Floor beam, F. B (C3, C4):-

Beam size = 10″ x 12″

Effective depth, d = 12″ -2.5″ = 9.5″

Self weight of beam            =  x 150                = 125 Ib/ft

All main wall weight          =    ( 10′ – ) x 120 = 450 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″

Self weight of beam F. B-3 =  x 150 = 156.25 Ib/ft

All main wall weight          =    ( 10′ – ) x 120 = 437.5 Ib/ft

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″

Self weight of beam F. B-10 =  x 150           = 156.25 Ib/ft

All main wall weight          =   ( 10 – ) x 120 = 437.5 Ib/ft

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″

Self weight of beam  =  x 150                        = 156.25 Ib/ft

All main wall weight          =    ( 10 – ) x 120 = 437.5 Ib/ft

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″

Self weight of beam  =  x 150                 = 156.25 Ib/ft

All main wall weight          =    (10 – ) x 120 = 437.5 Ib/ft

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″

Self weight of beam,  F.B-7 =  x 150                = 125 Ib /ft

All main wall weight          =    ( 10′ – ) x 120 = 450 Ib/ft

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″

Self weight of beam  F.B-4 =  x 150 = 125/ft

All main wall weight          =    ( 10′ – ) x 120 = 450 Ib/ft

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

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

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

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

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″

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

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)