Design of beam

Unit – 4
Design of Beams
In this chapter, it is intended to learn the method of designing the beams using the
principles developed in previous chapters. Design consists of selecting proper materials,
shape and size of the structural member keeping in view the economy, stability and
aesthetics. The design of beams are done for the limit state of collapse and checked for the
other limit states. Normally the beam is designed for flexure and checked for shear,
deflection, cracking and bond.
Design procedure
The procedure for the design of beam may be summarized as follows:
1. Estimation of loads
2. Analysis
3. Design
1. Estimation of loads
The loads that get realized on the beams consists of the following:
a. Self weight of the beam.
b. Weight of the wall constructed on the beam
c. The portion of the slab loads which gets transferred to the beams. These slab
loads are due to live loads that are acting on the slab dead loads such as self
weight of the slab, floor finishes, partitions, false ceiling and some special fixed
loads.
The economy and safety of the beams achieved depends on the accuracy with
which the loads are estimated.
The dead loads are calculated based on the density whereas the live loads are
taken from IS: 875 depending on the functional use of the building.
2. Analysis
For the loads that are acting on the beams, the analysis is done by any standard
method to obtain the shear forces and bending moments.
3. Design
a. Selection of width and depth of the beam.
The width of the beam selected shall satisfy the slenderness limits specified in
IS 456 : 2000 clause 23.3 to ensure the lateral stability.
b. Calculation of effective span (le) (Refer clause 22.2, IS 456:2000)
c. Calculation of loads (w)

d. Calculation of critical moments and shears.
The moment and shear that exists at the critical sections are considered for the
design. Critical sections are the sections where the values are maximum. Critical
section for the moment in a simply supported beam is at the point where the shear
force is zero. For continuous beams the critical section for the +ve bending
moment is in the span and –ve bending moment is at the support. The critical
section for the shear is at the support.
e. Find the factored shear (Vu) and factored moment (Mu)
f. Check for the depth based on maximum bending moment.
Considering the section to be nearly balanced section and using the equation
Annexure G, IS 456-2000 obtain the value of the required depth drequired. If the
assumed depth “d” is greater than the “drequired”, it satisfies the depth criteria based
on flexure. If the assumed section is less than the” drequired”, revise the section.
g. Calculation of steel.
As the section is under reinforced, use the equation G.1.1.(b) to obtain the
steel.
h. Check for shear.
i. Check for developmental length.
j. Check for deflection.
k. Check for Ast min, Ast max and distance between the two bars.
Anchorage of bars or check for development length
In accordance with clause 26.2 IS 456 : 2000, the bars shall be extended (or
anchored) for a certain distance on either side of the point of maximum bending moment
where there is maximum stress (Tension or Compression). This distance is known as the
development length and is required in order to prevent the bar from pulling out under tension
or pushing in under compression. The development length (Ld) is given by
‫ܮ‬ௗ =
∅ ߪ௦
4 ܼ௕ௗ
where, ∅ = Nominal diameter of the bar
ߪ௦= Stress in bar at the section considered at design load
Zbd= Design bond stress given in table 26.2.1.1 (IS 456 : 2000)
Table 26.2.1.1: Design bond stress in limit state method for plain bars in tension shall be as
below:
Grade of concrete M 20 M 25 M 30 M 35 M 40 and above
Design bond stress 1.2 1.4 1.5 1.7 1.9
߬௕ௗ N/mm2

Note: Due to the above requirement it can be concluded that no bar can be bent up or
curtailed upto a distance of development length from the point of maximum moment.
Due to practical difficulties if it is not possible to provide the required embedment or
development length, bends hooks and mechanical anchorages are used.
Flexural reinforcement shall not be terminated in a tension zone unless any one of the
following condition is satisfied:
a. The shear at the cut-off points does not exceed two-thirds that permitted,
including the shear strength of web reinforcement provided.
b. Stirrup area in excess of that required for shear and torsion is provided along each
terminated bar over a distance from the cut-off point equal to three-fourths the
effective depth of the member. The excess stirrup area shall be not less than
0.4bs/fy, where b is the breadth of the beam, s is the spacing and fy is the
characteristic strength of reinforcement in N/mm2
. The resulting spacing shall not
exceed d/8ߚ௕ where ߚ௕ is the ratio of the area of bars cut-off to the total area of
bars at the section, and d is the effective depth.
c. For 36 mm and smaller bars, the continuing bars provide double the area required
for flexure at the cut-off point and the shear does not exceed three-fourths that
permitted.
Positive moment reinforcement:
a. At least one-third the positive moment reinforcement in simple members and one-
fourth the positive moment reinforcement in continuous members shall extend
along the same face of the member into the support, to a length equal to Ld/3.
b. When a flexural member is part of the primary lateral load resisting system, the
positive reinforcement required to be extended into the support as described in (a)
shall be anchored to develop its design stress in tension at the face of the support.
c. At simple supports and at points of inflection, positive moment tension
reinforcement shall be limited to a diameter such that Ld computed for fd by 26.2.1
IS 456:2000 does not exceed.
‫ܯ‬ଵ
ܸ
+ ‫ܮ‬଴
where, M1 = moment of resistance of the section assuming all reinforcement at
the section to be stressed to fd;
fd = 0.87fy in the case of limit state design and the permissible stress ߪ௦௧ in the
case of working stress design;
V = shear force at the section due to the design loads;
L0 = sum of the anchorage beyond the centre of the support and the equivalent
anchorage value of any hook or mechanical anchorage at simple support; and

