration. For conversion, the formula listed in
Appendix B of the Wind Code can be used (as confirmed by some experts that
the formula can be used for downward conversion from 1-in-50 year to
1-in-10 year return periods). The 10 minutes mean speed can also be taken as
identical to that of hourly mean speed (also confirmed by the experts.) Or
alternatively, as a conservative approach, the factor 1 − 0.62 I 1.27 ln (t / 3600)
−0.11
 h 
can be applied where I is the turbulence intensity I = 0.087 
 500 

taken at top of the building and t = 600 sec;

(ii) The damping ratio recommended in the Wind Code which is 2% is for

B-2

Appendix B

ultimate design. A lower ratio may need to be considered for serviceability
check including acceleration. Nevertheless, a 10-year return period at damping
ratio 2% should be accepted which is the general practice by the Americans.
The worked examples follow are therefore based on damping ratio of 2%,
though the readers can easily work out the same for damping ratio of 1%
under the same principle.

The procedures for estimation of acceleration are demonstrated by 3 worked examples
that follow :

Worked Example B-1

For the 40-storey building shown in Figure B-1 which has been analyzed by ETABS,
the acceleration of the top residential floor in the for wind in X-direction is to be
computed.

Y

X

Figure B-1 – 40 storeys building for Worked Example B-1

Data : Building height h = 121.05 m;
Building plan width and depth are b = d = 43 m;
Lowest building natural frequencies for the respective motion can be
obtained with reference to the modal participating mass ratios as revealed by
dynamic analysis in ETABS or other softwares:
na1 = 0.297 Hz for rotation about Z axis (torsional)

B-3

Appendix B

na 2 = 0.3605 Hz for translation along Y-direction
na1 = 0.3892 Hz for translation along X-direction
For wind in X-direction :
−0.11 −0.11
 h   121.05 
I h = 0.1055  = 0.1055  = 0.1021 ; g v = 3.7
 90   90 
g f = 2 ln(3600na ) = 2 ln(3600 × 0.3892) = 3.8066 ;
0.25 0.25
h  121.05 
Lh = 1000  = 1000  = 1865.35 ;
 10   10 
1 1
B= = = 0.6989 ;
36h + 64b
2 2
36 ×121.05 2 + 64 × 432
1+ 1+
Lh 1865.35
0.11 0.11
 h   121.05 
Vh = V g   = 59.5  = 50.905 m/sec;
 500   500 
n L 0.3892 × 1865.35
N= a h = = 14.262 ;
Vh 50.905
1 1
S= = = 0.1019
 3.5na h   4na b   3.5 × 0.3892 ×121.05   4 × 0.3892 × 43 
1 +  1 +  1 +  1 + 
 Vh   Vh   50.905  50.905 
0.47 N 0.47 × 14.262
E= = = 0.0793 ;
(2 + N ) 2 5/6
(2 + 14.262 ) 2 5/6

2
g f SE 3.8066 2 × 0.1019 × 0.0793
G = 1 + 2I h g v B + = 1 + 2 × .1021 3.7 × 0.6989 +
2 2

ζ 0.02
=1.8155;
2
g f SE 3.8066 2 × 0.1019 × 0.0793
Gres = 2 I h = 2 × .1021 = 0.494 ;
ζ 0.02
∴ Gres / G = 0.272

Deflection (translation and rotation) of the centre of the top floor calculated
in accordance with Appendix G of HKWC2004 is
X-direction : 0.069m
Y-direction : 0.00061m
Z-direction : 0.000154rad
For this symmetrical layout, the Y-deflection and Z-rotation are small and
can be ignored.

Procedures :

(i) Conversion from 50 years return period to 10 years return period is by the
factor listed in Appendix B of HKWC2004. The factor is

B-4

Appendix B

 5 + ln (R )   5 + ln 10 
2 2

  =  = 0.6714
 5 + ln 50   5 + ln 50 

(ii) Conversion from hourly mean wind speed to 10 minutes mean wind speed is
by the factor
1 − 0.62 I 1.27 ln(t / 3600 ) = 1 − 0.62 × 0.10211.27 ln (600 / 3600) = 1.061
(iii) So the displacements converted to contain only the dynamic resonant
component and to 10 years return period, 10 minutes wind speed can be
obtained by multiplying the deflections obtained in accordance with Appendix
G of HKWC2004 by the aggregate factor of 0.272 × 0.6714 × 1.061 = 0.1938
(iv) The X-deflections for calculation of accelerations is therefore
0.1938 × 0.069 = 0.0134 m;
(v) The acceleration of the centre of the block in X-direction is therefore

(2πna 3 )2 × 0.0134 = (2π × 0.3892)2 × 0.0134 = 0.801 m/sec2 as the fundamental

frequency for X-translation is na 3 = 0.3892 Hz listed in the data.

Worked Example B-2

The acceleration of the block in Worked Example B-1 is redone by the Australian
Code AS/NSZ 1170.2:2002 Appendix G2 :
Total dead load is 539693 kN and total live load is 160810 kN
Using full dead load and 40% live load for mass computation :
Mass per unit height is

m0 =
(539693 + 160810 × 0.4) ÷ 9.8 × 103 = 509.165 ×103 kg/m
121.05
Overturning moment at 50 years return period is 1114040 kNm when wind is blocing
in the X-direction. When the moment is converted to contain only the dynamic
resonant component and to 10 years return period, 10 minutes wind speed, it becomes

M b = 0.272 × 0.6714 × 1.061 × 1114040 = 215857 kNm, the factors are quoted from
ˆ

Worked Example 1.
So the acceleration in the X-direction is
3 ˆ 3 × 215857 × 10 3
a= Mb = = 0.087 m/sec2.
m0 h 2
506.165 × 10 × 121.05
3 2

which is greater than that in Worked Example B-1.

B-5

Appendix B

Worked Example B-3

Another worked example for finding acceleration in Y-direction is demonstrated for a
building shown in Figure B-2 where torsional effect is significant. The building
suffers significant torsion where the displacement and acceleration of Point A (at
distance 30m in the X-direction and 1.6m in the Y-direction from the centre of rigidity
of the building) is most severe. The provision in the Australian Code should be quite
limited in this case and therefore not used.

C, centre of rigidity

1.6m A

30m

Figure B-2 – Layout of Building where torsional effect is significant

The first 3 fundamental frequencies are listed as follows. They can be read from
dynamic analysis of the building by ETABS with reference to the modal participating
mass ratios or other softwares. The dynamic resonant component factor

2 I h g f SE / ζ
2
are also calculated for the respective direction of motion whilst the

dynamic magnification factor G for wind in Y-direction is calculated to be 1.8227.

Fundamental Frequency Circular 2
g f SE Gres
Direction Periods (sec) f frequency Gres = 2 I h G
(Hz) ω = 2πf (Hz) ζ
Y-direction 2.6598 na = 0.376 ω1 = 2.3623 0.512 0.2809
Z-rotation 1.8712 na = 0.5344 ω 2 = 3.3578 0.3185 0.1747
X-direction 1.5652 na = 0.6389 ω3 = 4.014 0.3036 0.1665

Table B-1 – Fundamental Frequencies of Worked Example B-2

B-6

Appendix B

The displacements of centre of rigidity of the building at the top residential floor as
per analysis in accordance with Appendix G of the Wind Code 2004 after application
of the dynamic magnification factor, G is as follows in Table B-2. The corrected
values after discount for (i) Gresonant / G ; (ii) 10 minutes duration (factor 1.06); and (iii)
10 years return period (0.6714) are also listed.

Displacement at Centre of Rigidity
before adjustment (read from after adjustment for (i), (ii), (iii)
ETABS output)
X-displacement 0.0262m 0.00311m
Y-displacement 0.119m 0.0238m
Z-rotation 0.00196rad 0.000244rad

Table B-2 – Displacement of Worked Example B-2

The acceleration of the building at its centre of rigidity in X-direction, Y-direction and
The acceleration of the respective directions are :
X-direction : ω3 ∆ x = 4.014 2 × 0.00311 = 0.0501 m/sec2;
2

Y-direction : ω1 2 ∆ y = 2.36232 × 0.0238 = 0.1328 m/sec2;
Z-direction : ω 2 2θ z = 3.3578 2 × 0.000244 = 0.00275 rad/sec2;

The linear acceleration at point A will be the vector sum of that in the X and
Y-directions, each of which in turn comprises linear component equal to that in the
centre of rigidity and a component being magnified by the torsional effect.

Linear acceleration due to Z-rotation acceleration is 0.00275 × 30 = 0.0825 m/sec2.
Total acceleration in Y-direction is taken as the square root of sum of squares of the
direct linear Y acceleration and that induced by rotation. The reason why the
acceleration is not taken as algebraic sum of both is because they do not occur at the
same frequency. So the total acceleration in Y-direction is

0.1328 2 + 0.0825 2 = 0.1563 m/sec2.

Similarly, linear acceleration in the X-direction due to Z-rotation acceleration is
0.00275 × 1.6 = 0.0044 m/sec2;

Total acceleration in X-direction is 0.05012 + 0.0044 2 = 0.0503 m/sec2.

The vector sum of the acceleration of point A is therefore

B-7

Appendix B

0.15632 + 0.0503 2 = 0.1642 > 0.15 m/sec2 as required in Cl. 7.3.2 of the Code.

So provisions should be made to reduce the acceleration.

Thus it can be seen that, though the deflection complies with the limit of H/500, the
acceleration exceeds the limit of 0.15m/sec2. However, compliance with the former
may be adequate as per Code requirements.

Limitations of the two approaches

It should be borne in mind that the approaches described above are simplified ones.
As the approaches are very much based on the assumed natural frequency of the
building or arrival of such value by empirical method (the Australian Code), it follows
that they should be used in care when the dynamic behaviour of the building is
complicated such as having significant cross wind effects or coupling of building
modes is significant.

B-8

Appendix C

Derivation of Basic Design
Formulae of R.C. Beam sections
against Flexure

Appendix C

Derivation of Basic Design Formulae of R.C. Beam sections against
Bending

The stress strain relationship of a R.C. beam section is illustrated in Figure C-1.

ε ult = 0.0035
d’

x

d neutral axis

Stress Diagram Strain Diagram

Figure C-1 – Stress Strain diagram for Beam

In Figure C-1 above, the symbols for the neutral axis depth, effective depth, cover to
compressive reinforcements are x, d, and d’, as used in BS8110 and the Code.

To derive the contribution of force and moment by the concrete stress block, assume
the parabolic portion of the concrete stress block be represented by the equation
σ = Aε 2 + Bε (where A and B are constants) (Eqn C-1)
dσ
So = 2 Aε + B (Eqn C-2)
dε
dσ
As = Ec ⇒ B = Ec where Ec is the tangential Young’s Modulus of
dε ε =0
concrete listed in Table 3.2 of the Code.

dσ B E
Also = 0 ⇒ 2 Aε 0 + B = 0 ⇒ A = − =− c (Eqn C-3)
dε ε =ε 0 2ε 0 2ε 0

f cu
As σ = 0.67 when ε = ε 0
γm
f cu Ec Ec 0.67 f cu Ec 1.34 f cu
∴ 0.67 − =− ⇒ = ⇒ ε0 = (Eqn (C-4)
γ mε 0 2
ε0 2ε 0 γ mε 0 2 Ec γ m
(accords with 3.14 of the Concrete Code Handbook)

C-1

Appendix C

Ec 1.34 f cu
A=− where ε 0 =
2ε 0 Ec γ m
Ec 2
So the equation of the parabola is σ =− ε + Ecε for ε ≤ ε 0
2ε 0
Consider the linear strain distribution

xε 0 / ε ult
ε ult = 0.0035
ε = ε0

x
u
h

Figure C-2 – Strain diagram across concrete section

u
At distance u from the neutral axis, ε = ε ult
x
ε0
So stress at u from the neutral axis up to x is
ε ult
Eε Eε
2 2
Ec 2 E  u  u
σ =− ε + Ecε = − c  ε ult  + Ec  ε ult  = − c ult2 u 2 + c ult u (Eqn C-5)
2ε 0 2ε 0  x  x 2ε 0 x x
Based on (Eqn C-5), the stress strain profiles can be determined. A plot for grade 35 is
included for illustration :


18
16
14
12
Stress (MPa)

10
8
6
0.3769 where
4
ε0 = 0.001319
2
0
0 0.2 0.4 0.6 0.8 1

Figure C-3 – Stress strain profile of grades 35
C-2

Appendix C

Sectional Design of rectangular Section to rigorous stress strain profile

Making use of the properties of parabola in Figure C-4 offered by the parabolic
section as Fc1 given by

centre of mass

b
2
Area = ab
3

a

3
a
8

Figure C-4 – Geometrical Properties of Parabola

2 ε0 f 1.34ε 0 f cu
Fc1 = b x0.67 cu = bx (Eqn C-6)
3 ε ult γm 3γ m ε ult
and the moment exerted by Fc1 about centre line of the whole section

h  ε  3 ε0  h  5 ε0 
M c1 = Fc1  − x1 − 0
 ε − x
 8 ε  = Fc1  2 − x1 − 8 ε
 
 (Eqn C-7)
2  ult  ult    ult 

The force by the straight portion is

0.67 f cu  ε0  0.67 f cu bx  ε 
Fc 2 = 
x− x ε b =
 1 − 0
 ε 
 (Eqn C-8)
γm  ult  γm  ult 

The moment offered by the constant part about the centre line of the whole section is

h  ε  x
M c 2 = Fc 2  − 1 − 0
 ε  
2 (Eqn C-9)
2  ult  

The compressive force by concrete as stipulated in (Eqn C-6) and (Eqn C-8) is

1.34ε 0 f cu 0.67 f cu bx  ε  0.67 f cu bx  ε 
Fc = Fc1 + Fc 2 = bx + 1 − 0
 ε =
 3 − 0
 ε 

3γ m ε ult γm  ult  3γ m  ult 

For singly reinforcing sections, moment by concrete about the level of the tensile steel
is, by (Eqn C-7) and (C-9)

C-3

Appendix C

  5 ε 0    ε0  x
M = M c1 + M c 2 = Fc1 d − x1 −  8 ε  + Fc 2 d − 1 − ε  2 
  

  ult    
 ult   
1.34ε 0 f cu   5 ε 0  0.67 f cubx  ε   ε  x 
= 
bx d − x1 −  +
 1 − 0   d − 1 − 0  
 ε   ε 2
3γ mε ult 
  8 ε ult   γm  ult  
  ult  

M 1.34ε 0 f cu x  x  5 ε 0  0.67 f cu x  ε  1  ε  x 
⇒ 2 =  8 ε  + γ
1 − 1 −  1 − 0  1 − 1 − 0  
 ε   
bd 3γ m ε ult d  d 
 ult   m d  2  ε ult  d 
ult   
0.67 f cu x  1 ε 0   1 1 ε 0 1  ε0   x
2
M  
⇒ = 1 −
  + − +
 −  ε   d 

bd 2
γ m d  3 ε ult   2 3 ε ult 12  ult   
   
0.67 f cu  1 1 ε 0 1  ε   x 
2
0.67 f cu  1 ε 0  x M
2

⇒ − + −  0    +
ε  1 −  − =0
γ m  2 3 ε ult 12  ult   d  γ m  3 ε ult  d bd 2
 
 
(Eqn C-10)
x
which is a quadratic equation in .
d
x
As is limited to 0.5 for singly reinforcing sections for grades up to 45 under
d
moment distribution not greater than 10% (Clause 6.1.2.4 of the Code), by (Eqn C-10),
M
will be limited to K ' values listed as
bd 2 f cu

K ' = 0.154 for grade 30
K ' = 0.152 for grade 35
K ' = 0.151 for grade 40
K ' = 0.150 for grade 45

which are all smaller than 0.156 under the simplified stress block.

x
However, for 45 < f cu ≤ 70 where is limited to 0.4 for singly reinforcing
d
sections under moment distribution not greater than 10% (Clause 6.1.2.4 of the Code),
M
again by (Eqn 3-1) will be limited to
bd 2 f cu
K ' = 0.125 for grade 50
K ' = 0.123 for grade 60
K ' = 0.121 for grade 70

which are instead greater than 0.120 under the simplified stress block. This is because
at concrete grade > 45, the Code has limited the rectangular stress block to 0.8 times
of the neutral axis depth.

