CSI ETABS & SAFE MANUAL: Slab Analysis and Design to EC2

CSI ETABS & SAFE MANUAL
Part‐III: Model Analysis & Design of Slabs
According to Eurocode 2
AUTHOR: VALENTINOS NEOPHYTOU BEng (Hons), MSc
REVISION 2: August, 2014

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ABOUT THIS DOCUMENT
This document presents an example of analysis design of slab using ETABS.
This example examines a simple single story building, which is regular in plan
and elevation. It is examining and compares the calculated ultimate moment
from CSI ETABS & SAFE with hand calculation. Moment coefficients were
used to calculate the ultimate moment. However it is good practice that such
hand analysis methods are used to verify the output of more sophisticated
methods.
Also, this document contains simple procedure (step-by-step) of how to
design solid slab according to Eurocode 2.The process of designing elements
will not be revolutionised as a result of using Eurocode 2.
Due to time constraints and knowledge, I may not be able to address the
whole issues.
Please send me your suggestions for improvement. Anyone interested to
share his/her knowledge or willing to contribute either totally a new section
about ETABS or within this section is encouraged.
For further details:
My LinkedIn Profile:
http://www.linkedin.com/profile/view?id=125833097&trk=hb_tab_pro_top
Email: valentinos_n@hotmail.com
Slideshare Account:http://www.slideshare.net/ValentinosNeophytou

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TABLE OF CONTENTS
1. SLAB MODELING .................................................................................... 4
2. THEORETICAL CALCULATION OF ULTIMATE MOMENTS ......... 5
3. DESIGN OF SLAB ACCORDING TO EUROCODE 2 ........................... 7
4. WORKED EXAMPLE : ANALYSIS AND DESIGN OF RC SLAB
USING CSI ETABS AND SAFE .............................................................. 11
5. ANALYSIS RESULTS ............................................................................. 17
6. DESIGN THE SLAB FOR FLEXURAL USING MOMENT CAPACITY
VALUES .................................................................................................... 19
ANNEX A - EXAMPLE OF HOW TO DETERMINE THE DESIGN BENDING
MOMENT USING MOMENT COEFFICIENTS...…………………….22
ANNEX B - EXAMPLE OF HOW TO DETERMINE THE MOMENT CAPACITY
OF RC SLAB………………………………………………………..…….28
ANNEX C - EXAMPLE OF DESIGN SLAB PANEL WITH TWO
DISCONTINUOUS EDGES…………..…..………………………..…….32
ANNEX D - EXAMPLE OF DESIGN SLAB PANEL WITH ONE
DISCONTINUOUS EDGES………………………………………..…….48
ANNEX E - EXAMPLE OF DESIGN INTERIOR PANEL SLAB..…………..…….65

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1. SLAB MODELING
1.1 ASSUMPTIONS
In preparing this document a number of assumptions have been made to avoid over
complication; the assumptions and their implications are as follows.
a) Element type : SHELL
b) Meshing (Sizing of element) : Size= min{Lmax/10 or l000mm}
c) Element shape : Ratio= Lmax/Lmin = 1 ≤ ratio ≤ 2
d) Acceptable error : 20%
1.2 INITIAL STEP BEFORE RUN THE ANALYSIS
a) Sketch out by hand the expected results before carrying out the analysis.
b) Calculate by hand the total applied loads and compare these with the sum of
the reactions from the model results.

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2. THEORETICAL CALCULATION OF ULTIMATE MOMENTS
Maximum moments of two-way slabs
If ly/lx<2: Design as a Two-way slab
If lx/ly> 2: Deisgn as a One-way slab
Note: lx is the longer span
ly is the shorter span
2 in
Msx= asxnlx
direction of span lx
Maximum moment of Simply supported (pinned) two-way slab
n: is the ultimate load m2
2 in
Msy= asynlx
direction of span ly
Bending moment coefficient for simply supported slab
ly/lx 1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0
asx 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118
asy 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029
Maximum moment of Restrained supported (fixed) two-way slab
2 in
Msx= asxnlx
direction of span lx
2 in
Msy= asynlx
direction of span ly
Bending moment coefficient for two way rectangular slab supported by beams
(Manual of EC2 ,Table 5.3)
Type of panel and moment
considered
Short span coefficient for value of Ly/Lx Long-span coefficients for all
1.0 1.25 1.5 1.75 2.0 values of Ly/Lx
Interior panels
Negative moment at continuous edge 0.031 0.044 0.053 0.059 0.063 0.032
Positive moment at midspan 0.024 0.034 0.040 0.044 0.048 0.024
One short edge discontinuous
One long edge discontinuous
Two adjacent edges discontinuous

6
Maximum moment of Simply supported (pinned)
one-way slab
(Manual of EC2, Table 5.2)
L: is the effective span
Maximum moments of one-way slabs
If ly/lx<2: Design as a Two-way slab
If lx/ly> 2: Deisgn as a One-way slab
Note: lxis the longer span
lyis the shorter span
MEd= 0.086FL
F: is the total ultimate
load =1.35Gk+1.5Qk
Note: Allowance has been made in the coefficients in
Table 5.2 for 20% redistribution of moments.
Maximum moment of continuous supported one-way
slab
Uniformly distributed loads
End support condition Moment
End support support MEd =-0.040FL
End span MEd =0.075FL
Penultimate support MEd= -0.086FL
Interior spans MEd =0.063FL
Interior supports MEd =-0.063FL
F: total design ultimate load on span
Note: Allowance has been made in the coefficients in
Table 5.2 for 20% redistribution of moments.

Check of the amount of reinforcement provided above the “minimum/maximum amount of
7
3. DESIGN OF SLAB ACCORDING TO EUROCODE 2
FLEXURAL DESIGN
(EN1992-1-1,cl. 6.1)
Determine design yield strength of reinforcement
푓푦푑 =
푓푦푘
훾푠
Determine K from:
퐾 =
푀퐸푑
푏푑2푓푐푘
퐾′ = 0.6훿 − 0.18훿2 − 0.21
K<K′ (no compression reinforcement required)
Obtain lever arm z:푧 =
푑
2
1 + 1 − 3.53퐾 ≤ 0.95푑
K>K′(then compression reinforcement required –
not recommended for typical slab)
Obtain lever arm z:푧 =
푑
2
1 + 1 − 3.53퐾′ ≤ 0.95푑
δ=1.0 for no redistribution
δ=0.85 for 15% redistribution
δ=0.7 for 30% redistribution
퐴푠.푟푒푞 =
푀퐸푑
푓푦푑 푧
퐴푠푥 .푟푒푞 =
푀퐸푑 ,푠푥
푓푦푑 푧
퐴푠푦 .푟푒푞 =
푀퐸푑 ,푠푦
푓푦푑 푧
Area of steel reinforcement required:
One way solid slab Two way solid slab
For slabs, provide group of bars with area As.prov per meter width
Spacing of bars (mm)
75 100 125 150 175 200 225 250 275 300
Bar
Diameter
(mm)
8 670 503 402 335 287 251 223 201 183 168
10 1047 785 628 524 449 393 349 314 286 262
12 1508 1131 905 754 646 565 503 452 411 377
16 2681 2011 1608 1340 1149 1005 894 804 731 670
20 4189 3142 2513 2094 1795 1571 1396 1257 1142 1047
25 6545 4909 3927 3272 2805 2454 2182 1963 1785 1636
32 10723 8042 6434 5362 4596 4021 3574 3217 2925 2681
For beams, provide group of bars with area As. prov
Number of bars
1 2 3 4 5 6 7 8 9 10
Bar
Diameter
(mm)
8 50 101 151 201 251 302 352 402 452 503
10 79 157 236 314 393 471 550 628 707 785
12 113 226 339 452 565 679 792 905 1018 1131
16 201 402 603 804 1005 1206 1407 1608 1810 2011
20 314 628 942 1257 1571 1885 2199 2513 2827 3142
25 491 982 1473 1963 2454 2945 3436 3927 4418 4909
32 804 1608 2413 3217 4021 4825 5630 6434 7238 8042
퐴푠,푚푖푛 =
(CYS NA EN1992-1-1, cl. NA 2.49(1)(3))
0.26푓푐푡푚 푏푑
푓푦푘
reinforcement “limit
≥ 0.0013푏푑 ≤ 퐴푠,푝푟표푣 ≤ 퐴푠,푚푎푥 = 0.04퐴푐

