Crank - Slide Forces 2

[gatz]

We're trying to come up with a way to calculate how much resultant force is produced by turning a crank.
[This looks like an engine crank and piston arrangement]
I've looked on the 'net and can't seem to find a definitive answer.

We have a diagram of a 1" radius crank going from 180deg on the extreme bottom (BTC) to 0deg (TDC). To this 1" radius is attached a 7.500" rod, that in turn is pinned to a vetical slide.
The center of the crank circle is coincident with the slide axis.
If we impart a force of 450 ft lbs to the crank, what will we get as measured in the direction of the slide for the angular rotation variable? (neglect friction)
What formula(e) is used?

Thanks in advance,
Gatz

 

[ivymike]

why don't you resolve the torque into x&y components for your rod (versus angle), and plot whichever of those is along the axis of your slider? You only need to know how to sum forces and how to calculate the angle of the rod vs crank angle. You're going to have a hard time reacting your moment near TDC.

 

[gatz]

That may be our problem,.... how to set up the force vectors.
We realize that the force near 0 and then again at 180 is approaching infinity, but it's what we need to know.
The force at at some angle less than 90deg would be approximatley, but not as much as the 450 lbs.
If it's easier to visualize, liken this to a mechanical press and ram arrangement.

 

[zekeman]

If you call the crank arm R, the connecting rod L, the crank angle @,and the distance from the slide pivot to the crank center, I get the y component,Fv at the slide froom
Fy=f1[cot@/*R/y-1/sin@]
where f1 is the crank force perpendicular to the crank arm; it is derived from the law of cosines, viz,
L^2=R^2+y^2-2Rycosine@ by taking a first differential and using the energy equation
Fvdy=f1Rd@
Fv=f1Rd@/dy
To get the horizontal component,Fh, you make use of the fact that the resultant force is colinear with the rod and therefore
Fh=Fv*tan# where # is vertical angle(gotten from law of sines) that the rod makes with y direction.

 

[zekeman]

Correction the first equation should be
Fv=......
not Fy=.....

 

[gatz]

zekeman, you lost me on the 1st sentence and the 1st eq.
Is the distance from the slide pivot to the crank center: "d" ?

In the 1st equation, if y = Fv, what is the y variable in the equation?
(after correction from your 2nd post...)

Fv=f1[cot@/*R/ y -1/sin@]

Thnx, Gatz

 

[wes616]

http://iel.ucdavis.edu/design/fourbar/

Wes C.
------------------------------
When they broke open molecules, they found they were only stuffed with atoms. But when they broke open atoms, they found them stuffed with explosions...

 

[zekeman]

Gatz,
The first sentence should mean that the force in the "y" direction is Fv and the distance from the crank center to the slide center is y, which is a variable depending on the position of the crank which explains the first equation. By the way, the first equation should read
Fv=f1[cot@*R/y-1/sin@]
removing that / sign.
Now wherever you see a "d" it is the differential, so that
dy means differential of y and d@ is the differential of the crank angle, which are shown in the derivation only.
I think I will show an example to make this clear.
If the crank angle is 20 degrees, then we first get "y" from the law of cosines, my second eq.
L^2=R^2+y^2-2Rycos@
L=7.5, R=1.0. Evaluating
7.5^2=1+y^2-2ycos20
56.25=1+y^2-1.879y The useful solution for this quadratic is
y=8.4316
and from
Fv=f1[cot@*R/y-1/sin@]=450[cot20*1/8.4316-1/sin20]=-1169 lb
Law of sines we get vertical angle # from
L/sin20=R/sin#: sin#=sin20(R/L)=04046 ; #=2.32deg
And Fh,the horiz component of force at the slide is
Fh=Fvtan#=47.36 lb
The -1169lb means that the vertical force on slide is downward and the the total force is sqrt(1169^2+47.36^2)= 1170 lb
Hope this clarifies things.

 

[gatz]

zekeman, it got a little more involved that we first thought, but we were able to get through it.
A coupla things.... we got 2.614° for # angle.
sin#=[sin(20)* (1"/7.500")]
sin#=[.34202 * .13333]
sin#=[.0456]
# = 2.614

Because we know 2 of the 3 angles that the crank arm, rod & vertical line make, we can now find the large angle which is 180-(20+2.6137)=157.3863
Then, for "y" dim, rather than quadratic,
we used the Law of Sines; ie...
y=[(sin(157.3863) * 7.500) / sin(20)] = 8.4138
Results were the same.

I set this up in Excel and now we have a way to calculate forces for any number of variables.

If it's not too much trouble, could you explain the derivation of the first formula?
Fv=f1[cot@*R/y-1/sin@]

We sure appreciate your help, and many, many thanks.
Gatz

 

[zekeman]

Indeed, your method of getting y is easier and I guess that my arithmetic is not so hot either.
The derivation involves taking the first differential of the cosine eq or recalling
L^2=R^2+y^2-2Rycos@
Then a first differential is
0=0+2y(dy)-2Rcos@(dy)+2Rysin@(d@) from whih we get
(1) R(d@/dy=[Rcos@-y]/ysin@ or (R/y)cot@-1/sin@
Recalling the energy eq I showed in my first post
(2) Fv(dy)= f1Rd@ from which we get
(2') R(d@/dy)=Fv/f1
Substituting in (1) it is clear ( to me?) that
Fv/f1=(R/y)cot@-1/sin@
While this is correct, there is a far easier method to getting the solution to the frictionless case you have here by using force vectors to place the point where the rod, crank and crank force act in equilibrium. Frictionless implies that forces from the link and rod act along their centerlines so that you can constuct a force vector diagram that has the crank force perpendicular to the dirction of the link centerline and the rod force also acting in the direction of its centerline. The resulting vector shows that f1 and force along crank form a right angle and the force vector F ( the one you are looking for)along the rod is at an angle =@+#. Invoking your favorite law of sines on the force vector diagram yields
F/f1=1/sin(@+#).This agrees with the solution of the sample problem of F=450/sin(22.6)=1171
Put this in your Excel worksheet.
There is another method to get this by breaking the rod and replacing it with an external force F and using f1 the other external force and setting the sum of moments about the crank pivot to zero. You would get the same answer. Talk about beating something to death!