One way to avoid pushing the frame rails apart would be to make the unit functional by itself, with a crossmember, and bolt that between the frame rails... something like a 2" square tubing with 3/8" walls. that way there are no outward forces on the frame. I would add significant reinforcing where the hinges mount to the square tube though.

 

apology accepted I probably got a jumped a little too quick on that one too. I really do appreciate the help though, thanks again. And thats part of the problem, the prof isn't very clear on what he wants. We don't even know if we need these calculations, but we're going with the btter safe than sorry route. The class is called engineering graphics 103 and so far we've learned about autoCAD, word and excel in the class.

Furball: I mentioned in one my other posts that it is attached to a crossmember. It has 'dog ears' welded to it for the two pivot points. I'll try to get a snapshot of the autocad drawing and upload it in a little bit to give you a better idea of whats up under the bumper.

 

http://www.pitt.edu/~jbc11/carjack.jpg

Take a look at that pic to know what I'm talking about here. I made it in paint in about 30 secs, so the size & everything is outta wack, had to link to it.

Anyways, to figure out just the compressive loads in the bars, you need to know the length of the bars, the height the car is from the ground, and the weight of the car. The picture labels the lengths of the bars, and the height is labeled as "h". For a factor of safety, let's use the entire weight of the car, not just the front axle rating. To find the loads in each of the bars, you would use a static situation where horizontal & vertical forces are both zero. The vertical force equation is:
weight of car= left load * (h/35.5) + right load * (h/36.5)

And the horizontal force equation is:
left load * (s1/35.5) = right load * (s2/36.5)

After some algebra:
http://www.pitt.edu/~jbc11/math.jpg

What I found is that with a pressure of 2700 psi on a piston with diameter 1.75 in, you can put out about 6491 lbs of force. If your load is less than that (which it should be, since this worked), then you've at least proved this puts out enough force to lift the car. If your load comes out higher than this, then my math is probably wrong and you should come smack me .

On a side note, I'm really not a big fan of only having one contact point, just because it makes things really unstable. If you want my idea on how to keep relatively the same system, pm me & I'll see if I can make a clear pic to send you.

Hope I helped.

 

here is the image of the autocad drawing we made:
is isn't exactly what the pictures of the first post show, but it will soon be modified to look very much like this


I'll see what I can come up with for max height and weight of car.

 

I think I'd go for a more stable platform. I'd use a longer "foot" like you have in the center, and shorten my "inactive arm". Use a slightly larger ram for better stability as well. With a 1.75" bore you're looking at about a 1.25" shaft.. That's a little weak I think. I might even go as far as using four individual rams each with it's own foot similar to an outrigger on a crane or back-hoe. You would apply pressure to each cylinder to raise the car.

 

interesting ideas, but what we're cost limited, what you see is what you get at this point. If we had the money there would be a ram at each wheel with its own platform, but we don't.
The guy that is making it was talking about building the foot to be 8-10inches.

 

I'd also suggest running through the calculations at the point where the front tires first lift off the ground. The more horizontal the cylinder the greater the force it will need to apply, but the springs help lift the car, so this will be the point where the greatest force is required. (For this class, you could probably get away with neglecting the work the springs do to lift the car, if you make a note of it on your calculations.)

Also, while I see the idea behind using the weight of the car to gain a factor of safety, the point of this exercise is basically to impress the professor more than to engineer a car lifting system. So, I'd use the front axle rating as the weight, and then try to do a little research on what a typical FS is for something along these lines (it'll be high, given the risk involved in a failure, as well as the corrosive environment the device is stored/used in) and use that.

 

Even at 8 to 10 inches it's going to be very unstable. I wouldn't even want to change a tire on that small of a foot. The slightest jerk and it may (will) fall. I'd go for closer to 18 inches. And about 10 inches wide. give it a nice platform.

 

I understand what you're saying, but it just isn't going to 'fall'. It will sway side to side as it sits now, there is no way that you'd be able to overcome a 10inch wide platform and have it fall over. you'd literally have to lift the car up and push it over.. and even then I don't think it'd tip.

 

Well, if you need help remembering the basic trig functions this is what my teacher taught us.

Some old hippie caught a high trippin on acid.
Sin 0pposite/Hypotnuse Cos Adjacent/Hyp Tan Opp/Adj
or S O/H C A/H T O/A


say the angle your trying to find is the bottom right, your adj side would be the bottom, your opposite would be all the way to the left, and your hyp. is always opp. of the 90* angle.

|\
| \ Hyp
| \
| .\
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