Explanation Given information: The diameter of the bolt ( d b ) is 20 mm . The inner diameter of the tube ( D t ) is 50 mm . The outer diameter of the tube ( d t ) is 60 mm . The bolt and tube are made of A-36 steel. The Young’s modulus of steel ( E ) is 200 GPa . The length of the bolt ( L b ) is 160 mm . The length of the tube ( L t ) is 150 mm . The force applied to the bolt ( P ) is 40 kN . Calculation: Show the free body diagram of the left portion of the cut assembly as in Figure 1. Refer Figure 1. Calculate the horizontal forces by applying the equation of equilibrium: Sum of horizontal forces is equal to 0. ∑ F x = 0 F b + F t − 40 × 10 3 = 0 F b + F t = 40 × 10 3 (1) Calculate the area of the bolt ( A b ) using the formula: A b = π 4 d b 2 (2) Substitute 20 mm for d b in Equation (2). A b = π 4 ( 20 ) 2 = 314.159 mm 2 Calculate the area of the tube ( A t ) using the formula: A t = π 4 ( D s 2 − d s 2 ) (3) Substitute 60 mm for D t and 50 mm for d t in Equation (3). A t = π 4 ( 60 2 − 50 2 ) = 863.938 mm 2 The deformation in the bolt is equal to the deformation in the tube. It is expressed as follows: δ b = δ t P b L b A b E = P t L t A t E P b L b A b = P t L t A t (4) Here, P b is the force in the bolt and P t is the force in the tube. Substitute 160 mm for L b , 150 mm for L t , 314.159 mm 2 for A b , and 863.938 mm 2 for A t in Equation (4)

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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The term stress indicates the measure of internal forces that develop due to an external force's application on an object. When a specific external force is applied to an object, internal stress generates in the material. The value of stress can obtain b…

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Chapter 4.5, Problem 4.45P

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the normal stress in the tube and bolt (σt and σb) .

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Chapter 4.5, Problem 4.44P

Chapter 4.5, Problem 4.46P

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the normal stress in the tube and bolt ( σ t and σ b ) .

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