--- title: "Check of a solid brick wall under gravity load" subject: "Design of Timber and Masonry Structures" chapter: 5 key: "semester_5/design_of_timber_and_masonry_structures:solid-masonry-wall-gravity-check" source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/solid-masonry-wall-gravity-check site: BCT Engineering --- # Check of a solid brick wall under gravity load Design of Timber and Masonry Structures · Chapter 5: Design of Masonry Walls for Gravity Loads - Asked in: year not given, 6 marks - Video: [Masonry Wall Design - Part 1 || Interior Cross Wall Design || IS 1905:1987 || Nepali Language](https://www.youtube.com/watch?v=X41MEWk5MMY) — Sumit Paudel - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/solid-masonry-wall-gravity-check - Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-5 ## Question A single-storey load-bearing brick wall is 230 mm thick and 3.2 m high (clear height between floor and roof slab). The RC roof slab bears over the full wall thickness on both sides and gives lateral and rotational restraint at the top, and the wall is fixed at the base. The basic compressive stress is $f_b = 1.0$ N/mm². The slab transfers a working load of 120 kN per metre length of wall at the top, and the unit weight of masonry is 19 kN/m³. Using the IS 1905 method with $k_s$ from the table (SR 10 → 0.89, SR 12 → 0.84), check the wall and find the maximum load per metre it can carry. ## Short answer Using H_eff=0.75×3.2=2.4 m, SR=2400/230=10.43, k_s=0.879, k_a=1.0, f_c=0.879 N/mm². Self-weight 13.98 kN/m, total load 133.98 kN/m gives stress 0.583 N/mm² < 0.879, safe. Maximum total load capacity = 202 kN/m. ## Answer #### Step 1: Slenderness Lateral and rotational restraint at both ends gives $H_{eff} = 0.75H$: $$ \begin{aligned} H_{eff} &= 0.75 \times 3200 = 2400\ \text{mm} \\ SR &= \frac{2400}{230} = 10.43 \end{aligned} $$ $SR < 27$, so OK. #### Step 2: Stress reduction factor Interpolate between $SR = 10$ ($k_s = 0.89$) and $SR = 12$ ($k_s = 0.84$): $$ k_s = 0.89 - \frac{0.43}{2}\times 0.05 = 0.879 $$ #### Step 3: Permissible stress Per metre length, area $A = 0.23 \times 1.0 = 0.23$ m² $> 0.2$ m², so $k_a = 1.0$; for bricks $k_p = 1.0$. $$ f_c = 1.0 \times 0.879 \times 1.0 \times 1.0 = 0.879\ \text{N/mm}^2 $$ #### Step 4: Actual stress at base $$ \begin{aligned} W_{wall} &= 19 \times 0.23 \times 3.2 = 13.98\ \text{kN/m} \\ P &= 120 + 13.98 = 133.98\ \text{kN/m} \\ f &= \frac{133\,980}{230 \times 1000} = 0.583\ \text{N/mm}^2 \end{aligned} $$ $0.583 < 0.879$, so the wall is **safe** (utilisation 66 percent). #### Step 5: Maximum load $$ P_{max} = f_c\,A = 0.879 \times 230 \times 1000 = 202\,170\ \text{N/m} \approx 202\ \text{kN/m} $$ Allowable additional slab load $= 202.2 - 14.0 = 188\ \text{kN/m}$. **Answer: Actual stress 0.58 N/mm² < 0.879 N/mm², safe. Maximum total load = 202 kN/m (slab load about 188 kN/m).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A 230 mm wall 3.2 m high has slab restraint at top and is fixed at the base. What is SR? A. 3200/230 = 13.9 B. 0.75 x 3200/230 = 10.43 C. 2 x 3200/230 = 27.8 D. 1.5 x 3200/230 = 20.9 Answer: B. Full restraint at both ends gives H_eff = 0.75H = 2400 mm, so SR = 2400/230 = 10.43. (Key point: Slenderness ratio) 2. SR = 10.43 lies between SR 10 (0.89) and SR 12 (0.84). What is f_c with f_b = 1.0 N/mm^2? A. 0.879 N/mm^2, with k_a = k_p = 1.0 B. 0.89 N/mm^2, ignoring interpolation C. 1.0 N/mm^2 D. 0.84 N/mm^2 Answer: A. Interpolation gives k_s = 0.879; per-metre area is 0.23 m^2 > 0.2 so k_a = 1.0 and k_p = 1.0 for bricks. (Key point: Permissible stress) 3. The wall carries 120 kN/m from the slab; its own weight is 19 x 0.23 x 3.2 = 13.98 kN/m. The base stress is: A. 0.52 N/mm^2 (slab load only) B. 0.88 N/mm^2 C. 0.58 N/mm^2, below 0.879, so the wall is safe D. 1.2 N/mm^2 Answer: C. P = 133.98 kN/m over 230 x 1000 mm^2 gives 0.583 N/mm^2, which is less than 0.879 N/mm^2. (Key point: Actual stress at base) 4. What is the maximum total load per metre, and the allowable slab load? A. 202 kN/m total; about 188 kN/m slab load after subtracting self-weight B. 202 kN/m slab load C. 120 kN/m total D. 133.98 kN/m total Answer: A. P_max = f_c x A = 0.879 x 230 x 1000 = 202 kN/m; subtracting 14 kN/m of self-weight leaves about 188 kN/m. 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