Floor beam, F. B (C3, C4):-
Beam size = 10″ x 12″
Effective depth, d = 12″ -2.5″ = 9.5″
a) Load Calculation:-
Self weight of beam = x 150 = 125 Ib/ft
All main wall weight = ( 10′ – ) x 120 = 450 Ib/ft
Total Dead load = 125 + 450 = 575 Ib/ft
Total Live load = 40 x .83 = 33.2Ib/ft
Factored Load = 1.2 x 575 + 1.6 x 33.2 = 743.12Ib/ft
b) Moment Calculation and ‘d ‘check.
(-) Moment at Int. support = = = 4.59 k-ft
(+) Ve moment at Mid span = = = 3.16 k-ft
Mu = 4.59 k-ft
b = 0.85 x ßl x x
= 0.85 x 0.85 x x
= 0.037
max = 0.75 ρb = 0.037 x .75 = 0.028
Mu = bd fy (1- 0.59 x
or, 4.59 x 12= 0.9 x 0.028 x 10 x d x 40 x(1- 0.59 x 0.028 x
= 2.65″ < 9.5″
So, ok
c) Reinforcement Calculation:
Steel for Int. Support:-
= = 56.509 <200 psi
= 0.005
As = bd = 0.005 x 10 x 9.5 = 0.475 in
As (min) = 0.0018 x b x t = 0.0018 x 10 x 12 = 0.216 in
Use 2 # 5 Straight bars.
Steel for mid span:-
= = 46.69 < 200 psi
= 0.005
As = bd = 0.005 x 10 x 9.5 = 0.475in
Use ,2 # 5 Straight bar.
Floor beam, (A1, B1):-
Beam size = 15″ x 10″
Effective depth, d = 15″ -2.5″ = 12.5″
a) Load Calculation:
Self weight of beam F. B-3 = x 150 = 156.25 Ib/ft
All main wall weight = ( 10′ – ) x 120 = 437.5 Ib/ft
Load coming from Slabs:-
S5- Dead load = 185 x ( + 0.5’)x 0.71 = 881.36 Ib/ft
Total, Dead load = 156.25 + 437.5 + 881.36 = 1475.11 Ib/ft
Live load = 40 x ( + 0.5’) x 0.71
= 190.56 Ib/ft
Factored load, (F. B-3) = 1.2 x D.L + 1.6 L.L
= 1.2 x 1475.11 + 1.6 x 190.56
= 2075.028 Ib/ft
=2.075 k/ft
b) Moment Calculation and ‘d check.
(-) Ve Moment at Ext. support = = 32.50 k-ft
(+) Ve moment at Mid span = = 37.14 k-ft
(-) Ve Moment at Int. support = = 51.997 52 k-ft
Mu = 52 k-ft
b = 0.85 x ßl x
= 0.85 x 0.85 x x
= 0.037
max = 0.75 b = 0.75 x 0.037 = 0.0278
Mu = bd fy (1- 0.59 x
Or, d =
= = 8.93″ < 12.5″
So, ok
c) Reinforcement Calculation:
Steel for Ext. Support:-
R= = = 277.33 Psi >200 psi
= 0.0071
min = = = 0.0050
min = = = 0.0041
As = bd = 0.0071 x 10 x 12.5 = 0.89 inch
Use 2 # 5 Straight bar’s + 1 # 5 Ext: top.
Ap = 0.96inch
Steel for mid span:-
R= = 316.93 psi > 200 psi
= 0.0077
min = = 0.0050
min = = 0.0041
As = bd = 0.0077 x 10 x 12.5 = 0.962 in
Use 2 # 5 Straight bars + 1 # 5 Ext: Bottom.
Ap = 0.965in
Steel for Int. Support:-
R= = = 443.73 psi > 200 psi
= 0.013
As = bd = 0.013 x 10 x 12.5 = 1.625 in
Use 2 # 5 Straight bars + 2 # 6 Ext: top.
