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8/11/2019 Slab Design Me
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Design of slab
Given Data:-
1. No of storey = 52. Plinth height = 0.6 m
3. Length along shorter direction = 5.60 m
4. Length along longer direction = 6.00 m
5. Cover for mild exposure = 15.00 mm
6. Bars used = 10.00 mm (assummed)
7. Bars used in edge strip of slab = 8.00 mm (assummed)
8. Characteristic Strength of concrete, fck= 20.00 N/mm2
9. Characteristic Strength ofsteel, fy= 415.00 N/mm2
Loading
i) Loading on slab
a) Self wt due to dead load = 24000 N/m2
b) Loading due to 25 mm flooring
c) 10 mm thick ceiling plaster 250 N/m2
d) 10 mm thick turfelting 250 N/m2
ii) Load due to partition wall on slab = 1000 N/m2
iii) Live load on floor slab = 4000 N/m2
iv) Live load on roof = 1500 N/m2
v) Basic wind pressure = 1500 N/m2
Ref Step
1. Thickness of slab and durability consideration
lx = 5.6 m
Cover for mild exposure= 15.00 mm
Basic value of span / eff depth = 26
d= 215.3846 mm
Provided d = 110 mmOverall depth = 130 mm
2. Calculation of design load
Dead Load = 3620 N/m2
Live load = 1500 N/m2
Design load = (Factored Load) 1.5 (D.L + L.L) = 7680 N/m2
3. Calculation of moment
ly/lx = 1.071
Case -I - Two adjacent edges discontinous
x (+ve moment at mid span) = 0.0386x (-ve moment at continous edge) = 0.0513
y (+ve moment at mid span) = 0.035
y (-ve moment at continous edge) = 0.047
Longer direction:
8429.57 N-m (+ve moment)
11319.71 N-m (-ve moment)
Shorter direction:
Calculation
IS 456-
2000, Cl
23.2.1
IS 456-
2000,Annex D-
1.1, Table
26
IS 456-
2000,
Table 18
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9296.61 N-m (+ve moment)
12345.70 N-m (-ve moment)
Case -II - Interior Panels
x (+ve moment at mid span) = 0.027
x (-ve moment at continous edge) = 0.0355
y(+ve moment at mid span) = 0.024
y (-ve moment at continous edge) = 0.032
Longer direction:
5780.28 N-m (+ve moment)
7707.03 N-m (-ve moment)
Shorter direction:
6502.81 N-m (+ve moment)
8549.99 N-m (-ve moment)
Case -III - One short edge discontinous
x (+ve moment at mid span) = 0.031
x (-ve moment at continous edge) = 0.0413y (+ve moment at mid span) = 0.028
y (-ve moment at continous edge) = 0.037
Longer direction:
6743.65 N-m (+ve moment)
8911.26 N-m (-ve moment)
Shorter direction:
7466.19 N-m (+ve moment)
9946.89 N-m (-ve moment)
Case -III - One long edge discontinous
x (+ve moment at mid span) = 0.0316x (-ve moment at continous edge) = 0.0420
y (+ve moment at mid span) = 0.028
y (-ve moment at continous edge) = 0.037
Longer direction:
6743.65 N-m (+ve moment)
8911.26 N-m (-ve moment)
Shorter direction:
7610.70 N-m (+ve moment)
10115.48 N-m (-ve moment)
Maxm moment = 12345.70 N-m
Check depth for maxm B.M
Mmax= 0.138fckbd2
d = 66.88 < 110 mm OK
4. Check for shear
Maxm design shear, V = 0.5wlx= 21504 N
Shear Strss, c = V/bd = 0.195 N/mm2
Table C of
SP 16
Table 12.3
of 'RCC
Design' by
Ver hese &
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< Safe minm shear stress, 0.28 N/mm2
5. Area of steel
Case -I - Two adjacent edges discontinous
Middle strip (For shorter direction)
Width of middle strip = 4.5 m
Steel at midspan
M = 8429.57 N-m d = 110 mm
M/bd2
= 0.697
% steel = 0.203%
Steel provided > 0.12% which is minimum
Ast = 223.3 mm2
Spacing of 10 mm dia bar= 351.72 mm
Provided spacing : 330 mm c/c < 3d or 450 mm
Steel at support
M = 11319.71 N-m d = 110 mm
M/bd2
= 0.936
% steel = 0.288%
Steel provided > 0.12% which is minimum
Ast = 316.25 mm2
Spacing of 10 mm dia bar= 248.35 mm
Provided spacing : 225 mm c/c < 3d or 450 mm
Middle strip (For longer direction)
Steel at midspan
Width of middle strip = 4.2 m
M = 9296.61 N-m d = 100.00 mm
M/bd2
= 0.930
% steel = 0.274%Steel provided > 0.12% which is minimum
Ast = 273.5 mm2
Spacing of 10 mm dia bar= 287.17 mm
Provided spacing : 275 mm c/c < 3d or 450 mm
Steel at support
M = 12345.70 N-m d = 100 mm
M/bd2
= 1.235
% steel = 0.371%
Steel provided > 0.12% which is minimum
Ast = 371 mm2
Spacing of 10 mm dia bar= 211.70 mm
Provided spacing : 200 mm c/c < 3d or 450 mm
Edge strip (For shorter direction)
Width of edge strip = 0.75 m
Minm reinforcement @ 0.12% of cross sectional area is provided
Ast = 156 mm2
(Considering 1 m in width)
IS 456-
2000,
Annex D-
SP 16
Table 2
IS 456
Table 19
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No of 8 mm dia bar used = 3 nos
Edge strip (For longer direction)
Width of edge strip = 0.7 m
Minm reinforcement @ 0.12% of cross sectional area is provided
Ast = 156 mm2
(Considering 1 m in width)
No of 8 mm dia bar used = 3 nos
Torsional reinforcement
Area of steel (at corner where two adjacent edges discontinus) =
= 3/4 th Amount of reinforcement required for maxm +ve midspan BM =
167.475 mm2
Reinforcement is to be provided for a distance of lx/5 = 1.12 m
in 4 layers at mutually perpendicular direction both at top & bottom
Spacing of reinforcement of 8 mm dia = 300.1372 mm
Provided spacing : 300 mm c/c
Area of corner steel = 83.7375 mm2
Spacing of reinforcement of 8 mm dia = 600.2745 mm
Provided spacing : 600 mm c/c
6. Check of deflection
Basic value of span / eff depth = 26
% of steel provided along lx direction = 0.203%
Modification factor = 1.6
Allowable value of span / eff depth = 41.6
d= 134.62 mm
> provided i.e. 110 mm, OK7. Check cracking
Steel > 0.12%, minm required, OK.
