Explanation Given information: The tie rods are made up of A-36 steel rod. The length of each rods ( L ) is 0.75 m . The cross-sectional area of each rods ( A ) is 125 mm 2 . The young’s modulus of steel ( E ) is 200 GPa . The lead of the screw is 1.5 mm . Assumption: Apply the sign convention for calculating the equations of equilibrium as below: For the horizontal forces equilibrium condition, take the force acting towards right side as positive ( → + ) and the force acting towards left side as negative ( ← − ) . For the vertical forces equilibrium condition, take the upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive. Calculation: Show the free body diagram of the turnbuckle as in Figure 1. Refer to Figure 1: Here, T A B is force of the rod AB , T C D is force of the rod CD , and T E F is force of the rod EF . Calculate the moment about point E by applying the equation of equilibrium: Sum of moments about point E is equal to 0. ∑ M E = 0 − ( T A B × 0.5 ) + ( T C D × 0.5 ) = 0 0.5 T A B = 0.5 T C D T A B = T C D Therefore, T = T A B = T C D (1) Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 T E F − 2 T = 0 T E F = 2 T (2) Show the free body diagram of the assembly as in Figure 2. Refer Figure 2. Here, δ A / B is deformation at AB and δ E / F is deformation at EF . The rod EF shortens 1.5 mm causing AB (and DC ) to elongate. Therefore, 0.0015 = δ A / B + δ E / F 0

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Chapter 4.5, Problem 4.40P

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Find the forces in the rods AB, CD, and EF (TAB, TCD, and TEF).

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Chapter 4.5, Problem 4.39P

Chapter 4.5, Problem 4.41P

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Find the forces in the rods AB , CD , and EF ( T A B , T C D , and T E F ) .

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