---
title: "Beam reinforcement for given loads, 6 m span"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:beam-reinforcement-for-given-loads-6"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-reinforcement-for-given-loads-6
site: BCT Engineering
---

# Beam reinforcement for given loads, 6 m span

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

- Asked in: 2067 Asar (old course), 12 marks
- Video: [Reinforcement Arrangement of Beam - Reinforcement Design of Beam](https://www.youtube.com/watch?v=Gee_U5yvT5Q) — Engr Sami 
- 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-reinforcement-for-given-loads-6
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

Design the reinforcement required for a simple rectangular beam having effective span length of 6 m. The beam is carrying 8 kN/m load from 120 mm thick slab. Consider the width of beam 250 mm and overall depth of beam to be 450 mm. For loading calculation, consider live load on floor: 5 kN/m, floor finish: 3 kN/m, partition wall: 10 kN/m. M20 concrete and Fe415 are used.

## Short answer

Doubly reinforced beam required as Mu (194.5 kNm) > Mulim (116 kNm). Provide 4-25 mm bars (1963 mm²) in tension and 4-16 mm bars (804 mm²) in compression. Effective depth 410 mm. Deflection check satisfied with allowable L/d 23.1 > actual 14.6.

## Answer

**Given:** $L=6$ m, $b=250$ mm, $D=450$ mm, M20, Fe415. Assume effective cover 40 mm, $d=450-40=410$ mm, $d'=40$ mm.

**Step 1: Loads**

| Item | kN/m |
|---|---|
| Load from 120 mm slab | 8 |
| Live load | 5 |
| Floor finish | 3 |
| Partition wall | 10 |
| Self weight $0.25\times0.45\times25$ | 2.812 |
| **Total (service)** | **28.812** |

$$
w_u=1.5\times28.812=43.22\ \text{kN/m},\qquad M_u=\frac{w_uL^2}{8}=\frac{43.22\times6^2}{8}=194.48\ \text{kN·m}
$$

**Step 2: Limiting moment**

$$
M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times250\times410^2=115.96\ \text{kN·m}<M_u
$$

The beam must be **doubly reinforced**.

**Doubly reinforced design**

$$
M_{u2}=M_u-M_{u,lim}=194.48-115.96=78.53\ \text{kN·m}
$$

Limiting depth $x_{u,max}=0.48d=196.8$ mm. Strain in compression steel (similar triangles):

$$
\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{40}{196.8}\right)=0.00279
$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=352.1$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$
A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{78.53\times10^6}{(352.1-8.92)(410-40)}=619\ \text{mm}^2
$$

$$
A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=981\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=588\ \text{mm}^2
$$

$$
A_{st}=A_{st1}+A_{st2}=1569\ \text{mm}^2
$$

**Provide:** tension 4-25 mm dia ($1963$ mm$^2$, two layers) and compression 4-16 mm dia ($804$ mm$^2$). Use 8 mm stirrups at suitable spacing ($\le16\phi_c$, 300 mm).

**Deflection check:** $p_t=1.92\%$, $f_s=0.58f_y\dfrac{A_{st,req}}{A_{st,prov}}=192.4$ N/mm$^2$, $k_t\approx0.98$ (Fig. 4); $p_c=0.78\%$, $k_c=1.18$:

$$
\left(\frac{L}{d}\right)_{allow}=20\times0.98\times1.18=23.1>\frac{6000}{410}=14.63
$$

Safe.

**Answer:** $A_{st}=1569$ mm$^2$ (4-25 mm) and $A_{sc}=619$ mm$^2$ (4-16 mm).

## Check yourself

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

1. How is the load assembled?
   A. Only live load
   B. Self weight excluded
   C. Slab load, live, finish, partition and self weight are summed then x1.5
   D. Only partition wall
   Answer: C. w = 28.812 kN/m, wu = 43.22 kN/m. (Key point: Load total)

2. Why?
   A. Mu < Mu,lim
   B. Span is 6 m
   C. Cover large
   D. Mu = 194.48 > Mu,lim = 115.96 kN·m for fixed size
   Answer: D. Excess moment Mu2 = 78.53 kN·m requires Asc. (Key point: Doubly reinforced)

3. Which formula gives Asc?
   A. Mu2/((fsc - fcc)(d - d'))
   B. Mu2/(0.87fy d)
   C. Mu/(fy b)
   D. Mu2/(fck b d)
   Answer: A. Couple of compression steel with net stress fsc - fcc and lever arm d - d'. (Key point: Asc)

4. kt 1.50 and kc 1.18 give allowable 35.5. Result?
   A. Unsafe
   B. Actual 14.63 < 35.5, safe
   C. Equal
   D. Cannot decide
   Answer: B. Allowable L/d exceeds actual. (Key point: Deflection)

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