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SHEAR FORCE DESIGN
(EN1992-1-1,cl 6.2)
Maximum moment of Simply supported (pinned)
(Manual of EC2, Table 5.2)
MEd= 0.4F
one-way slab
F: is the total ultimate
load =1.35Gk+1.5Qk
Maximum shear force of continuous supported
one-way slab
Uniformly distributed loads
End support condition Moment
End support support MEd =0.046F
Penultimate support MEd= 0.6F
Interior supports MEd =0.5F
F: total design ultimate load on span
Determine design shear stress, vEd
vEd=VEd/b·d
Reinforcement ratio, ρ1 (EN1992-1-1, cl 6.2.2(1))
ρ1=As/b·d
푘 = 1 +
200
푑
Design shear resistance
≤ 2,0with 푑 in mm
푉푅푑 .푐 =
0.18
훾푐
푘 100휌1푓푐푘
1
3 + 푘1휎푐푝 푏푑
푉푅푑 .푐 .푚푖푛 = 0.0035 푓푐푘 푘1.5 + 푘1휎푐푝 푏푑
Alternative value of design shear resistance, VRd.c (Concrete centre) (ΜΡa)
ρI =
Effective depth, d (mm)
As/(bd)
≤200 225 250 275 300 350 400 450 500 600 750
0.25% 0.54 0.52 0.50 0.48 0.47 0.45 0.43 0.41 0.40 0.38 0.36
0.50% 0.59 0.57 0.56 0.55 0.54 0.52 0.51 0.49 0.48 0.47 0.45
0.75% 0.68 0.66 0.64 0.63 0.62 0.59 0.58 0.56 0.55 0.53 0.51
1.00% 0.75 0.72 0.71 0.69 0.68 0.65 0.64 0.62 0.61 0.59 0.57
1.25% 0.80 0.78 0.76 0.74 0.73 0.71 0.69 0.67 0.66 0.63 0.61
1.50% 0.85 0.83 0.81 0.79 0.78 0.75 0.73 0.71 0.70 0.67 0.65
1.75% 0.90 0.87 0.85 0.83 0.82 0.79 0.77 0.75 0.73 0.71 0.68
≥2.00% 0.94 0.91 0.89 0.87 0.85 0.82 0.80 0.78 0.77 0.74 0.71
k 2.000 1.943 1.894 1.853 1.816 1.756 1.707 1.667 1.632 1.577 1.516
Table derived from: vRd.c=0.12k(100ρIfck)1/3≥0.035k1.5fck
0.5
where k=1+(200/d)0.5≤0.02
If VRdc≥VEd≥VRdc.min, Concrete strut is adequate in resisting shear
stress
Shear reinforcement is not required in slabs

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DESIGN FOR CRACKING
(EN1992-1-1,cl.7.3)
Asmin<As.prov
Minimum area of reinforcement steel
within tensile zone
(EN1992-1-1,Eq. 7.1)
퐴푠.푚푖푛 =
푘푘푐 푓푐푡 ,푒푓푓 퐴푐푡
휎푠
Chart to calculate unmodified steel stress σsu
(Concrete Centre - www.concretecentre.com)
Crack widths have an influence on the durability of the RC member. Maximum crack width
sizes can be determined from the table below (knowing σs, bar diameter, and spacing).
Maximum bar diameter and maximum spacing to limit crack widths
(EN1992-1-1,table7.2N&7.3N)
σs
(N/mm2)
Maximum bar diameter and spacing for
maximum crack width of:
0.2mm 0.3mm 0.4mm
160 25 200 32 300 40 300
200 16 150 25 250 32 300
240 12 100 16 200 20 250
280 8 50 12 150 16 200
300 6 - 10 100 12 150
Note. The table demonstrates that cracks widths can be reduced if;
 σs is reduced
 Bar diameter is reduced. This mean that spacing is reduced if As.provis to be the
same.
 Spacing is reduced
kc=0.4 for bending k=1 for web
width < 300mm or k=0.65for web >
800mm fct,eff= fctm = tensile strength after 28
days Act=Area of concrete in tension=b (h-
(2.5(d-z))) σs=max stress in steel
immediately after crack initiation
휎푠 = 휎푠푢
퐴푠.푟푒푞
퐴푠.푝푟표푣
1
훿
or 휎푠 = 0.62
퐴푠.푟푒푞
퐴푠.푝푟표푣
푓푦푘

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DESIGN FOR DEFLECTION
(EN1992-1-1,cl.7.4)
Simplified Calculation approach
푙
푑
= 퐾 11 + 1.5 푓푐푘
휌0
휌
+ 3.2 푓푐푘
휌0
휌
− 1
1.5
푖푓휌 ≤ 휌0
푙
푑
= 퐾 11 + 1.5 푓푐푘
휌0
휌 − 휌′ +
1
12
푓푐푘
휌,
휌0
푖푓휌 > 휌0
Span/effective depth ratio
(EN1992-1-1, Eq. 7.16a and 7.16b)
The effect of cracking complicacies the deflection calculations of the RC member under
service load. To avoid such complicate calculations, a limit placed upon the span/effective
depth ration.
Note: The span-to-depth ratios should ensure that deflection is limited to span/250
Structural system modification factor
(CY NA EN1992-1-1,NA. table 7.4N)
The values of K may be reduced to account for long span as follow:
 In beams and slabs where the span>7.0m, multiply by leff/7
Type of member K
Cantilever 0.4
Flat slab 1.2
Simply supported 1.0
Continuous end
span
1.3
Continuous interior
span
1.5
Reference reinforcement
ratio
(EN1992-1-1,cl. 7.4.2(2))
휌0 = 0.001 푓푐푘
Tension reinforcement ratio
(EN1992-1-1,cl. 7.4.2(2))
휌 =
퐴푠.푟푒푞
푏푑

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4. WORKED EXAMPLE : ANALYSIS AND DESIGN OF RC SLAB USING
CSI ETABS AND SAFE
4.1 DIMENSIONS:
Depth of slab, h: h=170mm
Length in longitudinal direction, Ly: Ly=5m
Length in transverse direction, Lx: Lx=5m
Number of slab panels: N=3x3
4.2 LOADS:
Dead load:
Self weight, gk.s: gk.s=4.25kN/m2
Extra dead load, gk.e: gk.e=2.00kN/m2
Total dead load, Gk: Gk=6.25kN/m2
Live load:
Live load, qk: gk=2.00kN/m2
Total live load, Qk: Qk=2.00kN/m2
4.3 LOAD COMBINATION:
Total load on slab: 1.35Gk+1.5Qk=
ULS: 1.35*6.25+1.5*2.00=11.4kN/m2
Total load on slab: 1.35Gk+1.5Qk=
SLS: 1.00*6.25+1.00*2.00=8.25kN/m2

12
4.4 LAYOUT OF MODEL:
Figure 1: Layout of the model

13
4.5 PROCEDURE FOR EXPORTING ETABS MODEL TO SAFE
A very useful and powerful way to start a model in SAFE is to import the model
from ETABS. Floor slabs or basemats that have been modeled in ETABS can be
exported from ETABS.
From that form, the appropriate floor load option can be selected, along with the
desired load cases. After the model has been exported as an .f2k text file, the same
file can then be imported into SAFE using the File menu > Import command.
Using the export and import steps will complete the transfer of the slab geometry,
section properties, and loading for the selected load cases. The design strips need
to be added to the imported model since design strips are not defined as part of the
ETABS model.
ETABS: File > Export > Storey as SAFE
Text File commands saves the specified story level as a SAFE.f2k text input file.
You can later import this file/model into SAFE.
Figure 2: Load to Export to SAFE
Notes:
Model must be analyzed and locked to export.
The export floor loads only option is for individual floor plate design.
The export floor loads and loads from above is used to design foundation.
The export floor loads plus Column and Wall Distortions is necessary only when
displacement compatibility could govern and needs to be checked floor slab
design.(Effects punching shear and flexural reinforcement design).

