Question

Determine the factor of safety for a 9 foot long hollow steel

column 3.5 inches on a side that has a wall thickness of 0.225
inches and is loaded with a 22 kip load. Use the steel E of 29 *10
^6 psi and assume the column is pin connected at each end.

Answer 1

factor of safety for A36 structural steel is 0.82

Step-by-step explanation:

given data:

side of column = 3.5 inches

wall thickness = 0.225 inches

load P = 22 kip

Length od column = 9 ft

we know that critical stress is given as

`\sigma_(cr) = (\pi^2 E)/((l/r)^2)`

where

r is radius of gyration
`= \sqrt{\fra{I}{A}}`

Here I is moment od inertia
`= (b_1^2)/(12) - (b_2^2)/(12)`

`I == (3.5^2)/(12) - (3.05^2)/(12) = 5.294 in^4`

For hollow steel area is given as
`A = b_1^2 -b_2^2`

`A = 3.5^2 -3.05^2 = 2.948 in^2`

critical stress
`\sigma_(cr) = = (\pi^2* 29* 10^6)/(((9*12)/(1.34))^2)`

`\sigma_(cr) =  44061.56 lbs/inc^2`

considering Structural steel A36

so A36
`\sigma_y = 36ksi`

factor of safety
`= (yield\ stress)/(critical\  stress)`

factor of safety =
`(36*10^3)/(44061.56) = 0.82`

factor of safety for A36 structural steel is 0.82