Spacing of steel < 3d, OK.
Diameter of steel < 200/8 i.e. 25 mm, OK.
.
IS 456-
2000,
Annex D-
1.8 & D-1.9
Amount of reinforcement at corner where one edge is continous & other
discontinous is to be provided equal to half of that provided earlier.
IS 456-
2000, Cl
23.2.1 &
Fig. 3
IS 456-
2000, Cl
25.5.2.1
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Design of slab
Given Data:-
1. Length along shorter direction = 5.60 m
2. Length along longer direction = 6.00 m
3. Cover for mild exposure = 15.00 mm
4. Bars used = 10.00 mm (assummed)
5. Bars used in edge strip of slab = 8.00 mm (assummed)
Loadingi) Loading on slab
a) Self wt due to dead load = 24000 N/m2
b) Loading due to 25 mm flooring 750 N/m2
c) 10 mm thick ceiling plaster 250 N/m2
d) 10 mm thick turfelting 250 N/m2
ii) Load due to partition wall on slab = 1000 N/m2
iii) Live load on floor slab = 4000 N/m2
iv) Live load on roof = 1500 N/m2
Ref Step1. Thickness of slab and durability consideration
lx = 5.6 m
Cover for mild exposure= 15.00 mm
Basic value of span / eff depth = 26
d= 215.3846 mm
Provided d = 140 mm (assummed)
Overall depth = 160 mm
2. Calculation of design load
Dead Load = 5840 N/m2
Live load = 4000 N/m
2
Design load = (Factored Load) 1.5 (D.L + L.L) = 14760 N/m2
3. Calculation of moment
ly/lx = 1.071
Case -I - Two adjacent edges discontinousx (+ve moment at mid span) = 0.0386
x (-ve moment at continous edge) = 0.0515
y (+ve moment at mid span) = 0.035
y (-ve moment at continous edge) = 0.0467
Longer direction:
16200.58 N-m (+ve moment)21600.77 N-m (-ve moment)
Shorter direction:
17866.92 N-m (+ve moment)
23822.56 N-m (-ve moment)
Case -II - Interior Panels
x (+ve moment at mid span) = 0.027
x (-ve moment at continous edge) = 0.0360
Calculation
IS 456-
2000, Cl
23.2.1
IS 456-
2000,Table 18
IS 456-
2000,
Annex D-
1.1, Table
26
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y (+ve moment at mid span) = 0.024
y (-ve moment at continous edge) = 0.0320
Longer direction:
11108.97 N-m (+ve moment)
14811.96 N-m (-ve moment)
Shorter direction:
12497.59 N-m (+ve moment)
16663.45 N-m (-ve moment)
Case -III - One short edge discontinous
x (+ve moment at mid span) = 0.031
x (-ve moment at continous edge) = 0.0413
y (+ve moment at mid span) = 0.028
y (-ve moment at continous edge) = 0.0373
Longer direction:
12960.46 N-m (+ve moment)
17280.61 N-m (-ve moment)
Shorter direction:14349.08 N-m (+ve moment)
19132.11 N-m (-ve moment)
Case -III - One long edge discontinous
x (+ve moment at mid span) = 0.0316
x (-ve moment at continous edge) = 0.0421
y (+ve moment at mid span) = 0.028
y (-ve moment at continous edge) = 0.0373
Longer direction:
12960.46 N-m (+ve moment)
17280.61 N-m (-ve moment)
Shorter direction:
14626.81 N-m (+ve moment)
19502.41 N-m (-ve moment)
Maxm moment = 23822.56 N-m
Check depth for maxm B.M
Mmax= 0.138fckbd2
d = 92.91 < 140 mm OK
4. Check for shear
Maxm design shear, V = 0.5wlx= 41328 NShear Strss, c = V/bd = 0.295 N/mm
2
< Safe minm shear stress, 0.28 N/mm2
5. Area of steel
Case -I - Two adjacent edges discontinous
Middle strip (For shorter direction)
Width of middle strip = 4.5 m
Table C of
SP 16
Table 12.3
of 'RCCDesign' by
Verghese &
IS 456
Table 19
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Steel at midspan
M = 16200.58 N-m d = 140 mm
M/bd2
= 0.827 % steel = 0.241%
Steel provided > 0.12% which is minimum
Ast = 337.40 mm2
Spacing of 10 mm dia bar= 232.78 mm
Provided spacing : 225 mm c/c < 3d or 450 mm
Steel at support
M = 21600.77 N-m d = 140 mm
M/bd2
= 1.102 % steel = 0.328%
Steel provided > 0.12% which is minimum
Ast = 458.53006 mm2
Spacing of 10 mm dia bar= 171.29 mm
Provided spacing : 150 mm c/c < 3d or 450 mm
Middle strip (For longer direction)
Steel at midspan
Width of middle strip = 4.2 m
M = 17866.92 N-m d = 130.00 mm
M/bd2
= 1.057 % steel = 0.313%
Steel provided > 0.12% which is minimum
Ast = 407.14531 mm2
Spacing of 10 mm dia bar= 192.90 mm
Provided spacing : 175 mm c/c < 3d or 450 mm
Steel at support
M = 23822.56 N-m d = 130 mm
M/bd2
= 1.410 % steel = 0.429%
Steel provided > 0.12% which is minimum
Ast = 557.12436 mm2
Spacing of 10 mm dia bar= 140.97 mm
Provided spacing : 125 mm c/c < 3d or 450 mm
Edge strip (For shorter direction)
Width of edge strip = 0.75 m
Minm reinforcement @ 0.12% of cross sectional area is provided
Ast = 192 mm2
(Considering 1 m in width)
No of 8 mm dia bar used = 3 nos
Edge strip (For longer direction)Width of edge strip = 0.7 m
Minm reinforcement @ 0.12% of cross sectional area is provided
Ast = 192 mm2
(Considering 1 m in width)
No of 8 mm dia bar used = 3 nos
Torsional reinforcement
Area of steel (at corner where two adjacent edges discontinus) =
IS 456-
2000,
Annex D-
1.7
IS 456-
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= 3/4 th Amount of reinforcement required for maxm +ve midspan BM =
253.0488 mm2
Reinforcement is to be provided for a distance of lx/5 = 1.12 m
in 4 layers at mutually perpendicular direction both at top & bottom
Spacing of reinforcement of 8 mm dia = 198.6395 mm
Provided spacing : 175 mm c/c
Area of corner steel = 126.5244 mm2
Spacing of reinforcement of 8 mm dia = 397.279 mm
Provided spacing : 375 mm c/c
6. Check of deflection
Basic value of span / eff depth = 26
% of steel provided along lx direction = 0.241%
Modification factor = 1.4
Allowable value of span / eff depth = 36.4
d= 153.85 mm> provided i.e. 140 mm, OK
7. Check cracking
Steel > 0.12%, minm required, OK.
