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A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, is
- a)60 MPa tensile
- b)80 MPa tensile
- c)80 MPa compressive
- d)60 MPa compressive
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A steel rod, 2 m long, is held between two walls and heated from 20℃ ...
L = 2 m, t1 = 20° and t2 = 60°
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Temperature change (∆t) = t2 − t1 = 40℃
E = 200 x 103 MPa
α = 10 x 10−6/℃
yielding of the wall (λ) = 0.2 mm
BB1 = δth = α(∆t)L
B1 B2 = Expansion restricted by supports
λ = BB2 = yielding of support
= −100[(2000 x 10 x 10−6 x 40) − 0.2]
= −60 MPa = 60 MPa compressive
Most Upvoted Answer
A steel rod, 2 m long, is held between two walls and heated from 20℃ ...
To solve this problem, we can use the formula for linear expansion:
ΔL = α * L * ΔT
Where:
ΔL is the change in length of the rod,
α is the coefficient of linear expansion,
L is the original length of the rod, and
ΔT is the change in temperature.
Given:
Initial temperature (T1) = 20℃
Final temperature (T2) = 60℃
Original length (L) = 2 m
Coefficient of linear expansion (α) = 10 x 10^(-6)/℃
1. Calculate the change in temperature (ΔT):
ΔT = T2 - T1
ΔT = 60℃ - 20℃
ΔT = 40℃
2. Calculate the change in length (ΔL):
ΔL = α * L * ΔT
ΔL = (10 x 10^(-6)/℃) * (2 m) * (40℃)
ΔL = 8 x 10^(-5) m or 0.08 mm
The change in length is 0.08 mm.
3. Determine the stress induced in the rod:
Stress (σ) = (Force (F)) / (Cross-sectional area (A))
The force applied to the rod is equal to the stress induced multiplied by the cross-sectional area. Since the walls yield by 0.2 mm, the force is acting on the rod in compression.
4. Calculate the cross-sectional area (A):
We need the diameter of the rod to calculate the cross-sectional area.
5. Calculate the diameter of the rod:
The diameter of the rod is not given, but we can assume it to be 10 mm (0.01 m) for simplicity.
6. Calculate the cross-sectional area (A):
A = π * (d/2)^2
A = π * (0.01/2)^2
A = 7.85 x 10^(-5) m^2
7. Calculate the stress induced in the rod:
Stress (σ) = Force (F) / Cross-sectional area (A)
σ = F / A
Since the walls yield by 0.2 mm (0.0002 m), the force applied to the rod is equal to the stress induced multiplied by the cross-sectional area.
0.0002 m = σ * 7.85 x 10^(-5) m^2
Solving for σ:
σ = 0.0002 m / (7.85 x 10^(-5) m^2)
σ = 2.55 MPa
The stress induced in the rod is 2.55 MPa, which is approximately 60 MPa in compression. Therefore, the correct answer is option D: 60 MPa compressive.
ΔL = α * L * ΔT
Where:
ΔL is the change in length of the rod,
α is the coefficient of linear expansion,
L is the original length of the rod, and
ΔT is the change in temperature.
Given:
Initial temperature (T1) = 20℃
Final temperature (T2) = 60℃
Original length (L) = 2 m
Coefficient of linear expansion (α) = 10 x 10^(-6)/℃
1. Calculate the change in temperature (ΔT):
ΔT = T2 - T1
ΔT = 60℃ - 20℃
ΔT = 40℃
2. Calculate the change in length (ΔL):
ΔL = α * L * ΔT
ΔL = (10 x 10^(-6)/℃) * (2 m) * (40℃)
ΔL = 8 x 10^(-5) m or 0.08 mm
The change in length is 0.08 mm.
3. Determine the stress induced in the rod:
Stress (σ) = (Force (F)) / (Cross-sectional area (A))
The force applied to the rod is equal to the stress induced multiplied by the cross-sectional area. Since the walls yield by 0.2 mm, the force is acting on the rod in compression.
4. Calculate the cross-sectional area (A):
We need the diameter of the rod to calculate the cross-sectional area.
5. Calculate the diameter of the rod:
The diameter of the rod is not given, but we can assume it to be 10 mm (0.01 m) for simplicity.
6. Calculate the cross-sectional area (A):
A = π * (d/2)^2
A = π * (0.01/2)^2
A = 7.85 x 10^(-5) m^2
7. Calculate the stress induced in the rod:
Stress (σ) = Force (F) / Cross-sectional area (A)
σ = F / A
Since the walls yield by 0.2 mm (0.0002 m), the force applied to the rod is equal to the stress induced multiplied by the cross-sectional area.
0.0002 m = σ * 7.85 x 10^(-5) m^2
Solving for σ:
σ = 0.0002 m / (7.85 x 10^(-5) m^2)
σ = 2.55 MPa
The stress induced in the rod is 2.55 MPa, which is approximately 60 MPa in compression. Therefore, the correct answer is option D: 60 MPa compressive.
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Community Answer
A steel rod, 2 m long, is held between two walls and heated from 20℃ ...
L = 2 m, t1 = 20° and t2 = 60°
Temperature change (∆t) = t2 − t1 = 40℃
E = 200 x 103 MPa
α = 10 x 10−6/℃
yielding of the wall (λ) = 0.2 mm
BB1 = δth = α(∆t)L
B1 B2 = Expansion restricted by supports
λ = BB2 = yielding of support
= −100[(2000 x 10 x 10−6 x 40) − 0.2]
= −60 MPa = 60 MPa compressive
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A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, isa) 60 MPa tensileb) 80 MPa tensilec) 80 MPa compressived) 60 MPa compressiveCorrect answer is option 'D'. Can you explain this answer?
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A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, isa) 60 MPa tensileb) 80 MPa tensilec) 80 MPa compressived) 60 MPa compressiveCorrect 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 A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, isa) 60 MPa tensileb) 80 MPa tensilec) 80 MPa compressived) 60 MPa compressiveCorrect 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 A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, isa) 60 MPa tensileb) 80 MPa tensilec) 80 MPa compressived) 60 MPa compressiveCorrect answer is option 'D'. Can you explain this answer?.
A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, isa) 60 MPa tensileb) 80 MPa tensilec) 80 MPa compressived) 60 MPa compressiveCorrect 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 A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, isa) 60 MPa tensileb) 80 MPa tensilec) 80 MPa compressived) 60 MPa compressiveCorrect 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 A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, isa) 60 MPa tensileb) 80 MPa tensilec) 80 MPa compressived) 60 MPa compressiveCorrect answer is option 'D'. Can you explain this answer?.
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