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What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?
(Answer up to two decimal places)
Correct answer is '11.91'. Can you explain this answer?
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Most Upvoted Answer
What will be the largest allowable diameter (in mm) of a 3 m long stee...
We have L = 3 m, Φ = 30π/180 = 523.6 x 10-3 rad, τ = 80 x 106 Pa
G = 77 x 109 Pa
From Torsional equation,
Φ = TL/GJ ⇒ T = GJΦ/L
Hence, d = 2c = 2 x 5.953 ≅ 11.91 mm
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Community Answer
What will be the largest allowable diameter (in mm) of a 3 m long stee...
To find the largest allowable diameter of the steel rod, we can use the torsion formula:
θ = (T * L) / (G * J)
Where:
θ is the angle of twist (in radians)
T is the torque applied (in Nm)
L is the length of the rod (in meters)
G is the shear modulus or modulus of rigidity (in Pa)
J is the polar moment of inertia (in m^4)
In this case, we are given that the rod is twisted through 30°, which we can convert to radians:
θ = 30° * (π/180) = 0.5236 radians
We are also given that the length of the rod is 3 meters, the shear modulus is 77 GPa (77 * 10^9 Pa), and we need to find the largest allowable diameter.
To find the polar moment of inertia (J) for a solid round rod, we can use the formula:
J = (π/32) * d^4
Where d is the diameter of the rod.
Rearranging the torsion formula, we can solve for the torque (T):
T = (θ * G * J) / L
Substituting the expression for J, we get:
T = (θ * G * (π/32) * d^4) / L
To find the largest allowable diameter, we need to find the maximum torque that the rod can handle. Assuming the maximum allowable torque is T_max, we can rearrange the equation to solve for d:
d^4 = (32 * T_max * L) / (θ * G * π)
Taking the fourth root of both sides, we get:
d = [(32 * T_max * L) / (θ * G * π)]^(1/4)
Substituting the given values, we have:
d = [(32 * T_max * 3) / (0.5236 * 77 * 10^9 * π)]^(1/4)
Calculating this expression will give us the largest allowable diameter of the steel rod in mm.
θ = (T * L) / (G * J)
Where:
θ is the angle of twist (in radians)
T is the torque applied (in Nm)
L is the length of the rod (in meters)
G is the shear modulus or modulus of rigidity (in Pa)
J is the polar moment of inertia (in m^4)
In this case, we are given that the rod is twisted through 30°, which we can convert to radians:
θ = 30° * (π/180) = 0.5236 radians
We are also given that the length of the rod is 3 meters, the shear modulus is 77 GPa (77 * 10^9 Pa), and we need to find the largest allowable diameter.
To find the polar moment of inertia (J) for a solid round rod, we can use the formula:
J = (π/32) * d^4
Where d is the diameter of the rod.
Rearranging the torsion formula, we can solve for the torque (T):
T = (θ * G * J) / L
Substituting the expression for J, we get:
T = (θ * G * (π/32) * d^4) / L
To find the largest allowable diameter, we need to find the maximum torque that the rod can handle. Assuming the maximum allowable torque is T_max, we can rearrange the equation to solve for d:
d^4 = (32 * T_max * L) / (θ * G * π)
Taking the fourth root of both sides, we get:
d = [(32 * T_max * L) / (θ * G * π)]^(1/4)
Substituting the given values, we have:
d = [(32 * T_max * 3) / (0.5236 * 77 * 10^9 * π)]^(1/4)
Calculating this expression will give us the largest allowable diameter of the steel rod in mm.
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What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?(Answer up to two decimal places)Correct answer is '11.91'. Can you explain this answer?
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What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?(Answer up to two decimal places)Correct answer is '11.91'. 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 What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?(Answer up to two decimal places)Correct answer is '11.91'. 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 What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?(Answer up to two decimal places)Correct answer is '11.91'. Can you explain this answer?.
What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?(Answer up to two decimal places)Correct answer is '11.91'. 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 What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?(Answer up to two decimal places)Correct answer is '11.91'. 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 What will be the largest allowable diameter (in mm) of a 3 m long steel rod (G = 77 GPa) if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa?(Answer up to two decimal places)Correct answer is '11.91'. Can you explain this answer?.
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