--- title: "Reduction factors in permissible stress of masonry" subject: "Concrete Technology and Masonry Structure" chapter: 8 key: "semester_5/concrete_technology_and_masonry_structure:reduction-factors-permissible-stress" source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/reduction-factors-permissible-stress site: BCT Engineering --- # Reduction factors in permissible stress of masonry Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads - Asked in: 2082 Bhadra, 4 marks; 2081 Bhadra, 3 marks (asked 2 times) - 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/reduction-factors-permissible-stress - Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8 ## Question What is the concept (role) of using reduction factors (coefficients) in calculating permissible stress in masonry structure? ## Short answer Permissible compressive stress is obtained by multiplying basic compressive stress by reduction factors: ks accounts for slenderness and load eccentricity, ka for small cross‑sectional area, and kp for unit shape. They adjust allowable stress to real wall conditions including buckling, bending, workmanship, and unit geometry. ## Answer The permissible compressive stress in masonry cannot be taken equal to the strength of the unit or prism, because a real wall differs from the tested specimen. IS 1905 therefore starts from the **basic compressive stress** $f_b$ (Table 8; for a short, axially loaded member built of units of standard shape) and multiplies it by reduction factors: $$ f_c = f_b \times k_s \times k_a \times k_p $$ #### Role of each factor (IS 1905 cl. 5.4.1) - **Stress reduction factor $k_s$ (Table 9):** accounts for the slenderness ratio and the eccentricity of the load. A tall, slender wall buckles at a lower stress, and an eccentric load causes bending. $k_s$ falls from 1.0 (SR $\le 6$) to as low as 0.15 (SR 27 with $e/t = 1/3$). - **Area reduction factor $k_a$:** for small sections (area $A < 0.2$ m$^2$, such as columns and piers) the effect of local weak spots and poor workmanship is larger; $k_a = 0.7+1.5A$ ($A$ in m$^2$). - **Shape modification factor $k_p$ (Table 10):** $f_b$ is based on the crushing strength of the unit, but for the same unit strength, masonry built with taller units has fewer bed joints and is stronger. Units with $h/w \le 0.75$ have $k_p = 1$; taller units get a factor above 1 (up to 1.8 for tall, low-strength units). The factor of safety is already built into $f_b$. The reduction factors adjust it so that the allowed stress depends on how slender the member is, how it is loaded, how large its section is and what shape its units are. In design they are also used with the 25 percent increase allowed when $e/t > 1/24$ (cl. 5.4.1.4). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can permissible stress not equal prism strength? A. The prism is always weaker B. A real wall differs from the tested prism in slenderness, load and shape C. Mortar has no effect D. Codes forbid testing Answer: B. Reduction factors account for these differences. (Key point: Why reduce) 2. What does k_s account for? A. Area of section B. Shape of unit C. Mortar colour D. Slenderness ratio and eccentricity Answer: D. Tall walls buckle and eccentric loads cause bending. (Key point: k_s) 3. When does the area reduction factor apply? A. For all walls B. For area over 1 m^2 C. When area is less than 0.2 m^2, as for columns and piers D. For timber floors Answer: C. k_a = 0.7 + 1.5A. (Key point: k_a) 4. What is k_p for units with height/width <= 0.75? A. 1.0 B. 1.8 C. 0.7 D. 0.2 Answer: A. Taller units get a factor above 1. 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