Ap = 1.65in
d) Stirrup design:-
Vv = 1.15 x = 18.89 K at support face
Critical shear strength:
Vu = 18.89 – x 2.075 = 16.73 k
Concrete shear strength:
Vc = 2 x 10 x = 13.70k
Vc = 0.75 x 13.70 = 10.275
Vs = (Vu – Vc) = 16.73 – 10.275 = 6.45
4 bwd = 4 x 0.75 x 10 x = 20.54 > Vc
Smax = = 6.25″
Smax = 24″
Use, 2 legs # 3 @ 6″ c / c throught the beam length.
Floor beam (B1, C1):-
Beam size = 10″ x 15″
Effective depth, d = 15″ -2.5″ = 12.5″
a) Load Calculation:-
Self weight of beam F. B-10 = x 150 = 156.25 Ib/ft
All main wall weight = ( 10 – ) x 120 = 437.5 Ib/ft
Load coming from Slabs:-
S1, Dead load = 185 x ( x 0.5) x 0.71 = 881.36 Ib/ft
Total, Dead load = 156.25 + 437.5 + 881.36 = 1475.11 Ib/ft
Live load = 40 x ( + 0.5) x 0.71
= 190.56 Ib/ft
Factored load, (F. B-10) = 1.2 D.L + 1.6 L.L
= 1.2 x 1475.11 + 1.6 x 190.56
= 2075.028 Ib/ft
=2.075 k/ft
b) Moment Calculation and ‘d’ check:-
(-) Ve Moment at Ext. support = = 29.83 k-ft
(+) Ve moment at Mid span = = 34.1 k-ft
(-) Ve Moment at Int. support = = 47.43 k-ft
Mu = 47.43 k-ft
b = 0.85 x ßl x
= 0.85 x 0.85 x x
= 0.0371
max = 0.75 b = 0.75 x 0.0371 = 0.0278
Mu = bd fy (1- 0.59 x
Or, 47.43 x 12 = 0.90 x 0.0278 x 10 x d x 40 x
= = 5.11″ < 12.5″
So, ok
c) Reinforcement Calculation:
Steel for Ext. Support:-
R= = 254.55 psi > 200 psi
= 0.0065
min = = 0.0050
min = = 0.0041
As = bd = 0.0065 x 10 x 12.5 = 0.812 in
Use 2 # 5 Straight bar’s + 1 # 5 Ext: top.
Ap = 0.96in
Steel for mid span:-
R= = 290.99 psi > 200 psi
= 0.0070
min = = 0.0050
As = bd = 0.0070 x 10 x 12.5 = 0.875 in
Use 2 # 5 Straight bars + 1 # 5 Ext: Bottom bar.
Ap = .96 in
Steel for Int. Support:-
R= = = 407.3 Psi > 200 psi
= 0.011
As = 0.011 x 10 x 12.5 = 1.375 in
Use 2 # 5 Straight bars + 2 # 6 Ext: top.
Ap = 1.65in
d) Stirrup design:-
Vv = 1.15 x = 14.7 K at support face
Critical shear strength:
Vu = 14.7 – x 2.075 = 12.53 k
Concrete shear strength,
Vc = 2 x 10 x = 13.70k
Vc = 0.75 x 13.70 = 10.275
Vs = (Vu – Vc) = 12.53 – 10.275 = 2.26
4 bwd = 4 x 0.75 x 10 x = 20.54 > Vs
Smax = = 6.25″
Smax = 24″
Use, 2 legs # 3 @ 6″ c / c throught the beam length.
