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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?

Step-by-Step

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}}