Spacing of steel < 3d, OK.
Diameter of steel < 200/8 i.e. 25 mm, OK.
IS 456-
2000, Cl
25.5.2.1
2000,
Annex D-
1.8 & D-1.9
Amount of reinforcement at corner where one edge is continous & other
discontinous is to be provided equal to half of that provided earlier.
IS 456-
2000, Cl
23.2.1 &
Fig. 3
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Assumption
1. No axial load is coming on beam
3. In case of horizontal loading, point of inflexon occurs at mid span.
4. Maximum +ve B.M. occurs at the centre and maxmimum -ve B.M. occurs at the support.
5. Stiffness of column (I/L) is assummed to be same at each storey.
Ref Step
1. Calculation of load on beam
In larger direction
Overall depth of beam = 500 mm (assummed)
Width of beam = 300 mm (assummed)
Diameter of bar provided in longer direction = 25 mm
Diameter of bar provided in shorter direction = 25 mm
Diameter of 2 lgged stirrup = 8 mm
Spacing of 2 lgged stirrup = 200 mmCover of reinforcement = 25 mm
Height of each storey, h = 3.9 m
Dead Load
For weight of wall coming on beam, 25 to 30% should be deducted for opening
30% of weight of wall per meter on floor = 300 N/m2
Hence, dead load of floor slab on beam = 5540 N/m2
Dead load on beam from slab = 99276.8 N
(Bothside of beam is considered)
Eqivalent slab load, Wsl= 16546.13 N/m
Self weight of beam = 3600.00 N/m
Hence, total D.L., Wd(factored)= 30219.20 N/m
Live load
Live load of floor slab on beam = 4000 N/m2
L.L. on beam from slab = 71680.00 N
(Bothside of beam is considered)
Eqivalent live load on beam per meter = 11946.67 N/m
Hence, total L.L., Wl(factored)= 17920.00 N/m
In shorter direction
Overall depth of beam = 450 mm (assummed)
Width of beam = 300 mm (assummed)Dead Load
Hence, dead load of floor slab on beam = 5540 N/m2
Dead load on beam from slab = 86867.2 N
(Bothside of beam is considered)
Eqivalent slab load, Wsl= 15512 N/m
Self weight of beam = 3240.00 N/m
Hence, total D.L., Wd(factored) = 28128.00 N/m
2. Point of inflexon i.e. the point of zero moment occur at 1/10th span span frm the support in vertical loading.
Calculation
IS 456-
2000,
Cl 24.5
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Live load
L.L. on beam from slab = 62720 N
(Bothside of beam is considered)
Eqivalent live load on beam per meter = 11200 N/m
Hence, total L.L., Wl(factored) = 16800.00 N/m
2. Calculation of B.M. and S.F. on beam in longer direction.
Wd = 30219.20 N/m
Wl = 17920.00 N/m
Span Moment
B.M. near middle of end span = 155169.6 N-m
B.M. at middle of interior span = 99088.8 N-m
Support Moment
B.M. at support next to the end support = -180469.12 N-m
B.M. at other interior support = -162337.60 N-m
Shear forces
At end support = 120910.08 NAt support next to the end support (outer side) = 173301.12 N
