---
title: "Limit state design of beam, 45 kN/m"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:limit-state-design-of-beam-45"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/limit-state-design-of-beam-45
site: BCT Engineering
---

# Limit state design of beam, 45 kN/m

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

- Asked in: 2065 Kartik (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/limit-state-design-of-beam-45
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

Design a simply supported reinforced concrete beam of rectangular section at limit state of collapse in flexure and shear. The beam is subjected to a design load of 45 kN/m (including self weight). Clear span of the beam is 6 m. Take M20 grade of concrete and Fe415 steel as the materials of beam.

## Short answer

Singly reinforced beam 300×600 mm with effective depth 550 mm. Factored moment 223 kNm < Mulim 250 kNm. Provide 5-20 mm bars (1571 mm²) for Ast=1355 mm². Shear design: nominal shear stress 0.67 N/mm² > τc 0.61 N/mm²; provide minimum 8 mm 2-legged stirrups at 300 mm c/c throughout.

## Answer

**Given:** $w=45$ kN/m (design load, i.e. factored, including self weight), clear span 6 m, M20, Fe415. The 45 kN/m is taken as the design (factored) load.

**Assumptions:** support width 300 mm; $b=300$ mm, $D=600$ mm, effective cover 50 mm, $d=550$ mm.

**Step 1: Effective span:** least of $c/c=6+0.3=6.3$ m and clear $+d=6+0.55=6.55$ m, so $L_{eff}=6.30$ m.

**Step 2: Flexure**

$$
M_u=\frac{w_uL^2}{8}=\frac{45\times6.30^2}{8}=223.26\ \text{kN·m}
$$

$$
M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times550^2=250.40\ \text{kN·m}>M_u
$$

so the section is singly reinforced.

$$
M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)\ \Rightarrow\ A_{st}=1355\ \text{mm}^2
$$

Provide 5-20 mm dia bars ($1571$ mm$^2$); minimum steel $0.85bd/f_y=338$ mm$^2$ OK; $x_u=262.6$ mm $<0.48d=264$ mm OK.

**Step 3: Shear**

Support shear (at face) $V_u=\dfrac{w_u\times6}{2}=135.0$ kN. At the critical section $d$ from the face:

$$
V_{u,d}=135.0-45\times0.55=110.2\ \text{kN},\qquad \tau_v=\frac{110.2\times10^3}{300\times550}=0.67\ \text{N/mm}^2<2.8
$$

$p_t=\dfrac{100\times1571}{300\times550}=0.95\%$ gives $\tau_c=0.61$ N/mm$^2$ (IS 456 Table 19). As $\tau_v>0.61$:

$$
V_{us}=V_{u,d}-\tau_cbd=110.2-100.4=9.9\ \text{kN}
$$

Using 8 mm 2-legged stirrups ($A_{sv}=100.5$ mm$^2$):

$$
s_v=\frac{0.87f_yA_{sv}d}{V_{us}}=\frac{0.87\times415\times100.5\times550}{9.9\times10^3}=2027\ \text{mm}
$$

Limits: $\le0.75d=412$ mm, $\le300$ mm, $\le\dfrac{0.87f_yA_{sv}}{0.4b}=302$ mm. The spacing required for $V_{us}$ is much larger than these limits, so the maximum spacing (minimum shear reinforcement) governs. Provide **8 mm 2-legged stirrups at 300 mm c/c** throughout the span.

**Answer:** Beam 300 x 600 mm, $A_{st}=1355$ mm$^2$ (5-20 mm), stirrups 8 mm 2L @ 300 mm c/c.

## Check yourself

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

1. How is 45 kN/m interpreted?
   A. Service load needing x1.5
   B. Live load only
   C. Dead load only
   D. As factored load already including self weight, so no 1.5 factor is applied
   Answer: D. The problem says design load including self weight, so wu = 45. (Key point: Meaning of design load)

2. Why singly reinforced?
   A. Mu = 223.26 kN·m < Mu,lim = 250.40 kN·m
   B. Mu > Mu,lim
   C. Because shear is low
   D. Span is 6 m
   Answer: A. Section capacity suffices without compression steel. (Key point: Singly reinforced)

3. What is Vu,d?
   A. Support shear itself
   B. wuL/4
   C. Support shear less wu x d
   D. Zero
   Answer: C. 135 - 45 x 0.55 = 110.2 kN at d from the face. (Key point: Critical shear)

## Related in Design of RCC Structure

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- [Beam flexure and deflection, 4.5 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-flexure-and-deflection-4-5) — Design of RCC Structure, chapter 4, asked 2×
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