   # - 11 Determine the maximum permissible length L m a x for a steel pipe column that is fixed at the base and free at the top and must support an axial load P = 40 kips (see figure). The column has an outside diameter d = 4.0 in., wall thickness t = 0.226 in., E = 29,000 ksi, and σ y = 42 ksi. ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347 ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 11, Problem 11.9.11P
Textbook Problem
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## -11 Determine the maximum permissible length Lmaxfor a steel pipe column that is fixed at the base and free at the top and must support an axial load P = 40 kips (see figure). The column has an outside diameter d = 4.0 in., wall thickness t = 0.226 in., E = 29,000 ksi, and σ y = 42 ksi. To determine

The value of the maximum permissible length Lmax for a steel pipe column.

### Explanation of Solution

Given Information:

The outside diameter of column, do = 4.0 in.

The thickness of the wall of column, t = .226 in.

The allowable load, P = 40 k

The value of the young modulus, E = 29000 ksi

The value of maximum stress, sy = 36 ksi

Concept Used:

For fixed free end, K=2

di=do2tA=π4(do2di2)I=π64(do4di4)r=IA( L r)max=200

Here, r = radius of gyration.

Lc=r 2 π 2 E σ y Lc=critical length of column in ft

i=1,2...4n1i=NA          ifKLir>KLcr&n1i=53+3( K L i r )8( K L c r ) ( K L i r )38 ( K L c r )3ifKLirKLcrn2i=2312          ifKLir>KLcr&n2i=NA          ifKLirKLcr

σallow=σy[1 n 1 i (1 ( K L i r ) 2 2 ( K L c r ) 2 )]ifKLirKLcrσallow=σy[ ( K L c r ) 22 n 2 i ( K L i r ) 2]ifKLir>KLcrPallow=σallow×A

Calculation:

di=4.02(.226)=3.548in.A=π4(42 3.5482)=2.6795in.2I=π64(44 3.5484)=4.7877in.4r= 4.7877 2.6795=1.3367in.( KL r)max=200Lmax=200×1.3367/2=133.67in

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