---
title: "Corner stresses of 400 x 800 column (200 kN)"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:column-stress-400x800-200kn"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/column-stress-400x800-200kn
site: BCT Engineering
---

# Corner stresses of 400 x 800 column (200 kN)

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

- Asked in: 2070 Chaitra, 8 marks
- Video: [EX 04 | BENDING STRESSES | RECTANGULAR COLUMN | ECCENTRIC LOADING ANYWHERE IN COLUMN](https://www.youtube.com/watch?v=Fe-ftpSHkV8) — Structural Funda by:- KheNiSiR
- 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-400x800-200kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

A column section 400 mm $\times$ 800 mm carries a load of 200 kN acting at 160 mm from the 800 mm face and 350 mm from the 400 mm face. Determine the stress intensities at all four corners.

## Short answer

Eccentricities40 mm and50 mm; A =320000 mm², Z₁ =21,33×10⁶ mm³, Z₂ =42,67×10⁶ mm³. Direct stress0.625 N/mm², bending stresses0.375 and0.234 N/mm². Corner stresses: A =1.234, B =0.766, C =0.484, D =0.016 N/mm² (compression positive).

## Answer

Distances are measured from the faces. The 160 mm distance is across the 400 mm side (from the 800 mm face) and the 350 mm distance is along the 800 mm side (from the 400 mm face). Hence $e_1 = 200-160 = 40$ mm and $e_2 = 400-350 = 50$ mm.

#### Data

Section 400 mm $\times$ 800 mm, $P=200$ kN. The load acts 160 mm from the 800 mm face and 350 mm from the 400 mm face.

#### Eccentricities

$$
\begin{aligned} e_{1} &= 40\ \text{mm (across the 400 mm side)}\\ e_{2} &= 50\ \text{mm (along the 800 mm side)} \end{aligned}
$$

#### Section properties

$$
\begin{aligned} A &= 400\times800 = 320000\ \text{mm}^2\\ Z_1 &= \frac{800\times400^2}{6} = 21333333\ \text{mm}^3\\ Z_2 &= \frac{400\times800^2}{6} = 42666667\ \text{mm}^3 \end{aligned}
$$

#### Stresses

$$
\begin{aligned} \sigma_0 &= \frac{P}{A} = \frac{200\times10^3}{320000} = 0.6250\ \text{N/mm}^2\\ \sigma_{b1} &= \frac{Pe_1}{Z_1} = \frac{200\times10^3\times40}{21333333} = 0.3750\ \text{N/mm}^2\\ \sigma_{b2} &= \frac{Pe_2}{Z_2} = \frac{200\times10^3\times50}{42666667} = 0.2344\ \text{N/mm}^2 \end{aligned}
$$

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

| Corner | Position | Working | Stress |
|---|---|---|---|
| A | near the 800 face and near the 400 face | $0.6250 + 0.3750 + 0.2344$ | **1.2344** |
| B | near the 800 face, far from the 400 face | $0.6250 + 0.3750 - 0.2344$ | **0.7656** |
| C | far from the 800 face, near the 400 face | $0.6250 - 0.3750 + 0.2344$ | **0.4844** |
| D | far from both faces | $0.6250 - 0.3750 - 0.2344$ | **0.0156** |

Check: $\sigma_0$ = average of the four corners = 0.6250. All four corners are in compression, because the load lies inside the kernel ($6e_1/400+6e_2/800 = 0.975 < 1$).

**Answer: A = 1.234, B = 0.766, C = 0.484, D = 0.016 N/mm²** (compression positive); maximum 1.234 N/mm² at A.

## Check yourself

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

1. The load is 160 mm from the 800 mm face. What is e1 across the 400 mm side?
   A. 160 mm
   B. 200 - 160 = 40 mm
   C. 400 - 160 = 240 mm
   D. 800 - 160 = 640 mm
   Answer: B. e is the distance from the centroid, half of 400 mm is 200 mm, so 40 mm. (Key point: Eccentricities)

2. How are the corner stresses obtained?
   A. sigma = P/A only
   B. sigma = P/A x (1 +/- e1 +/- e2)
   C. sigma = P/A +/- P e1/Z1 +/- P e2/Z2
   D. sigma = P/Z1 +/- P/Z2
   Answer: C. Direct stress combined with biaxial bending stresses. (Key point: Stress formula)

3. What is the stress at corner A?
   A. 0.625 + 0.375 + 0.2344 = 1.234 N/mm^2
   B. 0.625 + 0.375 - 0.2344 = 0.766
   C. 0.625 N/mm^2
   D. 0.016 N/mm^2
   Answer: A. Corner A is nearest the load in both directions. (Key point: Maximum stress)

4. Why are all corners in compression?
   A. Because P/A is larger than 1 N/mm^2
   B. Because e1 and e2 are equal
   C. Because corner D is the average stress
   D. The load lies inside the kernel: 6e1/400 + 6e2/800 = 0.975 < 1
   Answer: D. Within the kernel the minimum stress stays positive (D = 0.016 N/mm^2). (Key point: Tension check)

## Close variants set in other exams

- [Corner stresses of 400 x 800 column (250 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-stress-400x800-250kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
- [Corner stresses of 500 x 900 column (300 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-stress-500x900-300kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×

## Related in Concrete Technology and Masonry Structure

- [Column 500 x 900: stress intensities, critical corner](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-stress-500x900-critical) — Concrete Technology and Masonry Structure, chapter 8
- [Corner stresses of 450 x 750 column (150 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-stress-450x750-150kn) — Concrete Technology and Masonry Structure, chapter 8

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