I have a dynamic model which undergoes two distinct stages. The system starts out with certain initial conditions and once a specific point is reached in stage 1, these ending conditions are used as the initial conditions for stage 2, and the remainder of the solution is calculated. I am attempting to solve for the steady-state equilibrium point by taking the derivative of the potential energy with respect to each generalized coordinate, setting equal to zero, and solving. For stage 1:

$$ V_{im} = \frac{1}{2}k_1y_1^2 + \frac{1}{2}k_2(y_2-y_1)^2 + k_r \theta_o^2+m_1gy_1 \\ +m_2g(y_1+y_2)+2m_3g(y_1-l \sin{\theta_o}) +\frac{1}{4}k_{1n}y_1^4+\frac{1}{4}k_{2n}(y_2-y_1)^4 $$ and for stage 2:

$$ V_{de} = \frac{1}{2}k_1y_1^2+\frac{1}{2}k_2(y_2-y_1)^2+k_r\theta^2 + m_1gy_1 \nonumber\\ + m_2g(y_1+y_2) + 2m_3g(y_1-l\sin{\theta}) + \frac{1}{4}k_{1n}y_1^4 + \frac{1}{4}k_{2n}(y_2-y_1)^4 $$ where $ \theta_o$ was a constant while $ \theta$ is variable. Still, these reveal the same two derivatives:

$$ \frac{\partial V}{\partial y_1} = k_1y_1+k_2(y_1-y_2)+g(m_1+m_2+2m_3) +k_{1n}y_1^3-k_{2n}(y_2-y_1)^3=0 $$ and

$$ \frac{\partial V}{\partial y_2} = k_2(y_2-y_1)+m_2g +k_{2n}(y_2-y_1)^3= 0 $$

Therefore, I would assume that both stages would have the same steady-state position, but this is not always the case once I run the entire simulation numerically in MATLAB (details not shown). Sometimes the steady-state of stage 1 matches the calculated value and not stage 2, and sometimes it is the opposite way around. What could be causing this discrepancy?

For comparison, stage 1 and stage 2 position graphs as described are below:

enter image description here enter image description here

The values of the constants are the same throughout the simulations.

  • $\begingroup$ You have a cubic equation, so it has three roots. When finding zero points, you have to test them for minima, saddle points or maxima. This is not always easy. But any good book on generalized coordinate methods should cover it. $\endgroup$
    – Phil Sweet
    Apr 28, 2021 at 22:15
  • $\begingroup$ Hi Phil, that is a good point and I hadn't thought of that yet. Even so, the equilibrium values for stage 2 don't solve the equations which is quite strange to me. I would expect, if anything, for it to be another root, but it isn't. Do you think it could be something else? $\endgroup$
    – alcopo63q
    Apr 29, 2021 at 16:07


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.