--- title: "Deflection control, sufficient stiffness method" subject: "Design of RCC Structure" chapter: 8 key: "semester_7/design_of_rcc_structure:deflection-control-sufficient-stiffness-method" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/deflection-control-sufficient-stiffness-method site: BCT Engineering --- # Deflection control, sufficient stiffness method Design of RCC Structure · Chapter 8: Limit States of Serviceability: Deflection and Cracking - Asked in: 2079 Baisakh, 4 marks; 2065 Shrawan (old course), 7 marks; 2065 Kartik (old course), 6 marks (asked 3 times) - Video: [How to determine Modification factor as per IS:456-2000 ?](https://www.youtube.com/watch?v=ny7VEAuwM3s) — cwill structures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/deflection-control-sufficient-stiffness-method - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-8 ## Question Explain how the deflection of an RC flexural member is controlled by the empirical (sufficient stiffness) method as per IS 456. ## Short answer Deflection controlled by limiting span/effective depth ratio per IS 456 cl. 23.2.1. Basic l/d: cantilever 7, simply supported 20, continuous 26. Modified by factors kt (tension steel), kc (compression steel), kf (flanged beams). Permissible l/d = basic × kt × kc × kf. For spans >10m, basic value multiplied by 10/span. Actual l/d must not exceed permissible. ## Answer #### Idea of the method Calculating the actual deflection takes a long time, so IS 456 cl. 23.2.1 allows deflection to be controlled indirectly. If the ratio of span to effective depth of a beam, slab or cantilever does not exceed a permitted value, the deflection will be within the limits of cl. 23.2 (final deflection including creep, shrinkage and temperature $\le \text{span}/250$, and the part occurring after finishes are applied $\le \text{span}/350$ or 20 mm, whichever is less). This is the **empirical, or sufficient stiffness (span/depth ratio)**, method. #### Procedure 1. Take the **basic span/effective depth ratio**: | Member | Basic $l/d$ | |---|---| | Cantilever | 7 | | Simply supported | 20 | | Continuous | 26 | 2. Modify it by factors: - $k_t$: **modification factor for tension steel** from Fig. 4 of IS 456, which depends on $p_t$ and the service stress $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}}$. Less steel and a lower stress give a larger factor (up to 2.0). - $k_c$: **modification factor for compression steel** from Fig. 5, based on $p_c = 100A_{sc}/bd$ (maximum 1.5). - $k_f$: **reduction factor for flanged beams** from Fig. 6, based on the ratio $b_w/b_f$ only (0.8 when $b_w/b_f \le 0.3$). For flanged beams, $p_t$ and $p_c$ for Figs. 4 and 5 are worked out on the area $b_f d$. 3. For spans above 10 m (other than cantilevers), the basic value is multiplied by $10/\text{span (m)}$; for cantilevers longer than 10 m, deflection must be calculated. 4. The **permissible ratio** is $$ \left(\frac{l}{d}\right)_{perm} = \left(\frac{l}{d}\right)_{basic} \times k_t \times k_c \times k_f $$ 5. Compare it with the **actual** $l/d$. If actual $\le$ permissible, deflection is satisfied; if not, increase $d$ (or the steel) and check again. #### Notes - For two-way slabs, the check is made on the shorter span. - This method is a conservative approximation of Annex C (the detailed method, which adds short-term, creep and shrinkage deflections). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the empirical method, deflection is controlled by A. Keeping span/effective depth within a permitted value B. Limiting bar diameter C. Calculating creep and shrinkage in full D. Limiting concrete cover Answer: A. If l/d is not more than permitted, deflection limits are satisfied. (Key point: Idea) 2. Basic l/d values for cantilever, simply supported and continuous members are A. 10, 20 and 30 B. 26, 20 and 7 C. 7, 20 and 26 D. 20, 7 and 26 Answer: C. IS 456 gives 7, 20 and 26 respectively. (Key point: Basic ratios) 3. The tension steel factor kt depends on A. The span only B. pt and service stress fs = 0.58 fy Ast,req/Ast,prov C. The bar diameter only D. fck only Answer: B. More steel and lower stress give a larger kt (up to 2.0). (Key point: Factor kt) 4. The permissible l/d is A. kt × kc only B. Basic l/d + kt + kc + kf C. Basic l/d × kt × kc × kf D. Basic l/d / kt Answer: C. Modification factors multiply the basic ratio; compare with the actual l/d. 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