VIPSolutions logo ✨ VIPSolutions

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.

Question:

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.

This is AI generated Response.
Asked by: varrisanju5@gmail.com
Created at: 2025-11-12 08:37:52

Final Answer:

Rendered Raw Text
Did you like this answer?
# 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.

📢 Instructions:

  • Ask anything relatd to above question only.
  • For new Question goto :

Study Help, Fast Answers

VIPSolutions focuses on clear, step‑by‑step explanations so you can learn quickly. Whether you need an AI‑generated walkthrough or a short hint to get unstuck, each solution is organized for fast reading and easy review later.

Search similar questions, compare approaches, and bookmark the best answers for revision. Our goal is simple: quick, reliable study help that feels natural—not noisy.