ES 202 - Theory of structures-IV
Dr. Beena K. P.
Dept. of Civil Engineering
Theory of structures -IV
Module –III(6 hrs-15%)
Slabs
1. Classification (one way slab and two way slabs)
2. Design of one way slab under flexure and shear ( simply
supported and continuous slabs)
Slabs
Classification
One way slab
simply supported
continuous
ly
Two way slab Unit width
𝑙𝑦
> 2 − 𝑜𝑛𝑒 𝑤𝑎𝑦 𝑠𝑙𝑎𝑏 lx
𝑙𝑥
𝑙𝑦
≤ 2 − 𝑡𝑤𝑜 𝑤𝑎𝑦 𝑠𝑙𝑎𝑏
𝑙𝑥
Slabs
Slabs
Slabs
Slabs
One way slab Two way slab
Supported by beams only in two Supported by beams in all four
sides – load transfer only in two sides- load transfer on all four
edges edges
Ratio of longer span to shorter Ratio of longer span to shorter
one is greater than 2 one is less than or equal to 2
Main reinforcement is provided Main reinforcement is provided
only in one direction (along in both directions
shorter span)
One way slab – Simply Supported
Reinforcement consists of
Main (longitudinal) reinforcement or moment steel
Temperature steel or distribution reinforcement to
minimise cracks and shrinkage due to temperature
Analysis of slab is similar to that of a beam of 1m width
One way slab – Simply Supported
Effective span – Clause 22.2, Pg 34, IS 456
Effective Span =
Clear Span + Effective Depth of Slab or beam
OR
Centre to centre of supports
whichever is less
Design of One way slab – Simply Supported
Pbm 1. Design a simply supported roof slab for a room 8m X
3.5m clear in size if the superimposed load is 5kN/m2. Use
M15 and Fe 415
lx=3.5m ly =8m
ly=8m Unit width,1m
fck =15 N/mm2
fy=415 N/mm2 lx = 3.5m
Design of One way slab – Simply Supported
𝑙𝑦 8
= = 2.286
𝑙𝑥 3.5
> 2 − ℎ𝑒𝑛𝑐𝑒 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑠𝑖𝑔𝑛𝑒𝑑 𝑎𝑠 𝑜𝑛𝑒 𝑤𝑎𝑦 𝑠𝑙𝑎𝑏
Clear span = 3.5m
Considering unit width perpendicular to the span
b=1000 mm
3.5m
Design of One way slab – Simply Supported
General steps
1. Calculate the loads
2. Find the maximum factored bending moment and shear force
3. Fix the depth of the section, effective depth and overall depth,
using Mu,lim expression for flexure.
2
𝑥𝑢,𝑚𝑎𝑥 0.42𝑥𝑢,𝑚𝑎𝑥
𝑀𝑢,𝑙𝑖𝑚 = 0.36𝑓𝑐𝑘 𝑏𝑑 1−
𝑑 𝑑
4. Find the area of steel required
𝑓 𝐴𝑠𝑡
𝑀𝑢 = 0.87𝑓𝑦 𝐴𝑠𝑡 𝑑 1 − 𝑓𝑦 𝑏𝑑
𝑐𝑘
5. Check for shear and deflection
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
the depth of slab from deflection criteria
l/d ratio for simply supported slab = 20,
clause 23.2.1, pg 37, IS 456
𝑙 3500
‘d’ should be grater than = = 175𝑚𝑚
20 20
‘d’ is the minimum depth
Cover to reinforcement
Assuming a clear cover of 15mm and 10mm dia steel for reinforcing
bars
Effective cover= 15+10/2=20mm
Total (Overall) depth = 175+20=195mm
Design of One way slab – Simply Supported
Load calculations
Superimposed load = 5kN/m2( dead load + live load)
Superimposed load along the span =5 X 1= 5kN/m
Self weight = 1X 0.195 X 25 = 4.875 kN/m
Total load/m = 5+4.875=9.875 kN/m
