Explanation Given information: The modulus of elasticity of the concrete ( E ) is 29 kN/mm 2 . The maximum intensity of the pile ( p 0 ) is 180 kN/m . The axial force of the pile ( P ) is 1500 kN . The diameter of the pile ( d ) is 300 mm . The length of the pile ( L ) is 12 m . Calculation: Calculate the cross-sectional area of the pile A ( y ) : A ( y ) = π 2 d 2 (1) Substitute 300 mm for d in Equation (1). A ( y ) = π 4 × 300 2 = 22.5 × 10 3 π mm 2 Show the free body diagram of the entire pile as in Figure 1. Refer Figure 1. Here, y is distance from the bottom to neutral axis and p ( y ) is intensity of the pile at y distance. Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 F + ( 1 2 × 180 × 12 ) − 1500 = 0 F − 420 = 0 F = 420 kN Hence, the resisting bearing force F for equilibrium is 420 kN _ . Calculate the intensity of the pile at y distance p ( y ) : p ( y ) = p 0 L y (2) Substitute 180 kN/m for p 0 and 12 m for L in Equation (2). p ( y ) = 180 12 y = 15 y kN/m Show the free body diagram of the section of the pile as in Figure 2. Refer Figure 2. Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 ( 1 2 × 15 y ) y + 420 − P ( y ) = 0 7

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Chapter 4.2, Problem 4.30P

To determine

the maximum intensity p0 kN/m for equilibrium.

To find: the corresponding elastic shortening of the pile (δ) .

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Chapter 4.2, Problem 4.29P

Chapter 4.5, Problem 4.31P

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the maximum intensity p 0 kN/m for equilibrium. To find: the corresponding elastic shortening of the pile ( δ ) .

the maximum intensity p0 kN/m for equilibrium. To find: the corresponding elastic shortening of the pile (δ) .

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Chapter 4 Solutions

the maximum intensity p0 kN/m for equilibrium.

To find: the corresponding elastic shortening of the pile (δ) .