https://file.aiquickdraw.com/mathbot/file/1728918350464ub8rtbv4.jpg

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.

The wooden pile shown in the figure has a diameter of 120 mm and is subjected to a load of P = 65 kN. Along the length of the pile and around its perimeter, soil supplies a constant frictional resistance of w = 3.15 kN/m. The length of the pile is L = 6.0 m and its elastic modulus is E = 8.0 GPa. Calculate (a) the force FB needed at the base of the pile for equilibrium and (b) the magnitude of the downward displacement at A relative to B.

Learn how to calculate the equilibrium force at the base of a wooden pile subjected to an external load of 65 kN and a frictional resistance of 3.15 kN/m. Also, find the downward displacement at the top relative to the base using key engineering concepts such as axial deformation and elastic modulus. This solution is suitable for civil engineering students.

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