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A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.
- a)0.34
- b)0.38
- c)0.36
- d)0.32
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A tensile specimen with 12 mm initial diameter and 50 mm initial lengt...
Given,
d1 = 12 mm
l1 = 50 mm p = 90 kN
d2 = 10 mm
= 0.36
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Community Answer
A tensile specimen with 12 mm initial diameter and 50 mm initial lengt...
To calculate the true strain along the length of the tensile specimen, we can use the equation:
ε_true = ln(Lf/Li)
Where:
ε_true is the true strain,
Lf is the final length of the specimen, and
Li is the initial length of the specimen.
We are given that the initial length of the specimen is 50 mm, and the final length is not provided. However, we can calculate the final length using the change in diameter.
Let's assume the final diameter is Df. The change in diameter can be calculated using the equation:
ΔD = Di - Df
Where:
ΔD is the change in diameter,
Di is the initial diameter, and
Df is the final diameter.
We are given that the initial diameter is 12 mm and the final diameter is 10 mm. Therefore:
ΔD = 12 mm - 10 mm
ΔD = 2 mm
Since the material is assumed to be incompressible, we know that the volume remains constant. Hence, we can use the equation:
π/4 * Di^2 * Li = π/4 * Df^2 * Lf
Where:
Di is the initial diameter,
Li is the initial length,
Df is the final diameter, and
Lf is the final length.
Substituting the known values, we have:
π/4 * (12 mm)^2 * 50 mm = π/4 * (10 mm)^2 * Lf
Simplifying the equation:
(1440 mm^3) * 50 mm = (100 mm^3) * Lf
72000 mm^4 = 100 mm^3 * Lf
Lf = 72000 mm^4 / 100 mm^3
Lf = 720 mm
Now we can substitute the values of Lf and Li into the equation for true strain:
ε_true = ln(720 mm/50 mm)
ε_true = ln(14.4)
Using a calculator, the natural logarithm of 14.4 is approximately 2.664. Therefore, the true strain along the length of the tensile specimen is:
ε_true ≈ 2.664
So, the correct answer is option 'a) 0.34'.
ε_true = ln(Lf/Li)
Where:
ε_true is the true strain,
Lf is the final length of the specimen, and
Li is the initial length of the specimen.
We are given that the initial length of the specimen is 50 mm, and the final length is not provided. However, we can calculate the final length using the change in diameter.
Let's assume the final diameter is Df. The change in diameter can be calculated using the equation:
ΔD = Di - Df
Where:
ΔD is the change in diameter,
Di is the initial diameter, and
Df is the final diameter.
We are given that the initial diameter is 12 mm and the final diameter is 10 mm. Therefore:
ΔD = 12 mm - 10 mm
ΔD = 2 mm
Since the material is assumed to be incompressible, we know that the volume remains constant. Hence, we can use the equation:
π/4 * Di^2 * Li = π/4 * Df^2 * Lf
Where:
Di is the initial diameter,
Li is the initial length,
Df is the final diameter, and
Lf is the final length.
Substituting the known values, we have:
π/4 * (12 mm)^2 * 50 mm = π/4 * (10 mm)^2 * Lf
Simplifying the equation:
(1440 mm^3) * 50 mm = (100 mm^3) * Lf
72000 mm^4 = 100 mm^3 * Lf
Lf = 72000 mm^4 / 100 mm^3
Lf = 720 mm
Now we can substitute the values of Lf and Li into the equation for true strain:
ε_true = ln(720 mm/50 mm)
ε_true = ln(14.4)
Using a calculator, the natural logarithm of 14.4 is approximately 2.664. Therefore, the true strain along the length of the tensile specimen is:
ε_true ≈ 2.664
So, the correct answer is option 'a) 0.34'.
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Question Description
A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.a)0.34b)0.38c)0.36d)0.32Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.a)0.34b)0.38c)0.36d)0.32Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.a)0.34b)0.38c)0.36d)0.32Correct answer is option 'C'. Can you explain this answer?.
A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.a)0.34b)0.38c)0.36d)0.32Correct answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.a)0.34b)0.38c)0.36d)0.32Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.a)0.34b)0.38c)0.36d)0.32Correct answer is option 'C'. Can you explain this answer?.
Solutions for A tensile specimen with 12 mm initial diameter and 50 mm initial length is subjected to a load of 90 kN. After some time the diameter is 10 mm. Assuming it as incompressible material, Calculate the true strain along the length.a)0.34b)0.38c)0.36d)0.32Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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