---
title: "Falling mass on bar with spring, 130 cm"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:falling-mass-bar-spring-130cm"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/falling-mass-bar-spring-130cm
site: BCT Engineering
---

# Falling mass on bar with spring, 130 cm

Theory of Structures I · Chapter 2: Analysis by the Strain Energy Method

- Asked in: 2073 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/falling-mass-bar-spring-130cm
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

A bar of 3 cm diameter and of length 130 cm is supported rigidly in the vertical position at the top and is provided with a hollow falling mass and a collar at the bottom which supports a spring 10 cm long. Find the stress developed in the bar if the falling mass is 4 kg and it falls from a height of 1.15 m measured from the collar top. Take stiffness of the spring $k = 40$ kN/m and $E = 210\ \text{GN/m}^2$.

## Short answer

A 4kg mass falls 1.15m onto a spring (k=40kN/m, free length 10cm) in series with a bar (3cm diameter, 130cm long, E=210GPa). Using energy balance: loss of potential energy = strain energy in spring and bar. Maximum force P = 1854.9N, stress in bar σ = P/A = 2.62 N/mm².

## Answer

#### Given data
Bar: $d = 3$ cm, $L = 130$ 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 1.15 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:**

$$
h' = 1.15 - 0.10 = 1.05\ \text{m}
$$

#### Section properties and stiffnesses
$$
A = \frac{\pi d^2}{4} = 706.86\ \text{mm}^2 = 7.0686\times10^{-4}\ \text{m}^2
$$

$$
\frac{1}{k_{bar}} = \frac{L}{AE} = \frac{1.30}{7.0686\times10^{-4}\times2.1\times10^{11}} = 8.758\times10^{-9}\ \text{m/N},\qquad \frac{1}{k} = 2.500\times10^{-5}\ \text{m/N}
$$

Total flexibility of spring + bar: $c = 2.5009\times10^{-5}$ 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:

$$
W\,(h' + \delta) = \tfrac12 P\delta,\qquad \delta = cP
$$

$$
W(h' + cP) = \tfrac12 cP^2 \;\Rightarrow\; P^2 - 2WP - \frac{2Wh'}{c} = 0
$$

$$
P = W + \sqrt{W^2 + \frac{2Wh'}{c}} = 39.24 + \sqrt{1539.8 + 3.2950\times10^{6}} = 1854.9\ \text{N}
$$

Spring compression $= P/k = 46.37$ mm; bar elongation $= PL/AE = 0.01624$ mm (very small).

#### Stress in the bar
$$
\sigma = \frac{P}{A} = \frac{1854.9}{706.86} = 2.624\ \text{N/mm}^2
$$

**Answer: maximum stress in the bar $\approx 2.62$ N/mm$^2$** (maximum load 1.855 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 0.056 N/mm$^2$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Spring and bar are in series, so:
   A. Stiffnesses add
   B. The same force passes through both and flexibilities add
   C. The bar carries all the force
   D. Spring carries none
   Answer: B. Series elements share force. (Key point: Series action)

2. Free fall before touching the spring is:
   A. 1.15 m
   B. 1.25 m
   C. 1.15 - 0.10 = 1.05 m
   D. 0.10 m
   Answer: C. Spring is 10 cm long. (Key point: Free fall)

3. The energy equation is:
   A. W h' = P
   B. W cP = P^2
   C. W(h' - cP) = c P
   D. W(h' + cP) = (1/2) c P^2
   Answer: D. δ = cP. (Key point: Energy balance)

4. Maximum stress in the bar is about:
   A. 2.62 N/mm^2
   B. 0.056 N/mm^2
   C. 26.2 N/mm^2
   D. 1.85 N/mm^2
   Answer: A. 1854.9 / 706.86. (Key point: Result)

## Close variants set in other exams

- [Falling mass on bar with spring, 125 cm](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Afalling-mass-bar-spring-125cm) — Theory of Structures I, chapter 2, asked 2×

## Related in other subjects

- [0.45 kg collar and spring on circular rod](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Acollar-spring-circular-rod-0-45kg) — Applied Mechanics (Dynamics), chapter 2