--- title: "Allowable load on 30 x 60 cm column" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:column-30x60-allowable-load" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/column-30x60-allowable-load site: BCT Engineering --- # Allowable load on 30 x 60 cm column Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2080 Baisakh, 7 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-30x60-allowable-load - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question Determine the allowable axial load on a column 30 cm $\times$ 60 cm constructed with brick in cement mortar using 200 $\times$ 100 $\times$ 100 mm brick. The height of the column is 3.2 m. The roof consists of RCC beam and slab. ## Short answer Slenderness ratio =10.67≤12, ka =0.970, ks =0.873, kp =1.10. Permissible stress =0.74×0.873×0.970×1.10 =0.690 N/mm². Allowable total axial load at base =0.690×0.18 =124.1 kN (roof load about113.2 kN after self-weight) for7,5 N/mm² bricks in1∶5 cement mortar. ## Answer #### Data and assumptions - Column 300 mm $\times$ 600 mm, height 3.2 m, supporting an RCC beam and slab roof, so it is laterally supported in both directions and $h_{ef}=H=3.2$ m (IS 1905 cl. 4.3.2, Note 2). - Brick $200\times100\times100$ mm, so height/width $=1.0$ and $k_p$ comes from Table 10. The brick strength and mortar are not given, so **7.5 N/mm$^2$ brick with cement mortar 1:5 (M1)** is assumed as the basic case; other grades are tabulated below. #### Step 1: Factors $$ \begin{aligned} SR &= \frac{h_{ef}}{t} = \frac{3.2}{0.30} = 10.67\ (<12)\\ A &= 0.30\times0.60 = 0.18\ \text{m}^2\ (<0.2)\ \Rightarrow\ k_a=0.7+1.5\times0.18 = 0.970\\ k_s &= 0.873\ \text{(Table 9, } e=0)\\ k_p &= 1.10\ \text{(7.5 N/mm}^2\text{, } h/w=1.0) \end{aligned} $$ #### Step 2: Allowable load $$ \begin{aligned} f_c &= f_b k_s k_a k_p = 0.74\times0.873\times0.970\times1.10 = 0.690\ \text{N/mm}^2\\ P &= f_c A = 0.690\times0.18\times10^3 = 124.1\ \text{kN} \end{aligned} $$ Self weight $=0.18\times3.2\times19 = 10.9$ kN, so the load that can be applied from the roof is $124.1 - 10.9 = 113.2$ kN. #### Step 3: Allowable total load for other grades (kN) | Brick (N/mm²) | $k_p$ | M2 (1:6) | M1 (1:5) | H2 (1:4) | H1 (1:3) | |---|---|---|---|---|---| | 5 | 1.20 | 80.5 | 91.5 | 91.5 | 91.5 | | 7.5 | 1.10 | 99.0 | 124.1 | 124.1 | 125.8 | | 10 | 1.10 | 135.9 | 161.0 | 161.0 | 167.7 | **Answer: allowable axial load = 124.1 kN at the base (about 113.2 kN of roof load after deducting self weight), for 7.5 N/mm$^2$ brick in 1:5 cement mortar.** With 5 N/mm$^2$ brick and 1:6 mortar the capacity is only 80.5 kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is SR for the 300 x 600 mm column of height 3.2 m with RCC roof? A. 5.33, using 0.60 m B. 8.0, using 0.75H C. 10.67, using h_ef = 3.2 m and t = 0.30 m D. 21.3, using 2H Answer: C. Supported in both directions gives h_ef = H; the least dimension 0.30 m is used. (Key point: Slenderness ratio) 2. What is k_a for A = 0.18 m^2? A. 0.7 + 1.5 x 0.18 = 0.970 B. 1.0 because A is large C. 0.7 + 0.18 = 0.88 D. 0.7 x 1.5 x 0.18 = 0.19 Answer: A. A < 0.2 m^2 so k_a = 0.7 + 1.5A. (Key point: Area factor) 3. What is the allowable load on the column? A. P = 0.690 x 300 x 600 / 1000 = 12.4 kN B. P = 0.690 x 180000 = 124.1 kN C. P = 113.2 kN total at base D. P = 80.5 kN Answer: B. f_c = 0.74 x 0.873 x 0.970 x 1.10 = 0.690 N/mm^2, times A = 0.18 m^2. (Key point: Allowable load) 4. How much roof load can be applied? A. 124.1 + 10.9 = 135 kN B. 124.1 kN, self weight ignored C. 10.9 kN D. Total 124.1 minus self weight 10.9 = 113.2 kN Answer: D. Self weight consumes part of the capacity at the base. 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