Concept explainers
A column has both ends pinned and has a length of 32 in. It is made of SAE 1040 HR steel and has a circular shape with a diameter of 0.75 in. Determine the critical load.
The critical load that can be applied to a column of circular cross section whose both ends are pinned.
Answer to Problem 1P
The critical load of the column is given by 4490 lb
Explanation of Solution
Given Information:
Cross section of the column: Circular
Diameter of the column, D = 0.75 in
Length of the column, L = 32 in
End conditions: Both ends pinned
Material = SAE 1040 HR steel
So, Young’s Modulus, E =
Yield strength, Sy= 42000 psi
Concept used:
For long Column Euler’s Formula used
For short Column Euler’s Formula used
Here,
Pcr is Critical load
E is Young’s Modulus
Imin is Minimum Moment of Inertia
Le is effective length (Based on the end conditions of Column)
For both ends pinned condition, Le = L
Sy is yield stress
Sr Slenderness ratio =
r is radius of gyration
A cross section area
Calculation:
For solid round section,
r = D/4
= 0.75/4
= 0.188 in
Slenderness ratio,
Compute Column constant for the above material
Sr>Cc, so Column is long, Use Euler’s Formula
Moment of Inertia,
From Equation (1)
Thus, critical load is 4490 lb.
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