Floor beam, F. B (A2, B2):-
Beam size = 10″ x 15″
Effective depth, d = 15″ – 2.5″ = 12.5″
a) Load Calculation:-
Self weight of beam = x 150 = 156.25 Ib/ft
All main wall weight = ( 10 – ) x 120 = 437.5 Ib/ft
Load coming from Slab:-
S4 & S5, Dead load = 185 x ( ) x ( ) = 1664 Ib/ft
Total, Dead load = 156.25 + 437.5 + 1664 = 2257.75 Ib/ft
Live load = 40 x ( ) x ( )
= 359.78 Ib/ft
Factored load = 1.2 D.L + 1.6 L.L
= 1.2 x 2257.75 + 1.6 x 359.78
= 3284.95 Ib/ft
=3.28 k/ft
b) Moment Calculation and ‘d’ check:-
(-) Ve Moment at Ext. support = = 51.44 k-ft
(+) Ve moment at Mid span = = 58.79 k-ft
(-) Ve Moment at Int: support = = 82.32 k-ft
Mu = 85.02 k-ft
b = 0.85 x ßl x
= 0.85 x 0.85 x x
= 0.037
max = 0.75 b = 0.75 x 0.037 = 0.0278
Mu = bd fy (1- 0.59 x
Or, d =
= = 11.423″ < 12.5″
So, ok
c) Reinforcement Calculation:
Steel for Ext. Support:-
R= = = 453.46
= 0.0125
min = = 0.0050
min = = 0.0041
As = bd = 0.0125 x 10 x 12.5 = 1.56 in
Use 2 # 6 Straight bars + 2 # 6 Ext: top bars.
Ap = 1.56 inch
Steel for mid span:-
R= = = 518.23
= 0.014
min = = 0.0050
min = = 0.0041
As = bd = 0.014 x 10 x 12.5 = 1.75 in
Use 2 # 5 Straight bars + (2 # 6 + 1 # 5)Ext: Bottom bar’s.
Ap = 1.96 in
Steel for Int. Support:-
R= = = 725.504
= 0.022
As = bd = 0.022 x 10 x 12.5 = 2.75 in
Use 2 # 5 Straight bar’s + (2 # 8 + 1 # 7)Ext: top.
Ap = 2.82in
d) Stirrup design:-
Vv = 1.15 x = 30.88 at support face
Critical shear at `d’ distance,
Vu = 30.80 – x 3.393 = 27.35 k
Concrete shear strength:-
Vc = 2 wd = 2 x 10 x = 13.69 k
Vc = 0.75 x 13.69 = 10.27 k
Vs = (Vu – Vc) = 27.35 – 10.27 = 17.08 kip
4 bwd = 4 x 0.75 x 10 x = 20.54 > Vs
Smax = = = 6.25″
Smax = 24″
Use, 2 legs # 3 @ 4″ c / c
S = = = 4.8″ 4″ c / c
Floor beam, F B (B2, C2):-
Beam size = 15″ x 10″
Effective depth, d = 15″ – 2.5″ = 12.5″
a) Load Calculation:-
Self weight of beam = x 150 = 156.25 Ib/ft
All main wall weight = (10 – ) x 120 = 437.5 Ib/ft
Load coming from Slab:-
S1 & S2, Dead load = 185 x ( ) x ( ) = 1664.00 Ib/ft
Total, Dead load = 156.25 + 437.5 + 1664 = 2257.75 Ib/ft
Live load = 40 x ( ) x ( )
= 359.78 Ib/ft
Factored load = 1.2 D.L + 1.6 L.L
= 1.2 x 2257.75 + 1.6 x 359.78
= 3284.95 Ib/ft
=3.28 k/ft
b) Moment Calculation and ‘d ‘ check.
(-) Ve Moment at Ext. support = = 47.15 k-ft
(+) Ve moment at Mid span = = 53.89 k-ft
(-) Ve Moment at Int: support = = 75.44 k-ft
Mu = 75.44 k-ft
b = 0.85 x 0.85 x x
= 0.037
max = 0.75 b = 0.75 x 0.037 = 0.0278
Mu = bd fy (1- 0.59 x
Or, d =
= 10.76″ < 12.5″
So, ok
c) Reinforcement Calculation:
Steel for Ext. Support:-
R= = = 402.45
= 0.012
min = = 0.0050
min = = 0.0041
As = bd = 0.012 x 10 x 12.5 = 1.5 in
Use 2 # 6 Straight bars + 2 # 6 Ext: top bars.
Ap = 1.63 in
Steel for mid span:-
R= = = 460
min = 0.012
min = = 0.0050
min = = 0.0041
As = bd = 0.012 x 10 x 12.5 = 1.50 in
Use 2 # 6 Straight bar’s + 2 # 6 Ext: Bottom bar.
Ap = 1.65 in
Steel for Int. Support:-
R= = = 643.75
min = 0.014
As = bd = 0.014 x 10 x 12.5 = 1.75 in
Use 2 # 5 Straight bars + (2 # 8 + 1 # 5)Ext: top.
Ap = 1.96in
d) Stirrup design:-
Vv = = 24.87 at support face
Critical shear at `d’ distance,
Vu = 24.87 – x 3.28 = 21.45 k
Concrete shear strength,
Vc = 2 bwd = 2 x 10 x = 13.69k
Vc = 0.75 x 13.69 = 10.27k
Vs = (Vu – Vc) = 24.45 – 10.27 = 11.18Kip
4 bwd = 4 x 0.75 x 10 x = 20.54 > Vs
Smax = = = 6.25″
Smax = 24″
Use, 2 legs # 3 @ 5″ c / c
S = = = 7.37″ > 5″ c / c, ok.
Floor beam, F. B (A3, D3):-
Beam size = 10″ x 12″ = 0.89
= 82
Effective depth, d = 12″ – 2.5″ = 9.5″
a) Load Calculation:-
Self weight of beam, F.B-7 = x 150 = 125 Ib /ft
All main wall weight = ( 10′ – ) x 120 = 450 Ib/ft
Load coming from Slabs:-
S3 & S4, Dead load = 185 x ( ) x ( ) = 1804.29 Ib/ft
Total, Dead load = 125 + 450 + 1804.29 = 2379.29 Ib/ft
Live load = 40 x ( ) x ( )
= 210.60 Ib/ft
Factored load = 1.2 x 2379.29 + 1.6 x 210.60
= 3192.11 k/ft
=3.19 k/ft
b) Moment Calculation and ‘d’ check.
(-) Ve Moment at Ext. support = = = 19.94 k-ft
(+) Ve moment at Mid span = = 22.79 k-ft
(-) Ve Moment at Int: support = = 31.90 k-ft
Mu = 31.90 k-ft
b = 0.85 x ßl x
= 0.85 x 0.85 x x
= 0.037
max = 0.75 b = 0.75 x 0.037 = 0.028
Mu = bd fy (1- 0.59 x
Or, 31.90 x 12 = 0.9 x .028 x 12 x d x 40 (1- 0.59 x .028 x
= 5.62″ 5.5” < 9.5″
So, ok
c) Calculation of Reinforcement
Steel for Ext. Support:-
= = 244.99 > 200 psi
= 0.007
As = bd = 0.007 x 12 x 9.5 = 0.798 in
Use 2 # 6 bar Straight.
Steel for mid span:-
= = 280.57 > 200 psi
= 0.008
As = bd = 0.008 x 12 x 9.5 = 0.912 in
Use 2 # 5 bar Straight + 1 # 5Ext.
Steel for Int. Support:-
= = 392.7 > 200 psi
= 0.011
As = bd = 0.011 x 12 x 9.5 = 1.254 in
Use 2 # 6 bar Straight + 1 # 6 bar.
Floor beam, F B (B3, C3):-
Beam size = 10″ x 12″
Effective depth, d = 12″ – 2.5 = 9.5″
a) Load Calculation:-
Self weight of beam F.B-4 = x 150 = 125/ft
All main wall weight = ( 10′ – ) x 120 = 450 Ib/ft
Load coming from Slabs:-
m = = = 0.74
S2, Dead load = 185 x ( ) x .24 = 249.75 Ib/ft
Total, Dead load = 125 + 450 + 249.75 = 824.75 Ib/ft
Live load = 40 x ( ) x .24 = 54.00 Ib/ft
Factored load F,B-4 = 1.2 x 824.75 + 1.6 x 54
= 1076.10 Ib/ft
=1.07 k/ft
b) Moment Calculation and ‘d’ check.