At support next to the end support (inner side) = 164235.36 N
At other interior support = 155169.60 N
Hence, maxm +ve B.M. = 155169.6 N-m
Maxm -ve B.M. = -180469.12 N-m
Maxm S.F. = 173301.12 N
3. Calculation of B.M. and S.F. on beam in shorter direction.
Wd = 28128 N/m
Wl = 16800 N/m
Span Moment
B.M. near middle of end span = 126192.64 N-m
B.M. at middle of interior span = 80657.92 N-m
Support Moment
B.M. at support next to the end support = -146748.07 N-m
B.M. at other interior support = -132046.51 N-m
Shear forces
At end support = 105342.72 N
At support next to the end support (outer side) = 150958.08 N
At support next to the end support (inner side) = 143082.24 N
At other interior support = 135206.40 N
Hence, maxm +ve B.M. = 126192.64 N-m
Maxm -ve B.M. = -146748.07 N-m
Maxm S.F. = 150958.08 N
4. Analysis of frame with lateral loads (Considering wind load on longer side of the frame)
(Considering wind on larger side of the frame)
Total surface area = 482.4 Sqm
IS 456-
2000,
Cl
22.5.1,
Table
12
IS 456-
2000,
Cl
22.5.1,
Table
12
IS 456-
2000,
Cl
22.5.1
&
22.5.2,
Table
13
IS 456-2000,
Cl
22.5.1
&
22.5.2,
Table
13
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Nos of bay: 4
Effective surface area (after 20% reduction for opening) = 385.92 Sqm
Basic wind pressure = 1500 N/m2
Total wind load = 578880 N
Wind load/bay = 144720 N
Wind load/bay/each storey = 28944 N
Hence, P = 28944 N
(assummed to be acting on junction)
Calculation of column shear
Assumption
From Portal method, we have, shear in each interior column =
2 X Shear in each exterior column of same storey (x) = 2x
5 storied building is being designed
At Gr floor
x + 2x + 2x + x = 2 x P/2 + 4 x P =
or, 6x = 5P = 144720 N
x = 24120 N
At 1st floor
x + 2x + 2x + x = 1 x P/2 + 4 x P =or, 6x = 4.5P = 130248 N
x = 21708 N
At 2nd floor
x + 2x + 2x + x = 1 x P/2 + 3 x P =
or, 6x = 3.5P = 101304 N
x = 16884 N
At 3rd floor
x + 2x + 2x + x = 1 x P/2 + 2 x P =
or, 6x = 2.5P = 72360 N
x = 12060 N
At 4th floor
x + 2x + 2x + x = 1 x P/2 + 1 x P =
or, 6x = 1.5P = 43416 N
x = 7236 N
At roof
x + 2x + 2x + x = 1 x P/2 =
or, 6x = 0.5P = 14472 N
x = 2412 N
Calculation of column moments
Assumption
Moments at centre is zero. So, point of contraflexure lies at midspan i.e. at mid height
End moment in column = Shear in column X h/2, where h is the storey height.
Moment (end) in exterior columnsAt Gr floor - 1st Floor
M (exterior) = 42330.6 N-m
At 1st floor - 2nd Floor
M (exterior) = 32923.8 N-m
At 2nd floor - 3rd Floor
M (exterior) = 23517.00 N-m
At 3rd floor - 4th Floor
M (exterior) = 14110.2 N-m
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At 4th floor - roof
M (exterior) = 4703.4 N-m
Moment in interior columns
= 2 X Moment in exterior column of same storey.
Calculation of beam moments
At roof
M = 4703.4 N-m
At 4th floor
M = 18813.6 N-m
At 3rd floor
M = 37627.2 N-m
At 2nd floor
M = 56440.8 N-m
At 1st floor
M = 75254.4 N-m
At Gr floor
M = 49566.6 N-mAxial load in interior column = 0, due to wind load
5. Analysis of frame with lateral loads (Considering wind load on shorter side of the frame)
Total surface area = 337.68 Sqm
Nos of bay: 3
Effective surface area (after 20% reduction for opening) = 270.144 Sqm
Basic wind pressure = 1500 N/m2
Total wind load = 405216 N
Wind load/bay = 135072 N
Wind load/bay/each storey = 27014.4 N
Hence, P = 27014.4 N
(assummed to be acting on junction)
Calculation of column shear
Assumption
From Portal method, we have, shear in each interior column =
2 X Shear in each exterior column of same storey (x) = 2x
5 storied building is being designed
At Gr floor
x + 2x + 2x + 2x + x = 2 x P/2 + 4 x P =
or, 8x = 5P = 135072 N
x = 16884 N
At 1st floor
x + 2x + 2x + 2x + x = 1 x P/2 + 4 x P =or, 8x = 4.5P = 121564.8 N
x = 15195.6 N
At 2nd floor
x + 2x + 2x + 2x + x = 1 x P/2 + 3 x P =
or, 8x = 3.5P = 94550.4 N
x = 11818.8 N
At 3rd floor
x + 2x + 2x + 2x + x = 1 x P/2 + 2 x P =
Beam moment acts in opposite direction to that of column at that joint so that sum of column
moment is equal to the sum of beam moment.
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or, 8x = 2.5P = 67536 N
x = 8442 N
At 4th floor
x + 2x + 2x + 2x + x = 1 x P/2 + 1 x P =
or, 8x = 1.5P = 40521.6 N
x = 5065.2 N
At roof
x + 2x + 2x + 2x + x = 1 x P/2 =or, 8x = 0.5P = 13507.2 N
x = 1688.4 N
Calculation of column moments
Assumption
Moments at centre is zero. So, point of contraflexure lies at midspan i.e. at mid height
End moment in column = Shear in column X h/2, where h is the storey height.
Moment (end) in exterior columns
At Gr floor - 1st Floor
M (exterior) = 29631.42 N-m
At 1st floor - 2nd Floor
M (exterior) = 23046.66 N-mAt 2nd floor - 3rd Floor
M (exterior) = 16461.9 N-m
At 3rd floor - 4th Floor
M (exterior) = 9877.14 N-m
At 4th floor - roof
M (exterior) = 3292.38 N-m
Moment in interior columns
= 2 X Moment in exterior column of same storey.
Calculation of beam moments
At roof
M = 3292.38 N-m
At 4th floor
M = 13169.52 N-m
At 3rd floor
M = 26339.04 N-m
At 2nd floor
M = 39508.56 N-m
At 1st floor
M = 52678.08 N-m
At Gr floor
M = 36385.02 N-mAxial load in interior column = 0, due to wind load
6. Determination of design moment
Beam (interior) in larger direction (at 3rd floor)
+ve moment due to vertical load = 155169.6 N-m
-ve moment due to vertical load = 180469.12 N-m
End(-ve) moment due to wind load: 26339.04 N-m
Design +ve moment at mid span = 155169.6 N-m
Beam moment acts in opposite direction to that of column at that joint so that sum of column
moment is equal to the sum of beam moment.
