Explanation Given information: The mass of the collar D is m = 35 kg . The modulus of elasticity of the steel rod is E = 200 GPa . The maximum stress in the rod is σ all = 250 MPa . The length of the rod AB is L A B = 2 m . The length of the rod BC is L B C = 1.5 m . The diameter of rod AB is d A B = 40 mm The diameter of rod BC is d B C = 30 mm Calculation: Consider the acceleration due to gravity as g = 9.81 m / s 2 . Calculate the weight of the collar ( m ) as shown below. W = m g Substitute 9.81 m / s 2 for g and 35 kg for m . W = 35 kg × 9.81 m / s 2 = 343.35 kg ⋅ m / s 2 × 1 N 1 kg ⋅ m / s 2 = 343.35 N Calculate the cross sectional area A of the rod as shown below. A = π d 2 4 (1) For rod AB . Substitute 40 mm for d in Equation (1). A A B = π × 40 2 4 = 1 , 256.64 mm 2 × ( 10 − 3 m 1 mm ) 2 = 1.2566 × 10 − 3 m 2 For rod BC . Substitute 30 mm for d in Equation (1). A B C = π × 30 2 4 = 706.86 mm 2 × ( 10 − 3 m 1 mm ) 2 = 0.7069 × 10 − 3 m 2 The rod BC has the minimum area A min = 0.7069 × 10 − 3 m 2 . Calculate the load applied to the rod ( P m ) as shown below. P m = σ all A min Substitute 250 MPa for σ all and 0.7069 × 10 − 3 m 2 for A min . P m = 250 MPa × 10 6 N / m 2 1 MPa × 0.7069 × 10 − 3 m 2 = 176 , 725 N × 1 kN 1 , 000 N = 176.725 kN Calculate the strain energy ( U ) as shown below. U = P 2 L 2 A E (2) For rod AB . Substitute 176 , 725 N for P , 2 m for L , 1.2566 × 10 − 3 m 2 for A , and 200 GPa for E in Equation (2). U A B = ( 176 , 725 N ) 2 ( 2 m ) 2 × 1

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 11.5, Problem 45P

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The distance h for which the maximum stress in the rod is 250 MPa.

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Chapter 11.5, Problem 44P

Chapter 11.5, Problem 46P

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The distance h for which the maximum stress in the rod is 250 MPa .

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The distance h for which the maximum stress in the rod is 250 MPa.

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