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An axial load of 100kN is applied onto a bar of length 3 m and cross-sectional area of 100 mm2. If E = 200 kN/mm2 the total strain energy in the bar is equal to
- a)750 N-mm
- b)7500 N-mm
- c)75,000 N-mm
- d)7,50,000 N-mm
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
An axial load of 100kN is applied onto a bar of length 3 m and cross-...
Axial load, P = 100 × 103N
View all questions of this test
Length, L = 3000 mm 2
Cross section area, A = 100 mm E = 200 × 103 N/mm2
U =
= 
= 
= 750000 Nmm
Most Upvoted Answer
An axial load of 100kN is applied onto a bar of length 3 m and cross-...
To calculate the total strain energy in the bar, we can use the formula:
Strain Energy = (1/2) * Force * Displacement
1. Calculate the force:
Given that the axial load applied on the bar is 100 kN, the force can be calculated as follows:
Force = 100 kN = 100,000 N
2. Calculate the displacement:
The displacement can be calculated using the formula:
Displacement = Length * Strain
To find the strain, we can use Hooke's Law formula:
Strain = Stress / Young's Modulus
Given that the cross-sectional area of the bar is 100 mm^2 and the applied load is 100 kN, the stress can be calculated as follows:
Stress = Force / Area = 100,000 N / 100 mm^2 = 1,000 N/mm^2
Now, we can calculate the strain:
Strain = Stress / Young's Modulus = 1,000 N/mm^2 / 200 kN/mm^2 = 0.005
Next, we can calculate the displacement:
Displacement = Length * Strain = 3 m * 0.005 = 0.015 m
3. Calculate the strain energy:
Now, we can calculate the strain energy using the formula:
Strain Energy = (1/2) * Force * Displacement
= (1/2) * 100,000 N * 0.015 m
= 1,500 N-m
However, the given options are in N-mm, so we need to convert the strain energy to N-mm:
1 N-m = 1,000 N-mm
Therefore, the strain energy in N-mm is:
Strain Energy = 1,500 N-m * 1,000 N-mm / 1 N-m = 1,500,000 N-mm
The correct answer is option 'D' which is 7,50,000 N-mm.
Strain Energy = (1/2) * Force * Displacement
1. Calculate the force:
Given that the axial load applied on the bar is 100 kN, the force can be calculated as follows:
Force = 100 kN = 100,000 N
2. Calculate the displacement:
The displacement can be calculated using the formula:
Displacement = Length * Strain
To find the strain, we can use Hooke's Law formula:
Strain = Stress / Young's Modulus
Given that the cross-sectional area of the bar is 100 mm^2 and the applied load is 100 kN, the stress can be calculated as follows:
Stress = Force / Area = 100,000 N / 100 mm^2 = 1,000 N/mm^2
Now, we can calculate the strain:
Strain = Stress / Young's Modulus = 1,000 N/mm^2 / 200 kN/mm^2 = 0.005
Next, we can calculate the displacement:
Displacement = Length * Strain = 3 m * 0.005 = 0.015 m
3. Calculate the strain energy:
Now, we can calculate the strain energy using the formula:
Strain Energy = (1/2) * Force * Displacement
= (1/2) * 100,000 N * 0.015 m
= 1,500 N-m
However, the given options are in N-mm, so we need to convert the strain energy to N-mm:
1 N-m = 1,000 N-mm
Therefore, the strain energy in N-mm is:
Strain Energy = 1,500 N-m * 1,000 N-mm / 1 N-m = 1,500,000 N-mm
The correct answer is option 'D' which is 7,50,000 N-mm.
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An axial load of 100kN is applied onto a bar of length 3 m and cross-sectional area of 100 mm2. If E = 200 kN/mm2 the total strain energy in the bar is equal toa) 750 N-mmb) 7500 N-mmc) 75,000 N-mmd) 7,50,000 N-mmCorrect answer is option 'D'. Can you explain this answer?
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An axial load of 100kN is applied onto a bar of length 3 m and cross-sectional area of 100 mm2. If E = 200 kN/mm2 the total strain energy in the bar is equal toa) 750 N-mmb) 7500 N-mmc) 75,000 N-mmd) 7,50,000 N-mmCorrect answer is option 'D'. 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 An axial load of 100kN is applied onto a bar of length 3 m and cross-sectional area of 100 mm2. If E = 200 kN/mm2 the total strain energy in the bar is equal toa) 750 N-mmb) 7500 N-mmc) 75,000 N-mmd) 7,50,000 N-mmCorrect answer is option 'D'. 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 An axial load of 100kN is applied onto a bar of length 3 m and cross-sectional area of 100 mm2. If E = 200 kN/mm2 the total strain energy in the bar is equal toa) 750 N-mmb) 7500 N-mmc) 75,000 N-mmd) 7,50,000 N-mmCorrect answer is option 'D'. Can you explain this answer?.
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