--- title: "Wall with openings: stress, brick, mortar" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:wall-with-openings-single-storey" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-with-openings-single-storey site: BCT Engineering --- # Wall with openings: stress, brick, mortar Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2069 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-with-openings-single-storey - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question The external wall of a single storeyed house is 230 mm thick and has door and window openings as shown in the figure. The plinth level is 1500 mm above the top of foundation footing and floor-to-ceiling height is 2800 mm. The one-way RCC slab of 3500 mm clear span bears on the walls and is 115 mm thick. Determine the maximum stress in the wall and calculate the strength of the bricks and grade of mortar required for the wall. Live load = 1.5 kN/m$^2$; lintel level = 2000 mm. [Figure: plan of the 230 mm wall with segments of 230 mm, 2000 mm, 1000 mm (opening), 1800 mm and 230 mm.] ## Short answer Maximum average stress 0.115 N/mm² at top of footing (extreme fibre 0.136 N/mm²). Required basic compressive stress 0.126 N/mm²; use bricks with crushing strength not less than 3.5 N/mm² and cement mortar M2 (1:6). ## Answer #### Data and assumptions - Wall 230 mm thick, modular bricks ($k_p=1$). Total length 5.26 m ($230+2000+1000+1800+230$ mm), with a 1.0 m wide opening up to lintel level 2.0 m. Solid length above plinth $=5.26-1.00=4.26$ m; below plinth level the wall is continuous over the full 5.26 m. - Ceiling height 2.8 m; 115 mm one-way RCC slab of 3.5 m clear span bearing on this wall. Finish and roofing assumed 1.0 kN/m$^2$ (not given). Live load 1.5 kN/m$^2$. Eccentricity of the slab load assumed $e/t=1/12$ (IS 1905 App. A-2). - The load over the opening is carried by the lintel to the solid wall on either side (IS 1905 cl. 5.3.2, 5.3.3), so above the plinth the solid parts carry the whole load. - Two sections are checked: just above plinth level (solid wall only) and at the top of the footing (full length, including the plinth wall). - Brick masonry 19 kN/m$^3$, RCC 25 kN/m$^3$. #### Step 1: Loads on the wall $$ \begin{aligned} w &= 25\times0.115+1.0+1.5 = 5.375\ \text{kN/m}^2\\ \text{Slab reaction per m} &= w\times\frac{3.5}{2} = 9.41\ \text{kN/m}\\ \text{Slab load on whole wall} &= 9.41\times5.26 = 49.5\ \text{kN}\\ \text{Wall above lintel (over opening)} &= 0.23\times1.0\times(2.8-2.0)\times19 = 3.50\ \text{kN}\\ \text{Solid wall self weight} &= 0.23\times4.26\times2.8\times19 = 52.1\ \text{kN}\\ N_{plinth} &= 105.1\ \text{kN}\\ \text{Plinth wall (full length)} &= 0.23\times5.26\times1.5\times19 = 34.5\ \text{kN}\\ N_{footing} &= 105.1+34.5 = 139.6\ \text{kN} \end{aligned} $$ #### Step 2: Maximum stress in the wall $$ \begin{aligned} \text{Above plinth: } \sigma &= \frac{105.1\times10^3}{0.23\times4.26\times10^6} = 0.107\ \text{N/mm}^2\\ \text{At top of footing: } \sigma &= \frac{139.6\times10^3}{0.23\times5.26\times10^6} = 0.115\ \text{N/mm}^2 \end{aligned} $$ Bending from the slab eccentricity ($e = t/12 = 19.2$ mm on the 49.5 kN slab load): $M = 49.5\times0.0192 = 0.95$ kNm. At the footing, $Z = 5260\times230^2/6 = 46.4\times10^6$ mm$^3$, so $\sigma_b = 0.95\times10^6/46.4\times10^6 = 0.020$ N/mm$^2$. **Maximum stress in the wall: 0.115 N/mm$^2$ (average) at the top of the footing, or about $0.115+0.020 = 0.136$ N/mm$^2$ at the extreme fibre.