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## Q. 2.2

A 200-kg machine is attached to the end of a cantilever beam of length L = 2.5 m, elastic modulus $E = 200 \times 10^9 N/m^2$, and cross-sectional moment of inertia $1.8\times 10^–6 m^4$. Assuming the mass of the beam is small compared to the mass of the machine, what is the stiffness of the beam?

## Verified Solution

From Table D.2 the deflection equation for a cantilever beam with a concentrated unit load at z=L is

$\omega \left(z\right) =\frac{1}{EI}\left(-\frac{1}{6}z^3+\frac{1}{2}z^2 \right)$          (a)

The deflection at the end of the beam is

$\omega \left(L\right)=\frac{1}{EI}\left(-\frac{L^3}{6}+\frac{1}{2}L^2 \right)=\frac{L^3}{3EI}$          (b)

The stiffness of the cantilever beam at its end is

$k=\frac{3EI}{L^3}=\frac{3\left(200\times 10^9 N/m^2\right)\left(1.8\times 10^-6 m^4\right) }{\left(2.5m\right)^3 } = 6.91\times 10^4 N/m$          (c)

 T A B L E    D . 2 The deflection, y(z), of a uniform beam of elastic modulus E and cross-sectional moment of inertia I due to a unit concentrated load applied at z = a is $y\left(z\right)=\frac{1}{EI} \left[\frac{1}{6}\left(z-a\right)^3 u \left(z-a\right)+ \frac{1}{6}\sum\limits_{i=1}^{n}{R_{i}\left(z-z_{i} \right)^3 u\left(z-z_{i} \right)+C_{1} \frac{z^3}{6}+C_{2} \frac{z^2}{2}+C_3z+C_4 }\right]$ where $R_{i}$ is the reaction at an intermediate support located at $z = z_{i}$. The forms of the constants and the intermediate reactions for common beams are given as follows. $C_{1}=-1$                     $C_{3}=0$ $C_{2}=a$                        $C_{4}=0$ $C_{1}=\frac{a}{L}-1$                     $C_{3}=\frac{aL}{6}\Bigl(1-\frac{a}{L} \Bigr)\Bigl(2-\frac{a}{L}\Bigr)$ $C_{2}=0$                             $C_{4}=0$ $C_{1}=\frac{1}{2}\left(1-\frac{a}{L} \right)\left[\left(\frac{a}{L} \right)^{2}-2\frac{a}{L} -2 \right]$               $C_{3}=0$ $C_{2}=\frac{1}{2}a\left(1-\frac{a}{L} \right)\left(2-\frac{a}{L} \right)$                            $C_{4}=0$ $C_{1}=-\left(1-\frac{a}{L} \right)^{2}\left(1+\frac{2a}{L} \right)$                     $C_{3}=0$ $C_{2}=a\left(1-\frac{a}{L} \right)^{2}$                                        $C_{4}=0$ $C_{1}= -\frac{3}{2} +\frac{3a}{2z_{1}}+\frac{1}{2}\left(1-\frac{a}{z_{1}} \right)^{3}u\left(z_{1}-a\right)$                                 $C_{3}=0$ $C_{2}=\frac{z_{1}}{2} \left(1-\frac{a}{z_{1}} \right)\left[1-\left(1-\frac{a}{z_{1}}\right)^{3} u\left(z_{1}-a\right) \right)$                          $C_{4}=0$ $R_{1}=\frac{1}{2}-\frac{3a}{2z_{1}}-\frac{1}{2} \left(1-\frac{a}{z_{1}}\right)^{3} u\left(z_{1}-a\right)$ $C_{1}=\frac{a}{z_{1}}-1$                     $C_{3}=-\left(1-\frac{a}{z_{1}}\right)\frac{z^{2}_{1} }{6}\left[\left(1-\frac{a}{z_{1}}\right)^{2}u\left(z_{1}-a\right)-1 \right]$ $C_{2}=0$                                $C_{4}=0$ $R_{1}=-\frac{a}{z_{1}}$