14
Figure 3: Load cases selection
Figure 4: Load combination selection

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4.6 DRAW DESIGN STRIPS
Use the Draw menu > Draw Design Strips command to add design strips to the
model. Design strips are drawn as lines, but have a width associated with them.
Design strips are typically drawn over support locations (e.g., columns), with a
width equal to the distance between midspan in the transverse direction.
Design strips determine how reinforcing will be calculated and positioned in the
slab. Forces are integrated across the design strips and used to calculate the
required reinforcing.
Typically design strips are positioned in two principal directions: Layer A and
Layer B.
Select the Auto option. The added design strips will automatically adjust their
width to align with adjacent strips.
Figure 5: Design strip for x direction

16
Figure 6: Design strip for y direction
Figure 7: Model after drawing design strip

17
5. ANALYSIS RESULTS
Figure 8: Maximum hogging and Sagging moment at Short span direction Lx
Figure 9: Maximum Shear Force at Short span direction Lx

Figure 10: Maximum hogging and Sagging moment at Long span direction
18
Ly
Figure 11: Maximum Shear Force at Short span direction Ly

6. DESIGN THE SLAB FOR FLEXURAL USING MOMENT CAPACITY
19
VALUES
SAFE: Display > Show slab forces/stresses

20
Figure 12: Bending moment for M11 (Mx – direction) contours displayed
The figure above indicates that the proposed bending reinforcements are adequate to
resist the design moment (hogging & sagging moments).

21
Figure13: Bending moment for M22 (My – direction) contours displayed
The figure above indicates that the proposed bending reinforcements are adequate to
resist the design moment (hogging & sagging moments).

ANNEX A - EXAMPLE OF HOW TO DETERMINE THE DESIGN
22
BENDING MOMENT USING MOMENT COEFFICIENTS

CALUCLATIION
SHEET
BEAM FLEXURAL AND SHEAR
CAPACITY CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
BENDING MOMENT COEFFICIENTS FOR TWO-WAY SPANNING RECTANGULAR SLABS
(Table 5.3, Manual to EC2 - IStrucTE)
GEOMETRICAL DATA:
Shorter effective span of panel (clear span): lx  5000mm
Longer effective span of panel: ly  5000mm
Type of panel and moment considered: Slab_type:= "Interior panel"
Slab_type:= "One short edge discontinuous"
Slab_type:= "One long edge discontinuous"
Slab_type:= "Two adjacent edges discontinuous"
Slab_type  "Two adjacent edges discontinuous"
Ratio of Ly/Lx: Ratio
ly
lx
  1
LOADINGS:
Characteistic permanent action: Gk 6.25kN m 2  
Characteistic variable action: Qk 2kN m 2  
PARTIAL FACTOR FOR LOADS:
Permanent action (dead load) - Ultimate limit state (ULS): γGk.ULS  1.35
Variable action (live load) - Ultimate limit state (ULS): γQk.ULS  1.50
Permanent action (dead load) - Ultimate limit state (SLS): γGk.SLS  1.00
Variable action (live load) - Ultimate limit state (SLS): γQk.SLS  1.00
DESIGN LOADS:
Ultimate design load (ULS): FEd.ULS γGk.ULSGk  γQk.ULSQk 11.438 kN m 2    
Ultimate design load (SLS): FEd.SLS γGk.SLSGk  γQk.SLSQk 8.25 kN m 2    
MOMENT COEFFICIENT:
Short span - Bending moment coefficient for negative moment (hogging moment) at
continuous edge
SEISMIC ASSESSMENT OF
EXISTING RC BUILDING
Page 23 of 27

CALUCLATIION
SHEET
CAPACITY CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
βsx.support 0.031
lx
ly
if  1.0  Slab_type = "Interior panel"
0.044 1.0
ly
lx
if   1.25  Slab_type = "Interior panel"
0.053 1.25
ly
lx
0.059 1.5
ly
lx
0.063 1.75
ly
lx
0.039
lx
ly
if  1.0  Slab_type = "One short edge discontinuous"
0.050 1.0
ly
lx
if   1.25  Slab_type = "One short edge discontinuous"
0.058 1.25
ly
lx
0.063 1.5
ly
lx
0.067 1.75
ly
lx
0.039
lx
ly
if  1.0  Slab_type = "One long edge discontinuous"
0.059 1.0
ly
lx
if   1.25  Slab_type = "One long edge discontinuous"
0.073 1.25
ly
lx
0.082 1.5
ly
lx
0.089 1.75
ly
lx
0.047
lx
ly
if  1.0  Slab_type = "Two adjacent edges discontinuous"
0.066 1.0
ly
lx
if   1.25  Slab_type = "Two adjacent edges discontinuous"
l

Page 24 of 27

CALUCLATIION
SHEET
CAPACITY CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
0.078 1.25
ly
lx
0.087 1.5
ly
lx
0.093 1.75
ly
lx
Short span - Bending moment coefficient for positive moment (sagging moment) at
continuous edge
βsx.midspan 0.024
lx
ly
if  1.0  Slab_type = "Interior panel"
0.034 1.0
ly
lx
0.040 1.25
ly
lx
0.044 1.5
ly
lx
0.048 1.75
ly
lx
0.029
lx
ly
if  1.0  Slab_type = "One short edge discontinuous"
0.038 1.0
ly
lx
0.043 1.25
ly
lx
0.047 1.5
ly
lx
0.050 1.75
ly
lx
0.030
lx
ly
if  1.0  Slab_type = "One long edge discontinuous"
0.045 1.0
ly
lx
0.055 1.25
ly
lx
l

Page 25 of 27

CALUCLATIION
SHEET
CAPACITY CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
0.062 1.5
ly
lx
0.067 1.75
ly
lx
0.036
lx
ly
if  1.0  Slab_type = "Two adjacent edges discontinuous"
0.049 1.0
ly
lx
0.059 1.25
ly
lx
0.065 1.5
ly
lx
0.070 1.75
ly
lx
Long span - Bending moment coefficient for negative moment (hogging moment) at
continuous edge
βsy.support 0.032 if Slab_type = "Interior panel"
0.037 if Slab_type = "One short edge discontinuous"
0.037 if Slab_type = "One long edge discontinuous"
0.045 if Slab_type = "Two adjacent edges discontinuous"

Long span - Bending moment coefficient for positive moment (sagging moment) at
continuous edge
βsy.midspan 0.024 if Slab_type = "Interior panel"
0.028 if Slab_type = "One short edge discontinuous"
0.028 if Slab_type = "One long edge discontinuous"
0.034 if Slab_type = "Two adjacent edges discontinuous"

