--- title: "Safe load on a brick masonry column" subject: "Design of Timber and Masonry Structures" chapter: 5 key: "semester_5/design_of_timber_and_masonry_structures:brick-masonry-column-safe-load" source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/brick-masonry-column-safe-load site: BCT Engineering --- # Safe load on a brick masonry column Design of Timber and Masonry Structures · Chapter 5: Design of Masonry Walls for Gravity Loads - Asked in: year not given, 6 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/brick-masonry-column-safe-load - Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-5 ## Question A brick masonry column of cross-section 340 mm × 460 mm and height 3.5 m is laterally restrained at the top and bottom but not rotationally (effective height = actual height). The basic compressive stress is $f_b = 1.1$ N/mm². Using IS 1905 with $k_s$ from SR 10 → 0.89, SR 12 → 0.84, find the safe axial load. Compare it with that of a 460 mm × 460 mm column. ## Short answer SR based on least lateral dimension: 3500/340=10.29, k_s=0.883, area <0.2 m² gives k_a=0.935, f_c=0.907 N/mm², safe axial load=142 kN. For 460×460 column: SR=7.61, k_s=0.96, k_a=1.0, f_c=1.056 N/mm², load=223 kN. ## Answer #### Slenderness (column) Slenderness is based on the **least lateral dimension**: $$ H_{eff} = 3500\ \text{mm}, \qquad SR = \frac{3500}{340} = 10.29 $$ $$ k_s = 0.89 - \frac{0.29}{2}\times 0.05 = 0.883 $$ #### Area factor $A = 0.34 \times 0.46 = 0.1564$ m² $< 0.2$ m² $$ k_a = 0.7 + 1.5\times0.1564 = 0.935 $$ $k_p = 1.0$ for brick. #### Permissible stress and safe load $$ \begin{aligned} f_c &= f_b\,k_s\,k_a\,k_p = 1.1 \times 0.883 \times 0.935 \times 1.0 = 0.907\ \text{N/mm}^2 \\ P &= f_c A = 0.907 \times 340 \times 460 = 141\,900\ \text{N} \approx 142\ \text{kN} \end{aligned} $$ #### Comparison with 460 × 460 mm column For $SR < 10$ the full IS 1905 table is needed (SR 6 → 1.00, SR 8 → 0.95). $SR = 3500/460 = 7.61$, so $k_s = 0.95 + (8 - 7.61)/2 \times 0.05 = 0.96$; $A = 0.2116$ m² $> 0.2$, so $k_a = 1.0$. $$ f_c = 1.1 \times 0.96 \times 1.0 = 1.056\ \text{N/mm}^2, \qquad P = 1.056 \times 460 \times 460 = 223\,400\ \text{N} = 223\ \text{kN} $$ | Column | $SR$ | $k_s$ | $k_a$ | $f_c$ (N/mm²) | Safe load (kN) | |---|---|---|---|---|---| | 340 × 460 | 10.29 | 0.883 | 0.935 | 0.907 | 142 | | 460 × 460 | 7.61 | 0.960 | 1.000 | 1.056 | 223 | The larger column carries 57 percent more load because of its larger area and lower slenderness. The self-weight of the column ($19 \times 0.1564 \times 3.5 = 10.4$ kN) should be deducted from the safe load to get the superimposed load, about 131 kN for the 340 × 460 column. **Answer: Safe axial load ≈ 142 kN for 340 × 460 mm (about 131 kN superimposed after self-weight) and ≈ 223 kN for 460 × 460 mm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 340 x 460 mm column, slenderness ratio is based on which dimension? A. The larger dimension 460 mm B. The least lateral dimension 340 mm, giving SR = 10.29 C. The average of the two dimensions D. The diagonal of the section Answer: B. Slenderness uses the least lateral dimension, so SR = 3500/340 = 10.29. (Key point: Slenderness ratio) 2. Why is an area factor k_a = 0.935 applied to the 340 x 460 column? A. Its area 0.1564 m^2 is less than 0.2 m^2, so k_a = 0.7 + 1.5A B. Its height exceeds 3 m C. Its area is more than 0.2 m^2 D. Bricks always need k_a = 0.935 Answer: A. For sections under 0.2 m^2, k_a = 0.7 + 1.5A reduces the permissible stress. (Key point: Area factor) 3. How are the permissible stress and safe load obtained? A. f_c = f_b + k_s + k_a; P = f_c / A B. f_c = f_b x k_s x k_a x k_p; P = f_c x A C. f_c = f_b / (k_s k_a); P = f_c x A D. f_c = f_b x k_s only; P = f_c x A Answer: B. The permissible stress is the basic stress times the factors, and load is stress times area (about 142 kN). (Key point: Permissible stress and load) 4. Why does the 460 x 460 mm column carry about 57 percent more load? A. It has lower slenderness (7.61), k_a = 1.0 and larger area B. It has a higher basic stress C. Its effective height is smaller D. Self-weight is ignored for it Answer: A. Lower SR gives higher k_s, area over 0.2 m^2 gives k_a = 1, and area is greater, giving 223 kN versus 142 kN. 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