Flitched timber–steel beam capacity

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.

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).