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A circular bar 20 mm in diameter and 200 mm long is subjected to a force of 20 kN. Find the strain in the bar if the value of E = 80 GPa.
- a)7.9 × 10-2
- b)7.9 × 10-4
- c)3.2 × 10-2
- d)3.2 × 10-3
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A circular bar 20 mm in diameter and 200 mm long is subjected to a fo...
E = 80 GPa = 80 × 103 MPa = 80 × 103 N/mm2
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Stress = Load/Area
Strain = Stress/E
Most Upvoted Answer
A circular bar 20 mm in diameter and 200 mm long is subjected to a fo...
E = 80 GPa = 80 × 103 MPa = 80 × 103 N/mm2
Stress = Load/Area
Strain = Stress/E
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Community Answer
A circular bar 20 mm in diameter and 200 mm long is subjected to a fo...
Given data:
Diameter of bar, d = 20 mm
Length of bar, L = 200 mm
Force applied, F = 20 kN = 20,000 N
Young's modulus, E = 80 GPa = 80 × 10^9 N/m^2
We know that the strain in a material is defined as the ratio of the change in length to the original length. In this case, the bar is subjected to an axial force, which will cause it to elongate. The strain can be calculated using the formula:
Strain = ΔL / L
where ΔL is the change in length and L is the original length.
To find ΔL, we can use the formula for axial stress:
Stress = F / A
where A is the cross-sectional area of the bar. For a circular bar, the area is given by:
A = πd^2 / 4
Substituting the values, we get:
A = π(20 mm)^2 / 4 = 314.16 mm^2
Now, the stress can be calculated as:
Stress = F / A = 20,000 N / 314.16 mm^2 = 63.66 N/mm^2
Using the formula for axial strain:
Strain = Stress / E
Substituting the values, we get:
Strain = 63.66 N/mm^2 / (80 × 10^9 N/m^2) = 7.9575 × 10^-4
Therefore, the strain in the bar is 7.9 × 10^-4.
Hence, option B is the correct answer.
Diameter of bar, d = 20 mm
Length of bar, L = 200 mm
Force applied, F = 20 kN = 20,000 N
Young's modulus, E = 80 GPa = 80 × 10^9 N/m^2
We know that the strain in a material is defined as the ratio of the change in length to the original length. In this case, the bar is subjected to an axial force, which will cause it to elongate. The strain can be calculated using the formula:
Strain = ΔL / L
where ΔL is the change in length and L is the original length.
To find ΔL, we can use the formula for axial stress:
Stress = F / A
where A is the cross-sectional area of the bar. For a circular bar, the area is given by:
A = πd^2 / 4
Substituting the values, we get:
A = π(20 mm)^2 / 4 = 314.16 mm^2
Now, the stress can be calculated as:
Stress = F / A = 20,000 N / 314.16 mm^2 = 63.66 N/mm^2
Using the formula for axial strain:
Strain = Stress / E
Substituting the values, we get:
Strain = 63.66 N/mm^2 / (80 × 10^9 N/m^2) = 7.9575 × 10^-4
Therefore, the strain in the bar is 7.9 × 10^-4.
Hence, option B is the correct answer.
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A circular bar 20 mm in diameter and 200 mm long is subjected to a force of 20 kN. Find the strain in the bar if the value of E = 80 GPa.a)7.9 × 10-2b)7.9 × 10-4c)3.2 × 10-2d)3.2 × 10-3Correct answer is option 'B'. Can you explain this answer?
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A circular bar 20 mm in diameter and 200 mm long is subjected to a force of 20 kN. Find the strain in the bar if the value of E = 80 GPa.a)7.9 × 10-2b)7.9 × 10-4c)3.2 × 10-2d)3.2 × 10-3Correct answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A circular bar 20 mm in diameter and 200 mm long is subjected to a force of 20 kN. Find the strain in the bar if the value of E = 80 GPa.a)7.9 × 10-2b)7.9 × 10-4c)3.2 × 10-2d)3.2 × 10-3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular bar 20 mm in diameter and 200 mm long is subjected to a force of 20 kN. Find the strain in the bar if the value of E = 80 GPa.a)7.9 × 10-2b)7.9 × 10-4c)3.2 × 10-2d)3.2 × 10-3Correct answer is option 'B'. Can you explain this answer?.
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