Question 2 of 6: Overhung diving board — impact loading, principal stress and safety factor (30 marks)

Question 2: Overhung diving board — impact loading, principal stress and safety factor (30 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. An overhung board pinned at its heel, resting on a roller 0.7 m along, with the diver at the free end 2 m from the pin.

Find. The largest principal stress produced by the landing, and the static factor of safety against the ultimate stress.

Approach. Convert the fall into an equivalent static load through the energy-based impact factor, carry that dynamic force through beam statics to the largest bending moment (which occurs at the roller, not at the pin), then convert the moment to a surface bending stress — which, because the free surface carries no shear, is the largest principal stress.

1. Weight of the falling diver. The impacting body is the 100 kg person: $$W = mg = 100\times 9.81 = 981\ \text{N}$$ The board's own 29 kg does not fall — it is already deflected and is, in any case, embedded in the measured static deflection. Its effect is quantified in the callout below.
2. Impact factor from the energy balance. Equating the work done by the falling weight through the total drop $\left(h_j+\delta_{\max}\right)$ to the strain energy stored in the board, with the board treated as a linear spring of stiffness $k = W/\delta_{st}$, gives the standard result $$n = 1+\sqrt{1+\frac{2h_j}{\delta_{st}}}$$ Substituting the measured static deflection, $$n = 1+\sqrt{1+\frac{2\times 250}{131}} = 1+\sqrt{4.8168} = 3.195$$ Using the measured $\delta_{st}$ rather than a computed one is the point of the question: it absorbs the real end fixity, the taper and the material's true modulus, none of which are given.
3. Equivalent static (dynamic) force at the tip. $$F_{dyn} = nW = 3.195 \times 981$$ $$\boxed{F_{dyn} = 3134\ \text{N}}$$
4. Support reactions under the dynamic load. Taking moments about the pin $A$ for the roller reaction $R_B$ at 0.7 m with the load at 2.0 m, $$R_B = F_{dyn}\frac{2.0}{0.7} = 3134\times 2.857 = 8954\ \text{N}\ (\uparrow)$$ $$R_A = R_B - F_{dyn} = 8954-3134 = 5820\ \text{N}\ (\downarrow)$$ The pin therefore pulls down on the heel of the board: it is a hold-down, which is exactly how a real diving-board fulcrum assembly is anchored.
5. Largest bending moment. On an overhung beam loaded only at the free tip, the bending moment grows linearly from zero at the tip to a maximum over the roller and then falls linearly back to zero at the pin. The peak (hogging) moment is therefore $$M_{\max}=F_{dyn}\times\left(2.0-0.7\right)=3134\times 1.3$$ $$\boxed{M_{\max}=4.074\ \text{kN}\cdot\text{m at the roller}}$$ Checking from the other side, $R_A\times 0.7 = 5820\times 0.7 = 4.074\ \text{kN}\cdot\text{m}$ — the same value, as it must be.
6. Section modulus of the rectangular board. Bending is about the strong-in-plan but thin-in-depth axis, so the depth in the formula is the 32 mm thickness: $$S=\frac{bh^{2}}{6}=\frac{305\times 32^{2}}{6}=\frac{312\,320}{6}=52\,053\ \text{mm}^{3}$$
7. Largest principal stress. At the extreme fibre the transverse shear stress is zero and the surface is traction-free, so the bending stress is a principal stress and the other two principal stresses are zero: $$\sigma_1=\frac{M_{\max}}{S}=\frac{4.074\times 10^{6}\ \text{N}\cdot\text{mm}}{52\,053\ \text{mm}^{3}}$$ $$\boxed{\sigma_1 = 78.3\ \text{MPa (tension on the top surface over the roller)}}$$
8. Static factor of safety. Comparing with the longitudinal ultimate stress, $$n_s=\frac{S_{ut}}{\sigma_1}=\frac{130}{78.3}$$ $$\boxed{n_s = 1.66}$$ The board survives the specified dive with a 66 % margin on ultimate stress — adequate for a one-off overload check, though a fibre-reinforced board in repeated service would also need a fatigue assessment, which this question does not ask for.