25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1) · December 2013
Question 5 of 6: Overhung Diving Board — Impact Stress
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 · 98-Mar-B1 Advanced Machine Design.
Open-book, 3 hours, 100 marks. The paper requires all of Part I (Problems 1 and 2)
plus any three of the four Part II problems (3–6). All six problems are solved in full below.
Reference texts. R. G. Budynas & J. K. Nisbett, Shigley’s
Mechanical Engineering Design, 10th ed. (shafts & critical speed §7; bolted joints
§8; lubrication & journal bearings §12; brakes §16); R. C. Juvinall &
K. M. Marshek, Fundamentals of Machine Component Design (bearings, brakes, impact);
R. C. Hibbeler, Mechanics of Materials (bending, impact loading).
Check: this paper, although listed under Applied Thermodynamics and Heat Transfer, is headed “98-Mar-B1” with printed title “Advanced Machine Design”; it is a general mechanical machine-design paper — stress/yield criteria, shaft deflection/critical speed, bolted-joint preload, journal-bearing sizing, impact loading, and drum brakes — with zero thermodynamics or heat-transfer content, and it is solved as the exam actually printed.
Given. Rectangular board $b\times h = 300\times35$ mm; pin support at the back,
roller (fulcrum) 0.7 m forward, free end (load point) 2.0 m from the pin; person mass
60 kg jumps $h_j = 0.25$ m; static tip deflection under the person
$\delta_{st} = 0.10$ m; board self-weight 25 kg.
Overhung diving board: pin at the back, roller fulcrum at 0.7 m,
person’s weight $P$ at the free end 2.0 m from the pin.
Find. The largest principal stress in the board caused by the diver landing after a
0.25 m jump.
Approach. Convert the fall into a dynamic amplification (impact factor) from energy
conservation using the given static deflection, apply the amplified force at the tip, find the maximum
bending moment at the fulcrum, and divide by the section modulus; at the extreme fibre the bending stress is
itself the principal stress.
Static weight. $W = mg = 60(9.81) = 588.6$ N.
Impact factor. Equating the energy of the jump to the strain energy of an equivalent
static deflection,
$$n = 1+\sqrt{1+\dfrac{2h_j}{\delta_{st}}} = 1+\sqrt{1+\dfrac{2(0.25)}{0.10}} = 1+\sqrt{6} = \boxed{3.449}.$$
Dynamic force. $F_{dyn} = nW = 3.449(588.6) = 2030$ N.
Maximum bending moment. The critical section is the roller (fulcrum), where the
overhang from the fulcrum to the tip is $2.0-0.7 = 1.3$ m, so
$$M_{\max} = F_{dyn}(1.3) = 2030(1.3) = \boxed{2640\ \text{N}\cdot\text{m}}.$$
Largest principal stress. At the top/bottom fibre the transverse shear is zero, so the
bending stress is the maximum principal stress,
$$\sigma_1 = \dfrac{M_{\max}}{S} = \dfrac{2640}{6.125\times10^{-5}} = \boxed{43.1\ \text{MPa}}.$$
Check: the 25-kg board self-weight is not needed for the impact
amplification — the given static deflection already embodies the board’s flexibility under the
diver. Its dead-weight bending stress ($\sim$3 MPa) simply adds to, and is small beside, the 43 MPa impact
stress.
