Overhung diving board — impact stress and safety factor: 22-Mec-B1 Advanced Machine Design, May 2013, Q2

Question 2: Overhung diving board — impact 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. Board cross-section $b=305\text{ mm}$ wide by $h=32\text{ mm}$ thick; pinned support at the wall end, roller support $0.7\text{ m}$ from the pin, free (diving) end $2.0\text{ m}$ from the pin; diver mass $60\text{ kg}$ ($W=588.6\text{ N}$); jump height $h_j=0.25\text{ m}$; static tip deflection under the diver $\delta_{st}=0.10\text{ m}$; board mass $25\text{ kg}$; $S_{ut}=200\text{ MPa}$.

Find. (a) the largest principal (bending) stress under the impact landing; (b) the static factor of safety on $S_{ut}$.

Diving board as a beam with a pin at the wall, a roller at $0.7\text{ m}$, and the diver load $P$ at the $2.0\text{ m}$ free end. The maximum bending moment occurs at the roller.

Approach. Treat the landing as a suddenly-applied (impact) load: obtain the impact factor from energy conservation using the given static deflection, scale the diver’s weight to the dynamic force, find the maximum bending moment at the roller support, and convert to bending stress through the section modulus. At the outer fibre the state is uniaxial, so the largest principal stress equals the bending stress.

Section modulus of the board. Bending is about the axis with the $32\text{ mm}$ dimension as depth: $$S = \frac{b\,h^2}{6} = \frac{0.305\,(0.032)^2}{6} = 5.21\times10^{-5}\text{ m}^3.$$
Impact (dynamic magnification) factor. The diver rises $h_j$ and falls back, delivering energy $W(h_j+\delta_{dyn})=\tfrac12 k\,\delta_{dyn}^2$ with $k=W/\delta_{st}$. Solving gives the classic falling-weight factor $$n = 1 + \sqrt{1 + \frac{2h_j}{\delta_{st}}} = 1 + \sqrt{1 + \frac{2(0.25)}{0.10}} = 1+\sqrt{6} = 3.45.$$
Dynamic force at the free end. $$F_{dyn} = n\,W = 3.45 \times 588.6\text{ N} = 2.03\times10^{3}\text{ N}.$$
Maximum bending moment. With the pin at $0$ and the roller at $0.7\text{ m}$, the overhang from the roller to the load is $2.0-0.7 = 1.3\text{ m}$; taking the free-body of the overhang, the moment is greatest at the roller: $$M_{max} = F_{dyn}\,(2.0-0.7) = 2.03\times10^{3}\times1.3 = 2.64\times10^{3}\text{ N}\cdot\text{m}.$$
Largest principal stress. At the top/bottom fibre the transverse shear is zero, so the bending stress is a principal stress: $$\sigma_1 = \frac{M_{max}}{S} = \frac{2.64\times10^{3}}{5.21\times10^{-5}}.$$ $$\boxed{\sigma_1 \approx 50.7\text{ MPa}.}$$
Static factor of safety. $$n_s = \frac{S_{ut}}{\sigma_1} = \frac{200}{50.7}.$$ $$\boxed{n_s \approx 3.9.}$$

Question 2 — results
Quantity	Symbol	Value
Impact factor	$n$	$3.45$
Dynamic tip force	$F_{dyn}$	$2.03\text{ kN}$
Max bending moment (at roller)	$M_{max}$	$2.64\text{ kN}\cdot\text{m}$
Largest principal stress	$\sigma_1$	$50.7\text{ MPa}$
Static safety factor	$n_s$	$3.9$

Check / assumptions: The energy method uses the diver’s own static deflection ($\delta_{st}=0.10\text{ m}$) as the reference; the board’s 25-kg self-weight is therefore not needed for the impact factor (its constant bending stress adds only a small offset and is neglected, as is common for this energy formulation). The impact factor assumes a rigid, non-rebounding contact and no energy loss.

Topic: Energy (impact) loading of a beam and the resulting bending stress.

Key relations: $k=W/\delta_{st}$; impact factor $n=1+\sqrt{1+2h/\delta_{st}}$; $F_{dyn}=nW$; $\sigma=M/S$, $S=bh^2/6$.

Why this works: A weight dropped onto an elastic member does work equal to the strain energy stored at maximum deflection. Equating the two and solving the quadratic yields a magnification factor that depends only on the ratio of drop height to static deflection — independent of the actual stiffness. Because the maximum stress scales linearly with the maximum force, the same factor multiplies the static stress.

Common pitfall: Placing the maximum moment at the pin instead of at the roller (the overhang controls), or using the total 2 m as the moment arm instead of the $1.3\text{ m}$ overhang; also forgetting the leading “$1+$” in the impact factor, which understates the force by nearly a factor of two.

Source: Hibbeler, Mechanics of Materials, §14 (impact/energy methods); Norton, Machine Design, §3.9 (impact loading).

Question 2 of 6: Overhung diving board — impact stress and safety factor

Question 2: Overhung diving board — impact stress and safety factor (30 marks)