--- title: "Brick column 35 cm x 50 cm allowable load" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:column-35x50-axial-eccentric" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/column-35x50-axial-eccentric site: BCT Engineering --- # Brick column 35 cm x 50 cm allowable load Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2082 Bhadra, 12 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/column-35x50-axial-eccentric - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question A brick in cement mortar column of size 35 cm $\times$ 50 cm is axially loaded with $P$ kN. The height of the column is 4 m and is fully restrained at top and bottom. Determine the allowable value of $P$. Also, find the allowable compressive stress and load when the load is applied at an eccentricity of 110 mm about the major axis of bending. Take size of brick 200 $\times$ 100 $\times$ 150 mm. ## Short answer Assuming brick 7.5 N/mm² and mortar M1 (1:5), axially loaded column: allowable stress 0.864 N/mm², P = 151.2 kN. With 110 mm eccentricity about major axis: allowable max stress 1.040 N/mm², load ≈ 76.4 kN. ## Answer #### Data and assumptions - Column 350 mm $\times$ 500 mm, height 4 m, fully restrained at top and bottom. The strengths of brick and mortar are not given, so **brick 7.5 N/mm$^2$ and cement mortar 1:5 (grade M1) are assumed** (IS 1905 Table 8: $f_b=0.74$ N/mm$^2$). - Brick $200\times100\times150$ mm: height/width $=150/100=1.5$, so the shape factor is $k_p=1.30$ (Table 10, 7.5 N/mm$^2$). #### Step 1: Slenderness ratio (IS 1905 cl. 4.3, 4.6.2) With full restraint at both ends, $h_{ef}=0.75H = 0.75\times4 = 3.0$ m. For a column the greater slenderness ratio governs, so the least dimension (350 mm) is critical: $$ SR=\frac{3.0}{0.35}=8.57\ (<12,\ \text{OK}) $$ #### Step 2: Axially loaded column $$ \begin{aligned} A &= 0.35\times0.5 = 0.175\ \text{m}^2 \ (<0.2)\ \Rightarrow\ k_a = 0.7+1.5A = 0.9625\\ k_s &= 0.933\ \text{(Table 9, } e=0,\ SR=8.57)\\ f_c &= f_b k_s k_a k_p = 0.74\times0.933\times0.9625\times1.30 = 0.864\ \text{N/mm}^2\\ P &= f_c A = 0.864\times0.175\times10^3 = 151.2\ \text{kN} \end{aligned} $$ #### Step 3: Eccentricity 110 mm about the major axis Bending about the major axis means the eccentricity is measured along the 500 mm side: $e/D = 110/500 = 0.22$. This exceeds $1/6$ (middle-third limit $=83$ mm), so part of the section would be in tension; the tension area is neglected (cl. 5.4.1.4 b, 5.4.4). The stress on the compressed part is triangular, over a depth of $$ 3\left(\frac{D}{2}-e\right)=3\times(250-110)=420\ \text{mm},\qquad A_{eff}=350\times420=0.147\ \text{m}^2 $$ $k_s=0.899$ (Table 9, $SR=8.57$, $e/D=0.22$, interpolated). Since $e/D > 1/6$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 b): $$ \begin{aligned} f_c &= 0.74\times0.899\times0.9625\times1.30 = 0.832\ \text{N/mm}^2\\ f_{c,\text{allow}} &= 1.25\times0.832 = 1.040\ \text{N/mm}^2\ \text{(maximum stress at the compressed edge)}\\ P_{ecc} &= \tfrac{1}{2}\,f_{c,\text{allow}}\,A_{eff} = \tfrac{1}{2}\times1.040\times0.147\times10^3 = 76.4\ \text{kN} \end{aligned} $$ **Answer: axially loaded: allowable stress 0.864 N/mm$^2$, $P = 151.2$ kN. With 110 mm eccentricity: allowable (maximum) compressive stress 1.040 N/mm$^2$, allowable load $\approx 76.4$ kN** (for brick 7.5 N/mm$^2$, mortar 1:5). Without the 25 percent increase these become 0.832 N/mm$^2$ and 61.2 kN. All values change in proportion to $f_b$ if other brick and mortar are used. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A 4 m column fully restrained at top and bottom has what effective height? A. 0.75H = 3.0 m B. H = 4.0 m C. 2H = 8.0 m D. 0.5H = 2.0 m Answer: A. With full restraint at both ends the effective height is 0.75H = 3.0 m. (Key point: Effective height) 2. Which dimension is used for the slenderness ratio of the 350 x 500 mm column and what is SR? A. Larger dimension 500 mm, SR = 6.0 B. Least dimension 350 mm, SR = 3.0/0.35 = 8.57 C. Least dimension 350 mm, SR = 4.0/0.35 = 11.4 D. Mean dimension 425 mm, SR = 7.06 Answer: B. For a column the greater slenderness governs, so the least dimension is used with h_ef = 3.0 m. (Key point: Slenderness ratio) 3. How is the permissible compressive stress of the column obtained? A. f_c = f_b + k_s + k_a + k_p B. f_c = f_b x k_s / (k_a x k_p) C. f_c = f_b x k_s x k_a x k_p D. f_c = P / A only Answer: C. Basic stress is multiplied by the stress reduction, area reduction and shape factors. (Key point: Axial permissible stress) 4. Why is the effective area reduced when e = 110 mm along the 500 mm side? A. The brick shape factor changes with eccentricity B. The column becomes more slender with eccentricity C. Mortar strength drops when the load is eccentric D. e/D = 0.22 exceeds 1/6, so part of the section is in tension and is neglected, leaving a 420 mm effective depth Answer: D. Beyond the middle third, tension area is ignored and the effective depth is 3(D/2 - e) = 420 mm. 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