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• # Scissor Lift calculation

Discussion in 'Calculations' started by nabilishes, Jan 23, 2011.

1. ### ValmikiMember

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Apologies HMCK, I had failed to scroll down on your sketch, so only glimpsed your R1+R2 equation.

Still avoiding the equation, would your sketch get the same result if you were looking at a pin jointed triangle, where the total weight (M+Ma) were supported on the apex, and the holding force equated to the tension in the horizontal member of the triangle???

2.

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4. ### hmck57Member

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Valmiki -

If I'm understanding your question correctly, you're asking about three members pinned together in a triangular frame, loaded at one apex. In this case you have a truss - the free body diagram and reactions can be seen here: http://twitpic.com/3yplo2. As such there won't be any horizontal reaction load (at least as I've drawn it) except as might be due to a small amount of friction in the roller at joint 1. The only way there would be a horizontal reaction would be due to a horizontal load.

5. ### hmck57Member

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Valmiki -

No apology necessary!

Guy - just FYI, I submitted a Feedback request to Engineer's Edge on their scissors jack equation and referenced this forum. We'll see if anything comes of it. Thanks for double checking my work.

6. ### Kelly BrambleNew Member

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Hello folks!

I received a feedback post on Engineers Edge from Hal McKinnon today in regards to the Scissor Lift equation and image posted on Engineers Edge website - http://www.engineersedge.com/mechanics_ ... r-lift.htm.

On a quick review of Hal's proof, I believe Hal is correct. Looking in Engineers Edge backend, we do have another scissor lift image showing the actuator load point at the center scissor pivot pin loading horizontal. I suspect our images where reversed.

Anyway, the webpage has been put our â€œreview listâ€ and will be corrected as required.

Appreciate the feedback!

7. ### StevenMember

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Hello.

I have different answers to you!

From the principle of virtual work:

F*l*SIN(f)*df=W*2*l*COS(f)*df+Wg*l*COS(f)*df

F=(2W+Wg)/Tan(f)

8. ### ValmikiMember

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Hi HMCK57,

Yes, the idea I was trying to picture was that the tension in the bottom (horizontal) member of the pin jointed triangle ref your free body diagram for the triangle) would equate to the actuator force when the koad M+Ma was applied vertically to the apex of the pin jointed triangle.

Now, to represent the scissor more accurately, then I would place a second identical but inverted pin jointed triangle with its own apex balanced on the apex of the original pin jointed triangle.

Again placing the entire load on top of the whole balancing structure, the tension in the horizontal link of the upper inverted triangle would be equal to the tension in the horizontal link of the original lower pin jointed triangle.

Since both these tension forces have to be created or restrained by the actuator, it would appear that the required actuator force would be doubled (I know that this is not entirely true as the load applied to the top inverted triangle would only be M and not M+MA, but for a mental picture I'm ignoring this for the time being).

My point was that equating the scissor to one pin jointed triangle with the load applied to it's apex would give a tension in the base horizontal link of half the actuator force, and not the whole actuator force as for a scissor arrangement.

Could not the actuator force be arrived at by treating the scissor as a pair of pin jointed triangles, and summing the tensions in the horizontal links (using M+Ma as the load to get the tension in the bottom triangle, and M as the load in the upper inverted pin jointed triangle)?

Would this give a result which matches your scissor free body diagram?

For multiple scissors, the same principle would apply, of simply summing the tensions (you'd have to apportion all the Ma loads accordingly).

(sorry about not representing my words with sketches, but I've only got dial up here, so attempting to do so is pretty futile for me).

I would be interested if you could find the time to reply.

9. ### StevenMember

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Hello, guys.

My former post has a mistake.
So I repost my solution again.
From the principle of virtual work:

?W=F*?x-W*?y1-Wg*?y2
?x=2*l*COS(?)
?y1=2*l*SIN(?)
?y2=l*SIN(?)
Make derivative calculation on the ?W.

we can get
F*2*l*SIN(?)*??=W*2*l*COS(?)*??+Wg*l*COS(?)*??

F=(2W+Wg)/2Tan(?)

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