Falling mass on bar with spring, 125 cm

Given data

Bar: $d = 2$ cm, $L = 125$ cm, $E = 210$ GPa. Falling mass $m = 4$ kg ($W = mg = 39.24$ N). Spring: $k = 40$ kN/m $= 40\,000$ N/m, free length 10 cm. The mass falls 1.0 m measured from the top of the collar.

The spring (on the collar) and the bar act in series: the same force $P$ passes through both. The mass first falls freely to the top of the spring, then compresses the spring while stretching the bar.

Free fall before touching the spring:

$$h' = 1.0 - 0.10 = 0.90\ \text{m}$$

Section properties and stiffnesses

$$A = \frac{\pi d^2}{4} = 314.16\ \text{mm}^2 = 3.1416\times10^{-4}\ \text{m}^2$$ $$\frac{1}{k_{bar}} = \frac{L}{AE} = \frac{1.25}{3.1416\times10^{-4}\times2.1\times10^{11}} = 1.895\times10^{-8}\ \text{m/N},\qquad \frac{1}{k} = 2.500\times10^{-5}\ \text{m/N}$$

Total flexibility of spring + bar: $c = 2.5019\times10^{-5}$ m/N. If the maximum force is $P$, the total shortening/extension is $\delta = cP$.

Energy balance

Loss of potential energy = strain energy stored in the spring and bar:

$$W\,(h' + \delta) = \tfrac12 P\delta,\qquad \delta = cP$$ $$W(h' + cP) = \tfrac12 cP^2 \;\Rightarrow\; P^2 - 2WP - \frac{2Wh'}{c} = 0$$ $$P = W + \sqrt{W^2 + \frac{2Wh'}{c}} = 39.24 + \sqrt{1539.8 + 2.8231\times10^{6}} = 1719.9\ \text{N}$$

Spring compression $= P/k = 43.00$ mm; bar elongation $= PL/AE = 0.03259$ mm (very small).

Stress in the bar

$$\sigma = \frac{P}{A} = \frac{1719.9}{314.16} = 5.475\ \text{N/mm}^2$$

Answer: maximum stress in the bar $\approx 5.47$ N/mm$^2$ (maximum load 1.720 kN). The spring absorbs almost all the energy, which is why the stress is small. For comparison, the same weight applied statically would cause a stress of only 0.125 N/mm$^2$.