25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1) · December 2016
Question 2 of 6: Overhung Diving Board — Impact Stress and Safety Factor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 98-Mar-B1, December 2016. Open book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; three of the four Part II problems (3–6) are required. All six problems are solved below.
Given. The board is pinned at one end and rests on a roller 0.7 m away; it overhangs a further 1.3 m to the free tip (overall span 2.0 m) where the diver lands.
Given data
Quantity
Symbol
Value
Cross-section (width × depth)
$b\times h$
$305\times32\ \text{mm}$
Diver mass
$m$
$100\ \text{kg}$ ($W=981\ \text{N}$)
Jump height at tip
$h_j$
$0.25\ \text{m}$
Static deflection under diver
$\delta_{st}$
$0.131\ \text{m}$
Board self-weight
—
$29\ \text{kg}$ (not used)
Ultimate stress
$S_{ut}$
$130\ \text{MPa}$
Find. The largest principal stress produced by the diver landing after a 0.25 m jump, and the static safety factor against the ultimate stress.
Overhung diving board: pin at the heel, roller 0.7 m along, 1.3 m overhang to the diver. The hogging moment peaks at the roller.
Approach. Treat the landing as an impact (energy) problem: the diver falls back through the jump height, so an impact factor scales up the static load; then find the peak bending moment at the roller support, convert to surface bending stress (which is the largest principal stress because transverse shear vanishes at the extreme fibre), and finally the safety factor.
Section modulus of the rectangular board. $S=\dfrac{b\,h^2}{6}=\dfrac{0.305\times0.032^2}{6}=5.205\times10^{-5}\ \text{m}^3.$
Impact (dynamic magnification) factor. Equating the diver’s kinetic-plus-potential energy at impact to the strain energy stored in the board gives the standard result, using the measured static deflection $\delta_{st}$ under the diver’s weight:
$$n = 1+\sqrt{1+\frac{2h_j}{\delta_{st}}}=1+\sqrt{1+\frac{2(0.25)}{0.131}}=1+\sqrt{4.817}=3.19.$$
The board’s own 29 kg weight is a static bias that does not enter the impact factor and is not needed.
Dynamic force at the tip. $F_{dyn}=n\,W=3.19\times(100\times9.81)=3.13\times10^{3}\ \text{N}.$
Peak bending moment. For a load on the overhang, the largest moment is hogging over the roller, equal to the force times the overhang $L_o=2.0-0.7=1.3\ \text{m}$:
$$M_{\max}=F_{dyn}\,L_o=3134\times1.3=4.07\times10^{3}\ \text{N}\cdot\text{m}.$$
Largest principal stress. At the top/bottom fibre the transverse shear is zero, so the bending stress is the largest principal stress:
$$\boxed{\ \sigma_{\max}=\frac{M_{\max}}{S}=\frac{4074}{5.205\times10^{-5}}=78.3\ \text{MPa}\ }$$
Static safety factor. Against the ultimate longitudinal stress,
$$n_{s}=\frac{S_{ut}}{\sigma_{\max}}=\frac{130}{78.3}=1.66.$$
The board survives the landing with a modest margin.
