Give step-by-step solution with explanation and final answer:
7
.
1
6
A building column carries a factored ultimate vertical downward load of
3
2
0
k
.
It is to be supported on a
3
ft deep,

Question

Give step-by-step solution with explanation and final answer:
7
.
1
6
 ﻿A building column carries a factored ultimate vertical downward load of 
3
2
0
 ﻿k 
.
 ﻿It is to be supported on a 
3
 ﻿ft deep, square footing. The soil beneath this footing has the following properties: 
γ
=
2
0
.
5
k
N
m
3
,
c
'
=
5
kPa, and 
φ
'
=
3
6
°
.
 ﻿The groundwater table is at a depth of 
5
 ﻿ft below the ground surface. Using LRFD
,
 ﻿with a resistance factor of 
0
.
5
 
,
 ﻿compute the required footing width.
Show all work and break down each part of answer

VIPSolutions · Accepted Answer

# Problem Breakdown and Solution

Given Data:
- Factored ultimate load, $ P_u = \, \text{kN} \- Footing depth, \( D_f = 3\, \text{ft} $
 Soil properties:
  - Unit weight, $ \gamma = 20.5\, \text{kN/m}^3)
  - Cohesion, \( c' = 5\, \text{kPa} $
  - Friction angle, $ \phi' = 36^\circ $
- Groundwater table depth, $ 5\, \text{ft} $ below ground surface
- Resistance factor, $ \phi_{LRFD} = .5 $
- Goal: Find the required footing width $ B $

---

# Step 1: Convert units to SI (meters) for consistency

| Quantity | Value | Conversion | SI value |
|------------|--------|--------------|-----------|
| Depth of footing, $ D_f $ | 3 ft | $ 1\, \text{ft} = .3048\, \text{m} $ | $ 3 \times .3048 = .9144\, \text{m} $ |
| Groundwater depth | 5 ft | $ 5 \times .3048 = 1.524\, \text{m} $ |
| Soil unit weight, $ \gamma $ | 20.5 kN/m³ | Already SI |
| Load, $ P_u $ | 320 kN | Already SI |

---

# Step 2: Calculate the total vertical load per unit width

Since the load is given as a factored ultimate load $ P_u $, the actual vertical load is:

$$
Q_{Load} = P_u = 320\, \text{kN}
$$

---

# Step 3: Determine the effective stress parameters

**Assumption:** The footing is square, and the load is centered.

---

# Step 4: Compute the passive and active earth pressures

### a) Effective vertical stress at foundation level:

$$
\sigma_v = \gamma \times D_f
$$

$$
\sigma_v = 20.5\, \text{kN/m}^3 \times .9144\, \text{m} = 18.75\, \text{kPa}
$$

### b) Adjust for groundwater:

- Since groundwater is at 1.524 m (~5 ft) and footing is at .9144 m, the groundwater is **above** the footing, meaning the soil is saturated at the surface but not under the footing.

- For this calculation, approximate that the effective stress is:

$$
\sigma' = \sigma_v - u
$$

where $ u $ is the pore water pressure at the footing.

- Pore pressure at footing level:

$$
u = \gamma_{water} \times \text{height of water above the footing}
$$

- The water table is at 1.524 m below ground surface, footing at .9144 m:

$$
\text{Water above footing} = 1.524 - .9144 = .6096\, \text{m}
$$

- Water unit weight:

$$
\gamma_{water} \approx 9.81\, \text{kN/m}^3
$$

- Pore water pressure:

$$
u = 9.81\, \text{kN/m}^3 \times .6096\, \text{m} \approx 5.98\, \text{kPa}
$$

- Effective vertical stress:

$$
\sigma' = 18.75 - 5.98 \approx 12.77\, \text{kPa}
$$

---

# Step 5: Compute the ultimate bearing capacity $q_u$

Using Terzaghi's bearing capacity equation for strip footing (approximate for square footing):

$$
q_u = c' N_{c} + \sigma' N_{q} + .5 \gamma B N_{\gamma}
$$

Where:

- $ c' = 5\, \text{kPa} $
- $ \phi' = 36^\circ $
- $ N_c, N_q, N_{\gamma} $ are bearing capacity factors dependent on $ \phi' $
- $ B $ is the footing width (unknown, to be found)

---

# Step 6: Calculate bearing capacity factors $ N_c, N_q, N_{\gamma} $

Using standard values for $ \phi' = 36^\circ $:

$$
N_q = e^{\pi \tan \phi'} \times \tan^2 \left(45^\circ + \frac{\phi'}{2}\right)
$$

$$
N_c = \frac{N_q - 1}{\tan \phi'}
$$

$$
N_{\gamma} = 2 (N_q + 1) \tan \phi'
$$

Calculate step-by-step:

### a) $ N_q $:

$$
\pi \tan 36^\circ \approx 3.1416 \times .7265 = 2.283
$$
$$
e^{2.283} \approx 9.80
$$
$$
45^\circ + 18^\circ = 63^\circ
$$
$$
\tan 63^\circ \approx 1.9626
$$
$$
N_q = 9.80 \times (1.9626)^2 = 9.80 \times 3.850 = 37.7
$$

### b) $ N_c $:

$$
N_c = \frac{N_q - 1}{\tan 36^\circ} = \frac{37.7 - 1}{.7265} = \frac{36.7}{.7265} \approx 50.5
$$

### c) $ N_{\gamma} $:

$$
N_{\gamma} = 2 (N_q + 1) \tan 36^\circ = 2 (37.7 + 1) \times .7265 = 2 \times 38.7 \times .7265 \approx 56.2
$$

---

# Step 7: Express ultimate bearing capacity $ q_u $

$$
q_u = c' N_c + \sigma' N_q + .5 \gamma B N_{\gamma}
$$

$$
q_u = 5 \times 50.5 + 12.77 \times 37.7 + .5 \times 20.5 \times B \times 56.2
$$

Calculate each term:

- $ c' N_c = 5 \times 50.5 = 252.5\, \text{kPa} $
- $ \sigma' N_q = 12.77 \times 37.7 \approx 481.5\, \text{kPa} $
- The third term:

$$
.5 \times 20.5 \times B \times 56.2 \approx 577.3 B
$$

Thus,

$$
q_u = 252.5 + 481.5 + 577.3 B
$$

---

# Step 8: Apply LRFD resistance factor

Design bearing capacity:

$$
q_{allow} = \phi_{LRFD} \times q_u = .5 \times q_u
$$

The actual load per unit area (assuming unit width) must satisfy:

$$
Q_{u} = q_{allow} \times B^2
$$

Given $ P_u = 320\, \text{kN} $, and assuming the load is applied over a square footing of width $ B $:

$$
P_u = q_{allow} \times B^2
$$

Expressed in terms of known quantities:

$$
320 = .5 \times (252.5 + 481.5 + 577.3 B) \times B^2
$$

Simplify:

$$
320 = \frac{1}{2} \times (734 + 577.3 B) \times B^2
$$
$$
640 = (734 + 577.3 B) \times B^2
$$

---

# Step 9: Solve for $ B $

Rewrite as a cubic equation:

$$
640 = 734 B^2 + 577.3 B^3
$$

Bring all to one side:

$$
577.3 B^3 + 734 B^2 - 640 = 
$$

Use numerical methods (trial or quadratic approximation):

### Trial $ B = 1\, \text{m} $:

$$
577.3 \times 1 + 734 \times 1 - 640 = 577.3 + 734 - 640 = 671.3 > 
$$
Too high, try \( B =

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