To determine The average normal stress developed in the rods. Explanation Given information: The rods AB and CD consist of red brass C83400 copper. The rod EF consists of stainless 304 steel alloy. The diameter of copper rods AB and CD ( d A B and d C D ) is 30 mm . The diameter of steel alloy rod EF ( d E F ) is 40 mm . The length of copper rods AB and CD ( L A B and L C D ) is 300 mm The lengthof steel alloy rod EF ( L E F ) is 450 mm . The forces on AB and CD rods ( P ) is 40 kN . The Young’s modulus of steel ( E st ) is 101 GPa . The Young’s modulus of copper ( E co ) is 193 GPa . The stiffnessof the supports at A , C and F ( k ) is 200 MN/m Calculation: Show the free body diagram of the assembly as in Figure 1. Refer Figure 1. The assembly is symmetry. Therefore, the forces in the rod AB and CD are F A B = F C D = F . Calculate the horizontal forces by applying the equation of equilibrium: Sum of horizontal forces is equal to 0. ∑ F x = 0 F + F + F E F − 2 ( 40 × 10 3 ) = 0 2 F + F E F − 80 × 10 3 = 0 (1) Calculate the area ( A A B ) of AB rod using the formula: A A B = π 4 d A B 2 (2) Substitute 30 mm for d A B in Equation (2). A A B = π 4 ( 30 ) 2 = 706.858 mm 2 Calculate the area ( A E F ) of rod EF using the formula: A E F = π 4 d E F 2 (3) Substitute 40 mm for d E F in Equation (3). A E F = π 4 ( 40 ) 2 = 1256.637 mm 2 Show the free body diagram of the assembly as in Figure 2. Refer to Figure 2. Here, δ F is deformation at F , δ P is deformation at load, and δ E F is deformation at EF Calculate support reactions at the rigid supports AB , CD , and EF using the method of superposition: δ F = δ P − δ E F F E F k = ( P L A B A A B E s t + P k ) − ( F E F L E F A E F E c u + F E F 2 L A B A A B E s t + F E F 2 k ) (4) Substitute 200 × 10 3 N/mm for k , 40 × 10 3 N for P , 300 mm for L A B , 450 mm for L E F , 706.858 mm 2 for A A B , 1256.637 mm 2 for A E F , 101 × 10 3 N/mm 2 for E st , and 193 × 10 3 N/mm 2 for E cu in Equation (4)

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ISBN: 9789810694364

Author: Russell C Hibbeler

Publisher: Pearson Education

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When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 4.5, Problem 4.44P

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The average normal stress developed in the rods.

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Chapter 4.5, Problem 4.43P

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