--- title: "Multi-storey wall: stresses at each level" subject: "Design of Timber and Masonry Structures" chapter: 5 key: "semester_5/design_of_timber_and_masonry_structures:three-storey-wall-base-stress-check" source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/three-storey-wall-base-stress-check site: BCT Engineering --- # Multi-storey wall: stresses at each level Design of Timber and Masonry Structures · Chapter 5: Design of Masonry Walls for Gravity Loads - Asked in: year not given, 8 marks - 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/three-storey-wall-base-stress-check - Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-5 ## Question A three-storey load-bearing brick wall has a uniform thickness of 230 mm in all storeys; each storey height is 3.0 m with RC slabs at each floor and roof level giving lateral and rotational restraint at both ends of the wall in each storey. The roof slab transfers 15 kN/m and each of the two floor slabs transfers 25 kN/m (working loads, including live load). Unit weight of masonry is 19 kN/m³. The basic compressive stress for the masonry is $f_b = 0.5$ N/mm². Find the stress at the base of each storey and check against the permissible stress (take $k_s$ for SR 8 → 0.95, SR 10 → 0.89). If the ground storey fails, redesign its thickness as 345 mm. ## Short answer Stresses at base: top 0.122 N/mm², middle 0.288 N/mm², ground 0.454 N/mm² (permissible 0.448). Ground storey fails; redesign to 345 mm thick wall reduces stress to 0.321 N/mm² (< 0.49 N/mm²). ## Answer #### Step 1: Permissible stress Each storey is fixed at both ends: $H_{eff} = 0.75 \times 3000 = 2250$ mm. $$ SR = \frac{2250}{230} = 9.78, \quad k_s = 0.89 + \frac{0.22}{2}\times 0.06 = 0.897 $$ Area per metre $= 0.23$ m² $> 0.2$, so $k_a = 1.0$, $k_p = 1.0$: $$ f_c = 0.5 \times 0.897 = 0.448\ \text{N/mm}^2 $$ #### Step 2: Self-weight per storey $$ W_{s} = 19 \times 0.23 \times 3.0 = 13.11\ \text{kN/m} $$ #### Step 3: Loads and stresses at each base level | Level (base of) | Roof | Floors | Wall self-weight | Total (kN/m) | Stress (N/mm²) | Permissible | Check | |---|---|---|---|---|---|---|---| | Top storey | 15 | 0 | 13.11 | 28.11 | 0.122 | 0.448 | Safe | | Middle storey | 15 | 25 | $2 \times 13.11 = 26.22$ | 66.22 | 0.288 | 0.448 | Safe | | Ground storey | 15 | 50 | $3 \times 13.11 = 39.33$ | 104.33 | 0.454 | 0.448 | Slightly over | Stress $= \text{load}/(230 \times 1000)$. The ground storey stress exceeds the permissible value by about 1 percent, so it is not acceptable. #### Step 4: Redesign of the ground storey with 345 mm wall Self-weight of the 345 mm ground wall $= 19 \times 0.345 \times 3.0 = 19.67$ kN/m. Load at base: $$ P = 15 + 50 + 2 \times 13.11 + 19.67 = 110.89\ \text{kN/m} $$ $$ \begin{aligned} SR &= \frac{2250}{345} = 6.52, \quad k_s \approx 0.99 \\ f_c &= 0.5 \times 0.99 = 0.49\ \text{N/mm}^2 \\ f &= \frac{110\,890}{345 \times 1000} = 0.321\ \text{N/mm}^2 < 0.49 \end{aligned} $$ A 345 mm wall at the ground storey is safe (stress ratio 0.65). The 230 mm upper wall should sit centrally on the 345 mm wall (57.5 mm projection on each side) so that the load stays concentric. If it were flush with one face, the eccentricity of 57.5 mm would add large bending stress and the check above would no longer hold. **Answer: Stresses at base: top 0.122, middle 0.288, ground 0.454 N/mm² (permissible 0.448). The ground storey needs a 345 mm wall (stress 0.321 N/mm² < 0.49).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What effective height is used for each storey with slab restraint (fixed) at both ends? A. H_eff = 1.0 x 3000 = 3000 mm B. H_eff = 0.75 x 3000 = 2250 mm C. H_eff = 2.0 x 3000 = 6000 mm D. H_eff = 0.5 x 3000 = 1500 mm Answer: B. Lateral and rotational restraint at both ends gives 0.75H, so SR = 2250/230 = 9.78. (Key point: Effective height) 2. What is the permissible stress for the 230 mm wall? A. 0.5 N/mm^2, since k_s is ignored B. 0.448 N/mm^2 = f_b x k_s with k_s = 0.897, k_a = k_p = 1 C. 0.95 N/mm^2 D. 0.89 N/mm^2 Answer: B. k_s is interpolated between 0.95 and 0.89 for SR 9.78 (0.897); the area is above 0.2 m^2 so k_a = 1; f_c = 0.448. (Key point: Permissible stress) 3. What load acts at the base of the ground storey? A. 15 + 25 + 13.11 kN/m B. 15 + 50 + 3 x 13.11 = 104.33 kN/m, i.e. roof, two floors and three storeys of wall C. Only 3 x 13.11 kN/m D. 15 + 25 + 3 x 13.11 kN/m Answer: B. Roof load, both floor loads and the self-weight of all three storeys act at the ground storey base. (Key point: Load accumulation at the base) 4. What is the result for the ground storey and the remedy? A. 0.454 > 0.448 N/mm^2, so unsafe; use a 345 mm wall which gives 0.321 < 0.49 B. 0.454 < 0.448, safe C. Unsafe by a large margin; replace by a column D. Safe because the excess is only 1 percent Answer: A. The stress slightly exceeds the permissible value, so it is rejected, and thickening to 345 mm reduces the stress and raises k_s. 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