Explanation Given information: The joint is subjected to an axial force of P = 9 kN . The diameter of the bolt is 6 mm. Calculation: The cross sectional dimension of the shaded shear plane is 100 mm × 100 mm . Find the cross sectional area of the shaded shear plane as shown below. A P = 100 × 100 = 10 , 000 mm 2 Find the cross sectional area of the bolt as shown below. A b = π d b 2 4 Here, d b is the diameter of the bolt. Substitute 6 mm for d b . A b = π × 6 2 4 = 28.274 mm 2 Sketch the Free Body Diagram of the shear plane of the bolt as shown in Figure 1. Refer to Figure 1. Find the internal loadings as shown below. Apply the Equations of Equilibrium as shown below. Summation of forces along y direction is Equal to zero. ∑ F y = 0 4 V b − 9 = 0 4 V b = 9 V b = 2.25 kN Find the average shear stress developed in each shear plane of the bolt ( τ avg ) a − a as shown below. ( τ avg ) a − a = V b A b Here, V b is the shear force and A b is the cross sectional area of the bolt. Substitute 2.25 kN for V b and 28.274 mm 2 for A b . ( τ avg ) a − a = 2.25 kN × 1 , 000 N 1 kN 28.274 = 79.6 N / mm 2 Convert the unit from N / mm 2 to MPa

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 1.5, Problem 1.52P

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The average shear stress developed in each shear plane of the bolt and the member.

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Chapter 1.5, Problem 1.51P

Chapter 1.5, Problem 1.53P

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