Concept explainers
A wide-flange steel column (E = 30 × l06 psi) of W12 × 87 shape (see figure) h as a length L = 28 ft. It is supported only at the ends and may buckle in any direction.
Calculate the allowable load Pallowbased upon the critical load with a factor of safety n = 2.5. Consider the following end conditions: (a) pinned-pinned, (b) fixed-free, (c) fixed-pinned, and (d) fixed-fixed.
i.
The allowable load for the pinned-pinned end condition.
Answer to Problem 11.4.5P
The allowable load for the pinned-pinned condition is 252.842 ksi
Explanation of Solution
Given:
E=30 ×106 psi
L= 28 ft
Factor of of safety = 2.5
Column type: W12×87
Concept Used:
Calculation:
Conclusion:
The allowable load for the pinned-pinned condition is 252.842 ksi
ii.
The allowable load for the fixed-free end condition.
Answer to Problem 11.4.5P
The allowable load for the fixed-free end condition is 63.206 ksi
Explanation of Solution
Given:
E=30 ×106 psi
L= 28 ft
Factor of of safety = 2.5
Column type: W12×87
Concept Used:
Calculation:
Conclusion:
The allowable load for the fixed-free end condition is 63.206 ksi
iii.
The allowable load for the fixed-pinned end condition.
Answer to Problem 11.4.5P
The allowable load for the fixed-pinned end condition is 517.28 ksi
Explanation of Solution
Given:
E=30 ×106 psi
L= 28 ft
Factor of of safety = 2.5
Column type: W12×87
Concept Used:
Calculation:
Conclusion:
The allowable load for the fixed-pinned end condition is 517.28 ksi
iv.
The allowable load for the fixed-fixed end condition.
Answer to Problem 11.4.5P
The allowable load for the fixed-fixed end condition is 1011.296 ksi
Explanation of Solution
Given:
E=30 ×106 psi
L= 28 ft
Factor of of safety = 2.5
Column type: W12×87
Concept Used:
Calculation:
Conclusion:
The allowable load for the fixed-fixed end condition is 1011.296 ksi
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Chapter 11 Solutions
Mechanics of Materials (MindTap Course List)
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- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning