Concept explainers
Solve the preceding problem for a steel pipe column (E = 210 GPa) with length L = 1.2 m, inner diameter d2= 36 mm, and outer diameter d2=40 mm.
i.
The critical load for the pinned-pinned end condition.
Answer to Problem 11.4.4P
The critical load for the pinned-pinned condition is 62.201 kN
Explanation of Solution
Given:
E=210 GPa
L= 1.2 m
d1= 36 mm
d2= 40 mm
Concept Used:
Calculation:
Conclusion:
The critical load for the pinned-pinned condition is 62.201 kN
ii.
The critical load for the fixed-free end condition.
Answer to Problem 11.4.4P
The critical load for the fixed-free end condition is 15.550 kN
Explanation of Solution
Given:
E=210 GPa
L= 1.2 m
d1= 36 mm
d2= 40 mm
Concept Used:
Calculation:
Conclusion:
The critical load for the fixed-free end condition is 15.550 kN
iii.
The critical load for the fixed-pinned end condition.
Answer to Problem 11.4.4P
The critical load for the fixed-pinned end condition is 127.304 kN
Explanation of Solution
Given:
E=210 GPa
L= 1.2 m
d1= 36 mm
d2= 40 mm
Concept Used:
Calculation:
Conclusion:
The critical load for the fixed-pinned end condition is 127.304 kN
iv.
The critical load for the fixed-fixed end condition.
Answer to Problem 11.4.4P
The critical load for the fixed-fixed end condition is 248.804 kN
Explanation of Solution
Given:
E=210 GPa
L= 1.2 m
d1= 36 mm
d2= 40 mm
Concept Used:
Calculation:
Conclusion:
The critical load for the fixed-fixed end condition is 248.804 kN
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Chapter 11 Solutions
Mechanics of Materials (MindTap Course List)
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