Structural Concrete - Hassoun

436 Chapter 12 Slender Columns
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Direction of
analysis
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Figure 12.6 Example 12.4.
Use f c ′ = 5ksi, f y = 60 ksi, and the ACI Code requirements. Assume an exterior column load of
two-thirds the interior column load, a corner column load of one-third the interior column load.
Solution
1. Calculate the factored forces using load combinations.
For gravity loads,
P u = 1.2D + 1.6L = 1.2(300)+1.6(100) =520K
M u,top = M 1ns = 1.2M D + 1.6M L = 1.2(32)+1.6(20) =70.4K⋅ ft
M u,bottom = M 2ns = 1.2M D + 1.6M L = 1.2(54)+1.6(36) =122.4K⋅ ft
For gravity plus wind load,
P u =(1.2D + 1.0L + 1.6W)
=[1.2(300)+1.0(100)+0] =460K
M u,top = 1.2M D + 1.0M L + 1.6M w
= 1.2(32)+1.0(20)+1.6(50) =138.4K⋅ ft (top total)
M u,tns = 1.2M D + 1.0M L = 1.2(32)+1.0(20) =58.4K⋅ ft (top nonsway)
M u,ts = 1.6M w = 1.6 × 50 = 80K ⋅ ft (top sway)
M u,bottom = 1.2M D + 1.0M L + 1.6M w
= 1.2(54)+1.0(36)+1.6(50) =180.8K⋅ ft (bottom total)
M u,bns = 1.2M D + 1.0M L = 100.8K⋅ ft (bottom nonsway)
M u,bs = 1.6M w = 80K ⋅ ft (bottom sway)
Other combinations are not critical.
Check for minimum e:
e = M u
P u
122.4 × 12
e for gravity loads, e = = 2.82in.
520
180.8 × 12
e for gravity plus wind loads, e = = 4.72in.
460
e min = 0.6 + 0.03h = 0.6 + 0.03(18) =1.14in.
e > e min
safe