Question

-1-1.5" clearance 1st layer: As = 4 in? (calculate di), then Mn,1 2nd layer: As = 2 in? (calculate d2), then M1,2 3rd layer: As = 2 in? (calculate d3), then Mn,3 4th layer: As = 4 in? (calculate d4), then Mn,4 Summation of o Mn,1 to 0M2,4 shall be at least equal to Mn = 485 k-ft.

Calculate the moment capacity of the column, using the area of steel provided in the section

Answer 1

Given:

overall depth, D=24"

therefore, d=24"-1.5" = 22.5" (d'=1.5")

center of the column, c=24"/2 = 12"

as per ACI

$\beta _{1}$=0.85

$\phi$=0.65

Es=29000 ksi

a=$\beta _{1}$c=0.85x12" = 10.2"

i) 1st layer

As=4 in2

total 9no of bars

As/9= (($\pi$/4)*d12)

d1=0.7522 in

Mn = Y Cc + y cs+yT

$f_{s}^{'}$=0.003Es((c-d')/c) = 0.003x29000x((12"-1.5")/12") = 76.125 ksi

fs=0.003Es((d-c)/c) = 0.003x29000x(22.5"-12")/12") = 76.125 ksi

cc=0.85$f_{c}^{'}$$\beta _{1}$cb = 0.85x4x0.85x10x24 = 693.6 k (assume $f_{c}^{'}$ =4 ksi)

cs=$A_{s}^{'}$$f_{s}^{'}$ = 4x60 = 240 k (assume $f_{s}^{'}$ =60 ksi)

Ts=Asfs= 4x60 = 240 k

ys=c-(a/2) = 12"-(10.2"/2) = 6.9"

ys'=c-d' = 12"-1.5" = 10.5"

ys=d-c = 22.5"- 12" = 10.5"

Mn=(6.9x693.6)+(10.5x240)+(10.5x240) = 9825.84 k-in = 818.82 k-ft

$\phi$Mn=0.65x818.82 = 532.233 k-ft

ii) 2nd layer

As=2 in2

total 9no of bars

As/9= (($\pi$/4)*d12)

d1=0.0.5318 in

Mn = Y Cc + y cs+yT

$f_{s}^{'}$=0.003Es((c-d')/c) = 0.003x29000x((12"-1.5")/12") = 76.125 ksi

fs=0.003Es((d-c)/c) = 0.003x29000x(22.5"-12")/12") = 76.125 ksi

cc=0.85$f_{c}^{'}$$\beta _{1}$cb = 0.85x5.8x0.85x10x24 = 1005.72 k (assume $f_{c}^{'}$ =5.8 ksi)

cs=$A_{s}^{'}$$f_{s}^{'}$ = 2x80 = 160 k (assume $f_{s}^{'}$ =80 ksi)

Ts=Asfs= 2x80 = 160 k

ys=c-(a/2) = 12"-(10.2"/2) = 6.9"

ys'=c-d' = 12"-1.5" = 10.5"

ys=d-c = 22.5"- 12" = 10.5"

Mn=(6.9x1005.72)+(10.5x160)+(10.5x160) = 10299.468 k-in = 858.286 k-ft

$\phi$Mn=0.65x818.82 = 557.89 k-ft

iii) 3rd layer

As=2 in2

total 9no of bars

As/9= (($\pi$/4)*d12)

d1=0.0.5318 in

Mn = Y Cc + y cs+yT

$f_{s}^{'}$=0.003Es((c-d')/c) = 0.003x29000x((12"-1.5")/12") = 76.125 ksi

fs=0.003Es((d-c)/c) = 0.003x29000x(22.5"-12")/12") = 76.125 ksi

cc=0.85$f_{c}^{'}$$\beta _{1}$cb = 0.85x5.8x0.85x10x24 = 1005.72 k (assume $f_{c}^{'}$ =5.8 ksi)

cs=$A_{s}^{'}$$f_{s}^{'}$ = 2x80 = 160 k (assume $f_{s}^{'}$ =80 ksi)

Ts=Asfs= 2x80 = 160 k

ys=c-(a/2) = 12"-(10.2"/2) = 6.9"

ys'=c-d' = 12"-1.5" = 10.5"

ys=d-c = 22.5"- 12" = 10.5"

Mn=(6.9x1005.72)+(10.5x160)+(10.5x160) = 10299.468 k-in = 858.286 k-ft

$\phi$Mn=0.65x818.82 = 557.89 k-ft

iv) 4th layer

As=4 in2

total 9no of bars

As/9= (($\pi$/4)*d12)

d1=0.7522 in

Mn = Y Cc + y cs+yT

$f_{s}^{'}$=0.003Es((c-d')/c) = 0.003x29000x((12"-1.5")/12") = 76.125 ksi

fs=0.003Es((d-c)/c) = 0.003x29000x(22.5"-12")/12") = 76.125 ksi

cc=0.85$f_{c}^{'}$$\beta _{1}$cb = 0.85x4x0.85x10x24 = 693.6 k (assume $f_{c}^{'}$ =4 ksi)

cs=$A_{s}^{'}$$f_{s}^{'}$ = 4x60 = 240 k (assume $f_{s}^{'}$ =60 ksi)

Ts=Asfs= 4x60 = 240 k

ys=c-(a/2) = 12"-(10.2"/2) = 6.9"

ys'=c-d' = 12"-1.5" = 10.5"

ys=d-c = 22.5"- 12" = 10.5"

Mn=(6.9x693.6)+(10.5x240)+(10.5x240) = 9825.84 k-in = 818.82 k-ft

$\phi$Mn=0.65x818.82 = 532.233 k-ft