Summary of moment coefficient:
Short span - Moment coefficient - support: βsx.support  0.047
Short span - Moment coefficient - midspan: βsx.midspan  0.036
Long span - Moment coefficient - support: βsy.support  0.045
Long span - Moment coefficient - midspan: βsy.midspan  0.034
Page 26 of 27

CALUCLATIION
SHEET
CAPACITY CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
BENDING MOMENT RESULTS:
Note: Bending moment per unit width.
Short span - Bending moment at support: MEd.sx.sup βsx.supportFEd.ULS lx
  2  13.439kN
  2  10.294kN
Short span - Bending moment at midspan: MEd.sx.mid βsx.midspanFEd.ULS lx
  2  12.867kN
Long span - Bending moment at support: MEd.sy.sup βsy.supportFEd.ULS lx
  2  9.722kN
Long span - Bending moment at midspan: MEd.sy.mid βsy.midspanFEd.ULS lx
Page 27 of 27

ANNEX B - EXAMPLE OF HOW TO DETERMINE THE MOMENT
28
CAPACITY OF RC SLAB

CALUCLATIION
SHEET
BEAM FLEXURAL CAPACITY
CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
REINFORCED CONCRETE SOLID SLAB DESIGN TO EUROCODE 2
Note: The following colour key is a guide to using the full calculation page.
INPUT DTATA
COMPUTED OUTPUT
DATA TO BE CHECKED
STANDARD DATA
Figure 1: Analysis of rectangular section - stress strain
ASSUMPTIONS:
GEOMETRICAL DATA:
Concrete cover: cnom  25mm
Breadth of the section (assumed 1m strip): b  1m
Depth of the section: h  170mm
Longitudinal diameter (tension zone - bottom): dt  10mm
Longitudinal diameter (compression zone - top): dc  12mm
Spacing of steel reinforcement: sp  200mm
   392.699mm2
Area of steel reinforcement provided: As.prov.t π
2
4
dt
 m
sp
   565.487mm2
Area of steel reinforcement provided: As.prov.c π
2
4
dc
 m
sp
Effective depth of the section. d: d h  cnom
dt
2
   140mm
Effective depth of the section. d2: d2 cnom
dc
2
   31mm
MATERIAL PROPERTIES:
Mean characteristic compressive
SLAB DESIGN TO EUROCODE 2 Page 29 of 31

CALUCLATIION
SHEET
CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
cylinder strength of concrete
(Laboratory results): fck 30N mm 2  
Characteristic yield strength of
steel reinforcement:
fyk 500N mm 2  
PARTIAL SAFETY FACTOR (CYS NA EN1992-1-1,Table 2.1):
Partial factor for reinforcement
steel (NA CYS EN 1992-1-1:2004, Table 2.1)): γs  1.15
Partial factor for concrete
(NA CYS EN 1992-1-1:2004, Table 2.1)): γc  1.5
DESIGN STRENGTHS OF MATERIAL(EN1992-1-1,cl.3.1.6):
Design yield strength of reinforcement
(EN1992-1-1,Fig.3.8): fyd
fyk
γs
434.783 N mm 2    
Coefficient value for compressive strength
(NA CYS EN 1992-1-1:2004, cl. NA 2.8): αcc  1
Design value of concrete compressive strength
fcd
(EN 1992-1-1:2004, Equation 3.15):
αccfck
γc
20 N mm 2    
RECTANGULAR STRESS BLOCK FACTORS:
Factor, λ λ  0.8 if fck  50MPa
 0.8
(EN1992-1-1,Eq.3.19&3.20)
0.8
fck  50MPa
400MPa



if fck  50MPa
Factor, η η  1.0 if fck  50MPa
 1
(EN1992-1-1,Eq.3.21&3.22)
1.0
fck  50MPa
200MPa



if fck  50MPa
BENDING MOMENT CAPACITY (AT MIDSPAN) FOR A SINGLY REINFORCED SECTION
Figure 2: Detail of reinforcement slab at midspan
For equilibrium, the ultimate design moment, must be balanced by the moment of resistance
of the section (figure 1):
Fc  Fst
Fst  fydAs.prov.t  170.739kN
Fc  fcdbλx  kN
Therefore depth of stress block is:

CALUCLATIION
SHEET
CHECK
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
s
fydAs.prov.t
fcdb
  8.537mm
x
s
  10.671mm
λ
To ensure rotation of the plastic hinge with sufficient yielding of the tension steel and also to
allow for other factors such as the strain hardening of the steel, EC2 limit the depth of neutral
axis to:
Check  if (x  0.45d"PASS" "FAIL" )  "PASS"
z d
s
2
   135.732mm
Moment capacity: MRd  fydAs.prov.tz  23.175kNm
BENDING CAPACITY (AT SUPPORTS) OF SECTION WITH COMPRESSION
REINFORCEMENT AT ULTIMATE LIMIT STATE
Figure 3: Detail of reinforcement slab at support
For equilibrium, the ultimate design moment, must be balanced by the moment of resistance
of the section (figure 1):
Fst  Fc  Fsc
Fsc  fydAs.prov.c  245.864kN
Fst  fydAs.prov.t  170.739kN
Fc  fcdbλx
Therefore depth of stress block is:
s
fydAs.prov.c  As.prov.t
  3.756mm
fcdb
x
s
  10.671mm
λ
Check  if (x  0.45d"PASS" "FAIL" )  "PASS"
To ensure rotation of the plastic hinge with sufficient yielding of the tension steel and also to
allow for other factors such as the strain hardening of the steel, EC2 limit the depth of neutral
axis to:
Moment capacity: MRd. fcdbs d
s
2
 

   fydAs.prov.cd  d2  37.176kNm

32
ANNEX C - EXAMPLE OF DESIGN SLAB PANEL WITH TWO
DISCONTINUOUS EDGES

CALUCLATIION
SHEET
REINFORCED CONCRETE
SOLID SLAB DESIGN TO
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
INPUT DTATA ASSUMPTIONS:
1. Fire resistance 1hour (REI 60).
2. Exposure class of concrete XC1.
3. No redistribution of bending moment made.
COMPUTED OUTPUT
DATA TO BE CHECKED
STANDARD DATA
GEOMETRICAL DATA:
Structural_system:= "Simply supported"
"End span of continuous slab"
"Interior span"
"Flat slab"
"Cantilever"
Structural system:
Structural_system  "End span of continous slab"
Depth of slab: h  170mm
Strip width: b  1000mm
Type of slab:
Type_slab "Two way slab"
ly
lx
 if  2.0
 "Two way slab"
"One way slab"
ly
lx
if  2.0
ANALYSIS & LOADING RESULTS:
TWO DISCONTINOUS EDGE Page 33 of 48

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Figure 1: Bending moment diagram for x - direction
Figure 2: Bending moment diagram for y - direction

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Figure 3: Shear force diagram for x - direction

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Figure 4: Shear force diagram for y - direction
Loads:
Quasi-permanent value of variable action: ψ2  0.3
Short span:
Design bending moment at short span - continuous support: Mx.1  21.14kNm
Design bending moment at short span - middle: Mx.m  12.35kNm
Design shear force at short span - continous support: Vx.1  21kN
Design shear force at short span - discontinous support: Vx.2  13kN
Long span:
Design bending moment at long span - continous support: My.1  10.52kNm
Design bending moment at long span - middle: My.m  11.86kNm
Design shear force at long span - continous support: Vy.1  18kN

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Design shear force at long span - discontinous support: Vy.2  13kN
STEEL REINFORCEMENT PROPERTIES:
Bars diameter for short/long span-midspan: ϕy.p  10mm
steel reinforcement: fyk 500N mm 2   
CONCRETE PROPERTIES:
Characteristic compressive cylinder
strength of concrete: fck 30N mm 2   
Mean value of compressive sylinder
strength
(EN 1992-1-1:2004, table 3.1): fctm 0.3
fck
MPa


