Question 7 of 8: Composite Wood-Steel Beam

Question 7: Composite Wood-Steel Beam (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. Whether the section survives the 70 kN load in BOTH materials, and the maximum load it can carry.

Approach. Use the transformed-section method (steel widths × $n=E_s/E_w$) to get the composite $\bar y$ and $I_{tr}$, then apply $\sigma_{wood}=My/I_{tr}$ and $\sigma_{steel}=nMy/I_{tr}$ at each extreme fibre.

1. Transform to wood-equivalent, $n=E_s/E_w=20$. The section is symmetric top to bottom about the wood's own mid-depth, so the transformed centroid sits there: $\bar y = 150$ mm from the wood's bottom fibre.
2. Transformed moment of inertia: $$I_{tr} = 2.523\times10^9\ \text{mm}^4$$ with $y_{wood}=\pm150$ mm to the wood's own extreme fibres and $y_{steel}=\pm190$ mm to the outer steel fibres.
3. (a) Applied moment and stress check at the 70 kN load: $$M = \frac{PL}{4} = \frac{70(6)}{4} = 105\ \text{kN}\cdot\text{m}$$ $$\sigma_{wood} = \frac{My_{wood}}{I_{tr}} = \frac{(105\times10^6)(150)}{2.523\times10^9} = \pm6.24\ \text{MPa} \;(<10\ \text{MPa allow})$$ $$\sigma_{steel} = \frac{nMy_{steel}}{I_{tr}} = \frac{20(105\times10^6)(190)}{2.523\times10^9} = \pm158.1\ \text{MPa} \;(<240\ \text{MPa allow})$$ $$\boxed{\text{Both materials stay within their allowable stress} \Rightarrow \text{the beam CAN support the 70 kN load}}$$
4. (b) Maximum load without failure. Find the moment that first exhausts either material's allowable, then convert back through $M=PL/4$: $$M_{wood,allow} = \frac{10\,I_{tr}}{150} = 168.2\ \text{kN}\cdot\text{m}\qquad M_{steel,allow} = \frac{240\,I_{tr}}{20(190)} = 159.4\ \text{kN}\cdot\text{m}$$ Steel governs (lower capacity): $$\boxed{P_{max} = \frac{4\,M_{steel,allow}}{L} = \frac{4(159.4)}{6} = 106.2\ \text{kN}}$$