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An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square : (σcirtical)circular’ is
- a)1 : 4
- b)3 : 4
- c)4 : 3
- d)4 : 1
Correct answer is option 'C'. Can you explain this answer?
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An a aluminium column of square cross-section (10 mm × 10 mm) an...
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An a aluminium column of square cross-section (10 mm × 10 mm) an...
Explanation:
Given Data:
- Square cross-section column: Side length = 10 mm, Length = 300 mm
- Circular cross-section column: Diameter = 10 mm, Length = 300 mm
Formula:
- The critical buckling load for a column can be calculated using Euler's formula:
P_critical = (π^2 * E * I) / (K * L)^2
Where,
P_critical = Critical load
E = Modulus of elasticity
I = Moment of inertia
K = Column effective length factor
L = Length of the column
Calculation:
1. For the square cross-section column:
- Moment of inertia (I) for square cross-section = (side length)^4 / 12
I_square = (10 mm)^4 / 12 = 833.33 mm^4
- Effective length factor (K) for both ends pinned = 1
- Critical stress for square column: σ_critical_square = P_critical / A
A = Cross-sectional area = (side length)^2 = 10 mm^2
2. For the circular cross-section column:
- Moment of inertia (I) for circular cross-section = π * (diameter)^4 / 64
I_circular = π * (10 mm)^4 / 64 = 491.87 mm^4
- Critical stress for circular column: σ_critical_circular = P_critical / A
A = π * (diameter/2)^2 = 78.54 mm^2
Ratio of critical stresses:
- σ_critical_square : σ_critical_circular = (P_critical / A)_square : (P_critical / A)_circular
- Substituting the values and simplifying, we get:
σ_critical_square : σ_critical_circular = (π^2 * E * I_square) / (K * L)^2 : (π^2 * E * I_circular) / (K * L)^2
σ_critical_square : σ_critical_circular = I_square : I_circular
σ_critical_square : σ_critical_circular = 833.33 mm^4 : 491.87 mm^4
σ_critical_square : σ_critical_circular = 4 : 3
Therefore, the ratio of critical stresses for the square and circular columns is 4 : 3.
Given Data:
- Square cross-section column: Side length = 10 mm, Length = 300 mm
- Circular cross-section column: Diameter = 10 mm, Length = 300 mm
Formula:
- The critical buckling load for a column can be calculated using Euler's formula:
P_critical = (π^2 * E * I) / (K * L)^2
Where,
P_critical = Critical load
E = Modulus of elasticity
I = Moment of inertia
K = Column effective length factor
L = Length of the column
Calculation:
1. For the square cross-section column:
- Moment of inertia (I) for square cross-section = (side length)^4 / 12
I_square = (10 mm)^4 / 12 = 833.33 mm^4
- Effective length factor (K) for both ends pinned = 1
- Critical stress for square column: σ_critical_square = P_critical / A
A = Cross-sectional area = (side length)^2 = 10 mm^2
2. For the circular cross-section column:
- Moment of inertia (I) for circular cross-section = π * (diameter)^4 / 64
I_circular = π * (10 mm)^4 / 64 = 491.87 mm^4
- Critical stress for circular column: σ_critical_circular = P_critical / A
A = π * (diameter/2)^2 = 78.54 mm^2
Ratio of critical stresses:
- σ_critical_square : σ_critical_circular = (P_critical / A)_square : (P_critical / A)_circular
- Substituting the values and simplifying, we get:
σ_critical_square : σ_critical_circular = (π^2 * E * I_square) / (K * L)^2 : (π^2 * E * I_circular) / (K * L)^2
σ_critical_square : σ_critical_circular = I_square : I_circular
σ_critical_square : σ_critical_circular = 833.33 mm^4 : 491.87 mm^4
σ_critical_square : σ_critical_circular = 4 : 3
Therefore, the ratio of critical stresses for the square and circular columns is 4 : 3.
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An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square: (σcirtical)circular’isa)1 : 4b)3 : 4c)4 : 3d)4 : 1Correct answer is option 'C'. Can you explain this answer?
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An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square: (σcirtical)circular’isa)1 : 4b)3 : 4c)4 : 3d)4 : 1Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square: (σcirtical)circular’isa)1 : 4b)3 : 4c)4 : 3d)4 : 1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square: (σcirtical)circular’isa)1 : 4b)3 : 4c)4 : 3d)4 : 1Correct answer is option 'C'. Can you explain this answer?.
An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square: (σcirtical)circular’isa)1 : 4b)3 : 4c)4 : 3d)4 : 1Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square: (σcirtical)circular’isa)1 : 4b)3 : 4c)4 : 3d)4 : 1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An a aluminium column of square cross-section (10 mm × 10 mm) and length 300 mm has both ends pinned. This is replaced by a circular cross-section (of diameter 10 mm) column of the same length and made of the same material with the same end conditions. The ratio of critical stresses for these two columns corresponding to Euler’s critical load, (σcritical)square: (σcirtical)circular’isa)1 : 4b)3 : 4c)4 : 3d)4 : 1Correct answer is option 'C'. Can you explain this answer?.
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