To determine The average normalstress in the concrete. The average normal stress in the steel. Explanation Given information: The dimension of square Pipe ( a ) is 300 mm The length of Pipe ( L ) is 500 mm . The young’s modulus of concrete ( E c ) is 24 GPa . The young’s modulus of steel ( E st ) is 200 GPa . An axial loadof the Pipe ( P ) is 80 kN . The outer diameter of the pipe ( D ) is 80 mm . The inner diameter of the pipe ( d ) is 70 mm . Calculation: Show the free body diagram of the Pipe as in Figure 1. Refer Figure 1. Here, P con is axial force of the concrete and P st is axial force of the steel. Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 P con + P st − 80 = 0 (1) Calculate the area concrete ( A con ) using formula: A con = π 4 d 2 (2) Substitute 70 mm for d in Equation (2). A con = π 4 × 70 2 = 3 , 848.45 mm 2 Calculate the area steel ( A st ) using formula: A st = π 4 ( D 2 − d 2 ) (3) Substitute 80 mm for D and 70 mm for d in Equation (3). A st = π 4 × ( 80 2 − 70 2 ) = 1 , 178.097 mm 2 The steel bows and concrete are bonded, δ st = δ con P st L A st E st = P con L A con E c P st A st E st = P con A con E c (4) Substitute 1 , 178.097 mm 2 for A st , 3 , 848.45 mm 2 for A con , 24 GPa for E c , and 200 GPa for E st in Equation (4). P st 1 , 178.097 × 200 = P con 3 , 848

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Chapter 4.5, Problem 4.33P

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The average normalstress in the concrete.

The average normal stress in the steel.

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Chapter 4.5, Problem 4.32P

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