Force calculation for opening guard on machine

1. Jul 18, 2016

Jeremy Sawatzky

Hello people smarter than me.

I am working on a project that utilizes a gas shock to open a guard that protects machine components. A very common scenario. I have nailed down my positions for where the shock is mounted, pivot point, shock length, stroke, guard CG etc...

I have tried to do the math and my boss has tried to help me as well however I think his math may be incorrect. Before I order the shocks I want to be sure that I know what I need. I have attached 2 photos. one with the guard open and one closed. For simplicity sake assume I am using 1 gas shock.

We would like the guard to stay shut by gravity alone and obviously stay open.

Guard weight: 87.4lbs

http://imgur.com/a/47EFI

Thanks so much.

2. Jul 19, 2016

Jeremy Sawatzky

Anyone want to have a go?

3. Jul 27, 2016

Jeremy Sawatzky

This would be super helpful!

4. Jul 27, 2016

Mech_Engineer

You basically need to balance the net moment being applied about the machine guard's pivot point and calcuclating:
1. Open case: want to see if the shock's moment applied is larger than the gravity moment applied, this means the door will stay open
2. Closed case: see if the shock's moment applied is smaller than the gravity moment applied, meaning the door will stay shut
For the open case:
1. Calculate perpendicular component of gravity applied at hinge to find moment M_g (force x distance)
2. Calculate perpendicular component of piston force at hing to find moment M_p (force x distance)
3. For this case, M_p needs to be greater than M_g with some safety factor

For the closed case:
1. Calculate perpendicular component of gravity applied at hinge to find moment M_g (force x distance)
2. Calculate perpendicular component of piston force at hing to find moment M_p (force x distance)
3. For this case, M_p needs to be less than M_g with some safety factor
Does this make sense?

5. Jul 27, 2016

Jeremy Sawatzky

It does sort of, however my math instructions have been all over the map. My boss is hard to understand sometimes and I really doubt that his math is correct.

I need someone to do the math on their end and then I can compare it to mine. I come up with needing a 70lb gas shock to keep the door open and will also stay shut when closed. BUT... my common sense says that this will not be enough to keep the door open. This is due to the angle of the shock when the door is open.

If you could run through the math that would be super helpful.

6. Jul 27, 2016

Mech_Engineer

Can you post your work so we can check to see if there are any glaring errors?

7. Jul 27, 2016

Jeremy Sawatzky

Okay. This is a mess of different instruction I have received.

OPEN
(Gravity Force) 15.109" x 87.4lbs which equals 1320.53 inch lbs. - I have no idea if this conversion is correct. My boss said it was...

(Shock Force) 90° - 56.09° = 33.91° And again this is what my boss told me to do. 33.91 SINE = 0.557% efficiency which means the 70lb shock is applying 39lbs of force vertically

Im lost beyond here, honestly I am so mixed up with all the different instructions I am getting from here, different websites, my boss etc. I really need someone to post the correct math because I am 99% certain this is all wrong.

Thanks.

8. Jul 27, 2016

Mech_Engineer

No conversion is needed, force multiplied by distance gives you a moment (also called torque in some cases). You might later need to convert those resulting units (in-lbf) to something more standard like N*m, but it works for now.

You need to balance the moment caused by the weight of the cover and the moment applied by the cylinder. Let's break the problem into steps:

First, to calculate the moment applied by the weight of the cover. You already have the correct distance measured in your diagram, so you just need the "normal force," which would be the component of gravity which is acting perpendicular to the line between the cover's pivot and the C.G. See here:

Next: to calculate the moment applied by the gas cylinder, you need to find the length of the arm between the cover's hinge and the end of the gas cylinder. Once you have that length, you can calculate what that force needs to be by taking the moment caused by the cover, and divide by the distance between the pivot and the end of the gas cylinder. See here:

Making sense so far?

9. Jul 27, 2016

Jeremy Sawatzky

Ok so 17.144 x 87.4 = 1498.38

Then 1498.38 Divided by 34.257 (the missing dimension) = 43.74.

So the next step is determining the force of the cylinder at that angle. How do I do that?

