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Please help to implement following differential equation numerical solution in mathcad
Numerical solution for single degree of freedom elastic system.
Spring stiffness = k = 2000 lbf/ft
Weight = W = 64 lbf
Force function:
F(t) = 500t+50 if t <=0.1
=-1250t+225 if 0.1 <=t<= 0.14
= 50 if t>=0.14
Acceleration function:
A(t) = ½ F(t) – 1000 *Y
Say time step = dt = 0.02
Initial Condition:
Displacement:
Y0 = 0 that is displacement is zero when t = 0
Y1 = ½ * A0 *dt^2 that is first time step which t = 0.02 this gives displacement but in terms of Acceleration at t =
Acceleration:
A0 = 0 when t = 0
Solution:
Y (t+dt) = 2 * Y (t) – Y(t-dt) + A(t) * dt^2
For t – 0 to 0.4 with dt = 0.02
Return:
Y, A
Never cross post.
Why have you posted this twice? Just a pointer - for a collab to help they will have manually type all the data into Mathcad, which takes times. It would be much easier if the question/query is accompanied with a worksheet.
Mike
Never cross post.
Now the worksheet has been postyed in the other thread, even though all the comments are in this one. Very irritating!
my bad. This was my first post and so was an mistake Hope that i am not adding to the mess here by attaching the sheet.
Alan Stevens posted a ODE solution but i was hoping to solve it numerically based on the original post. I am attaching my workseet
Again apologise for the inconvenince.
Thanks
Weight = W = 64 lbf
Weight is not forcce, it's mass.
Acceleration function:
A(t) = ½ F(t) – 1000 *Y
Acceleration is not force.
There are a number of ways to do this in Mathcad. Odesolve, Rkadapt, etc.
Let's see what you've tried. Post a sheet.
f.kohlhepp wrote:
Weight = W = 64 lbf
Weight is not forcce, it's mass.
Weight is indeed force.
Richard Jackson wrote:
Weight is indeed force.
Sorry! You're right of course.
Should have said that for this application we don't want force, we want mass. Acceleration is force/mass (most of the time.)
Should have said that for this application we don't want force, we want mass. Acceleration is force/mass (most of the time.)
True