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Design -ve moment at support = 206808.16 N-m
Axial load on interior column =
Beam (interior) in shorter direction (at 3rd floor)
+ve moment due to vertical load = 126192.64 N-m
-ve moment due to vertical load = 146748.07 N-m
End(-ve) moment due to wind load: 37627.2 N-m
Design +ve moment at mid span = 126192.64 N-mDesign -ve moment at support = 184375.27 N-m
Axial load on interior column =
Maxm S.F. = 173301.12 N
7. Design of flexural reinforcement in longer direction.For an under-reinforced section,Mmax= 0.138fckbd
Effective depth, d = 433 mm
Effective depth, d (provided) = 462.5 mm > reqd
At mid span
Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd
2
= 2.418 Pt= 0.804%
Ast(reqd) = 1205.77 mm2
No of 25 mm bars provided at bottom of mid span = 3
At support
Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd2= 3.223
Pt= 1.183%
Ast(reqd) = 1774.65 mm2
No of 25 mm bars provided at top of support = 4
8. Design of flexural reinforcement in shorter direction.
For an under-reinforced section,Mmax= 0.138fckbd2
Effective depth, d = 390 mm
Effective depth, d (provided) = 412.5 mm > reqd
At mid span
Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd2= 2.472
Pt= 0.826%
Ast(reqd) = 1115.74 mm2
No of 25 mm bars provided at bottom of mid span = 3
At support
Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd2= 3.612
Pt= 1.417%
Ast(reqd) = 1913.23 mm2
No of 25 mm bars provided at top of support = 4
8. Design of shear reinforcement.
Design shear force, V = 173301.12 N
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Shear stress, v= V/bd = 1.155 N/mm2
Design shear stress of cocrete, c= *0.85 (0.8 fck)(15) -1+/6
where = max(1,0.8fck/0.89pt)
Pt= 1.183%
= 1.96
c= 0.661 N/mm2
< v
Hence, shear reinforcement necessary.Shear reinforcement provided for shear force, Vs = V - cbd
Vs = 74198.4 N
Using vertical stirrups of 2 legged 8 mm c/c (Fe 250 grade)
Asv= 100.53 mm2
Sv= 0.87 fyAsvd/Vs Sv= 226.25 mm
Provided spacing = 200 mm
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Assumption
1. Typical interior column is considered. (Project work is consulted)
3. In case of horizontal loading, point of inflexon occurs at mid span.
4. Maximum +ve B.M. occurs at the centre and maxmimum -ve B.M. occurs at the support.5. Stiffness of column (I/L) is assummed to be same at each storey.
Ref Step
1. Calculation of load on column
(a) Column(interior, next to end support) at 3rd floor, Total no: 6
At 3rd floor, S.F. of beam at next to end support (longer direction)
Outer Side: 173301.12 N
Inner Side: 164235.36 N
Avg S.F. at c.g. of column = 168768.24 N
Load = 337536.48 N
S.F. of beam at next to end support (shorter direction)Outer Side: 150958.08 N
Inner Side: 143082.24 N
Avg S.F. at c.g. of column = 147020.16 N
Load = 294040.32 N
Load coming on column from roof slab and beam at roof level
Roof slab = 258048 N
Roof beam (longer side) = 23976 N
Roof beam (Shorter side) = 19353.6 N
Total: 301377.6 N
Load coming on column from 4th floor (assummed as same from 3rd floor)
i.e. 631576.8 N
Deduction of live load @ 10% for each floor from top:
From position of the column, total deduction: 20.00%
Total deduction = 40320 N
Design axial load on column : 1524.21 KN
(b) Column(exterior) at 3rd floor in longer side, Total no: 6
S.F. of beam at end support (longer direction): 120910.08 N
Load = 241820.16 N
S.F. of beam at end support (shorter direction): 105342.72 N
Load = 210685.44 N
Load coming on column from roof slab and beam at roof level
Roof slab = 129024 N
Roof beam (longer side) = 11988 NRoof beam (Shorter side) = 19353.6 N
Total: 160365.6 N
Load coming on column from 4th floor (assummed as same from 3rd floor)
i.e. 452505.6 N
Deduction of live load @ 10% for each floor from top:
From position of the column, total deduction: 20.00%
Total deduction = 38707.2 N
Design axial load on column : 1026.67 KN
2. Point of inflexon i.e. the point of zero moment occur at 1/10th span span frm the support in vertical
loading.
Calculation
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Column(exterior) at 3rd floor in shorter side, Total no: 4
S.F. of beam at end support (longer direction): 120910.08 N
Load = 241820.16 N
S.F. of beam at end support (shorter direction): 105342.72 N
Load = 210685.44 N
Load coming on column from roof slab and beam at roof level
Roof slab = 129024 N
Roof beam (longer side) = 23976 NRoof beam (Shorter side) = 9676.8 N
Total: 162676.8 N
Load coming on column from 4th floor (assummed as same from 3rd floor)
i.e. 452505.6 N
Deduction of live load @ 10% for each floor from top:
From position of the column, total deduction: 20.00%
Total deduction = 38707.2 N
Design axial load on column : 1028.98 KN
(d) Column(corner) at 3rd floor, Total no: 4
S.F. of beam at end support (longer direction): 120910.08 N
Load = 241820.16 NS.F. of beam at end support (shorter direction): 105342.72 N
Load = 210685.44 N
Load coming on column from roof slab and beam at roof level
Roof slab = 64512 N
Roof beam (longer side) = 11988 N
Roof beam (Shorter side) = 9676.8 N
Total: 86176.8 N
Load coming on column from 4th floor (assummed as same from 3rd floor)
i.e. 452505.6 N
Deduction of live load @ 10% for each floor from top:
From position of the column, total deduction: 20.00%
Total deduction = 19353.6 N
Design axial load on column : 971.83 KN
2. Calculation of bending moment on column
B.M. due to dead load & live load = 0
B.M. due to wind load (on larger side) = 47034 N-m
B.M. due to wind load (on shorter side) = 32923.8 N-m
Design B.M.= 47034 N-m
3. Provided section of column, size & no. of reinforcement:
400 mm X 400 mm
Main bar: 4 nos - 25 mm
Tie: 8 mm Cover: 40 mm
M25 fck= 25.00 N/mm2
4. Short Column or Slender Column ?
Unsupported length, ly = 3400 mm (for buckling about y-axis)
Unsupported length, lx 3450 mm (for buckling about x-axis)
Lateral dimension, Dx = 400 mm
Lateral dimension, Dy = 400 mm
Grade of concrete:
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ley = kyx ly = 4488 mm
Buckling with respect to major (local x-) axis:
jt(Ic/ hs)= 1.09E+06 mm3
jt(Ib/ lb)= 8.14E+05 mm3
1= 2 = 0.47
Referring to Fig. 27 of the Code, ky= 1.39
lex = kxx lx = 4795.5 mm
Alternative: Effective Lengths by Formulas
Buckling with respect to minor (local y-) axis:
1= 2= 1.05
ky = 1.32
ley = 4471.26 mm
Buckling with respect to major (local x-) axis:1= 2= 1.34
kx = 1.40
lex = 4841.70 mm
ley/Dy = 11.22 < 12
lex/Dx = 11.99 < 12
The column should be designed as a short column.