at a point of inflection, L0 is limited to the effective depth of the members or
12∅ , whichever is greater; and
∅ = diameter of bar.
The value of M1/V in the above expression may be increased by 30 percent when the ends of
the reinforcement are confined by a compressive reaction.
Negative moment reinforcement:
At least one third of the total reinforcement provided for negative moment at the
support shall extend beyond the point of inflection for a distance not less than the effective
depth of the member of 12߮ or one-sixteenth of the clear span whichever is greater.
Anchorage of bars
Anchoring of bars is done to provide the development length and maintain the
integrity of the structure.
Anchoring bars in tension:
a. Deformed bars may be used without end anchorages provided development length
requirement is satisfied. Hooks should normally be provided for plain bars in tension.
b. Bends and hooks – shall conform to IS 2502
1. Bends – The anchorage value of bend shall be taken as 4 times the diameter of the
bar for each 450
bend subject to a maximum of 16 times the diameter of the bar.
2. Hooks – The anchorage value of a standard U-type hook shall be equal to 16 times
the diameter of the bar.
Anchoring bars in compression:
The anchorage length of straight bar in compression shall be equal to the development
length of bars in compression as specified in clause 26.2.1 of IS 456:2000. The projected
length of hooks, bends and straight lengths beyond bends if provided for a bar in
compression, shall only be considered for development length.
Mechanical devices for anchorage:
Any mechanical or other device capable of developing the strength of the bar without
damage to concrete may be used as anchorage with the approval of the engineer-in-charge.
Anchoring shear reinforcement:
a. Inclined bars – The development length shall be as for bars in tension; this length
shall be measured as under:
1. In tension zone, from the end of the sloping or inclined portion of the bar, and
2. In the compression zone, from the mid depth of the beam.
b. Stirrups – Not withstanding any of the provisions of this standard, in case of
secondary reinforcement, such as stirrups and transverse ties, complete development
lengths and anchorages shall be deemed to have been provided when the bar is bent

through an angle of at least 900
round a bar of at least its own diameter and is
continued beyond the end of the curve for a length of at least eight diameters, or when
the bar is bent through an angle of 1350
and is continued beyond the end of the curve
for a length of at least six bar diameters or when the bar is bent through an angle of
1800
and is continued beyond the end of the curve for a length of at least four bar
diameters.
Reinforcement requirements
1. Minimum reinforcement:
The minimum area of tension reinforcement shall be not less than that given by the
following:
‫ܣ‬௦
ܾ݀
=
0.85
݂௬
where, As = minimum area of tension reinforcement.
b = breadth of beam or the breadth of the web of T-beam,
d = effective depth, and
fy = characteristic strength of reinforcement in N/mm2
2. Maximum reinforcement – The maximum area of tension reinforcement shall not
exceed 0.04bD
Compression reinforcement:
The maximum area of compression reinforcement shall not exceed 0.04bD.
Compression reinforcement in beams shall be enclosed by stirrups for effective lateral
restraint.
Pitch and diameter of lateral ties:
The pitch of shear reinforcement shall be not more than the least of the following
distances:
1. The least lateral dimension of the compression members;
2. Sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied;
and
3. 300 mm.
The diameter of the polygonal links or lateral ties shall be not less than one-fourth of
the diameter of the largest longitudinal bar, and in no case less than 16 mm.

Slenderness limits of beams to ensure lateral stability
A beam is usually a vertical load carrying member. However, if the length of the
beam is very large it may bend laterally. To ensure lateral stability of a beam the following
specifications have been given in the code.
A simply supported or continuous beam shall be so proportioned that the clear
distance between the lateral restraints does not exceed 60b or
ଶହ଴௕మ
ௗ
whichever is less,
where d is the effective depth of the beam and b the breadth of the compression face midway
between the lateral restraints.
For a cantilever, the clear distance from the free end of the cantilever to the lateral
restraint shall not exceed 25b or
ଵ଴଴௕మ
ௗ
whichever is less.
Problems:
1. Design a singly reinforced SSB of clear span 5m to support a working live load of 25
kN/m run. Use Fe 415 steel and M 20 grad concrete. Assume the support thickness as
230 mm.
Step 1 (a):Fixing up the depth of the section.
Taking
௅
ௗ
= 20, for SSB [Refer 23.2.1, pg 37]
݀ =
௅
ଶ଴
=
ହ
ଶ଴
= 0.25 m = 250 mm
Providing a cover of 25 mm, overall depth D = 250 + 25 = 275 mm
Dimensions of the section.
Width b = 230 mm
depth d = 250 mm
Step 1 (b): Check for lateral stability/lateral buckling
Refer page 39, clause 23.3
Allowable l = 60b or
ଶହ଴ ௕మ
ௗ
Allowable l = 60b = 13800 mm = 13.8 m
Or
ଶହ଴ ௕మ
ௗ
= 52900 mm = 52.9 m
Allowable l = Lesser of the two values
= 13.8 m
Actual l of the beam (5m) < Allowable value of l. Hence ok