0.667



MPa 2.9 N mm 2    
PARTIAL SAFETY FACTORS:
(NA CYS EN 1992-1-1:2004, Table 2.1)): γc  1.5
(EN1992-1-1,Fig.3.8): fyd
fyk
γs
434.783 N mm 2    
(NA CYS EN 1992-1-1:2004, cl. NA 2.8): αcc  1
fcd
(EN 1992-1-1:2004, Equation 3.15):
αccfck
γc
20 N mm 2    
CONCRETE COVER TO REINFORCEMENT:
Allowance in design for deviation
(Assuming no measurement of cover)
(EN1992-1-1,cl.4.4.1.3(3):
Δcdev  10mm
Minimum cover due to bond
(Diameter of bar)
(EN1992-1-1,Table 4.2):
cmin.b  ϕy.p  10mm
Minimum cover due to environmental
condition (Condition :XC1)
("How to design to Eurocode 2",Table 8):
cmin.dur  15mm
Minimum concrete cover
(EN1992-1-1,Eq.4.2):
cmin  maxcmin.bcmin.dur10mm  15mm
Nominal cover
(EN1992-1-1,Eq.4.1):
cnom  cmin  Δcdev  25mm

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
FIRE DESIGN CHECK:
Minimum slab thickness
(EN1992-1-2,Table 5.8):
hs.min  80mm
Fire_resistance  if h  hs.min"OK" "NOT OK"   "OK"
Axis distance to top and bottom
reinforcement, a
(EN1992-1-2,Table 5.8):
amin  20mm
Minimum distance to top and bottom
reinforcement:
aprov cnom
ϕy.p
2
   30mm
Fire_resistance  if aprov  amin"OK" "NOT OK"   "OK"
REINFORCEMENT DESIGN AT MID-SPAN IN SHORT SPAN DIRECTION:
Actual bar size: ϕx.m  10mm
Actual bar spacing: sx.m  200mm
   392.699mm2
Area of reinforcement provided: Asx.m π
2
4
ϕx.m
 m
sx.m
dx.m h  cnom
ϕx.m
2
   140mm
Values for Klim
(Assumed no redistribution):
K
Mx.m
b dx.m
  0.021 Klim  0.22
 2 f ck
Compression  if K  Klim"NOT REQUIRED" "REQUIRED"   "NOT REQUIRED"
Level arm:
z min
dx.m
2
1  1  3.53K


0.95dx.m


  133mm
Area of reinforcement required for
bending:
Asx.p.m
Mx.m
fydz
  213.571mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.m0.0013bdx.m


  211.102m
Maximum reinforcement
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdx.m 5.6 10   3mm2
Check_steel_1  if Asx.p.m  Asx.m  As.min  Asx.m  As.max"OK" "NOT OK"   "OK"
Ratio_1
maxAs.minAsx.p.m
  0.544
Asx.m
Stress in the
reinforcement
(IStrucTE EC2 Manual)
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.p.m
Asx.m
1




 141.617 N mm 2    

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Maximum spacing (for wk=0.3mm)
(EN1992-1-1,Table 7.3N:
smax 300mm if σs  160MPa
  300mm
275mm if 160MPa  σs  180MPa
Maximum spacing of bars
(EN1992-1-1,cl.9.3.1.1(3):
smax.  min3h400mmsmax  300mm
Spacing_1  if sx.m  smax."OK" "NOT OK"   "OK"
Ratio_s_1
sx.m
smax
  0.667
REINFORCEMENT DESIGN AT CONTINUOUS SUPPORT IN SHORT SPAN DIRECTION:
Actual bar size: ϕx.1  12mm
Actual bar spacing: sx.1  200mm
   565.487mm2
Area of reinforcement provided: Asx.1 π
2
4
ϕx.1
 m
sx.1
dx.1 h  cnom
ϕx.1
2
   139mm
Values for Klim
K
Mx.1
b dx.1
  0.036 Klim  0.22
 2 f ck
Level arm:
z min
dx.1
2
1  1  3.53K


0.95dx.1


  132.05mm
bending:
Asx.n.1
Mx.1
fydz
  368.209mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.10.0013bdx.1


  209.594mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdx.1 5.56 10   3mm2

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Check_steel_2  if Asx.n.1  Asx.1  As.min  Asx.1  As.max"OK" "NOT OK"   "OK"
Ratio_2
maxAs.minAsx.n.1
  0.651
Asx.1
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.n.1
Asx.1
1




 169.552 N mm 2    
(EN1992-1-1,Table 7.3N:
  275mm
(EN1992-1-1,cl.9.3.1.1(3):
Spacing_2  if sx.1  smax."OK" "NOT OK"   "OK"
Ratio_s_2
sx.1
smax
  0.727
REINFORCEMENT DESIGN AT MID-SPAN IN LONG SPAN DIRECTION:
Actual bar size: ϕy.m  10mm
Actual bar spacing: sy.m  200mm
   392.699mm2
Area of reinforcement provided: Asy.m π
2
4
ϕy.m
 m
sy.m
dy.m h  cnom
ϕy.m
2
   140mm
Values for Klim
K
My.m
b dy.m
  0.02 Klim  0.22
 2 f ck

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Level arm:
z min
dy.m
2
1  1  3.53K


0.95dy.m


  133mm
bending:
Asy.p.m
My.m
fydz
  205.098mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdy.m0.0013bdy.m


  211.102mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdy.m 5.6 10   3mm2
Check_steel_3  if Asy.p.m  Asy.m  As.min  Asy.m  As.max"OK" "NOT OK"   "OK"
Ratio_3
maxAs.minAsy.p.m
  0.538
Asy.m
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asy.p.m
Asy.m
1




 135.998 N mm 2    
(EN1992-1-1,Table 7.3N:
  0.3m
(EN1992-1-1,cl.9.3.1.1(3):
Spacing_3  if sy.m  smax."OK" "NOT OK"   "OK"
Ratio_s_3
sy.m
smax
  0.667
REINFORCEMENT DESIGN AT CONTINUOUS SUPPORT IN LONG SPAN DIRECTION:
Actual bar size: ϕy.1  10mm
Actual bar spacing: sy.1  200mm
   392.699mm2
Area of reinforcement provided: Asy.1 π
2
4
ϕy.1
 m
sy.1

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
dy.1 h  cnom
ϕy.1
2
   140mm
Values for Klim
K
My.1
b dy.1
  0.018 Klim  0.22
 2 f ck
Level arm:
z min
dy.1
2
1  1  3.53K


0.95dy.1


  133mm
bending:
Asy.n.1
My.1
fydz
  181.925mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdy.10.0013bdy.1


  211.102mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdy.1 5.6 10   3mm2
Check_steel_4  if Asy.n.1  Asy.1  As.min  Asy.1  As.max"OK" "NOT OK"   "OK"
Ratio_4
maxAs.minAsy.n.1
  0.538
Asy.1
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asy.n.1
Asy.1
1




 120.632 N mm 2    
(EN1992-1-1,Table 7.3N:
  300mm
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_4
sy.m
smax
  0.667

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
SHEAR CAPACITY CHECK AT SHORT SPAN CONTINUOUS SUPPORT:
Effective depth factor
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dx.1


0.5
 


  2
Reinforcement ratio: ρ1 min 0.02
Asx.1
bdx.1



4.068 10 3   
Minimum shear resistance
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdx.1


N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
VRd.c.x.1 max VRd.c.min
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdx.1


