1. Assuming general shear failure and using the Terzaghi equation with a FOS of 4 find the allowable gross bearing load the following footing can sustain: 3ft continuous footing set 3ft below grade in a 28deg phi soil with a c' of 400psf and a unit wt of 110psf .
1. Assuming general shear failure and using the Terzaghi equation with a FOS of 4 find the allowable gross bearing load the following footing can sustain: 3ft continuous footing set 3ft below grade in a 28deg phi soil with a c' of 400psf and a unit wt of 110psf .
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![### Bearing Capacity Calculation Using Terzaghi's Equation
1. **Problem Statement:**
- **Scenario:** Assume general shear failure.
- **Task:** Using the Terzaghi equation with a Factor of Safety (FOS) of 4, find the allowable gross bearing load the following footing can sustain.
- **Footing Details:** A 3 ft continuous footing set 3 ft below grade.
- **Soil Properties:** 28° phi (friction angle) soil, cohesion (c') of 400 psf, and a unit weight (γ) of 110 psf.
### Calculation Steps:
1. **Terzaghi’s Bearing Capacity Equation:**
\[ q_{ult} = c' N_c + \gamma D_f N_q + 0.5 \gamma B N_\gamma \]
Where:
- \( q_{ult} \) = ultimate bearing capacity
- \( c' \) = cohesion of soil
- \( γ \) = unit weight of soil
- \( D_f \) = depth of footing
- \( B \) = width of continuous footing
- \( N_c, N_q, N_γ \) = bearing capacity factors dependent on the angle of internal friction (φ)
2. **Bearing Capacity Factors for φ = 28° (approx.):**
- \( N_c \approx 22.5 \)
- \( N_q \approx 12.0 \)
- \( N_γ \approx 15.7 \)
3. **Plugging in the Values:**
\[ q_{ult} = (400 \times 22.5) + (110 \times 3 \times 12.0) + (0.5 \times 110 \times 3 \times 15.7) \]
4. **Calculate Each Term:**
\[ 400 \times 22.5 = 9000 \]
\[ 110 \times 3 \times 12.0 = 3960 \]
\[ 0.5 \times 110 \times 3 \times 15.7 = 2590.5 \]
5. **Sum of Terms:**
\[ q_{ult} = 9000 + 3960 + 2590.5 = 15550.5 \text{ psf} \]
6. **Applying Factor of Safety:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24dbfb66-daf6-4b8e-a748-b5c870ef5592%2Fd1306f25-5977-4532-a861-423625b3e5e9%2Fdo19jx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Bearing Capacity Calculation Using Terzaghi's Equation
1. **Problem Statement:**
- **Scenario:** Assume general shear failure.
- **Task:** Using the Terzaghi equation with a Factor of Safety (FOS) of 4, find the allowable gross bearing load the following footing can sustain.
- **Footing Details:** A 3 ft continuous footing set 3 ft below grade.
- **Soil Properties:** 28° phi (friction angle) soil, cohesion (c') of 400 psf, and a unit weight (γ) of 110 psf.
### Calculation Steps:
1. **Terzaghi’s Bearing Capacity Equation:**
\[ q_{ult} = c' N_c + \gamma D_f N_q + 0.5 \gamma B N_\gamma \]
Where:
- \( q_{ult} \) = ultimate bearing capacity
- \( c' \) = cohesion of soil
- \( γ \) = unit weight of soil
- \( D_f \) = depth of footing
- \( B \) = width of continuous footing
- \( N_c, N_q, N_γ \) = bearing capacity factors dependent on the angle of internal friction (φ)
2. **Bearing Capacity Factors for φ = 28° (approx.):**
- \( N_c \approx 22.5 \)
- \( N_q \approx 12.0 \)
- \( N_γ \approx 15.7 \)
3. **Plugging in the Values:**
\[ q_{ult} = (400 \times 22.5) + (110 \times 3 \times 12.0) + (0.5 \times 110 \times 3 \times 15.7) \]
4. **Calculate Each Term:**
\[ 400 \times 22.5 = 9000 \]
\[ 110 \times 3 \times 12.0 = 3960 \]
\[ 0.5 \times 110 \times 3 \times 15.7 = 2590.5 \]
5. **Sum of Terms:**
\[ q_{ult} = 9000 + 3960 + 2590.5 = 15550.5 \text{ psf} \]
6. **Applying Factor of Safety:**
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