Step 2: Effective span
Referring class 22.2 page 34,
Effective span le = clear span + d
Or le = clear span +
ଵ
ଶ
support thickness +
ଵ
ଶ
support thickness
= clear span +
௧ೞ
ଶ
+
௧ೞ
ଶ
Whichever is lesser.
le = 5000 + 250 mm = 5250 mm
Or le = 5000 +
ଶଷ଴
ଶ
+
ଶଷ଴
ଶ
= 5230 mm
Therefore le = 5230 mm
Step 3: Calculation of loads:
Consider 1m length of the beam
a. Dead load = (0.23 x 0.275 x 1m x 25 kN/m3
) = 1.58 kN/m
b. Live load = 25 kN/m
Total working load w = 26.58 kN/m
Factored load = 26.58 x 1.5
Wu = 39.87 kN/m ≈ 40 kN/m
Factored moment Mu =
ௐೠ × ௟೐
మ
଼
=
ସ଴ ×ହ.ଶଷమ
଼
= 136.76 kN-m
Factored shear =
ସ଴ ×ହ.ଶଷ
ଶ
= 104.6 kN
Step 4: Check for depth based on flexure or bending moment consideration
Assuming the section to be nearly balanced, and equating Mu to Mulim,
Mu = Mulim = 136.76 kN-m
Using the equation G 1.1 (c), Annexure G IS 456-2000
‫ܯ‬௨௟௜௠ = 0.36
௫ೠ೘ೌೣ
ௗ
ቀ1 − 0.42
௫ೠ೘ೌೣ
ௗ
ቁ ܾ݀ଶ
݂௖௞
136.76 × 10଺
= 0.36 × 0.48 ሺ1 − 0.42 × 0.48ሻ230݀ଶ
× 20
d = 464.21 mm

Assumed depth d is less than the required depth of 464 mm. Hence revise the section
Assume
d = 500 mm
b = 230 mm
Loads:
Dead load = 0.23 x 0.525 x 1 x 25 = 2.875 kN/m
Live load = 25 kN/n
Total working load = 27.875 kN/m
Factored load = 27.875 x 1.5 = 41.8 ≈ 42 kN/m
ௐೠ × ௟೐
మ
଼
=
ସଶ ×ହ.ଶଷమ
଼
= 143.6 kN-m
Factored shear =
ସଶ ×ହ.ଶଷ
ଶ
= 109.83 kN
Check for depth based on flexure
Using the equation G 1.1 (c)
‫ܯ‬௨௟௜௠ = 0.36
௫ೠ೘ೌೣ
ௗ
ቀ1 − 0.42
௫ೠ೘ೌೣ
ௗ
ቁ ܾ݀ଶ
݂௖௞
143.6 × 10଺
= 0.36 × 0.48 ሺ1 − 0.42 × 0.48ሻ230݀ଶ
× 20
d = 475.68 mm
Assumed depth is greater than the required depth of 475.68 mm.
Required ‘d’ = 476 mm and Assumed ‘d’ = 500 mm
Hence ok.
Therefore we shall continue with d = 500 mm and D = 525 mm
Check whether the section is under reinforced
Actual moment acting Mu = 143.6 kM-m
Using equation G 1.1 (c)
‫ܯ‬௨௟௜௠ = 0.36
௫ೠ೘ೌೣ
ௗ
ቀ1 − 0.42
௫ೠ೘ೌೣ
ௗ
ቁ ܾ݀ଶ
݂௖௞

‫ܯ‬௨௟௜௠ = 0.36 × 0.48 ሺ1 − 0.42 × 0.48ሻ230 × 500ଶ
× 20
= 158.66 kN-m
Mu < Mulim
Hence the section is under reinforced
Step 5: Calculation of steel:
Since the section is under reinforced we have,
Using equation G 1.1 (b)
‫ܯ‬௨ = 0.87݂௬ ‫ܣ‬௦௧ ݀ ቀ1 −
஺ೞ೟௙೤
௕ௗ ௙೎ೖ
ቁ
143.6 × 10଺
= 0.87 × 415 × ‫ܣ‬௦௧ × 500 ቀ1 −
஺ೞ೟×ସଵହ
ଶଷ଴ ×ହ଴଴ ×ଶ଴
ቁ
Solving the quadratic equation, Ast = 960.33 mm2
≈ 960 mm2
Choosing 8 mm diameter bars,
Area of 1 bar =
గ
ସ
× 8ଶ
= 50.265 mm2
Therefore number of bars of 8mm required = 19.10 = 20 bars
Distance between any two bars
Minimum distance between two bars is greater of the following:
a. Size of the aggregate + 5 mm
20 mm + 5 mm
b. Size of the bar (whichever is greater)
Therefore minimum distance = 25 mm
Distance between bars =
ଶଷ଴ିଶ×ଶହିଶ×଼
ଵଽ
= 8.63
1.63 < 25. Therefore 8 mm dia bars cannot be provided.
Let us choose 16 mm dia bars.
Area of 1 bar =
గ
ସ
× 16ଶ
= 201.06 mm2
Therefore number of bars of 16 mm required = 4.77 = 5 bars
Distance between bars =
ଶଷ଴ିଶ×ଶହିଶ×଼ିହ×ଵ଺
ସ
= 21 mm

