Answer to Question #162293 in Mechanical Engineering for Zawad

Question #162293

A hollow circular column is made of AISI C1015, as rolled and its Outer and inner diameter is 4.5 in and 4.021 in.

I = 7.2 in^4, k= 1.5 in, L= 15 ft, effective length= 0.65*L, N=2.5.

Single column area= 3.174 in^2

total load,F= 200 kips

Determine minimum number of column and equivalent stress.


1
Expert's answer
2021-02-10T01:11:38-0500

Outer diameter=do=4.5 in, inner diameter=di=4.021 in, I= 7.2 in4, k= 1.5 in= radius of gyration, L=15 ft, effective Length=0.65 L, N=2.5, single column area= 3.174 in2,Total load= F=200 kips

section modulus=Z= "\\frac{I}{y}= \\frac{7.2}{2.25}=3.2 in^3"

Critical load for one coloumn

P="\\frac{\\pi^2EI}{L^2}=\\frac{3.14^2\\times 27557\\times 7.2}{180^2}=60.377 kips"


(i) Number of column="\\frac{200}{60.377}=3.31" , number of column will be 4


(ii) Now value of equivalnt stress= "\\frac{60.377}{3.174}=19.022 ksi"


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Comments

AKM Jahangir Alam
31.05.21, 12:40

Good

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