Effective span = clear span + effective depth of slab
= 3.5 +0.175=3.675m
Design of One way slab – Simply Supported
Maximum Bending moment and Shear Force
𝑤𝑙 2 9.875𝑋3.6752
Working moment = = =16.671 kNm
8 8
Factored Moment = 1.5 X 16.671=25.01kNm
𝑤𝑙
Maximum Factored Shear Force=
2
1.5𝑋9.875𝑋3.675
= = 𝟐𝟕. 𝟐𝟐𝒌𝑵
2
Design of One way slab – Simply Supported
Depth of slab from flexure criteria
𝑥𝑢,𝑚𝑎𝑥 0.42𝑥𝑢,𝑚𝑎𝑥
𝑀𝑢,𝑙𝑖𝑚 = 0.36𝑓𝑐𝑘 𝑏𝑑 2 1−
𝑑 𝑑
25.01 𝑋 106 = 0.36 ∗ 15 ∗ 1000 ∗ 𝑑 2 ∗ 0.48 1 − 0.42 ∗ 0.48
d=109.92mm ≈110mm
Provide an effective depth of 150mm
Overall depth =150+15+10/2=170mm
Design of One way slab – Simply Supported
Area of tension steel
𝑓𝑦 𝐴𝑠𝑡
𝑀𝑢 = 0.87𝑓𝑦 𝐴𝑠𝑡 𝑑 1 − 𝑓𝑐𝑘 𝑏𝑑
415∗𝐴𝑠𝑡
25.01𝑋106 = 0.87 ∗ 415𝐴𝑠𝑡 ∗ 150 1 − 15∗1000∗150
𝑨𝒔𝒕 =510 mm2
𝜋∗102
Area of 10mm bar = =78.54mm2
4
1000∗78.54
spacing of bars, = =156.25mm
510
Provide 10mm bar @150mm c/c
Bend alternate bars @L/7=3500/7=500mm from the face of
support where moment reduces to less than half its maximum
value
Design of One way slab – Simply Supported
Temperature Reinforcement (Pg 48, section 26.5.2.1)
- 0.12% of the gross concrete area (tor steel)
- which will be provided in the longitudinal direction
% steel = 0.12*1000*170=204mm2
Providing 6mm dia bars
𝜋∗62
Area of 6mm dia bars = =28.27mm2
4
1000∗28.27
spacing of bars, = =138.57mm
204
Provide 6mm bar @130mm c/c
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
Check for shear (Pg 73, Table 19, same as that for beam)
100𝐴𝑠𝑡
% tension steel = - spacing doubled at support
𝑏𝑑
1000
100∗78.5∗ 300
= = 0.174%
1000∗150
Shear strength of concrete for 0.174% of steel
τc=0.294N/mm2
For 170mm thick slab, k=1.25(Clause 40.2.1.1, Pg 72)
τc’=kτc=1.25*0.294=0.37N/mm2
𝑉𝑢 27.22∗103
Nominal Shear stress, 𝜏𝑣 = =
𝑏𝑑 1000∗150
= 𝟎. 𝟏𝟑𝟗 𝑵ൗ𝒎𝒎𝟐 <τc’ =0.37N/mm2, hence safe in shear.
Only minimum shear reinforcement is required
Provide 8mm two legged stirrups @ 300 mm c/c
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
Check for Deflection ( Pg 37,38)
100𝐴𝑠𝑡
% tension steel at mid span =
𝑏𝑑
100∗78.5∗1000
= = 0.35%
1000∗150∗150
𝑓𝑠 = 0.58𝑓𝑦 = 0.58 ∗ 415 = 240.7 𝑁/𝑚𝑚2
from Fig 4, Pg 38, IS 456
Modification factor for l/d ratio , γ=1.3, clause
23.2.1(c)
Modified l/d ratio = 1.3 X20=26
𝑙 3670
Actual , = = 24.5 < 26, ℎ𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒 𝑖𝑛 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑑 150
Design of One way slab – Simply Supported
6Φ@130mmc/c
175mm
10Φ@150mmc/c
10Φ@300mmc/c 3.5m
Detailing of reinforcement in slab
Design of One way slab – Simply Supported
Pbm 2) Design a R C Slab for a room having inside dimension
3m X 7m. The thickness of supporting wall is 300mm. The
slab carries 75mm thick lime concrete at its top. Live load in
slab is 2kN/m2. Assume the slab to be simply supported at its
end. Use M15 concrete and Fe 250
ly =7m
lx=3m
Unit width,1m
ly=7m