Minimum distance required = 25 mm
Therefore 16 mm dia cannot be used.
Let us choose 25 mm dia bars.
Area of 1 bar =
గ
ସ
× 25ଶ
= 490.890 mm2
Therefore number of bars of 25mm required = 1.95 = 2 bars
Distance between the bars =
ଶଷ଴ିଶ×ଶହିଶ×଼ିଶ×ଶହ
ଵ
= 114 mm
Check for Ast min
‫ܣ‬௦௧ ௠௜௡ =
଴.଼ହ௕ௗ
଴.଼଻௙೤
‫ܣ‬௦௧ ௠௜௡ =
଴.଼ହ ×ଶଷ଴ ×ହ଴଴
଴.଼଻ ×ସଵହ
= 270.7 mm2
Check for Ast max
Ast max = 0.04 x b x D = 4830 mm2
Ast provided = 982 mm2
Ast min < Ast < Ast max
Hence ok.
Check for shear
Factored load = 42 kN/m
Support reaction =
௪௟
ଶ
=
ସଶ ×ହ
ଶ
= 105 kN
Vu = 105 kN
߬௩ =
௏ೠ
௕ௗ
= 0.913 N/mm2
ܲ௧ =
ଵ଴଴஺ೞ೟
௕ௗ
=
ଵ଴଴×ଽ଼ଶ
ଶଷ଴×ହ଴଴
= 0.8539
From table 19, IS 456-2000 page 73
߬௖ = 0.58 N/mm2
From table 20, IS 456-2000 page 73
߬௖ ௠௔௫ = 2.8 N/mm2

߬௖ < ߬௩ < ߬௖ ௠௔௫
Hence design of shear reinforcement is required
Selecting 2 leg vertical stirrups of 8 mm diameter, Fe 415 steel,
‫ܣ‬௦௩ = 2 ×
గ
ସ
× 8ଶ
= 100 mm2
Vc = Shear force taken up by the concrete
=
ఛ೎௕ௗ
ଵ଴଴଴
=
଴.ଶ଼×ଶଷ଴×ହ଴଴
ଵ଴଴଴
= 66.7 kN
Vu = 105 kN
Vus = Vu - Vc
= 105 – 66.7 = 38.3 kN
Vus =
଴.଼଻×௙೤×஺ೞೡ×ௗ
ௌೡ
from clause 40.4
38.3 x 103
=
଴.଼଻×ସଵହ×ଵ଴଴×ହ଴଴
ௌೡ
Sv = 471.3 mm
Check for maximum spacing
Maximum spacing = 0.75d or 300mm whichever is lesser
Maximum spacing = 375 or 300mm
Therefore maximum spacing allowed = 300mm
Let us provide 8 mm dia 2-leg vertical stirrups at a spacing of 300 mm.
Check for Asv min:
Asv provided = 100 mm2
‫ܣ‬௦௩ ௠௜௡ =
଴.ସ௕ௌೡ
଴.଼଻௙೤
= 76.44 mm2
Asv provided > Asv min
Hence ok.
Check for deflection:
Allowable
௟
ௗ
= Basic
௟
ௗ
x Mt x Mc x Mf

Basic
௟
ௗ
= 20 as the beam is simply supported
To determine Mt
fs = 0.58 × 415 ×
ଽ଺଴
ଽ଼ଶ
= 235.3 N/mm2
from fig 4, Mt = 1
To determine Mc
From fig 5, Mc = 1 [since there is no compression reinforcement]
To determine Mf
௕ೢ
௕೑
= 1 [since it is rectangular section bw = bf]
Therefore allowable l/d = 20 x 1 x 1 x 1 = 20
Actual l/d =
ହଶଷ଴
ହ଴଴
= 10.46 < Allowable l/d.
Hence ok.
2. Design a cantilever beam of clear span 2m subjected to a factored live load of 30 kN/m
run. Use M 20 grade concrete and Fe 415 steel. The cantilever forms the end of a
continuous beam. Support thickness = 230mm
Step 1 (a):Fixing up the depth of the section.
Taking
௅
ௗ
= 7, for cantilever [Refer 23.2.1, pg 37]
݀ =
௅
଻
=
ଶ଴଴଴
଻
= 285.7 mm
However provide d = 450 mm
Providing a cover of 25 mm, overall depth D = 450 + 25 = 475 mm
Dimensions of the section.
Width b = 230 mm
Depth d = 450 mm
Step 1 (b): Check for lateral stability/lateral buckling
Refer page 39, class 23.3
Allowable l = 25b or
ଵ଴଴ ௕మ
ௗ
Allowable l = 25b = 5750 mm = 5.75 m

Or
ଵ଴଴ ௕మ
ௗ
= 11750 mm = 11.75 m
Allowable l = Lesser of the two values
= 5.75 m
Actual l of the beam (5m) < Allowable value of l.
Hence ok
Step 2: Effective span
Referring class 22.2 page 34,
Effective span le = clear span + d
Or le = clear span +
ଵ
ଶ
support thickness
= clear span +
௧ೞ
ଶ
Whichever is lesser.
Le = 2 m + 450 mm = 2450 mm
Or Le = 2 m +
ଶଷ଴
ଶ
= 2115 mm
Therefore le = 2115 mm
Step 3: Calculation of loads
Consider 1m length of the beam
a. Dead load = (0.23 x 0.475 x 1m x 25 kN/m3
)x1.5 = 4.096 ≈ 4.1 kN/m
b. Factored live load = 30 kN/m
Total Factored load Wu = 34.1 kN/m ≈ 35 kN/m
ௐೠ × ௟೐
మ
ଶ
=
ଷହ ×ଶ.ଵଵହమ
ଶ
= 78.28 kN-m
Factored shear = 35 x 2.115 = 74.025 kN
Step 4: Check for depth based on flexure or bending moment consideration
Assuming the section to be nearly balanced, and equating Mu to Mulim,

