Composite Timber/Steel Beam — Bending and Glue-Joint Shear: 04-BS-6, December 2019, Q7

Question 7: Composite Timber/Steel Beam — Bending and Glue-Joint Shear (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.

Quantity	Value
Timber	144 mm × 240 mm, on top
Steel strap	120 mm × 10 mm, glued to bottom face
$M$ (positive/sagging)	20 kN·m
$V$	15 kN
$E_{wood}$	7 GPa
$E_{steel}$	200 GPa

Find. (1) $\sigma_{wood,max}$; (2) $\sigma_{steel,max}$; (3) glue-joint shear stress.

Transformed section: 144x240 mm timber over a 120x10 mm steel strap; bending-stress distribution at right.

Approach. Use the transformed-section method (modular ratio $n=E_s/E_w$) to locate the neutral axis and combined $I$ of the whole section, then apply the flexure formula to each material (multiplying the steel result by $n$) and the shear-flow formula at the glue interface.

Modular ratio and centroid. $n=200/7=28.57$. Measuring $y$ from the bottom of the composite section: wood centroid $y_w=10+120=130\text{ mm}$ ($A_w=144\times240=34{,}560\text{ mm}^2$); steel centroid $y_s=5\text{ mm}$ ($A_s=120\times10=1200\text{ mm}^2$). $\bar y=\dfrac{A_wy_w+nA_sy_s}{A_w+nA_s}=\dfrac{34{,}560(130)+28.57(1200)(5)}{34{,}560+28.57(1200)}=\boxed{67.75\text{ mm}}$.
Transformed moment of inertia. $I_w=\dfrac{144(240)^3}{12}+A_w(130-\bar y)^2=165.9\times10^6+34{,}560(62.25)^2$. $I_s=n\left[\dfrac{120(10)^3}{12}+A_s(5-\bar y)^2\right]$. Summing: $I_{tot}=\boxed{435.1\times10^6\text{ mm}^4}$.
Part (1) — wood stress. Top-fibre distance $c_{top}=250-\bar y=182.25\text{ mm}$ (compression, sagging moment). $\sigma_{wood}=Mc_{top}/I=20\times10^6(182.25)/435.1\times10^6=\boxed{8.38\text{ MPa (compression, top fibre)}}$ — this exceeds the tension value at the wood/steel interface ($2.65\text{ MPa}$), so the TOP fibre governs.
Part (2) — steel stress. Bottom-fibre distance $=\bar y=67.75\text{ mm}$ (tension). $\sigma_{steel}=n\,Mc_{bot}/I=28.57(20\times10^6)(67.75)/435.1\times10^6=\boxed{89.0\text{ MPa (tension)}}$.
Part (3) — glue-joint shear. $Q_{wood}=A_w(y_w-\bar y)=34{,}560(62.25)=2.151\times10^6\text{ mm}^3$ (first moment of the wood about the composite NA, taken above the glue line). Shear plane width = strap width $b_s=120\text{ mm}$ (the narrower of the two contact widths). $\tau_{glue}=VQ/(Ib_s)=15{,}000(2.151\times10^6)/(435.1\times10^6\times120)=\boxed{0.618\text{ MPa}}$.

Final results
Quantity	Value
$\bar y$ (from bottom)	67.75 mm
$I_{tot}$ (transformed)	$435.1\times10^6\text{ mm}^4$
$\sigma_{wood,max}$	8.38 MPa (compression, top)
$\sigma_{steel,max}$	89.0 MPa (tension, bottom)
$\tau_{glue}$	0.618 MPa

Topic: Transformed-section bending of a two-material composite beam

Key relations: $n=E_2/E_1$; transform areas by $n$ (keep true dimensions for own-axis $I$ and parallel-axis terms); flexure $\sigma_1=Mc/I_{tr}$, $\sigma_2=nMc/I_{tr}$; shear flow $\tau=VQ/(Ib)$ using the ACTUAL contact width at the interface.

Why this works: Transforming the stiffer material's area by $n=E_s/E_w$ converts the composite cross-section into an equivalent SINGLE-material (wood) section, so ordinary flexure and shear-flow formulas apply directly — the factor $n$ is then reintroduced only when converting the transformed-section strain back to the real steel stress.

Common pitfall: Checking the wood stress only at the glue interface (a much smaller value here) and missing that the FAR (top) fibre governs; and using the full timber width instead of the narrower steel-strap width for the glue-joint shear plane.

Source: Hibbeler, Mechanics of Materials, 10th ed., §6.5–6.6 Composite Beams / Transformed Section, §7.3 Shear Flow

Question 7 of 8: Composite Timber/Steel Beam — Bending and Glue-Joint Shear

Question 7: Composite Timber/Steel Beam — Bending and Glue-Joint Shear (20 marks)