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A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2?
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A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both e...
**Given:**
- Length of the strut (L) = 1.2 m
- Diameter of the strut (d) = 30 mm = 0.03 m
- Axial thrust at the ends (P) = 20 kN
- Transverse point load at the center (W) = 1.8 kN
- Modulus of elasticity (E) = 208 GPa = 208 GN/m²
**To find:**
The maximum stress induced in the strut.
**Assumptions:**
- The strut is made of a homogeneous and isotropic material.
- The strut is perfectly cylindrical and has a uniform cross-section.
- The strut is subjected to pure axial loading and the point load does not induce any bending moment.
**Analysis:**
1. Calculating the area of the strut:
- The diameter of the strut is given as 30 mm, which means the radius (r) is half of the diameter.
- r = 0.03 / 2 = 0.015 m
- The cross-sectional area (A) of the strut can be calculated using the formula: A = πr²
- A = π(0.015)² = 0.00070686 m²
2. Calculating the axial stress due to the axial thrust:
- The axial stress (σ₁) can be calculated using the formula: σ₁ = P / A
- σ₁ = 20 kN / 0.00070686 m² = 28.3 MN/m²
3. Calculating the bending stress due to the transverse point load:
- Since the strut is hinged at both ends, it is free to rotate. Therefore, the transverse point load does not induce any bending moment.
- Hence, the bending stress can be considered zero (σ₂ = 0).
4. Calculating the total stress:
- The total stress (σ) induced in the strut can be calculated using the formula: σ = σ₁ + σ₂
- σ = 28.3 MN/m² + 0 = 28.3 MN/m²
**Answer:**
The maximum stress induced in the strut is 28.3 MN/m², which is approximately equal to 28.3 MPa.
- Length of the strut (L) = 1.2 m
- Diameter of the strut (d) = 30 mm = 0.03 m
- Axial thrust at the ends (P) = 20 kN
- Transverse point load at the center (W) = 1.8 kN
- Modulus of elasticity (E) = 208 GPa = 208 GN/m²
**To find:**
The maximum stress induced in the strut.
**Assumptions:**
- The strut is made of a homogeneous and isotropic material.
- The strut is perfectly cylindrical and has a uniform cross-section.
- The strut is subjected to pure axial loading and the point load does not induce any bending moment.
**Analysis:**
1. Calculating the area of the strut:
- The diameter of the strut is given as 30 mm, which means the radius (r) is half of the diameter.
- r = 0.03 / 2 = 0.015 m
- The cross-sectional area (A) of the strut can be calculated using the formula: A = πr²
- A = π(0.015)² = 0.00070686 m²
2. Calculating the axial stress due to the axial thrust:
- The axial stress (σ₁) can be calculated using the formula: σ₁ = P / A
- σ₁ = 20 kN / 0.00070686 m² = 28.3 MN/m²
3. Calculating the bending stress due to the transverse point load:
- Since the strut is hinged at both ends, it is free to rotate. Therefore, the transverse point load does not induce any bending moment.
- Hence, the bending stress can be considered zero (σ₂ = 0).
4. Calculating the total stress:
- The total stress (σ) induced in the strut can be calculated using the formula: σ = σ₁ + σ₂
- σ = 28.3 MN/m² + 0 = 28.3 MN/m²
**Answer:**
The maximum stress induced in the strut is 28.3 MN/m², which is approximately equal to 28.3 MPa.
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A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2?
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A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2? 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 A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2?.
A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2? 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 A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical strut of length 1.2 m and dia 30 mm, is hinged at both ends. It is subjected to axial thrust of 20 KN at it's ends and a transverse point load of 1.8 KN at the centre. If E=208 GN/ m^2 for the material, the maximum stress induced would be: A 223 MN /m^2 B 253 MN/m^2 C 323 MN/m^2 D 353MN/m^2?.
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