C-4

Appendix C

x
With the analyzed by (Eqn C-9), the forces in concrete
d
1.34ε 0 f cu 0.67 f cu bx  ε  F 0.67 f cu  1 ε 0  x
Fc = Fc1 + Fc 2 = bx + 1 − 0  ⇒ c =
 ε  bd 1 − 
3γ m ε ult γm  ult  γ m  3 ε ult  d
 
can be calculated which will be equal to the required force to be provided by steel,
thus
A 0.67 f cu  1 ε 0  x Ast 1 0.67 f cu  1 ε 0  x
0.87 f y st = 1 −
 3 ε  d ⇒ bd = 0.87 f
 1 − 
bd γm  ult  y γ m  3 ε ult  d
 
(Eqn C-11)

M
When exceeds the limited value for single reinforcement. Compression
bd 2 f cu
reinforcements at d ' from the surface of the compression side should be added. The
compression reinforcements will take up the difference between the applied moment
and K 'bd 2

 M 
 2 
Asc  d '   M  Asc  bd f − K '  f cu

1 −  =  − K ' ⇒ =
cu
0.87 f y   (Eqn C-12)
bd  d   bd 2 f cu  bd  d'
0.87 f y 1 − 
 d

And the same amount of steel will be added to the tensile steel.

 M 
 2 
 1 ε0   bd f − K '  f cu
Ast 1 0.67 f cu  
= 1 − 3 ε η +
 cu
 (Eqn C-13)
bd 0.87 f y γ m  ult   d'
0.87 f y 1 − 
 d

x
where η is the limit of ratio which is 0.5 for grade 45 and below and 0.4 for
d
grades up to and including 70.

Furthermore, there is a limitation of lever arm ratio not to exceed 0.95 which requires
0.67 f cu  1 1 ε 0 1  ε 0   x 
2
 1 ε0 x
2
0.67 f cu
− + −      + 1 − 
γ m  2 3 ε ult 12  ε ult   d 
  γm  3ε

d

 ult
≤ 0.95
0.67 f cu  1 ε 0  x
1 − 
γ n  3 ε ult  d
 

C-5

Appendix C

 1 ε0 
0.051 −
 3ε 

x  
⇒ ≥
ult
(Eqn C-14)
d 1 1 ε 1 ε 
2

 − 0
+  0
 
 
 2 3 ε ult 12  ε ult
  


Thus the lower limits for the neutral axis depth ratios are 0.112, 0.113, 0.114 and
0.115 for grades 30, 35, 40, 45 respectively. Thus for small moments acting on beam
x M
with not fulfilling (Eqn C-14), Ast = (Eqn C-15)
d 0.87 f y × 0.95d

As illustration for comparison between the rigorous and simplified stress block
M
approaches, plots of against steel percentages for grade 35 is plotted as
bd 2

Comparison of Reinforcement Ratios for grade 35 according to the Rigorous
and Simplfied Stress Block (d'/d = 0.1)
Ast/bd - Rigorous Stress Approach Ast/bd - Simplified Stress Approach
Asc/bd - Rigorous Stress Approach Asc/bd - Simplified Stress Approach
14
13
12
11
10
9
8
2
M/bd

7
6
5
4
3
2
1
0
0 0.5 1 1.5 2 2.5 3 3.5 4

It can be seen that the differences are very small, maximum error is 1%.

However, for high grade concrete where the K ' values are significantly reduced in
the rigorous stress block approach (mainly due to the switching of upper limits of the
neutral axis depth ratios from 0.5 to 0.4 and 0.33 for high grade concrete), the
differences are much more significant for doubly reinforced sections, as can be seen
from the Design Charts enclosed in this Appendix that compressive steel ratios
increased when concrete grade switches from grade 45 to 50 as the neutral axis depth
ratio changes from 0.5 to 0.4.

C-6

Appendix C

Determination of reinforcements for Flanged Beam Section – T- or L-Sections

For simplicity, only the simplified stress block in accordance with Figure 6.1 of the
Code is adopted in the following derivation. The symbol η is used to denote the
ratio of the length of the simplified stress block to the neutral axis depth. Thus
η = 0.9 for f cu ≤ 45 ; η = 0.8 for 45 < f cu ≤ 70 ; η = 0.72 for 70 < f cu ≤ 100 .

The exercise is first carried out by treating the width of the beam as beff and analyze

the beam as if it is a rectangular section. If η of neutral axis depth is within the
x hf
depth of the flange, i.e. η ≤ , the reinforcement so arrived is adequate for the
d d
x hf
section. The requirement for η ≤ is
d d
x K hf
η = 1− 1− ≤ (Eqn C-15)
d 0.225 d
1
The lever arm z = d − ηx
2

z 1 x 1 K  K
⇒ = 1 − η = 1 − 1 − 1 −

 = 0.5 + 0.25 −
 (Eqn C-16)
d 2 d 2 0.225  0.9

x hf
If, however, η > , the section has to be reconsidered with reference to Figure
d d
C-5.

0.67 f cu
beff γm

hf ηx
x
d

bw

Figure C-5 – Analysis of a T or L beam section

For singly reinforced sections, taking moment about the level of the reinforcing steel,

C-7

Appendix C

 h f  0.67 f cu η 
M=
0.67 f cu
(b − bw )h f  d −  +
  
bw (ηx ) d − x 
γm eff
 2  γm  2 
M 0.67 f cu  beff  h f  1 h f  0.67 f cu  x  η x 
⇒ = 
 b − 1 d  1 − 2 d  + γ
   η 1 −  (Eqn C-17)
bw d 2
γm  w    m  d  2 d 
Mf 0.67 f cu h f  beff  1 h f 
Putting =  − 11 −
 2 d   (Eqn C-18)
bw d 2 γ m d  bw   
The equation is in fact the contribution of the moment of resistance of the section by
the flange, (Eqn C-17) becomes

0.67 f cu η 2  x  0.67 f cu x M − M f
2

  − η + =0 (Eqn C-19)
γm 2 d  γm d bw d 2

x
which is a quadratic equation for solution of where M f can be predetermined
d
x
by (Eqn C-18). Provided ≤ ϕ where ϕ = 0.5 for f cu > 45 ; 0.4 for f cu > 70
d
and 0.33 for f cu > 100 , single reinforcement be provided by the following equation
which is derived by balancing the steel force and the concrete force.

 beff  hf x
0.87 f y Ast =
0.67 f cu
γm
[(b
eff ]
− bw )h f + bwηx ⇒
Ast
=
0.67 f cu
bw d γ m 0.87 f y

 − 1 + η 
 d
 bw  d

(Eqn C-20)

x
If = ϕ , the maximum moment of resistance by concrete is reached which is (by
d
taking moment about the tensile steel level)
0.67 f cu   hf  1 
(beff − bw )h f  d −  + bwηϕd  d − ϕd 

M c max = M f max + M b max =  
γm   2   2 
Mc 0.67 f cu  beff  hf  1 hf   1 
⇒ K '= = 
 − 1  1 −
d  2 d   + ηϕ 1 − ϕ  (Eqn C-21)
bw d 2
γm  bw     2 
and tensile steel required will be, by (Eqn C-20)
Ast ,bal 0.67 f cu  beff  hf 
= 
 b − 1 d + ηϕ 
 (Eqn C-22)
bw d γ m 0.87 f y  w  

If the applied moment exceeds M c , the “excess moment” will be taken up
compressive steel Asc with cover to reinforcement c ' .
( )
0.87 f y Asc d − d ' = M − M c

C-8

Appendix C

Asc M − Mc 1  M Mc 
⇒ = =  − 
bw d bw d (1 − d ' / d )0.87 f y 0.87 f y (1 − d ' / d )  bw d
2 2
bw d 2 
 M  beff  hf  1 hf  1 
 + ηϕ 1 − ϕ  
1 0.67 f cu
=  − 
 − 1 1 −
d  2 d   
0.87 f y (1 − d ' / d )  bw d

2
γm  bw     2   
(Eqn C-23)
The total tensile steel will be
Ast A A
= st ,bal + sc
bw d bw d bw d

The followings are stated for f cu ≤ 45 where η = 0.9 and ϕ = 0.5 which is most
commonly used in flexural members:

For η × neutral axis depth below flange, (Eqn C-19) can be written as :
x M −Mf
2
x
0.1809 f cu   − 0.402 f cu + =0 (Eqn C-24)
d  d bw d 2
Mf 0.67 f cu h f  beff  1 h f 
where 0.9 x ≥ h f and = 
 b − 11 − 2 d
 

bw d 2
γm d  w  
Ast 0.67 f cu  beff  hf x
=  b − 1 d + 0.9 d 
  (Eqn C-24)
bw d γ m 0.87 f y  w  

For double reinforcements where x > 0.5d by (Eqn C-24), substituting η = 0.9 and
ϕ = 0.5 into (Eqn C-23)
Asc 1  M 0.67 f cu  beff  hf  1 hf  
=  − 
 − 1 1 −
d  2 d  + 0.3375 

bw d 0.87 f y (1 − d ' / d )  bw d

2
γm  bw    
(Eqn C-25)
By (Eqn C-22),
Ast A A 0.67 f cu  beff  hf  Asc
= st ,bal + sc =  b − 1 d + 0.45 + b d
  (Eqn C-26)
bw d bw d bw d γ m 0.87 f y  w   w

C-9


14

12

10

8
2
M/bd

6

4

2

0
0 0.5 1 1.5 2 2.5 3 3.5 4

Chart C-1


14

12

10

8
2
M/bd

6

4

2

0
0 0.5 1 1.5 2 2.5 3 3.5 4

Chart C-2


14

12

10

8
2
M/bd

6

4

2

0
0 0.5 1 1.5 2 2.5 3 3.5 4

Chart C-3


14

12

10

8
2
M/bd

6

4

2

0
0 0.5 1 1.5 2 2.5 3 3.5 4

Chart C-4

Appendix D

Underlying Theory and Design
Principles for Plate Bending
Element

Appendix D

Underlying Theory and Design Principles for Plate Bending Element

By the finite element method, a plate bending structure is idealized as an assembly of
discrete elements joined at nodes. Through the analysis, “node forces” at each node of
an element, each of which comprises two bending moments and a shear force can be
obtained, the summation of which will balance the applied load at the node. Figures
D-1a and D-1b illustrates the phenomena.

FZ at FZ at Z
Node 4 Node 3
FZ at Y
Node 1
3 MX at X
Node 2
FZ at Note :
1 Node 2
2 MY at MX, MY, FZ represents
Node 3 respectively the bending
moments about X, Y axes
MY at and the force in the Z axis
MX at MX at Node 2 at the nodes of the plate
Node 1 Node 2
bending element

Figure D-1a – Diagrammatic illustration of the Node Forces at the four Nodes of a
Plate Bending Element 1234.

FA, external load
applied at the
common node
For equilibrium, F1,
F4 F2, F3 and F4 which
are node vertical
F1 shear of 4 elements
F3 at the common node
will sum up to
balance the
externally applied
F2
force FA such that
F1+ F2+ F3 + F4 =
FA. Balancing of
moments is similar.

Figure D-1b – Diagrammatic illustration of balancing of Node shear forces at a
common node to 2 or more adjoining elements. The four elements joined at the
common node are displaced diagrammatically for clarity.

D-1

Appendix D

The finite element method goes further to analyze the “stresses” within the discrete
elements. It should be noted that “stresss” is a terminology of the finite element
method which refer to bending moments, twisting moments and shear forces per unit
width in plate bending element. They represent the actual internal forces within the
plate structure in accordance with the plate bending theory. R.H. Woods (1968) has
developed the famous Wood-Armer Equations to convert the bending moments and
twisting moments (both are moments per unit width) at any point to “design
moments” in two directions for structural design purpose.

Outline of the plate bending theory

Apart from bending moment in two mutually perpendicular directions as well known
by engineers, a twisting moment can be proved to be in existence by the plate bending
theory. The bending and twisting moments constitutes a “moment field” which
represents the actual structural behaviour of a plate bending structure. The existence
of the twisting moment and its nature are discussed in the followings. Consider a
triangular element in a plate bending structure with two of its sides aligning with the
global X and Y directions as shown in Figure D-2 where moments M X and M Y
(both in kNm per m width) are acting respectively about X and Y. A moment M B
will generally be acting on the hypotenuse making an angle of θ with the X-axis as
shown to achieve equilibrium. However, as the resultant of M X and M Y does not
necessarily align with M B , so there will generally be a moment acting in the
perpendicular direction of M B to achieve equilibrium which is denoted as M T . The
vector direction of M T is normal to the face of the hypotenuse. So instead of
“bending” the element like M X , M Y and M B which produces flexural stresses, it
“twists” the element and produce shear stress in the in-plane direction. The shear
stress will follow a triangular pattern as shown in Figure D-2 for stress-strain
compatibility. M T is therefore termed the “twisting moment”. Furthermore, in order
to achieve rotational equilibrium about an axis out of plane, the shear stress will have
to be “complementary”. As the hypotenuse can be in any directions of the plate
structure, it follows that at any point in the plate bending structure, there will
generally be two bending moments, say M X and M Y in two mutually perpendicular
directions coupled with a complementary twisting moment M XY as indicated in
Figure 11a. The phenomenon is in exact analogy to the in-plane stress problem where
generally two direct stresses coupled with a shear stress exist and these components
vary with directions. The equations relating M B , M T with M X , M Y , M XY and
θ derived from equilibrium conditions are stated as follows:

D-2

Appendix D

MB =
1
(M X + M Y ) + 1 (M X − M Y ) cos 2θ + M XY sin 2θ
2 2
(Eqn D-1)
M T = (M X − M Y ) sin 2θ − M XY sin 2θ
1
2

In addition, if θ is so varied that M T vanishes when θ = φ , then the element will
be having pure bending in the direction. The moments will be termed the “principal
moments” and denoted as M 1 , M 2 , again in exact analogy with the in-plane stress
problem having principal stresses at orientations where shear stresses are zero. The
1 2 M XY
angle φ can be worked out by φ = tan −1 (Eqn D-2)
2 (M X − M Y )

Plate Structure

MT

MY
MB
Y θ
Complementary shear stress pattern
MX
X

Figure D-2 – Derivation and nature of the “Twisting Moment”

D-3

Appendix D

MXY Y
X

MX M1 M2
MY

MXY MXY
MY M1
M2 φ
MX

MXY

Figure D-3a – General co-existence Figure D-3b – Principal moment in a
of bending moments and twisting plate bending structure
moment in a plate bending structure

Again, as similar to the in-plane stress problem, one may view that the plate bending
structure is actually having principal moments “bending” in the principal directions
which are free of “twisting”. Theoretically, it will be adequate if the designer designs
for these principal moments in the principal directions which generally vary from
point to point. However, practically this is not achievable for reinforced concrete
structures as we cannot vary the directions of the reinforcing steels from point to point
and from load case to load case.

The “stress” approach for design against flexure would therefore involve formulae for
providing reinforcing steels in two directions (mostly in orthogonal directions)
adequate to resist the “moment field” comprising the bending moments and twisting
moments. The most popular one is the “Wood Armer” Equations by Woods (1968),
the derivation of which is based on the “normal yield criterion” which requires the
provided reinforcing steels at any point to be adequate to resist the normal moment
which is the bending moment M B in any directions as calculated from (Eqn D-1).
The effects of the twisting moments have been taken into account in the formulae.
The Wood Armer Equations are listed as follows.

For bottom steel reinforcement provisions:

∗ ∗
Generally M X = M X + M XY ; M Y = M Y + M XY ;
2
∗ ∗ ∗ M XY
If M X < 0 , then M X = 0 and M Y = M Y +
MX

D-4

Appendix D
2
∗ ∗ ∗ M XY
If M Y < 0 , then M Y = 0 and M X = M X +
MY

For top steel reinforcement provisions:

∗ ∗
Generally M X = M X − M XY ; M Y = M Y − M XY ;
2
∗ ∗ ∗ M
If M X > 0 , then M X =0 and M = M Y − XY
Y
MX
2
∗ ∗ ∗ M
If M > 0 , then M = 0
Y Y and M X = M X − XY
MY
(Eqn D-3)

The equations have been incorporated in the New Zealand Standard NZS 3101:Part
2:1995 as solution approach for a general moment field.

The “stress” approach is therefore based on the actual structural behaviour of the plate
bending structure which is considered as a direct and realistic approach. The approach
is particularly suitable for structures analyzed by the finite element method which
produces a complete set of components of the internal forces of the plate bending
structures including twisting moments, Q max . Design has to cater for all these
components to ensure structural adequacy.