Using the equation G 1.1 (c)
‫ܯ‬௨௟௜௠ = 0.36
௫ೠ೘ೌೣ
ௗ
ቀ1 − 0.42
௫ೠ೘ೌೣ
ௗ
ቁ ܾ݀ଶ
݂௖௞
78.28 × 10଺
= 0.36 × 0.48 ሺ1 − 0.42 × 0.48ሻ230݀ଶ
× 20
d = 222 mm
dassumed > drequired
Hence ok.
Step 5: Calculation of steel
Since the section is under reinforced we have,
Using equation G 1.1 (b)
‫ܯ‬௨ = 0.87݂௬ ‫ܣ‬௦௧ ݀ ቀ1 −
஺ೞ೟௙೤
௕ௗ ௙೎ೖ
ቁ
78.28 × 10଺
= 0.87 × 415 × ‫ܣ‬௦௧ × 450 ቀ1 −
஺ೞ೟×ସଵହ
ଶଷ଴ ×ସହ଴ ×ଶ଴
ቁ
Solving the quadratic equation, Ast = 540.33 mm2
≈ 540 mm2
Choosing 16 mm diameter bars,
Area of 1 bar =
గ
ସ
× 16ଶ
= 201.06 mm2
Therefore number of bars of 8mm required = 2.69= 3 bars
Distance between any two bars
Minimum distance between two bars is greater of the following:
a. Size of the aggregate + 5 mm
20 mm + 5 mm
b. Size of the bar (whichever is greater)=16mm
Therefore minimum distance = 25 mm
Distance between the bars =
ଶଷ଴ିଶ×ଶହିଶ×ଵ଺ିଶ×଼
ଶ
= 58mm
Distance provided = 58mm > Minimum distance 25mm
Hence ok.

Check for Ast min
‫ܣ‬௦௧ ௠௜௡ =
଴.଼ହ௕ௗ
଴.଼଻௙೤
‫ܣ‬௦௧ ௠௜௡ =
଴.଼ହ ×ଶଷ଴ ×ସହ଴
଴.଼଻ ×ସଵହ
= 243.66 mm2
Ast provided = 3 x
గ
ସ
x 162
= 603.18 mm2
> Ast min
Hence ok.
Check for Ast max
Ast max = 0.04 x b x D = 4370 mm2
Ast provided = 603.18 mm2
Ast min < Ast < Ast max
Hence ok.
Check for shear
Vu = 74.025 kN
߬௩ =
௏ೠ
௕ௗ
= 0.715 N/mm2
ܲ௧ =
ଵ଴଴஺ೞ೟
௕ௗ
=
ଵ଴଴×଺଴ଷ.ଵ଼
ଶଷ଴×ସହ଴
= 0.58
From table 19,
߬௖ = 0.51 N/mm2
From table 20,
߬௖ ௠௔௫ = 2.8 N/mm2
߬௖ < ߬௩ < ߬௖ ௠௔௫
Hence design of shear reinforcement is required
Selecting 2 leg vertical stirrups of 8 mm diameter, Fe 415 steel,
‫ܣ‬௦௩ = 2 ×
గ
ସ
× 8ଶ
= 100 mm2
Vc = Shear force taken up by the concrete
=
ఛ೎௕ௗ
ଵ଴଴଴
=
଴.ହଵ×ଶଷ଴×ସହ଴
ଵ଴଴଴
= 52.78 kN

Vu = 74.025 kN
Vus = Vu - Vc
= 74.025 – 52.785 = 21.24 kN
Vus =
଴.଼଻×௙೤×஺ೞೡ×ௗ
ௌೡ
21.24 x 103
=
଴.଼଻×ସଵହ×ଵ଴଴×ସହ଴
ௌೡ
Sv = 764 mm
Check for maximum spacing
Maximum spacing = 0.75d or 300mm whichever is lesser
Maximum spacing = 337.5 or 300mm
Therefore maximum spacing allowed = 300mm
Let us provide 8 mm dia 2-leg vertical stirrups at a spacing of 300 mm.
Check for Asv min:
Asv provided = 100 mm2
‫ܣ‬௦௩ ௠௜௡ =
଴.ସ௕ௌೡ
଴.଼଻௙೤
= 76.44 mm2
Asv provided > Asv min
Hence ok.
Check for deflection:
Allowable
௟
ௗ
= Basic
௟
ௗ
x Mt x Mc x Mf
Basic
௟
ௗ
= 7 as the beam is cantilever
From fig 4, Mt = 1.2
From fig 5, Mc = 1
From fig 6,
௕ೢ
௕೑
= 1 [Since it is rectangular section bw = bf]
Therefore allowable l/d = 7 x 1.2 x 1 x 1 = 8.4
Actual l/d =
ଶଵଵହ
ସହ଴
= 4.7 < Allowable l/d. Hence ok.

3. Design a reinforced concrete beam of rectangular section using the following data:
Effective span = 5 m
Width of beam = 250 mm
Overall depth = 500 mm
Service load (DL+LL) = 40 kN/m
Effective cover = 50 mm
Materials : M20 grade concrete and Fe 415 steel
a. Data
b = 250 mm fck = 20 N/mm2
D = 500 mm fy = 415 N/mm2
d = 450 mm Es = 2 x 105
N/mm2
d’= 50 mm
le = 5 m
w = 40 kN/m and Wu = 40 x 1.5 =60 kN/m
b. Ultimate moments and shear forces
Mu =
ௐೠ × ௟೐
మ
଼
=
଺଴ ×ହమ
଼
= 187.5 kN-m
Vu = Factored shear =
ܹ‫ ݑ‬× ݈݁
ଶ
= 150 kN
c. Determination of Mulim and fsc
‫ܯ‬௨௟௜௠ = 0.36
௫ೠ೘ೌೣ
ௗ
ቀ1 − 0.42
௫ೠ೘ೌೣ
ௗ
ቁ ܾ݀ଶ
݂௖௞
‫ܯ‬௨௟௜௠ = 0.36 × 0.48 ሺ1 − 0.42 × 0.48ሻ250 × 450ଶ
× 20
= 140 kN.m
Since Mu > Mu lim, design a doubly reinforced section
(Mu – Mu lim) = 187.5-140 = 47.5 kN.m
݂௦௖ = ߳௦௖ × ‫ܧ‬௦
Where, ∈௦௖= ቄ
଴.଴଴ଷହሺ௫ೠ ೘ೌೣିௗᇲሻ
௫ೠ ೘ೌೣ
ቅ
fsc = ቄ
଴.଴଴ଷହሺ௫ೠ ೘ೌೣିௗᇲሻ
௫ೠ ೘ೌೣ
ቅ ‫ܧ‬௦

