---
title: "Design beam 300x700 with shear, 7 m span"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:design-beam-300x700-with-shear-7"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-beam-300x700-with-shear-7
site: BCT Engineering
---

# Design beam 300x700 with shear, 7 m span

Design of RCC Structure · Chapter 4: Design of beams: Behavior in Flexure

- Asked in: 2066 Bhadra (old course), 15 marks
- Video: [Design of Singly Reinforced Beam | Limit State Method | RCC structure | Mumbai University | IOE, TU](https://www.youtube.com/watch?v=Zop0L689DO8) — Parash Joshi - Civil Construction and Tutor
- 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-beam-300x700-with-shear-7
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

Design a simply supported rectangular beam with the effective span of 7 m. The size of beam is required to be limited to 300 mm x 700 mm. Design shear reinforcement also. Take a live load of 70 kN/m. Use M20 concrete and Fe415 grade steel.

## Short answer

Beam is doubly reinforced due to Mu (691 kNm) > Mulim (350 kNm). Provide 8-25 mm bars (3927 mm²) in tension and 6-20 mm bars (1885 mm²) in compression. Shear reinforcement: 10 mm 2-legged stirrups at 240 mm c/c near supports for 2.13 m, then 300 mm c/c in central zone.

## Answer

**Given:** $L=7$ m (effective), $b=300$ mm, $D=700$ mm, live load 70 kN/m, M20, Fe415. Assume effective cover 50 mm: $d=650$ mm, $d'=50$ mm. Support width assumed 300 mm.

**Step 1: Loads and moment**

$$
\begin{aligned}
\text{Self weight}&=0.30\times0.70\times25=5.25\ \text{kN/m},\quad w=70+5.25=75.25\ \text{kN/m}\\
w_u&=1.5w=112.88\ \text{kN/m},\qquad M_u=\frac{w_uL^2}{8}=\frac{112.88\times7^2}{8}=691.36\ \text{kN·m}
\end{aligned}
$$

**Step 2: Limiting moment**

$$
M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times650^2=349.74\ \text{kN·m}<M_u
$$

The beam is **doubly reinforced**.

**Doubly reinforced design**

$$
M_{u2}=M_u-M_{u,lim}=691.36-349.74=341.62\ \text{kN·m}
$$

Limiting depth $x_{u,max}=0.48d=312.0$ mm. Strain in compression steel (similar triangles):

$$
\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{312.0}\right)=0.00294
$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=353.4$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$
A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{341.62\times10^6}{(353.4-8.92)(650-50)}=1653\ \text{mm}^2
$$

$$
A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1867\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=1577\ \text{mm}^2
$$

$$
A_{st}=A_{st1}+A_{st2}=3444\ \text{mm}^2
$$

**Provide:** tension 8-25 mm dia ($3927$ mm$^2$, in two layers), compression 6-20 mm dia ($1885$ mm$^2$).

**Step 3: Shear design**

Support shear $V_u=\dfrac{w_uL}{2}=395.1$ kN. Critical section at $d$ from the face of support, i.e. $0.15+0.65=0.80$ m from the support centre:

$$
V_{u,d}=w_u\left(\frac{L}{2}-0.80\right)=304.8\ \text{kN}
$$

$$
\tau_v=\frac{V_{u,d}}{bd}=\frac{304.8\times10^3}{300\times650}=1.56\ \text{N/mm}^2<\tau_{c,max}=2.8\ \text{N/mm}^2\ \text{(OK)}
$$

Taking all tension bars to the support, $p_t=\dfrac{100\times3927}{300\times650}=2.01\%$, so $\tau_c=0.79$ N/mm$^2$ (IS 456 Table 19). As $\tau_v>\tau_c$, shear reinforcement is required:

$$
V_{us}=V_{u,d}-\tau_cbd=304.8-154.3=150.5\ \text{kN}
$$

Using 10 mm 2-legged stirrups ($A_{sv}=157$ mm$^2$):

$$
s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times157\times650}{150.5\times10^3}=245\ \text{mm}
$$

Limits: $s_v\le0.75d=487$ mm, 300 mm, and $\dfrac{0.87f_yA_{sv}}{0.4b}=473$ mm. Provide **10 mm 2-legged stirrups at 240 mm c/c** near the supports for a length of about 2.13 m (where $V_u>\tau_cbd$). In the remaining central zone provide the minimum: 10 mm at 300 mm c/c. Compression steel is tied by these stirrups (spacing $\le 16\phi$ of compression bar).

**Answer:** $A_{st}=3444$ mm$^2$ (8-25 mm), $A_{sc}=1653$ mm$^2$ (6-20 mm); stirrups 10 mm 2L at 240 mm c/c.

## Check yourself

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

1. Why is compression steel needed?
   A. Mu is smaller than Mu,lim
   B. Shear is high
   C. Mu = 691.36 kN·m exceeds Mu,lim = 349.74 kN·m for the fixed 300 x 700 size
   D. Span is 7 m
   Answer: C. Size is limited, so excess moment Mu2 is resisted by compression steel. (Key point: Doubly reinforced)

2. Where is design shear taken?
   A. At the support centre
   B. At midspan
   C. At 2d from the face
   D. At distance d from the face of the support
   Answer: D. Shear is checked at d from the face (0.15 + 0.65 = 0.80 m from support centre) and Vu,d = 304.8 kN. (Key point: Critical shear section)

3. Why are stirrups designed for Vus = Vu,d - tau_c b d?
   A. Concrete carries tau_c b d; stirrups carry the balance
   B. Stirrups carry whole Vu
   C. Concrete carries none
   D. Vus is moment
   Answer: A. tau_v = 1.56 > tau_c = 0.79, so steel takes Vus = 150.5 kN via sv = 0.87fyAsv d/Vus. (Key point: Shear steel)

4. What limits stirrup spacing?
   A. sv <= 2d only
   B. sv <= 0.75d, 300 mm and 0.87fyAsv/(0.4b)
   C. sv <= 600 mm
   D. No limit
   Answer: B. Computed 245 mm is within limits, so 240 mm c/c is provided near supports. (Key point: Spacing limits)

## Related in Design of RCC Structure

- [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4
- [Shear reinforcement, 300x500 beam, 370 kN](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Ashear-reinforcement-300x500-beam-370-kn) — Design of RCC Structure, chapter 5
- [Shear design for 230x450 beam, 5 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Ashear-design-for-230x450-beam-5) — Design of RCC Structure, chapter 5
- [Beam reinforcement for given loads, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-reinforcement-for-given-loads-6) — Design of RCC Structure, chapter 4
- [Design T-beam with 150 mm slab](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-t-beam-with-150-mm) — Design of RCC Structure, chapter 4
- [Bent-up bars shear, 550 mm effective depth](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abent-up-bars-shear-550-mm) — Design of RCC Structure, chapter 5, asked 3×

## Related in other subjects

- [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7
- [Design of a simply supported timber beam](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Adesign-timber-floor-beam) — Design of Timber and Masonry Structures, chapter 3
- [Beam 6 m span with 100 kN secondary beam load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-20knm-100kn-midspan) — Design of Steel and Timber Structure, chapter 7
- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7