  76.743k
Shear_1  if Vx.1  VRd.c.x.1"NO SHEAR REQUIRED" "SHEAR REQUIRED" 
Shear_1  "NO SHEAR REQUIRED"
Ratio1
Vx.1
VRd.c.x.1
  0.274
SHEAR CAPACITY CHECK AT SHORT SPAN DISCONTINUOUS SUPPORT:
Flexural reinforcement at
As.req  Asx.m0.25  98.175mm2
discontinuous support
EN1992-1-1,cl.9.3.1.2(2):
Bar spacing: sx.2  sx.m  200mm
   251.327mm2
2
4
ϕx.2
 m
sx.2
Effective depth:
dx.2 h  cnom
ϕx.2
2
   141mm
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dx.2


0.5
 


  2
Asx.2
bdx.2



1.782 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdx.2


N mm 2    54.06kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdx.2


  59.143k

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Ratio2
Vx.2
VRd.c.x.2
  0.22
SHEAR CAPACITY CHECK AT LONG SPAN CONTINUOUS SUPPORT:
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dy.1


0.5
 


  2
Asy.1
bdy.1



2.805 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdy.1


N mm 2    53.677kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
VRd.c.y.1 max VRd.c.min
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdy.1


  68.294kN
Shear_3  if Vy.1  VRd.c.y.1"NO SHEAR REQUIRED" "SHEAR REQUIRED" 
Ratio3
Vy.1
VRd.c.y.1
  0.264
SHEAR CAPACITY CHECK AT LONG SPAN DISCONTINUOUS SUPPORT:
As.req  Asy.m0.25  98.175mm2
EN1992-1-1,cl.9.3.1.2(2):
Bar spacing: sy.2  sy.m  200mm
   251.327mm2
2
4
ϕy.2
 m
sy.2
Effective depth:
dy.2 h  cnom
ϕy.2
2
   141mm
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dy.2


0.5
 


  2
Asy.2
bdy.2



1.782 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdy.2


N mm 2    54.06kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdy.2


  59.143kN

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Ratio4
Vy.2
VRd.c.y.2
  0.22
BASIC SPAN-TO-DEPTH DEFLECTION RATIO CHECK:
Reference reinforcement ratio: ρo 0.001
fck
MPa


0.5
 5.477 10 3   
Required compression reinforcement
(at mid-span - short span): ρc  0
Required tension reinforcement
(at mid-span - short span):
ρt max 0.0035
Asx.m
bdx.m



3.5 10 3   
Structural system factor
(EN1992-1-1,Table 7.4N):
Kδ 1.0 if Structural_system = "Simply supported"
  1.3
1.3 if Structural_system = "End span of continous slab"
1.5 if Structural_system = "Interior span"
1.2 if Structural_system = "Flat slab"
0.4 if Structural_system = "Cantilever"
Basic limit span-to-depth ratio
(EN1992-1-1,Eq.7.16a&7.16b):

Limx.bas Kδ 11 1.5
fck
MPa


0.5
 
 40.689
ρo
ρt
  3.2
fck
MPa


0.5

ρo
ρt
 1


1.5
 


ρt ρo  if
Kδ 11 1.5
fck
MPa


0.5

ρo
ρt  ρc
  1
12
fck
MPa


0.5

ρc
ρo
 





if ρt  ρo
Actual span to effective depth ratio: Ratioact
lx
dx.m
  35.714
Deflection  if Ratioact  Limx.bas"OK" "NOT OK"   "OK"
Ratio
Ratioact
Limx.bas
  0.878
CALCULATION SUMMARY RESULTS:
Short span - Bending capacity: PASS/FAIL: Ratio:
Check bending capacity at midspan: Check_steel_1  "OK" Ratio_1  0.544
Spacing at midspan reinforcement: Spacing_1  "OK" Ratio_s_1  0.667
Check bending capacity at support 1: Check_steel_2  "OK" Ratio_2  0.651
Spacing at support 1 reinforcement: Spacing_2  "OK" Ratio_s_2  0.727

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Long span - Bending capacity: PASS/FAIL: Ratio:
Short span - Shear capacity: PASS/FAIL: Ratio:
Check shear capacity at support 1: Shear_1  "NO SHEAR REQUIRED" Ratio1  0.274
Long span - Shear capacity: PASS/FAIL: Ratio:
Deflection: PASS/FAIL: Ratio:
Check deflection of panel: Deflection  "OK" Ratio  0.878
RENFORCEMENT SUMMARY:
Short span:
Midspan in short span direction: ϕx.m  10mm at C/C sx.m  200mm
Continuous support 1 in short span direction: ϕx.1  12mm at C/C sx.1  200mm
Discontinuous support 2 in short span direction: ϕx.2  8mm at C/C sx.2  200mm
Long span:
Midspan in short span direction: ϕy.m  10mm at C/C sy.m  200mm
Continuous support 1 in long span direction: ϕy.1  10mm at C/C sy.1  200mm
Discontinuous support 2 in long span direction: ϕy.2  8mm at C/C sy.2  200mm

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
ϕy.2  8mmsy.2  200mm
ϕx.2  8mmsx.2  200mm
ϕx.1  12mmsx.1  200mm
ϕx.m  10mmsx.m  200mm
ϕy.m  10mm sy.m  200mm
ϕy.1  10mmsy.1  200mm

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
mm2

48
ANNEX D - EXAMPLE OF DESIGN SLAB PANEL WITH ONE
DISCONTINUOUS EDGES

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
COMPUTED OUTPUT
DATA TO BE CHECKED
STANDARD DATA
GEOMETRICAL DATA:
"Interior span"
"Flat slab"
"Cantilever"
Structural system:
Structural_system  "End span of continous slab"
Type of slab:
ly
lx
 if  2.0
 "Two way slab"
"One way slab"
ly
lx
if  2.0
ONE DISCONTINUOUS EDGE Page 49 of 64

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Loads:
Short span:
Design bending moment at short span - continuous support: Mx.1  21kNm
Design bending moment at short span - middle: Mx.m  7kNm
Long span:
Design bending moment at long span - continous support: My.1  20kNm
Design bending moment at long span - middle: My.m  12kNm

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
strength
(EN 1992-1-1:2004, table 3.1): fctm 0.3
fck
MPa


0.667



MPa 2.9 N mm 2    
(NA CYS EN 1992-1-1:2004, Table 2.1)): γc  1.5
(EN1992-1-1,Fig.3.8): fyd
fyk
γs
434.783 N mm 2    
(NA CYS EN 1992-1-1:2004, cl. NA 2.8): αcc  1
fcd
(EN 1992-1-1:2004, Equation 3.15):
αccfck
γc
20 N mm 2    
(EN1992-1-1,cl.4.4.1.3(3):
Δcdev  10mm
(Diameter of bar)
(EN1992-1-1,Table 4.2):
(EN1992-1-1,Eq.4.2):

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Nominal cover
(EN1992-1-1,Eq.4.1):
FIRE DESIGN CHECK:
(EN1992-1-2,Table 5.8):
hs.min  80mm
reinforcement, a
(EN1992-1-2,Table 5.8):
amin  20mm
reinforcement:
aprov cnom
ϕy.p
2
   30mm
   392.699mm2
2
4
ϕx.m
 m
sx.m
dx.m h  cnom
ϕx.m
2
   140mm
Values for Klim
K
Mx.m
b dx.m
  0.012 Klim  0.22
 2 f ck
Level arm:
z min
dx.m
2
1  1  3.53K


0.95dx.m


  133mm
bending:
Asx.p.m
Mx.m
fydz
  121.053mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.m0.0013bdx.m


