Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 11, Problem 11.9.24P
To determine

The allowable axial load.

Expert Solution & Answer
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Answer to Problem 11.9.24P

The allowable axial load at L=0.6m is 96.03kN .

The allowable axial load at L=0.8m is 83.68kN .

The allowable axial load at L=1m is 60.77kN .

The allowable axial load at L=1.2m is 42.19kN .

Explanation of Solution

Given:

The outside diameter of the pipe is d2=80mm and inside diameter of the pipe is d1=72mm .

Concept Used:

Write the expression for the allowable axial load.

  Pa=σaA(1)

The allowable axial load is Pa , the allowable stress is σa, and the area of cross-section is A .

Write the expression for the area of cross-section.

  A=π4(d22d12)

Substitute π4(d22d12) for A in equation (1).

  Pa=σa×π4(d22d12)(2)

The equation for the aluminum alloy (6061-T6) column allowable stress as follows:

  σa=20.20.126(kLr)ksikLr66 ....... (3)

  σa=51000( kL r)2ksikLr66(4)

The radius of gyration is r and the constant is k .

Since the column is fixed at the base and free at the top so,

  k=2

Write the expression for the radius of gyration.

  r=IA(5)

The area moment of inertia is I .

Write the equation of area moment of inertia as follows:

  I=π64(d24d14)

Substitute π64(d24d14) for I and π4(d22d12) for A in equation (5).

  r= π 64( d 2 4 d 1 4 ) π 4( d 2 2 d 1 2 )r=116( d 2 2 + d 1 2 )( d 2 2 d 1 2 )( d 2 2 d 1 2 )r=14( d 2 2+ d 1 2)(6)

Calculation:

As per the given problem

  d1=72mm , d2=80mm , L=0.6m , 0.8m , 1m and 1.2m .

Convert the diameter into inch .

  d1=(72mm)(1in25.4mm)=2.835ind2=(80mm)(1in25.4mm)=3.150in

Substitute 2.835in for d1 and 3.150in for d2 in equation (6).

  r=14( 2.835 2+ 3.150 2)=1.059in

When L=0.6m ,

  kLr=2×(0.6m)( 39.37in 1m)1.059=44.612

Since kLr66 , so

Substitute 44.612 for kLr in equation (3).

  σa=20.2(0.126×44.612)ksi=14.58ksi

Substitute 14.58ksi for σa , 2.835in for d1 and 3.150in for d2 in equation (2).

  Pa=14.58×103×π4(3.1522.8352)=(21588.48lbf)(0.00444822kN1lbf)=96.03kN

When L=0.8m ,

  kLr=2×0.8×( 39.37in 1m)1.059=59.483

Since kLr66 , so

Substitute 59.483 for kLr in equation (3).

  σa=20.2(0.126×59.483)ksi=12.705ksi

Substitute 12.705ksi for σa , 2.835in for d1 and 3.150in for d2 in equation (2).

  Pa=12.705×103×π4(3.1522.8352)=(18812.186lbf)(0.00444822kN1lbf)=83.68kN

When L=1m ,

  kLr=2×1×( 39.37in 1m)1.059=74.35

Since kLr66 , so

Substitute 74.35 for kLr in equation (4).

  σa=51000( 74.35)2ksi=9.22589ksi

Substitute 9.22589ksi for σa , 2.835in for d1, and 3.150in for d2 in equation (2).

  Pa=9.22589×103×π4(3.1522.8352)=(13660.7lbf)(0.00444822kN1lbf)=60.77kN

When L=1.2m ,

  kLr=2×1.2×( 39.37in 1m)1.059=89.223

Since kLr66 , so

Substitute 89.223 for kLr in equation (4).

  σa=51000( 89.223)2ksi=6.406ksi

Substitute 6.406ksi for σa , 2.835in for d1, and 3.150in for d2 in equation (2).

  Pa=6.406×103×π4(3.1522.8352)=(9485.309lbf)(0.00444822kN1lbf)=42.19kN

Conclusion:

The allowable axial load at L=0.6m is 96.03kN .

The allowable axial load at L=0.8m is 83.68kN .

The allowable axial load at L=1m is 60.77kN .

The allowable axial load at L=1.2m is 42.19kN .

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Chapter 11 Solutions

Mechanics of Materials (MindTap Course List)

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