--- title: "Brick and mortar: 20 cm wall, 3.4 m" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:wall-eccentric-20cm-100kn-3-4m" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-eccentric-20cm-100kn-3-4m site: BCT Engineering --- # Brick and mortar: 20 cm wall, 3.4 m Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2079 Bhadra, 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-eccentric-20cm-100kn-3-4m - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question A wall 20 cm thick, using modular bricks, carries at the top a load of 100 kN/m having resultant eccentricity of 1/12. The wall is 5 m long between cross walls and is of 3.4 m clear height between RCC slabs at top and bottom. What should be the required strength of brick and grade of mortar? Assume that joints are not raked. ## Short answer Required bricks of minimum 7.5 N/mm² with M2 (1:6) mortar. Effective height 2.55 m gives slenderness ratio 12.75, ks=0.784; eccentricity ratio 1/12 allows 25% increase. Base stress 0.565 N/mm² gives required basic compressive stress 0.576 N/mm². ## Answer #### Data and assumptions - Wall 20 cm thick, modular bricks, joints not raked. Load 100 kN/m at top, $e/t=1/12$. - Wall 5 m long between cross walls, clear height 3.4 m between RCC slabs (slab thickness not given, $H=3.4$ m). - RCC slabs give lateral and rotational restraint (Table 4 Sl. 1). - 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 $H = 3.40$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.40 = 2.550$ 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: $$ SR = \frac{h_{ef}}{t} = \frac{2.550}{0.200} = 12.75\ (<27,\ \text{Table 7, cement mortar}) $$ #### Step 2: Design load per metre run at the base $$ \begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.200\times3.40\times19 = 12.92\ \text{kN/m}\\ P &= 100.00 + 12.92 = 112.92\ \text{kN/m} \end{aligned} $$ Stress on the wall: $$ \sigma = \frac{P}{A} = \frac{112.92\times10^3}{200\times10^3\ \text{mm}^2/\text{m}} = 0.565\ \text{N/mm}^2 $$ #### Step 3: Reduction factors - Stress reduction factor $k_s$ (Table 9, $SR=12.75$, $e/t=0.083$, by linear interpolation) $= 0.784$ - Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.20 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) - Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25 #### Step 4: Required basic compressive stress $$ \sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.565}{0.784\times1.0\times1.0\times1.25} = 0.576\ \text{N/mm}^2 $$ (Without the 25 percent increase the requirement would be $f_b \ge 0.720$ N/mm$^2$.) #### Step 5: Strength of brick and grade of mortar (IS 1905 Table 8) Smallest standard unit strength giving $f_b \ge 0.576$ N/mm$^2$: | Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) | |---|---|---| | M2 (cement:sand 1:6) | 7.5 | 0.59 | | M1 (cement:sand 1:5) | 7.5 | 0.74 | | H2 (cement:sand 1:4) | 7.5 | 0.74 | | H1 (cement:sand 1:3) | 7.5 | 0.75 | **Answer: use bricks of crushing strength not less than 7.5 N/mm$^2$ with mortar M2 (cement:sand 1:6)** ($f_b = 0.59$ N/mm$^2$ $\ge$ 0.576). 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. For RCC slabs at top and bottom, h_ef = 0.75 x 3.40 = 2.55 m. What is SR (t = 0.20 m)? A. 17.00 B. 12.75 C. 6.38 D. 8.50 Answer: B. SR = h_ef/t = 2.55/0.20 = 12.75, less than 27 so acceptable (height governs over 5 m length). (Key point: Effective height and SR) 2. Why is self weight added to the 100 kN/m load? A. Because the load acts at the top and the wall weight adds to it at the base: P = 100 + 12.92 = 112.92 kN/m B. Self weight is never considered in walls C. Because it acts as the eccentric load D. To account for slab dead load Answer: A. Design is at the base, where the wall weight 0.2 x 3.4 x 19 = 12.92 kN/m adds to the top load. (Key point: Design load) 3. Why is f_b = 0.565/(0.784 x 1.25) used rather than 0.565/0.784? A. e/t = 1/12 exceeds 1/24, so a 25 percent increase in permissible stress is allowed B. Because the bricks are modular C. Because SR is above 20 D. Because k_a is 1.25 Answer: A. IS 1905 allows 25 percent more stress when e/t > 1/24; this lowers required f_b from 0.720 to 0.576 N/mm^2. 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