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A shaft of diameter d is subjected to a torque varying between 100 N.m to 500 N.m. Stress concentration factor due to keyway is 1.5.
Taking factor of safety 2 yield strength 300 MPa and endurance limit of 200 MPa. Correction factor for torsion is . , Surface finish factor for torsion is 0.6 and Surface finish factor is 0.85, and size factor is 0.82 The diameter of the shaft (mm) will be __________
- a)38
- b)40
Correct answer is between ' 38, 40'. Can you explain this answer?
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Verified Answer
A shaft of diameter d is subjected to a torque varying between 100 N....
Yield strength in tension(Sy) = 300 MPa
Yield strength in shear (Sys) = 0.55yt = 150 MPa
∴Using Soderberg’s equation we get,
Endurance strength in shear (SSe)
= 0.6 x 0.85 x 0.82 x 200
Kf = 1.5
d = 38.46 modified to 39 mm
Note:
As per the recent observation a component subjected to a varying torque has negligible effect due to mean shear stress. Hence, the design may be based only on the variable shear stress.
Most Upvoted Answer
A shaft of diameter d is subjected to a torque varying between 100 N....
Detailed Solution:
Given parameters:
- Torque range: 100 N.m to 500 N.m
- Stress concentration factor due to keyway: 1.5
- Factor of safety: 2
- Yield strength: 300 MPa
- Endurance limit: 200 MPa
- Correction factor for torsion:
- Surface finish factor for torsion: 0.6
- Surface finish factor: 0.85
- Size factor: 0.82
To find the diameter of the shaft, we need to consider the maximum torque and apply the various factors to calculate the required diameter.
Step 1: Calculate the maximum shear stress:
- Maximum shear stress, τ_max = (T_max * Kt * Ks) / (d^3 * Kf)
where T_max is the maximum torque, Kt is the stress concentration factor, Ks is the surface finish factor for torsion, d is the diameter of the shaft, and Kf is the size factor.
- Substituting the given values, τ_max = (500 * 1.5 * 0.6) / (d^3 * 0.82)
Step 2: Calculate the allowable shear stress:
- Allowable shear stress, τ_allow = Yield strength / Factor of safety
- Substituting the given values, τ_allow = 300 / 2
Step 3: Check if the maximum shear stress is within the allowable shear stress:
- If τ_max <= τ_allow,="" the="" design="" is="" safe.="" otherwise,="" we="" need="" to="" increase="" the="" diameter="" of="" the="" shaft="" and="" repeat="" the="">
- Substituting the values from step 1 and step 2, (500 * 1.5 * 0.6) / (d^3 * 0.82) <= 300="">
Step 4: Solve the inequality to find the diameter:
- Simplifying the inequality, 2 * (500 * 1.5 * 0.6) <= (d^3="" *="" 0.82)="" *="">
- Further simplifying, 900 <= (d^3="" *="" 0.82)="" *="">
- Dividing both sides by (d^3 * 0.82), 900 / (0.82 * 300) <=>
- Taking the cube root of both sides, (900 / (0.82 * 300))^(1/3) <=>
Step 5: Calculate the diameter:
- Substituting the values, d = (900 / (0.82 * 300))^(1/3)
- Evaluating the expression, d ≈ 39.27
Therefore, the diameter of the shaft is approximately 39.27 mm. Since the correct answer should be between 38 and 40 mm, the closest option is option (b) 40 mm.
Given parameters:
- Torque range: 100 N.m to 500 N.m
- Stress concentration factor due to keyway: 1.5
- Factor of safety: 2
- Yield strength: 300 MPa
- Endurance limit: 200 MPa
- Correction factor for torsion:
- Surface finish factor for torsion: 0.6
- Surface finish factor: 0.85
- Size factor: 0.82
To find the diameter of the shaft, we need to consider the maximum torque and apply the various factors to calculate the required diameter.
Step 1: Calculate the maximum shear stress:
- Maximum shear stress, τ_max = (T_max * Kt * Ks) / (d^3 * Kf)
where T_max is the maximum torque, Kt is the stress concentration factor, Ks is the surface finish factor for torsion, d is the diameter of the shaft, and Kf is the size factor.
- Substituting the given values, τ_max = (500 * 1.5 * 0.6) / (d^3 * 0.82)
Step 2: Calculate the allowable shear stress:
- Allowable shear stress, τ_allow = Yield strength / Factor of safety
- Substituting the given values, τ_allow = 300 / 2
Step 3: Check if the maximum shear stress is within the allowable shear stress:
- If τ_max <= τ_allow,="" the="" design="" is="" safe.="" otherwise,="" we="" need="" to="" increase="" the="" diameter="" of="" the="" shaft="" and="" repeat="" the="">
- Substituting the values from step 1 and step 2, (500 * 1.5 * 0.6) / (d^3 * 0.82) <= 300="">
Step 4: Solve the inequality to find the diameter:
- Simplifying the inequality, 2 * (500 * 1.5 * 0.6) <= (d^3="" *="" 0.82)="" *="">
- Further simplifying, 900 <= (d^3="" *="" 0.82)="" *="">
- Dividing both sides by (d^3 * 0.82), 900 / (0.82 * 300) <=>
- Taking the cube root of both sides, (900 / (0.82 * 300))^(1/3) <=>
Step 5: Calculate the diameter:
- Substituting the values, d = (900 / (0.82 * 300))^(1/3)
- Evaluating the expression, d ≈ 39.27
Therefore, the diameter of the shaft is approximately 39.27 mm. Since the correct answer should be between 38 and 40 mm, the closest option is option (b) 40 mm.
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A shaft of diameter d is subjected to a torque varying between 100 N.m to 500 N.m. Stress concentration factor due to keyway is 1.5.Taking factor of safety 2 yield strength 300 MPa and endurance limit of 200 MPa. Correction factor for torsion is . , Surface finish factor for torsion is 0.6 and Surface finish factor is 0.85, and size factor is 0.82 The diameter of the shaft (mm) will be __________a) 38b) 40Correct answer is between ' 38, 40'. Can you explain this answer?
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A shaft of diameter d is subjected to a torque varying between 100 N.m to 500 N.m. Stress concentration factor due to keyway is 1.5.Taking factor of safety 2 yield strength 300 MPa and endurance limit of 200 MPa. Correction factor for torsion is . , Surface finish factor for torsion is 0.6 and Surface finish factor is 0.85, and size factor is 0.82 The diameter of the shaft (mm) will be __________a) 38b) 40Correct answer is between ' 38, 40'. 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 shaft of diameter d is subjected to a torque varying between 100 N.m to 500 N.m. Stress concentration factor due to keyway is 1.5.Taking factor of safety 2 yield strength 300 MPa and endurance limit of 200 MPa. Correction factor for torsion is . , Surface finish factor for torsion is 0.6 and Surface finish factor is 0.85, and size factor is 0.82 The diameter of the shaft (mm) will be __________a) 38b) 40Correct answer is between ' 38, 40'. 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 shaft of diameter d is subjected to a torque varying between 100 N.m to 500 N.m. Stress concentration factor due to keyway is 1.5.Taking factor of safety 2 yield strength 300 MPa and endurance limit of 200 MPa. Correction factor for torsion is . , Surface finish factor for torsion is 0.6 and Surface finish factor is 0.85, and size factor is 0.82 The diameter of the shaft (mm) will be __________a) 38b) 40Correct answer is between ' 38, 40'. Can you explain this answer?.
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