5.
Minimum Eccentricities
ex,min= 20.233 (> 20.0 mm) < 0.5Dx= 20.00 mm
ey,min= 20.133 (> 20.0 mm) < 0.5Dy= 20.00 mm
Design load (factored) = 1524.21 KN
Primary moments for design
Mux = 47.03 KN-M
Muy = 32.92 KN-M
Corresponding (primary) eccentricities:
ex= 30.86 > ex,min= 20.233 mm
ey= 21.60 > ey,min= 20.133 mm
The primary eccentricities should not be less than the minimum eccentricities.
Primary moments for design:
Mux = 47.03 KN-M
IS 456
2000 Cl.
25.4
Ratio of the unsupported length (l) to the least lateral dimension (d) of a column i.e. l/d 60
Note 2: Considering the effective lengths given by the Code charts, the slenderness ratios of
the column are obtained as follows:
Note 1: The effective lengths predicted by the two different methods are fairly close.
2000
Annex E
Fig 27
IS 456
2000 Cl.
25.3.1
As the column is unbraced and bent in double
curvature, Mu= M2+ Mafor unbraced columns, where
M2is the higher of the two end moments
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Muy = 32.92 KN-M
Additional moments
Without modification factor, additional eccentricities
eax= Dx(lex/Dx)2/2000 29.30 mm
eay= Dy(ley/Dy)2/2000 24.99 mm
Assuming modification factors kax= kay 0.5 (to be verified later),
additional moments:Max= Pu(kax eax) 22.33 KN-M
May= Pu(kay eay) 19.05 KN-M
Total factored moments
Mux = Mux+ Max= 69.37 KN-M
Muy = Muy+ May= 51.97 KN-M
Trial section as slender column
Designing for a resultant uniaxial moment with respect to the minor axis,
99.68 KN-M
Pu= 1524.21 KN
0.38
0.06
Assuming 25 for main bars, 8 ties and 40 mm clear cover, d' = 60.5 mm
d'/D = 0.15 0.15
with "Reinforcement Distributed Equally on Four Sides" and referring chart of SP 16
p/fck= 0.04 preqd = 1.00 As,reqd= 1600 mm2
As,provided= 1963.50 mm2
> 1600.00 mm2
Hence, OK
pprovided= 1.23% p/fck= 0.049
Check additional moments
Assuming a clear cover of 40mm d' = 60.5 mm
d'/Dx= 0.15 0.15
d'/Dy= 0.15 0.15
Corresponding to p/fck= 0.049
For d'/Dx= 0.15 Pub,x/fckbD = 0.11
Pub,x= 440.00 KN
For d'/Dy= 0.15 Pub,y/fckbD = 0.11
Pub,y= 440.00 KN
Puz
= 0.45fck
Ag+ (0.75f
y 0.45f
ck)A
s P
uz=
2389.05 KN Modification factors:
The actual (revised) total moments are obtained as:
Mux = Mux+ Max= 66.85 KN-M
Referring to Charts of SP :16, the ultimate loads Pub,x & Pub,yat balanced failure can be
determined by considering the stress level fyd= 0.87fy
0.444
0.444
Pu/fckbD =
Mu/fckbD2=
Mu1.15(Mux2+Muy
2)
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Muy = Muy+ May= 49.83 KN-M
Check safety under biaxial bending
Referring to the design Charts in SP : 16, uniaxial moment capacities corresponding to
Pu/fckbD = 0.38 & p/fck= 0.049
For d'/Dx= 0.15 Mux1/fckbD2= 0.09
For d'/Dy= 0.15 Muy1/fckbD2= 0.09
Mux1= 144.00 KN-M > Mux = 66.85 KN-MMuy1= 144.00 KN-M > Muy = 49.83 KN-M
Pu/Puz= 0.638 which lies between 0.2 & 0.8
n= 1.73
6. Transverse reinforcement
Minimum diameter, t = 6.25 mm
Maximum spacing,st= 300 mm
Provided 8 ties @ 300 c/c
Trial section as short column
Designing for a resultant uniaxial moment with respect to the minor axis,
66.02 KN-M
Pu= 1524.21 KN
0.38
0.04
Assuming 25 for main bars, 8 ties and 40 mm clear cover, d' = 60.5 mm
d'/D = 0.15 0.15
with "Reinforcement Distributed Equally on Four Sides" and referring chart of SP 16p/fck= 0.01 preqd = 0.8 As,reqd= 1280 mm
2
As,provided= 1963.50 mm2
> 1280.00 mm2
Hence, OK
pprovided= 1.23% p/fck= 0.049
Check safety under biaxial bending
Referring to the design Charts in SP : 16, uniaxial moment capacities corresponding to
Pu/fckbD = 0.38 & p/fck= 0.049
For d'/Dx= 0.15 Mux1/fckbD2= 0.07
For d'/Dy= 0.15 Muy1/fckbD2= 0.07
Mux1= 112.00 KN-M > Mux = 47.03 KN-M
Muy1= 112.00 KN-M > Muy = 32.92 KN-M
Pu/Puz= 0.638 which lies between 0.2 & 0.8
n= 1.73
Transverse reinforcement
< 1 Hence, safe0.42
0.34 < 1 Hence, safe
IS 456
2000 Cl.26.5.3.2
Mu1.15(Mux2+Muy
2)
