--- title: "Brick column design for 80 kN axial load" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:brick-column-80kn-roof" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/brick-column-80kn-roof site: BCT Engineering --- # Brick column design for 80 kN axial load Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2080 Bhadra, 8 marks - Video: [Design of Brick Masonry structures](https://www.youtube.com/watch?v=86w9uX41e_I) — Arunachala Civil - 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/brick-column-80kn-roof - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question Design a brick column using locally available bricks to carry an axial load of 80 kN from a roof consisting of RCC beams and slabs. The height of the brick column is 3.5 m. Take size of brick 200 $\times$ 100 $\times$ 100 mm. ## Short answer Trial sections checked for slenderness and stress. Adopt a 400 mm × 400 mm column using bricks of minimum crushing strength7.5 N/mm² in cement mortar1∶5 (M1). The permissible stress is0.710 N/mm² against actual0.567 N/mm². ## Answer #### Data and assumptions - Brick $200\times100\times100$ mm (height/width $=1.0$, so $k_p>1$ from Table 10). Load 80 kN axial, height 3.5 m. RCC beams and slab give lateral support in both directions, so the effective height is the actual height $h_{ef}=H=3.5$ m (IS 1905 cl. 4.3.2, Note 2). - Column in cement mortar; self weight added (19 kN/m$^3$). Maximum slenderness ratio 12 (Table 7). #### Step 1: Trial sections For each size: $SR=H/t$, $k_a=0.7+1.5A$ (A < 0.2 m$^2$), $k_s$ from Table 9 ($e=0$), $P=80+$ self weight, $\sigma=P/A$, and the unit strength needed so that $\sigma\le f_b k_sk_ak_p$ ($k_p$ taken at the unit strength): | Section (mm) | SR | $k_a$ | $k_s$ | $P$ (kN) | $\sigma$ (N/mm²) | Min. unit strength (N/mm²) for each mortar | |---|---|---|---|---|---|---| | 300 $\times$ 300 | 11.67 | 0.835 | 0.848 | 86.0 | 0.955 | M2: 30; M1: 25; H2: 20; H1: 17.5 | | 400 $\times$ 400 | 8.75 | 0.940 | 0.927 | 90.6 | 0.566 | M2: 10; M1: 7.5; H2: 7.5; H1: 7.5 | A 300 mm square column needs very strong bricks (17.5 N/mm$^2$ or more), so it is not practical for locally made bricks. #### Step 2: Adopt 400 mm $\times$ 400 mm column $$ \begin{aligned} SR &= \frac{3500}{400} = 8.75\ (<12)\\ A &= 0.16\ \text{m}^2,\ k_a=0.7+1.5\times0.16=0.940\\ k_s &= 0.927\\ \text{Self weight} &= 0.16\times3.5\times19 = 10.64\ \text{kN}\\ P &= 80+10.64 = 90.64\ \text{kN}\\ \sigma &= \frac{90.64\times10^3}{160000} = 0.567\ \text{N/mm}^2 \end{aligned} $$ For 7.5 N/mm$^2$ bricks in 1:5 mortar (M1): $f_b=0.74$, $k_p=1.1$: $$ f_c = 0.74\times0.927\times0.940\times1.1 = 0.710\ \text{N/mm}^2 \ \ge\ 0.567 $$ (5 N/mm$^2$ bricks give only 0.523 N/mm$^2$, not enough.) **Answer: provide a 400 mm $\times$ 400 mm brick column (two bricks by two bricks) built with bricks of crushing strength not less than 7.5 N/mm$^2$ in cement mortar 1:5 (M1)**, with all courses broken-bonded and the column tied to the RCC beam above. Allowable stress 0.710 N/mm$^2$ against actual 0.567 N/mm$^2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the effective height taken as the actual height 3.5 m? A. The column is fully restrained giving 0.75H B. RCC beams and slab give lateral support in both directions C. Brick size is 200 x 100 x 100 D. The column is unsupported so 2H is used Answer: B. Lateral support in both directions gives h_ef = H per IS 1905 cl. 4.3.2. (Key point: Effective height with RCC roof) 2. Why is a 300 x 300 mm column rejected? A. Its SR exceeds 12 B. It cannot carry 80 kN under any bricks C. It needs bricks of 17.5 N/mm^2 or more, impractical for local bricks D. Its area is more than 0.2 m^2 Answer: C. Its SR is 11.67 (OK) but the high stress 0.955 N/mm^2 demands very strong units. (Key point: Why not 300 x 300) 3. What must be added to the 80 kN roof load? A. Self weight of column = 0.16 x 3.5 x 19 = 10.64 kN B. Nothing, self weight is ignored C. Half the self weight D. Self weight of the RCC slab only Answer: A. Design load P = 80 + 10.64 = 90.64 kN. (Key point: Self weight) 4. What is adopted? A. 400 x 400 mm with 5 N/mm^2 bricks B. 300 x 300 mm with 7.5 N/mm^2 bricks C. 400 x 400 mm with 10 N/mm^2 bricks in M2 only D. 400 x 400 mm with bricks of at least 7.5 N/mm^2 in 1:5 mortar (f_c 0.710 > 0.567) Answer: D. 5 N/mm^2 bricks give only 0.523 N/mm^2, insufficient, while 7.5 N/mm^2 gives 0.710 N/mm^2. 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