** (Just above plinth level: 0.107 average, about 0.132 at the extreme fibre.) #### Step 3: Slenderness and permissible stress Above plinth: the solid parts (2.23 m and 2.03 m long) are longer than $4t = 0.92$ m, so they act as walls, not as columns between openings. With the RCC slab at the top and the plinth floor at the bottom (full restraint), $H=2.8+0.115/2=2.858$ m: $$ \begin{aligned} h_{ef} &= 0.75\times2.858 = 2.143\ \text{m}\\ SR &= \frac{2.143}{0.23} = 9.32\ (<27) \end{aligned} $$ $k_s=0.894$ (Table 9, $e/t=1/12$), $k_a=1.0$ (section area greater than 0.2 m$^2$), $k_p=1$, and a 25 percent increase applies ($e/t>1/24$): $$ f_b \ge \frac{0.107}{0.894\times1.25} = 0.096\ \text{N/mm}^2 $$ At the top of the footing (full-length wall, conservatively taking the full height from footing to slab): $H = 1.5+2.8+0.0575 = 4.36$ m, $h_{ef}=0.75\times4.36=3.27$ m, $SR=14.2$ ($<27$), $k_s = 0.734$: $$ f_b \ge \frac{0.115}{0.734\times1.25} = 0.126\ \text{N/mm}^2 $$ The section at the top of the footing governs (required $f_b = 0.126$ N/mm$^2$). #### Step 4: Brick and mortar (IS 1905 Table 8) | Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) | |---|---|---| | M2 (cement:sand 1:6) | 3.5 | 0.35 | | M1 (cement:sand 1:5) | 3.5 | 0.35 | | H2 (cement:sand 1:4) | 3.5 | 0.35 | | H1 (cement:sand 1:3) | 3.5 | 0.35 | **Answer: maximum stress about 0.115 N/mm$^2$ (average; about 0.136 N/mm$^2$ at the extreme fibre) at the top of the footing; required $f_b = 0.126$ N/mm$^2$, so the lowest class of brick (3.5 N/mm$^2$) with M2 mortar (1:6, $f_b=0.35$) is adequate.** For durability and practical reasons, adopt bricks of not less than 5 N/mm$^2$ with 1:6 cement mortar. Provide lintels with a bearing of at least 230 mm (or 1/10 of the span) on each side. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the load above the 1.0 m opening treated? A. Carried by the lintel to the solid wall on either side B. Ignored entirely C. Carried by the sill D. Carried by the footing directly below the opening Answer: A. IS 1905 cl. 5.3.2-5.3.3: the lintel transfers load to the solid parts, so only the solid length (4.26 m) resists it. (Key point: Load over opening) 2. Average stress is N/A with N = 105.1 kN and A = 0.23 x 4.26 m^2. The result is: A. 0.107 N/mm^2 B. 0.161 N/mm^2 C. 0.5 N/mm^2 D. 0.03 N/mm^2 Answer: A. 105.1e3/(0.23 x 4.26e6) = 0.107 N/mm^2 (extreme fibre 0.161 with e/t = 1/12). (Key point: Average stress) 3. The effective height used is 0.75H + 0.25H1. What does H1 represent? A. Slab thickness B. Height of the opening (2.0 m) C. Wall length D. Plinth height Answer: B. For a wall with openings, cl. 4.3.3 a gives h_ef = 0.75 x 2.857 + 0.25 x 2.0 = 2.643 m, SR = 11.49. (Key point: Effective height with opening) 4. Required f_b = 0.104 N/mm^2. The brick adopted is: A. Lowest class 3.5 N/mm^2 is adequate; practice adopts 5 N/mm^2 with 1:6 mortar B. 20 N/mm^2 with H1 C. 10 N/mm^2 with M1 minimum D. Wall cannot be built in brick Answer: A. Table 8 gives 0.35 for the lowest class, far above 0.104; 5 N/mm^2 is adopted for durability. 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