--- title: "Design simply supported beam, 6 m span" subject: "Design of RCC Structure" chapter: 4 key: "semester_7/design_of_rcc_structure:design-simply-supported-beam-6-m" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-simply-supported-beam-6-m site: BCT Engineering --- # Design simply supported beam, 6 m span Design of RCC Structure · Chapter 4: Design of beams: Behavior in Flexure - Asked in: 2068 Baisakh (old course), 15 marks - Video: [Design of simply supported beam](https://www.youtube.com/watch?v=oPUImjzyu8k) — Parag Pal - 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/design-simply-supported-beam-6-m - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4 ## Question A beam of 6 m span is simply supported and carrying 24 kN/m live load and 3 kN/m dead loads excluding self weight. The beam is made of M20 concrete and Fe415 steel. Design the beam. Shear design is not required. ## Short answer Beam size 300 x 600 mm, singly reinforced. Required tension steel 1275 mm² (5-20 mm). Limiting moment 250.4 kN·m > 212.62 kN·m; deflection check safe (L/d = 10.9 < 23.8). ## Answer **Assumptions:** effective span 6 m; $b=300$ mm, $D=600$ mm ($\approx L/10$), effective cover 50 mm, so $d=550$ mm. M20, Fe415. Live load 24 kN/m and dead load 3 kN/m (excluding self weight). **Step 1: Loads** $$ \begin{aligned} \text{Self weight}&=0.30\times0.60\times25=4.50\ \text{kN/m}\\ w&=24+3+4.50=31.50\ \text{kN/m},\qquad w_u=1.5w=47.25\ \text{kN/m} \end{aligned} $$ **Step 2: Moment** $$ M_u=\frac{w_uL^2}{8}=\frac{47.25\times6^2}{8}=212.62\ \text{kN·m} $$ **Step 3: Check depth for a singly reinforced section** $$ M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times550^2=250.40\ \text{kN·m}>M_u $$ A singly reinforced section is adequate ($d_{req}=\sqrt{M_u/(0.138f_{ck}b)}=507$ mm $<550$ mm). **Step 4: Tension steel** $$ M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=1275\ \text{mm}^2 $$ Provide 5-20 mm dia bars, $A_{st,prov}=1571$ mm$^2$ ($p_t=0.95\%$). Minimum steel $0.85bd/f_y=338$ mm$^2$ is satisfied; even with the provided steel, $x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b}=262.6$ mm $<0.48d=264$ mm. **Step 5: Deflection check** $$ f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=195.4\ \text{N/mm}^2,\quad k_t\approx1.19\ \text{(Fig. 4)},\quad \left(\frac{L}{d}\right)_{allow}=20k_t=23.8>\frac{6000}{550}=10.91 $$ The beam is **safe** in deflection. Provide 2-12 mm hanger bars and 8 mm stirrups (shear design not required). **Answer:** Beam 300 x 600 mm with 5-20 mm bars at the bottom ($A_{st}=1275$ mm$^2$ required). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is beam self weight added to the loads? A. It is a live load B. It is ignored in limit state C. It is a dead load not included in the given 3 kN/m D. It is already in the 24 kN/m Answer: C. Loads are stated as excluding self weight, so 0.3 x 0.6 x 25 = 4.5 kN/m is added. (Key point: Why self weight) 2. Why is a singly reinforced section adequate? A. Mu is greater than Mu,lim B. Because Fe415 is used C. Because span is 6 m D. Mu = 212.62 kN·m is less than Mu,lim = 250.40 kN·m Answer: D. When Mu <= Mu,lim, tension steel alone suffices; d required (507 mm) < d provided (550 mm). (Key point: Singly reinforced decision) 3. How is Ast found for singly reinforced section? A. Solve Mu = 0.87 fy Ast d (1 - Ast fy/(b d fck)) B. Ast = Mu/(fy d) C. Ast = 0.85 b d/fy D. Ast = 0.138 fck b d Answer: A. Standard IS 456 Annex G flexure formula solved for Ast; minimum steel 0.85bd/fy is then checked. (Key point: Ast equation) 4. Which controls (L/d)allow = 23.1 vs actual 10.91? A. Actual exceeds allowable, so unsafe B. Actual L/d is less than allowable, so safe C. They are unrelated D. Allowable is 7 Answer: B. Basic 20 x kt 1.15 = 23.1 is greater than 6000/550 = 10.91. 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