Design against shear

As an alternative to checking or designing punching shear in slab in accordance with
6.1.5.7 of the Code by which the punching shear load created by column (or pile in
pile cap) is effectively averaged over a perimeter, more accurate design or checking
can be carried out which is based on finite element analysis by which an accurate
shear stress distribution in the slab structure can be obtained. The finite element
analysis outputs the “shear stresses” (shear force per unit width) in accordance with
the general plate bending theory at the “X-face” and “Y-face” of an element which are
∂M Y ∂M XY ∂M X ∂M XY
respectively Q XZ = − + and QYZ = − , as
∂x ∂y ∂y ∂y
diagrammatically illustrated in Figure D-4. It can be easily shown that the maximum
shear after “compounding” these two components will occur in a plane at an

 Q XZ 
orientation θ = tan −1 
 
 on plan and the value of the maximum shear is
 QYZ 

D-5

Appendix D

Qmax = Q XZ + QYZ
2 2
as per the illustration in the same Figure. Thus one can view

that both Q XZ and QYZ are components of the actual shears in a pre-set global axis
system. The actual shear stress is Qmax , the action of which tends to produce shear
failure at the angle θ on plan as shown in Figure D-3. So the designer needs to
check or design for Qmax at the spot. There is no necessity to design for Q XZ and
QYZ separately.

Plate structure Out-of plane shear QXZ, QYZ and Qθ

Qθ
Z Q XZ
θ
Y X Plan
QYZ
The triangular element formed from
θ the rectangular element by cutting
into half. By varying the angle θ,
Q XZ QYZ the maximum shear on the
hypotenuse is obtained.

Derivation
A rectangular element extracted
from a plate structure showing For vertical equilibrium, Qθ can be expressed as
shear per unit width in the X and Qθ × 1 = Q XZ sin θ + QYZ cos θ
Y directions For Qθ to be maximum set
dQθ Q
= 0 ⇒ Q XZ cos θ − QYZ sin θ = 0 ⇒ tan θ = XZ
dθ QYZ
Q max Q XZ 2 QYZ 2
⇒ Qθ = Qmax = +
θ X Q XZ 2 + QYZ 2 Q XZ 2 + QYZ 2

= Q XZ 2 + QYZ 2

Q max
Potential shear failure at the
orientation θ = tan −1 (QXZ / QYZ )

Figure D-4 – Diagrammatic illustration of shear “stresses” in the X and Y faces
of an element in Plate bending structure, potential shear failure and Derivation of
the magnitude and orientation of the design shear stress

Following the usual practice of designing against shear in accordance with the Code,
if the Qmax does not exceed allowable shear strength of concrete based on vc (the

D-6

Appendix D

design concrete shear stress with enhancement as appropriate) no shear
reinforcements will be required. Otherwise, reinforcements will be required to cater
for the difference.

The “stress” approach for shear design based on Qmax can best be carried out by
graphical method, so as to avoid handling the large quantity of data obtainable in the
finite element analysis. An illustration of the method for a raft footing is indicated in
Figure D-5 as :
(i) an enveloped shear stress (shear force per unit width) contour map of a structure
due to applied loads is first plotted as shown in Figure D-5(a);
(ii) the concrete shear strength contour of the structure which is a contour map
indicating the shear strength of the concrete structure after enhancement of the
design concrete shear stress (vc) to closeness of supports in accordance with
established code requirements (say BS8110) is plotted as shown in Figure
D-5(b);
(iii) locations where the stresses exceed the enhanced strengths be reinforced by
shear links as appropriate in accordance with established code requirements as
shown in Figure D-5(c).

Figure D-5a – Stress contour of enveloped shear “stresses” of a raft footing due to
applied load

D-7

Appendix D

Figure D-5b – Strength contour of the raft footing with enhancement

Figure D-5c – Arrangement of shear reinforcements

D-8

Appendix E

Moment Coefficients for three
side supported Slabs

Bending Coefficients in the plate of the indicated support conditions and length
breadth ratio are interpolated from Table 1.38 of “Tables for the Analysis of
Plates, Slabs and Diaphragms based on Elastic Theory”

b
=2
a
0.0933 0.1063 0.1104

0.0842 0.0956 0.0993

0.0721 0.0814 0.0844

Simply supported
Simply supported

b
0.0516 0.0574 0.0594

0.0196 0.0207 0.0211

Fixed support

a

Coefficients for bending along X direction (+ve : sagging; –ve : hogging)
Moments at various points is c. f . × q × a 2 where q is the u.d.l.

E-1

b
=2
a
0 0 0

0.0116 0.0132 0.0138

0.0158 0.0180 0.0188

Simply supported
Simply supported

b
0.0136 0.0155 0.0161

-0.0020 -0.0029 -0.0032

-0.0607 -0.0698 -0.0729

Fixed support

a

Coefficients for bending along Y direction (+ve : sagging; –ve : hogging)
Moments at various points is c. f . × q × b 2 where q is the u.d.l.

E-2

Appendix F

Derivation of Design Formulae
for Rectangular Columns to
Rigorous Stress Strain Curve of
Concrete

Appendix F
Derivation of Design Formulae for Rectangular Columns to Rigorous
Stress Strain Curve of Concrete

(I) Computing stress / force contribution of concrete stress block

Assuming the parabolic portion of the concrete stress block as indicated in Fig. 3.8 of
HKCoP2004 be represented by the equation
σ = Aε 2 + Bε (where A and B are constants) (Eqn F-1)
dσ
So = 2 Aε + B (Eqn F-2)
dε
dσ
As = Ec ⇒ B = Ec where Ec is the tangential Young’s Modulus of
dε ε =0
concrete listed in Table 3.2 of the Code.

dσ B E
Also = 0 ⇒ 2 Aε 0 + B = 0 ⇒ A = − =− c (Eqn F-3)
dε ε =ε 0 2ε 0 2ε 0

f cu
As σ = 0.67 when ε = ε 0
γm
f cu Ec Ec 0.67 f cu Ec 1.34 f cu
∴ 0.67 − =− ⇒ = ⇒ ε0 = (Eqn (F-4)
γ mε 0 2
ε0 2ε 0 γ mε 0 2 Ec γ m
(accords with 3.14 of the Concrete Code Handbook)
E 1.34 f cu
A=− c where ε 0 =
2ε 0 Ec γ m
Ec 2
So the equation of the parabola is σ =− ε + Ecε for ε ≤ ε 0
2ε 0
Consider the linear strain distribution

xε 0 / ε ult
ε ult = 0.0035
ε = ε0

x
u

h

Figure F-1 – Strain diagram across concrete section

F-1

Appendix F
u
At distance u from the neutral axis, ε = ε ult
x
ε0
So stress at u from the neutral axis up to x is
ε ult
2
Eε Eε
2
E E  u  u
σ = − c ε 2 + Ecε = − c  ε ult  + Ec  ε ult  = − c ult2 u 2 + c ult u (Eqn F-5)
2ε 0 2ε 0  x  x 2ε 0 x x
Based on (Eqn F-5), the stress strain profiles for grade 35 within the concrete
compression section are plotted in Figure F2 for illustration.


18
16
14
12
Stress (MPa)

10
8
6
4 0.3769 where
ε0 = 0.001319
2
0
0 0.2 0.4 0.6 0.8 1

Figure F-2 – Stress strain profile of grades 35

By the properties of parabola as shown in Figure F-3, we can formulate total force
offered by the parabolic section as Fc1 given by

centre of mass

b
2
Area = ab
3

a

3
a
8

Figure F-3 – Geometrical Properties of Parabola

F-2

Appendix F
2 ε0 f 1.34ε 0 f cu
Fc1 = b x0.67 cu = bx (Eqn F-6)
3 ε ult γm 3γ mε ult
and the moment exerted by Fc1 about centre line of the whole section

h  ε  3 ε0  h  5 ε0 
M c1 = Fc1  − x1 − 0
 ε − x
 8 ε  = Fc1  − x1 −
 
 (Eqn F-7)
2  ult  ult   2  8 ε ult 

The force by the straight portion is

0.67 f cu  ε  0.67 f cu bx  ε 
Fc 2 = x− x 0
 b =
 1 − 0
 ε 
 (Eqn F-8)
γm  ε ult  γm  ult 

The moment offered by the constant part about the centre line of the whole section is

h  ε  x
M c 2 = Fc 2  − 1 − 0
 ε  
2 (Eqn F-9)
2  ult  

Thus if full section of concrete in compression exists in the column section
Fc Fc1 Fc 2 0.67 f cu  1 ε0 x
= + = 1 −
 3ε 
h (Eqn F-10)
bh bh bh γm  ult
M c1 1.34ε 0 f cu  h  5 ε 0  1 1.34ε 0 f cu  x   1  x  5 ε 0 
= bx  − x1 −
  2 =
    −  1 − 
bh 2 3γ m ε ult  2  8 ε ult  bh
  3γ m ε ult  h   2  h  8 ε ult 
  
M c 2 0.67 f cu bx  ε  h  ε  x 1 0.67 f cu  x  ε 0   ε  x
= 1 − 0   − 1 − 0   2 =
 ε  2  ε  2 bh  1 −  1 − 1 − 0  
bh 2 γm  ult  
  ult  
 2γ m  h  ε ult    ε ult  h 
    
M c M c1 + M c 2 1.34ε 0 f cu  x   1  x  5 ε 0  0.67 f cu  x  ε   ε  x 
= =    −  1 −  8 ε  + 2γ
  1 − 0  1 − 1 − 0  
 ε   ε h
bh 2
bh 2
3γ mε ult  h   2  h  ult   m  h  ult    ult  

0.67 f cu  x  1 1 ε 0  1 1 ε 0  x
2
 1ε  
=   − + − + −  0    (Eqn F-11)
γ m  h  2 6 ε ult  2 3 ε ult 12  ε ult


  h
   

(II) Derivation of Basic Design Formulae of R.C. column sections

Cases 1 to 7 with different stress / strain profile of concrete and steel across the
x
column section due to the differences in the neutral axis depth ratios, , are
h
investigated. The section is reinforced by continuous reinforcements Ash along its
length h idealized as continuum and reinforcements at its end faces Asb with cover
d'.

Pursuant to the derivation of the stress strain relationship of concrete and steel, the

F-3

Appendix F
stress strain diagram of concrete and steel for Cases 1 to 7 are as follows, under the
definition of symbols as :

b: width of the column
h: length of the column
x: neutral axis depth of the column
Asb : total steel area at the end faces of the column
d' : concrete cover to the centre of the end face steel
Ash : total steel area along the length of the column

Case 1 (a) – where (i) x/h < 7/3(d’/h) for d’/h ≤ 3/14; and (ii) x/h < 7/11(1 – d’/h) for
d’/h > 3/14

Pursuant to the derivation of the stress strain relationship of concrete and steel, the
stress strain diagram of concrete and steel for Case 1(a) is as indicated in Figure
F-1(a) :

It should be noted that Fsc1 is in elastic whilst Fst1 is in plastic range as d’/h < 3/14

Steel compressive force in the portion steel elastic zone by Asb is
 x − d' 7  d'
Fsc1 =   × 0.87 f y × 0.5 Asb = 1 −  × 0.87 f y × 0.5 Asb (Eqn F-12)
 4x / 7  4 x
Steel compressive force in the portion steel plastic zone by Ash is
A  3x  3 x
Fsc 2 = 0.87 f y × sh   = 0.87 f y × Ash   (Eqn F-13)
h  7  7 h
Steel compressive force in the portion steel elastic zone by Ash is
A  4x  1 2 x
Fsc 3 = 0.87 f y × sh   × = 0.87 f y × Ash   (Eqn F-14)
h  7  2 7 h
Steel tensile force in the portion steel plastic zone by Asb is
Fst1 = 0.87 f y × 0.5 Asb (Eqn F-15)
Steel tensile force in the portion steel plastic zone by Ash is
A  11x   11 x 
Fst 2 = 0.87 f y × sh  h −  = 0.87 f y × Ash 1 −  (Eqn F-16)
h  7   7 h
Steel tensile force in the portion steel elastic zone by Ash is
Ash  4 x  1 2 x
Fst 3 = 0.87 f y ×   × = 0.87 f y × Ash   (Eqn F-17)
h  7  2 7 h
To balance the external load N u
Fc1 + Fc 2 + Fsc1 + Fsc 2 + Fsc 3 − Fst1 − Fst 2 − Fst 3 = N u
⇒ Fc1 + Fc 2 + Fsc1 + Fsc 2 − Fst1 − Fst 2 = N u

F-4

Appendix F
0.67 f cu  ε x
 3 − 0  + 0.87 f y  2 − 1 sh +  −
N x A 3 7 d' h  A
⇒ u =  ε h    0.87 f y sb
bh 3γ m  ult   h  bh  8 8 h x  bh
(Eqn F-18)

d'

xε 0 / ε ult εs = 0.002

d'
ε = ε0 ε ult = 0.0035

4x/7 4x/7 3x/7

x
u

h

Strain diagram across whole section

Fc1 Fc2 0.67 f cu
γm

Concrete stress Block

Fsc1

h–11x/7 4x/7
Fsc2 0.87fy
Fsc3

Fst3
Fst2 4x/7 3x/7

Steel stress Block
Fst1

Figure F-1(a) – Concrete and steel stress strain relation for Case 1(a)

Re-arranging (F-18)
 0.67 f cu  1 ε 0  Ash  x   N u
2
 Ash 3 Asb  x
 1 −
 3 ε  + 2 × 0.87 f y bh  h  −  bh + 0.87 f y  bh − 8 bh  h

 γm 
 ult    
  
7 A d'
− 0.87 f y sb = 0 (Eqn F-19)
8 bh h

F-5

Appendix F
x
(Eqn F-19) can be used for solve for
h

To balance the external load M u
M c1 + M c 2 + M sc1 + M sc 2 + M sc 3 + M st1 + M st 2 + M st 3 = M u
h   h 3x   h 3x 4 x  h 
M c1 + M c 2 + Fsc1  − d '  + Fsc 2  −  + Fsc 3  − −  + Fst1  − d ' 
2   2 14   2 7 21  2 
 h 1  11x   h  11x  4 x 
+ Fst 2  −  h −  + Fst 3  −  h − − = Mu
2 2  7  2  7  21  
h   h 3x   h 13x  h 
⇔ M u = M c1 + M c 2 + Fsc1  − d '  + Fsc 2  −  + Fsc 3  −  + Fst1  − d ' 
2   2 14   2 21  2 
 11x   29 x h 
+ Fst 2   + Fst 3  −  (Eqn F-20)
 14   21 2 
where
M c1 1.34ε 0 f cu  h  5 ε 0  1 1.34ε 0 f cu  x   1  x  5 ε 0 
= bx  − x1 − 8 ε  bh 2 = 3γ ε
    −  1 − 

bh 2
3γ m ε ult 2 
 ult  
 m ult  h   2  h  8 ε ult 
 
(Eqn F-21)
M c 2 0.67 f cu bx  ε  h  ε  x 1 0.67 f cu  x  ε   ε  x
= 1 − 0   − 1 − 0   2 =
 ε  2    1 − 0  1 − 1 − 0  
bh 2 γm  ult  
  ε ult  2  bh
 2γ m  h  ε ult    ε ult  h 
    
(Eqn F-22)
M sc1 7  d '  Asb  h  7  d' h  Asb  1 d ' 
2
= 1 −  × 0.87 f y × 0.5 2  − d '  = 1 −  × 0.87 f y × 0.5  − 
bh 4 x bh  2  4 h x bh  2 h 
(Eqn F-23)
M sc 2  3 x  h 3x  1 Ash  3 x  1 3 x 
2
= 0.87 f y × Ash   −  2 = 0.87 f y   −  (Eqn F-24)
bh  7 h  2 14  bh bh  7 h  2 14 h 
M sc 3  2 x  h 13x  1 Ash  2 x  1 13 x 
2
= 0.87 f y × Ash   −  2 = 0.87 f y   −  (Eqn F-25)
bh  7 h  2 21  bh bh  7 h  2 21 h 
M st1 0.87 f y × 0.5 Asb  h  Asb  1 d ' 
=  − d '  = 0.87 f y × 0.5  −  (Eqn F-26)
bh 2 bh 2 2  bh  2 h 
M st 2 A  11 x  11x  Ash  11 x  11 x 
2
= 0.87 f y × sh 1 −
2
  = 0.87 f y × 1 −   (Eqn F-27)
bh bh  7 h  14  bh  7 h  14 h 
M st 3  2 x  29 x h  1 A  2 x  29 x 1 
= 0.87 f y × Ash   −  2 = 0.87 f y × sh   − 
bh 2  7 h  21 2  bh bh  7 h  21 h 2 
(Eqn F-28)
Summing up
M c1 + M c 2 0.67 f cu  x  1 1 ε 0  1 1 ε 0 1  ε0   x
2
 
=   − + − + −   ε   h
 (Eqn F-29)
bh 2
γ m  h  2 6 ε ult  2 3 ε ult 12  ult   
   
M sc1 + M st1 7  d ' h  A  1 d' A  1 d'
2
= 1 −  × 0.87 f y × 0.5 sb  −  + 0.87 f y × 0.5 sb  − 
bh 4 h x bh  2 h  bh  2 h 

F-6

Appendix F
Asb  1 d '  11 7 d ' h 
= 0.87 f y  −  − 
bh  2 h  8 8 h x 
(Eqn F-30)
M sc 2 + M st 2 A  3 x  1 3 x  Ash  11 x  11 x 
2
= 0.87 f y sh   −  + 0.87 f y × 1 −  
bh bh  7 h  2 14 h  bh  7 h  14 h 
Ash  x  65  x  
2

= 0.87 f y   −    (Eqn F-31)
bh  h  49  h  
 
M sc 3 + M st 3 A  2 x  1 13 x  Ash  2 x  29 x 1 
2
= 0.87 f y sh   −  + 0.87 f y ×   − 
bh bh  7 h  2 21 h  bh  7 h  21 h 2 
2
Ash 32  x 
= 0.87 f y   (Eqn F-32)
bh 147  h 
Ms  Asb  1 d '  11 7 d ' h  Ash  x  163  x  2  
 
Total 2
= 0.87 f y   −  − +   −     (Eqn F-33)
bh  bh  2 h  8 8 h x  bh  h  147  h   
  
0.67 f cu  x  1 1 ε 0  1 1 ε 0 1  ε0   x
2
Mu  
∴ 2 =   − + − + −   ε   h

bh γ m  h  2 6 ε ult  2 3 ε ult 12  ult   
   
 A  1 d '  11 7 d ' h  A  x  163  x   

2

+ 0.87 f y  sb  −  −  + sh   −    (Eqn F-34)
 bh  2 h  8 8 h x  bh  h  147  h   
  

Case 1 (b) – 7/11(1 – d’/h) ≤ x/h < 7/3(d’/h) where d’/h > 3/14

Case 1(b) is similar to Case 1(a) except that both Fsc1 and Fst1 are in the elastic range
as d’/h > 3/14.