= ቄ
଴.଴଴ଷହ[ሺ଴.ସ଼ ×ସହ଴ሻିହ଴]
଴.ସ଼ ×ସହ଴
ቅ 2 × 10ହ
= 538 N/mm2
But fsc ≯ 0.87fy = (0.87 x 415) = 361 N/mm2
Therefore fsc = 361 N/mm2
steel Asc = ቂ
ሺெೠିெೠ ೗೔೘ሻ
௙ೞ೎ሺௗିௗᇲሻ
ቃ
= ቂ
ሺସ଻.ହ ×ଵ଴లሻ
ଷ଺ଵ ×ସ଴଴
ቃ = 329 mm2
Provide 2 bars of 16mm diameter (Asc = 402 mm2
)
‫ܣ‬௦௧ଶ = ൬
஺ೞ೎௙ೞ೎
଴.଼଻௙೤
൰ = ቀ
ଷଶଽ ×ଷ଺ଵ
଴.଼଻×ସଵହ
ቁ = 329 mm2
‫ܣ‬௦௧ଵ = ൤
଴.ଷ଺௙೎ೖ௕௫ೠ ೗೔೘
଴.଼଻௙೤
൨
= ቂ
଴.ଷ଺×ଶ଴×ଶହ଴×଴.ସ଼×ସହ଴
଴.଼଻×ସଵହ
ቃ = 1077 mm2
Total tension reinforcement = Ast = (Ast1 + Ast2)
= (1077 + 329)
= 1406 mm2
Provide 3 bars of 25mm diameter (Ast = 1473 mm2
)
d. Shear reinforcements
߬௩ = ሺܸ௨ ܾ݀⁄ ሻ = (150 x 103
) / (250 x 450) = 1.33 N/mm2
ܲ௧ =
ሺଵ଴଴஺ೞሻ
௕ௗ
=
ଵ଴଴ ×ଵସ଻ଷ
ଶହ଴ ×ସହ଴
= 1.3
Referring table 19 of IS : 456 – 2000 ,
߬௖ = 0.68 N/mm2
߬௖௠௔௫ = 2.8 N/mm2
for M20 concrete from table 20 of IS 456-2000
Since ߬௖ < ߬௩ < ߬௖௠௔௫ , shear reinforcements are required.
Vus = [Vu – (߬௖bd)]
= [150-(0.68 x 250 x 450)10-3
] = 73.5 kN

Using 8 mm diameter 2 legged stirrups,
ܵ௩ =
0.87×݂‫ݕ‬×‫ܣ‬‫ݒݏ‬×݀
ܸ‫ݏݑ‬
=
଴.଼଻×ସଵହ×ଶ×ହ଴×ସହ଴
଻ଷ.ହ×ଵ଴య
= 221mm
Maximum spacing is 0.75d or 300 mm whichever is less
ܵ௩ > 0.75݀ = ሺ0.75 × 450ሻ = 337.5 mm
Adopt a spacing of 200 mm near supports gradually increasing to 300 mm towards the centre
of the span.
e. Check for deflection control
(l/d)actual = (5000/450) = 11.1
(l/d)allowable= [(l/d)basic x Mt x Mc x Mf]
Pt = 1.3 and Pc = [(100 x 402) / (250 x 450)] = 0.35
Refer Fig 4, Mt = 0.93
Fig 5, Mc = 1.10
Fig 6, Mf = 1.0
(l/d)allowable=[(20 x 0.93 x 1.10 x 1] = 20.46
(l/d)actual < (l/d)allowable
Hence deflection control is satisfied.
f. Reinforcement details

4. A tee beam slab floor of an office comprises of a slab 150 mm thick spanning between
ribs spaced at 3m centres. The effective span of the beam is 8 m. Live load on floor is
4 kN/m2
. Using M-20 grade concrete and Fe-415 HYSD bars, design one of the
intermediate tee beam.
a. Data
L = 8 m spacing of the tee beam = 8 m
Df = 150 mm fck = 20 N/mm2
Live load on slab = 4 kN/m2
fy = 415 N/mm2
b. Cross sectional dimensions
Assume
௟
ௗ
= 16
Therefore d = 500mm and D = 550 mm
Hence the tee beam parameters are:
d = 500 m
D = 550 mm
bw = 300 mm Df = 150 mm
c. Loads
Self weight of slab = (0.15 x 25 x 3) = 11.25 kN/m
Floor finish = (0.6 x 3) = 1.80
Self weight of rib = (0.3 x 0.4 x 25) = 3.00
Plaster finishes = ………………. = 0.45
Total dead load = = 16.50 kN/m
Live load = = 4.00 kN/m
Design ultimate load Wu = 1.5(16.50+4.0) = 30.75 kN/m
d. Ultimate moments and shear forces:
Mu =
ௐೠ ×௟మ
଼
=
ଷ଴.଻ହ ×଼మ
଼
= 246 kN-m
Vu =
ܹ‫ ݑ‬× ݈
ଶ
= 123 kN
e. Effective width of flange
i. bf = [(L0/6) + bw + 6 Df]
= [(8/6) + 0.3 + (6 x 0.15)]

