--- title: "Shear design of T-beam on 300x600 section" subject: "Design of RCC Structure" chapter: 5 key: "semester_7/design_of_rcc_structure:shear-design-of-t-beam-on" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/shear-design-of-t-beam-on site: BCT Engineering --- # Shear design of T-beam on 300x600 section Design of RCC Structure · Chapter 5: Design for Shear and Torsion - Asked in: 2078 Bhadra, 8 marks - Video: [Design for Shear Reinforcement in Beams - Problems](https://www.youtube.com/watch?v=9esi2xeKCNM) — Stan Academy - 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/shear-design-of-t-beam-on - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5 ## Question A simply supported normal T beam of 6 m clear span with service load of 40 kN/m. It is reinforced with 4 numbers of 20 mm diameter bars at support. Design the shear reinforcement near the support considering the shear contribution of 2 numbers of 20 mm dia bars near support. The beam has cross section of 300 mm x 600 mm overall. Use M20 concrete and Fe 415 steel. ## Short answer At d from support, Vud=147 kN and Vus=77.2 kN. Bent-up bars give 160.4 kN but are limited to 38.6 kN; balance is 38.6 kN. Provide 8 mm 2-legged stirrups at 300 mm c/c near the support. ## Answer **Assumptions:** the 40 kN/m service load includes self weight; web $b_w=300$ mm, $D=600$ mm, effective cover 50 mm, $d=550$ mm. Of the four 20 mm bars, 2 are bent up at 45$^\circ$ near the support and 2 continue straight. M20, Fe415. For a T-beam, shear is calculated on the web width. **Step 1: Design shear** $$ w_u=1.5\times40=60.0\ \text{kN/m},\quad V_u=\frac{60.0\times6}{2}=180.0\ \text{kN} $$ At $d$ from the face: $V_{u,d}=180.0-60.0\times0.55=147.0$ kN, so $\tau_v=\dfrac{147.0\times10^3}{300\times550}=0.891$ N/mm$^2<2.8$, OK. **Step 2: Concrete strength.** Straight bars continuing to support: 2-20, $A_{st}=628.3$ mm$^2$: $$ p_t=\frac{100\times628.3}{300\times550}=0.381\%,\quad\tau_c=0.423\ \text{N/mm}^2,\quad V_c=\tau_cb_wd=69.8\ \text{kN} $$ $V_{us}=147.0-69.8=77.2$ kN is to be carried by shear reinforcement. **Step 3: Bent-up bars (2-20 at 45$^\circ$)** $$ V_{us,bent}=0.87f_yA_{sb}\sin\alpha=0.87\times415\times628.3\times0.7071=160.4\ \text{kN} $$ As bent bars may carry only half of the shear reinforcement load (cl. 40.4), the usable value is $0.5\times77.2=38.6$ kN. **Step 4: Stirrups for the balance** $=77.2-38.6=38.6$ kN. With 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$): $$ s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.5\times550}{38.6\times10^3}=517\ \text{mm} $$ Maximum spacing limits: $\dfrac{0.87f_yA_{sv}}{0.4b}=302$ mm, $0.75d=412$ mm and 300 mm, so the limit of 300 mm governs. **Answer:** Provide 8 mm 2-legged vertical stirrups at 300 mm c/c near the support, in addition to the 2 bars bent up at 45$^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which width is used for shear in a T-beam? A. Flange width B. Average of flange and web C. Overall depth D. Web width bw = 300 mm Answer: D. Shear is resisted by the web. (Key point: Web width) 2. What Ast gives tau_c? A. 2-20 mm straight bars (628.3 mm2), pt = 0.381% B. All four bars (1256.6 mm2) C. Only the bent bars D. 1.2% Answer: A. The two bent bars no longer act as tension steel at the support, so tau_c = 0.423. (Key point: Straight bars) 3. How much shear are the bent bars credited with? A. 160.4 kN B. 38.6 kN, half of Vus = 77.2 kN C. 77.2 kN D. None Answer: B. The 50% limit (cl. 40.4) applies even though 160.4 kN is available. (Key point: Bent-up credit) 4. Why is 300 mm c/c provided though 517 mm is calculated? A. 517 mm exceeds beam depth B. Fe415 needs 300 mm C. Maximum spacing limits (300 mm) govern D. Bent bars need it Answer: C. Spacing must not exceed the least of 302, 412 and 300 mm. 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