Problem 1. Find the nominal axial compression and design capacity of a rectangular column sixe is 16" * 24* and reinforced with 12 - 20 mm bars. Given fy = 60 ksi and fc'= 4ksi?
Solution:
Nominal axial compressive force for ties column
Pn = 0.8 { 0.85fc' (Ag - Ast) + Asfy}
Here, fc'= 4 ksi
Ag= (16 * 24)
= 384 inch2
Ast= 12 * π/4 * 0.7872
= 5.84 in2
fy = 60 ksi
Pn = 0.8 { 0.85 * 4*(384 - 5.84) + 5.84*60}
= 1308.9 k
Design capacity = φPn = (0.65 * 1308.9) k
= 850.78k (Ans.)
Problem 2. Determine the allowable compressive stress in the following loadin the column with column size 10"*10" and use 10 -3/4 " bar, given fy = 60 ksi and fc'= 3ksi?
Solution:
We Know,
Pn = 0.8 { 0.85fc' (Ag - Ast) + Asfy}
Here, fc'= 3 ksi
Ag= (10 * 10)
= 100 inch2
Ast= 10 * π/4 * (3/4)2
= 1.104 in2
fy = 60 ksi
Allowable load = 0.65 *0.8 { 0.85 * 3*(100 -1.104) + 1.104*60}
= 165.6 kip (Ans.)
Problem 3.The value of fc' = 4000 psi, fy= 60000 psi and steel area is 2%. Find the design ultimate axial stress of the column with zero eccentricity?
Solution:
Axial ultimate load,
Pu = φ0.8 { 0.85fc' (Ag - Ast) + Asfy}
Here, Ag= (12 * 12)
Ast= 0.02 * 144
=2.88 in2
∴Pu = 0.65* 0.8 { 0.85 *4 (144 - 2.88) + 2.88*60}
= 339.4 k
Axial Stress = {339.4/ (12*12)}
= 2.36 kip/inch2 (Answer.)
very nice
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