= 2.53 m
= 2530 mm
ii. Centre to centre of ribs = (3-0.3) = 2.7 m
Hence the least of i and ii is bf = 2530 mm
f. Moment capacity of flange
Muf = 0.36 fck bf Df (d-0.42 Df)
= 0.36 x 20 x 2530 x 150 (500-0.42 x 150)
= 1194 x 106
N.mm
= 1194 kNm
Since Mu < Muf , xu < Df
Hence the section is considered as rectangular with b = bf
g. Reinforcements
‫ܯ‬௨ = 0.87݂௬ ‫ܣ‬௦௧ ݀ ൬1 −
஺ೞ೟௙೤
௕೑ௗ ௙೎ೖ
൰
246 × 10଺
= 0.87 × 415 ‫ܣ‬௦௧ × 500 ቀ1 −
ସଵହ஺ೞ೟
ଶହଷ଴×ହ଴଴×ଶ଴
ቁ
Solving, Ast = 1417 mm2
Provide 3 bars of 25 mm diameter (Ast = 1473 mm2
) and two hanger bars of 12 mm diameter
on the compression face.
h. Shear reinforcements
߬௩ = ሺܸ௨ ܾ௪݀⁄ ሻ = (123 x 103
) / (300 x 500) = 0.82 N/mm2
ܲ௧ =
ሺଵ଴଴஺ೞ೟ሻ
௕ೢௗ
=
ଷ଴଴×ହ଴଴
= 0.98
Referring table 19 of IS : 456 – 2000 ,
߬௖ = 0.60 N/mm2
Since ߬௩ > ߬௖ , shear reinforcements are required.
Vus = [Vu – (߬௖bwd)]
= [123-(0.60 x 300 x 500)10-3
] = 33 kN
Using 8 mm diameter 2 legged stirrups,

ܵ௩ =
0.87×݂‫ݕ‬×‫ܣ‬‫ݒݏ‬×݀
ܸ‫ݏݑ‬
=
଴.଼଻×ସଵହ×ଶ×ହ଴×ହ଴଴
ଷଷ×ଵ଴య
= 547mm
Maximum spacing is 0.75d or 300 mm whichever is less
ܵ௩ > 0.75݀ = ሺ0.75 × 500ሻ = 375 mm
Hence provide 8 mm diameter 2 legged stirrups at 300 mm centres throughout the length of
the beam.
i. Check for deflection control
ܲ௧ =
௕ೢௗ
=
ଷ଴଴×ହ଴଴
= 0.98
൬
௕ೢ
௕೑
൰ = ቀ
ଷ଴଴
ଶହଷ଴
ቁ = 0.118
Refer Fig 4, and read out Mt = 2.00
Fig 5, and read out Mc = 1.00
Fig 6, and read out Mf = 0.80
(L/d)max= [(L/d)basic x Mt x Mc x Mf]
= [16 x 2 x 1 x 0.8] = 25.6
(L/d) provided = (8000/500) = 16 < 25.6
Hence deflection control is satisfied
j. Reinforcement details

5. Design a L beam for an office floor to suit the following data:
Data
Clear span = L = 8m
Thickness of flange = Df = 150 mm
Live load on the slab = 4 kN/m2
Spacing of beams = 3 m
fck = 20 N/mm2
fy = 415 N/mm2
L-beams are monolithic with R.C columns
Width of column = 300 mm
Cross sectional dimensions
Since L-beam is subjected to bending, torsion and shear forces, assume a trial section having
span/depth ratio of 12.
Therefore ‘d’ = (8000/12) = 666 mm
Adopt d = 700 mm
D = 750 mm
bw = 300 mm
Effective span
Effective span is least of
i. Centre to centre of supports = (8+0.3) = 8.3 m
ii. Clear span + effective depth = (8+0.7) = 8.7 m
Therefore L = 8.3 m
Loads
Self weight of slab = (0.15 x 25 x0.5 x 3) = 11.25 kN/m
Floor finish = (0.6 x0.5x 3) = 1.80
Self weight of rib = (0.3 x 0.6 x 25) = 3.00
Live load = (4 x 0.5 x 3) = 6.00 kN/m
Total working load = w = 17 kN/m

Effective flange width
Effective flange width bf is least of the following values:
i. bf = (L0/12) + bw + 3Df
= (8000/12) + 300 + (3x150) = 1442
ii. bf = bw + 0.5 times the spacing between the ribs
= 300 + (0.5x2700) = 1650 mm
Therefore bf = 1442 mm
Ultimate bending and shear force
At support section:
Mu = 1.5 (17 x 8.32
) / 12 = 147 kN.m
Vu = 1.5 (0.5 x 17 x 8.3) = 106 kN
At centre of the span section:
Mu = 1.5 (17 x 8.32
) / 24 = 73 kN.m
Torsional moments at support section
Torsional moment is produced due to dead load of slab and live load on it.
(working load/m – rib self weight) = (17-4.50) = 12.50 kN/m
Therefore total ultimate load on slab = 1.5 (12.50 x 8.3) = 156 kN
Total ultimate shear force = (0.5 x 156) = 78 kN
Distance of centroid of shear force from the centre line of the beam
= (0.5 x 1442 – 150 ) = 571 mm = 0.571 m
Ultimate torsional moment = Tu = 78 x 0.571 = 44.5 kN.m
Equivalent bending moment and shear force
According to IS : 456 – 2000, clause 41.4.2, at support section, the equivalent bending
moment is compared as:
Mel = (Mu + Mt)
Where, Mt = Tu = ቂ
ଵାሺ஽ ௕⁄ ሻ
ଵ.଻
ቃ = 44.5 ቂ
ଵାሺ଻ହ଴ ଷ଴଴⁄ ሻ
ଵ.଻
ቃ
= 92 kN.m
Therefore Mel = (147 + 92 ) = 239 kN.m

