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A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6 per oC and E = 2×105 N/mm2)
- a)0
- b)144 N/mm2
- c)10.57 kN/mm2
- d)210 N/mm2
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A bar of uniformly tapering section with 75 mm and 150 mm diameter, is...
Expansion of bar due to temperature only,
ΔL = αtL = 12 x 10-6 x 30 x 5.5 = 1.98 x 10-3 m
Since (6 – 5.998 = 0.002 m) of expansion is allowed, but expansion in bar due to temperature change is 0.00198 m, which is less than 0.002 m. Hence there will not be any stress inside the bar.
ΔL = αtL = 12 x 10-6 x 30 x 5.5 = 1.98 x 10-3 m
Since (6 – 5.998 = 0.002 m) of expansion is allowed, but expansion in bar due to temperature change is 0.00198 m, which is less than 0.002 m. Hence there will not be any stress inside the bar.
Most Upvoted Answer
A bar of uniformly tapering section with 75 mm and 150 mm diameter, is...
To calculate the stress induced in the bar due to a temperature change, we can use the formula:
Stress = Young's modulus * Coefficient of thermal expansion * Temperature change
Given:
Length of the bar (L) = 5.998 m
Diameter at one end (d1) = 75 mm = 0.075 m
Diameter at the other end (d2) = 150 mm = 0.150 m
Distance between the end supports (D) = 6 m
Temperature change (ΔT) = 30°C
Coefficient of thermal expansion (α) = 12 x 10^-6 (per °C)
First, let's calculate the average diameter of the bar:
Average diameter (d_avg) = (d1 + d2) / 2
= (0.075 + 0.150) / 2
= 0.1125 m
Next, let's calculate the change in length of the bar due to the temperature change:
Change in length (ΔL) = α * L * ΔT
= (12 x 10^-6) * 5.998 * 30
= 0.021594 m
Now, let's calculate the original cross-sectional area (A1) and the final cross-sectional area (A2) of the bar:
Original cross-sectional area (A1) = π * (d1/2)^2
= π * (0.075/2)^2
= 0.004418 m^2
Final cross-sectional area (A2) = π * (d2/2)^2
= π * (0.150/2)^2
= 0.017678 m^2
Next, let's calculate the change in cross-sectional area (ΔA):
Change in cross-sectional area (ΔA) = A2 - A1
= 0.017678 - 0.004418
= 0.01326 m^2
Finally, let's calculate the stress induced in the bar:
Stress = (ΔL / L) / ΔA
= (0.021594 / 5.998) / 0.01326
= 0.002884 / 0.01326
= 0.2181 MPa
Therefore, the stress induced in the bar for a temperature change of 30°C is 0.2181 MPa.
Stress = Young's modulus * Coefficient of thermal expansion * Temperature change
Given:
Length of the bar (L) = 5.998 m
Diameter at one end (d1) = 75 mm = 0.075 m
Diameter at the other end (d2) = 150 mm = 0.150 m
Distance between the end supports (D) = 6 m
Temperature change (ΔT) = 30°C
Coefficient of thermal expansion (α) = 12 x 10^-6 (per °C)
First, let's calculate the average diameter of the bar:
Average diameter (d_avg) = (d1 + d2) / 2
= (0.075 + 0.150) / 2
= 0.1125 m
Next, let's calculate the change in length of the bar due to the temperature change:
Change in length (ΔL) = α * L * ΔT
= (12 x 10^-6) * 5.998 * 30
= 0.021594 m
Now, let's calculate the original cross-sectional area (A1) and the final cross-sectional area (A2) of the bar:
Original cross-sectional area (A1) = π * (d1/2)^2
= π * (0.075/2)^2
= 0.004418 m^2
Final cross-sectional area (A2) = π * (d2/2)^2
= π * (0.150/2)^2
= 0.017678 m^2
Next, let's calculate the change in cross-sectional area (ΔA):
Change in cross-sectional area (ΔA) = A2 - A1
= 0.017678 - 0.004418
= 0.01326 m^2
Finally, let's calculate the stress induced in the bar:
Stress = (ΔL / L) / ΔA
= (0.021594 / 5.998) / 0.01326
= 0.002884 / 0.01326
= 0.2181 MPa
Therefore, the stress induced in the bar for a temperature change of 30°C is 0.2181 MPa.
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A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer?
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A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer?.
A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer?.
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A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6peroC and E = 2×105N/mm2)a)0b)144 N/mm2c)10.57 kN/mm2d)210 N/mm2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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