A wooden beam, 250mm x 150mm has a steel strap 10mm x 150mm fixed at the top and the bottom. The beam is subjected to a bending moment of 5kNm around the horizontal axis. - $E_s = 200 GPa \ \ \ and \ \ \ E_w = 20 GPa$

$M = 5kNm = 5 × 10^6 Nmm \ \ \ E_s = 200 GPa$

$E_w = 20 GPa$

Converting the flitched beam to an equivalent section of wood.

The modular ratio is given by m = $\frac{E_s}{E_w} = \frac{200}{20} = 10$

The dimension of steel parallel to the neutral axis = 150mm

∴Equivalent Dimension=150 × m = 150 × 10 = 1500mm

The new equivalent shape of the wood is that of an I – section. It has a web and a flange.

The moment of inertia of the equivalent section of the wood wrt the neutral axis is given by:

$$I_W = I_{Web} + 2I_{Flange}$$

$$I_W = [\frac{bd^3}{12}]_{Web} + 2[\frac{bd^3}{12} + Ay^2]_{Flange}$$

$$I_W = [\frac{[150] × 250^3}{12}]_{Web} + 2 [\frac{[1500] × 10^3}{12} + (15000)(130)^2]_{Flange}$$

$I_W = 702.562 × 10^6 mm^4$

From the Flexural equation, we have: $\frac{f}{y} = \frac{M}{I}$

For the Equivalent section of wood: $\frac{f_w}{y} = \frac{M}{I_w}$

$f_w = \frac{(5 × 10^6)}{702.562 × 10^6}$

$f_w = 0.889 N/mm^2$

Since m = 10, the stress in steel at the same layer is given by:

$f_s = 10 × 0.889$

$f_s = 8.89 N/mm^2$

The extreme fibers of steel are at a distance of y = 135. Assuming that the variation of stress is of a linear nature, we have:

$$[f_s]_{y=135} = \frac{[f_s]_{y = 125}}{125} × 135$$

$$[f_s]_{y=135} = \frac{8.89}{125} × 135$$

$$[f_s]_{y=135} = 9.6 N/mm^2$$