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A solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What should
be the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?
Take σy = 210 N/mm2 and μ = 0.25
be the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?
Take σy = 210 N/mm2 and μ = 0.25
______________
Correct answer is between '32,33'. Can you explain this answer?
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Verified Answer
A solid shaft is subjected to a bending moment of 300 Nm and a twistin...
M = 300 N.m
T = 225 N.m
T = 225 N.m
With a factor of safety, the condition to be satisfied is
Most Upvoted Answer
A solid shaft is subjected to a bending moment of 300 Nm and a twistin...
The yield stress of the material as 300 MPa.
Solution:
From the given data, we can calculate the maximum principal stress and maximum shear stress as follows:
Maximum bending stress:
σ_b = M_b * c / I
where M_b = 300 Nm (bending moment)
c = distance from the neutral axis to the outermost fiber (assumed as half of the diameter, d/2)
I = π/4 * d^4 (moment of inertia for a solid circular shaft)
Substituting the values, we get:
σ_b = 32 * M_b / π * d^3
Maximum twisting stress:
τ_t = M_t * d / J
where M_t = 225 Nm (twisting moment)
J = π/32 * d^4 (polar moment of inertia for a solid circular shaft)
Substituting the values, we get:
τ_t = 16 * M_t / π * d^3
Now, according to the maximum principal strain theory, the maximum principal strain should be less than or equal to the yield strain divided by the factor of safety:
ε_max = σ_max / E + τ_max / G <= σ_yield="" (e="" *="">
where E = 200 GPa (Young's modulus of the material)
G = 80 GPa (shear modulus of the material)
σ_yield = 300 MPa (yield stress of the material)
FS = 2 (factor of safety)
Substituting the expressions for σ_max and τ_max, we get:
32 * M_b / π * d^3 / E + 16 * M_t / π * d^3 / G <= σ_yield="" (e="" *="">
Simplifying and solving for d, we get:
d >= (16 * M_t * G + 32 * M_b * E)^(1/3) / (π * (E * G * σ_yield / FS)^(1/3))
Substituting the given values, we get:
d >= (16 * 225 * 80e9 + 32 * 300 * 200e9)^(1/3) / (π * (200e9 * 80e9 * 300e6 / 2)^(1/3))
d >= 29.4 mm
Therefore, the minimum diameter required for the shaft with a factor of safety of 2 according to the maximum principal strain theory is 29.4 mm.
Solution:
From the given data, we can calculate the maximum principal stress and maximum shear stress as follows:
Maximum bending stress:
σ_b = M_b * c / I
where M_b = 300 Nm (bending moment)
c = distance from the neutral axis to the outermost fiber (assumed as half of the diameter, d/2)
I = π/4 * d^4 (moment of inertia for a solid circular shaft)
Substituting the values, we get:
σ_b = 32 * M_b / π * d^3
Maximum twisting stress:
τ_t = M_t * d / J
where M_t = 225 Nm (twisting moment)
J = π/32 * d^4 (polar moment of inertia for a solid circular shaft)
Substituting the values, we get:
τ_t = 16 * M_t / π * d^3
Now, according to the maximum principal strain theory, the maximum principal strain should be less than or equal to the yield strain divided by the factor of safety:
ε_max = σ_max / E + τ_max / G <= σ_yield="" (e="" *="">
where E = 200 GPa (Young's modulus of the material)
G = 80 GPa (shear modulus of the material)
σ_yield = 300 MPa (yield stress of the material)
FS = 2 (factor of safety)
Substituting the expressions for σ_max and τ_max, we get:
32 * M_b / π * d^3 / E + 16 * M_t / π * d^3 / G <= σ_yield="" (e="" *="">
Simplifying and solving for d, we get:
d >= (16 * M_t * G + 32 * M_b * E)^(1/3) / (π * (E * G * σ_yield / FS)^(1/3))
Substituting the given values, we get:
d >= (16 * 225 * 80e9 + 32 * 300 * 200e9)^(1/3) / (π * (200e9 * 80e9 * 300e6 / 2)^(1/3))
d >= 29.4 mm
Therefore, the minimum diameter required for the shaft with a factor of safety of 2 according to the maximum principal strain theory is 29.4 mm.
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A solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What shouldbe the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?Take σy = 210 N/mm2 and μ = 0.25______________Correct answer is between '32,33'. Can you explain this answer?
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A solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What shouldbe the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?Take σy = 210 N/mm2 and μ = 0.25______________Correct answer is between '32,33'. 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 solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What shouldbe the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?Take σy = 210 N/mm2 and μ = 0.25______________Correct answer is between '32,33'. 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 solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What shouldbe the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?Take σy = 210 N/mm2 and μ = 0.25______________Correct answer is between '32,33'. Can you explain this answer?.
A solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What shouldbe the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?Take σy = 210 N/mm2 and μ = 0.25______________Correct answer is between '32,33'. 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 solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What shouldbe the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?Take σy = 210 N/mm2 and μ = 0.25______________Correct answer is between '32,33'. 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 solid shaft is subjected to a bending moment of 300 Nm and a twisting moment of 225 Nm. What shouldbe the diameter of the shaft with a factor of safety of 2 according to the maximum principal strain theory?Take σy = 210 N/mm2 and μ = 0.25______________Correct answer is between '32,33'. Can you explain this answer?.
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