Calculating the forces on a crankshaft assembly

[inamukak]

Hi everyone,

So I'm working on this project where I have to determine the total force that is being applied to an assembly that is attached to a crank shaft which is in turn attached to a motor that's driving it. I looked up the equations for the acceleration and velocity of a crank shaft (or a piston at the end of it) and tried to calculate the values based on that but the numbers I'm getting are huge. I'm also dumbing it down to a very basic model ignoring friction, vibration and other factors. The following are the values that I have right now:

m = 2995 lbs
Motor rpm = 358

I converted the motor rpm to get the angular velocity of the crank, which I got as w= 37.47 rad/s
I then used this value to determine the acceleration based on a formula I looked up online:

a = -r(w^2)*(cos(theta) + ((cos (2*theta))/n)

where
n = l/r
l = 8 in
r = 0.3125 in

I'm getting an acceleration of around 450 in/s^2 (if I consider 1 as the maximum value of cosine). Now if I use F = ma here, I get a force of approximately 1.35 million lbs. Just looking at the number makes me think it's wrong. So I just wanted to check what factors I should be considering here or what I'm doing wrong and the correct way of doing this. Any help would be appreciated.

Thank you!

 

[jrmichler]

Take a hard look at the units in F = ma. Especially m. The units absolutely, positively, must cancel out correctly.

 

[inamukak]

Take a hard look at the units in F = ma. Especially m. The units absolutely, positively, must cancel out correctly.

The units for mass are in pounds, which I got from Solidworks. The acceleration units are in/s^2. So it does come out to be lb-in/s^2 which is equivalent to lbf. Is that correct or am I wrong there?

 

[Dr.D]

Forget about looking up equations and work this problem from first principles for yourself. You will learn a lot more, and you will understand what you have. Thus far you have said nothing about the piston mass, the connecting rod mass, the connecting rod mass moment of inertia, all of which are important factors.

 

[jack action]

Double check your equation.

First, the parenthesizes don't match. Second, somehow, w^2 must divide r, such that you get the unit in/s². By doing so, you will get acceleration of around 2 X 10^4 in/s² which should give you a force of about 0.7 lb instead of 1.35 millions lb.

 

[jrmichler]

So it does come out to be lb-in/s^2 which is equivalent to lbf. Is that correct or am I wrong there?

Wrong. In the English system of units, one pound force is the force to accelerate an object weighing one pound at an acceleration of 386 inches per second squared. I get the same acceleration that you did, you just need to properly calculate F = ma.

Hint: The unit of mass in English units is the slug.
2nd hint: Pounds force vs pounds mass vs slugs is confusing, and you need to wrap your mind around what they are.
3rd hint: If all else fails, translate your problem into metric, solve it, then translate back into English units.

 

[inamukak]

Wrong. In the English system of units, one pound force is the force to accelerate an object weighing one pound at an acceleration of 386 inches per second squared. I get the same acceleration that you did, you just need to properly calculate F = ma.

Hint: The unit of mass in English units is the slug.
2nd hint: Pounds force vs pounds mass vs slugs is confusing, and you need to wrap your mind around what they are.
3rd hint: If all else fails, translate your problem into metric, solve it, then translate back into English units.

Thank you for the reply! I actually did end up doing that, converting to SI and checking the results...when they didn't match, I looked up to see if lbf = lb-in/s^2 and it wasn't...like you mentioned, I had to divide my answer by 386 to get the right force. Now it seems the value is low, but I didn't get a chance to dig deeper into this. I will try to run some FEA on this and compare my results with that and go from there.

Thank you again!

 

[inamukak]

Double check your equation.

First, the parenthesizes don't match. Second, somehow, w^2 must divide r, such that you get the unit in/s². By doing so, you will get acceleration of around 2 X 10^4 in/s² which should give you a force of about 0.7 lb instead of 1.35 millions lb.

Thank you for replying! Wouldn't dividing w^2 by r give me units of rad^2/ (in-s^2)?

 

[Dr.D]

In the English system of units, one pound force is the force to accelerate an object weighing one pound at an acceleration of 386 inches per second squared. I get the same acceleration that you did, you just need to properly calculate F = ma.

Hint: The unit of mass in English units is the slug.
2nd hint: Pounds force vs pounds mass vs slugs is confusing, and you need to wrap your mind around what they are.
3rd hint: If all else fails, translate your problem into metric, solve it, then translate back into English units.

I have to disagree with this approach. The first statement is true, but it confuses much more than helps. The hint is correct; in the FPS system, the unit of mass is the slug. In the IPS system (inch-pound-second), the unit of mass is the lb-s^2/in.

What I really disagree with is the suggestion to convert to SI, solve, and then re-convert the result. This is a poor idea involving two conversions with opportunities for error (round off error if nothing else) all for nothing. Newton's Second Law always works correctly if a proper, consistent system of units is used. SI is no more or less consistent than either FPS or IPS units; they all work perfectly.

 

[jack action]

Thank you for replying! Wouldn't dividing w^2 by r give me units of rad^2/ (in-s^2)?

My bad! I went over the equation too quickly. You are right, it would be the wrong unit and it is correct in the actual form.