Concept explainers
An aluminum pipe column (E = 10,400 ksi) with a length L = 10.0 ft has inside and outside diameters d1= 5.0 in. and d2= 6.0 in., respectively (sec figure). The column is supported only at the ends and may buckle in any direction.
Calculate the critical load Pcrfor the following end conditions: (a) pinned-pinned, (b) fixed-free, (c) fixed-pinned, and (d) fixed-fixed.
i.
The critical load for the pinned-pinned end condition.
Answer to Problem 11.4.3P
The critical load for the pinned-pinned condition is 235 ksi
Explanation of Solution
Given:
E=10400 ksi
L= 10 ft
d1= 5 in
d2= 6 in
Concept Used:
Calculation:
Conclusion:
The critical load for the pinned-pinned condition is 235 ksi
ii.
The critical load for the fixed-free end condition.
Answer to Problem 11.4.3P
The critical load for the fixed-free end condition is 58.75 ksi
Explanation of Solution
Given:
E=10400 ksi
L= 10 ft
d1= 5 in
d2= 6 in
Concept Used:
Calculation:
Conclusion:
The critical load for the fixed-free end condition is 58.75 ksi
iii.
The critical load for the fixed-pinned end condition.
Answer to Problem 11.4.3P
The critical load for the fixed-pinned end condition is 480.96 ksi
Explanation of Solution
Given:
E=10400 ksi
L= 10 ft
d1= 5 in
d2= 6 in
Concept Used:
Calculation:
Conclusion:
The critical load for the fixed-pinned end condition is 480.96 ksi
iv.
The critical load for the fixed-fixed end condition.
Answer to Problem 11.4.3P
The critical load for the fixed-fixed end condition is 940 ksi
Explanation of Solution
Given:
E=10400 ksi
L= 10 ft
d1= 5 in
d2= 6 in
Concept Used:
Calculation:
Conclusion:
The critical load for the fixed-fixed end condition is 940 ksi
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Chapter 11 Solutions
Mechanics of Materials (MindTap Course List)
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