(a) The steps involved in modeling the system of Figure 2.11(a) by the system of Figure 2.8 are shown in Figure 2.12. Equation (2.28) is used to replace the two parallel springs by an equivalent spring of stiffness 3k. The three springs on the left of the mass are then in series, and Equation (2.31) is used to obtain an equivalent stiffness.
If the mass in Figure 2.11(a) is given a displacement x to the right, then the spring on the left of the mass will increase in length by x, while the spring on the right of the mass will decrease in length by x. Thus, each spring will exert a force to the left on the mass. The spring forces add; the springs behave as if they are in parallel. Hence Equation (2.28) is used to replace these springs by the equivalent spring shown in Figure 2.12(c). (b) The deflection of the simply supported beam due to a unit load at x 2 m is calculated using Table D.2

\omega \left(z=2m\right) =\omega \left(\frac{2L}{3} \right) =\frac{4l^3}{243EI}           (a)

from which the equivalent stiffness is obtained

k_1=\frac{243EI}{4L^3}=\frac{243\left(210\times 10^9 N/m^2\right)\left(5\times 10^{-4}m^4\right) }{4\left(3m\right)^{3} }=2.36\times 10^{8}N/m          (b)

The displacement of the block of mass m equals the displacement of the beam at the location where the spring is attached plus the change in length of the spring. Hence the beam and spring act as a series combination. Equation (2.31) is used to calculate their equivalent stiffness

k_eq=\frac{1}{\frac{1}{2.36\times 10^{8}N/m} +\frac{1}{1\times 10^{8}N/m} } =7.03\times 10^{7}N/m         (c)

(c) The aluminum core of shaft AB is rigidly bonded to the steel shell. Thus the angular rotation at B is the same for both materials. The total resisting torque transmitted t section BC is the sum of the torque developed in the aluminum core and the torque developed in the steel shell. Thus the aluminum core and steel shell of shaft AB behave as two torsional springs in parallel. The resisting torque in shaft AB is the same as the resisting torque in shaft BC. The angular displacement at C is the angular displacement of B plus the angular displacement of C relative to B. Thus shafts AB and BC behave as two torsional springs in series. In view of the preceding discussion and using Equations (2.28) and (2.31), the equivalent stiffness of shaft AC is

k_eq=\frac{1}{\frac{1}{k_{t_{AB_{al}}}} +\frac{1}{k_{t_{AB_{st}}}}+\frac{1}{k_{t_{BC}}} }           (d)

where the torsional stiffness of a shaft is k_t=JG/ L and

k_{t_{AB_{al}}}=\frac{\frac{\pi}{32}\left(0.04m\right)^{4}\left(40\times 10^{9}\frac{N}{m^{2}}\right) }{0.3m} =3.35\times 10^{4 }\frac{N.m}{rad}         (e)

k_{t_{AB_{st}}}=\frac{\frac{\pi}{32}\left[\left(0.05m\right)^{4}- \left(0.04m\right)^{4}\right] \left(80\times 10^{9}\frac{N}{m^{2}}\right) }{0.3m} =9.66\times 10^{4 }\frac{N.m}{rad}          (f)

k_{t_{BC}}=\frac{\frac{\pi}{32}\left[\left(0.06m\right)^{4}- \left(0.036m\right)^{4}\right] \left(80\times 10^{9}\frac{N}{m^{2}}\right) }{0.2m} =4.43\times 10^{5}\frac{N.m}{rad}          (g)

Substitution of these values into the equation for k_{eq} gives

k_{t,eq}= 1.01\times 10^{5}N.m/rad          (h)

d) Under the assumption that the rate of taper of the bar is small the following mechanics of materials equation is used to calculate the change in length of the bar due to a unit load applied at its end:

\Delta =\int_{0}^{L}{\frac{dz}{AE} }         (i)

The area varies linearly over the length of the bar A=\left(h_{1}-\frac{h_{1}-h_{2}}{L}z \right)bThe change in
length is

\Delta =\frac{1}{bE}\int_{0}^{L}{\frac{dz}{h_{1}-\frac{h_[1]-h_{2}}{L}z } }= \frac{1}{bE}\left(\frac{-1}{h_[1]-h_{2}} \right)In \left(h_{1}- \frac{h_[1]-h_{2}}{L} \right)\mid ^{L}_{0} =\frac{L}{bE\left(h_[1]-h_{2}\right) } In \left(\frac{h_{1}}{h_{2}} \right)

=\frac{2m}{\left(0.013m\right)\left(210\times 10^{9}N/m^{2}\right)\left(0.025m-0.02m\right) } In\frac{0.025m}{0.02m}

=3.27\times 10^{8}m/N          (j)

Thus, the equivalent stiffness of the shaft is

k_{eq}= \frac{1}{\Delta } =\frac{1}{3.27\times 10^{-8}m/N} = 3.06\times 10^{7}N/m          (k)

k_{eq}=\sum\limits_{i=1}^{n}{k_{i}}          (2.28)
k_{eq}=\frac{1}{ \sum\limits_{i=1}^{n}\frac{1}{k_{1}} }         (2.31)

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}}