Re Figure F-1(b), the various components of stresses in concrete and in steel are
identical to that of Case 1(b) except that by Fst1 , the stress of which is
h − x − d'
0.87 f y
4x / 7
F h − x − d' A 7 h d ' h  Asb
So the st1 = − 0.87 f y × 0.5 sb = − × 0.87 f y  − 1 − 
bh 4x / 7 bh 8 x h x  bh
Fst1 Fsc1 7  h A
+ = × 0.87 f y  2 −  sb
bh bh 8  x  bh
N u 0.67 f cu  ε x A Asb
 3 − 0  + 0.87 f y  2 − 1 sh +  −
x 7 7 h
∴ =  ε     0.87 f y (Eqn F-35)
bh 3γ m  ult  h  h  bh  4 8 x  bh
 0.67 f cu  1 ε0  A  x
2
  7 Asb Ash  N u  x
⇒ 1 −
 3ε   + 0.87 f y × 2 sh   + 0.87 f y 
  − − 
 γm
  ult  bh  h  
  4 bh bh  bh  h
7 Asb
− 0.87 × =0 (Eqn F-36)
8 bh

F-7

Appendix F
M sc1 + M st1 7  d ' h  Asb  1 d '  7 h d ' h  Asb  1 d ' 
2
= 1 −  × 0.87 f y  −  + 0.87 f y ×  − 1 −   − 
bh 8 h x bh  2 h  8 x h x  bh  2 h 
7 A  1 d '  d'  h
= 0.87 f y × sb  − 1 − 2  (Eqn F-37)
8 bh  2 h  hx
M u 0.67 f cu  x  1 1 ε 0   1 ε 0 1 1  ε 0   x 
2
 
∴ 2 =   −
 2 6 ε  +  3 ε − 2 − 12  ε   h 
    
bh γ m  h  ult   ult  ult   
   
 7 Asb  1 d ' 
 d '  h A  x  163  x   
2

+ 0.87 f y   − 1 − 2  + sh   −    (Eqn F-38)
 8 bh  2 h 
 h  x bh  h  147  h   
 

d'

xε 0 / ε ult εs = 0.002

d' ε = ε0
ε ult = 0.0035

4x/7 4x/7 3x/7

x
u

h


Fc1 Fc2 0.67 f cu
γm


Fsc1
h–11x/7 4x/7

Fsc2 0.87fy
Fsc3

Fst3
Fst2 4x/7 3x/7

Fst1
Steel stress Block

Figure F-1(b) – Concrete and steel stress strain relation for Case 1(b)

F-8

Appendix F
Case 2 – 7/3(d’/h) ≤ x/h < 7/11(1 – d’/h)

There are two sub-cases to be considered in Case 2,
d' 3 d' 3
i.e. Case 2(a) – ≥ and Case 2(b) – <
h 14 h 14
d' 3 7 d' 1 7  d' 1
For Case 2(a), where ≥ . However, ≥ and 1 −  < . So this case
h 14 3 h 2 11  h 2
d' 3
doesn’t exist. For Case 2(b), where < both Asc1 and Ast1 are in the plastic
h 14
zone as shown in Figure F-2.

d'
εs = 0.002
xε 0 / ε ult
d' ε ult = 0.0035
ε = ε0

4x/7

x
u

h


Fc1 Fc2 0.67 f cu
γm

Fsc1

h–11x/7 4x/7
0.87fy
Fsc3 Fsc2

Fst3
Fst2
4x/7 3x/7

Fst1
Steel stress Block

Figure F-2 – Concrete and steel stress strain relation for Case 2(b)

F-9

Appendix F
The various components of stresses in concrete and steel are identical to that of Case
2(a) except that of Asc1 where
 1 d'
Fsc1 = 0.87 f y × 0.5 Asb and M sc1 = 0.87 f y × 0.5 Asb  − 
2 h 
It can be seen that Fst1 and Fsc1 are identical but opposite in direction, so cancel out.
By formulation similar to the above,
 1 ε0  x
N u 0.67 f cu  x  Ash
= 1 −
 3 ε  h + 0.87 f y  2 h − 1 bh
 (Eqn F-39)
bh γm  ult   
Nu A
+ 0.87 f y sh
x bh bh
⇒ = (Eqn F-40)
h  0.67 f cu  1 ε 0  A 
 1 −
  + 2 × 0.87 f y sh 

 γ m  3 ε ult  bh 

M u 0.67 f cu  x  1 1 ε 0  1 1 ε 0  x
2
 1ε  
=   − + − + −  0   
bh 2
γ m  h  2 6 ε ult  2 3 ε ult 12  ε ult


  h
   
 A  1 d '  A  x  163  x  2  
 
+ 0.87 f y  sb  −  + sh   −    (Eqn F-41)
 bh  2 h  bh  h  147  h   
  

Case 3 – where 7/3(d’/h) ≤ x/h < 7/11 for d’/h > 3/14 and
7/11(1 – d’/h) ≤ x/h < 7/11 for d’/h < 3/14

The concrete / steel stress / strain diagram is worked out as indicated in Figure F-3 :

The components of stresses are identical to Case 2 except that Fst1 become elastic
which is
0.87 f y (h − x − d ') (h − x − d ') 7 A = 0.87 f  h − d ' − 1 7 A
Fst1 = × 0.5 Asb = 0.87 f y sb y  sb
4x / 7 x 8  x x 8
F  h d ' h  7 Asb
⇒ st1 = 0.87 f y  − − 1
bh  x h x  8 bh
F F A  h d ' h  7 Asb A  11 7 h 7 d ' 
∴ sc1 + st1 = 0.87 f y × 0.5 sb − 0.87 f y  − − 1 = 0.87 f y sb  − + 
bh bh bh  x h x  8 bh bh  8 8 x 8 x 
(Eqn F-42)
N 0.67 f cu  1 ε 0  x Asb  11 7 h 7 d '   x  Ash
∴ u = 1 −
 3 ε  h + 0.87 f y bh  8 − 8 x + 8 x  + 0.87 f y  2 h − 1 bh

bh γm  ult     
(Eqn F-43)
Re-arranging (Eqn F-43)
 0.67 f cu  1 ε 0  A  x
2
  A 11 Ash  N u  x 
 1 −
  + 2 × 0.87 f y sh   + 0.87 f y  sb
   − −  
 γ m  3 ε ult 
 bh  h  
  bh 8 bh  bh  h 

F-10

Appendix F
7 Asb  d ' 
+ 0.87 f y  − 1 = 0 (Eqn F-44)
8 bh  h 

d'
εs = 0.002
xε 0 / ε ult
ε ult = 0.0035
d'
ε = ε0

4x/7

x
u

h

Strain diagram across concrete section

Fc1 Fc2 0.67 f cu
γm

Fsc1

h–11x/7 4x/7
Fsc2 0.87fy
Fsc3

Fst3
Fst2
4x/7 3x/7

Fst1
Steel stress Block

Figure F-3 – Concrete and steel stress strain relation for Case 3

To balance the external moment M u , all components are identical to Case 2 except
that by Fst1 which is
M st1 7  h d '  h  1 A 7  h d '  1 d ' 
2
= 0.87 f y Asb  − − 1 − d '  2 = 0.87 f y sb  − − 1 − 
bh 8  x x  2  bh bh 8  x x  2 h 
(Eqn F-45)

F-11

Appendix F
0.67 f cu  x  1 1 ε 0   1 ε 0 1 1  ε 0   x 
2
Mu  
∴ =   − 
 2 6 ε  +  3 ε − 2 − 12  ε   h 
   
bh 2
γ m  h  ult   ult  ult   
   
 1 d '  7 h 7 d ' h 3  Asb  x  163  x  2  Ash 
 
+ 0.87 f y  −  − −  +   −     (Eqn F-46)
 2 h  8 x 8 h x 8  bh  h  147  h   bh 
   
A A x
So, by pre-determining the steel ratios for sb and sh , we can solve for by
bh bh h
(Eqn F-44) under the applied load N u . The moment of resistance M u can then be
obtained by (Eqn F-46). The section is adequate if M u is greater than the applied
moment.

Case 4 – where x ≤ h < 11x/7, i.e. 7/11 ≤ x/h < 1

The concrete / steel stress / strain diagram is worked out as in Figure 3-4. The stress
components are identical to Case 3 except that Fst2 vanishes and Fst3 reduces as
indicated in Figure F-4 :
Steel tensile force in the portion steel elastic zone by Ash is

(h − x )× 1 h − x = 0.87 f y × 7 (h − x ) Ash = 0.87 f y × 7  h + x − 2  Ash
2
Ash
Fst 3 = 0.87 f y ×  
h 2 4x / 7 8 hx 8x h 
(Eqn F-47)
To balance the external load N u
N u 0.67 f cu bx  1 ε 0  A 11 7 h 7 d ' 
⇒ = 1 −
 3ε  + 0.87 f y sb  −
  + 
bh γm  ult  bh  8 8 x 8 x 
5x 7h x 
+ 0.87 f y × Ash − 0.87 f y ×  + − 2  Ash
7h 8x h 
N u 0.67 f cu  1 ε0 x A 11 7 h 7 d '  Ash  9 x 7 h 7 
⇒ = 
1 − 3 ε  + 0.87 f y sb  −
h  +  + 0.87 f y − − + 
bh γm  
ult bh  8 8 x 8 x  bh  56 h 8 x 4 
(Eqn F-48)
 0.67 f cu  ε  9 A  x
2
  A 11 Ash 7  N u  x 
⇔  3 − 0  − × 0.87 f y sh   + 0.87 f y  sb
 + −  
 3γ m   ε ult  56
 bh  h    bh 8 bh 4  bh  h 
7  A  d'  A 
+ 0.87 f y  sb  − 1 − sh  = 0 (Eqn F-49)
8  bh  h  bh 
To balance the external load M u about the centre of the column section
M u M c M sc1 + M st1 M sc 2 M sc 3 M st 3
= + + + +
bh 2 bh 2 bh 2 bh 2 bh 2 bh 2
M M M +M F 2  h 3 x  F 3  h 3x 4 x  F 3  h  h − x 
⇒ u = c + sc1 2 st1 + sc2  −  + sc2  − −  + st 2  − 
2 2

bh bh bh bh  2 14  bh  2 7 21  bh  2  3 
(Eqn F-50)

F-12

Appendix F
M M M
Total Moment 2
= c + s , i.e.
2
bh bh bh 2
  x 
 x  1 1 ε 0   1 ε 0 1 1  ε 0
2
Mu 0.67 f cu  
=   − + − −    
bh 2 γm  h  2 6 ε ult   3 ε ult 2 12  ε ult
   

  h 

 
 1 d '   7  h  7  d '  h  3  A
 7 h 9 x 9  x   Ash 
2

+ 0.87 f y  −     −    −  sb +  − +    
 2 h   8  x  8  h  x  8  bh  48 x 112 h 392  h   bh 
   
(Eqn F-51)

d'
εs = 0.002
xε 0 / ε ult
d' ε ult = 0.0035
ε = ε0
0.0035(h − x )
ε1 = 4x/7
x

x
u

h


Fc1 Fc2 0.67 f cu
γm


h–x
Fsc2 0.87fy
Fsc3
Fst3

4x/7 3x/7

Steel stress Block

Figure 3-4 – Concrete and steel stress strain relation for Case 4

F-13

Appendix F
Case 5 – where x>h>(1 – ε0/εult)x, i.e 1 ≤ x/h < 1/(1 – ε0/εult)

The concrete / steel stress / strain diagram is worked out as follows. It should be noted
that the neutral axis depth ratio is greater than unity and hence becomes a hypothetical
concept :

d'
εs = 0.002
xε 0 / ε ult
ε ult = 0.0035
ε = ε0 d'

0.0035(h − x )
ε1 = 4x/7
x

x
u

h


Fc1 Fc2 0.67 f cu
γm


Fsc2 0.87fy
Fsc3

4x/7 3x/7

Steel stress Block

Figure F-5 – Concrete and steel stress strain relation for Case 5

Concrete compressive stresses and forces
xε 0 / ε ult
Eε Eε
2

Fc1 = ∫ σ bdu where σ = − c ult2 u 2 + c ult u (Eqn F-52)
x−h
2ε 0 x x

F-14

Appendix F
xε 0 / ε ult xε 0 / ε ult xε 0 / ε ult
 Ec ε ult 2 Ec ε ult 
2
Eε b
2
Eε b
Fc1 = ∫h − 2ε 0 x 2 u + x u bdu = − 2cε 0ult 2 x∫hu du + c xult ∫ u du
2

x−   x − x−h

Fc1  E c ε ult  ε 0  E c ε ult  ε 0  x  Eε 
2 2 3 2

⇒ =  −
− 1   +  E c ε ult − c ult 
− 1
bh  2  ε ult 2
   6ε 0  ε ult 3
 

 h  2ε 0  
 Ecε ult 2 Ecε ult  h Ecε ult 2  h  2
+ 
 2ε − 2  x − 6ε  x  (Eqn F-53)
 0  0  
0.67 f cu bx  ε  F 0.67 f cu  ε x
Fc 2 = 1 − 0  ⇔ c 2 =
 ε  1 − 0  (Eqn F-54)
γm  ult  bh γ m  ε ult  h
 
Fc Fc1 Fc 2  E c ε ult  ε 0  Eε 2ε 3  x  Eε 
2 2

= + =  − 1 − c ult  0 3 − 1 +  E c ε ult − c ult 
bh bh bh  2  ε ult 2  6ε 0  ε ult  h  2ε 0 
      
 E c ε ult 2 E c ε ult  h E c ε ult 2  h  2 0.67 f cu  ε x
+ −  −   + 1 − 0 
 ε h
 2ε 2 x 6ε 0  x  γm 
 0  ult 

Fc  E c ε ult  ε 0  Eε 2ε 3  Eε  ε  x  Eε 
2 2

=  − 1 − c ult  0 3 − 1 + c 0 1 − 0  +  E c ε ult − c ult 
bh  2  ε ult 2
 



6ε 0  ε ult 
 2  ε ult  h 
   2ε 0 

Eε 2 Eε  h Ecε ult 2  h  2
+  c ult − c ult
 2ε
 −
x   (Eqn F-55)
 0 2  6ε 0  x 

Steel compressive force in the portion steel plastic zone by Asb is
F A
Fsc1 = 0.87 f y × 0.5 Asb ⇔ sc1 = 0.87 f y × 0.5 sb (Eqn F-56)
bh bh
0.87 f y ( x − h + d ') (x − h + d ') 7 A = 0.87 f 1 − h + d '  7 A
Fsc1' = × 0.5 Asb = 0.87 f y sb y  sb
4x / 7 x 8  x x 8
Fsc1'  h d '  7 Asb  h d ' h  7 Asb  h  d '  7 Asb
⇒ = 0.87 f y 1 − +  = 0.87 f y 1 − +  = 0.87 f y 1 − 1 − 
bh  x x  8 bh  x h x  8 bh  x h  8 bh

