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Chapter 2

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