FVE2 - Machine on simply supported beam

A machine of mass $m=500[\mathrm{kg}]$ is mounted on a simply supported beam of length $l=2[\mathrm{m}]$ having a regular cross section (depth = 0.1 [m], width = 1.2 [m]\), and Young's modulus $E=2.06\times {10}^{11}\left[\frac{\mathrm{N}}{[m{]}^2}\right]$. To reduce the vertical deflection of the beam, a spring of stiffness $k$ is attached at mid-span. Calculate the stiffnes $k$ needed to reduce the deflection of the beam by 25, 50, and 75 %.

Explanation

To solve this example first the moment of inertia of the cross section of the beam must be calculated using the expression $$I=\frac{wd}{l}$$ where $w$ and $d$ are dimensions of the cross section, and $l$ is the length of the beam.
Next step is to calculate the stiffness of the simply supported beam, for a load at the middle. The stiffness is calcualted using the following expression: $$k_1=\frac{48EI}{l^3}$$ where $E$ is Young's modulus, $I$ is the first moment of inertia of the beams cross-section, and $l$ is the length of the beam.
To calculate the deflection i.e. original deflection the following equation is required. $$\delta =\frac{G}{k_1}=\frac{mg}{k_1}$$ where $m,g,and{k}_1$ are mass, gravitational acceleration, and stiffness.
To calculate the required stiffness of the attached spring at mid-span to reduce the deflection by 25, 50 and 75% we will start by introducing the equivalent stiffness that is sum of the original stiffness $k_1$ and the new unknown stiffness. $$k_{eq}=k_1+k$$ To determine the unknown value $k$ simple use the equation: $$\frac{mg}{k_{eq}}=\frac{\delta }{4}$$ the right side of equation will be equal to $\frac{\delta }{4}$ in case the deflection has to be reduced by 25%, $\frac{\delta }{2}$ in case the deflection has to be reduced by 50%, and $\frac{3\delta }{4}$ if the deflection has to be reduced by 75%. From previous equation the $k_{eq}$ is determined.

Solution

The parameters that where given in the problem description. $$\begin{array}{rcl}m& =& 500[\mathrm{kg}]\\ l& =& 2[\mathrm{m}]\\ d& =& 0.1[\mathrm{m}]\\ w& =& 1.2[\mathrm{m}]\\ E& =& 2.06\times {10}^{11}\left[\frac{\mathrm{N}}{{\mathrm{m}}^2}\right]\\ g& =& 9.81\left[\frac{m}{s^2}\right]\end{array}$$ We need to calculate the moment of inertia of the cross section of the beam: $$\begin{array}{rcl}I& =& \frac{w{d}^3}{12}\\ I& =& \frac{1.2\cdot {0.1}^3}{12}\\ I& =& {10}^{-4}\left[{\mathrm{m}}^4\right]\end{array}$$ The stiffness of simply supported beam, for load at middle: $$\begin{array}{rcl}{k}_1& =& \frac{48EI}{l^3}\\ k_1& =& \frac{48(2.06\times {10}^{11})({10}^{-4})}{8}\\ k_1& =& 12.36\times {10}^7\left[\frac{\mathrm{N}}{\mathrm{m}}\right]\end{array}$$ The original deflection: $$\begin{array}{rcl}\delta & =& \frac{G}{k_1}=\frac{mg}{k_1}\\ \delta & =& \frac{500\cdot 9.81}{12.36\cdot {10}^7}\\ \delta & =& 396.8447\cdot {10}^{-7}[\mathrm{m}]\end{array}$$ If we add the spring of stiffness k to the system then the equivalent spring stiffness is equal to: $$k_{eq}=k+k_1$$

1. Determine the spring stiffness $k$ if the beam deflection is reduced by 25%.
$$\begin{array}{rcl}\frac{mg}{k_{eq}}& =& \frac{\delta }{4}\\ k_{eq}& =& \frac{4mg}{\delta }\\ k_{eq}& =& 4k_1=k+k_1\\ k& =& 3\cdot k=3\ast 12.36\cdot {10}^7\\ k& =& 37.08\cdot {10}^7\left[\frac{\mathrm{N}}{\mathrm{m}}\right]\end{array}$$
2. Determine the spring stiffness $k$ if the beam deflection is reduced by 50\% of its original value.
$$\begin{array}{rcl}\frac{mg}{k_{eq}}& =& \frac{\delta }{2}\\ k_{eq}& =& \frac{2mg}{\delta }\\ k_{eq}& =& 2k_1=k+k_1\\ k& =& k=12.36\cdot {10}^7\left[\frac{\mathrm{N}}{\mathrm{m}}\right]\end{array}$$
3. Determine the spring stiffness $k$ if the beam deflection is reduced by 75\% of its original value.
$$\begin{array}{rcl}\frac{mg}{k_{eq}}& =& \frac{3\delta }{4}\\ k_{eq}& =& \frac{4mg}{3\delta }\\ k_{eq}& =& \frac{4}{3}{k}_1=k+k_1\\ k& =& \frac{1}{3}k=\frac{1}{3}12.36\cdot {10}^7\\ k& =& 4.12\times {10}^7\left[\frac{\mathrm{N}}{\mathrm{m}}\right]\end{array}$$

Solution in Python

To solve this example in Python we will need two libraries i.e. numpy and matplotlib.

import numpy as np
import matplotlib.pyplot as plt

The first step is to define the mass, length, depth, widht, and Youg's modulus.

m = 500
l = 2
d = 0.1
w = 1.2
E = 2.06e11
g = 9.81

Now calculate the moment of inertia:

I = (w*d**3)/12
print(f'I = {I}')

After executing the code written so far the following output is obtained.

I = 0.00010

The stiffness of simply supported beam, for load at the middle:

k1 = (48*E*I)/l**3
print(f'k1 = {k1}')

The output of the previous print function is

k1 = 123600000

The original deflection

delta = (m*g)/k1
print(f'delta = {delta}')

The print function will show the value of the delta variable.

delta = 3.968446601941746e-05

Now the k will be calculated as:

k = [3*k1, k1, k1/3]
print(f'k = {k}')

Since we have to calculate the k for different cases of beam deflection reduction we have used already derived expression and created a list were all three values will be stored. The output of the last print function is given below.

k = [370800000.0000001, 123600000.00000004, 41200000.000000015]