Overhung Diving Board under Impact: 22-Mec-B1 Advanced Machine Design, December 2014, Q2

Question 2: Overhung Diving Board under Impact (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. The board is a beam pinned at the left end and simply supported on a roller 0.7 m to the right; it overhangs 1.3 m from the roller to the free tip (overall span 2.0 m), where the diver lands.

Given data
Cross-section (width × depth)	$b=305$ mm, $h=32$ mm
Diver mass	$m=70$ kg
Jump height (fall onto board)	$h_j=0.25$ m
Static tip deflection under diver	$\delta_{st}=10$ mm
Ultimate stress (longitudinal)	$S_{ut}=130$ MPa
Support geometry	pin @ 0, roller @ 0.7 m, tip @ 2.0 m

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

Overhung diving board: pin at the left end, roller at 0.7 m, diver load P at the 2.0 m free tip. The critical bending section is the roller (overhang 1.3 m).

Approach. Convert the landing into an equivalent static force through the impact (dynamic-magnification) factor built from the given static deflection, find the peak bending moment at the roller, then the surface bending stress (which is the largest principal stress since transverse shear is zero at the extreme fibre), and finally the safety factor.

Section modulus of the board. For a rectangle, $Z=\dfrac{bh^2}{6}=\dfrac{(0.305)(0.032)^2}{6}=5.205\times10^{-5}\ \text{m}^3$.
Impact (dynamic-magnification) factor. A mass falling a height $h_j$ onto a support that deflects $\delta_{st}$ under that same weight produces a load $n$ times the static weight, with $$n = 1+\sqrt{1+\dfrac{2h_j}{\delta_{st}}} = 1+\sqrt{1+\dfrac{2(0.25)}{0.010}} = 1+\sqrt{51} = 8.14.$$ The board’s own 25 kg weight is a distractor: the impact factor already embeds the board’s flexibility through $\delta_{st}$, and the board’s static self-weight stress is not part of the diver’s landing event.
Equivalent dynamic force at the tip. With diver weight $W=mg=(70)(9.81)=686.7\ \text{N}$, $$F_{dyn}=nW=(8.14)(686.7)=5.59\times10^{3}\ \text{N}.$$
Peak bending moment. The maximum moment on an overhanging beam occurs at the interior (roller) support and equals the tip force times the overhang length: $$M_{\max}=F_{dyn}\,(2.0-0.7)=(5590.7)(1.3)=7.27\times10^{3}\ \text{N}\cdot\text{m}.$$
Largest principal stress. At the top/bottom surface the state is uniaxial bending (shear stress is zero there), so the largest principal stress equals the bending stress: $$\boxed{\;\sigma_1=\frac{M_{\max}}{Z}=\frac{7267.9}{5.205\times10^{-5}}=1.396\times10^{8}\ \text{Pa}=139.6\ \text{MPa}\;}$$
Static factor of safety. $$n_s=\frac{S_{ut}}{\sigma_1}=\frac{130}{139.6}=0.93.$$

Check / engineering judgement: the factor of safety is below unity ($n_s\approx0.93$), so the board would be overstressed and could fail under this landing. This is the honest result of the given data — the very shallow $\delta_{st}=10$ mm makes the board stiff, which drives the impact factor up to 8.1. In practice a diving board is deliberately compliant (large $\delta_{st}$) precisely to keep the impact factor low; the numbers here flag that this board is too stiff and too highly stressed for the specified duty.

Problem 2 results
Quantity	Value
Impact factor $n$	8.14
Equivalent dynamic force $F_{dyn}$	5.59 kN
Peak moment at roller $M_{\max}$	7.27 kN·m
Largest principal stress $\sigma_1$	139.6 MPa
Static factor of safety $n_s$	0.93 (unsafe, $\lt 1$)

Topic. Impact (suddenly applied / dropped) loading of a beam via the energy-based dynamic-magnification factor.

Key relations. Impact factor $n=1+\sqrt{1+2h/\delta_{st}}$; bending stress $\sigma=M/Z$; $Z=bh^2/6$; factor of safety $n_s=S_{ut}/\sigma$.

Why this works. Equating the potential energy lost by the falling mass, $W(h+\delta_{max})$, to the strain energy stored in the beam, $\tfrac12 k\,\delta_{max}^2$ with $k=W/\delta_{st}$, yields $\delta_{max}=n\,\delta_{st}$ and an equivalent static force $nW$. Because stress is linear in load for an elastic beam, the same factor multiplies the static stress. The largest principal stress at the surface equals the bending stress because the transverse shear stress vanishes at the extreme fibre.

Common pitfall. Using $\delta_{st}$ in metres inconsistently, or adding the board’s self-weight stress to the impact stress — the self-weight is a decoy. Another trap is placing $M_{\max}$ at the tip; on an overhang the peak moment is at the interior support.

Source. Shigley §4–6 (deflection, energy methods, impact); Juvinall & Marshek §7 (impact loading).

Question 2 of 6: Overhung Diving Board under Impact

Question 2: Overhung Diving Board under Impact (30 marks)