fck =15 N/mm2
fy=250 N/mm2 lx = 3 m
Design of One way slab – Simply Supported
𝑙𝑦 7
= = 2.33
𝑙𝑥 3
> 2 − ℎ𝑒𝑛𝑐𝑒 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑠𝑖𝑔𝑛𝑒𝑑 𝑎𝑠 𝑜𝑛𝑒 𝑤𝑎𝑦 𝑠𝑙𝑎𝑏
Clear span = 3m
Considering unit width perpendicular to the span
b=1000 mm
3m
Design of One way slab – Simply Supported
the depth of slab from deflection criteria
l/d ratio for simply supported slab = 20,
clause 23.2.1, pg 37, IS 456
𝑙 3000
‘d’ should be grater than = = 150𝑚𝑚
20 20
‘d’ is the minimum depth
Giving allowance for the % steel provided, the effective depth can be
little bit lower, hence assume a depth of 110mm
Cover to reinforcement
Assuming a clear cover of 15mm and 10mm dia steel for reinforcing bars
Effective cover= 15+10/2=20mm
Total (Overall) depth = 110+20=130mm
Design of One way slab – Simply Supported
Load calculations
Unit wt of lime concrete = 20kN/m3
Dead wt of lime concrete =20X1X0.075=1.5kN/m
Self weight of slab= 1X 0.13 X 25 = 3.25 kN/m
Slab live load= 2kN/m2
Slab live load along the span =2 X 1= 2kN/m
Total load/m =1.5+3.25+2=6.75 kN/m
Effective span = clear span + effective depth of slab
= 3 +0.11=3.11m
Design of One way slab – Simply Supported
Maximum Bending moment and Shear Force
𝑤𝑙 2 6.75𝑋3.112
Working moment = = =8.16 kNm
8 8
Factored Moment = 1.5 X 8.16=12.24 kNm
𝑤𝑙
Maximum Factored Shear Force=
2
1.5𝑋6.75𝑋3.11
= = 𝟏𝟓. 𝟕𝟒𝒌𝑵
2
Design of One way slab – Simply Supported
Depth of slab from flexure criteria
𝑥𝑢,𝑚𝑎𝑥 0.42𝑥𝑢,𝑚𝑎𝑥
𝑀𝑢,𝑙𝑖𝑚 = 0.36𝑓𝑐𝑘 𝑏𝑑 2 1−
𝑑 𝑑
12.24 𝑋 106 = 0.36 ∗ 15 ∗ 1000 ∗ 𝑑 2 ∗ 0.53 1 − 0.42 ∗ 0.53
d=74.17 ≈80mm
An effective depth of 110mm provided is sufficient
Overall depth =110+15+10/2=130mm
Design of One way slab – Simply Supported
Area of tension steel
𝑓𝑦 𝐴𝑠𝑡
𝑀𝑢 = 0.87𝑓𝑦 𝐴𝑠𝑡 𝑑 1 − 𝑓𝑐𝑘 𝑏𝑑
250∗𝐴𝑠𝑡
12.24𝑋106 = 0.87 ∗ 250 ∗ 𝐴𝑠𝑡 ∗ 110 1 − 15∗1000∗110
𝑨𝒔𝒕 =558.93 mm2
𝜋∗102
Area of 10mm bar = =78.54mm2
4
1000∗78.54
spacing of bars, = =140.52mm
558.93
Provide 10mm bar @140mm c/c
Bend alternate bars @L/7=3000/7=428≈430mm from the face
of support where moment reduces to less than half its
maximum value
Design of One way slab – Simply Supported
Temperature Reinforcement (Pg 48, section 26.5.2.1)
- 0.15% of the gross concrete area
- which will be provided in the longitudinal direction
% steel = 0.15*1000*130=195mm2
Providing 6mm dia bars
𝜋∗62
Area of 6mm dia bars = =28.27mm2
4
1000∗28.27
spacing of bars, = =146.51mm
195
Provide 6mm bar @140mm c/c
Design of One way slab – Simply Supported
Check for shear (Pg 73, Table 19, same as that for beam)
100𝐴𝑠𝑡
% tension steel =
𝑏𝑑
1000
100∗78.5∗ 280
= = 0.255%
1000∗110
Shear strength of concrete for 0.255% of steel
τc=0.361N/mm2 for fck= 15N/mm2
For 130mm thick slab, k=1.3(Clause 40.2.1.1, Pg 72)
τc’=kτc=1.3*0.361=0.47N/mm2
𝑉𝑢 15.74∗103
Nominal Shear stress, 𝜏𝑣 = =
𝑏𝑑 1000∗110
= 0.143 𝑁Τ𝑚𝑚2 <τc’ =0.47N/mm2, hence safe in shear.
