Composite Timber-Steel-Aluminum Beam (Transformed Section): 04-BS-6, May 2018, Q5

Question 5: Composite Timber-Steel-Aluminum 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. Cross-section, bottom to top: steel plate $150\times20\text{ mm}$, timber core $150\times150\text{ mm}$, aluminum plate $150\times20\text{ mm}$ (all $150\text{ mm}$ wide). $M=20\text{ kN}\cdot\text{m}$. $E_{wood}=10\text{ GPa}$, $E_{steel}=200\text{ GPa}$, $E_{al}=70\text{ GPa}$.

Find. The maximum normal stress in each of the three materials, using the transformed-section method.

Composite cross-section: steel plate (bottom), timber core, aluminum plate (top), all 150 mm wide.

Actual bending stress vs. depth: jumps at each material interface (aluminum compression at top, steel tension at bottom).

Approach. Transform the steel and aluminum layers into equivalent wood by scaling their widths by the modular ratio $n=E_i/E_{wood}$, locate the transformed centroid and moment of inertia, then convert the wood-equivalent flexure stress back to the real stress in each material.

Modular ratios. $n_{steel}=E_{steel}/E_{wood}=200/10=20$; $n_{al}=E_{al}/E_{wood}=70/10=7$.
Transformed areas and centroid (y measured from the bottom of the steel). Steel: $A=20\times150\times20=60{,}000\text{ mm}^2$ at $\bar y=10$; wood: $A=150\times150=22{,}500\text{ mm}^2$ at $\bar y=95$; aluminum: $A=7\times150\times20=21{,}000\text{ mm}^2$ at $\bar y=180$. $\bar Y=\dfrac{\sum A_i\bar y_i}{\sum A_i}=\dfrac{6{,}517{,}500}{103{,}500}=\boxed{63.0\text{ mm}}$ (falls inside the timber layer).
Transformed moment of inertia. $I_{tr}=\sum\left[I_i+A_i(\bar y_i-\bar Y)^2\right]=170.4+65.3+288.3=\boxed{523.9\times10^{6}\text{ mm}^4}$ (wood-equivalent).
Part (i) — wood. The farthest wood fibre from the N.A. is at the top of the wood ($y=170\text{ mm}$, $c=170-63.0=107.0\text{ mm}$; the bottom of wood at $c=63.0-20=43.0\text{ mm}$ is closer). $\sigma_{wood}=\dfrac{Mc}{I_{tr}}=\dfrac{20\times10^6\times107.0}{523.9\times10^6}=\boxed{4.09\text{ MPa (compression, at the wood/aluminum interface)}}$.
Part (ii) — steel. Bottom fibre, $c=63.0\text{ mm}$. $\sigma_{steel}=n_{steel}\dfrac{Mc}{I_{tr}}=20\times\dfrac{20\times10^6\times63.0}{523.9\times10^6}=\boxed{48.1\text{ MPa (tension)}}$.
Part (iii) — aluminum. Top fibre, $c=190-63.0=127.0\text{ mm}$. $\sigma_{al}=n_{al}\dfrac{Mc}{I_{tr}}=7\times\dfrac{20\times10^6\times127.0}{523.9\times10^6}=\boxed{33.9\text{ MPa (compression)}}$.

Final results
Quantity	Value
Transformed centroid (from bottom)	63.0 mm
I_transformed (wood-equiv.)	523.9 × 10⁶ mm⁴
σ_wood,max	4.09 MPa (compression)
σ_steel,max	48.1 MPa (tension)
σ_aluminum,max	33.9 MPa (compression)

Topic: Composite (multi-material) beam bending: transformed-section method

Key relations: Modular ratio $n=E_i/E_{ref}$; transformed width $b_{tr}=nb$; flexure formula $\sigma_{tr}=M\bar y/I_{tr}$ then $\sigma_{actual}=n\,\sigma_{tr}$.

Why this works: Plane sections remain plane regardless of material, so STRAIN still varies linearly with $y$; since $\sigma=E\varepsilon$, the ACTUAL stress jumps at every material interface even though the transformed (wood-equivalent) stress is one continuous linear function of $y$ across the whole depth.

Common pitfall: Forgetting to multiply the transformed-section stress by $n$ to recover the real stress in steel/aluminum (reporting the wood-equivalent value directly), and forgetting that stiffer, farther-from-centroid materials (here steel) usually govern the design even with a modest area.

Source: Hibbeler, Mechanics of Materials, 10th ed., §6.6 Composite Beams (Transformed-Section Method)

Question 5 of 8: Composite Timber-Steel-Aluminum Beam (Transformed Section)

Question 5: Composite Timber-Steel-Aluminum Beam (Transformed Section) (20 marks)