Pu/fckbD =
Mu/fckbD2=
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Assumption
Bearing capacity of soil = 120 KN/m2
Depth of soil tested for bearing capacity = 1.5 m
Characteristic strength of concrete, fck= 20 N/mm2
Clear cover: 75 mm
Diameter of bar used = 25 mm
Ref Step
1. Calculation of load on footing (under the column designed).
Load on column at 3rd floor = 1524.211 KN
Includes load from roof = 301.378 KN
-do- from 4th floor = 631.577 KN
-do- from 3rd floor = 631.577 KN
Add: Load from 2nd floor = 631.577 KN
Add: Load from 1st floor = 631.577 KN
Total: 2787.365 KN
Deduction of live load for 1st & 2nd floor = 20%
Total deduction = 40.32 KN
Total load after deduction = 2747.045 KNSelf wt of column @ 6 KN/m
180.9 KN (factored)
Total axial load on footing = 2927.945 KN
Self wt of footing (assummed)= 10.00%
292.794 KN
Design axial load = 3220.739 KN (factored)
Design bending moment = 84.661 KN-M
(due to wind load on larger side on G.L.)
2. Design of isolated footing, eccentrically loaded
Size of footingResultant eccentricity of loading at footing base,
e = 26.29 mm
Assuming e < L/6 (i.e., L > 158 mm
where, L = Dimension along the plane of section of moment.
B = Bradth
q1 & q2= Foundation pressure
Now, soil pressure = 180 KN/m2
BL2 17.893 L 2.82204 0
BL2 17.893 L 2.82204 0
L2 4.473 L 0.70551 0
L = 4.626 m
Assume B = 4.00 m
L = 4.75 m
Projection (in the short direction) = 2175.00 mm
Calculation
Assuming an enhanced soil pressure under ultimate loads is equivalent to considering
allowable pressures at the serviceability limit state. A load factor of 1.5 is assumed here.
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Projection (in the long direction) = 1800.00 mm
Maxm soil pressure, q1= 175.14 KN/m2
Minm soil pressure, q2= 163.88 KN/m2
Thickness of footing based on shear
(a) One-way shear
Assume: d = 975 mmDepth at edge = 250 mm
172.30 KN/m2
0.172 N/mm2
The section at critical plane is trapezoidal.
Width at top = 3300 mm (1:2 slope is maintained)
Eff depth at critical plane, d' = 650.00 mm
Depth of N.A = k d' = 183.95 mm k = 0.283
Width of the section at N.A, b' = 3710.35 mm
Shear force at critical plane,V' = 861.82 KN
Shear stress, =V'/(b'd') = 0.357 N/mm2 > allowable c = 0.36
(for M 20 concrete with nominal pt = 0.25) N/mm2
(b) Two-way shear
Calculated shear stress, v= ksc ks= 1
c = 1.12 N/mm2
Width of footing at critical plane, ao(x-x dirn) = 1375 mm
Width of footing at critical plane, bo(y-y dirn) = 1375 mm
Depth of footing at critical plane, do= 812.5 mm
Pressure under column axis, q = 169.51 KN/m2
Punching shear force, F = 2900.25 KN
Punching shear stress, v= 0.6490 < c= 1.12 N/mm2
(c) Depth from bending
Section Y-Y
Moment about y-y axis, Myy = 1603.80 KN-M
Section X-X
Moment about y-y axis, Myy = 1304.40 KN-M
Width of equivalent rectangle for trapejoidal shaped footing, beqv= b + (B-b)/8
beqv= 943.75 mm
For an under-reinforced section,Mmax= 0.138fckbd2
deff= 784.68 mm < d (assummed) = 975 mm
D 1070 mm
effective depth (long span) = 982.5 mm
The critical section is located d away from the column face. Intensity of pressure
contributing to the factored one-way shear is
IS 456
2000
Table 19
The critical section is located d/2 from the periphery of the column all around. The
average pressure contributing to the factored two-way shear is
IS 456
2000 Cl
31.6.3.1
Hence, one-way shear governs the footing slab thickness and d 975 mm. Assuming a
clear cover of 75 mm and a bar diameter of 25 mm,
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effective depth (short span) = 957.5 mm
3. Check maximum soil pressure (at factored loads)
Assumption
Unit weights of concrete = 24.00 kN/m3
unit weights of soil = 18.00 kN/m3
qmax-gross
= 186.754 kN/m3
> 180 kN/m3
Hence O.K.
4. Design of flexural reinforcement
(a) long span
Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd2= 0.350
Pt= 0.099% < Pt(assummed for one way shear) , 0.25 %
Pt(reqd)= 0.250%
Ast(reqd) = 11667.19 mm2
No of 25 mm bars provided = 24Corresponding spacing = 190
Development length required = 1175 mm
Length available from face of the column = 2175 mm > Reqd length
Hence, OK.
Provided 24 no 25 mm bars @ 190 mm c/c along long span.
(a) Short span
Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd2= 0.356
Pt= 0.101% < Pt(assummed for one way shear) , 0.25 %
Pt(reqd)= 0.250%
Ast(reqd) = 9575.00 mm2
No of 25 mm bars provided = 20
Corresponding spacing = 225
Development length required = 1175 mm
Length available from face of the column = 1800 mm > Reqd length
Hence, OK.
Provided 20 no 25 mm bars @ 225 mm c/c along long span.
Placing of steel
(a) Reinforcements in long direction placed uniformly across the full width.
5. Transfer of forces at column base
Capacity =0.45 fck x 2 x Area = 2880 KN < 2927.945
Hence, dowels are needed.
At least 4 no bars equal to 0.5% of the area of supported column are extended to the footing.
Reinf in central brand width/ Total reinf in short direction = 2/(1), where is the ratio of
the long side to the short side.