Thanks so much.

10. Jul 27, 2016

Mech_Engineer

Finally, you need to decompose the cylinder's force into two components, like shown:

This should give you what you need to calculate the applied moment force from the cylinder, or to back-calculate the required froce from the cylinder given a moment from the weight of the cover. make sense?

Last edited: Jul 27, 2016
11. Jul 27, 2016

Jeremy Sawatzky

Yea that makes sense.

The missing angle is 27.57°

Is this a SINE calculation?

12. Jul 27, 2016

Jeremy Sawatzky

My mistake the missing angle is 12.76°

13. Jul 27, 2016

Mech_Engineer

Yes, take your calculated force (43.74 lbf) and divide by sin(angle) to find the force needed from the cylinder. I think you'll find the force is unexpectedly high, and the door will not want to stay open as a result...

Last edited: Jun 28, 2017
14. Jul 27, 2016

Jeremy Sawatzky

Ok so to confirm my math.

12.76 SIN is 0.22.

43.74/0.22 = 198.82. - Far higher than I expected.

Thanks very much for the help. I will try the calculations with the door in the shut position and report back.

15. Jul 27, 2016

Jeremy Sawatzky

actually, in your post with this written at the top (Finally, you need to decompose the cylinder's force into two components, like shown:)

Are the top and bottom lines parallel? Because if they aren't, how do I know where to position the bottom line to get the angle measurement?

Ive made these calculations assuming those lines are supposed to be parallel. My previous angle calculations were wrong. The correct angle is 20.52°.

I am coming up with a required cylinder force of 125lbs to keep the lid open.

And to keep the lid closed is <1,093lb.

So using a 150lb cylinder will keep the lid open and wont open it once it is closed.

Last edited: Jul 27, 2016
16. Jul 27, 2016

Mech_Engineer

The good news is, now you know the real numbers and can make design changes as needed. I personally think your geometry layout is not yet optimal. Knowing what you know now, it will be possible to improve the location of the cylinder to make it more effective. For example, if in the open position theta is 90 degrees, this means the cylinder's force would be 100% effective at holding the door open.

I would recommend re-locating the cylinder's mounting points (and perhaps changing its stroke and length) such that:
1. When the machine's cover is open the cylinder's force is acting perpendicular (or at least nearly perpendicular) to an imaginary line between the door's pivot joint and the cylinder's end point. This will ensure 100% (or nearly 100%) of the cylinder force is utilized for holding the door open, and you don't need an over-powered cylinder to hold the door due to sub-optimal angles.
2. When the machine's door is closed the cylinder should form a line which points at the door's pivot. In essence, when closed you want the angles to be such that the cylinder is 0% effective, so that regardless of the strength of the cylinder, it cannot open the door in the closed position.
3. Also keep in mind: the longer the distance between the door's pivot point and the end of the air cylinder, the lower the cylinder' force requirement. So take a look at the recommended changes above, but don't move the cylinder too close to the door's pivot point because this will also increase your force requirements. Practically speaking, I would keep the distance at least the same as the distance from the pivot to the cover's C.G.
Good luck.

17. Jul 27, 2016

Mech_Engineer

Yes, Line AB is parallel to Line CD, because Line BD should be perpendicular to both of them (I just didn't draw them very well). See updated picture:

18. Jul 27, 2016

Mech_Engineer

Another thought: make sure your cylinder is a little stronger (maybe 10% or 20% more) than needed so that it holds the door open easily, otherwise it may have some trouble if the door is heavier than expected, or the angles are slightly different than expected.

19. Jul 27, 2016

Jeremy Sawatzky

Thanks very very much for your help. My boss was miles off of the correct math.

With my updated math I actually came up with 125lb needed to hold open the door. So that is more reasonable.

Unfortunately the parts have already been cut for manufacturing and altering the design now would be costly.

I am happy to toss 2x 75lb cylinders in there. The door will take a little bit of force to open but once it reaches about a 1/3rd of the way the gas shock will take over completely according to my calculations. It will take approximately 7.4lbs of force to pull the door down if my math is correct so thats good.

Thanks!!