---
title: "Flitched timber–steel beam capacity"
subject: "Design of Timber and Masonry Structures"
chapter: 3
key: "semester_5/design_of_timber_and_masonry_structures:flitched-beam-moment-capacity"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/flitched-beam-moment-capacity
site: BCT Engineering
---

# Flitched timber–steel beam capacity

Design of Timber and Masonry Structures · Chapter 3: Structural Elements of Timber Structures

- Asked in: year not given, 8 marks
- Video: [Unit-7; Flitched Beam Design | Timber Structure | Design of Steel & Timber Structure PU PoU TU KU ](https://www.youtube.com/watch?v=jhYaYU0C94M) — Civil Engineering Solution Channel
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/flitched-beam-moment-capacity
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-3

## Question

A flitched beam is made of two timber planks, each 75 mm wide × 250 mm deep, with a steel plate 10 mm × 250 mm deep sandwiched between them (all bolted together to act as a unit). Span is 4.0 m, simply supported. Permissible stresses: timber 10 N/mm², steel 140 N/mm². $E_{timber} = 9000$ N/mm², $E_{steel} = 2 \times 10^5$ N/mm². Find the safe moment of resistance and the safe uniformly distributed load. Compare with the timber alone.

## Short answer

Convert steel to equivalent timber using modular ratio m=22.2. Equivalent I=4.847×10^8 mm⁴. Steel stress governs: MR=24.4 kN·m. For 4 m span, safe UDL=12.2 kN/m; deflection 9.3 mm < 11.1 mm, OK.

## Answer

#### Principle
In a flitched beam the timber and steel strains are equal at each fibre, so stresses are proportional to modulus. Convert steel to an equivalent timber section using the modular ratio

$$
m = \frac{E_s}{E_t} = \frac{2\times 10^5}{9000} = 22.2
$$

#### Equivalent moment of inertia (in timber units)
$$
\begin{aligned}
I_t &= 2 \times \frac{75 \times 250^3}{12} = 1.953 \times 10^8\ \text{mm}^4 \\
I_s &= \frac{10 \times 250^3}{12} = 1.302 \times 10^7\ \text{mm}^4 \\
I_{eq} &= I_t + m I_s = 1.953\times10^8 + 22.2 \times 1.302\times10^7 = 4.847\times10^8\ \text{mm}^4
\end{aligned}
$$

#### Moment of resistance
Extreme fibre distance $y = 125$ mm.

- From timber: $M_t = \dfrac{\sigma_t I_{eq}}{y} = \dfrac{10 \times 4.847\times10^8}{125} = 38.8$ kN·m
- From steel: the steel stress is $m$ times the equivalent timber stress, so $M_s = \dfrac{\sigma_s I_{eq}}{m\,y} = \dfrac{140 \times 4.847\times10^8}{22.2 \times 125} = 24.4$ kN·m

The lower value controls: $M_R = 24.4$ kN·m (steel reaches its limit first).

#### Safe load
$$
M = \frac{wL^2}{8} \;\Rightarrow\; w = \frac{8 \times 24.43}{4^2} = 12.2\ \text{kN/m}
$$

Deflection check (limit $L/360 = 11.1$ mm):
$$
\delta = \frac{5wL^4}{384 E_t I_{eq}} = \frac{5 \times 12.2 \times 4000^4}{384 \times 9000 \times 4.847\times10^8} = 9.3\ \text{mm} < 11.1\ \text{mm}
$$

#### Comparison with timber alone
Timber only: $M = 10 \times 1.953\times10^8 / 125 = 15.6$ kN·m, giving $w = 7.8$ kN/m.

| Beam | $M_R$ (kN·m) | Safe $w$ (kN/m) |
|---|---|---|
| Timber planks only | 15.6 | 7.8 |
| Flitched beam | 24.4 | 12.2 |

The steel plate raises the capacity by about 56 percent.

**Answer: $M_R = 24.4$ kN·m; safe UDL $= 12.2$ kN/m (deflection 9.3 mm, OK).**

## Check yourself

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

1. Why is the steel converted to equivalent timber using m = Es/Et = 22.2?
   A. Steel and timber carry equal stress
   B. Timber carries no load
   C. Strains are equal at each fibre, so stresses are proportional to modulus
   D. Steel and timber have the same E
   Answer: C. The two materials strain equally, so steel stress is m times the timber stress at the same fibre, and the steel is replaced by an equivalent timber area. (Key point: Modular ratio principle)

2. Which expression gives the equivalent moment of inertia in timber units?
   A. I_t + I_s
   B. I_t + m I_s
   C. I_t / m + I_s
   D. m (I_t + I_s)
   Answer: B. I_eq = I_t + m I_s, which is 1.953 x 10^8 + 22.2 x 1.302 x 10^7 = 4.847 x 10^8 mm^4. (Key point: Equivalent moment of inertia)

3. Timber alone allows 38.8 kN.m and steel allows 24.4 kN.m in the flitched beam. What is M_R?
   A. 38.8 kN.m, the larger value
   B. 31.6 kN.m, the average
   C. 63.2 kN.m, the sum
   D. 24.4 kN.m, because steel reaches its limit first
   Answer: D. The lower value controls, so M_R = 24.4 kN.m, where steel stress is m times the equivalent timber stress. (Key point: Governing material)

4. What is the safe UDL, and how does it compare with timber alone (7.8 kN/m)?
   A. 12.2 kN/m, about 56 percent more than timber alone
   B. 7.8 kN/m, no gain
   C. 24.4 kN/m, double
   D. 12.2 kN/m, about 22 percent more
   Answer: A. w = 8M/L^2 = 8 x 24.43/16 = 12.2 kN/m, versus 7.8 kN/m for timber alone, a rise of about 56 percent. (Key point: Safe UDL and comparison)

## Related in Design of Timber and Masonry Structures

- [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
- [Timber beam types and design criteria](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Atimber-beam-types-design-criteria) — Design of Timber and Masonry Structures, chapter 3

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