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A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter for
the remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar take
E = 105 N/mm2
the remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar take
E = 105 N/mm2
Correct answer is between '0.16,0.24'. Can you explain this answer?
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A straight bar 500 mm long is 30 mm in diameter for the first 350 mm l...
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A straight bar 500 mm long is 30 mm in diameter for the first 350 mm l...
Given:
- Length of the bar (L) = 500 mm
- Diameter of the bar (d1) = 30 mm for the first 350 mm length
- Diameter of the bar (d2) = 20 mm for the remaining length
- Axial pull force (P) = 20 kN = 20,000 N
- Young's modulus (E) = 105 N/mm²
To find:
- Extension of the bar
Assumptions:
- The bar is made of a homogeneous and isotropic material.
- The bar is subjected to axial tension, resulting in uniform stress and strain along its length.
- The bar remains in the elastic region and obeys Hooke's law.
Solution:
1. Calculate the cross-sectional area of the bar at both diameters:
- Area of the first section (A1) = π * (d1/2)² = π * (30/2)² = 2250 mm²
- Area of the second section (A2) = π * (d2/2)² = π * (20/2)² = 1000 mm²
2. Calculate the change in length for each section using the formula:
- Change in length (ΔL) = (P * L) / (A * E)
- Change in length for the first section (ΔL1) = (P * L1) / (A1 * E)
= (20,000 * 350) / (2250 * 105) = 0.3048 mm
- Change in length for the second section (ΔL2) = (P * L2) / (A2 * E)
= (20,000 * 150) / (1000 * 105) = 0.2857 mm
3. Total change in length (ΔL) = ΔL1 + ΔL2 = 0.3048 + 0.2857 = 0.5905 mm
4. Convert the change in length from mm to meters:
- ΔL = 0.5905 mm = 0.5905 * 10⁻³ m = 0.0005905 m
5. The extension of the bar is equal to the change in length:
- Extension of the bar = ΔL = 0.0005905 m = 0.59 mm
Conclusion:
The extension of the bar under an axial pull of 20 kN is approximately 0.59 mm.
- Length of the bar (L) = 500 mm
- Diameter of the bar (d1) = 30 mm for the first 350 mm length
- Diameter of the bar (d2) = 20 mm for the remaining length
- Axial pull force (P) = 20 kN = 20,000 N
- Young's modulus (E) = 105 N/mm²
To find:
- Extension of the bar
Assumptions:
- The bar is made of a homogeneous and isotropic material.
- The bar is subjected to axial tension, resulting in uniform stress and strain along its length.
- The bar remains in the elastic region and obeys Hooke's law.
Solution:
1. Calculate the cross-sectional area of the bar at both diameters:
- Area of the first section (A1) = π * (d1/2)² = π * (30/2)² = 2250 mm²
- Area of the second section (A2) = π * (d2/2)² = π * (20/2)² = 1000 mm²
2. Calculate the change in length for each section using the formula:
- Change in length (ΔL) = (P * L) / (A * E)
- Change in length for the first section (ΔL1) = (P * L1) / (A1 * E)
= (20,000 * 350) / (2250 * 105) = 0.3048 mm
- Change in length for the second section (ΔL2) = (P * L2) / (A2 * E)
= (20,000 * 150) / (1000 * 105) = 0.2857 mm
3. Total change in length (ΔL) = ΔL1 + ΔL2 = 0.3048 + 0.2857 = 0.5905 mm
4. Convert the change in length from mm to meters:
- ΔL = 0.5905 mm = 0.5905 * 10⁻³ m = 0.0005905 m
5. The extension of the bar is equal to the change in length:
- Extension of the bar = ΔL = 0.0005905 m = 0.59 mm
Conclusion:
The extension of the bar under an axial pull of 20 kN is approximately 0.59 mm.
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A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer?
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A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer? 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 A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer?.
A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer? 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 A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer?.
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A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer?, a detailed solution for A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer? has been provided alongside types of A straight bar 500 mm long is 30 mm in diameter for the first 350 mm length and 20mm diameter forthe remaining length. If the bar is subjected to an axial pull of 20 kN find the extension of the bar takeE = 105 N/mm2Correct answer is between '0.16,0.24'. Can you explain this answer? theory, EduRev gives you an
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