---
title: "Design 200x400 beam for bending, shear, torsion"
subject: "Design of RCC Structure"
chapter: 5
key: "semester_7/design_of_rcc_structure:design-200x400-beam-for-bending-shear"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-200x400-beam-for-bending-shear
site: BCT Engineering
---

# Design 200x400 beam for bending, shear, torsion

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

- Asked in: 2080 Bhadra, 10 marks
- 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-200x400-beam-for-bending-shear
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-5

## Question

Design a rectangular RC beam having cross sectional dimension 200 mm x 400 mm subjected to design bending moment 100 kN-m, design shear force 80 kN (at critical section) and design torsional moment 15 kN-m. Consider M25 grade concrete and Fe415 grade steel.

## Short answer

Equivalent shear is 200 kN, tau_ve=2.857 N/mm2 below 3.1; Me1=126.47 kN-m requires doubly reinforced section. Provide tension 4-20 mm (1225 mm2), compression 3-16 mm (420 mm2), and closed 10 mm 2-legged stirrups at 90 mm c/c.

## Answer

**Given:** $b=200$ mm, $D=400$ mm, $M_u=100$ kN·m, $V_u=80$ kN, $T_u=15$ kN·m, M25, Fe415. Assume effective cover 50 mm: $d=350$ mm. Centre-to-centre distances of the corner bars: $b_1=200-100=100$ mm, $d_1=400-100=300$ mm.

**Step 1: Equivalent shear and moment**

$$
\begin{aligned}
V_e&=V_u+1.6\frac{T_u}{b}=80+1.6\times\frac{15}{0.2}=200.0\ \text{kN},\qquad \tau_{ve}=\frac{200.0\times10^3}{200\times350}=2.857\ \text{N/mm}^2\\
M_t&=T_u\frac{1+D/b}{1.7}=15\times\frac{1+400/200}{1.7}=26.47\ \text{kN·m}\\
M_{e1}&=M_u+M_t=100+26.47=126.47\ \text{kN·m}
\end{aligned}
$$

$\tau_{ve}<\tau_{c,max}=3.1$ N/mm$^2$ (M25), so the section is acceptable. $M_t<M_u$, so there is no $M_{e2}$.

**Step 2: Longitudinal steel.** $M_{u,lim}=0.138f_{ck}bd^2=0.138\times25\times200\times350^2=84.50$ kN·m $<M_{e1}$, so the section is **doubly reinforced** for $M_{e1}$.

**Doubly reinforced design**

$$
M_{u2}=M_{e1}-M_{u,lim}=126.47-84.50=41.97\ \text{kN·m}
$$

Limiting depth $x_{u,max}=0.48d=168.0$ mm. Strain in compression steel (similar triangles):

$$
\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{168.0}\right)=0.00246
$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=344.0$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=11.15$ N/mm$^2$.

$$
A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{41.97\times10^6}{(344.0-11.15)(350-50)}=420\ \text{mm}^2
$$

$$
A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=838\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=387\ \text{mm}^2
$$

$$
A_{st}=A_{st1}+A_{st2}=1225\ \text{mm}^2
$$

Provide tension 4-20 mm ($1257$ mm$^2$) and compression 3-16 mm ($603$ mm$^2$) bars.

**Step 3: Transverse reinforcement.** $p_t=\dfrac{100\times1257}{200\times350}=1.80\%$, $\tau_c=0.79$ N/mm$^2$; $\tau_{ve}>\tau_c$.

$$
\frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1(0.87f_y)}+\frac{V_u}{2.5d_1(0.87f_y)}=1.385+0.295=1.680\ \text{mm}^2/\text{mm}
$$

For 10 mm 2-legged closed stirrups ($A_{sv}=157.1$ mm$^2$): $s_v=93$ mm. Limits: $x_1=125$ mm, $(x_1+y_1)/4=112$ mm, 300 mm.

**Answer:** Tension steel $A_{st}=1225$ mm$^2$ (4-20 mm), compression steel $A_{sc}=420$ mm$^2$ (3-16 mm), closed 10 mm 2-legged stirrups at 90 mm c/c.

## Check yourself

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

1. Why is the section doubly reinforced?
   A. tau_ve exceeds tau_c
   B. Mt exceeds Mu
   C. Torsion always needs compression steel
   D. Mu,lim = 84.5 kN.m is less than Me1 = 126.47 kN.m
   Answer: D. The excess Mu2 = 41.97 kN.m has to be taken by compression steel plus balancing tension steel. (Key point: Why doubly reinforced)

2. How is Asc obtained?
   A. Asc = Mu2/((fsc - fcc)(d - d'))
   B. Asc = Mu2/(0.87 fy d)
   C. Asc = 0.36 fck b xu/0.87fy
   D. Asc = Mu,lim/(fsc(d - d'))
   Answer: A. Asc = 41.97e6/((344 - 11.15)x300) = 420 mm2. (Key point: Compression steel formula)

3. How is total Ast formed?
   A. Ast1 only
   B. Ast = Ast1 (for Mu,lim) + Ast2 (balancing Asc) = 838 + 387
   C. Ast = Asc
   D. Ast1 - Ast2
   Answer: B. Total 1225 mm2, provided as 4-20 mm. (Key point: Total tension steel)

4. Why is the section acceptable though tau_ve = 2.857 is large?
   A. It is below tau_c
   B. Tau_c,max is 2.8 for M25
   C. It is below tau_c,max = 3.1 N/mm2 for M25
   D. Check is ignored for torsion
   Answer: C. M25 has tau_c,max = 3.1 N/mm2, so Ve = 200 kN is allowed. (Key point: Section check)

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