Fsc1 Fsc1' A  h  d '  7 Asb 11 7 h  d '  Asb
+ = 0.87 f y × 0.5 sb + 0.87 f y 1 − 1 −  = 0.87 f y  − 1 −  
bh bh bh  x h  8 bh 8 8 x h  bh
(Eqn F-57)

Steel compressive force in the portion steel plastic zone by Ash is
Ash  3 x  3 x F  3 x  Ash
Fsc 2 = 0.87 f y ×   = 0.87 f y × Ash   ⇔ sc 2 = 0.87 f y   (Eqn F-58)
h  7  7 h bh  7 h  bh
Steel compressive force in the portion steel elastic zone by Ash is
x − h Ash  3x   x − h  Ash  3x  1
Fsc 3 = 0.87 f y × ×  h −  + 0.87 f y 1 − × h − ×
4x / 7 h  7   4x / 7  h  7  2
F  7 7 h 33 x  Ash
⇒ sc 3 = 0.87 f y  − −  (Eqn F-59)
bh  4 8 x 56 h  bh

F-15

Appendix F
N F F F F F
As N u = Fc + Fsc1 + Fsc1' + Fsc 2 + Fsc 3 ⇔ u = c + sc1 + sc1' + sc 2 + sc 3
bh bh bh bh bh bh
Nu   1 ε 02 1 ε 0 1 1 ε ult  9 Ash  x
⇒ 
=  Ecε ult  − + − +  − 0.87 f y 
bh    6 ε ult
2
2 ε ult 2 6 ε 0  56 bh  h

 Ec ε ult
2
11 Asb 7 Ash 
+  Ec ε ult −
 + 0.87 f y + 0.87 f y 
 2ε 0 8 bh 4 bh 
 Ec ε ult 2 Ec ε ult 7 Asb  d'  7 Ash  h Ec ε ult
2
h
2

+ − + 0.87 f y  − 1 − 0.87 f y  −   (EqnF-60)
 2ε 0 2 8 bh h  8 bh  x 6ε 0  x

Re-arranging (Eqn F-60)
  1 ε 0 2 1 ε 0 1 1 ε ult  9 Ash  x 
3

 Ec ε ult  − 
 6 ε 2 + 2 ε − 2 + 6 ε  − 0.87 f y 56 bh  h 

  ult ult 0   

 Eε
2
11 Asb 7 Ash N u  x 
2

+  Ec ε ult − c ult + 0.87 f y + 0.87 f y −  
 2ε 0 8 bh 4 bh bh  h 

E ε 2
Eε A 7  d'  7 Ash  x E c ε ult
2

+  c ult − c ult + 0.87 f y sb  − 1 − 0.87 f y  − =0
 2ε 0
 2 bh 8  h  8 bh  h
 6ε 0
(Eqn F-61)
x
which is a cubic equation in .
h
Summing the Moments as follows :

Concrete compressive stresses and moments
xε 0 / ε ult
Ec ε ult 2 Ec ε ult
2
h 
M c1 = ∫ σ bdu  − x + u  where σ = − u + u
x−h 2  2ε 0 x 2 x
xε 0 / ε ult
 E c ε ult 2 2 E c ε ult   h 
M c1 = ∫h − 2ε 0 x 2 u + x u b 2 − x + u du
   
x−  
M c1  1  1 x   ε 2 x h  1  ε 0
3
 x  2 x h 
2
= E c ε ult   −   2 − 1 + 2 −  +  3 − 1  + 3 − 3 +
ε
0
h ε  h  
bh  2 2 h   ult
    x  3  ult
   h x 

E c ε ult  1  1 x   ε 0 x h  h   1  ε  x  2 h  h  
2 3 2 4 2
x
−   −   3 − 1 + 3 − 3 +    +  0 4 − 1  + 4 − 6 + 4 −    

2ε 0  3  2 h   ε ult
 h x  x   4  ε ult
 
 h 
 h x  x  


M c 2 0.67 f cu bx  ε  h  ε x 1 0.67 f cu  x  ε   ε  x
= 1 − 0
 ε   − 1 − 0
 2  ε   2 =
 2 bh  1 − 0  1 − 1 − 0
 ε   ε  
h
bh 2
γm  ult    
ult 2γ m  h  ult    ult  
M sc1 0.87 f y × 0.5 Asb  h  A  1 d'
2
= 2  − d '  = 0.87 f y × 0.5 sb  − 
bh bh 2  bh  2 h 
M sc1'  h d '  7 Asb  h 1 7  h d ' h  1 d '  Asb
2
= −0.87 f y 1 − +   − d '  = −0.87 f y 1 − +  − 
bh  x x  8 bh  2 h 8  x h x  2 h  bh

F-16

Appendix F
7   d '  h  1 d '  Asb
= −0.87 f y 1 +  − 1  − 
8   h  x  2 h  bh
 
M u M c1 M c 2 M sc1 M sc1' M sc 2 M sc 3
Total Moment = + + + + +
bh 2 bh 2 bh 2 bh 2 bh 2 bh 2 bh 2
Mu  1 ε 0 3 1 ε 0 2 1 ε 0 1 1 ε ult  x  2
⇒ 2 = E c ε ult  − + − + −  
bh  24 ε 3 6 ε 2 4 ε 6 24 ε 0  h 
 ult ult ult 
 1 ε 0 2 1 ε 0 1 1 ε ult  x  1 1 ε ult h 1 ε ult  h 
2

+ E c ε ult − + − + 
 + E c ε ult  −  + E c ε ult
x  
 12 ε ult

2
4 ε ult 4 12 ε 0  h
  12 12 ε 0  24 ε 0  x 
Asb  1 d '   7  d '  h 3  7 h 9 x 9  x   Ash
2

+ 0.87 f y ×  −  1 −  −  + 0.87 f y  − +   
bh  2 h   8 
 h  x 8  48 x 112 h 392  h   bh
 
(Eqn F-62)

Case 6 – where (1 – ε0/εult)x>h>3x/7, i.e. 1/(1 – ε0/εult) ≤ x/h < 7/3

Case 6 is similar to Case 5 except that Fc1 vanishes. The concrete / steel stress / strain
diagram is worked out as in Figure F-6 :

0.67 f cu
Referring to (Eqn F-55) by replacing Fc1 + Fc 2 by
γm
N u 0.67 f cu 11 7 h  d '  Asb  7 7 h 9 x  Ash
= + 0.87 f y  − 1 −   + 0.87 f y  − − 
bh γm  8 8 x h  bh  4 8 x 56 h  bh
(Eqn F-63)
2
9 Ash  x   N u 0.67 f cu  11 Asb 7 Ash  x 
Re-arranging ⇔ 0.87 f y   + − − 0.87 f y  +  
56 bh  h   bh γm  8 bh 4 bh  h 
7  d'  A 7 Ash 
− 0.87 f y   − 1 sb − =0 (Eqn F-64)
 8  h  bh 8 bh 
x
which is a quadratic equation in which can be solved
h

For Moment that can be provided by the section, similar to Case 5 except that
M c = 0 . So
Mu Asb  1 d '   7  d '  h 3  7 h 9 x 9  x   Ash
2

= 0.87 f y ×  −   1 −  −  + 0.87 f y  − +   
bh 2 bh  2 h   8  h  x 8  48 x 112 h 392  h   bh
 
(Eqn F-65)

F-17

Appendix F

d'
εs = 0.002

ε ult = 0.0035
d'

4x/7

x

h


Fc2 0.67 f cu
γm


Fsc2 0.87fy
Fsc3

4x/7 3x/7

Steel stress Block

Figure 3-6 – Concrete and steel stress strain relation for Case 6

Case 7 – where x/h ≥ 7/3

In this case, the concrete and steel in the entire column section are under ultimate
stress. The axial load will be simply
N u 0.67 f cu A A 
= + 0.87 f y  sb + sh  (Eqn F-66)
bh γm  bd bd 
and the moment is zero.
Mu
=0 (Eqn F-67)
bh 2

F-18

Appendix F

(III) Design formulae for 4-bar column sections for determination of
reinforcement ratios

b

h

It is the aim of the section of the Appendix to derive formulae for the determination of
Asb
against applied axial load and moment under a pre-determined sectional size. In
bh
A
the following derivations, sh are set to zero. The process involves :
bh
A
(i) For the 7 cases discussed in the foregoing, eliminate sb between equations
bh
Nu Mu A
obtained from balancing and 2
by making sb subject of formulae
bh bh bh
Nu
in the equation for balancing of substitute into the equation for balancing
bh
Mu x
of 2
. The equation obtained in a polynomial in which can be solved
bh h
by equations (if quadratic or cubic or even 4th power) or by numerical methods.
x
Solution in will be valid if the value arrived at agree with the
h
pre-determined range of the respective case;
x
(ii) Back substitute the accepted value of into the equation obtained by
h
Nu A
balancing to solve for sb .
bh bh

Case 1 (a) – where (i) x/h < 7/3(d’/h) for d’/h ≤ 3/14; and (ii) x/h < 7/11(1 – d’/h) for
d’/h > 3/14

Ash
Putting =0;
bh
N 0.67 f cu  1 ε0  x  3 7 d' h  Asb
(Eqn F-18) ⇒ u = 1 −
 3ε  + −
 h 8 8 h x 0.87 f y bh
bh γm  ult   

F-19

Appendix F
N u 0.67 f cu 1 ε0 x
− 1 −
 3ε 
h
Asb bh γm  ult 
⇒ 0.87 f y = (Eqn F-68)
bh  3 7 d' h 
 − 
8 8 h x

Substituting into (Eqn F-34)
M u 0.67 f cu  x  1 1 ε 0  1 1 ε 0 1  ε0   x
2
 
⇒ 2 =   − + − + −   ε   h

bh γ m  h  2 6 ε ult  2 3 ε ult 12  ult   
   
A  1 d '  11 7 d ' h 
+ 0.87 f y sb  −  − 
bh  2 h  8 8 h x 
0.67 f cu 3  1 1 ε 0 1  ε 0   x 
2 3

⇒ − + −    
γ m 8  2 3 ε ult 12  ε ult   h 
  

0.67 f cu  1 1 ε 0 29 d ' 3 d ' ε 0 7 d '  ε 0   x 
2 2

+ − + + − +    
γ m  2 6 ε ult 16 h 4 h ε ult 96 h  ε ult   h   

 N u 11  1 d '  3 M u 0.67 f cu 7  d '  2  1 ε 0  x
+  − − +    − 1
 bh 8  2 h  8 bh

2
γ m 8  h   3 ε ult  h
 
7 M u d ' N u 7  1 d '  d '
+ 2
−  −  =0 (Eqn F-69)
 8 bh h bh 8  2 h  h 

x A
Upon solving (Eqn F-69) for , back-substitution into (Eqn F-68) to calculate sb .
h bh

Case 1 (b) – 7/11(1 – d’/h) ≤ x/h < 7/3(d’/h) where d’/h > 3/14

Ash
Putting = 0 ; (Eqn F-35)
bh
N 0.67 f cu  1 ε 0  x  7 7 h  Asb
⇒ u = 1 −  + − 0.87 f y
bh γ m  3 ε ult  h  4 8 x 
  bh
N u 0.67 f cu  1 ε 0  x
− 1 − 
Asb bh γ m  3 ε ult  h
 
⇒ 0.87 f y = (Eqn F-70)
bh 7 7 h
 − 
4 8 x
A
Substituting into (Eqn F-38), again putting sh = 0 and simplifying,
bh
0.67 f cu 7  1 1 ε 0 1  ε 0   x  0.67 f cu  21 7 ε 0 7  ε 0   x 
2 3 2 2

⇒ − + −     +  − +    
γ m 4  2 3 ε ult 12  ε ult   h 
   γ m 16 12 ε ult 96  ε ult   h 
  
 
7  M u 0.67 f cu  1 1 ε 0   d '  d '   x 7  M u 1  d'  Nu 
2 2
 
−  2 +  −  1 − 2 + 2    +  2 + 1 − 2  =0
4  bh
 γ m  2 6 ε ult  
  h  h    h 8  bh
  2 h  bh  

F-20

Appendix F
(Eqn F-71)
x A
Upon solving (Eqn F-71) for , back-substitution into (Eqn F-70) to calculate sb .
h bh

Case 2 – 7/3(d’/h) ≤ x/h < 7/11 – 7/11(d’/h) and d’/h > 3/14

Ash
Using the equations summarized in Section 4 and setting = 0 in (Eqn F-39)
bh
x N u  0.67 f cu  1 ε 0 
= ÷ 1 −  (Eqn F-72)
h bh  γ m  3 ε ult 
  
x
Substituting obtained in (Eqn F-72), substituting into (Eqn F-41) and calculate
h
Asb
as
bh
Asb  M u 0.67 f cu  x   1  x  1 ε 0 1 ε 0  x  1  ε 0   x   
2
   1 d ' 
= 2 −    1 −  − +  −   ε   h   ÷ 0.87 f y  2 − h 
  
bh  bh γ m  h   2  h  6 ε ult 3 ε ult  h  12  ult     
  
(Eqn F-73)

Case 3 – where 7/3(d’/h) ≤ x/h < 7/11 for d’/h > 3/14 and
7/11(1 – d’/h) ≤ x/h < 7/11 for d’/h < 3/14
and Case 4 – where 7/11 ≤ x/h < 1

x Nu A
Using the equation relating and in (Eqn F-43) and setting sh = 0 .
h bh bh
N u 0.67 f cu  ε x
 3 − 0  + 0.87 f y sb  −
A 11 7 h 7 d ' 
=  h  + 
bh 3γ m  ε ult  bh  8 8 x 8 x 
N u 0.67 f cu  1 ε 0  x
− 1 − 
Asb bh γ m  3 ε ult  h
 
⇔ = (Eqn F-74)
bh  11 7 h 7 d ' h 
0.87 f y  − + 
 8 8 x 8 h x
A
Substituting sh = 0 , (Eqn F-46) and simplifying
bh
0.67 f cu 11  1 ε 0 1 1  ε 0   x 
2 3

⇔  − −    
γ m 8  3 ε ult 2 12  ε ult   h 
  

0.67 f cu  21 7 ε 0 13 d ' 7  ε 0  5 ε 0 d ' 7  ε 0  d '  x 
2 2 2

+  − − +   + −    
γ m 16 12 ε ult 16 h 96  ε ult  12 ε ult h 96  ε ult  h  h 
   
 
 0.67 f cu 7  d '  1 ε 0 d ' 1 d ' ε 0  3  1 d '  N u 11 M u  x 
+  − 11 −  − + −  −  − 2  
 γm 8 h  3 ε ult h 3 h ε ult  8  2 h  bh 8 bh  h 


F-21

Appendix F
7  1 d '  d '  N 7 M u  d' 
−  −  − 1 u − 2 
− 1 = 0 (Eqn F-75)
8  2 h  h  bh 8 bh  h 
 x
which is a cubic equation in  
h
 x 7  d' A
Upon solving   lying between 1 −  and 1, sb can be obtained by
h 11  h bh
back-substituting into (Eqn 5-7)
N u 0.67 f cu  ε0  x
Asb bh
−
3γ m 

3 − ε  h
ult 
= (Eqn F-76)
bh  11 7 h 7 d ' h 
0.87 f y  − + 
 8 8 x 8 h x

Case 5 – where 1 ≤ x/h < 1/(1 – ε0/εult)
Ash
Referring to Case 5 of Section 3 and setting = 0 in (Eqn F-60)
bd
Asb
Solving by
bd
N u  E c ε ult  ε 0  Eε 2  ε 03  Eε  ε  x  Eε 
2 2

=  − 1 − c ult  − 1 + c 0 1 − 0  +  E c ε ult − c ult 
bh  2  ε ult 2  6ε 0 ε 3  2   ε  h  2ε 0 
   ult  ult 
  
 E c ε ult 2 E c ε ult  h E c ε ult 2  h  2 11 7  d '  h  A
+ −  −   + 0.87 f y  −  − 1  sb
x
 2ε 0 2  6ε 0  x   8 8 h  x  bh
Asb  N u  1 1 ε 0 1 ε ult 1 ε 0  x  11 7  d '  h 
2

⇒ 0.87 f y = − − − + +  ÷ −  − 1
bh  bh  2 6 ε ult 2 6 ε 0 2 ε ult  h   8 8  h  x 
     
  Eε
2
 E ε 2 E ε  Eε
2 2

+ −  E c ε ult − c ult  −  c ult − c ult  h + c ult  h   ÷ 11 − 7  d ' − 1 h 
   
 
 