Only minimum shear reinforcement is required
Provide 8mm two legged stirrups @ 300 mm c/c
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
Design of One way slab – Simply Supported
Check for Deflection ( Pg 37,38)
100𝐴𝑠𝑡
% tension steel at mid span =
𝑏𝑑
100∗78.5∗1000
= = 0.51%
1000∗110∗140
𝑓𝑠 = 0.58𝑓𝑦 = 0.58 ∗ 250 = 145 𝑁/𝑚𝑚2
from Fig 4, Pg 38, IS 456
Modification factor, γ=1.9, clause 23.2.1(c)
𝑙 3110
= = 28.27 < 38, ℎ𝑒𝑛𝑐𝑒 𝑠𝑎𝑓𝑒 𝑖𝑛 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑑 110
Design of One way slab – Simply Supported
6Φ@140mmc/c
130mm
10Φ@140mmc/c
10Φ@280mmc/c 3m
Detailing of reinforcement in slab
Continuous slabs
A simply supported one way slab is supported on its two
sides by beams. The supports carry the loads along one
direction of the slab, therefore it is called a one-way slab.
Whereas, a continuous slab extends over three or
more support beams in a given direction. .
Continuous slabs
Span moments – sagging
tension at bottom fibres
Support moments – hogging
tension at top fibres
A B C D E
Moment and shear coefficients
C D
3m
One way reinforced continuous slabs
Pbm. Design a continuous one way roof slab having three
equal spans of 3m each with the following data. Imposed
load 2.5kN/m2, floor finish 3mm, Concrete grade M15, Steel
grade Fe415
A E B F C G D
3m 3m 3m
l/d <26 for continuous one way slab ( pg 37, clause 23.2.1)
3000/26= 115 mm
Hence provide 120mm thick slab
Owing to symmetry one half of the slab is considered.
One way reinforced continuous slabs
One way reinforced continuous slabs
Loads
Dead load of slab +finish =(0.12+0.03)*1*25=3.75kN/m
Imposed load =2.5*1=2.5kN/m
Factored dead load= 1.5 *3.75 = 5.625kN/m
Total factored dead load = 5.625*3=16.875kN
Factored imposed load = 1.5*2.5=3.75 kN/m
Total factored imposed load = 3.75*3= 11.25 kN
Moment coefficients for continuous beams
Bending Moment Coefficients (Pg 36, Table 12, IS 456)
E B F
Type of Load Span Moment (at near At support next to (at middle of
middle of end span) the end support interior span)
D.L. = 16.875 kN 1/12 -1/10 1/16
L L = 11.25 kN 1/10 -1/9 1/12
1/12*16.875*3 + -1/10*16.875*3 + 1/16*16.875*3 +
Total Moment 1/10*11.25*3 = -1/9*11.25*3 = 1/12*11.25*3 =
7.59 kNm 8.82 kNm 4.92 kNm
One way reinforced continuous slabs
For finding depth the largest moment is 8.82 kNm
2
𝑥𝑢,𝑚𝑎𝑥 0.42𝑥𝑢,𝑚𝑎𝑥
𝑀𝑢,𝑙𝑖𝑚 = 0.36𝑓𝑐𝑘 𝑏𝑑 1−
𝑑 𝑑
8.82X106 =0.36X15X1000X0.48(1-0.42X0.48)d2
d=65.3mm
Provide an overall depth of 110mm
Assume 8 mm bars with 20mm clear cover.
