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Business. There are three new inserts: 2A, 2B and 2C. Some of you are asking me where the lectures are posted Go to our website http://www.me.rochester.edu/courses/ME213/ Click on schedule Go to the day of the lecture (I am not posting in advance). Lecture 5: Friction and damping.
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Business There are three new inserts: 2A, 2B and 2C Some of you are asking me where the lectures are posted Go to our website http://www.me.rochester.edu/courses/ME213/ Click on schedule Go to the day of the lecture (I am not posting in advance)
Lecture 5: Friction and damping dry friction, nonlinear, not easily expressed analytically linear “viscous” damping, proportional to velocity
dry (sliding) friction f µW W Friction force opposite to the direction of motion Obviously the friction force cannot generate motion and so does not act to exceed the applied force, which would generate motion to the left
friction force no motion motion applied force
We have static and dynamic friction coefficients The former generally larger than the latter µW y2(t) y (There are more complicated models, but we won’t deal with them.)
Some comments We can measure the static and dynamic coefficients using an inclined plane
When it moves we’ll have This is a simple initial value problem. Integrating twice and applying the initial conditions I can replace the trigonometric functions and do a bit of algebra to give me
The equation of motion is where the coefficient of friction comes from the previous figure This is not amenable to simple mathematics I’ll talk about what we can do at the end of the lecture
If we add a thin layer of liquid, we’ll have viscous friction The resistance to motion will be proportional to the shear stress in the fluid For a thin layer the flow will be laminar and the shear stress will be proportional to the speed which we can generalize to
This will be our model for friction, and for damping in general Shock absorbers, screen door dampers, various hydraulic dampers all work this way. I’ll use it as a general approximation for all damping Now let’s look at the model analytically.
damper/dashpot damping constant y combine in the same way as springs
c5 c1 c2 c4 m c3 c6
c5 c7 c4 c3 m c6
c5 c4 c8 m c6
c8 c4 c9 m
c8 c10 m
mass-spring-damper system m c f k
natural frequency damping ratio General forced, damped one degree of freedom equation
unforced, a = 0 How do we solve this? Linear homogeneous ordinary differential equations with constant coefficients have exponential solutions!
The equation is not the entire problem; we need initial conditions
characteristic polynomial s is real if z > 1 overdamped s is complex if z < 1 underdamped s has a repeated root (-wn) if z = 1 critically damped
initial conditions combine (1) and (2) solve (3) to get (4)
This is a general expression for any value of z The most interesting and common case is the underdamped (z < 1) case.
In the underdamped case the denominators are imaginary s2 is the complex conjugate of s1 the two terms are complex conjugates and the expression is real as it must be
In the overdamped case the formula in the box is obviously real In the critically damped case, the two values of s are the same, and the differential equation is One solution is e-wt The other is t times this, as we can show
calculate y and its derivatives substitute into equation (1) all the terms cancel
Only the underdamped case has complex exponents (and associated complex coefficients). It’s also the most common case in engineering.
Let’s pursue this case if z < 1 the damped natural frequency
We can use that to rewrite the exponentials in terms of trigonometric functions
We can use this to write the general solution as avoiding explicit complex arithmetic damping ratio ONLY FOR THE UNDERDAMPED CASE
The connection between the two forms expand red for cosine blue for sine
How do you measure the decay rate — the damping factor? The log-decrement method: text pp. 40-43 The idea is that successive maxima are related by z Start with a little analysis