Question 19 A section of a rotating shaft is under an alternating bending moment of 70Nm and a torque of 35Nm. The yield strength of the material is 560 MPa and the ultimate tensile strength of the material is 700 MPa. The estimated actual endurance strength is 210 MPa. Stress concentration factor is 1.5 for well-rounded fillet. Design Factor = 2. The minimum diameter of the section of the shaft is: (Use mm as the unit for your answer) Given:

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter3: Torsion
Section: Chapter Questions
Problem 3.11.5P: A square tube section has side dimension of 20 in. arid thickness of 0.5 in. If the section is used...
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I'm not sure how to get the final result in mm.

Nm = Newton meters

MPa = MegaPascals

Question 19
A section of a rotating shaft is under an alternating bending moment of 70NM and a torque of
35NM. The yield strength of the material is 560 MPa and the ultimate tensile strength of the
material is 700 MPa. The estimated actual endurance strength is 210 MPa. Stress concentration
factor is 1.5 for well-rounded fillet. Design Factor = 2.
The minimum diameter of the section of the shaft is:
(Use mm as the unit for your answer)
Given:
M, = 70N · m; T = 25N · m; S, = 560MP%; Su = 700MP%; Si = 210MP%; K, = 1.5; N = 2
Solve:
[Use Appendix → Preferred Basic Sizes (D)]
1/3
32 x N
D =
(K, × M,
1/3
32 x 2
1.5 x 70Nm
25NM
210 x 10^6Pa,
560 x 10^6Pa.
Answer:
D =
Transcribed Image Text:Question 19 A section of a rotating shaft is under an alternating bending moment of 70NM and a torque of 35NM. The yield strength of the material is 560 MPa and the ultimate tensile strength of the material is 700 MPa. The estimated actual endurance strength is 210 MPa. Stress concentration factor is 1.5 for well-rounded fillet. Design Factor = 2. The minimum diameter of the section of the shaft is: (Use mm as the unit for your answer) Given: M, = 70N · m; T = 25N · m; S, = 560MP%; Su = 700MP%; Si = 210MP%; K, = 1.5; N = 2 Solve: [Use Appendix → Preferred Basic Sizes (D)] 1/3 32 x N D = (K, × M, 1/3 32 x 2 1.5 x 70Nm 25NM 210 x 10^6Pa, 560 x 10^6Pa. Answer: D =
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