Ve = Vu + 1.6 (Tu / b)
= 106 + 1.6 (44.5/0.3)
= 334 kN
Main longitudinal reinforcement
Support section is designed as rectangular section to resist the hogging equivalent bending
moment Mel = 239 kN.m
Mu lim = 0.138 fck bd2
= (0.138 x 20 x 300 x 7002
) 10-6
= 405.7 kN.m
Since Mel < Mu lim, the section is under reinforced.
‫ܯ‬௘௟ = 0.87݂௬ ‫ܣ‬௦௧ ݀ ቀ1 −
஺ೞ೟௙೤
௕ௗ ௙೎ೖ
ቁ
239 × 10଺
= 0.87 × 415 ‫ܣ‬௦௧ × 700 ቀ1 −
ସଵହ஺ೞ೟
ଷ଴଴×଻଴଴×ଶ଴
ቁ
Solving, Ast = 1056.3 mm2
Provide 3 bars of 22 mm diameter on the tension side (Ast = 1140 mm2
)
Area of steel required at centre of span to resist a moment of Mu = 73 kN.m will be less than
the minimum given by:
‫ܣ‬௦௧ ሺ୫୧୬ሻ = ൬
଴.଼ହ௕ೢௗ
௙೤
൰ =ቀ
଴.଼ହ×ଷ଴଴×଻଴଴
ସଵହ
ቁ
= 430 mmm2
Provide 2 bars of 20 mm diameter (Ast = 628 mm2
)
Side face reinforcement
According to clause 26.5.1.7 of IS : 456 code, side face reinforcement of 0.1 percent
of web area is to be provided for member subjected to torsion, when the depth exceeds 450
mm.
Therefore area of reinforcement = (0.001 x 300 x 750) = 225 mm2
Provide 10 mm diameter bars (4 numbers) two on each face as horizontal reinforcement
spaced 200 mm centres.

Shear reinforcements
߬௩௘ = ቀ
௏೐
௕ೢௗ
ቁ = ቀ
ଷସସ×ଵ଴య
ଷ଴଴×଻଴଴
ቁ = 1.63 N/mmm2
ܲ௧ =
௕ೢௗ
=
ଵ଴଴ ×ଵଵସ଴
ଷ଴଴×଻଴଴
= 0.542
From table 19 IS:456 read out,
߬௖ = 0.49 N/mmm2
< ߬‫݁ݒ‬
Hence shear reinforcements are required.
Using 10 mm diameter two legged stirrups with side covers of 25 mm and top and bottom
covers of 50 mm, we have b1 = 250 mm, d1 = 650 mm, Asv = (2 x 78.5) = 157 N/mmm2
The spacing Sv is computed using the equations specified in clause 451.4.3 of IS : 456-2000
code.
ܵ௩ = ቈ
଴.଼଻௙೤஺ೞೡௗభ
ቀ
೅ೠ
್భ
ቁା ቀ
ೇೠ
మ.ఱ
ቁ
቉
= ൥
଴.଼଻×ସଵହ×ଵହ଻×଺ହ଴
൬
రర.ఱ×భబల
మఱబ
൰ା ൬
భబల×భబయ
మ.ఱ
൰
൩
= 167 mm
OR
ܵ௩ = ቂ
଴.଼଻஺ೞೡ௙೤
ሺఛೡ೐ିఛ೎ሻ௡
ቃ
= ቂ
଴.଼଻×ଵହ଻×ସଵହ
ሺଵ.଺ଷି଴.ସଽሻଷ଴଴
ቃ
= 165 mm
Provide 10 mm diameter two legged stirrups at a minimum spacing given by clause 26.5.1.7
of IS 456
Adopt a minimum spacing based on shear and torsion computations computed as Sv = 160
mm.

Check for deflection control
ܲ௧ =
ଵ଴଴ ×ଵଵସ଴
ଷ଴଴×଻଴଴
= 0.54
ܲ௖ =
ଵ଴଴ ×଺ଶ଼
ଷ଴଴×଻଴଴
= 0.299
൬
௕ೢ
௕೑
൰ = ቀ
ଷ଴଴
ଵସସଶ
ቁ = 0.208
Refer Fig 4, and read out Mt = 1.20
Fig 5, and read out Mc = 1.10
Fig 6, and read out Mf = 0.80
(L/d)max= [(L/d)basic x Mt x Mc x Mf]
= [9.20 x 1.2 x 1.1 x 0.8] = 21.12
(L/d) provided = (8300/700) = 11.85 < 21.12
Hence deflection control is satisfied.
Reinforcement details

Design of beam

More Related Content

What's hot (20)

Similar to Design of beam (20)

Recently uploaded (20)

Design of beam