  211.102mm
Ratio_1
  0.538
Asx.m

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.p.m
Asx.m
1




 80.269 N mm 2    
(EN1992-1-1,Table 7.3N:
  300mm
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_1
sx.m
smax
  0.667
REINFORCEMENT DESIGN AT CONTINUOUS SUPPORT 1 IN SHORT SPAN DIRECTION:
   565.487mm2
2
4
ϕx.1
 m
sx.1
dx.1 h  cnom
ϕx.1
2
   139mm
Values for Klim
K
Mx.1
b dx.1
  0.036 Klim  0.22
 2 f ck
Level arm:
z min
dx.1
2
1  1  3.53K


0.95dx.1


  132.05mm
bending:
Asx.n.1
Mx.1
fydz
  365.771mm2

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.10.0013bdx.1


  209.594mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdx.1 5.56 10   3mm2
Ratio_2
  0.647
Asx.1
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.n.1
Asx.1
1




 168.429 N mm 2    
(EN1992-1-1,Table 7.3N:
  275mm
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_2
sx.1
smax
  0.727
   565.487mm2
2
4
ϕx.2
 m
sx.2

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
dx.2 h  cnom
ϕx.2
2
   139mm
Values for Klim
K
Mx.2
b dx.2
  0.036 Klim  0.22
 2 f ck
Level arm:
z min
dx.2
2
1  1  3.53K


0.95dx.2


  132.05mm
bending:
Asx.n.2
Mx.2
fydz
  365.771mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.20.0013bdx.2


  209.594mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdx.2 5.56 10   3mm2
Ratio_3
  0.647
Asx.2
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.n.2
Asx.2
1




 168.429 N mm 2    
(EN1992-1-1,Table 7.3N:
  275mm
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_3
sx.2
smax
  0.727

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
   392.699mm2
2
4
ϕy.m
 m
sy.m
dy.m h  cnom
ϕy.m
2
   140mm
Values for Klim
K
My.m
b dy.m
  0.02 Klim  0.22
 2 f ck
Level arm:
z min
dy.m
2
1  1  3.53K


0.95dy.m


  133mm
bending:
Asy.p.m
My.m
fydz
  207.519mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdy.m0.0013bdy.m


  211.102mm
Ratio_4
  0.538
Asy.m
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asy.p.m
Asy.m
1




 137.603 N mm 2    
(EN1992-1-1,Table 7.3N:
  0.3m

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_4
sy.m
smax
  0.667
REINFORCEMENT DESIGN AT CONTINUOUS SUPPORT IN LONG SPAN DIRECTION:
   565.487mm2
2
4
ϕy.1
 m
sy.1
dy.1 h  cnom
ϕy.1
2
   139mm
Values for Klim
K
My.1
b dy.1
  0.035 Klim  0.22
 2 f ck
Level arm:
z min
dy.1
2
1  1  3.53K


0.95dy.1


  132.05mm
bending:
Asy.n.1
My.1
fydz
  348.353mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdy.10.0013bdy.1


  209.594mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdy.1 5.56 10   3mm2
Ratio_5
  0.616
Asy.1
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk
 
 min
Asy.n.1
Asy.1
1




 160.409 N mm 2    

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
(EN1992-1-1,Table 7.3N:
  275mm
(EN1992-1-1,cl.9.3.1.1(3):
Spacing_5  if sy.1  smax."OK" "NOT OK"   "OK"
Ratio_s_5
sy.1
smax
  0.727
SHEAR CAPACITY CHECK AT SHORT SPAN CONTINUOUS SUPPORT 1:
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dx.1


0.5
 


  2
Asx.1
bdx.1



4.068 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdx.1


N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc

k 100ρ1
fck
MPa





0.333
  bdx.1


  76.743k
Ratio1
Vx.1
VRd.c.x.1
  0.287
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dx.2


0.5
 


  2

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Asx.2
bdx.2



4.068 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdx.2


N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdx.2


  76.743k
Ratio2
Vx.2
VRd.c.x.2
  0.235
SHEAR CAPACITY CHECK AT LONG SPAN CONTINUOUS SUPPORT:
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dy.1


0.5
 


  2
Asy.1
bdy.1



4.068 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdy.1


N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa




 0.333
  bdy.1


  76.743kN
Ratio3
Vy.1
VRd.c.y.1
  0.274
SHEAR CAPACITY CHECK AT LONG SPAN DISCONTINUOUS SUPPORT:
As.req  Asy.m0.25  98.175mm2
EN1992-1-1,cl.9.3.1.2(2):
Bar spacing: sy.2  sy.m  200mm
   251.327mm2
2
4
ϕy.2
 m
sy.2
Effective depth:
dy.2 h  cnom
ϕy.2
2
   141mm

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dy.2


0.5
 


  2
Asy.2
bdy.2



1.782 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdy.2


N mm 2    54.06kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdy.2


  59.143kN
Ratio4
Vy.2
VRd.c.y.2
  0.22
fck
MPa


0.5
 5.477 10 3   
ρt max 0.0035
Asx.m
bdx.m



3.5 10 3   
(EN1992-1-1,Table 7.4N):
  1.3
(EN1992-1-1,Eq.7.16a&7.16b):
Limx.bas Kδ 11 1.5
fck
MPa


0.5
 
 40.689
ρo
ρt
  3.2
fck
MPa


0.5

ρo
ρt
 1


1.5
 





if ρt  ρo
Kδ 11 1.5
fck
MPa


0.5

ρo
ρt  ρc
  1
12
fck
MPa


0.5

ρc
ρo
 





if ρt  ρo
lx
dx.m
  35.714

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Ratio
Ratioact
Limx.bas
  0.878
Short span:
Discontinuous support 2 in short span direction: ϕx.2  12mm at C/C sx.2  200mm
Long span:

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:IK
Discontinuous support 2 in long span direction: ϕy.2  8mm at C/C sy.2  200mm
ϕy.2  8mmsy.2  200mm
ϕx.2  12mmsx.2  200mm ϕx.1  12mmsx.1  200mm
ϕy.m  10mm sy.m  200mm
ϕy.1  12mmsy.1  200mm

65
ANNEX E - EXAMPLE OF DESIGN INTERIOR PANEL SLAB

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
COMPUTED OUTPUT
DATA TO BE CHECKED
STANDARD DATA
GEOMETRICAL DATA:
"Interior span"
"Flat slab"
"Cantilever"
Structural system:
Structural_system  "Interior span"
Type of slab:
ly
lx
 if  2.0
 "Two way slab"
"One way slab"
ly
lx
if  2.0
INTERIOR PANEL Page 66 of 82

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
Loads:
Short span:
Design bending moment at short span - middle: Mx.m  6kNm
Design shear force at short span - discontinous support: Vx.2  21kN
Long span:
Design bending moment at long span - middle: My.m  6kNm
strength
(EN 1992-1-1:2004, table 3.1): fctm 0.3
fck
MPa


0.667



MPa 2.9 N mm 2    

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
(NA CYS EN 1992-1-1:2004, Table 2.1)): γc  1.5
(EN1992-1-1,Fig.3.8): fyd
fyk
γs
434.783 N mm 2    
(NA CYS EN 1992-1-1:2004, cl. NA 2.8): αcc  1
fcd
(EN 1992-1-1:2004, Equation 3.15):
αccfck
γc
20 N mm 2    
(EN1992-1-1,cl.4.4.1.3(3):
Δcdev  10mm
(Diameter of bar)
(EN1992-1-1,Table 4.2):
(EN1992-1-1,Eq.4.2):
Nominal cover
(EN1992-1-1,Eq.4.1):
FIRE DESIGN CHECK:
(EN1992-1-2,Table 5.8):
hs.min  80mm
reinforcement, a
(EN1992-1-2,Table 5.8):
amin  20mm
reinforcement:
aprov cnom
ϕy.p
2
   30mm