The critical sections for moment are located at the faces of the column in both directions
(XX and YY)
IS 456
2000 Cl
34.3.1
(b) Reinforcement in short direction will be placed in a central band equal to the width of
the footing and portion of the reinforcement determined for central band as per ratiobelow,
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Building Construction Tips
1. Anti-Termite Treatment:
a) Heptachlor 0.5 %
b) Chlordane 1.0%
c) Chloropyrifos 1.0%
2. Hydrated Lime & Clay Pozzolana:
3. Use of Flyash:
To prevent attack of termites, any of the following chemicals may be used by mixing with water to
make a solution that can be sprayed @ 5 liters per Sq Meter.
May be used for all masonry mortar and plaster. 20 to 30% cement can be saved by using Lime
Pozzolana. Lime Pozzolana mortar is cheaper than cement sand mortar. It also improves workability.
Qty of cement can be saved up to 20%, if Flyash is used in conjunction with Cement in mortars and
plasters.
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Cubical Content with floor wise structural cost
Ground Floor
Qty Unit Mark
1. Total Area: 176.53 Sq M A
2. Floor to Floor Height: 3 M B
3. Plinth Height: 0.6 M C
4. Depth of Foundation: 1.2 M D
5. Number of Floors 2 E
6. Ht of Parapet 1 M F
Cubical Content: [A X {B X E + C + (D/2) + F/2}] = 1359.28
Rate: 3,250.00 per Cum =
ABSTRACT
1. Total Structural Cost: = 4,417,663.25
2. Internal S & P works: 15% of Structural Cost = 662,649.49
3. Internal Electrical Works: 20% of Structural Cost = 883,532.65
4. Anti Termite: 1% of Structural Cost = 44,176.63
5. Quality Control: 1% of Structural Cost = 44,176.63
6. Foundation Treatment: 6% of Structural Cost = 265,059.80
7. External Electrification: 6% of Structural Cost = 265,059.80Sub Total 6,582,318.24
Add: Contingency: 3% = 197,469.55
Add: Administrative Charge: 8.5% = 559,497.05
7,339,284.84
4,417,663.25
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Cu M
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Spacing of Stirrup
Beam: Simply Supported Beam
Dimension: Wide: 200 mm
Eff Depth: 500 mm
Loading: F= 25000 N/m
(Uniformly distributed load including its own weight)
Eff Span: l = 6.00 mDetermination the length over which vertical stirrups to be provided
cbc = 5 N/mm2
c = 0.35 N/mm2
c max= 1.6 N/mm2
sv = 140 N/mm2
fy = 250 N/mm2
Maxm Shear Force, V = 75000 N
Shear Stress, v= 0.75 N/mm2 > 0.35 N/mm
2 Shear reinf
< 1.6 N/mm2
Distance (x) from the end of the beam where permissible shear stress is developed is given by
x = 1.6 m
Shear to be resisted by vertical stirrups, Vs = V - c . b . d 40000 N
Shear Reinf: Dia of reinf: 6 mm (2 legged stirrup)
Asv 56.57 mm2
Spacing, Sv at support: Asv . sv. D / Vs = 99.00 mm
Maxm Spacing: 0.75 d = 375 mm
Asv/bsv 0.4/fy where Asv 31.68 mm2
Sv Asvfy/0.4 b 176.78 mm
Hence, 6 mm dia 2 legged stirrups near support at 95 mm c/c is adopted.
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Design of Beam
Data given:
Section of the Beam: Rectangular Grade of Concrete: M 30
Grade of Steel: Fe 250
b= 250 mm
d= 350 mm Analysis as underreinforced beaml= 4 mm
Ref Step
1 Depth of N.A. for balanced failure
SP 16
For M 30, Es= 200000 N/mm2
For Fe 250, fy = 250 N/mm2
x/d = 0.531
Balanced Percentage of Steel
IS 875 Assumed unit wt of RCC: 25 kN/m3
(Uniformly distributed load including its own weight)
Dead Load = 0.2 x 1 x 25 = 5 kN/m2
Live Load= 300 kg/m2
= 3 kN/m2
2 Factored Load, Moment and Shear
Factoted Load, w = 1.5 ( D.L + L. L)
= 12 kN/m2
Factored (design) moment, Mu = wL2/8
= 24 kN-m
Design Shear, Vu : wL/2 = 24 kN
3 Load for serviceability condition, ws
ws= 1.0 (D.L + L.L) = 8 kN/m2
Calculation
IS 456
Table 12
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Design of Doubly Reinforced Beam
Data given:
Section of the Beam: Rectangular fck = 30 N/mm2
fy = 415 N/mm2
b= 280 mmd= 510 mm Analysis as underreinforced beam
d' = 50 mm
Asc= 402 mm2
Ast= 2455 mm2
Determination of ultimate moment capacity
Ref Step
Method I Solution by Strain Compatibility
1. Choose a suitable NA depth
xu/d for Fe 415 = 0.48
x = 244.8 mm
x= 240 mm (adopted)
2. Check strain and stress in steel
Ec= 0.0035
Est= 0.0039
Yeild strain of Fe 415 = 0.0038 Steel yields
3. Total Tension in steel
T = 0.87 fyAst
T = 886.38 kN
4. Check strain in compression steel
Esc= 0.0035 (1-d'/x) 0.0028
Corresponding stress in steel = 350 N/mm2
Cs= 140.7 kN
5. Compression in concrete and its point of action
Calculation
Compression steel doesn't yield.
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Cc= 725.76 kN
Center of Cc= 0.42 x = 100.8 mm
6. Total Compression
C = 866.46 kN
7. Check T = C
T = 886.38 kN C = 866.46 kN
8. Ultimate moment i.e. moment of Csand Ccabout tension steel
Mu = 361.70 kNm
Method II Solution by formulae
1. Mu1for concrete failure as singly reinforced
= 301.51 kNm
2. Balanced steel for Fe 415, fck= 30
p = 1.43%
Ast
=2042.04
mm2
3. Compression in steel
d'/d = 0.10
fsc= 353 N/mm2
Cs= 141.91 kN
4. Value of Mu2due to compression steel
Mu2= 65.28 kNm
5. Total Muconsidering compression failure
Muc= Mu1+ Mu2 = 367 kNm
6. Additional tension steel available
Ast2= Ast- Ast1 = 412.96 mm2
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