(-) Ve Moment at Ext. support = = 15.38 k-ft
(+) Ve moment at Mid span = = 17.58 k-ft
(-) Ve Moment at Int: support = = 24.61 k-ft
Mu = 24.61 k-ft
b = 0.85 x ßl x
= 0.85 x 0.85 x x
= 0.037
max = 0.75 ρb = 0.75 x 0.037 = 0.028
Mu = bd fy (1- 0.59 x
Or, 24.61 x 12 = 0.9 x .028 x 12 x d x 40 (1- 0.59 x .028 x
= 5.59″ < 9.5″
So, ok
c) Calculation of Reinforcement
Steel for Ext. Support:-
= = 189.35 > 200 psi 190.00
= 0.008
min = = = 0.005
min = = = 0.0041
As = bd = 0.008 x 12 x 9.5 = 0.912 in
Use 2 # 6 Straight bars + 1 # 6 Ext: top.
Steel for mid span:-
= = 216.43 216 > 200 psi
= 0.006
As = bd = 0.006 x 12 x 9.5 = 0.684 in
Use 2 # 6 Straight bars.
Steel for Int. Support:-
= = 302.98 303 > 200psi
= 0.008
As = bd = 0.008 x 12 x 9.5 = 0.912 in
Use 2 # 7 Straight bar.
d) Stirrup design:-
Vv = 1.15 x = 8.82k at support face
Critical shear at `d’ distance,
Vu = 8.82 – x 1.07 = 7.97 k
Concrete shear strength,
Vc = 2 bwd = 2 x x 12 x = 12.48k
Vc = 0.75 x 12.48 = 9.37k
Vs = (Vu – Vc) = 7.97 – 9.37 = -1.40 kip
4 bwd = 4 x 0.75 x 10 x = 15.61 > Vs
Smax = = 4.75″ 4.5″
Smax = 24″
Use, 2 legs # 3 @ 4.5″ c / c Throught the beam length.
DESIGN OF COLUMN.
Design Data:–
Colomn height = 10′ c / c
Column type = Tied
Clear cover = 1.5″
= 0.80 for tied column
g = 0.02
= 0.65
Materials:
Fy = 40 ksi
f′c = 3 ksi
wc = 150 psf
COLUMN (C-1):
Assume column size = 10″ x 30″ = 300 inch
a) Load Calculation:-
Load from F. B, (B1,B2) = = 13.97 kip
Load from F. B (B2,B3)= = 6.3 kip
Load from F. B ( A2,B2) = = 25.96 kip
Load from F. B (B2,C2) = = 22.75 kip
Self weight of Column = x 150 x (10 – )/1000 = 2.734 k
Factored load = 1.2 x (13.97 + 6.3 + 25.96 + 22.75 + 2.734)
= 86.06 kip
Total load for 6 – story = 6 x 86.06 = 516.36 kip
b) Check for size:-
Pu = Ag [ .85f′c (1- g) + g fy]
Or, 516.36 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 40)]
Or, 516.36 = 0.52
Ag = =300.99 inch
Size of column = 10″ x 30.1″ = 301inch
Provided column size = 10″ x 30″ =300 in
c) Calculation of steel:-
Pu = [0.85 x f′c (Ag – Ast) + Ast fy]
Or, 516.36 = 0.80 x .65 [0.85 x 3 x (300- Ast) + 40 Ast]
Or, 516.36 = 0.52(765-2.55 Ast + 40 Ast)
Or, 516.36 = 398 + 19.474 Ast
Ast = = 6.1 in
Use 14 # 6 bars, Ap = 6.8 in
Tie bar Calculation / Spacing :
Use # 3 bar as Tie:-
i) S = 16 x = 12″ c / c
ii) S = 48 x = 18″ c / c
iii) S = Minimum thickness = 10″ c / c
So we use # 3 bar @ 10″ c / c
COLUMN (C-2):-
Assume column size = 10″ x 18″ = 180 inch
a) Load Calculation:-
Load from F. B (A1,A2) = = 8.35 kip
Load from F. B (A2,A3)= = 6.04 kip
