What is the ultimate allowable capacity in axial compression of a 40 ft long, 12" diameter, steel pipe pile driven into stiff clay with c' = 1500 psf and a unit weight of 120 lb/ft³? Provide your answer in kips and assume a factor of 2.5. State which procedure you use and any assumptions made.
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What is the ultimate allowable capacity in axial compression of a 40 ft long, 12" diameter, steel pipe pile driven into stiff clay with c' = 1500 psf and a unit weight of 120 lb/ft³? Provide your answer in kips and assume a factor of 2.5. State which procedure you use and any assumptions made.
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Asked by: varrisanju5@gmail.com
Created at: 2025-11-12 08:37:52
Asked by: varrisanju5@gmail.com
Created at: 2025-11-12 08:37:52
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# Ultimate Allowable Axial Compression Capacity of a Steel Pipe Pile in Stiff Clay
## Problem Restatement
- **Pile**: Steel pipe, 12" diameter, 40 ft long
- **Soil**: Stiff clay, \( c' = 150 \) psf (undrained shear strength)
- **Unit weight of clay (\( \gamma \))**: 120 lb/ft³
- **Factor of safety**: 2.5
- **Required**: Ultimate allowable axial compression capacity in **kips**
- **Assumptions**:
- Cohesion is constant over pile length.
- Pile is fully embedded; tip capacity is considered.
- No groundwater table is considered.
- Use the **α-method** (API, NAVFAC, or similar).
---
## Step 1: Compute Shaft (Skin) Resistance
### **α-Method for Clays**
\[
Q_{s,\text{ult}} = \alpha \cdot c' \cdot P \cdot L
\]
Where:
- \( \alpha \) = adhesion factor (typical for stiff clays: .7)
- \( c' \) = undrained shear strength = 150 psf
- \( P \) = perimeter of pile = \( \pi d \)
- \( L \) = embedded length = 40 ft
#### **Calculate Perimeter**
\[
d = 12\,\text{in} = 1\,\text{ft}
\]
\[
P = \pi d = \pi \times 1\,\text{ft} = 3.14\,\text{ft}
\]
#### **Plug in Values**
\[
Q_{s,\text{ult}} = .7 \times 150\,\text{psf} \times 3.14\,\text{ft} \times 40\,\text{ft}
\]
\[
Q_{s,\text{ult}} = .7 \times 150 \times 3.14 \times 40
\]
\[
= .7 \times 150 = 105
\]
\[
105 \times 3.14 = 3297
\]
\[
3297 \times 40 = 131,880\,\text{lbs}
\]
---
## Step 2: Compute End (Tip/Bearing) Resistance
\[
Q_{b,\text{ult}} = N_c \cdot c' \cdot A
\]
Where:
- \( N_c \) = bearing capacity factor for clay (typically 9)
- \( c' \) = 150 psf
- \( A \) = area of pile tip = \( \pi d^2 / 4 \)
#### **Calculate Area**
\[
A = \frac{\pi}{4} \times (1\,\text{ft})^2 = .785\,\text{ft}^2
\]
#### **Plug in Values**
\[
Q_{b,\text{ult}} = 9 \times 150\,\text{psf} \times .785\,\text{ft}^2
\]
\[
= 13,500 \times .785 = 10,598\,\text{lbs}
\]
---
## Step 3: Compute Total Ultimate Capacity
\[
Q_{\text{ult}} = Q_{s,\text{ult}} + Q_{b,\text{ult}}
\]
\[
Q_{\text{ult}} = 131,880 + 10,598 = 142,478\,\text{lbs}
\]
---
## Step 4: Apply Factor of Safety
\[
Q_{\text{allowable}} = \frac{Q_{\text{ult}}}{\text{FS}}
\]
\[
Q_{\text{allowable}} = \frac{142,478}{2.5} = 56,991\,\text{lbs}
\]
Convert to **kips** (\( 1\,\text{kip} = 1,000\,\text{lbs} \)):
\[
Q_{\text{allowable}} = \frac{56,991}{1,000} = \boxed{57\,\text{kips}}
\]
---
## **Summary Table**
| Capacity Type | Value (lbs) | Value (kips) |
|---------------|-------------|--------------|
| Shaft | 131,880 | 132 |
| Tip | 10,598 | 10.6 |
| **Ultimate** | 142,478 | 142 |
| **Allowable** | 56,991 | 57 |
---
## **Assumptions and Procedure**
- Used **α-method** for cohesive soils.
- Adhesion factor \( \alpha = .7 \) for stiff clay.
- \( N_c = 9 \) for tip resistance.
- Cohesion and unit weight are constant over the pile length.
- Neglected groundwater and negative skin friction.
- Factor of safety applied to total capacity.
---
## **Final Allowable Axial Compression Capacity**
\[
\boxed{57\ \text{kips}}
\]
for a 40 ft long, 12" diameter, steel pipe pile driven into stiff clay (\( c' = 150\,\text{psf} \)), using a factor of safety of 2.5.
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