---
title: "Lateral vibration of a simply supported beam"
subject: "Theory of Mechanism and Machine II"
chapter: 10
key: "semester_7/theory_of_mechanism_and_machine_ii:beam-lateral-vibration-simply-supported"
source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/beam-lateral-vibration-simply-supported
site: BCT Engineering
---

# Lateral vibration of a simply supported beam

Theory of Mechanism and Machine II · Chapter 10: Vibration of Continuous Systems

- Asked in: year not given, 5+3 marks
- Video: [Simply Supported Beam explained practically and analytically in detail](https://www.youtube.com/watch?v=2QZS685f0wg) — Vicky Dep
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/beam-lateral-vibration-simply-supported
- Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-10

## Question

(a) Derive the equation for the lateral (flexural) vibration of a uniform beam and obtain the natural frequencies of a simply supported beam.
(b) A simply supported steel beam of length 3 m has a rectangular section 50 mm wide and 100 mm deep (vibrating in the 100 mm deep direction). Find the first three natural frequencies. $E = 200$ GPa, $\rho = 7850$ kg/m³.

## Short answer

The Euler-Bernoulli equation is EI y'''' + rho A y'' = 0; for simply supported ends, sin(beta l)=0 gives omega_n = n^2 pi^2/l^2 sqrt(EI/(rho A)). With I=4.1667e-6 m^4 and A=0.005 m^2, the first three frequencies are 25.43 Hz, 101.72 Hz, and 228.88 Hz.

## Answer

#### (a) Euler-Bernoulli beam equation
Consider an element $dx$ of a beam with bending stiffness $EI$, mass per unit length $\rho A$, deflection $y(x,t)$, shear force $V$ and bending moment $M$.

```text
     V+dV
 M ->|====|<- M+dM
     V
```
Vertical equilibrium: $\;V - (V + dV) = \rho A\,dx\,\dfrac{\partial^2 y}{\partial t^2}$, that is $-\dfrac{\partial V}{\partial x} = \rho A\dfrac{\partial^2 y}{\partial t^2}$.

Moments (neglecting rotary inertia and shear deformation): $V = \dfrac{\partial M}{\partial x}$ and $M = EI\dfrac{\partial^2 y}{\partial x^2}$. Hence

$$
EI\frac{\partial^4 y}{\partial x^4} + \rho A\frac{\partial^2 y}{\partial t^2} = 0
$$

Put $y = Y(x)\sin\omega t$ and $\beta^4 = \dfrac{\rho A\omega^2}{EI}$:

$$
\frac{d^4Y}{dx^4} - \beta^4Y = 0 \;\Rightarrow\; Y = C_1\cos\beta x + C_2\sin\beta x + C_3\cosh\beta x + C_4\sinh\beta x
$$

**Simply supported ends:** $Y(0) = Y''(0) = 0 \Rightarrow C_1 = C_3 = 0$. At $x = l$, $Y(l) = Y''(l) = 0$ gives $C_4 = 0$ and

$$
\sin\beta l = 0 \;\Rightarrow\; \beta_nl = n\pi
$$

$$
\omega_n = (n\pi)^2\sqrt{\frac{EI}{\rho Al^4}} = \frac{n^2\pi^2}{l^2}\sqrt{\frac{EI}{\rho A}}, \qquad Y_n = \sin\frac{n\pi x}{l}
$$

#### (b) Numerical
$$
I = \frac{bh^3}{12} = \frac{0.05 \times 0.1^3}{12} = 4.1667\times10^{-6}\ \text{m}^4, \qquad A = bh = 0.005\ \text{m}^2
$$

$$
\sqrt{\frac{EI}{\rho A}} = \sqrt{\frac{200\times10^9 \times 4.1667\times10^{-6}}{7850 \times 0.005}} = 145.710\ \text{m}^2/\text{s}
$$

$$
\omega_1 = \frac{\pi^2}{3^2}\times 145.710 = 159.79\ \text{rad/s}, \qquad f_1 = 25.43\ \text{Hz}
$$

$$
f_2 = 4f_1 = 101.72\ \text{Hz}, \qquad f_3 = 9f_1 = 228.88\ \text{Hz}
$$

**Answer:** $f_1 = 25.43$ Hz, $f_2 = 101.72$ Hz, $f_3 = 228.88$ Hz.

## Check yourself

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

1. The Euler-Bernoulli equation for lateral vibration of a uniform beam is:
   A. EI y_xx + rho A y_tt = 0
   B. EI y_xxxx + rho A y_tt = 0
   C. EI y_xxxx = rho A y_t
   D. E y_xxxx = c^2 y_tt
   Answer: B. Combining V = dM/dx, M = EI y_xx and the dynamic equilibrium gives EI y'''' + rho A y_tt = 0. (Key point: Beam equation)

2. For simply supported ends, the conditions Y = Y'' = 0 at both ends lead to:
   A. cos(beta l) = 0
   B. cos(beta l) cosh(beta l) = 1
   C. sin(beta l) = 0, so beta l = n pi
   D. tan(beta l) = tanh(beta l)
   Answer: C. Both ends give C1 = C3 = C4 = 0 leaving sin(beta l) = 0. (Key point: Simply supported conditions)

3. The natural frequencies of the simply supported beam are proportional to:
   A. n
   B. n^2
   C. 2n - 1
   D. 1/n
   Answer: B. w_n = (n pi)^2 sqrt(EI/(rho A l^4)), so f2 = 4 f1 and f3 = 9 f1. (Key point: Frequency pattern)

4. For the 3 m steel beam (50 x 100 mm), the fundamental frequency is about:
   A. 159.8 Hz
   B. 25.43 Hz
   C. 101.7 Hz
   D. 6.4 Hz
   Answer: B. w1 = (pi^2/9) x 145.71 = 159.79 rad/s, f1 = 25.43 Hz (159.8 is in rad/s). (Key point: Numerical)

## Related in Theory of Mechanism and Machine II

- [Lateral vibration of a stretched string](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Astring-wave-equation-lateral-vibration) — Theory of Mechanism and Machine II, chapter 10
- [Rod longitudinal and shaft torsional frequencies](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Alongitudinal-rod-torsional-shaft-frequencies) — Theory of Mechanism and Machine II, chapter 10
- [Natural frequencies of a two-mass system](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Atwo-dof-natural-frequencies-mode-shapes) — Theory of Mechanism and Machine II, chapter 7
- [Undamped dynamic vibration absorber](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Adynamic-vibration-absorber-design) — Theory of Mechanism and Machine II, chapter 7
- [Vibration terms, effects and elements](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Avibration-terms-effects-notes) — Theory of Mechanism and Machine II, chapter 6
- [Natural frequency with series-parallel springs](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Aseries-parallel-spring-mass-frequency) — Theory of Mechanism and Machine II, chapter 6

## Related in other subjects

- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7
- [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4
- [Simply supported I-beam: stress at flange-web junction](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-beam-flange-web-junction-stress) — Strength of Materials, chapter 7