Explanation Given information: The diameter of the solid shaft is 40 mm. Calculation: Take the clockwise torque as positive and the counter clockwise torque as negative. Find the torque at different points on the given shaft: Torque at point A = − 15 N ⋅ m . Torque at point B = − 15 − 25 = − 40 N ⋅ m . Torque at point C = − 40 + 30 = − 10 N ⋅ m . Torque at point D = − 10 − 60 = − 70 N ⋅ m . Torque at point E = − 70 − 70 = 0 . Sketch the internal torque diagram as shown in Figure 1. Refer Figure 1. The segment DE is subjected to a larger internal torque. The value of maximum internal torque is T D E = 70 N ⋅ m . The torsion formula for maximum shear stress in the solid shaft ( τ max ) is shown below: τ max = T D E c J (1) Here, T D E is the maximum internal torque, J is the polar moment of inertia, and c is the outer radius of the solid shaft. The outer radius of the solid shaft is r . The diameter is 40 mm. and hence, the radius value is 20 mm. Substitute r for c and π 2 r 4 for J in Equation (1)

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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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…

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Chapter 5.3, Problem 5.16P

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Find the absolute maximum shear stress in the shaft.

Sketch the shear stress distribution along the radial line of the shaft.

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Chapter 5.3, Problem 5.15P

Chapter 5.3, Problem 5.17P

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Find the absolute maximum shear stress in the shaft. Sketch the shear stress distribution along the radial line of the shaft.