Quantcast Design of a Column -Cont.

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ex = Mx/p
= 3,816,000/636,000
= 6 in > 1.8 in
ey = My/P
= 0/636,000
= 0 < 1.8, use 1.8 in
(6) From Figure 40,
dx = dy = 18 - 1.5 - 0.5 - 0.875/2
= 15.56 in
Asx = 4 x 0.6
= 2.40 in2
Asy = 2 x 0.6
= 1.20 in2
Find value of m,
m = fdy/(O.85 f'dc)
= 66,000/(0.85 x 5,000)
= 15.53
Using Equation (58),
eb = 0.20 h + 1.54 As m/b
ebx = 0.20 x 18 + (1.54 x 2.40 x 15.53)/18
= 6.79 in > 6 in, use Equation (62)
eby = 0.20 x 18 + (1.54 x 1.2 x 15.53)/18
= 5.19 in > 1.8 in, use Equation (62).
(7) Axial load from Equation (62):
As fdy
bhf'dc
Pu = +
e/(2d - h) + 0.5
3he/d2 + 1.18
2.4 x 66,000
Px =
6/(2 x 15.56 - 18) + 0.5
18 x 18 x 5,000
+
3 x 18 x 6/(15.56)2 + 1.18
= 808,775 lb
1.2 x 66,000
Py =  
1.8/(2 x 15.56 - 18) + 0.5
18 x 18 x 5,000
+  
3 x 18 x 1.8/(15.56)2 + 1.18
= 1,148,662 lb
2.08-123





 


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