---
title: "Unit and mortar for 200 x 400 column"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:column-200x400-120kn"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/column-200x400-120kn
site: BCT Engineering
---

# Unit and mortar for 200 x 400 column

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

- Asked in: 2078 Bhadra, 8 marks
- 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-200x400-120kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

Find the minimum strength of masonry unit and mortar type for a column 200 mm $\times$ 400 mm size to support 120 kN load with an eccentricity of 50 mm. The clear height of the column is 3 m and fully restrained at both ends.

## Short answer

Effective height =0.75×3 =2.25 m, SR =11.25≤12, ka =0.82, ks =0.823 (eccentricity increase1.25). Required fb =1.557/(0.823×0.82×1.25) =1.846 N/mm². Minimum unit strength25 N/mm² with H1 mortar (1∶3) or30 N/mm² with H2 (1∶4).

## Answer

#### Data and assumptions

- Column 200 mm $\times$ 400 mm, clear height 3 m, fully restrained at both ends, so $h_{ef}=0.75H=2.25$ m (IS 1905 Table 4 Sl. 1). Load 120 kN with an eccentricity of 50 mm, taken along the 400 mm side ($e/D=0.125$).
- Units with height/width $\le0.75$ are assumed ($k_p=1$). Self weight is added, 19 kN/m$^3$.

#### Step 1: Slenderness, factors and stress

$$
\begin{aligned} SR &= \frac{2.25}{0.20} = 11.25\ (<12)\\ A &= 0.2\times0.4 = 0.08\ \text{m}^2\ \Rightarrow\ k_a = 0.7+1.5\times0.08=0.82\\ \frac{e}{D} &= \frac{50}{400}=0.125\ (1/24<e/D<1/6)\ \Rightarrow\ k_s = 0.823,\ \text{increase }1.25\\ \text{Self weight} &= 0.08\times3\times19 = 4.56\ \text{kN}\\ P &= 120+4.56 = 124.56\ \text{kN}\\ \sigma &= \frac{124.56\times10^3}{80000} = 1.557\ \text{N/mm}^2 \end{aligned}
$$

($k_s$ from Table 9: at $e/D=0.125$ it is 0.86 for $SR=10$ and 0.80 for $SR=12$; interpolating for $SR=11.25$ gives 0.823.)

#### Step 2: Required basic compressive stress

$$
f_b \ge \frac{\sigma}{k_s k_a k_p \times 1.25} = \frac{1.557}{0.823\times0.82\times1.0\times1.25} = 1.846\ \text{N/mm}^2
$$

#### Step 3: Select unit and mortar (IS 1905 Table 8)

| Mortar | Min. unit strength (N/mm²) | $f_b$ |
|---|---|---|
| M2 (cement:sand 1:6) | 40 | 1.90 |
| M1 (cement:sand 1:5) | 35 | 1.90 |
| H2 (cement:sand 1:4) | 30 | 1.85 |
| H1 (cement:sand 1:3) | 25 | 1.91 |

**Answer: required $f_b = 1.846$ N/mm$^2$; the column needs masonry units of at least 25 N/mm$^2$ crushing strength with H1 mortar (1:3), or 30 N/mm$^2$ units with H2 mortar (1:4).** Such strong units are uncommon in brickwork, so in practice enlarge the section (e.g. 300 mm $\times$ 400 mm, or 400 mm square) or use an RCC column.

## Check yourself

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

1. What is h_ef and SR for the 200 x 400 mm column, 3 m clear, fully restrained?
   A. h_ef = 3 m, SR = 15
   B. h_ef = 0.75 x 3 = 2.25 m, SR = 11.25
   C. h_ef = 2.25 m, SR = 5.6
   D. h_ef = 6 m, SR = 30
   Answer: B. Full restraint gives 0.75H and the least dimension 200 mm gives SR 11.25 < 12. (Key point: Effective height)

2. For e/D = 0.125 what is applied to the stress?
   A. No increase and k_s = 1
   B. The section is cracked and the tension area neglected
   C. k_s = 0.819 with a 25 percent increase (1/24 < e/D <= 1/6)
   D. k_s = 0.819 with a 25 percent reduction
   Answer: C. IS 1905 allows a 25 percent increase in permissible stress for e/D between 1/24 and 1/6. (Key point: Eccentricity effect)

3. Which formula gives the required basic stress?
   A. f_b >= sigma x k_s x k_a
   B. f_b >= P / (A x 1.25) only
   C. f_b >= sigma x 1.25 / (k_s k_a)
   D. f_b >= sigma / (k_s k_a k_p x 1.25) = 1.854 N/mm^2
   Answer: D. Permissible stress must be at least sigma, so f_b is sigma divided by all the factors. (Key point: Required f_b)

4. Which choice satisfies f_b = 1.854?
   A. 25 N/mm^2 units with H1 mortar (f_b 1.91), or 35 N/mm^2 with M1 or H2
   B. 5 N/mm^2 units with M2
   C. 7.5 N/mm^2 units with M1
   D. 10 N/mm^2 units with any mortar
   Answer: A. Only strong units give f_b above 1.854 in Table 8; hence enlarge the section in practice. (Key point: Selection)

## Related in Concrete Technology and Masonry Structure

- [Brick column 35 cm x 50 cm allowable load](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-35x50-axial-eccentric) — Concrete Technology and Masonry Structure, chapter 8
- [Corner stresses of 400 x 800 column (200 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acolumn-stress-400x800-200kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
- [Effect of unit and mortar strength on masonry](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Aunit-mortar-strength-effect) — Concrete Technology and Masonry Structure, chapter 8
- [Brick and mortar for 230 mm wall (100 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-230mm-100kn-5m) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
- [Brick and mortar for 250 mm wall (350 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-250mm-350kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
- [Load bearing wall: crushing strength and mortar](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-250cm-79-5kn-axial) — Concrete Technology and Masonry Structure, chapter 8

## Related in other subjects

- [Square column with 80 mm eccentricity](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Asquare-column-with-80-mm-eccentricity) — Design of RCC Structure, chapter 10
- [Column with battens, 1500 kN, 8 m, fixed ends](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Acolumn-rolled-i-1500kn-8m-fixed) — Design of Steel and Timber Structure, chapter 6
- [Braced column 450x350, 4.0 m height](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abraced-column-450x350-4-0-m) — Design of RCC Structure, chapter 10
- [Bar fixed at both ends: load shares, displacements](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acircular-bar-fixed-both-ends-50kn) — Strength of Materials, chapter 3