Explanation Given information: The rods AB , CD , and EF are made of the 304 stainless steel. The modulus of elasticity of the steel ( E ) is 193 kN/mm 2 . The diameter of the rods AB and CD ( d ) is 5 mm . The diameter of the rod EF ( d E F ) is 12 mm . The total weight the pipe and the fluid ( P ) is 4 kN . The length of the rod ( L ) is 0.75 m . Calculation: Draw the free body diagram of the rodas shown in Figure 1: Refer Figure 1: Here, F A B is force in the rod AB and F C D is force in the rod CD . Calculate the moment about point A by applying the equation of equilibrium: Sum of moments about point A is equal to 0. ∑ M A = 0 ( F C D × 0.5 ) − ( 4 × 0.25 ) = 0 0.5 F C D − 1 = 0 0.5 F C D = 1 F C D = 1 0.5 F C D = 2 kN Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. Σ F y = 0 F A B + F C D − 4 = 0 F A B + F C D = 4 (1) Calculate the force of the rod AB ( F A B ) : Substitute 2 kN for F C D in Equation (1). F A B + 2 = 4 F A B = 4 − 2 F A B = 2 kN Draw the free body diagram of the rod as shown in Figure 2: Refer Figure 2: Here, F E F is force in the rod EF . Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. Σ F y = 0 − F A B + F E F − F C D = 0 (2) Calculate the force of the rod EF ( F E F ) : Substitute 2 kN for F A B and 2 kN for F C D in Equation (2). − 2 + F E F − 2 = 0 F E F − 4 = 0 F E F = 4 kN Calculate the cross-sectional area of the rods AB and CD ( A ) : A = π 4 d 2 (3) Substitute 5 mm for d in Equation (3). A = π 4 × 5 2 = 6.25 π mm 2 Calculate the cross-sectional area of the rod EF ( A E F ) : A E F = π 4 d E F 2 (4) Substitute 12 mm for d E F in Equation (4). A E F = π 4 × 12 2 = 36 π mm 2 Calculate the displacement of rod AB ( δ A B ) : δ A B = F A B L A E (5) Substitute 2 kN for F A B , 0

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ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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Chapter 4.2, Problem 4.21P

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