Sandwich Composite Cantilever Beam: 04-BS-6, May 2016, Q1

Question 1: Sandwich Composite 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.

Quantity	Value
Span (cantilever), L	3 m
Applied load, P	50 kN, at the free end
Core (plastic)	240 mm × 200 mm, E_core = 100 GPa, allowable = 220 MPa
Faces (aluminum)	240 mm wide × 6 mm thick (each, top & bottom), E_face = 75 GPa, allowable = 260 MPa

Cantilever sandwich beam: elevation and cross-section.

Find. (a) Whether the beam survives the 50 kN load in both materials; (b) the maximum load P before either material reaches its allowable stress.

Approach. Transform the section to an equivalent single-material section (base = plastic core, transform factor $n=E_{face}/E_{core}$), then check the bending stress in each material at the fixed end (where $M$ is largest) against its own allowable.

Maximum bending moment. The fixed end carries the full cantilever moment: $$M = PL = 50(3) = 150\ \text{kN}\cdot\text{m} = 1.500\times10^{8}\ \text{N}\cdot\text{mm}$$
Transform the section. With $n=E_{face}/E_{core}=75/100=0.75$, the section is doubly symmetric (200 mm core + 2$\times$6 mm faces, all 240 mm wide), so the neutral axis stays at the geometric centre: $$I_{core}=\frac{240(200)^3}{12}=1.600\times10^{8}\ \text{mm}^4$$ $$I_{face,\,each}=\frac{240(6)^3}{12}+240(6)(103)^2=1.528\times10^{7}\ \text{mm}^4$$ $$I_{tr}=I_{core}+n\big(2\,I_{face}\big)=1.600\times10^{8}+0.75(2)(1.528\times10^{7})=1.829\times10^{8}\ \text{mm}^4$$
Stress in the core (edge at $y=100$ mm, real material — use $I_{tr}$ directly): $$\sigma_{core}=\frac{My}{I_{tr}}=\frac{(1.5\times10^{8})(100)}{1.829\times10^{8}}=82.0\ \text{MPa} < 220\ \text{MPa}\quad\boxed{\text{OK}}$$
Stress in the aluminum faces (edge at $y=106$ mm; recover the ACTUAL stress in the transformed material by re-multiplying by $n$): $$\sigma_{face}=n\,\frac{My}{I_{tr}}=0.75\,\frac{(1.5\times10^{8})(106)}{1.829\times10^{8}}=65.2\ \text{MPa} < 260\ \text{MPa}\quad\boxed{\text{OK}}$$ Both materials clear their allowables comfortably, so the beam supports the 50 kN load.
Maximum load before failure. Solve each material's stress equation for the moment that brings it exactly to its allowable, then convert to a load ($M=PL$): $$M_{core}=\frac{220(1.829\times10^{8})}{100}=4.024\times10^{8}\ \text{N}\cdot\text{mm}\;\Rightarrow\;P_{core}=\frac{M_{core}}{L}=134.1\ \text{kN}$$ $$M_{face}=\frac{260(1.829\times10^{8})}{0.75(106)}=5.983\times10^{8}\ \text{N}\cdot\text{mm}\;\Rightarrow\;P_{face}=\frac{M_{face}}{L}=199.4\ \text{kN}$$ $$\boxed{P_{max}=\min(134.1,\ 199.4)=134.1\ \text{kN}\ \text{(core governs)}}$$

Quantity	Value
Applied moment at fixed end	150 kN·m
Core stress (allowable 220 MPa)	82.0 MPa — OK
Face stress (allowable 260 MPa)	65.2 MPa — OK
(a) Beam supports the 50 kN load?	Yes — both materials well within allowable
(b) Maximum load P_max	134.1 kN (core reaches 220 MPa first)

Topic: Transformed-section method for a composite (sandwich) beam of two elastic moduli.

Key relations: $n=E_2/E_1$; $I_{tr}=I_1+n\,I_2$ (widths of the non-base material scaled by $n$); $\sigma_1=\dfrac{My}{I_{tr}}$ for the base material, $\sigma_2=n\dfrac{My}{I_{tr}}$ for the transformed material.

Why this works: Plane sections remain plane, so strain varies linearly with $y$ across the whole depth regardless of material. Stress is $E\varepsilon$, so scaling a layer's width by $n=E_2/E_1$ makes it contribute to the section's flexural stiffness exactly as it does physically while letting one formula ($My/I_{tr}$) describe the base-material stress; the true stress in the transformed layer is recovered by multiplying back by $n$.

Common pitfall: Reporting the transformed-material stress straight from $My/I_{tr}$ without re-applying the $n$ factor (understates the aluminum stress here since $n<1$), or transforming with $n$ inverted. A second trap in this particular problem: assuming the "sandwich" naming implies a soft core and stiff skins, when the given data is the reverse ($E_{core}>E_{face}$) — always transform with the E-values actually given, not the usual intuition.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 6 (composite/transformed sections).

Question 1 of 8: Sandwich Composite Cantilever Beam

Question 1: Sandwich Composite Cantilever Beam (20 marks)