Question 8 of 8: Composite Concrete–Steel Cantilever Beam

Question 8: Composite Concrete–Steel Cantilever 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. (a) Whether the beam supports w = 3 kN/m (check every material at every governing fibre). (b) The maximum w before any material fails.

Approach. Transform the steel plate into an equivalent width of concrete using $n=E_s/E_c$, locate the transformed section's centroid and moment of inertia, then evaluate $\sigma=M(y-\bar y)/I_t$ at every fibre where a different material or a different allowable (tension vs compression) applies — the top of the steel, the top of the concrete (just below the plate), and the bottom of the concrete.

1. Maximum moment. Cantilever fixed end, UDL only: $$M_{max}=\frac{wL^2}{2}=\frac{3(4)^2}{2}=24.0\ \text{kN}\cdot\text{m}\ \text{(hogging — TENSION on top, compression on bottom)}$$
2. Transformed section. $$n=\frac{E_s}{E_c}=\frac{210}{30}=7.0$$ Transform the steel plate to an equivalent concrete width $7(200)=1400$ mm (same 10 mm thickness), sitting on top of the 200×450 mm concrete block. Measuring y from the bottom: $$A_{concrete}=200(450)=90{,}000\ \text{mm}^2\ (\bar y=225\ \text{mm}), \qquad A_{steel,t}=1400(10)=14{,}000\ \text{mm}^2\ (\bar y=455\ \text{mm})$$ $$\bar y = \frac{90{,}000(225)+14{,}000(455)}{104{,}000}=\boxed{256.0\ \text{mm from the bottom}}$$ $$I_t = \left[\frac{200(450)^3}{12}+90{,}000(225-256.0)^2\right]+ \left[\frac{1400(10)^3}{12}+14{,}000(455-256.0)^2\right]=\boxed{2.160\times10^9\ \text{mm}^4}$$
3. Stress at every governing fibre ($\sigma=M(y-\bar y)/I_t$; steel stress is $n\times$ the transformed value). Bottom of concrete ($y=0$, maximum compression): $$\sigma_{bot}=\frac{24\times10^6(0-256.0)}{2.160\times10^9}=\boxed{-2.84\ \text{MPa (compression)}}\quad(<15\ \text{MPa allowable, OK})$$ Top of concrete, just under the plate ($y=450$ mm): $$\sigma_{interface}=\frac{24\times10^6(450-256.0)}{2.160\times10^9}=\boxed{+2.16\ \text{MPa (tension)}}\quad(<3\ \text{MPa allowable, OK — but the tightest check, at 72\% of capacity})$$ Top of steel ($y=460$ mm, actual stress $=n\times$ transformed): $$\sigma_{steel}=7\!\left[\frac{24\times10^6(460-256.0)}{2.160\times10^9}\right]= \boxed{+15.9\ \text{MPa (tension)}}\quad(<240\ \text{MPa allowable, OK})$$
4. (a) Conclusion. All three checks pass at w = 3 kN/m: $$\boxed{\text{The composite beam CAN support the loading shown; the governing limit is concrete tension at the steel–concrete interface (2.16 of 3 MPa allowable, 72\% utilized).}}$$
5. (b) Maximum w. Every stress scales linearly with M (hence with w); scaling each fibre's stress to its own allowable and taking the smallest resulting w: $$w_{bottom-conc}=3\!\left(\frac{15}{2.84}\right)=15.8\ \text{kN/m}, \qquad w_{interface}=3\!\left(\frac{3}{2.16}\right)=4.17\ \text{kN/m}, \qquad w_{steel}=3\!\left(\frac{240}{15.9}\right)=45.4\ \text{kN/m}$$ $$\boxed{w_{max}=4.17\ \text{kN/m, governed by concrete tension at the interface}}$$

Final Results.