Question 6 of 8: Composite Concrete Cantilever with a Top Steel Plate

Question 6: Composite Concrete Cantilever with a Top Steel Plate (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. Cantilever, fixed at the left end, length $L=4\text{ m}$, full-span UDL $w$ (to be found). Cross-section: concrete $200\times450\text{ mm}$ with a $10\text{ mm}$ steel plate bonded on TOP. $E_c=25\text{ GPa}$, $E_s=200\text{ GPa}$. Allowable stresses: concrete $3\text{ MPa}$ tension / $15\text{ MPa}$ compression; steel $240\text{ MPa}$.

Find. Part (a): the maximum UDL w the composite cantilever can carry. Part (b): why bonding is required for composite action.

Approach. Use the transformed-section method (steel transformed into equivalent concrete, n=Es/Ec) to find the composite centroid and Itr, identify that a full-span UDL on a cantilever produces HOGGING moment throughout (tension on top, compression on bottom), then find the moment capacity governed by each material's weakest allowable stress and take the smallest.

1. Transformed section (steel → equivalent concrete, $n=E_s/E_c=200/25=8$). Concrete: $A_c=200\times450=90{,}000\text{ mm}^2$ at $\bar y_c=225\text{ mm}$ (from the bottom). Steel (transformed): $A_{s,tr}=n\times200\times10=16{,}000\text{ mm}^2$ at $\bar y_s=455\text{ mm}$. Composite centroid $\bar Y=\dfrac{90{,}000(225)+16{,}000(455)}{106{,}000}=\boxed{259.7\text{ mm}}$ (from the bottom).
2. Transformed moment of inertia. $I_{tr}=\left[\tfrac{200(450)^3}{12}+90{,}000(225-259.7)^2\right]+\left[8\left(\tfrac{200(10)^3}{12}\right)+16{,}000(455-259.7)^2\right]=\boxed{2{,}237.5\times10^{6}\text{ mm}^4}$.
3. Which fibre governs (hogging, full-span UDL on a cantilever). A cantilever under a downward UDL bends concave-down (hogging) over its ENTIRE length, so the TOP fibre is in tension and the BOTTOM fibre is in compression, at every section. The weakest allowable stress is concrete's tension limit ($3\text{ MPa}$), acting at the top of the CONCRETE ($c=450-259.7=190.3\text{ mm}$ from the N.A., just below the steel interface).
4. Moment capacity from each limit. Concrete compression (bottom, $c=259.7\text{ mm}$): $M_1=15(2{,}237.5\times10^6)/259.7=129.2\times10^{6}\text{ N}\cdot\text{mm}$. Concrete tension (top of concrete, $c=190.3\text{ mm}$): $M_2=3(2{,}237.5\times10^6)/190.3=\boxed{35.28\times10^{6}\text{ N}\cdot\text{mm}}$ (the SMALLEST, so it governs). Steel tension (top, $c=200.3\text{ mm}$, transformed stress $\times n$): $M_3=240(2{,}237.5\times10^6)/(8\times200.3)=335.2\times10^{6}\text{ N}\cdot\text{mm}$. Concrete tension governs: $M_{max}=\boxed{35.28\text{ kN}\cdot\text{m}}$.
5. Part (a) — solve for w. For a cantilever with a full-span UDL, $M_{max}=wL^2/2$ at the fixed end, so $w_{max}=2M_{max}/L^2=2(35.28\times10^6)/(4{,}000)^2=\boxed{4.41\text{ kN/m}}$ — this is BELOW the figure's printed "5 kN/m," i.e. as printed the beam would exceed the concrete's tensile limit by about 13%.
6. Part (b) — why bonding is required. The transformed-section (composite) analysis above assumes the steel plate and concrete deform TOGETHER, i.e. plane sections remain plane across the whole depth with no relative slip at the interface. That requires the interface to transmit a horizontal shear flow $q=VQ/I$ between the two materials; without bonding (or equivalent mechanical connectors), the plate would simply slide relative to the concrete, each layer would bend about its OWN separate neutral axis, and the section would lose essentially all of the stiffness and strength benefit the transformed-section calculation above relies on.