Explanation Given information: The solid shaft is subjected to a torque T and a moment M . The diameter of the solid shaft is d . Calculation: The polar moment of inertia ( J ) of a solid shaft with diameter ( d ) is expressed as follows: J = π d 4 32 The moment of inertia ( I ) of a solid shaft with diameter ( d ) is expressed as follows: I = π d 4 64 Express the stress ( σ ) in the shaft using the relation: σ = M c I (1) Here, M is the moment and c is the outer radius of the shaft. Substitute d 2 for c and π d 4 64 for I in Equation (1). σ = M ( d 2 ) π d 4 64 = 32 M π d 3 Express the shear stress ( τ ) in the shaft using the relation: τ = T c J (2) Here, T is the torque applied to the shaft. Substitute d 2 for c and π d 4 32 for J in Equation (2). τ = T ( d 2 ) π d 4 32 = 16 T π d 3 Express the principal stresses ( σ 1 , σ 2 ) using the relation: σ 1 , 2 = σ x + σ y 2 ± ( σ x − σ y 2 ) 2 + τ x y 2 (3) Substitute 32 M π d 3 for σ x , 0 for σ y , and 16 T π d 3 for τ x y in Equation (3)

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

See similar textbooks

Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

Videos

Question

Chapter 10.7, Problem 10.88P

To determine

To show:

The maximum allowable shear stress is σallow=16πd3[M+M2+T2] according to maximum normal stress theory.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.7, Problem 10.87P

Chapter 10.7, Problem 10.89P

Students have asked these similar questions

A converging elbow (see the figure below) turns water through an angle of 135° in a vertical plane. The flow cross section diameter is 400 mm at the elbow inlet, section (1), and 200 mm at the elbow outlet, section (2). The elbow flow passage volume is 0.2 m³ between sections (1) and (2). The water volume flowrate is 0.1 m³/s and the elbow inlet and outlet pressures are 140 kPa and 90 kPa. The elbow mass is 11 kg. Calculate the (a) horizontal (x direction) and (b) vertical (z direction) anchoring forces required to hold the elbow in place. D₂- Section (1)- D₁ = 400 mm 135° 200 mm Section (2) (a) Fx= i 20809.96 N (b) Fz= i -120265 N

2: A rectangular aluminum block is loaded uniformly in three directions. The loadings are as follows:a 50 kN total resulting compressive load in the x-direction, a 200 kPa uniformly distributed tensile load in they-direction, and a 0.03 MN total resulting tensile load in the z-direction. The block has the following dimensions:L = 1 m, b = 20 cm, h = 350 mm. Use E = 70 GPa and ν = 0.25.Determine the strain in the x and y axes respectively. For the strain in the y-direction to be equal to 0, how much uniformly distributed load inthe surface of y-direction should be added? (+ for tensile, - for compressive) Answers: 5 -1.122 x10-5 / 3 decimals 6 5.102 x10-6 / 3 decimals 7 -0.357 MPa / 3 decimals

A spherical balloon with a diameter of 9 m is filled with water vapour at 200˚C and 200 kPa. Determine the mass of water in the balloon. The R value for water is 0.4615 kJ/kg∙K.

Chapter 10 Solutions

Mechanics of Materials

Ch. 10.3 - Prove that the sum of the normal strains in...Ch. 10.3 - The state of strain at the point on the arm has...Ch. 10.3 - Prob. 10.3P Ch. 10.3 - Prob. 10.4P Ch. 10.3 - 10-5. The state of strain at the point on the gear...Ch. 10.3 - Use the strain transformation equations and...Ch. 10.3 - Use the strain transformation equations and...Ch. 10.3 - Prob. 10.8P Ch. 10.3 - Use the strain transformation equations to...Ch. 10.3 - Use the strain- transformation equations to...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The maximum allowable shear stress is σ allow = 16 π d 3 [ M + M 2 + T 2 ] according to maximum normal stress theory.

Concept explainers

Videos

To show: The maximum allowable shear stress is σallow=16πd3[M+M2+T2] according to maximum normal stress theory.

Want to see the full answer?

Chapter 10 Solutions

To show:

The maximum allowable shear stress is σallow=16πd3[M+M2+T2] according to maximum normal stress theory.