Load from F. B ( A2,B2) = = 26.85 kip
Self weight of Column = x 150 x (10 – )/1000 = 1.670 k
Factored load = 1.2 x (26.85 + 8.35 + 1.670 + 6.04)
= 51.5 kip
Total load column for 6 – story = 6 x 51.50 = 308.96 kip
b) Check for size:-
Pu = 261.38 kip (story -1)
Pu = Ag [0.85f′c (1- g) + g fy]
Or, 308.96 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (.02 x 40)]
Or, Ag = = 180.10 inch
Size of column = 10″ x 18″
So, provided column size = 10″ x 18.2″ =182 in
c) Calculation of steel:-
Pu = [0.85 x f′c (Ag – Ast) + Ast fy]
Or, 308.96 = 0.80 x 0.65 [0.85 x 3 (180 – Ast) + 40 Ast]
Or, 308.96 = 0.52(459.25 – 2.55 Ast + 40 Ast)
Or, 308.96 = 19.474 Ast + 238.69
Ast = = 3.60 in
Ast 3.60 in
So, we provided, Use 8 # 6 bars, Ap = 3.89 in
Tie bar Calculation / Spacing:
Use # 3 bar as Tie:-
i) S = 16 x = 12″ c / c
ii) S = 48 x = 18″ c / c
iii) S = 10″ c / c
So we use # 3 bar @ 10″ c / c
COLUMN (C-3):-
Assume column size = 10″ x 18″ = 180 inch
a) Load Calculation:-
Load from F. B (C1,C2) = = 8.57 kip
Load from F. B (C2,C3)= = 7.26 kip
Load from F. B ( B2,C2) = = 22.75 kip
Self weight of Column = x 150 x (10 – )/1000 = 1.670 k
Factored load = 1.2 x (8.57 + 7.26 + 22.75 + 1.670)
= 48.30 kip
Total load column for 6 . story = 6 x 48.30 = 289.77 kip
b) Check for size:-
Pu = 289.77 kip (story -1)
Pu = Ag [0.85f′c (1- g) + g fy]
Or, 289.77 = 0.80 x 0.65 Ag [0.85 x 3 (1- 0.02) + (.02 x 40)]
Or, Ag = = 168.95 inch
Size of column = 10″ x 18″ = 180 inch
So we use ( 10″ x 18″) Size.
c) Calculation of steel:-
Pu = [0.85 f′c (Ag – Ast) + Ast fy]
Or, 289.77 = 0.80 x 0.65 x [0.85 x 3 (180 – Ast) + 40 Ast]
Or, 289.77 = 0.52(459 – 2.55 Ast + 40 Ast)
Or, 289.77 = 238.69 + 19.474 Ast
Ast = = 2.62 in
Ast 2.62 in
Use 6 # 6 bars, Ap = 2.92 inch
Tie bars:-
Use # 3 bar as Tie:-
i) S = 16 x = 12″ c / c
ii) S = 48 x = 18″ c / c
iii) S = 10″ c / c
So we use # 3 bar @ 10 c / c
COLUMN (C-4):-
Assume column size = 10″ x 15″ = 150 inch
a) Load Calculation:-
Load from F. B (B1,B2) = = 13.97 kip
Load from F. B (A1,B1)= = 16.42 kip
Load from F. B ( B1,C1) = = 15.69 kip
Self weight of Column = x 150 x (10 – )/1000 = 1.39 k
Factored load = 1.2 x (1.39 + 13.97 + 16.42 + 15.69)
= 56.96 kip
Total column load for 6 . story = 6 x 56.96 = 341.78 kip
= 342kip.
b) Check for size:-
Pu = 342 kip (story -1)
Pu = Ag [0.85f′c (1- g) + g fy]
Or, 342 = 0.80 x 0.65 Ag [0.85 x 3 x (1- 0.02) + (.02 x 40)]
Or, Ag = = 199.35 inch
Size of column = 10″ x 20″
So, Provided size we use = 10″ x 20″ = 200 inch
So, ok.
