Steel/Wood Composite Beam — Transformed-Section Bending Stress: 04-BS-6, Undated paper, Q7

Question 7: Steel/Wood Composite Beam — Transformed-Section Bending Stress

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.

Quantity	Value
Overall section	100 mm wide × 300 mm tall (symmetric)
Steel flanges (top & bottom)	100 mm × 10 mm each, full width
Steel web (centre)	20 mm wide × 280 mm tall
Wood planks (two, either side of web)	40 mm × 280 mm each
$E_{steel}$, $E_{wood}$	200 GPa, 7 GPa
Applied moment $M$	60 kN·m (positive, about horizontal axis)

Find. The maximum bending stress in the wood and in the steel, and whether the wood/steel interface needs bonding for composite action.

Composite section: 100x300mm overall, steel flanges top/bottom + web, wood planks either side.

Approach. By symmetry the neutral axis sits at mid-height; sum $E_iI_i$ (about that common axis) for each material to get the composite $EI$, then apply $\sigma_i=E_i\,y\,\kappa$ with $\kappa=M/EI$ at each material's extreme fibre.

Second moments of area about the (mid-height) neutral axis. $$I_{web}=\frac{20(280)^3}{12}=3.659\times10^7\text{ mm}^4,\qquad I_{wood}=\frac{2(40)(280)^3}{12}=1.463\times10^8\text{ mm}^4$$ Each flange (own centroid 145 mm from the NA, by parallel-axis): $$I_{flange}=\frac{100(10)^3}{12}+100(10)(145)^2=2.103\times10^7\text{ mm}^4\ \ (\times2\text{ flanges})$$
Composite flexural rigidity. $$EI=E_{steel}\big(I_{web}+2I_{flange}\big)+E_{wood}\,I_{wood}$$ $$=200{,}000(3.659\times10^7+2(2.103\times10^7))+7000(1.463\times10^8)=1.6755\times10^{13}\text{ N}\!\cdot\!\text{mm}^2$$ $$\kappa=\frac{M}{EI}=\frac{60\times10^6}{1.6755\times10^{13}}=3.581\times10^{-6}\ \text{mm}^{-1}$$
Extreme-fibre stresses. Wood reaches only to $y=140$ mm (it stops where the flange begins); steel extends to $y=150$ mm (outer flange face). $$\boxed{\sigma_{steel,max}=E_{steel}(150)\kappa=107.4\text{ MPa}}$$ $$\boxed{\sigma_{wood,max}=E_{wood}(140)\kappa=3.51\text{ MPa}}$$ (Cross-checked with the transformed-to-steel method, $n=E_{wood}/E_{steel}=0.035$, shrinking each wood strip to an equivalent $1.4$ mm of steel: identical results.)
Bonding requirement. The wood planks and the steel web run the FULL depth of the section side-by-side (parallel to the neutral axis), not stacked one above the other. Because plane sections remain plane, the bending strain at any height $y$ is $\varepsilon=y\kappa$ — identical for every fibre at that height regardless of which material it is (wood or steel), with no dependence on lateral (z) position. There is therefore no tendency for wood and steel to slide relative to each other along the beam's length at their shared vertical interface, so no bonding or shear connectors are required for composite action here — unlike a beam where dissimilar materials are stacked in LAYERS through the depth, which does need shear transfer (via $\tau=VQ/(Ib)$ on the horizontal interface) to enforce a common curvature.

Result	Value
Composite $EI$	$1.676\times10^{13}\ \text{N}\!\cdot\!\text{mm}^2$
$\sigma_{steel,max}$	107.4 MPa (at $y=150$ mm)
$\sigma_{wood,max}$	3.51 MPa (at $y=140$ mm)
Bonding needed?	No — materials are side-by-side (parallel to NA), not stacked; strain compatibility is automatic

Topic: Bending of a composite beam of dissimilar materials — the weighted-$EI$ (transformed-section) method.

Key relations: $\varepsilon(y)=y\kappa$ (plane sections remain plane, independent of material); $\sigma_i=E_i\varepsilon=E_iy\kappa$; $\kappa=M/\sum E_iI_i$; transformed width $b'=b(E_{other}/E_{ref})$ as an equivalent check.

Why this works: Curvature is a property of the whole cross-section, common to every fibre; each material simply "sees" that common curvature through its own stiffness $E_i$. Weighting each material's $I$ by its own $E$ (rather than converting geometry first) gives the same answer as the transformed-area method and is often faster when several materials are present.

Common pitfall: Assuming ALL composite sections require adhesive/mechanical bonding — the requirement is specifically for materials stacked through the DEPTH (where a horizontal shear flow $VQ/Ib$ must be transferred); side-by-side (parallel) arrangements like this one need no such connection for flexural compatibility.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 6 §6.6 (Bending Members Composed of Composite Materials).

Question 7 of 8: Steel/Wood Composite Beam — Transformed-Section Bending Stress

Question 7: Steel/Wood Composite Beam — Transformed-Section Bending Stress