Question 5 of 6: Overhung Diving Board — Largest Principal Stress under Impact
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 22-Mec-B1 Advanced Machine Design — National Exams, May 2015. Open book, 3 hours, 100 marks. Part I (Problems 1–2) compulsory; answer any three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Reference texts. R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design (10th ed.) — deflection §4, fatigue §6, shafts §7, bolted joints §8, journal bearings §12, brakes & clutches §16; R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design; R. L. Norton, Machine Design: An Integrated Approach; R. C. Hibbeler, Mechanics of Materials.
Question 5: Overhung Diving Board — Largest Principal Stress under Impact (20 marks)
Given. A board pinned at the back and resting on a roller fulcrum 0.7 m forward, projecting to a free end 2.0 m from the pin; rectangular section 305 mm wide × 32 mm thick; a 60-kg diver drops from a 25-cm jump onto the tip.
Given data
Quantity
Symbol
Value
Diver mass (weight)
$m$ ($W$)
60 kg (588.6 N)
Jump height
$h$
25 cm = 0.25 m
Static deflection (person standing)
$\delta_{st}$
8 cm = 0.08 m
Section (width × depth)
$b\times t$
305 mm × 32 mm
Pin–roller / pin–tip
—
0.7 m / 2.0 m
Board mass (self-weight)
—
25 kg (distractor)
Find. The largest principal stress in the board when the diver lands.
Overhung diving board: pin at the back, roller fulcrum at 0.7 m, diver’s dynamic weight $P$ at the free end 2.0 m from the pin.
Approach. Convert the jump into an impact factor from the given static deflection, apply the amplified load at the tip, find the maximum bending moment (at the roller, where the overhang is cantilevered), and divide by the weak-axis section modulus. At the outer fibre the bending stress is uniaxial, so it is itself the largest principal stress.
Impact (dynamic magnification) factor. For a mass falling height $h$ onto a structure whose static deflection under that mass is $\delta_{st}$,
$$n=1+\sqrt{1+\frac{2h}{\delta_{st}}}=1+\sqrt{1+\frac{2(0.25)}{0.08}}=1+\sqrt{7.25}=3.69.$$
Dynamic tip load. The diver’s static weight $W=60(9.81)=588.6$ N is amplified to
$$F_{dyn}=nW=3.69(588.6)=2.17\text{ kN}.$$
The board’s own 25-kg weight is already reflected in the measured static deflection and does not add to the impact load—it is a distractor.
Maximum bending moment. The tip load is reacted at the pin and roller; the overhang beyond the roller acts as a cantilever of length $2.0-0.7=1.3$ m, so the moment is greatest at the roller support:
$$M_{\max}=F_{dyn}\,(2.0-0.7)=2173(1.3)=2.83\text{ kN}\!\cdot\!\text{m}.$$
Section modulus (weak axis). The board bends about the axis through its 32-mm thickness:
$$S=\frac{b\,t^{2}}{6}=\frac{0.305(0.032)^{2}}{6}=5.21\times10^{-5}\text{ m}^3.$$
Largest principal stress. At the top/bottom fibre the transverse shear is zero, so the state is uniaxial bending and the bending stress is the largest principal stress:
$$\boxed{\sigma_1=\frac{M_{\max}}{S}=\frac{2825}{5.21\times10^{-5}}=54.3\text{ MPa}.}$$
Interpretation. A 54-MPa peak is safe for a fibreglass/aluminium diving board (yield well above 100 MPa); note how strongly the stiff board (small $\delta_{st}=8$ cm) drives the impact factor up—a softer board would cut the stress sharply.
