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Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…?
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Square cross-section wooden column of length 3140mm is pinned at both ...
**Given:**
Length of the column (L) = 3140 mm
Young's modulus of elasticity (E) = 12 GPa = 12 × 10^9 Pa
Allowable compressive stress (σ) = 12 MPa = 12 × 10^6 Pa
Axial compressive load (P) = 200 kN = 200 × 10^3 N
Factor of safety (FoS) = 2
**Calculations:**
1. **Euler's Buckling Load (Pcr):**
The Euler's buckling load is given by the formula:
Pcr = (π^2 * E * I) / (L^2)
where I is the moment of inertia of the cross-section.
2. **Minimum Cross-Section Area (A):**
The minimum cross-section area can be calculated using the formula:
A = Pcr / σ
3. **Moment of Inertia (I) for a Square Cross-Section:**
For a square cross-section, the moment of inertia (I) can be calculated using the formula:
I = (b^4) / 12
where b is the side length of the square cross-section.
Let's calculate each step in detail:
1. **Euler's Buckling Load (Pcr):**
Pcr = (π^2 * E * I) / (L^2)
To calculate I, we need to know the side length of the square cross-section. Let's assume b.
2. **Minimum Cross-Section Area (A):**
A = Pcr / σ
3. **Moment of Inertia (I) for a Square Cross-Section:**
I = (b^4) / 12
Now, let's substitute the formulas into each other and solve for A:
Pcr = (π^2 * E * ((b^4) / 12)) / (L^2)
A = ((π^2 * E * ((b^4) / 12)) / (L^2)) / σ
A = ((π^2 * (12 × 10^9) * ((b^4) / 12)) / (3140^2)) / (12 × 10^6)
A = (π^2 * (b^4) * (12 × 10^9)) / (3.14 * (3140^2) * (12 × 10^6))
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * (10^9 / 10^6)
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3
Length of the column (L) = 3140 mm
Young's modulus of elasticity (E) = 12 GPa = 12 × 10^9 Pa
Allowable compressive stress (σ) = 12 MPa = 12 × 10^6 Pa
Axial compressive load (P) = 200 kN = 200 × 10^3 N
Factor of safety (FoS) = 2
**Calculations:**
1. **Euler's Buckling Load (Pcr):**
The Euler's buckling load is given by the formula:
Pcr = (π^2 * E * I) / (L^2)
where I is the moment of inertia of the cross-section.
2. **Minimum Cross-Section Area (A):**
The minimum cross-section area can be calculated using the formula:
A = Pcr / σ
3. **Moment of Inertia (I) for a Square Cross-Section:**
For a square cross-section, the moment of inertia (I) can be calculated using the formula:
I = (b^4) / 12
where b is the side length of the square cross-section.
Let's calculate each step in detail:
1. **Euler's Buckling Load (Pcr):**
Pcr = (π^2 * E * I) / (L^2)
To calculate I, we need to know the side length of the square cross-section. Let's assume b.
2. **Minimum Cross-Section Area (A):**
A = Pcr / σ
3. **Moment of Inertia (I) for a Square Cross-Section:**
I = (b^4) / 12
Now, let's substitute the formulas into each other and solve for A:
Pcr = (π^2 * E * ((b^4) / 12)) / (L^2)
A = ((π^2 * E * ((b^4) / 12)) / (L^2)) / σ
A = ((π^2 * (12 × 10^9) * ((b^4) / 12)) / (3140^2)) / (12 × 10^6)
A = (π^2 * (b^4) * (12 × 10^9)) / (3.14 * (3140^2) * (12 × 10^6))
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * (10^9 / 10^6)
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3.14 * (3140^2) * 12) * 10^3
A = (π^2 * (b^4)) / (3
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Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…?
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Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…?.
Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Square cross-section wooden column of length 3140mm is pinned at both ends. For the wood, Young’s modulus of elasticity is 12 GPa and allowable compressive stress is 12 MPa. The column needs to support an axial compressive load of 200kN. Using a factor of safety of 2 in the computation of Euler’s buckling load, the minimum cross-section area (in mm2) of the column is…?.
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