The wooden pile shown in Figure P5.14 has a diameter of 100 mm and is subjected to a load of P = 75 kN. Along the length of the pile and around its perimeter, soil supplies a constant frictional resistance of w = 3.70 kN/m. The length of the pile is L = 5.0 m and its elastic modulus is E = 8.3 GPa. Calculate
(a) the force $F_{B}$ needed at the base of the pile for equilibrium.
(b) the magnitude of the downward displacement at A relative to B.

Question

The wooden pile shown in Figure P5.14 has a diameter of 100 mm and is subjected to a load of P = 75 kN. Along the length of the pile and around its perimeter, soil supplies a constant frictional resistance of w = 3.70 kN/m. The length of the pile is L = 5.0 m and its elastic modulus is E = 8.3 GPa. Calculate
(a) the force [latex]F_{B}[/latex] needed at the base of the pile for equilibrium.
(b) the magnitude of the downward displacement at A relative to B.

Accepted Answer

(a) Force at base of pile: Consider the entire pile length L and sum forces in the vertical direction.

[latex]\Sigma F_{y}=-F_{B}-w L+P=0[/latex]

The force [latex]F_{B}[/latex] required for equilibrium is thus

[latex]F_{B}=P-w L=75 kN -(3.70 kN / m )(5.0 m )=56.5 kN[/latex]

(b) Magnitude of the downward displacement at A: Consider the free-body diagram shown. The equilibrium equation for the pile is

[latex]\Sigma F_{y}=P-w y+F(y)=0[/latex]

The internal force F(y) in the pile can now be expressed as

[latex]F(y)=w y-P=(3.70 kN / m ) y-75 kN[/latex]

The cross-sectional area of the pile is

[latex]A=\frac{\pi}{4}(0.100 m )^{2}=7.8540 imes 10^{-3} m ^{2}[/latex]

The total deformation of the pile can be calculated with the following integral

[latex]\delta=\int_{0}^{L} \frac{F(y)}{A(y) E(y)} d y=\frac{1}{A E} \int_{0}^{L} F(y) d y[/latex]

Thus, the total deformation of the pile is

[latex]\begin{aligned}\delta &=\frac{1}{A E} \int_{0}^{L}((3.70 kN / m ) y-75 kN ) d y=\frac{1}{A E}\left[\frac{3.70 kN / m }{2} y^{2}-(75 kN ) y ight]_{0}^{5.0} \&=\frac{1}{\left(7.8540 imes 10^{-3} m ^{2} ight)\left(8.3 imes 10^{6} kN / m ^{2} ight)}\left[\frac{3.70 kN / m }{2}(5.0 m )^{2}-(75 kN )(5.0 m ) ight] \&=\frac{-328.750 kN - m }{\left(7.8540 imes 10^{-3} m ^{2} ight)\left(8.3 imes 10^{6} kN / m ^{2} ight)}=-5.04309 imes 10^{-3} m =-5.04 mm\end{aligned}[/latex]

Since F(y) was assumed to be a tension force, a positive result here would indicate an elongation of the pile. The negative sign obtained for δ indicates that the pile contracts in length. Therefore, the upper end of the pile moves downward by the same amount.

The magnitude of the downward displacement at A relative to B is thus

[latex]u_{A}=5.04 mm \downarrow[/latex]

Question 5.14: The wooden pile shown in Figure P5.14 has a diameter of 100 ...

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