---
title: "Falling mass on bar with spring, 125 cm"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:falling-mass-bar-spring-125cm"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/falling-mass-bar-spring-125cm
site: BCT Engineering
---

# Falling mass on bar with spring, 125 cm

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

- Asked in: 2071 Magh, 6 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-125cm
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

A bar of 2 cm diameter and of length 125 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 one metre measured from the collar top. Take $g = 9.81\ \text{m/s}^2$, stiffness of the spring $k = 40$ kN/m and $E = 210\ \text{GN/m}^2$.

## Short answer

A 4kg mass falls 1m onto a spring (k=40kN/m, free length 10cm) in series with a bar (2cm diameter, 125cm long, E=210GPa). Energy balance gives maximum force P = 1719.9N, stress in bar σ = P/A = 5.47 N/mm².

## Answer

#### 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$^2$.

## Check yourself

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

1. The mass falls 1.0 m measured from the top of the collar, and the spring is 10 cm long. What free-fall height h' is used before the mass touches the spring?
   A. 1.00 m
   B. 0.90 m
   C. 1.10 m
   D. 0.10 m
   Answer: B. The mass touches the spring top 10 cm above the collar, so it falls freely 1.0 - 0.10 = 0.90 m. (Key point: Free fall height)

2. Why are the spring and the bar combined by adding their flexibilities (1/k + L/AE)?
   A. They carry the same force P, so their deformations add
   B. They have the same deformation but different forces
   C. The spring is much stiffer than the bar
   D. Their strain energies are always equal
   Answer: A. In series the same force passes through both, so total deformation is the sum of the spring compression and bar extension, giving c = 1/k + L/AE. (Key point: Spring and bar in series)

3. Which energy equation gives the maximum force P?
   A. W h' = P delta
   B. W (h' + delta) = (1/2) P delta, with delta = cP
   C. W h' = (1/2) P^2
   D. W delta = P h'
   Answer: B. The loss of potential energy of the mass over the fall plus the extra deformation equals the strain energy (1/2)P delta stored in spring and bar. (Key point: Energy balance)

4. After finding P = 1719.9 N, how is the stress in the bar obtained and what is it?
   A. sigma = P/k = 43 MPa
   B. sigma = P/A = about 5.47 N/mm^2
   C. sigma = W/A = 0.125 N/mm^2
   D. sigma = P L / (A E) = 0.0326 N/mm^2
   Answer: B. Stress is the maximum force divided by the bar area 314.16 mm^2, giving about 5.47 N/mm^2; the static value 0.125 is only for the weight applied gradually. (Key point: Stress in bar)

## Close variants set in other exams

- [Falling mass on bar with spring, 130 cm](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Afalling-mass-bar-spring-130cm) — 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