c) Calculation of steel:-
Pu = [0.85 f′c (Ag – Ast) + Ast fy]
Or, 342 = 0.80 x 0.65 x [0.85 x 3 (200 – Ast) + Ast x 40]
Or, 342 = 0.52(510 – 2.55 Ast + 40 Ast)
Or, 342 = 265 + 19.474 Ast
Ast = = 3.95 in =4inch
So, we provided, Use 10 # 6 bars, Ap = 4.86 inch
Tie bar Calculation:-
Use # 3 bar’s as Tie:-
i) S = 16 x = 12″ c / c
ii) S = 48 x = 18″ c / c
iii) S = 10″ c / c
So, we use # 3 bar’s @ 10″ c / c
COLUMN (C-5):-
Assume column size = 10″ x 12″
a) Load Calculation:-
Load from F. B (A1,A2) = = 8.35 kip
Load from F. B (A1,B1)= = 16.42 kip
Self weight of Column = x 150 x (10 – )/1000 = 1.115 k
Factored load = 1.2 x (1.115 + 8.35 + 16.42)
= 31.062 31.50 kip
Total column load for 6 . story = 6 x 31.50 = 189 kip.
b) Check for size:-
Pu = 189 kip (story -1)
Pu = Ag [0.85f′c (1- g) + g fy]
Or, 189 = 0.80 x 0.65 Ag [0.85 x 3 x (1- 0.02) + (.02 x 40)]
Or, 189 = 1.71548 Ag
Or, Ag = = 110.20 inch
Size of column = 10″ x 11″
So, Provided size = 10″ x 12″ = 120 inch
c) Calculation of steel:-
Pu = [0.85 f′c (Ag – Ast) + Ast fy]
Or, 189 = 0.80 x 0.65 [0.85 x 3 (120 – Ast) + Ast x 40]
Or,189 = 0.52(306 – 2.55 Ast + 40 Ast)
Or, 189 = 19.474 Ast + 159.12
Ast = = 1.53 inch
So, we use 6 # 5 bars, Ap = 1.86 in
So, ok
Tie bar Calculation:-
Use # 3 bar as Tie:-
i) Spacing, S = 16 x = 10″ c / c
ii) S = 48 x = 18″ c / c
iii) Minimum width of Column, S = 10″ c / c
So, # 3 bar’s @ 10″ c / c
COLUMN (C-6):-
Assume column size = 10″ x 12″ = 120 inch
a) Load Calculation:-
Load from F. B, (B1,C1) = = 15.73 kip
Load from F. B (C1,C2)= = 8.6 kip
Self weight of Column = x 150 x (10 – )/1000 = 1.114 k
Factored load = 1.2 x (1.114 + 8.6 + 15.73)
= 25.41 kip
Total column load for 6 . story = 6 x 25.41 = 152.5 kip.
b) Check for size:-
Pu = 297.04 kip (story -1)
Pu = Ag [0.85f′c (1- g) + g fy]
Or, 152.52 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + (0.02 x 40)]
Or, 125.50 = 1.71548 Ag
Or, Ag = = 88.9 inch
Size of column = 10″ x 10″= 100 inch
So, Provided size = 10″ x 12″ = 120 inch
c) Calculation of steel:-
Pu = [0.85 f′c (Ag – Ast) + Ast fy]
Or, 152.50 = 0.80 x 0.65 [0.85 x 3 (120 – Ast) + Ast x 40]
Or,152.50 = 0.52(-2.25Ast + 306 + 40 Ast)
Or, 152.50 = 19.474 Ast + 159.12
Ast = = – 0.34 inch
So, we use 4 # 5 bars, Ap = 1.24 inch
Tie bar Calculation:
Use # 3 bar as Tie:-
i) Spacing, S = 16 x = 10″ c / c
ii) S = 48 x = 18″ c / c
iii) S = 10″ c / c
So, Use # 3 bar’s @ 10″ c / c
Some are parts:
Analysis and Design of Building Components (Part 1)
Analysis and Design of Building Components (Part 2)
Analysis and Design of Building Components (Part 3)