Let's go step by step to solve this problem. Here's a summary of the key information from the image:

• Diameter of the pile $D = 120 \text{ mm}$.
• Load at the top $P = 65 \text{ kN}$.
• Frictional resistance per unit length $w = 3.15 \text{ kN/m}$.
• Length of the pile $L = 6.0 \text{ m}$.
• Elastic modulus $E = 8.0 \text{ GPa}$.

We need to calculate:

1. Force $F_B$ at the base of the pile for equilibrium.
2. The downward displacement at point $A$ relative to point $B$.

Part (a): Force $F_B$ for equilibrium

For equilibrium, the forces acting on the pile must sum to zero. The total external forces acting downward are the applied load $P$ and the distributed frictional resistance force along the length of the pile.

The total frictional resistance force along the length is: $F_{\text{friction}} = w \times L$ Substitute $w = 3.15 \text{ kN/m}$ and $L = 6.0 \text{ m}$: $F_{\text{friction}} = 3.15 \times 6 = 18.9 \text{ kN}$

The equilibrium equation becomes: $P = F_B + F_{\text{friction}}$ Substitute $P = 65 \text{ kN}$ and $F_{\text{friction}} = 18.9 \text{ kN}$: $65 = F_B + 18.9$ Solving for $F_B$: $F_B = 65 - 18.9 = 46.1 \text{ kN}$

Part (b): Downward displacement at $A$ relative to $B$

The downward displacement $\delta_A$ at point $A$ relative to point $B$ can be found using the relationship for axial displacement under a distributed load and elastic modulus. For a pile subjected to both an axial load $P$ and a frictional load $w$, the displacement $\delta$ at any point can be computed by integrating the differential equation for axial deformation.

The formula for the downward displacement $\delta_A$ due to the load and friction along the pile is given by: $\delta_A = \frac{P \cdot L}{A \cdot E} + \frac{w \cdot L^2}{2 \cdot A \cdot E}$ Where:

• $A$ is the cross-sectional area of the pile.

The cross-sectional area $A$ of the pile is: $A = \frac{\pi D^2}{4}$ Substitute $D = 120 \text{ mm} = 0.12 \text{ m}$: $A = \frac{\pi (0.12)^2}{4} = 0.0113 \text{ m}^2$

Now, substitute values into the displacement equation: $\delta_A = \frac{65 \times 6}{0.0113 \times 8 \times 10^9} + \frac{3.15 \times 6^2}{2 \times 0.0113 \times 8 \times 10^9}$ Solving each term:

First term: $\frac{65 \times 6}{0.0113 \times 8 \times 10^9} = \frac{390}{9.04 \times 10^7} = 4.31 \times 10^{-3} \text{ m}$

Second term: $\frac{3.15 \times 6^2}{2 \times 0.0113 \times 8 \times 10^9} = \frac{113.4}{1.808 \times 10^8} = 6.27 \times 10^{-4} \text{ m}$

Thus, the total displacement $\delta_A$ is: $\delta_A = 4.31 \times 10^{-3} + 6.27 \times 10^{-4} = 4.93 \times 10^{-3} \text{ m} = 4.93 \text{ mm}$

Final Answers:

• (a) The force $F_B$ needed for equilibrium is 46.1 kN.
• (b) The downward displacement at $A$ relative to $B$ is 4.93 mm.

Let me know if you'd like further details or if you have any other questions!

Here are some related questions for you to explore:

1. What would happen to $F_B$ if the frictional resistance $w$ increased?
2. How does the elastic modulus $E$ influence the displacement $\delta_A$?
3. What is the significance of the pile's cross-sectional area in determining its displacement?
4. How would a shorter pile (smaller $L$) affect the equilibrium force $F_B$?
5. Can we neglect the frictional force for very small piles?

Tip: Understanding how material properties (like $E$) affect deformation is key in civil and structural engineering when designing piles or columns.