Effective depth , d= 110-20-8/2=86mm
One way reinforced continuous slabs
The area of tension steel
at B
𝑓 𝐴𝑠𝑡
𝑀𝑢 = 0.87𝑓𝑦 𝐴𝑠𝑡 𝑑 1 − 𝑓𝑦 , Annexure G1, Pg 96
𝑐𝑘 𝑏𝑑
415XAst )
8.82X106=0.87X415XAstX86(1- 15𝑋1000𝑋86
Ast required = 316 mm2/m
𝜋∗82
Area of 8mm bar = 4 =50.26mm2
1000∗50.26
spacing of bars, = 316 =159.05mm
spacing 159mm round to 150mm
Use 8mm bars @150mm c/c
1000∗50.26
Ast provided = =335.06mm2/m
150
One way reinforced continuous slabs
The area of tension steel
at E
𝑓 𝐴𝑠𝑡
𝑀𝑢 = 0.87𝑓𝑦 𝐴𝑠𝑡 𝑑 1 − 𝑓𝑦 , Annexure G1, Pg 96
𝑐𝑘 𝑏𝑑
415XAst )
7.59X106=0.87X415XAstX86(1- 15𝑋1000𝑋86
Ast required = 267 mm2/m
Use 8mm bars @180mm c/c
Ast provided = 280 mm2/m
One way reinforced continuous slabs
The area of tension steel
at F
𝑓 𝐴𝑠𝑡
𝑀𝑢 = 0.87𝑓𝑦 𝐴𝑠𝑡 𝑑 1 − 𝑓𝑦 , Annexure G1, Pg 96
𝑐𝑘 𝑏𝑑
415XAst )
4.92X106=0.87X415XAstX86(1- 15𝑋1000𝑋86
Ast required = 204 mm2/m
Use 8mm bars @240mm c/c
Ast provided = 210 mm2/m
One way reinforced continuous slabs
Check
i) The maximum spacing cannot exceed 3dor 300mm
whichever is less (pg 46, 26.3.3 b)
i.e 3X86 = 258mm, hence O K
ii) Minimum area of steel for HYSD bars (pg 48, 26.5.2.1)
=0.12% of total cross sectional area
=(0.12X1000X110)/100=132mm2/m, O K
One way reinforced continuous slabs
Temperature reinforcement
- 0.15% of the gross concrete area will be provided in the
longitudinal direction.
- =(0.15X110X1000)/100
- =165mm2/m
Use 6mm bars @ 180 mm c/c
One way reinforced continuous slabs
Check for shear
Shear Force Coefficients ( Table 13, IS 456)
Type of load At support (A) At support next
to the end
support (B)
D.L =16.875 kN 0.4 0.6
L.L =11.25 kN 0.45 0.6
Shear Force Maximum @B
= 0.6X16.875+0.6X11.25= 16.875kN
One way reinforced continuous slabs
𝑉𝑢 16.875𝑋1000
Nominal Shear stress, 𝜏𝑣 = = =0.196 N/mm2
𝑏𝑑 1000𝑋86
At B % of tension steel
100𝐴𝑠 100𝑋336
= =0.39%
𝑏𝑑 1000𝑋86
From Table 19, pg 73 for 0.39% steel and M15 concrete τc=0.41 N/mm2
From pg 72, clause 40.2.1.1, for solid slabs depth less than 150mm, k=1.3
τc’=1.3X0.41=0.533>0.196 hence safe.
One way reinforced continuous slabs
One way reinforced continuous slabs
One way reinforced continuous slabs
Check for deflection (pg 37, 23.2.1)
𝐿
< 26 and L<10m
𝑑
Deflection maximum will be at E,
Modification factor for deflection ( clause 23.2.1.c)
100𝐴𝑠 100𝑋280
% steel at E 𝑃𝑡 = 𝑏𝑑
= 1000𝑋86 = 0.32%
From Fig 4, Pg 38, IS 456 for Fe 415 i.e fy=415 Mpa,
fs = 0.58 X415X267/280=229.5
Modification Factor =1.35
Allowable L/d = 26X1.35=35.1
Actual L/d= 3000/86=34.88<35.1 Hence safe
One way reinforced continuous slabs
8Φ@150mmc/c
110 mm
6Φ@180mmc/c 8Φ@240mmc/c
8Φ@360mmc/c
8Φ@180mmc/c
Reinforcement Details
