Analysis and Design of Building Components (Part 3)

COLUMN (C-7):-

Assume column size = 10″ x 15″ = 150 sq. inch

a) Load Calculation:-

Load from F. B – A2,A3 =  = 15.19 kip

Load from F. B – A3,A4 =  = 8.25 kip

Load from F. B – A3,D3 =  = 15.95 kip

Self weight of Column =  x 150 x (10 – ) = 1.40 k

Factored load                 = 1.2 x 1.40 + 15.19 + 8.25 + 15.95

= 41.07 kip

Total column load for 6 – stored = 6 x 41.07 = 246.42 kip

 b) Check for size:-

P_u = 243.11 (kip) (story -1)                        / ρ_g = 0.02

P_u  = A_g [ .85f′c (1- ρ_g) + ρ_gf_y]

Or, 246.42 = 0.80 x 0.65 x Ag [0.85 x 3 (1- 0.02) + .02 x 40]

A_g = 143.64 inch

Size of column = 10″ x 15″

Provided size = 150 in

 c) Calculation of steel:-

P_u =  [0.85 x f′c (A_g – A_st) + A_st f_y]

Or, 150 = .80 x .65 x [.85 x 3 x (150- A_st) + A_st x 40]

A_st = 2.27 in

Use # 7 bars, Ap = 2.40 in

Tie bar:-

Use # 3 bar @ as Tie

S = 16 x Main bar dia = 16 x  = 14″ c / c

S = 48 x Tie bar dia = 48 x  = 18″ c / c

S = Minimum width of column = 10″ c /c

We use # 3 bar @ 10″ c / c

COLUMN (C-8):-

Assume column size = 10″ x 24″

a)  Load Calculation:-

Load from F. B – B2,B3 =  = 24.08 kip

Load from F. B – B3,B4 =  = 13.08 kip

Load from F. B – B3,C3 =  = 8.11 kip

Load from F. B – D3,B3 =  = 3.12 kip

Self weight of Column =  x 150 x (10 – ) = 2.25 k

Factored load                 = 24.08 + 13.08 + 8.11 + 1.2 x 2.25+20 k

= 67.97 kip

Total column load for 6 – story = 67.97 x 6 = 407.82 kip

b) Check for size:-

P_u = 407.82 (kip)

P_u  = A_g [ .85f’c (1- ρ_g) + ρ_gf_y]

Or, 407.82  = 0.80 x 0.65 x A_g [0.85 x 3 (1- .02) + .02 x 40]

A_g = 237.80 in

Size of column = 10″ x 24″

Provided size = 240 in

So, Ok

c) Calculation of steel:-

P_u =  [0.85 f′c (A_g – A_st) + A_st f_y]

Or, 407.82 = .80 x .65 x [.85 x 3 x (240- A_st) + A_st x 40]

A_st = 4.59 in

Use 8 # 7 bars st, Ap = 4.80 in

Tie bar:

Use # 3 bar  as Tie

S = 16 x  = 10″ c / c

S = 48 x  = 18″ c / c

S = Minimum width of column = 10″ c / c

We use # 3 bar @ 10″ c / c

COLUMN (C-9):-

Assume column size = 10″ x 15″ = 150 sq. inch

a) Load Calculation:

Load from F. B – B3, C3 =  = 7.81 kip

Load from F. B – C2, C3 =  = 6.81 kip

Load from F. B – C3, C4 =  = 3.06 kip

Load from stair Beam =  = 19.38 kip

Self weight of Column =  x 150 x (10 – )/1000 = 1.97 k

Factored load                 = 1.2 x 1.97 + 7.81 + 3.06 + 19.38 +20

= 59.42 kip

Total column load for 6. story = 6 x 59.42= 356.54 kip

b) Check for size:-

P_u = 116.232 kip

P_u  = Ag [ .85f′c (1- ρ_g) + ρ_gf_y]

Or, 356.54  = 0.80 x .65 x A_g [0.85 x 3 (1- .02) + .02 x 40]

A_g = 207.83 in

Size of column = 10″ x 21″

Provide, Ag = 10″ x 21″= 210 in

So,  ok.

 c) Calculation of steel:-

P_u =  [0.85 f′c (A_g – A_st) + A_st fy]

Or, 356.54  = .80 x .65 x [.85 x 3 x (210- A_st) + A_st x 40]

A_st = 4.00 in

Use 10 # 6 bar straight.

Tie bar:-

Use # 3 bar  as Tie

S = 16 x  = 12″ c / c

S = 48 x  = 18″ c / c

S = Minimum width of column = 10″ c / c

We use # 3 bar @ 10″ c / c

Check for lateral tie:-

S =  = 4.75″ < 6″

So, Ok

No latered ties are required.

 COLUMN (C-10):-

Assume column size = 10″ x 15″ = 150 sq. inch

a) Load Calculation:-

Load from F. B – A3, D3 =  = 15.95 kip

Load from F. B – D3, D4 =  = 9.29 kip

Load from F. B – D3, B3 =  = 13.08 kip

Self weight of Column = 150 x  x 9′ = 1.41 k

Factored load                 = 15.95 + 9.29 + 13.08 + 1.41 x1.2

= 40.00 kip

Total column load for 6- story = 6 x 40 = 240 kip

b) Check for size:-

Pu = 240 kip (Story -1)

P_u  = A_g [ .85 x f′c (1- ρg) + ρ_g f_y]

Or, 240  = .80 x .65 x Ag [.85 x 3 (1- 0.02) + .02 x 40]

A_g = 139.90 in

Size of column = 10″ x 14″

Provided size  = 10″ x 15″= 150 Sq. inch.

