Question 2 of 6: Overhung diving board — impact loading, principal stress and safety factor (30 marks)

Question 2: Overhung diving board — impact loading, principal 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. An overhung diving board carrying a diver who jumps and lands at the free end; the figure gives the support layout and the board cross-section, and the question gives the diver mass, the jump height, the measured static deflection and the material strength.

Find. (1) the largest principal stress in the board when the diver jumps 25 cm and lands back on the free end, and (2) the static safety factor against the 130 MPa longitudinal ultimate stress.

Approach. Convert the falling diver into an equivalent static tip force using the energy-based impact factor built on the given static deflection, then find the peak (hogging) bending moment at the roller and divide by the section modulus; at the extreme fibre the transverse shear is zero, so the bending stress is the largest principal stress.

Read the figure carefully before doing any statics. The 2 m dimension runs from the pin at the heel of the board all the way to the free end where the load $P$ acts, and the 0.7 m dimension locates the roller from that same pin. The overhang beyond the fulcrum is therefore $2.0 - 0.7 = 1.3\text{ m}$, not 2 m.

1. Static weight of the diver. The impact analysis is driven by the weight that falls, so start from the diver's own weight. $$W = m g = (60\text{ kg})(9.81\text{ m/s}^2) = 588.6\text{ N}$$
2. Impact (dynamic magnification) factor from an energy balance. A mass released from height $h_j$ above a structure that deflects $\delta_{st}$ under that same mass statically delivers a peak force $n$ times the static weight. Equating the work done by the weight through the total fall to the strain energy stored at maximum deflection, and treating the board as a linear spring, gives the standard result $$n = 1 + \sqrt{1 + \frac{2h_j}{\delta_{st}}} = 1 + \sqrt{1 + \frac{2(0.250)}{0.100}} = 1 + \sqrt{6}$$ $$\boxed{\;n = 3.449\;}$$ The 100 mm static deflection is the whole point of the question: it converts the board's unknown stiffness into a directly usable number, so no second-moment-of-area or deflection calculation is needed to get $n$.
3. Equivalent static impact force at the free end. Multiply the falling weight by the magnification factor. $$F_{dyn} = nW = 3.449 \times 588.6 = 2030\text{ N}$$
4. Support reactions. Take moments about the pin at $A\,(x=0)$ with the roller at $B\,(x=0.7\text{ m})$ and the load at $C\,(x=2.0\text{ m})$. $$R_B = \frac{F_{dyn}\,L}{L_1} = \frac{2030 \times 2.0}{0.7} = 5801\text{ N}\ (\uparrow), \qquad R_A = F_{dyn} - R_B = -3771\text{ N}\ (\downarrow)$$ The pin at the heel is pulled down with 3.77 kN, which is why a real diving board is bolted down at the heel rather than simply resting on its support.
5. Peak bending moment. On the overhang the moment grows linearly from zero at the free end to its maximum at the fulcrum; between the two supports it falls linearly back to zero at the pin. The maximum is therefore at the roller and is hogging (tension on the top surface): $$M_{\max} = F_{dyn}\,(L - L_1) = 2030\text{ N} \times 1.300\text{ m} = 2640\text{ N}\!\cdot\!\text{m} = 2.64\text{ kN}\!\cdot\!\text{m}$$
6. Section modulus of the rectangular board. Bending is about the strong axis, so the 32 mm dimension is the depth. $$S = \frac{b h^2}{6} = \frac{(305)(32)^2}{6} = 52\,053\text{ mm}^3$$
7. Bending stress, which is the largest principal stress. At the top and bottom extreme fibres the transverse shear stress is zero, so the element there is in pure uniaxial bending: the stress element is already principal, with $\sigma_1 = \sigma_{bend}$ and $\sigma_2 = \sigma_3 = 0$. $$\sigma_1 = \frac{M_{\max}}{S} = \frac{2.640 \times 10^6\text{ N}\!\cdot\!\text{mm}}{52\,053\text{ mm}^3}$$ $$\boxed{\;\sigma_1 = 50.7\text{ MPa (tension on the top fibre, at the roller)}\;}$$ As a cross-check, the diver standing statically produces $\sigma_{st} = (588.6)(1300)/52\,053 = 14.70$ MPa, and $14.70 \times 3.449 = 50.7$ MPa, confirming the impact factor was applied consistently.
8. Static safety factor against the longitudinal ultimate stress. The board is an anisotropic laminate whose quoted strength is the longitudinal (fibre-direction) ultimate, which is the direction of the bending stress. $$N = \frac{S_{ut}}{\sigma_1} = \frac{130}{50.7}$$ $$\boxed{\;N = 2.56\;}$$

The board is adequate for this diver and this jump, with a factor of about 2.6 on the ultimate stress. Note how sensitive the result is to the board's flexibility: the 100 mm static deflection is what keeps the impact factor down to 3.45. A stiffer board deflecting only 10 mm under the same diver would give $n = 1+\sqrt{51} = 8.14$ and a stress of roughly 120 MPa, taking the safety factor below 1.1. Compliance is a safety feature in impact-loaded structures, not a defect.