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A rod of material E = 200 x 103 MPa and α = 10-3 mm/mm/°C is fixed at both the ends. It is uniformly heated such that the increase in temperature is 30°C. The stress developed in the rod is
- a)6000 N/mm2 (tensile)
- b)6000 N/mm2 (compressive)
- c)2000 N/mm2 (tensile)
- d)2000 N/mm2 (compressive)
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A rod of material E = 200 x 103 MPa and α = 10-3 mm/mm/°C is...
σ = E ∝ T = 200 × 103 × 10-3 × 30= 6000 N/mm2
Stress will be compressive in nature as material is not allowed to expand.
Most Upvoted Answer
A rod of material E = 200 x 103 MPa and α = 10-3 mm/mm/°C is...
A rod of material E = 200 x 10^3 MPa and A = 10^-4 m^2 is subjected to a tensile force F = 10^4 N. What is the strain in the rod?
The strain in a material is defined as the change in length (ΔL) divided by the original length (L). It can be calculated using the formula:
strain = ΔL / L
To calculate the strain, we first need to find the change in length (ΔL). The change in length is equal to the force applied (F) divided by the product of the cross-sectional area (A) and the Young's modulus (E).
ΔL = F / (A * E)
Plugging in the given values, we have:
ΔL = 10^4 N / (10^-4 m^2 * 200 x 10^3 MPa)
Simplifying the units, we convert MPa to Pa by multiplying by 10^6:
ΔL = 10^4 N / (10^-4 m^2 * 200 x 10^9 Pa)
ΔL = 0.05 m
Now we can calculate the strain:
strain = ΔL / L
strain = 0.05 m / L
Since the original length (L) is not given, we cannot calculate the exact strain without this information. However, we can calculate the strain as a ratio of the change in length to the original length.
The strain in a material is defined as the change in length (ΔL) divided by the original length (L). It can be calculated using the formula:
strain = ΔL / L
To calculate the strain, we first need to find the change in length (ΔL). The change in length is equal to the force applied (F) divided by the product of the cross-sectional area (A) and the Young's modulus (E).
ΔL = F / (A * E)
Plugging in the given values, we have:
ΔL = 10^4 N / (10^-4 m^2 * 200 x 10^3 MPa)
Simplifying the units, we convert MPa to Pa by multiplying by 10^6:
ΔL = 10^4 N / (10^-4 m^2 * 200 x 10^9 Pa)
ΔL = 0.05 m
Now we can calculate the strain:
strain = ΔL / L
strain = 0.05 m / L
Since the original length (L) is not given, we cannot calculate the exact strain without this information. However, we can calculate the strain as a ratio of the change in length to the original length.
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A rod of material E = 200 x 103 MPa and α = 10-3 mm/mm/°C is fixed at both the ends. It is uniformly heated such that the increase in temperature is 30°C. The stress developed in the rod isa)6000 N/mm2 (tensile)b)6000 N/mm2 (compressive)c)2000 N/mm2 (tensile)d)2000 N/mm2 (compressive)Correct answer is option 'B'. Can you explain this answer?
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A rod of material E = 200 x 103 MPa and α = 10-3 mm/mm/°C is fixed at both the ends. It is uniformly heated such that the increase in temperature is 30°C. The stress developed in the rod isa)6000 N/mm2 (tensile)b)6000 N/mm2 (compressive)c)2000 N/mm2 (tensile)d)2000 N/mm2 (compressive)Correct answer is option 'B'. Can you explain this answer? 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 rod of material E = 200 x 103 MPa and α = 10-3 mm/mm/°C is fixed at both the ends. It is uniformly heated such that the increase in temperature is 30°C. The stress developed in the rod isa)6000 N/mm2 (tensile)b)6000 N/mm2 (compressive)c)2000 N/mm2 (tensile)d)2000 N/mm2 (compressive)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rod of material E = 200 x 103 MPa and α = 10-3 mm/mm/°C is fixed at both the ends. It is uniformly heated such that the increase in temperature is 30°C. The stress developed in the rod isa)6000 N/mm2 (tensile)b)6000 N/mm2 (compressive)c)2000 N/mm2 (tensile)d)2000 N/mm2 (compressive)Correct answer is option 'B'. Can you explain this answer?.
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