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
   392.699mm2
2
4
ϕx.m
 m
sx.m
dx.m h  cnom
ϕx.m
2
   140mm
Values for Klim
K
Mx.m
b dx.m
  0.01 Klim  0.22
 2 f ck
Level arm:
z min
dx.m
2
1  1  3.53K


0.95dx.m


  133mm
bending:
Asx.p.m
Mx.m
fydz
  103.759mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.m0.0013bdx.m


  211.102mm
Ratio_1
  0.538
Asx.m
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.p.m
Asx.m
1




 68.802 N mm 2    
(EN1992-1-1,Table 7.3N:
  300mm
(EN1992-1-1,cl.9.3.1.1(3):

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
Ratio_s_1
sx.m
smax
  0.667
   565.487mm2
2
4
ϕx.1
 m
sx.1
dx.1 h  cnom
ϕx.1
2
   139mm
Values for Klim
K
Mx.1
b dx.1
  0.036 Klim  0.22
 2 f ck
Level arm:
z min
dx.1
2
1  1  3.53K


0.95dx.1


  132.05mm
bending:
Asx.n.1
Mx.1
fydz
  365.771mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.10.0013bdx.1


  209.594mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdx.1 5.56 10   3mm2
Ratio_2
  0.647
Asx.1
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.n.1
Asx.1
1




 168.429 N mm 2    

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
(EN1992-1-1,Table 7.3N:
  275mm
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_2
sx.1
smax
  0.727
   565.487mm2
2
4
ϕx.2
 m
sx.2
dx.2 h  cnom
ϕx.2
2
   139mm
Values for Klim
K
Mx.2
b dx.2
  0.036 Klim  0.22
 2 f ck
Level arm:
z min
dx.2
2
1  1  3.53K


0.95dx.2


  132.05mm
bending:
Asx.n.2
Mx.2
fydz
  365.771mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdx.20.0013bdx.2


  209.594mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdx.2 5.56 10   3mm2

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
Ratio_3
  0.647
Asx.2
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asx.n.2
Asx.2
1




 168.429 N mm 2    
(EN1992-1-1,Table 7.3N:
  275mm
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_3
sx.2
smax
  0.727
   392.699mm2
2
4
ϕy.m
 m
sy.m
dy.m h  cnom
ϕy.m
2
   140mm
Values for Klim
K
My.m
b dy.m
  0.01 Klim  0.22
 2 f ck

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
Level arm:
z min
dy.m
2
1  1  3.53K


0.95dy.m


  133mm
bending:
Asy.p.m
My.m
fydz
  103.759mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdy.m0.0013bdy.m


  211.102mm
Ratio_4
  0.538
Asy.m
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asy.p.m
Asy.m
1




 68.802 N mm 2    
(EN1992-1-1,Table 7.3N:
  0.3m
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_4
sy.m
smax
  0.667
REINFORCEMENT DESIGN AT CONTINUOUS SUPPORT 1 IN LONG SPAN DIRECTION:
   565.487mm2
2
4
ϕy.1
 m
sy.1

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
dy.1 h  cnom
ϕy.1
2
   139mm
Values for Klim
K
My.1
b dy.1
  0.036 Klim  0.22
 2 f ck
Level arm:
z min
dy.1
2
1  1  3.53K


0.95dy.1


  132.05mm
bending:
Asy.n.1
My.1
fydz
  365.771mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdy.10.0013bdy.1


  209.594mm
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdy.1 5.56 10   3mm2
Ratio_5
  0.647
Asy.1
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asy.n.1
Asy.1
1




 168.429 N mm 2    
(EN1992-1-1,Table 7.3N:
  275mm
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_5
sy.1
smax
  0.727

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
REINFORCEMENT DESIGN AT CONTINUOUS SUPPORT 2 IN LONG SPAN DIRECTION:
   565.487mm2
2
4
ϕy.2
 m
sy.2
dy.2 h  cnom
ϕy.2
2
   139mm
Values for Klim
K
My.2
b dy.2
  0.036 Klim  0.22
 2 f ck
Level arm:
z min
dy.2
2
1  1  3.53K


0.95dy.2


  132.05mm
bending:
Asy.n.2
My.2
fydz
  365.771mm2
Minimum
reinforcement
(EN1992-1-1,Eq.9.1N)
:
As.min max 0.26
fctm
fyk
 bdy.20.0013bdy.2


  209.594mm2
(EN1992-1-1,cl.9.2.1.1(3)): As.max 0.04bdy.2 5.56 10   3mm2
Ratio_6
  0.647
Asy.2
Stress in the
reinforcement
σs
fyk
γs


ψ2Qk  Gk
1.5Qk  1.35Gk


 min
Asy.n.2
Asy.2
1




 168.429 N mm 2    
(EN1992-1-1,Table 7.3N:
  275mm

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
(EN1992-1-1,cl.9.3.1.1(3):
Ratio_s_6
sy.2
smax
  0.727
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dx.1


0.5
 


  2
Asx.1
bdx.1



4.068 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdx.1


N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdx.1


  76.743k
Ratio1
Vx.1
VRd.c.x.1
  0.274
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dx.2

 0.5
 


  2
Asx.2
bdx.2



4.068 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdx.2


N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdx.2


  76.743k
Ratio2
Vx.2
VRd.c.x.2
  0.274
SHEAR CAPACITY CHECK AT LONG SPAN CONTINUOUS SUPPORT 1:

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dy.1


0.5
 


  2
Asy.1
bdy.1



4.068 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdy.1


N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdy.1


  76.743kN
Ratio3
Vy.1
VRd.c.y.1
  0.274
SHEAR CAPACITY CHECK AT LONG SPAN CONTINUOUS SUPPORT 2:
(EN1992-1-1,cl.6.2.2): k min 2.0 1
200mm
dy.2


0.5
 


  2
Asy.2
bdy.2



4.068 10 3   
(EN1992-1-1,Eq.6.3N &6.2b): VRd.c.min 0.035k
fck
MPa


0.5



bdy.2

N mm 2    53.293kN
Shear resistance
(EN1992-1-1,
Eq.6.2a):
0.18MPa
γc


k 100ρ1
fck
MPa





0.333
  bdy.2


  76.743kN
Ratio4
Vy.2
VRd.c.y.2
  0.274
fck
MPa


0.5
 5.477 10 3   
ρt max 0.0035
Asx.m
bdx.m



3.5 10 3   

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
(EN1992-1-1,Table 7.4N):
  1.5
(EN1992-1-1,Eq.7.16a&7.16b):
Limx.bas Kδ 11 1.5
fck
MPa


0.5
 
 46.949
ρo
ρt
  3.2
fck
MPa


0.5

ρo
ρt
 1


1.5
 





if ρt  ρo
Kδ 11 1.5
fck
MPa


0.5

ρo
ρt  ρc
  1
12
fck
MPa


0.5

ρc
ρo
 





if ρt  ρo
lx
dx.m
  35.714
Ratio
Ratioact
Limx.bas
  0.761

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
Short span:
Long span:

CALUCLATIION
SHEET
REINFORCED CONCRETE
EUROCODE 2
Date:01/09/2014
Rev:B
Calculated by:VN
Checked by:VN
ϕy.2  12mmsy.2  200mm
ϕx.2  12mmsx.2  200mm ϕx.1  12mmsx.1  200mm
ϕy.m  10mm sy.m  200mm
ϕy.1  12mmsy.1  200mm

CSI ETABS & SAFE MANUAL: Slab Analysis and Design to EC2

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CSI ETABS & SAFE MANUAL: Slab Analysis and Design to EC2