Using SymPy and Jupyter to reduce 5 linear equations to a single equation in two variables.

Solving a variant of problem 6.3, Gray, Costanzo, and Plesha (Gray, G.L., Costanzo, F., and Plesha, M.E., Engineering Mechanics: Dynamics, McGraw Hill, New York, 2010.), where kinetics are added (\(I_1\), \(I_2\), \(I_3\)).

For my students: This is Wright State University, ME 3210, Quiz 2, Spring 2016

The problem statement is: A Moment \(M\) is applied to the rightmost gear (gear 3). Find the acceleration of the gear as a resulting moment presuming inertias \(I_1\), \(I_2\), \(I_3\).

Figure awaiting copyright approval. Dummy graphic in place for the moment.

Figure 1: Gear figure from Gray, Costanzo, and Plesha.

I am using Python 3.5 in Jupyter (formerly iPython). The original notebook is available at my Github examples repository.

 # import symbolic capability to Python
 from sympy import *
 # print things all pretty
 from sympy.abc import *
 init_printing()
 # Need to define variables as symbolic for sympy to use them.
 r_A, r_B, r_C, r_D, F_AB, F_DC, M= symbols("r_A r_B r_C r_D F_AB F_DC M", real = True)
 I_1, I_2, I_3 = symbols("I_1 I_2 I_3", real = True)
 theta_1, theta_2, theta_3 = symbols("theta_1 theta_2 theta_3", real = True)

Defining equations. In SymPy, and equation is an expression that is equal to zero. All terms must be brought to the same side in defining it.

1 2	EA = F_DC r_D - I_3 diff(theta_3(t),t,t) EA

\begin{equation*} F_{DC} r_{D} - I_{3} \frac{d^{2}}{d t^{2}} \theta_{3}{\left (t \right )} \end{equation*}

1 2	EB = r_B* F_AB - r_C* F_DC - I_2 * diff(theta_2(t),t,t) EB

\begin{equation*} F_{AB} r_{B} - F_{DC} r_{C} - I_{2} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )} \end{equation*}

1 2	EC = r_A (-F_AB) + M - I_1 diff(theta_1(t),t,t) EC

\begin{equation*} - F_{AB} r_{A} - I_{1} \frac{d^{2}}{d t^{2}} \theta_{1}{\left (t \right )} + M \end{equation*}

1 2	E1 = diff(theta_3(t),t,t)* r_D - diff(theta_2(t),t,t) *r_C E1

\begin{equation*} - r_{C} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )} + r_{D} \frac{d^{2}}{d t^{2}} \theta_{3}{\left (t \right )} \end{equation*}

1 2	E2 = r_B * diff(theta_2(t),t,t) - diff(theta_1(t),t,t) *r_A E2

\begin{equation*} - r_{A} \frac{d^{2}}{d t^{2}} \theta_{1}{\left (t \right )} + r_{B} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )} \end{equation*}

1	solve(EA,diff(theta_3(t),t,t))

\begin{equation*} \left [ \frac{F_{DC} r_{D}}{I_{3}}\right ] \end{equation*}

1	solve(EB,F_AB)[0]

\begin{equation*} \frac{1}{r_{B}} \left(F_{DC} r_{C} + I_{2} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )}\right) \end{equation*}

I'm going to do substitutions in stages, eliminating one variable at a time for clarity.

1 2	step1 = EC.subs(F_AB, solve(EB,F_AB)[0]) step1

\begin{equation*} - I_{1} \frac{d^{2}}{d t^{2}} \theta_{1}{\left (t \right )} + M - \frac{r_{A}}{r_{B}} \left(F_{DC} r_{C} + I_{2} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )}\right) \end{equation*}

1 2	step2 = step1.subs(F_DC,solve(EA,F_DC)[0]) step2

\begin{equation*} - I_{1} \frac{d^{2}}{d t^{2}} \theta_{1}{\left (t \right )} + M - \frac{r_{A}}{r_{B}} \left(I_{2} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )} + \frac{I_{3} r_{C}}{r_{D}} \frac{d^{2}}{d t^{2}} \theta_{3}{\left (t \right )}\right) \end{equation*}

1 2	step3 = step2.subs(diff(theta_1(t),t,t),solve(E2,diff(theta_1(t),t,t))[0]) step3

\begin{equation*} - \frac{I_{1} r_{B}}{r_{A}} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )} + M - \frac{r_{A}}{r_{B}} \left(I_{2} \frac{d^{2}}{d t^{2}} \theta_{2}{\left (t \right )} + \frac{I_{3} r_{C}}{r_{D}} \frac{d^{2}}{d t^{2}} \theta_{3}{\left (t \right )}\right) \end{equation*}

1 2	step4 = step3.subs(diff(theta_2(t),t,t),solve(E1,diff(theta_2(t),t,t))[0]) step4

\begin{equation*} - \frac{I_{1} r_{B} r_{D}}{r_{A} r_{C}} \frac{d^{2}}{d t^{2}} \theta_{3}{\left (t \right )} + M - \frac{r_{A}}{r_{B}} \left(\frac{I_{2} r_{D}}{r_{C}} \frac{d^{2}}{d t^{2}} \theta_{3}{\left (t \right )} + \frac{I_{3} r_{C}}{r_{D}} \frac{d^{2}}{d t^{2}} \theta_{3}{\left (t \right )}\right) \end{equation*}

Finally, the acceleration of the right most gear as a function of the input moment

1	solve(step4,diff(theta_3(t),t,t))

\begin{equation*} \left [ \frac{M r_{A} r_{B} r_{C} r_{D}}{I_{1} r_{B}^{2} r_{D}^{2} + I_{2} r_{A}^{2} r_{D}^{2} + I_{3} r_{A}^{2} r_{C}^{2}}\right ] \end{equation*}

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