|Feb22-10, 11:23 AM||#1|
Inertial forces in crank slider mechanism
I was little confused about taking inertial forces into significance while calculating piston effort in a piston crank slider arrangement. I read an analysis of various forces acting in a piston cylinder arrangement in a book yersterday, and while calculating piston effort(net force on piston while piston is accelerating) they took inertial force into account which was opposite to direction of piston movement.
Whats the need of inertial force here? Basically what it's significance? I am a bit confused!
I have never done any analysis before taking inertial forces into account and applying de alembert's principle.
Please anyone explain me. It's urgent! :)
|Jun29-11, 09:15 AM||#2|
Newton's Second Law (F=ma) is applicable only if you're working in the Inertial Frame of Reference (Earth or any other frame moving at a constant velocity wrt Earth). You don't even need D'Alembert Principle when working in this Frame.
However, when you're working in an Accelerating Non Inertial Frame of Reference, you apply D'Alembert Principle to negate the effect of a Virtual Force which appears to be working but is not there.
Consider the classic case of a Girl on Roller Skates standing in the aisle of a bus. When bus starts moving (Force on Bus only, not the Girl), the girl appears to be stationary if seen from the Earth. Because Earth is an inertial frame of reference, Newton's Second Law (F=ma) is applicable. No Force No Acceleration.
If an Accelerating Non Inertial Frame of Reference eg. Bus is used, to a man standing at the back of the bus, the girl would appear to accelerate towards him. This acceleration is due to a virtual force which is not present physically. This virtual force needs to be taken care of because the Net Physical Force on the Girl is Zero (Girl is not being pushed or Pulled inside the Bus), there should be no acceleration.
To find the actual Physical Force on the Girl, we need to subtract the Virtual Force (which is responsible for Girl's acceleration,af and is equal to the acceleration of the Frame of Reference, Bus). Using
F = m*(af + ar) or F - m*af = m*ar
where F = Net Physical Force on the Object
af=acceleration of the Frame of Reference (+/-)
ar=acceleration of Object relative to the Accelerating Frame (-/+)
m*af = Virtual Force
F=0 (which is what should be).
D'Alembert Principle is useless when working in an Inertial Frame of Reference (I don't think F=ma needs to be touched, it's as simple a principle as can be), D'Alembert Principle finds an application when working in an Accelerating Non Inertial Frame of Reference.
So, F-ma = 0 is hardly helpful (it's rather confusing), the only form in which it's helpful is F - m*af = m*ar.
While calculating Piston Effort, the Frame of Reference used is the system in which all the parts are operating and all the forces are internal to the system. Hence, Virtual Force is subtracted.
|Jun29-11, 09:28 AM||#3|
I'll be happy to explain it to you further if you want more eloboration on Piston Effort.
|Jun29-11, 09:38 AM||#4|
Inertial forces in crank slider mechanism
The forces acting on a piston as the gas pressure pushing down from above on the cumbustion stroke and forces dues to inertia becuase the piston in constantly experienceing acceleration.
At low speeds the pressure force greatly exceeds the inertia forces so they can be ignored. At higher engine speeds inertia loads begin to dominate.
You can get the inertia forces by using simple kinematics to calcualte the acceleration of the components at certain points throughout the cycle. Multiplying this by the mass of each component will give you the inertia force.
I only have a low RPM indicated torque output plot.
In this image you can see a peak in torque output at the about 20-30 deg aTDC due to the combustion force. Howver you can see that the green line moves up and down. This is due to inertia. As RPM increases these peaks will become more and more extreme.
I will try to post a second picture showing the inertia dominated cycle, if I can find it at home.
|Similar Threads for: Inertial forces in crank slider mechanism|
|Slider crank -- with a twist!||General Physics||0|
|Slider Crank Problem||Mechanical Engineering||3|
|crank and rocker mechanism||Mechanical Engineering||5|
|Proving a 2D slider crank mechanism fits the Gruebler's Equation||Engineering, Comp Sci, & Technology Homework||1|
|crank slider-4 bar linkage||Mechanical Engineering||8|