---
title: "Beam steel at midspan, limit state"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:beam-steel-at-midspan-limit-state"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-steel-at-midspan-limit-state
site: BCT Engineering
---

# Beam steel at midspan, limit state

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

- Asked in: 2074 Chaitra, 15 marks
- Video: [Design of a Singly Reinforced Beam By Limit State Method](https://www.youtube.com/watch?v=6IY3x9Fk08o) — 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/beam-steel-at-midspan-limit-state
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

A beam of rectangular section is 300 mm wide and 500 mm deep to the centre of tensile reinforcement. It has to carry a dead load of 45 kN/m excluding its self weight. Find the steel reinforcement required for the mid span section. The beam has a span of 7 m. Use M20 concrete and Fe 415 steel. Effective cover to compression steel = 40 mm. Use limit state method.

## Short answer

Mu=451.34 kNm. Doubly reinforced: tension steel Ast=2907 mm² (6-25 mm), compression steel Asc=1543 mm² (5-20 mm).

## Answer

**Given:** $b=300$ mm, $d=500$ mm, span $L=7$ m, dead load 45 kN/m (excluding self weight), M20, Fe415, effective cover to compression steel $d'=40$ mm. Limit state of collapse in flexure.

**Assumption:** overall depth $D=550$ mm ($d$ plus 50 mm cover); the effective span is taken as 7 m.

**Step 1: Loads**

$$
\begin{aligned}
\text{Self weight}&=0.30\times0.55\times25=4.12\ \text{kN/m}\\
w&=45+4.12=49.12\ \text{kN/m}\\
w_u&=1.5\times49.12=73.69\ \text{kN/m}
\end{aligned}
$$

**Step 2: Design moment at mid-span**

$$
M_u=\frac{w_uL^2}{8}=\frac{73.69\times7^2}{8}=451.34\ \text{kN·m}
$$

**Step 3: Limiting moment of resistance** (Fe415, $x_{u,max}=0.48d$)

$$
M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times300\times500^2=206.95\ \text{kN·m}
$$

Since $M_u>M_{u,lim}$ and the size is fixed, the section is **doubly reinforced**.

**Doubly reinforced design**

$$
M_{u2}=M_u-M_{u,lim}=451.34-206.95=244.39\ \text{kN·m}
$$

Limiting depth $x_{u,max}=0.48d=240.0$ mm. Strain in compression steel (similar triangles):

$$
\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{240.0}\right)=0.00292
$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=353.2$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$
A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{244.39\times10^6}{(353.2-8.92)(500-40)}=1543\ \text{mm}^2
$$

$$
A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1436\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=1471\ \text{mm}^2
$$

$$
A_{st}=A_{st1}+A_{st2}=2907\ \text{mm}^2
$$

**Provide:**
- Tension steel: 6-25 mm dia ($2945$ mm$^2$), arranged in two layers.
- Compression steel: 5-20 mm dia ($1571$ mm$^2$) at $d'=40$ mm.
- Stirrups (8 mm, 2-legged) at spacing not more than $16\phi_c$, 300 mm and $b$ to hold the compression bars (shear design separately).

**Checks:** minimum steel $0.85bd/f_y=307$ mm$^2$ OK; $p_t=1.96\%<4\%$ and $p_c=1.05\%$ OK.

**Answer:** Mid-span $A_{st}=2907$ mm$^2$ (6-25 mm) and $A_{sc}=1543$ mm$^2$ (5-20 mm).

## Check yourself

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

1. How is the design load wu obtained?
   A. The 45 kN/m is multiplied by 1.5 and self weight is ignored
   B. Self weight is multiplied by 1.5 only
   C. wu = w / 1.5
   D. Self weight (b x D x 25) is added to the 45 kN/m, then multiplied by 1.5
   Answer: D. Self weight is part of the dead load, so w = 45 + 4.12 and wu = 1.5w = 73.69 kN/m. (Key point: Self weight and load)

2. Mu = 451.34 kN·m and Mu,lim = 206.95 kN·m. What follows?
   A. Compression steel is required for Mu2 = Mu - Mu,lim
   B. Singly reinforced steel is enough with more bars
   C. The section is over-reinforced, so reduce the load
   D. Shear reinforcement replaces compression steel
   Answer: A. With size fixed and Mu > Mu,lim, the excess moment is taken by Asc and extra tension steel. (Key point: Why doubly reinforced)

3. How is total Ast composed?
   A. Ast = Ast1 only
   B. Ast = Ast1 (balances limiting concrete force) + Ast2 (balances compression steel)
   C. Ast = Asc + Ast1
   D. Ast = Ast2 - Ast1
   Answer: B. Ast1 = 0.36fck b xu,max/0.87fy and Ast2 = Asc(fsc-fcc)/0.87fy; their sum is the total tension steel. (Key point: Total tension steel)

4. How is fsc found?
   A. fsc = 0.87 fy always
   B. fsc = 0.446 fck
   C. Strain 0.0035(1 - d'/xu,max) is read on the Fe415 design stress-strain curve
   D. fsc = 0.0035 x Es regardless of d'
   Answer: C. Strain at the compression steel level from similar triangles gives fsc from the design curve (353.2 here); 0.87fy is reached only at larger strain. (Key point: Compression steel stress)

## Related in Design of RCC Structure

- [Limit state design of beam, 45 kN/m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Alimit-state-design-of-beam-45) — Design of RCC Structure, chapter 4
- [WSM tensile steel for 230x425 beam](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Awsm-tensile-steel-for-230x425-beam) — Design of RCC Structure, chapter 2
- [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
- [Beam design, 250x475 section](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-design-250x475-section) — Design of RCC Structure, chapter 4, asked 2×
- [Shear reinforcement of T-beam, 250 kN](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Ashear-reinforcement-of-t-beam-250) — Design of RCC Structure, chapter 5
- [L-beam reinforcement for given moment](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Al-beam-reinforcement-for-given-moment) — Design of RCC Structure, chapter 4

## Related in other subjects

- [Beam 4 m span, 45 kN/m and 100 kN midspan](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-4m-45knm-100kn) — Design of Steel and Timber Structure, chapter 7
- [Bending stresses in a T-section beam](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Acast-iron-t-beam-bending-stresses) — Strength of Materials, chapter 3
- [Splitting tensile strength of cylinder and cube (500 kN)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Asplitting-tensile-cylinder-cube-500kn) — Concrete Technology and Masonry Structure, chapter 5, asked 2×
- [Cantilever with self-weight: P and moment of resistance](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acantilever-self-weight-t-section) — Strength of Materials, chapter 7