---
title: "T-shaped foundation, stress at P"
subject: "Soil Mechanics"
chapter: 8
key: "semester_4/soil_mechanics:t-shaped-foundation-stress-p"
source: https://www.bctengineer.com/q/semester_4/soil_mechanics/t-shaped-foundation-stress-p
site: BCT Engineering
---

# T-shaped foundation, stress at P

Soil Mechanics · Chapter 8: Vertical Stresses Below Applied Loads

- Asked in: 2077 Chaitra, 4 marks
- Video: [Vertical stress at the Corner of Rectangular foundation || Most important | Soil Mechanics Numerical](https://www.youtube.com/watch?v=sl7hpOoSeZc) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/t-shaped-foundation-stress-p
- Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-8

## Question

A T-shaped foundation as shown in the figure is loaded with a uniform load of 120 kPa. Determine the vertical stress at the point P at a depth of 5 m. [Take $I_N$ = 0.0629 for m = 0.6 and n = 0.3; $I_N$ = 0.1431 for m = 0.6 and n = 1.0; and $I_N$ = 0.1069 for m = 0.6 and n = 0.6.] [Figure: T-shaped loaded area; top flange 9 m wide and 3 m deep; below it a stem 1.5 m deep and 3 m wide with 3 m of flange on each side (dimensions 3 m, 3 m, 3 m along the bottom); point P at the left corner where the stem meets the flange. Exact position of P is hard to read.]

## Short answer

Divide into three rectangles with corner at P: 3×3 m I=0.1069, 6×3 m I=0.1431, 3×1.5 m I=0.0629. Sum I=0.3129. Stress=120 kPa×0.3129=37.5 kPa.

## Answer

#### Method

Divide the T-shaped loaded area into rectangles that have P as a common corner, and use the corner stress $\Delta\sigma_z = q\,I_N$ for each. Depth $z = 5$ m, $q = 120$ kPa.

```text
 +--------+---------------------+
 |  (1)   |        (2)          |  3 m
 |  3x3   |        6x3          |
 +--------P-----+---------------+
          | (3)  |
          | 3x1.5|              1.5 m
          +------+
```

P lies at the left corner where the stem meets the flange (on the lower edge of the flange).

- Rectangle (1): left flange, 3 m × 3 m: $m = 0.6$, $n = 0.6$, $I_N = 0.1069$
- Rectangle (2): right flange, 6 m × 3 m: $m = 0.6$, $n = 6/5 = 1.2$, $I_N = 0.1431$ (the question prints $n = 1.0$, but 0.1431 is the correct factor for $n = 1.2$)
- Rectangle (3): stem, 3 m × 1.5 m: $m = 0.6$, $n = 0.3$, $I_N = 0.0629$

(The given values match the exact influence factors for these shapes.)

#### Stress at P

$$
\Delta\sigma_z = q\,(I_1 + I_2 + I_3) = 120\,(0.1069 + 0.1431 + 0.0629) = 120\times0.3129
$$

$$
\Delta\sigma_z = 37.5\ \text{kPa}
$$

**Answer: The vertical stress at P at 5 m depth $\approx 37.5$ kPa.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the T-shaped loaded area, the stress at P is found by
   A. treating it as a circle
   B. dividing it into rectangles with P as a common corner and adding q x I_N for each
   C. using a single rectangle of the overall size
   D. the 2:1 method only
   Answer: B. Each rectangle gives q I_N at the common corner P; superposition gives the total. (Key point: Method)

2. For the 3 m x 3 m left flange at z = 5 m, m and n are 0.6 and 0.6, so I_N is
   A. 0.0629
   B. 0.1431
   C. 0.1069
   D. 0.3129
   Answer: C. The given table value for m = n = 0.6 is 0.1069. (Key point: Rectangle (1))

3. The sum of influence factors for the three rectangles is
   A. 0.1069 + 0.1431 + 0.0629 = 0.3129
   B. 0.1069 x 0.1431 x 0.0629
   C. 0.1431 only
   D. 0.2500
   Answer: A. The factors are added: 0.3129. (Key point: Sum of factors)

4. With q = 120 kPa the vertical stress at P is about
   A. 12.6 kPa
   B. 37.5 kPa
   C. 120 kPa
   D. 75.0 kPa
   Answer: B. 120 x 0.3129 = 37.5 kPa. (Key point: Result)

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