---
title: "Beam with torsion, 4.5 m clear span"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:beam-with-torsion-4-5-m"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-with-torsion-4-5-m
site: BCT Engineering
---

# Beam with torsion, 4.5 m clear span

Design of RCC Structure · Chapter 5: Design for Shear and Torsion

- Asked in: 2079 Baisakh, 10 marks
- Video: [Difference between Shear and Torsion Reinforcement  explained in English](https://www.youtube.com/watch?v=MNFpiXHRLls) — Dreamchasers
- 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/beam-with-torsion-4-5-m
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

A simply supported beam has clear span 4.5 m, support width 200 mm is subjected to imposed load of 35 kN/m. Beam is also subjected to torsional moment 18 kNm, consider M20 concrete and Fe415 steel. Design for bending and shear.

## Short answer

Effective span is 4.7 m; Me1=211.25 kN-m, so Ast=1265 mm2, provide 5-20 mm bars. Add 1-12 mm side face bar on each face and closed 8 mm 2-legged stirrups at 100 mm c/c.

## Answer

**Assumptions:** the imposed load of 35 kN/m and the torsion of 18 kN·m are service values, factored by 1.5. Section $b=300$ mm, $D=600$ mm, effective cover 50 mm, $d=550$ mm. M20, Fe415.

**Step 1: Effective span** (IS 456 cl. 22.2): the lesser of $c/c=4.5+0.2=4.7$ m and clear span $+d=4.5+0.55=5.05$ m, so $L_{eff}=4.70$ m.

**Step 2: Loads**

$$
\begin{aligned}
\text{Self weight}&=0.30\times0.60\times25=4.500\ \text{kN/m}\\
w_u&=1.5(35+4.500)=59.25\ \text{kN/m},\qquad T_u=1.5\times18=27.0\ \text{kN·m}\\
M_u&=\frac{w_uL^2}{8}=\frac{59.25\times4.70^2}{8}=163.60\ \text{kN·m}\\
V_u&=\frac{w_u\times4.5}{2}=133.31\ \text{kN (face)},\quad V_{u,d}=133.31-59.25\times0.55=100.72\ \text{kN}
\end{aligned}
$$

**Step 3: Equivalent forces** (cl. 41.3)

$$
\begin{aligned}
M_t&=T_u\frac{1+D/b}{1.7}=27.0\times\frac{1+600/300}{1.7}=47.65\ \text{kN·m}\\
M_{e1}&=M_u+M_t=163.60+47.65=211.25\ \text{kN·m}\\
V_e&=V_{u,d}+1.6\frac{T_u}{b}=100.72+1.6\times\frac{27.0}{0.3}=244.7\ \text{kN}
\end{aligned}
$$

$\tau_{ve}=\dfrac{244.7\times10^3}{300\times550}=1.483$ N/mm$^2<2.8$, OK. $M_t<M_u$, so there is no $M_{e2}$.

**Step 4: Flexure.** $M_{u,lim}=0.138\times20\times300\times550^2=250.40$ kN·m $>M_{e1}$, so the section is singly reinforced.

$$
M_{e1}=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=1265\ \text{mm}^2
$$

Provide 5-20 mm bars ($1571$ mm$^2$), $p_t=0.95\%$. Minimum steel $0.85bd/f_y=338$ mm$^2$, OK. Provide 2-12 mm bars at the top corners.

**Side face bars:** $D=600$ mm $>450$ mm, so for torsion (cl. 26.5.1.7(d)) provide $0.1\%\times300\times600=180$ mm$^2$ (90 mm$^2$ per face): 1-12 mm bar on each face at mid-depth ($113$ mm$^2$), giving a spacing of about 250 mm $<300$ mm.

**Step 5: Stirrups.** $\tau_c=0.61$ N/mm$^2$; $\tau_{ve}>\tau_c$. With $b_1=200$ mm and $d_1=500$ mm:

$$
\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_{u}}{2.5d_1(0.87f_y)}=0.748+0.223=0.971\ \text{mm}^2/\text{mm}
$$

The code minimum is $\dfrac{(\tau_{ve}-\tau_c)b}{0.87f_y}=0.727$ mm$^2$/mm, so 0.971 governs. For 8 mm 2-legged closed stirrups ($A_{sv}=100.5$ mm$^2$): $s_v=104$ mm; limits $x_1=225$ mm, $(x_1+y_1)/4=188$ mm, 300 mm.

**Answer:** $M_u=163.60$ kN·m; $A_{st}=1265$ mm$^2$ (5-20 mm) for $M_{e1}=211.25$ kN·m; 1-12 mm side face bar on each face; closed 8 mm 2-legged stirrups at 100 mm c/c.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the effective span?
   A. 5.05 m
   B. 4.5 m
   C. The greater, 5.05 m
   D. Least of c/c (4.7 m) and clear + d (5.05 m), so 4.70 m
   Answer: D. IS 456 cl. 22.2 takes the lesser value. (Key point: Effective span)

2. How are the loads obtained?
   A. wu = 1.5(35 + 4.5) = 59.25 kN/m including self weight; Tu = 27 kN.m
   B. wu = 35 kN/m ignoring self weight
   C. wu = 1.5x35 only
   D. Tu = 18 kN.m
   Answer: A. Self weight 0.3x0.6x25 = 4.5 kN/m is added before factoring. (Key point: Factored loads)

3. Which shear is used with Ve?
   A. Vu at the support (133.31 kN)
   B. Vu,d at d from the face (100.72 kN)
   C. Vu at mid-span
   D. Zero
   Answer: B. Ve = 100.72 + 1.6x27/0.3 = 244.7 kN. (Key point: Design shear)

4. Beam is designed for which moment?
   A. Mu = 163.60 only
   B. Mt = 47.65 only
   C. Me1 = Mu + Mt = 211.25 kN.m
   D. Mu - Mt
   Answer: C. Me1 is below Mu,lim = 250.4, so singly reinforced with Ast = 1265 mm2. (Key point: Flexural design)

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