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With one fixed end and other free end, a column of length L buckles at load, P1. Another column of same length and same cross-section fixed at both ends buckles at load P2. Value of P2/P1 is
- a)1
- b)2
- c)4
- d)16
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
With one fixed end and other free end, a column of length L buckles at...
The maximum load at which the column tends to have lateral displacement or tends to buckle is known as buckling or crippling load. Load columns can be analysed with the Euler’s column formulas can be given as
- For both end hinged, n = 1
- For one end fixed and other free, n = ½
- For both end fixed, n = 2
- For one end fixed and other hinged, n = √2
Most Upvoted Answer
With one fixed end and other free end, a column of length L buckles at...
Explanation:
When a column is subjected to an axial compressive load, it may fail by buckling. Buckling is a sudden failure mode that occurs when the column's deflection becomes large enough to cause it to buckle out of its straight path. The load at which buckling occurs is known as the buckling load. The buckling load depends on the column's length, cross-sectional area, and material properties.
Let us consider two columns of the same cross-sectional area and material properties, but with different boundary conditions. Column A has one fixed end and one free end, while column B has both ends fixed. The length of both the columns is the same.
Column A:
- One fixed end and one free end.
- Buckles at load P1.
Column B:
- Both ends fixed.
- Buckles at load P2.
The ratio of the buckling loads of the two columns can be calculated using Euler's Buckling Formula:
Pcr = (π²EI)/(KL)²
Where,
Pcr is the critical buckling load
E is the modulus of elasticity of the material
I is the moment of inertia of the cross-sectional area
K is the effective length factor
L is the length of the column
For column A, the effective length factor K = 2 because one end is fixed and the other end is free. For column B, K = 1 because both ends are fixed.
Since the columns have the same cross-sectional area and material properties, E and I are constant for both columns. Therefore, we can write:
P1/P2 = (K2L)²/(K1L)²
= (1/2)²
= 1/4
Therefore, P2/P1 = 4
Hence, the correct option is D.
When a column is subjected to an axial compressive load, it may fail by buckling. Buckling is a sudden failure mode that occurs when the column's deflection becomes large enough to cause it to buckle out of its straight path. The load at which buckling occurs is known as the buckling load. The buckling load depends on the column's length, cross-sectional area, and material properties.
Let us consider two columns of the same cross-sectional area and material properties, but with different boundary conditions. Column A has one fixed end and one free end, while column B has both ends fixed. The length of both the columns is the same.
Column A:
- One fixed end and one free end.
- Buckles at load P1.
Column B:
- Both ends fixed.
- Buckles at load P2.
The ratio of the buckling loads of the two columns can be calculated using Euler's Buckling Formula:
Pcr = (π²EI)/(KL)²
Where,
Pcr is the critical buckling load
E is the modulus of elasticity of the material
I is the moment of inertia of the cross-sectional area
K is the effective length factor
L is the length of the column
For column A, the effective length factor K = 2 because one end is fixed and the other end is free. For column B, K = 1 because both ends are fixed.
Since the columns have the same cross-sectional area and material properties, E and I are constant for both columns. Therefore, we can write:
P1/P2 = (K2L)²/(K1L)²
= (1/2)²
= 1/4
Therefore, P2/P1 = 4
Hence, the correct option is D.
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With one fixed end and other free end, a column of length L buckles at load, P1. Another column of same length and same cross-section fixed at both ends buckles at load P2. Value of P2/P1isa)1b)2c)4d)16Correct answer is option 'D'. Can you explain this answer?
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