Connecting rod moment of inertia

Dear Maurizio:
Google, Advanced all the words: crankshaft connecting rod formula the exact phrase: moment of inertia 113 hits. A reasonable number, that a few minutes of reasearch can wade through. Your English seems to be good enough...
David A. Smith

total or at any instant? you need to sum more than one rod and ofset only if your time reference of your purpose is longer than the cycle of one rotation - or for curiosity - the flywheel is sized on other parameters, and anyway, it too only uses one rod and offset as its prime driver. do one rod and one offset - for total, times the number of pistons rotating within the time frame - and don't forget 1) the rotational mass moment of inertia since the rod rotates on the pin and shaft 2) the inertia of the crank offset 3) the connecting pin for rotational, use the pin as the reference point for linear, use the block. for mass moment, only rotational
rotational moment of inertia
Odd - since the mass moment is a cube function of length and straight-up on mass, and since the calculation is a fairly simple calculus problem.
Why would a professor assign a system set of inertials, which is fairly complex, and then set an erroneus assumption on a simple calculation?
Tell me it's not another professor who lacks real world experience..

Sum the contribution of each particle of the crank and con rod times its distance from the axis of rotation for each increment of rotation between 0 - 90 degrees.
Brian Whatcott Altus OK

they come in different flavors?

no, he's not a professor, he's an engineer :-) To further clarify the issue, I have to compute rotational moment of inertia about shaft axis (one cycle and one piston) of crankshaft+con rod system. Crankshaft alone is not a problem, the problem is the connecting rod, because its movement is a complex one, neither rotational nor linear. How much its rotational component affects crankshaft rotational moment of inertia, so what is moment of inertia of the whole crankshaft+con rod system about shaft axis ? The advice I got to simplify computation is to assume 1/3 con rod mass as it was in the pin, rotating about shaft axis, and 2/3 as it had linear alternate movement. So I computed moment of inertia about shaft axis of the crankshaft+assumed mass in the pin. I wonder if there is anything more precise than such rough computation.
Thanks
Maurizio
> >how can I calculate the moment of inertia of a crankshaft-connecting rod > >system ? > > total or at any instant? > you need to sum more than one rod and ofset only if your time reference of > your purpose is longer than the cycle of one rotation - or for curiosity > - the flywheel is sized on other parameters, and anyway, it too only uses > one rod and offset as its prime driver. > > do one rod and one offset - for total, times the number of pistons rotating > within the time frame - and don't forget > 1) the rotational mass moment of inertia since the rod rotates on the pin and > shaft > 2) the inertia of the crank offset > 3) the connecting pin > > for rotational, use the pin as the reference point > for linear, use the block. > for mass moment, only rotational > > >I've been told to assume 1/3 of connecting rod mass to roughly compute its > rotational moment of inertia > > Odd - since the mass moment is a cube function of length and straight-up on > mass, and since the calculation is a fairly simple calculus problem. > > Why would a professor assign a system set of inertials, which is fairly > complex, and then set an erroneus assumption on a simple calculation? > > Tell me it's not another professor who lacks real world experience..

none replies to my question, if you have found one please submit it.

I think you mean BOTH rotational and linear
All complex problems use the same approach in their solution - one part at a time, one reference point,dditive or multiplicative interaction
Thus the connecting rod is a mass acting at a distance from the axis of the crank. It is also a mass-moment multiplied by a lever arm which in turn acts thru an inefficient connection.
One step at a time. No easy way.
.

I believe the missing concept here is "reflected inertia" -- the apparent rotational inertia due to loads being driven by a shaft thru other mechanisms--a gear reduction or lead screw, for example. Or the slider-crank in this case, which is a much uglier problem if an exact solution is required.
Ned Simmons

I am familiar with the phenomena, although "reflected inertia" is not the term we used. (I once built a machine to put reflected impedance, as we called it, back into the toytoa IC engines. And I have had experience with, and designed for, the transfer of electric motor's mass moment "backwards" thru gear sets on cranes, booms, and hoists.)
Thinking out loud, for discussion - and don't bet the house on this, this is the internet and you get what you pay for - I do not believe that reflected inertia is the condition here, although with the piston driving the system rather than the crank driving the system, moving the reference point to the wrist pin would be more accurate than using the crank center and would minime calculations. In other words, the moving piston sees the inertia of the crank, flywheel and drive train at the pin; the crank "sees" no inertia, like the rod, it is part of it. In this approach, the torsional frequency of the crank and flywheel system is then used to compare the force along the stroke with the torsional cycle to determine if crank torsional deflection and attendant energy storage and release adds any "reflected impedance" to the peak impedance of the force chain seen by the piston. In lay terms, the combustion force pushes on the rod and winds up the crank and the crank unwinds at its natural frequency while input force is lower yet still inputting, potentially giving more force on the crank and rod than combustion alone gives. This is usually minimized by mismatching the frequency of the crank and the piston travel
With the disconitnuity when the rod shifts across the crank-rod bearing gap during the rotation, you can get jerk - But for the inertia problem, the discontinuity generally becomes moot because when the rod is not connected, it's inertia felt by the crank is nil.
imho

The rods will have different moments of inertia about different axes, normally you might be interested in it's moment about the little end, big end, or it's centre of gravity.
To get any of these, break down the design of the rod into primitive elements, ( cylinders, cubes,... ) use their inertia's ( from your reference book ) and the parallel axis theorem to add them together for your rod.
Two rods of the same weight can have different inertia's ( one could have heavy ends and a light beam compared with an other ).... so you need details of the rods designee. Another factor which affects the importance of inertia in an analysis is the angle through which a rod swings, which depends on the stoke and the rod length, a long rod will swing less, so the effects of inertia will be smaller in relation to it's mass.
If you have been given an "out " from these complications by being given an " add 1/3 to the mass of the rod and forget inertia " heave a sigh of relief and take it :-).
For a detailed analysis of the system I think you also need the position of the rod's centre of gravity.