2ε 0   2ε 0 2 x  6ε 0  x    8 8  h  x 
  
(Eqn F-77)
A
Substituting into (Eqn F-62) and again setting sh = 0
bd
Mu  1  1 x   ε 2
 x h  1  ε 0
3
 x  2 x h 
= E c ε ult   −   2 − 1 + 2 −  +  3 − 1  + 3 − 3 +  
0
 h x  3  ε ult  h 
 2  2 h   ε ult
2
bh      h x 


E c ε ult  1  1 x   ε 0
2 
2 3
x 2
  ε 4  2

−   −   −1  + 3 − 3 h +  h   + 1   0 − 1 x  + 4 x − 6 + 4 h −  h  
     
2ε 0  3  2 h   ε ult 3  h

  x  x   4   ε ult 4
 
 h 
 h x  x  


0.67 f cu  x  ε 0   ε 0  x  A 1 d' 7 d ' h 3
+  1 −
 ε   1 − 1 −
 ε h   + 0.87 f y sb ×  −   1 −  − 
   
2γ m  h   
ult   ult   bh  2 h   8  h  x 8
x
to solve for . (Eqn F-78)
h

F-22

Appendix F
Asb
Back-substituting into (Eqn 5-10) to solve for
bd

Case 6 – where (1 – ε0/εult)x > h > 3x/7 i.e. 1/(1 – ε0/εult) ≤ x/h < 7/3(1–d’/h)

Ash
Referring to (Eqn F-63) of Case 6 of and setting =0
bd
N u 0.67 f cu 11 7 h  d '  Asb
= + 0.87 f y  − 1 −  
bh γm 8 8 x h  bh
Asb  N u 0.67 f cu  11 7 h  d ' 
⇒ 0.87 f y = − ÷ − 1 −  
bh  bh
 γm 
 8 8 x h 
Ash
Substituting into (Eqn F-64) of Case 6 and again setting =0
bd
M A  1 d '  7  d '  h 3 
2
= 0.87 f y sb ×  −   1 −  − 
bh bh  2 h   8  h  x 8
7  d '   N u 0.67 f cu  1 d '  M 
1 −    −  −  + 
x 8 h   bh γ m  2 h  bh 2 

⇒ =
h 11 M  N u 0.67 f cu  1 d '  3 
 + −  −  
 8 bh 2  bh
 γ m  2 h  8 

x A
With determined, calculate sb by
h bh
N u 0.67 f cu 11 7 h  d '  Asb
= + 0.87 f y  − 1 −  
bh γm 8 8 x h  bh
N u 0.67 f cu
−
Asb bh γm
⇒ = (Eqn F-79)
bh 11 7 h  d ' 
0.87 f y  − 1 −  
8 8 x h 

Case 7 – where x/h ≥ 7/3

N u 0.67 f cu A A N 0.67 f cu  1
= 
+ 0.87 f y sb ⇒ sb =  u − 
 0.87 f (Eqn F-80)
bh γm bh bh  bh γm  y

Mu
=0 (Eqn F-81)
bh 2

F-23

Appendix F – Summary of Design Charts
for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9
2 2
M/bh N/mm

Chart F - 1

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5
2 2
M/bh N/mm

Chart F - 2

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13
2 2
M/bh N/mm

Chart F - 3

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5
2 2
M/bh N/mm

Chart F - 4

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5
2 2
M/bh N/mm

Chart F - 5

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5
2 2
M/bh N/mm

Chart F - 6

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
2 2
M/bh N/mm

Chart F - 7

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5
2 2
M/bh N/mm

Chart F - 8

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15
2 2
M/bh N/mm

Chart F - 9

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5
2 2
M/bh N/mm

Chart F - 10

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5
2 2
M/bh N/mm

Chart F - 11

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12
2 2
M/bh N/mm

Chart F - 12

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5
2 2
M/bh N/mm

Chart F - 13

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15
2 2
M/bh N/mm

Chart F - 14

for Columns

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17
2 2
M/bh N/mm

Chart F - 15

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
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M/bh N/mm

Chart F - 16

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5
2 2
M/bh N/mm

Chart F - 17

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14
2 2
M/bh N/mm

Chart F - 18

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5
2 2
M/bh N/mm

Chart F - 19

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17
2 2
M/bh N/mm

Chart F - 20

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
2 2
M/bh N/mm

Chart F - 21

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5
2 2
M/bh N/mm

Chart F - 22

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14
2 2
M/bh N/mm

Chart F - 23

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16
2 2
M/bh N/mm

Chart F - 24

for Columns

55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5
2 2
M/bh N/mm

Chart F - 25

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5
2 2
M/bh N/mm

Chart F - 26

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13
2 2
M/bh N/mm

Chart F - 27

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5
2 2
M/bh N/mm

Chart F - 28

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16
2 2
M/bh N/mm

Chart F - 29

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5
2 2
M/bh N/mm

Chart F - 30

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5
2 2
M/bh N/mm

Chart F - 31

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13
2 2
M/bh N/mm

Chart F - 32

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5
2 2
M/bh N/mm

Chart F - 33

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2
N/bh N/mm

35

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5
2 2
M/bh N/mm

Chart F - 34

for Columns

60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45
5% steel
40 6% steel
7% steel
2
N/bh N/mm

35 8% steel

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18

2 2
M/bh N/mm

Chart F - 35

Rectangular Column R.C. Design to Code of Practice for Structural Use of Concrete 2004 - 4 bar column

Project :
Column Mark Floor

fcu = 35 N/mm2 fy = 460 N/mm2 Ec = 23700 N/mm2
b = 400 h = 500 b' = 330.00 h' = 430.00 cover= 50 bar size = 40
Basic Load Case Mx
Load Case No. 1 2 3 4 5 6
Load Case D.L. L.L. Wx Wy W45 W135
h
Axial Load P (kN) 2304.7 582.1 -362.17 -545.1 56.92 82.09 My
Moment Mx (kNm) 29.13 32.11 47.1 -75.12 98.1 8.93
Moment My (kNm) -31.33 16.09 2.15 44.2 76.99 35.21
b
N Mx My N/bh M/bh2 d/h / d/b x/h / y/h Steel Steel area
(kN) (kNm) (kNm) (N/mm2) (N/mm2) (%) (mm2)
Load Comb 1 1.4D+1.6L 4157.9 92.158 -18.118 Mx' = 99.411 20.79 0.9941 0.86 1.1701 1.9429 3885.7
Load Comb 2 1.2(D+L+Wx) 3029.6 130.01 -15.708 Mx' = 140.12 15.148 1.4012 0.86 0.9702 0.8 1600
Load Comb 3 1.2(D+L-Wx) 3898.8 16.968 -20.868 My' = 25.447 19.494 0.3181 0.825 1.3301 1.1905 2380.9
Load Comb 4 1.2(D+L+Wy) 2810 -16.656 34.752 My' = 41.507 14.05 0.5188 0.825 1.0602 0.8 1600
Load Comb 5 1.2(D+L-Wy) 4118.3 163.63 -71.328 Mx' = 192.82 20.591 1.9282 0.86 1.0405 2.4722 4944.4
Load Comb 6 1.2(D+L+W45) 3532.5 191.21 74.1 Mx' = 231.22 17.662 2.3122 0.86 0.9448 2.0502 4100.4
Load Comb 7 1.2(D+L-W45) 3395.9 -44.232 -110.68 My' = 125.49 16.979 1.5686 0.825 0.979 1.4098 2819.6
Load Comb 8 1.2(D+L+W135) 3562.7 84.204 23.964 Mx' = 96.983 17.813 0.9698 0.86 1.0901 1.1681 2336.1
Load Comb 9 1.2(D+L-W135) 3365.7 62.772 -60.54 My' = 81.79 16.828 1.0224 0.825 1.04 0.9815 1963
Load Comb 10 1.4(D+Wx) 2719.5 106.72 -40.852 Mx' = 135.67 13.598 1.3567 0.86 0.9324 0.8 1600
Load Comb 11 1.4(D-Wx) 3733.6 -25.158 -46.872 My' = 54.208 18.668 0.6776 0.825 1.15 1.226 2452.1
Load Comb 12 1.4(D+Wy) 2463.4 -64.386 18.018 Mx' = 78.184 12.317 0.7818 0.86 0.9619 0.8 1600
Load Comb 13 1.4(D-Wy) 3989.7 145.95 -105.74 Mx' = 192.25 19.949 1.9225 0.86 1.0306 2.2936 4587.2
Load Comb 14 1.4(D+W45) 3306.3 178.12 63.924 Mx' = 215.64 16.531 2.1564 0.86 0.9327 1.663 3326.1
Load Comb 15 1.4(D-W45) 3146.9 -96.558 -151.65 My' = 186.88 15.734 2.336 0.825 0.8772 1.6942 3388.3
Load Comb 16 1.4(D+W135) 3341.5 53.284 5.432 Mx' = 56.433 16.708 0.5643 0.86 1.1401 0.8 1600
Load Comb 17 1.4(D-W135) 3111.7 28.28 -93.156 My' = 103.6 15.558 1.2949 0.825 0.9803 0.8449 1689.7
Load Comb 18 1.0D+1.4Wx 1797.7 95.07 -28.32 Mx' = 120.97 8.9883 1.2097 0.86 0.7146 0.8 1600
Load Comb 19 1.0D-1.4Wx 2811.7 -36.81 -34.34 My' = 49.26 14.059 0.6158 0.825 1.0401 0.8 1600
Load Comb 20 1.0D+1.4Wy 1541.6 -76.038 30.55 Mx' = 105.72 7.7078 1.0572 0.86 0.5779 0.8 1600
Load Comb 21 1.0D-1.4Wy 3067.8 134.3 -93.21 Mx' = 193.56 15.339 1.9356 0.86 0.923 1.2156 2431.2
Load Comb 22 1.0D+1.4W45 2384.4 166.47 76.456 Mx' = 226.37 11.922 2.2637 0.86 0.7836 0.8 1600
Load Comb 23 1.0D-1.4W45 2225 -108.21 -139.12 My' = 191.32 11.125 2.3914 0.825 0.7278 0.8 1600
Load Comb 24 1.0D+1.4W135 2419.6 41.632 17.964 Mx' = 55.564 12.098 0.5556 0.86 0.9858 0.8 1600
Load Comb 25 1.0D-1.4W135 2189.8 16.628 -80.624 My' = 88.722 10.949 1.109 0.825 0.8718 0.8 1600
Steel required = 2.4722 4944.4

P versus Mx and My of the Column Section
P - Mx P - My Actual Loads Mx control Actual Loads My control
6000

5000

4000
P (kN)

3000

2000

1000

0
0 100 200 300 400 500 600
M (kNm)

Rectangular Column R.C. Design to Code of Practice for Structural Use of Concrete 2004

Project :
Column Floor

b = 1500 h = 2000 b' = 1285.17 h' = 1684.31 cover= 50

Steel provided : 15 Y 40 (Along each long sides h, excluding corner bars)
12 Y 40 (Along each short sides b, excluding corner bars)
4 Y 40 (Corner bars)

Total Steel Area = 72885 mm2 Steel Percentage = 2.43 % Max. Ultimate Load = 76069 kN
Area of Steel per mm length for the long sides bars (including corner bars) = 21.36 mm2/mm
Area of Steel along long sides (excluding corner bars) = 37699 mm2
Area of Steel per mm length for the short sides bars (including corner bars) = 23.46 mm2/mm
Area of Steel along short sides (excluding corner bars) = 30159 mm2

Basic Load Case
Load Case D.L. L.L. Wx Wy W45 W135
Axial Load P (kN) 37872 1101 -3628.1 -2611.1 -5692.3 8209.2
Moment Mx (kNm) -291.3 -37.11 470.81 -3700 -1750.3 4892.9
Moment My (kNm) -31.33 16.09 5.17 2700 2764 -3520.2

P Mx My
Load Comb 1 1.4D+1.6L 54782 -467.2 -18.118 Mx' = 476.55 Mux = 16452 Section OK
Load Comb 2 1.2(D+L+Wx) 42413 170.88 -12.084 Mx' = 179.2 Mux = 23474 Section OK
Load Comb 3 1.2(D+L-Wx) 51121 -959.06 -24.492 Mx' = 973.01 Mux = 18743 Section OK
Load Comb 4 1.2(D+L+Wy) 43634 -4834.1 3221.7 Mx' = 6999.6 Mux = 22861 Section OK
Load Comb 5 1.2(D+L-Wy) 49900 4045.9 -3258.3 My' = 4639 Muy = 14871 Section OK
Load Comb 6 1.2(D+L+W45) 39936 -2494.5 3298.5 My' = 4352.2 Muy = 18945 Section OK
Load Comb 7 1.2(D+L-W45) 53598 1706.3 -3335.1 My' = 3865.6 Muy = 13130 Section OK
Load Comb 8 1.2(D+L+W135) 56618 5477.4 -4242.6 My' = 5801.2 Muy = 11590 Section OK
Load Comb 9 1.2(D+L-W135) 36916 -6265.5 4206 Mx' = 9507.3 Mux = 26076 Section OK
Load Comb 10 1.4(D+Wx) 47941 251.31 -36.624 Mx' = 273.77 Mux = 20575 Section OK
Load Comb 11 1.4(D-Wx) 58099 -1067 -51.1 Mx' = 1090.8 Mux = 14185 Section OK
Load Comb 12 1.4(D+Wy) 49365 -5587.8 3736.1 Mx' = 7805.2 Mux = 19771 Section OK
Load Comb 13 1.4(D-Wy) 56676 4772.2 -3823.9 My' = 5179.4 Muy = 11560 Section OK
Load Comb 14 1.4(D+W45) 45051 -2858.3 3825.7 My' = 4911.9 Muy = 16951 Section OK
Load Comb 15 1.4(D-W45) 60989 2042.6 -3913.4 My' = 4416.8 Muy = 9164 Section OK
Load Comb 16 1.4(D+W135) 64513 6442.2 -4972.2 My' = 6446.8 Muy = 7132 Section OK
Load Comb 17 1.4(D-W135) 41527 -7257.9 4884.4 Mx' = 10685 Mux = 23910 Section OK
Load Comb 18 1.0D+1.4Wx 32792 367.83 -24.092 Mx' = 387.89 Mux = 27896 Section OK
Load Comb 19 1.0D-1.4Wx 42951 -950.43 -38.568 Mx' = 976.72 Mux = 23206 Section OK
Load Comb 20 1.0D+1.4Wy 34216 -5471.3 3748.7 Mx' = 8512.2 Mux = 27278 Section OK
Load Comb 21 1.0D-1.4Wy 41527 4888.7 -3811.3 My' = 5808.5 Muy = 18343 Section OK
Load Comb 22 1.0D+1.4W45 29902 -2741.8 3838.3 My' = 5236.3 Muy = 22460 Section OK
Load Comb 23 1.0D-1.4W45 45841 2159.2 -3900.9 My' = 4707.8 Muy = 16626 Section OK
Load Comb 24 1.0D+1.4W135 49364 6558.7 -4959.6 Mx' = 9502.2 Mux = 19771 Section OK
Load Comb 25 1.0D-1.4W135 26379 -7141.3 4897 Mx' = 11689 Mux = 29921 Section OK

P versus Mx and My of the Column Section
80000
70000
60000
50000
P (kN)

40000
30000
20000
10000
0
0 5000 10000 15000 20000 25000 30000 35000
M (kNm)

Appendix G

Derivation of Design Formulae
for Walls to Rigorous Stress
Strain Curve of Concrete

Appendix G
Derivation of Design Formulae for Shear Walls to Rigorous Stress Strain
Curve of Concrete

b

h

As similar to the exercise in Appendix F for columns, the exercise in this Appendix is
A
repeated for walls by which sb are set to zero in the various cases 1 to 7, using the
bh
equations summarized in Appendix F.

