--- title: "Load bearing wall: crushing strength and mortar" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:wall-250cm-79-5kn-axial" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-250cm-79-5kn-axial site: BCT Engineering --- # Load bearing wall: crushing strength and mortar Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2068 Chaitra, 10 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/wall-250cm-79-5kn-axial - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question A load bearing brick masonry wall of a building is 250 cm [sic] thick, is laterally supported by RCC slabs at top and bottom, which are 13 cm thick each, and the clear height between slabs is 3.5 m. If the wall has an axial load of 79.5 kN/m at the base, inclusive of self weight, what should be the crushing strength of bricks and grade of mortar for the wall? The wall is 5 m long between cross walls and bricks used are of modular size. Assume suitable data if any required. ## Short answer Required basic compressive stress 0.366 N/mm²; bricks of crushing strength not less than 5 N/mm² with cement mortar M2 (1:6) are adequate. ## Answer #### Data and assumptions - "250 cm" is read as 250 mm (25 cm) thick, since a 2.5 m thick load-bearing wall is not meaningful. Modular bricks, $k_p=1$. - Axial load 79.5 kN/m at the base inclusive of self weight ($e=0$). Wall 5 m long between cross walls; clear height 3.5 m between 13 cm RCC slabs, so $H=3.5+0.13=3.63$ m. - Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method. - Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$. #### Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6) Height between slab centres $H = 3.63$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.63 = 2.723$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 5.00$ m. The lesser value governs: $$ SR = \frac{h_{ef}}{t} = \frac{2.723}{0.250} = 10.89\ (<27,\ \text{Table 7, cement mortar}) $$ #### Step 2: Design load per metre run at the base Load at the base (self weight already included) $P = 79.50$ kN/m. Stress on the wall: $$ \sigma = \frac{P}{A} = \frac{79.50\times10^3}{250\times10^3\ \text{mm}^2/\text{m}} = 0.318\ \text{N/mm}^2 $$ #### Step 3: Reduction factors - Stress reduction factor $k_s$ (Table 9, $SR=10.89$, $e/t=0.000$, by linear interpolation) $= 0.868$ - Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.25 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$ - Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10) - $e/t \le 1/24$: no increase #### Step 4: Required basic compressive stress $$ \sigma \le f_b\,k_s\,k_a\,k_p\times1.00 \Rightarrow f_b \ge \frac{0.318}{0.868\times1.0\times1.0\times1.00} = 0.366\ \text{N/mm}^2 $$ #### Step 5: Strength of brick and grade of mortar (IS 1905 Table 8) Smallest standard unit strength giving $f_b \ge 0.366$ N/mm$^2$: | Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) | |---|---|---| | M2 (cement:sand 1:6) | 5 | 0.44 | | M1 (cement:sand 1:5) | 5 | 0.50 | | H2 (cement:sand 1:4) | 5 | 0.50 | | H1 (cement:sand 1:3) | 5 | 0.50 | **Answer: use bricks of crushing strength not less than 5 N/mm$^2$ with mortar M2 (cement:sand 1:6)** ($f_b = 0.44$ N/mm$^2$ $\ge$ 0.366). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is '250 cm thick' treated? A. As 2.5 m thick B. As 250 mm, since 2.5 m is not meaningful for a load-bearing wall C. As 25 mm D. As 25 m Answer: B. The unit is read as 250 mm (a [sic] in the question). (Key point: Data interpretation) 2. Clear height 3.5 m between 13 cm slabs gives H of: A. 3.5 m B. 3.63 m C. 3.37 m D. 3.75 m Answer: B. H is centre to centre: 3.5 + 0.13 = 3.63 m; h_ef = 0.75H = 2.723 m. (Key point: H) 3. Why is no 25 percent increase applied? A. Bricks are modular B. Load is axial, e/t = 0, which is not greater than 1/24 C. Mortar is M2 D. SR is above 10 Answer: B. The increase needs e/t > 1/24; here e = 0, so f_b = 0.318/0.868 = 0.366 N/mm^2. 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