Question 7 of 8: Composite wood/steel beam — transformed section

Question 7: Composite wood/steel beam — transformed section (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) $\sigma_{wood,max}$, $\sigma_{steel,max}$; (b) $w$ at first failure; (c) whether bonding is required.

Approach. Transform the steel plate to an equivalent width of wood using the modular ratio $n=E_s/E_w$, find the transformed centroid and second moment of area, then apply the flexure formula (scaling the transformed-section stress by $n$ wherever it is evaluated in real steel).

1. Modular ratio and transformed width. $n=E_s/E_w=210/10=21$, so the steel plate transforms to an equivalent wood width $b_{s,t}=21(160)=3360$ mm (thickness unchanged, 10 mm).
2. Transformed centroid (measured from the bottom of the plate, $y$ up). Wood centroid $y_w=10+120=130$ mm, area $38\,400\ \text{mm}^2$; transformed-steel centroid $y_{s,t}=5$ mm, area $33\,600\ \text{mm}^2$: $$\bar y=\frac{38\,400(130)+33\,600(5)}{38\,400+33\,600}=71.67\ \text{mm from the bottom}$$
3. Transformed moment of inertia. $$I_{tr}=\left[\frac{160(240)^3}{12}+38\,400(130-71.67)^2\right]+\left[\frac{3360(10)^3}{12}+33\,600(5-71.67)^2\right]=4.646\times10^{8}\ \text{mm}^4$$
4. Design moment. $M_{max}=wL^2/8=10(4)^2/8=20$ kN·m.
5. Part (a): extreme stresses. Top of wood ($y=250$ mm, compression) uses the transformed stress directly ($n=1$ for wood); bottom of steel ($y=0$, tension) is scaled by $n=21$: $$\sigma_{wood,top}=\frac{M(250-\bar y)}{I_{tr}}=\frac{20\times10^6(178.33)}{4.646\times10^8}=\boxed{7.68\ \text{MPa (compression)}}$$ $$\sigma_{steel,bottom}=\frac{nM(0-\bar y)}{I_{tr}}=\frac{21(20\times10^6)(-71.67)}{4.646\times10^8}=\boxed{-64.8\ \text{MPa}\ (64.8\ \text{MPa tension})}$$
6. Part (b): failure load. Both stresses scale linearly with $w$; find the $w$ that brings each material to its own allowable and take the smaller: $$w_{wood}=10\times\frac{8}{7.68}=10.42\ \text{kN/m},\qquad w_{steel}=10\times\frac{240}{64.8}=37.04\ \text{kN/m}$$ $$\boxed{w_{fail}=10.42\ \text{kN/m}\ \text{(wood governs)}}$$ At the design load of 10 kN/m the wood is already at 96% of its allowable stress while the steel is at only 27% — the plate is oversized relative to the wood's capacity.
7. Part (c): is bonding required? Composite action (a single transformed section, one neutral axis, the flexure formula used above) is valid only if no slip occurs at the wood–steel interface, which requires the interface to transmit the horizontal shear flow $q=VQ/I_{tr}$. At the supports $V_{max}=wL/2=20$ kN and $Q_{interface}=A_{s,t}(\bar y-5)=33\,600(66.67)=2.24\times10^6\ \text{mm}^3$, giving $$q=\frac{20\,000(2.24\times10^6)}{4.646\times10^8}=96.4\ \text{N/mm}$$ Yes, the plate must be bonded (glued, or otherwise mechanically connected with sufficient capacity to resist ≈96 N/mm of shear flow along the span) — without it, the wood and steel would each bend about their own separate neutral axes (non-composite, "stacked" behaviour), each carrying far less moment than the bonded section computed above, and the stresses found in parts (a)/(b) would not apply.