Cases 1 to 3 – where x/h ≤ 7/11

By (Eqn F-18) or (Eqn F-32) or (Eqn F-36) of Appendix F
N u 0.67 f cu  ε x  x A
=  3 − 0  + 0.87 f y  2 − 1 sh
 h
bh 3γ m  ε ult   h  bh
N u 0.67 f cu  ε x
− 3 − 0
 ε 
h
Ash bh 3γ m  
= ult
(Eqn G-1)
bh  x 
0.87 f y  2 − 1
 h 
Asb
Substituting into (Eqn F-31) or (Eqn F-34) or (Eqn F-38) and putting =0
bh
M u 0.67 f cu  x  1  x  1 ε 0 1 ε 0  x  1  ε 0   x  Ash  x  163  x  
2 2
 
=    1 −  − +  −     + 0.87 f y   −   
bh 2 γ m  h  2  h  6 ε ult 3 ε ult  h  12  ε ult   h 
  bh  h  147  h  
 
 
0.67 f cu  16 131 ε 0 1  ε 0   x   0.67 f cu  1 1 ε 0 1  ε 0   N u 163  x 
2 3 2 2
   
⇒  + −     +   − +   −  
γ m 147 441 ε ult 6  ε ult   h   γ m  2 3 ε ult 12  ε ult   bh 147  h 
     
   
N 2M u 0.67 f cu  1 1 ε 0  x M u
+ u − −  −  + =0 (Eqn G-2)
 bh bh
2
γ m  2 6 ε ult  h bh 2
 
x A
Upon solving , back substituting into (Eqn G-1) to calculate sh
h bh

Case 4 – where 7/11 < x/h ≤ 1

G-1

Appendix G
Asb
Referring to (Eqn F-39) of Appendix F and setting =0
bh
0.67 f cu  ε0  x Ash  9 x 7 h 7  N u
3γ m   3 − ε  h + 0.87 f y bh  − 56 h − 8 x + 4  = bh
 
ult   
N u 0.67 f cu  ε x
− 3 − 0 
 ε h
A bh 3γ m  ult 
⇒ sh = (Eqn G-3)
bh  9 x 7 h 7
0.87 f y  − − + 
 56 h 8 x 4 
A
Substituting into (Eqn F-41) with sb = 0
bh
M u 0.67 f cu  x  1  x  1 ε 0 1 ε 0  x  1  ε 0   x 
2
 
=    1 −  − +  −   ε   h 
  
bh 2
γ m  h  2  h  6 ε ult 3 ε ult  h  12  ult  
 
7 h 9 x 9  x   Ash
2

+ 0.87 f y  − +   
 48 x 112 h 392  h   bh
 
  1 ε 0 1 1  ε 0
2
 x  7 9 x 7 h 
M  7 9 x 7 h  0.67 f cu  x  1 1 ε 0  
⇒ u − − =   − +
 3 ε − 2 − 12  ε    − − 
bh 2  4 56 h 8 x  γm  h  2 6 ε ult   ult
 
 h  4 56 h 8 x 
   ult   
7 h 9 x 9  x    N u 0.67 f cu
2
 ε  x
+ − +    − 3 − 0
  
h
 48 x 112 h 392  h    bh
  3γ m  ε ult  
0.67 f cu  45 3  ε 0   x 
2
9 ε0
4

⇒ − + −    
γ m  784 196 ε ult 224  ε ult   h 
  

 0.67 f  7 7 ε 7  ε 0   9 N u  x 
2 3

+ cu
 − 0
+   −  
 γ m  8 12 ε ult 48  ε ult   392 bh  h 
   
 
 9 M 0.67 f cu 7  2 ε0 1  ε 0   9 N u  x 
2 2

+ − u
+ − 1.5 + −   +  
 56 bh
2
γm 8  3 ε ult 12  ε ult   112 bh  h 
  
  
7 M u 0.67 f cu 7  ε 0  x  7 M u 7 Nu 
+ + 3 −  + − − =0 (Eqn G-4)
4 bh 2
3γ m 12  ε ult  h  8 bh 2 48 bh 
 
x A
Upon solving , back-substitute into (Eqn G-3) to solve for sh
h bh

Case 5 – where 1 < x/h ≤ 1/(1 – ε0/εult)
A
Referring to (Eqn F-52) and setting sb = 0
bd
Nu  1 1 ε0 2
1 ε ult 1 ε 0  x  1 ε ult 
= Ec ε ult − − + +  + Ec ε ult 1 −
 2ε 

bh  2 6 ε ult
2
6 ε 0 2 ε ult  h  0 

G-2

Appendix G
 1 ε ult 1  h E c ε ult  h 
2 2
 7 7 h 9 x  Ash
+ E c ε ult 
2 ε 
−  −   + 0.87 f y  − −  (Eqn G-5)
 0 2 x 6ε 0  x   4 8 x 56 h  bh
A A
Substituting for 0.87 f y sh into (Eqn F-57), again setting sb = 0
bh bd
M  1 ε0 3
1 ε0
2
1 ε 0 1 1 ε ult  x 
2

= Ec ε ult  −
 24 ε + − + −  
bh 2
 ult
3
6 ε ult 2
4 ε ult 6 24 ε 0  h 
 1 ε0 2
1 ε 0 1 1 ε ult  x  1 1 ε ult  h 1 ε ult h
2

+ E c ε ult − + − +  + E c ε ult  −
 12 12 ε  x + E c ε ult 24 ε
  
 12 ε ult

2
4 ε ult 4 12 ε 0  h   0  0  x
7 h 9 x 9  x   Ash
2

+ 0.87 f y  − +   
 48 x 112 h 392  h   bh
 
M 7 7 h 9 x  1 ε 0 3 1 ε 0 2 1 ε 0 1 1 ε ult  7 7 h 9 x  x  2
⇒ 2 − −  = E c ε ult  −

+ − + −  − −
 4 8 x 56 h  h 
bh  4 8 x 56 h   24 ε ult
3
6 ε ult 2 4 ε ult 6 24 ε 0   
 1 ε 0 2 1 ε 0 1 1 ε ult  7 7 h 9 x  x  1 1 ε ult  7 7 h 9 x  h
+ E c ε ult − + − +  − −  + E c ε ult  −
 12 12 ε  4 − 8 x − 56 h  x

 12 ε ult

2
4 ε ult 4 12 ε 0  4 8 x 56 h  h
  0  
1 ε ult  7 7 h 9 x  h  Nu  7 h 9 x 9  x 
2 2

+ E c ε ult  − −   +  − +   
24 ε 0  4 8 x 56 h  x  bh  48 x 112 h 392  h  
 
 1 1 ε0 1 ε ult 1 ε 0  x  7 h 9 x 9  x 
2 2

− E c ε ult − − + +   − +   
 2 6 ε ult

2
6 ε 0 2 ε ult  h  48 x 112 h 392  h  
  
 1 ε ult  7 h 9 x   1 ε ult 1  h  7 h 9 x 9  x 
2 2
9  x
− E c ε ult 1 −
 
 48 x − 112 h + 392  h   − E c ε ult 
 −  
 − +   
 2 ε0 
   
  2 ε 0 2  x  48 x 112 h 392  h  
 
E c ε ult h  7 h 9 x 9  x 
2 2 2

+    − +   
6ε 0  x   48 x 112 h 392  h  
 
 3 ε 03 9 ε0
2
45 ε 0 3 9 ε ult  x 
3

⇒ E c ε ult  − + − +  
 448 ε 3 392 ε 2 1568 ε 196 3136 ε 0  h 
 ult ult ult 
  −7 ε0 3
7 ε0
2
7 ε0 79 289 ε ult  9 N u  x 
2

+  E c ε ult  + − + − +  
  96 ε 3 24 ε 2 16 ε 294 4704 ε 0  392 bh  h 
  ult ult ult  
  7 ε0 3
7 ε0
2
21 ε 0 289 1229 ε ult  9 M 9 N u  x 
+  E c ε ult  − + − + + −  
  193 ε 3 24 ε 2 32 ε 588 9408 ε 0  56 bh 2 112 bh  h 
  ult ult ult  
  7 ε0 2
7 ε0 5 281 ε ult  7 M   E c ε ult 125 7 M
2
7 Nu  h
+  E c ε ult  − + − −  + − + + 


 72 ε 2 24 ε
 ult ult 21 7056 ε 0  4 bh 2  
   ε 0 1344 8 bh 2 48 bh  x 
Eε E c ε ult 7  h 
2 2 2 3
7 h
+ c ult   −   =0 (Eqn G-6)
ε 0 96  x  ε 0 576  x 
x x
which is in fact an equation of 6th power in . is to be solved by numerical
h h

G-3

Appendix G
x
method. By back-substituting into (Eqn G-5)
h
Nu  1 1 ε 02 1 ε ult 1 ε 0  x  1 ε ult 
= Ec ε ult − − + +  + Ec ε ult 1 −
 2ε 

bh  2 6 ε ult
2
6 ε 0 2 ε ult  h  0 
1 ε 1h E ε h
2 2
 7 7 h 9 x  Ash
+ Ec ε ult  ult −  − c ult   + 0.87 f y  −
  − 
 2 ε0 2  x 6ε 0  x   4 8 x 56 h  bh
Nu  1 1 ε 0 2 1 ε ult 1 ε 0  x  1 ε ult   1 ε ult 1  h Ec ε ult 2  h 
2
− E c ε ult − − + + 1 −
 − Ec ε ult   − Ec ε ult 
 2 ε − 2 x  +  
Ash bh  2 6 ε ult

2
6 ε 0 2 ε ult  h
  2 ε0   0  6ε 0  x 
=
bh 7 7 h 9 x
0.87 f y  − − 
 4 8 x 56 h 

Case 6 – where 1/(1 – ε0/εult) < x/h ≤ 7/3

Asb
Consider (Eqn F-58) and substituting =0
bh
N u 0.67 f cu
−
N u 0.67 f cu  7 7 h 9 x  Ash Ash bh γm
= + 0.87 f y  − −  ⇒ =
bh γm  4 8 x 56 h  bh bh 7 7 h 9 x
0.87 f y  − − 
 4 8 x 56 h 
(Eqn G-7)
Substituting into (Eqn F-60)
Ash  7 h 9  x 
2
Mu 9 x
= 0.87 f y  − +   
bh 2 bh  48 x 112 h 392  h  
 
9  N u 0.67 f cu  x   9 M u 9  N u 0.67 f cu  x 
3 2

⇒ 
 −   + 
 h  −  −  
392  bh γm   56 bh 2 112  bh
 γ m  h 

7 M u x  7  N u 0.67 f cu  7 M u 
− +  − + =0 (Eqn G-8)
4 bh 2 h  48  bh
 γ m  8 bh 2 

x A
Solving the cubic equation for and substituting into (Eqn G-7) to solve for sh
h bh

Case 7 – where x/h > 7/3

Asb
By (Eqn F-61) and setting =0
bh
N u 0.67 f cu A A N 0.67 f cu  1
= + 0.87 f y sh ⇒ sh =  u −
 bh 
 0.87 f (Eqn G-9)
bh γm bh bh  γm  y

Mu
=0
bh 2

G-4

Appendix G – Summary of Design Charts for Walls

50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9
2 2
M/bh N/mm

Chart G - 1


50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5
2 2
M/bh N/mm

Chart G - 2


Design Chart of Rectangular Shear Wall with Uniform Vertical reinforcements to Code of Practice
50
0.4% steel
1% steel
45
2% steel
3% steel
40
4% steel
5% steel
35 6% steel
7% steel
30 8% steel
2
N/bh N/mm

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5
2 2
M/bh N/mm

Chart G - 3


55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
2 2
M/bh N/mm

Chart G - 4


55
0.4% steel
1% steel
50
2% steel

45 3% steel
4% steel
40 5% steel
6% steel
35 7% steel
8% steel
2
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
2 2
M/bh N/mm

Chart G - 5


60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2

35
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5
2 2
M/bh N/mm

Chart G - 6


60
0.4% steel
55 1% steel
2% steel
50 3% steel
4% steel
45 5% steel
6% steel
40
7% steel
8% steel
2

35
N/bh N/mm

30

25

20

15

10

5

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5
2 2
M/bh N/mm

Chart G - 7

Shear Wall R.C. Design to Code of Practice for Structural Use of Concrete 2004 - Uniform Reinforcements

Project :
Wall Mark Floor

b = 200 h = 2000 b' = 165.00 h' = 1500.00 cover= 25 bar size = 20

Basic Load Case My Mx
Load Case D.L. L.L. Wx Wy W45 W135 b
Axial Load P (kN) 3304.7 1582.1 -362.17 -245.1 56.92 82.09
Moment Mx (kNm) 29.13 32.11 2047.1 -1275.1 1098.1 888.93 h
Moment My (kNm) -31.33 16.09 2.15 44.2 206.5 35.21

N Mx My N/bh M/bh2 d/h / d/b x/h / y/b Steel Steel area
(kN) (kNm) (kNm) (N/mm2) (N/mm2) (%) (mm2)
Load Comb 1 1.4D+1.6L 7200 1500 100 Mx' = 1866.2 18 2.3328 - 0.8901 2.3359 9343.5
Load Comb 2 1.2(D+L+Wx) 5429.6 2530 -15.708 Mx' = 2607.8 13.574 3.2597 - 0.7152 2.3059 9223.4
Load Comb 3 1.2(D+L-Wx) 6298.8 -2383 -20.868 Mx' = 2473.2 15.747 3.0915 - 0.7772 2.5637 10255
Load Comb 4 1.2(D+L+Wy) 5570 -1456.7 34.752 Mx' = 1624.9 13.925 2.0311 - 0.8365 1.0928 4371.1
Load Comb 5 1.2(D+L-Wy) 6158.3 1603.6 -71.328 Mx' = 1918.9 15.396 2.3986 - 0.8314 1.7861 7144.5
Load Comb 6 1.2(D+L+W45) 5932.5 1391.2 229.51 My' = 306.62 14.831 3.8328 0.825 0.7508 2.7599 11040
Load Comb 7 1.2(D+L-W45) 5795.9 -1244.2 -266.09 My' = 336.52 14.49 4.2065 0.825 0.7255 3.0079 12032
Load Comb 8 1.2(D+L+W135) 5962.7 1140.2 23.964 Mx' = 1249.5 14.907 1.5618 - 0.9185 0.9046 3618.4
Load Comb 9 1.2(D+L-W135) 5765.7 -993.23 -60.54 Mx' = 1277.8 14.414 1.5972 - 0.9032 0.8122 3248.8
Load Comb 10 1.4(D+Wx) 4119.5 2906.7 -40.852 Mx' = 3150.7 10.299 3.9384 - 0.6025 2.5143 10057
Load Comb 11 1.4(D-Wx) 5133.6 -2825.2 -46.872 Mx' = 3068 12.834 3.835 - 0.6673 2.7996 11199
Load Comb 12 1.4(D+Wy) 4283.4 -1744.4 18.018 Mx' = 1849.7 10.709 2.3121 - 0.6995 0.7417 2966.7
Load Comb 13 1.4(D-Wy) 4969.7 1826 -105.74 Mx' = 2387.4 12.424 2.9842 - 0.7031 1.7913 7165.1
Load Comb 14 1.4(D+W45) 4706.3 1578.1 245.24 My' = 350.54 11.766 4.3818 0.825 0.6532 2.626 10504
Load Comb 15 1.4(D-W45) 4546.9 -1496.6 -332.96 My' = 435.07 11.367 5.4384 0.825 0.62 3.4305 13722
Load Comb 16 1.4(D+W135) 4741.5 1285.3 5.432 Mx' = 1315.1 11.854 1.6439 - 0.8274 0.4 1600
Load Comb 17 1.4(D-W135) 4511.7 -1203.7 -93.156 Mx' = 1731.6 11.279 2.1645 - 0.7373 0.6743 2697.1
Load Comb 18 1.0D+1.4Wx 2797.7 2895.1 -28.32 Mx' = 3093.4 6.9942 3.8667 - 0.5051 2.1902 8760.9
Load Comb 19 1.0D-1.4Wx 3811.7 -2836.8 -34.34 Mx' = 3050.1 9.5293 3.8126 - 0.5843 2.2839 9135.8
Load Comb 20 1.0D+1.4Wy 2961.6 -1756 30.55 Mx' = 1966 7.4039 2.4576 - 0.5307 0.607 2428.1
Load Comb 21 1.0D-1.4Wy 3647.8 1814.3 -93.21 Mx' = 2405.2 9.1196 3.0065 - 0.5941 1.3272 5308.9
Load Comb 22 1.0D+1.4W45 3384.4 1566.5 257.77 My' = 381.82 8.461 4.7727 0.825 0.5596 2.3848 9539.2
Load Comb 23 1.0D-1.4W45 3225 -1508.2 -320.43 My' = 442.13 8.0625 5.5266 0.825 0.5456 2.9149 11660
Load Comb 24 1.0D+1.4W135 3419.6 1273.6 17.964 Mx' = 1390.7 8.5491 1.7384 - 0.6436 0.4 1600
Load Comb 25 1.0D-1.4W135 3189.8 -1215.4 -80.624 Mx' = 1755.3 7.9744 2.1941 - 0.5706 0.4 1600
Steel required = 3.4305 13722

Plot of P (kN) versus M (kNm)
14000

12000

10000
P (kN)

8000

6000

4000

2000

0
0 500 1000 1500 2000 2500 3000 3500 4000 4500

M (kNm)

Appendix H

Estimation of Support Stiffnesses
of vertical supports to Transfer
Structures

Appendix H

Estimation of Support Stiffnesses of vertical supports to Transfer Structures

Simulation of Support Stiffness in Plate Bending Structure

For support stiffness, we are referring to the force or moment required to produce unit
vertical movement or unit rotation at the top of the support which are denoted by K Z ,
K θX , K θY for settlement stiffness along the Z direction, and rotational stiffnesses about
X and Y directions. These stiffnesses are independent parameters which can interact only
through the plate structure. Most softwares allow the user either to input numerical values
or structural sizes and heights of the support (which are usually walls or columns) by
which the softwares can calculate numerical values for the support stiffnesses as follows :

(i) For the settlement stiffness K Z , the value is mostly simply AE L where A is
the cross sectional of the support which is either a column or a wall, E is the
Young’s Modulus of the construction material and L is the free length of the
column / wall. The AE L simply measures the ‘elastic shortening’ of the
column / wall.

