---
title: "Dynamic multiplier for falling mass"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:dynamic-multiplier-falling-mass"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/dynamic-multiplier-falling-mass
site: BCT Engineering
---

# Dynamic multiplier for falling mass

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

- Asked in: 2077 Chaitra, 3 marks; 2072 Asoj, 4 marks (asked 2 times)
- 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/dynamic-multiplier-falling-mass
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

What is a dynamic multiplier? Derive the expression for the dynamic multiplier when a mass falls on the mid-span of a simply supported beam.

## Short answer

Dynamic multiplier = δ_max / δ_st. For a weight W falling from height h onto a simply supported beam, static deflection δ_st = WL^3/(48EI). Energy balance gives multiplier = 1 + sqrt(1 + 2h/δ_st).

## Answer

The **dynamic multiplier** (impact factor) is the ratio of the maximum dynamic deflection (or stress) to the static deflection (or stress) produced by the same load:

$$
\text{Dynamic multiplier} = \frac{\delta_{max}}{\delta_{st}}
$$

#### Mass falling on the mid-span of a simply supported beam
Let a weight $W$ fall from a height $h$ on to the mid-span of a simply supported beam of span $L$ and flexural rigidity $EI$. Let $\delta$ be the maximum mid-span deflection and neglect the beam's weight and energy losses.

Static deflection of the beam under $W$ at mid-span:

$$
\delta_{st} = \frac{WL^3}{48EI}
$$

So the beam stiffness at mid-span is $k = \dfrac{W}{\delta_{st}} = \dfrac{48EI}{L^3}$, and the equivalent dynamic load is $P = k\delta$.

**Energy balance:** loss of potential energy of the falling weight = strain energy stored in the beam.

$$
W(h + \delta) = \tfrac12 P\delta = \tfrac12 k\delta^2 = \tfrac12\frac{W}{\delta_{st}}\delta^2
$$

$$
\delta^2 - 2\delta_{st}\delta - 2\delta_{st}h = 0
$$

$$
\delta = \delta_{st}\left[1 + \sqrt{1 + \frac{2h}{\delta_{st}}}\right]
$$

Therefore the dynamic multiplier is

$$
\boxed{\frac{\delta}{\delta_{st}} = 1 + \sqrt{1+\frac{2h}{\delta_{st}}}}
$$

The maximum bending moment and stress are multiplied by the same factor, since the structure is linear. If $h = 0$ (suddenly applied load) the multiplier is 2. If $h \gg \delta_{st}$, then $\delta \approx \sqrt{2h\,\delta_{st}}$.

## Check yourself

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

1. Dynamic multiplier is:
   A. Ratio of static to dynamic deflection
   B. Ratio of height to deflection
   C. Ratio of maximum dynamic deflection to static deflection for the same load
   D. Ratio of E to G
   Answer: C. It is the impact factor. (Key point: Definition)

2. The energy balance for a falling weight is:
   A. W h = k δ
   B. W δ = k δ^2
   C. W(h - δ) = (1/2) k δ^2
   D. W(h + δ) = (1/2) k δ^2
   Answer: D. Potential energy lost includes extra fall δ. (Key point: Energy balance)

3. The dynamic multiplier is:
   A. 1 + sqrt(1 + 2h/δst)
   B. 1 + sqrt(1 + h/δst)
   C. sqrt(1 + 2h/δst)
   D. 2 + sqrt(2h/δst)
   Answer: A. Solving the quadratic in δ. (Key point: Result)

4. For a suddenly applied load (h = 0) the multiplier is:
   A. 1
   B. 2
   C. 4
   D. 1/2
   Answer: B. 1 + sqrt(1) = 2. (Key point: Special case)

## Related in Theory of Structures I

- [Beam mid-span deflection with temperature gradient](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-midspan-defl-temperature) — Theory of Structures I, chapter 3
- [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4
- [ILD at mid-span: BM and SF under given loads](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-mid-span-bm-sf-50-10-30) — Theory of Structures I, chapter 5
- [ILD and BM at mid span for triangular load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-triangular-load-mid-span-c) — Theory of Structures I, chapter 5
- [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×
- [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

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