---
title: "Ultimate moment of L-beam, doubly reinforced"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:ultimate-moment-of-l-beam-doubly"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/ultimate-moment-of-l-beam-doubly
site: BCT Engineering
---

# Ultimate moment of L-beam, doubly reinforced

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

- Asked in: 2074 Chaitra, 10 marks
- Video: [Design Moment Strength of Doubly RC Section Example 1 (1/2)- Reinforced Concrete Design](https://www.youtube.com/watch?v=6t5sWk3mFk4) — structurefree
- 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/ultimate-moment-of-l-beam-doubly
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

A L-beam of effective and flange width as 925 mm, effective depth as 450 mm, depth of flange as 100 mm, breadth of rib as 250 mm is reinforced with 4-20 mm bars as tension reinforcement and 3-16 mm dia bars as compression reinforcement. Find the ultimate moment of resistance of the section at limit state of collapse. Use M20 grade concrete mix and Fe415 grade steel.

## Short answer

Assume neutral axis in flange; force equilibrium yields x_u = 53.20 mm < 100 mm. Ultimate moment of resistance is 192.27 kN·m.

## Answer

**Given:** $b_f=925$ mm, $d=450$ mm, $D_f=100$ mm, $b_w=250$ mm; tension steel 4-20 mm, $A_{st}=1256.6$ mm$^2$; compression steel 3-16 mm, $A_{sc}=603.2$ mm$^2$; M20, Fe415. Effective cover to compression steel is not given; assume $d'=40$ mm.

**Step 1: Neutral axis.** Assume $x_u\le D_f$ so the flange acts as a rectangle of width 925 mm. Force equilibrium $C=T$:

$$
0.36f_{ck}b_fx_u+A_{sc}(f_{sc}-f_{cc})=0.87f_yA_{st}
$$

where $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$ and $f_{sc}$ depends on the strain $\varepsilon_{sc}=0.0035\left(1-\dfrac{d'}{x_u}\right)$.

Total tension $T=0.87\times415\times1256.6=453.7$ kN.

Trial and error (checking $f_{sc}$ from the Fe415 design curve each time; here the strain is in the elastic range, so $f_{sc}=E_s\varepsilon_{sc}$):

$$
x_u=53.20\ \text{mm},\quad \varepsilon_{sc}=0.0035\left(1-\frac{40}{53.20}\right)=0.000868,\quad f_{sc}=2\times10^5\times0.000868=173.7\ \text{N/mm}^2
$$

Check: $C=0.36\times20\times925\times53.20+603.2(173.7-8.92)=354.3+99.4=453.7$ kN $=T$.

$x_u=53.20$ mm $<D_f=100$ mm (flange) and $<x_{u,max}=0.48d=216.0$ mm, so the assumption is correct and the section is under-reinforced.

**Step 2: Ultimate moment** (about the tension steel)

$$
\begin{aligned}
M_u&=0.36f_{ck}b_fx_u(d-0.42x_u)+A_{sc}(f_{sc}-f_{cc})(d-d')\\
&=0.36\times20\times925\times53.20(450-0.42\times53.20)+603.2(173.7-8.92)(450-40)\\
&=151.52+40.75=192.27\ \text{kN·m}
\end{aligned}
$$

**Answer:** $M_u=192.27$ kN·m.

## Check yourself

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

1. Which equation is used to find xu for the doubly reinforced L-beam (NA in flange)?
   A. 0.36 fck bf xu = Asc fsc only
   B. 0.87 fy Asc = 0.36 fck bw xu
   C. 0.36 fck bf xu + Asc(fsc - fcc) = 0.87 fy Ast
   D. xu = 0.48d
   Answer: C. Compression is the concrete block plus net compression steel force; it must equal tension in steel. (Key point: Force equilibrium)

2. Why is xu found by trial and error?
   A. Because bf is unknown
   B. Because Fe415 has no yield point
   C. Because Ast changes with xu
   D. fsc depends on strain, which depends on xu
   Answer: D. Compression steel strain 0.0035(1 - d'/xu) depends on xu, and fsc is read from the stress-strain curve, so the equation is iterated. (Key point: Why trial and error)

3. Which checks confirm the result xu = 53.18 mm?
   A. xu < Df = 100 mm and xu < xu,max = 216 mm
   B. xu > Df and xu > xu,max
   C. xu = bw
   D. Ast > Asc
   Answer: A. NA in the flange validates rectangular treatment, and xu < xu,max confirms under-reinforcement. (Key point: Check of assumption)

4. What are the two parts of Mu?
   A. Tension steel couple plus shear capacity
   B. Concrete compression about the tension steel plus compression steel couple Asc(fsc - fcc)(d - d')
   C. Concrete moment about NA plus Ast x fy
   D. Flange moment only
   Answer: B. Mu = 0.36fck bf xu (d - 0.42xu) + Asc(fsc - fcc)(d - d'), giving 151.47 + 40.80 = 192.28 kN·m. (Key point: Ultimate moment terms)

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