Strictly speaking, the expression AE L is only correct if the column / wall is
one storey high and restrained completely from settlement at the bottom. However,
if the column / wall is more than one storey high, the settlement stiffness
estimation can be very complicated. It will not even be a constant value. The
settlement of the support is, in fact, ‘interacting’ with that of others through the
structural frame linking them together by transferring the axial loads in the
column / wall to others through shears in the linking beams. Nevertheless, if the
linking beams (usually floor beams) in the structural frame are ‘flexible’, the
transfer of loads from one column / wall through the linking beams to the rest of
the frame will generally be negligible. By ignoring such transfer, the settlement
stiffness of a column / wall can be obtained by ‘compounding’ the settlement
stiffness of the individual settlement stiffness at each floor as
1 1
KZ = =
L1 L L L Li
+ 2 + 3 + ...... n
A1 E1 A2 E 2 A3 E3 An E n
∑AE
i i

H-1

Appendix H

(ii) For the rotational stiffness, most of the existing softwares calculate the numerical
4 EI 3EI
values either by or , depending on whether the far end of the supporting
L L
column / wall is assumed fixed or pinned (where I is the second moment of area
of the column / wall section). However, apart from the assumed fixity at the far
4 EI 3EI
end, the formulae or are also based on the assumption that both ends of
L L
the column / wall are restrained from lateral movement (sidesway). It is obvious
that the assumption will not be valid if the out-of-plane load or the structural
layout is unsymmetrical where the plate will have lateral movements. The errors
may be significant if the structure is to simulate a transfer plate under wind load
which is in the form of an out-of-plane moment tending to overturn the structure.
Nevertheless, the errors can be minimized by finding the force that will be
required to restrain the slab structure from sideswaying and applying a force of
the same magnitude but of opposite direction to nullify this force. This magnitude
of this restraining force or nullifying force is the sum of the total shears produced
in the supporting walls / columns due to the moments induced on the walls /
columns from the plate analysis. However, the analysis of finding the effects on
the plate by the “nullifying force” has to be done on a plane frame or a space
frame structure as the 2-D plate bending model cannot cater for lateral in-plane
loads. This approach is adopted by some local engineers and the procedure for
analysis is illustrated in Figure H-1.

Lateral force, S , to
prevent sidesway

S1 M U1 S2 MU2 S3 MU3
h1
h2 h3
S1 S3
S2

Figure H-1 Diagrammatic illustration of the restraining shear or nullifying shear

In addition, the followings should be noted :

H-2

Appendix H

Note : 1. If the wall / column is prismatic and the lower end is restrained from
rotation, the moment at the lower end will be M Li = 0.5M Ui (carry-over
from the top); if the lower end is assumed pinned, the moment at it will be
zero;
M + M Li
2. The shear on the wall / column will be S i = Ui where M Ui is
hi
obtained from plate bending analysis and the total restraining shear is
S = ∑ Si

H-3

Appendix I

Derivation of Formulae for
Rigid Cap Analysis

Appendix I

Derivation of Formulae for Rigid Cap Analysis

Underlying Principles of the Rigid Cap Analysis

The “Rigid Cap Analysis” method utilizes the assumption of “Rigid Cap” in the
solution of pile loads beneath a single cap against out-of-plane loads, i.e. the cap is a
perfectly rigid body which does not deform upon the application of loads. The cap
itself may settle, translate or rotate, but as a rigid body. The deflections of a
connecting pile will therefore be well defined by the movement of the cap and the
location of the pile beneath the cap, taking into consideration of the connection
conditions of the piles. Consider a Pile i situated from a point O on the pile cap as

shown in Figure I-1 with settlement stiffness K iZ

Y
Pile i

+ve M Y

yi +ve M X

O X
xi

Figure I-1 – Derivation of Pile Loads under Rigid Cap

As the settlement of all piles beneath the Cap will lie in the same plane after the
application of the out-of-plane load, the settlement of Pile i denoted by ∆ iZ
can be defined by bO + b1 xi + b2 y i which is the equation for a plane in
‘co-ordinate geometry’ where bO , b1 and b2 are constants.

The upward reaction by Pile i is K iZ (bO + b1 xi + b2 y i )

Summing all pile loads :
Balancing the applied load P = ∑ K iZ (bO + b1 xi + b2 y i )
⇒ P = bO ∑ K iZ + b1 ∑ K iZ xi + b2 ∑ K iZ y i

I-1

Appendix I

Balancing the applied Moment M X = −∑ K iZ (bO + b1 xi + b2 y i ) y i
⇒ M X = −bO ∑ K iZ y i − b1 ∑ K iZ xi y i − b2 ∑ K iZ y i
2

Balancing the applied Moment M Y = ∑ K iZ (bO + b1 xi + b2 yi )xi
⇒ M Y = bO ∑ K iZ xi + b1 ∑ K iZ xi + b2 ∑ K iZ xi y i
2

It is possible to choose the centre O such that
∑ K iZ xi = ∑ K iZ yi = ∑ K iZ xi yi = 0 .
So the three equations become
P
P = bO ∑ K iZ ⇒ bO =
∑ K iZ
−MX
M X = −b2 ∑ K iZ y i ⇒ b2 =
2

∑K iZ yi
2

MY
M Y = b1 ∑ K iZ xi ⇒ b1 =
2

∑K iZ xi
2

The load on Pile i is then
P = ∑ K iZ (bO + b1 xi + b2 y i )
 P MY MX 
= K iZ  + xi − yi 
 ∑ K iZ ∑ K x 2 ∑ K iZ yi 
2
 iZ i 
PK iZ M Y K iZ M X K iZ
= + xi − yi
∑ K iZ ∑ K iZ xi 2 ∑ K iZ yi 2
To effect ∑K iZ xi = ∑ K iZ y i = ∑ K iZ xi y i = 0 , the location of O and the

orientation of the axes X-X and Y-Y must then be the “principal axes” of the
pile group.

Conventionally, designers may like to use moments along defined axes instead of
moments about defined axes. If we rename the axes and U-U and V-V after translation
and rotation of the axes X-X and Y-Y such that the condition

∑K iZ u i = ∑ K iZ vi = ∑ K iZ u i vi = 0 can be satisfied, then the pile load become

PK iZ M U K iZ M Y K iZ
PiZ = + u + v
∑ K iZ ∑ K iZ ui 2 i
∑ K iZ vi
2 i

If all piles are identical, i.e. all K iZ are equal, then the formula is reduced
P MU MV
PiZ = + u + v where N is the number of piles.
N ∑ ui 2 i
∑ vi
2 i

Or if we do not wish to rotate the axes to U and V , then only

∑K iZ xi = ∑ K iZ y i = 0 and the moment balancing equations becomes

I-2

Appendix I

M X = −bO ∑ K iZ y i − b1 ∑ K iZ xi y i − b2 ∑ K iZ y i
2

⇒ M X = −b1 ∑ K iZ xi y i − b2 ∑ K iZ y i
2

M Y = bO ∑ K iZ xi + b1 ∑ K iZ xi + b2 ∑ K iZ xi y i
2
and

⇒ M Y = b1 ∑ K iZ xi + b2 ∑ K iZ xi y i
2

Solving
P
P = bO ∑ K iZ ⇒ bO =
∑ K iZ
M X ∑ K iZ xi yi + M Y ∑ K iZ y i
2

b1 =
− (∑ K iZ xi y i ) +
2
(∑ K iZ xi
2
∑K iZ yi
2
)
− M Y ∑ K iZ xi y i − M X ∑ K iZ xi
2

b2 =
− (∑ K iZ xi y i ) +
2
(∑ K iZ xi
2
∑K iZ yi
2
)

So the pile load becomes

M X ∑ K iZ xi y i + M Y ∑ K iZ y i
2
PK iZ
PiZ = +
( )
K iZ xi
∑ K iZ − (∑ K iZ xi yi )2 + ∑ K iZ xi 2 ∑ K iZ yi 2
− M Y ∑ K iZ xi y i − M X ∑ K iZ xi
2

(∑ K )
+ K iZ y i .
− (∑ K iZ xi y i ) + ∑ K iZ yi
2 2 2
iZ xi

If all piles are identical, i.e. all KiZ are equal, then the formula is reduced

M X ∑ xi y i + M Y ∑ y i − M Y ∑ xi y i − M X ∑ xi
2 2
P
PiZ = + xi +
( ) ( )
yi
N − (∑ xi y i ) + ∑ xi 2 ∑ y i 2
2
− (∑ xi y i ) + ∑ xi ∑ y i
2 2 2

For a symmetrical layout where ∑x y i i = 0 , the equation is further reduced to

P MY −MX
PiZ = + x + yi
N ∑ xi 2 i
∑ yi 2

I-3

Appendix J

Mathematical Simulation of
Curves related to Shrinkage and
Creep Determination

Appendix J

Simulation of Curves for Shrinkage and Creep Determination
Simulation of K j values

Figure 3.5 of the Code is expanded and intermediate lines are added for reading more
accurate values. The intermediate values are scaled off from the expanded figure and
listed as follows (he = 50 mm which is seldom used is ignored) :

he = 100 mm he = 200 mm he = 400 mm he = 800 mm
Days Kj Days Kj Days Kj Days Kj
2 0.09 6 0.09 16.6 0.065 60 0.065
3 0.108 7 0.095 20 0.08 70 0.075
4 0.125 8 0.1 30 0.115 80 0.084
5 0.145 9 0.105 40 0.145 90 0.092
6 0.165 10 0.112 50 0.165 100 0.099
7 0.185 11 0.12 60 0.185 200 0.17
8 0.2 12 0.13 70 0.2 300 0.22
9 0.213 13 0.138 80 0.22 400 0.265
10 0.225 14 0.145 90 0.235 500 0.31
20 0.33 20 0.18 100 0.25 600 0.35
30 0.4 30 0.23 200 0.375 700 0.386
40 0.45 40 0.275 300 0.46 800 0.42
50 0.5 50 0.31 400 0.54 900 0.45
60 0.543 60 0.345 500 0.6 1000 0.48
70 0.57 70 0.37 600 0.64 2000 0.73
80 0.6 80 0.4 700 0.67 3000 0.83
90 0.625 90 0.425 800 0.7 4000 0.888
100 0.645 100 0.445 900 0.72 5000 0.923
200 0.775 200 0.61 1000 0.74 6000 0.95
300 0.827 300 0.7 2000 0.87 7000 0.97
400 0.865 400 0.75 3000 0.935 8000 0.98
500 0.892 500 0.79 4000 0.97
600 0.91 600 0.81 5000 0.99
700 0.927 700 0.84
800 0.937 800 0.855
900 0.945 900 0.87
1000 0.955 1000 0.883
1500 0.975 2000 0.955

J-1

Appendix J

Curves are plotted accordingly in Microsoft Excel as shown :

Simulation of Kj Values


1

0.9

0.8

0.7

0.6
Kj

0.5

0.4

0.3

0.2

0.1

0
1 10 100 1000 10000

Time since loading, Days

These curves are divided into parts and polynomial equations (x denote days) are
simulated by regression done by the Excel as follows :
(i) Effectiveness thickness he = 100 mm
for 2 ≤ x ≤ 10
Kj = –1.5740740764×10-6x6 + 7.1089743699×10-5x5 – 1.2348646738×10-3x4 +
1.0396454943×10-2x3 – 4.4218106746×10-2x2 + 1.0785366750×10-1x –
1.4422222154×10-2;
for 10 < x ≤ 100
Kj = –8.2638888726×10-12x6 + 2.9424679436×10-9x5 – 4.1646100361×10-7x4 +
2.9995170408×10-5x3 – 1.1964688098×10-3x2 + 3.0905446162×10-2x +
9.3000049487×10-3
for 100 < x ≤ 1000
Kj = –9.9999999553×10-18x6 + 3.7871794729×10-14x5 – 5.7487179303×10-11x4
+ 4.4829720169×10-8x3 – 1.9268813492×10-5x2 + 4.6787198128×10-3x +
3.3059999890×10-1
(ii) Effectiveness thickness he = 200 mm
for 1 ≤ x ≤ 10
Kj = –5.5555555584×10-7x6 + 1.9230769236×10-5x5 – 2.3632478631×10-4x4 +
1.1888111887×10-3x3 – 1.8372455154×10-3x2 + 5.1966197721×10-3x +
5.0666667394×10-2
for 10 < x ≤ 100
Kj = –6.0905886799×10-12x6 + 2.0287559012×10-9x5 – 2.6706836340×10-7x4 +

J-2

Appendix J

1.7840233064E×10-5x3 – 6.6454331705×10-4x2 + 1.7736234727×10-2x -
1.3696178365×10-2
for 100 < x ≤ 1000
Kj = –4.1666665317×10-19x6 + 4.6185897038×10-15x5 – 1.2899038408×10-11x4
+ 1.6179152071×10-8x3 – 1.0631842073×10-5x2 + 3.8848713316×10-3x +
1.4793333214×10-1
(iii) Effectiveness thickness he = 400 mm
for 1 ≤ x ≤ 16.6
Kj = 1.4187214466×10-6x4 – 3.5464080361×10-5x3 + 3.3384218737×10-4x2 –
2.2688256448×10-5x + 2.7836053347×10-2
for 16.6 < x ≤ 100
Kj = –1.5740740764×10-6x6 + 7.1089743699×10-5x5 – 1.2348646738×10-3x4 +
1.0396454943×10-6x3 – 4.4218106746×10-2x2 + 1.0785366750×10-1x –
1.4422222154×10-2
for 100 < x ≤ 1000
Kj = –9.3749999678×10-18x6 + 3.1193910157×10-4x5 – 4.0436698591×10-11x4
+ 2.6279902314×10-8x3 – 9.8112164735×10-6x2 + 2.8475810022×10-3x +
4.1166665811×10-2
for 1000 < x ≤ 5000
Kj = –8.3333333334×10-16x4 + 1.4166666667×10-11x3 – 9.6666666667×10-8x2
+ 3.3333333333×10-4x + 4.9000000000×10-1
(iv) Effectiveness thickness he = 800 mm
for 3 ≤ x ≤ 60
Kj = 9.5889348301×10-12x5 – 1.5604725262×10-8x4 + 1.8715280898×10-6x3 –
7.5635030550×10-5x2 + 1.8805930655×10-3x + 1.4981311831×10-2
for 60 < x ≤ 100
Kj = –5.4210108624×10-20x4 + 1.3010426070×10-17x3 – 5.0000000012×10-6x2
+ 1.6500000000×10-3x – 1.6000000000×10-2
for 100 < x ≤ 1000
Kj = –3.9583333158×10-18x6 + 1.4818910202×10-14x5 – 2.1967147366×10-11x4
+ 1.6383442558×10-8x3 – 6.5899851301×10-6x2 + 1.8249511657×10-3x –
3.1900000544×10-2

Simulation of K m values

Values of Figure 3.2 of the Code for Ordinary Portland Cement are read, Excel chart
is plotted and polynomial equations are simulated as :

J-3

Appendix J

Simulation of Km Values for Portland Cement

Ordinary Portland Cement Rapid Hardening Portland Cement

2

1.8

1.6

1.4

1.2
Km

1

0.8

0.6

0.4

0.2

0
1 10 100 1000

Age of Concrete at Time of Loading (Days)

for 1 ≤ x ≤ 7
Km = 8.3333333333×10-3x2 – 1.3333333333×10-1x + 1.925
for 7 < x ≤ 28
Km = 7.3129251701×10-4x2 – 4.4642857143×10-2x + 1.6766666667
for 28 < x ≤ 90
Km = 3.8967199783×10-5x2 – 8.6303876389×10-3x + 1.2111005693
for 90 < x ≤ 360
Km = 2.3662551440×10-6x2 – 1.9722222222×10-3x + 9.0833333333×10-1

J-4

Concrete design

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