---
title: "Corner stresses of 450 x 750 column (150 kN)"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:column-stress-450x750-150kn"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/column-stress-450x750-150kn
site: BCT Engineering
---

# Corner stresses of 450 x 750 column (150 kN)

Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads

- Asked in: 2079 Baisakh, 6 marks
- Video: [Stresses in concrete and steel bar for composite sections/simple stress and strain](https://www.youtube.com/watch?v=_jV2JYrQylg) — Prof.Dr.Pravin Patil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/column-stress-450x750-150kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

A column section 450 mm $\times$ 750 mm carries a load of 150 kN acting with eccentricity of 50 mm along the 750 mm side and 45 mm along the 450 mm side from the centroid. Calculate the stress intensities at all four corners.

## Short answer

Compute stresses by superimposing direct stress (P/A = 0.444 N/mm²) and bending stresses from eccentricities in both axes. Corner stresses: A = 0.889 N/mm², B = 0.533 N/mm², C = 0.356 N/mm², D = 0.000 N/mm² (compression).

## Answer

#### Data

Section 450 mm $\times$ 750 mm, $P=150$ kN. The eccentricity of the load from the centroid is 50 mm along the 750 mm side and 45 mm along the 450 mm side.

#### Eccentricities

$$
\begin{aligned} e_{1} &= 45\ \text{mm (across the 450 mm side)}\\ e_{2} &= 50\ \text{mm (along the 750 mm side)} \end{aligned}
$$

#### Section properties

$$
\begin{aligned} A &= 450\times750 = 337500\ \text{mm}^2\\ Z_1 &= \frac{750\times450^2}{6} = 25312500\ \text{mm}^3\\ Z_2 &= \frac{450\times750^2}{6} = 42187500\ \text{mm}^3 \end{aligned}
$$

#### Stresses

$$
\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{150\times10^3}{337500} = 0.4444\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{150\times10^3\times45}{25312500} = 0.2667\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{150\times10^3\times50}{42187500} = 0.1778\ \text{N/mm}^2 \end{aligned}
$$

Corner stresses $\sigma=\sigma_0\pm\sigma_{b1}\pm\sigma_{b2}$ (positive = compression, N/mm²):

| Corner | Position | Working | Stress |
|---|---|---|---|
| A | toward the load in both directions | $0.4444 + 0.2667 + 0.1778$ | **0.8889** |
| B | toward the load across the 450 side, away from it along the 750 side | $0.4444 + 0.2667 - 0.1778$ | **0.5333** |
| C | away from the load across the 450 side, toward it along the 750 side | $0.4444 - 0.2667 + 0.1778$ | **0.3556** |
| D | away from the load in both directions | $0.4444 - 0.2667 - 0.1778$ | **0.0000** |

Check: $\sigma_0$ = average of the four corners = 0.4444. Corner D has zero stress because the load lies exactly on the edge of the kernel ($6e_1/450+6e_2/750 = 0.6+0.4 = 1.000$), so no tension develops.

**Answer: A = 0.889, B = 0.533, C = 0.356, D = 0.000 N/mm²** (compression positive); maximum 0.889 N/mm² at A.

## Check yourself

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

1. What is sigma_0?
   A. 150/450 = 0.333
   B. P/A = 150000/337500 = 0.444 N/mm^2
   C. 0.889 N/mm^2
   D. 0.178 N/mm^2
   Answer: B. Direct stress is P over area. (Key point: Direct stress)

2. Which stress results from the 45 mm eccentricity across the 450 mm side?
   A. P e1/Z1 = 0.2667 N/mm^2 with Z1 = 750 x 450^2/6
   B. 0.1778 N/mm^2
   C. 0.4444 N/mm^2
   D. 0.0889 N/mm^2
   Answer: A. 0.1778 is from e2 = 50 mm; e1 uses Z1. (Key point: Bending stress)

3. What is corner D and what does that say about the load position?
   A. 0.889 N/mm^2, centroid
   B. -0.2 N/mm^2, outside kernel
   C. 0.000 N/mm^2, load is on the kernel edge (sum of 6e/b = 1)
   D. 0.533 N/mm^2, inside kernel
   Answer: C. 0.4444 - 0.2667 - 0.1778 = 0 and 6e1/450 + 6e2/750 = 1.000. (Key point: Corner D)

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