Ok.

c) Calculation of steel:-

P_u =  [0.85 f′c (A_g – A_st) + A_st f_y]

Or, 240  = .80 x 0.65 [.85 x 3 (150- A_st) + A_st x 40]

A_st = 2.11 in

Use 4 # 7 bar st, Ap = 2.40in

Tie bar:

Use # 3 bar  as Tie

S = 16 x  = 14″ c / c

S = 48 x  = 18″ c / c

S = Minimum width of column = 10″ c / c

We use # 3 bar @ 10″ c / c

Table:- 4.4.   Details of sectional dimensions and reinforcement arrangement of all Columns (C – 1 ~ C – 10):

Figure Plan of Grade beam

 Design of grade beams:

Design Data:–

Materials:-

F_y = 40 ksi

f′c = 3 ksi

w_c = 150 pcf

w _brick = 120 pcf

Thickness of wall on GB = 5″

Clear cover                       = 3″

 Grade Beam GB-1:

Size  = 10″ x 15″

Effective depth (d) = 15-3 = 12 inch

a) Load Calculation:-

Self weight of beam GB-1 =  x 150 = 156.25 Ib/ft

All main wall weight          =  ( 10- )x120 = 430.00 Ib/ft

Total Dead load                  = 156.25 + 450 = 606.25 Ib/ft

Factored load GB-1            = 1.2 DL    = 1.2 x 606.25

= 727.5 Ib/ft

= 0.73 k/ft

b) Moment Calculation and ‘d’ check:-

(-) Ve moment at Ext. support =  = 7.04 k-ft

(+) Ve moment at Mid span     =  = 8.04 k-ft

(-) Ve moment at Int. support =  = 11.26 k-ft

M_u = 11.26 k-ft

_b = 0.85ß₁ x  x

= 0.85 x 0.85 x x

= 0.037

_max = 0.75 _b = 0.75 x 0.037 = 0.028

M_u       = bd  f_y (1-0.59  x

Or 11.26  x 12 = 0.9 x 0.028×12 x d x 40 x (1-0.59 x .028 x

= 3.78″ < 12″

So, ok

 c) Reinforcement Calculation:

Mid span steel:-

= = 62.03 < 200 psi

_min = =  = 0.005

_min = =  = 0.0041

A_s = bd = 0.005x 12 x 12 = 0.72 inch

Use 2 # 6  hanger bar.

Steel for Int. Support:-

= = 86.88 psi < 200 psi

_min  = 0.005

_min = 0.0041

A_s = bd = 0.005x 12 x 12 = 0.72 inch

Use 2 # 6 hanger bar.

Steel for Ext. Support:-

= = 54.32 psi < 200 psi

A_s = bd = 0.005x 12 x 12 = 0.72 inch

Use 2 # 6  hanger bar.

d) Stirrup design:-

V_v = 1.15 x  = 4.86 K

V_u = 4.86 – ( ) x.73 = 4.13 k

V_c = 2 b_w d = 2  x 10 x 12 = 13.14 k

 Shear strength of concrete,

V_c = .75 x 13.14 = 9.86 K

Since, V_c > V_d; no shear reinforcement is required.

Use # 3 bar @  =  = 6″ c / c throughout beam length.

 Grade Beam GB-2:

Size  = 12″ x 15″

Effective depth, (d) = 15-3 = 12 inch

a) Load Calculation:-

Self weight of beam GB-2 =  x 150 = 156.25 Ib/ft

All main wall weight          =  ( 10- ) x120 = 450.00 Ib/ft

Total Dead load                  = 156.25 + 450.00 = 606.25 Ib/ft

Factored load GB-2            = 1.2 DL    = 1.2 x 606.25

= 727.5 Ib/ft

= 0.73 k/ft

 b) Moment Calculation and d check:-

(-) Ve moment at Ext. support =  = 11.43 k-ft

(+) Ve moment at Mid span     =  = 13.07k/ft

(-) Ve moment at Int. support =  = 18.79 k-ft

Now, M_u = 18.79 k-ft

_b = .85x ₁   x  x

= 0.85 x 0.85 x x

= 0.037

_max = 0.75 x 0.037  = 0.028

M_u       = bd  fy (1-0.59  x

Or 18.79  x 12 = 0.9 x .028×12 x d x 40 x (1-0.59 x .028 x

= 4.88″ < 12″

So, ok

c) Reinforcement Calculation:

Mid span steel:-

= = 100.84 < 200 psi

_min = =  = 0.005

_min = =  = 0.0041

A_s = bd = .005x 12 x 12 = 0.72

Use 2 # 6 bar hanger

Steel for Int. Support:-

= = 144.98 psi < 200 psi

_min  = 0.005

A_s =  0.005x 12 x 12 = 0.72 inch

Use 2 # 6 bar hanger.

Steel for Ext. Support:-

= = 88.19 psi < 200 psi

_min  = 0.005

A_s = 0.005x 12 x 12 = 0.72 inch

Use 2 # 6  bar hanger .

d) Stirrup design:-

V_u = 1.15 x  = 6.29 K

V_u = 6.29 – ( ) x .73 = 5.57 k

Vc = 2 b_w d = 2  x 10 x 12 = 13.14 k

Shear strength of concrete,

V_c = 0.75 x 13.14 = 9.86 K

Since, V_c > V_d; no shear reinforcement is required.

Use # 3 bar @  =  = 6″ c / c throughout beam length.

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Figure Reinfrocement Grade Beam ( GB-2 )

Table : Details of sectional dimensions and reinforcement arrangement of all grade beams ( GB- 1  GB – 2).

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)