Scissor Lift Mechanism 2

[JLKDS]

Hello,

I am designing a scissor lift mechanism with one arm 3 times the length of the other arms, with the remaining arms being symmetric. The mechanism is loaded at the end of the long arm. I am trying to determine what would be the force acting upwards at point A but it's proving to be a bit beyond my capabilities. Any help would be appreciated.

Many thanks

 

[JLKDS]

Sorry I didn't attach the image correctly. Here is a diagram of the proposed mechanism:

 

[renderu]

Let's call the bottom two supports point A and B. Do sum of moments about A. Now do sum of moments about B. And finally sum of forces in Y equal zero. You now have 3 equations with 3 unknowns.

 

[renderu]

Actually disregard that. I end up with zero=zero. That is not the way to go.

 

[jlnsol]

F will start at 5 times W and grow while pushing up.

 

[JLKDS]

Thanks for your responses guys. jlnsol, wouldn't F start at W when the arms are vertical and then increase to 5W when in the horizontal when A moves up?

 

[TheTick]

There is a simple way to "cheat" your solution: If you can sketch this mechanism in CAD (e.g. SolidWorks or Inventor), simply take the ratio of vertical movement from of load and point A. This is the net mechanical advantage.

For instance: if W moves 3 mm for every 1 mm of movement of A, your mechanical advantage is 1:3, and it will take 3N of force to hold 1N of weight.

The mechanical advantage will change as the mechanism extends and collapses. Take measurements at multiple heights and make a graph (in Excel?) to see the effect of height on mechanical advantage.

p.s. Yes, I can do the math. However, this method seems to work to help people understand the net results.

Looks like you would have to pull down at A to keep the mechanism from collapsin under its own weight.

 

[zekeman]

This "mechanism" will collapse as shown.
To make it work, remove F and instead put a horizontal cylinder or jack screw between the first two end pivots.
Look at you car jack to see it.

 

[TheTick]

Not true, Zeke. If A is constrained to ground (or w/ matching force), the mechanism does not collapse.

 

[rb1957]

lets say the thing is in static equilibrium.

it's "easy" to calc the external reactions with two applied loads, W and F ... sum moments at one support (which will tell you the reaction at the other support), then sum Fy.

but this isn't going to answer the question "what's F ?" the force required to balance W. for this i'd use energy/work ... the work done by F (F*dy) is equal to the work done by the weight (W*Dy). which now i read tick's post is what he's done for you ... but i think he's been beguiled by your sketch ... i think the mechanical advantage is 2:1, but a rough "finger" measurement of your sketch suggests that the 3L should be dimensioned to the joint ?? (which would make it 3:1)

to analyze the structure, the links are two force members (axial load only) and three force members (the three forces intersect at a point). so the top short link is a two force member, so you know the direction of the force (axial). the top long link is a three force member, you know the direction of two forces, so you know the direction of the 3rd force. and so on

Quando Omni Flunkus Moritati

 

[rb1957]

i'm not really convinced by that ... that says that F required to balance W is a fixed amount (3W) ... and a little more F will raise W, a little less will lower it,

am i right in thinking that you can delete the top diamond, or add another diamond above the existing one, without significantly changing the problem ??

Quando Omni Flunkus Moritati

 

[jlnsol]

F is roughly 5 times W for all positions between flat horizontal and up vertical. (end positions to be avoided here) Your sketch shows F up but in reality F will be down and the reaction force in B will be up and greater than F.

 

[rb1957]

if A moves up 1, then the joint above will move down 1, and the joints at the LH and RH ends of the upper diamond will move down 2, and the joint at the top of the diamond (the culcrum for the top lever) will move down 3, and the weight will move down 3+2 ...

Quando Omni Flunkus Moritati

 

[zekeman]

Tick,
You are right, never thought one would design a scissor mechanism with
negative mechanical advantage, since you need 5X the weight to lift it.
This is a first.
Rb,
Kinematics
You do it this way, nothing fancy
Geometry
Y1=5Lsin@ at the weight
y2=Lsin@ at the bottom pull point
dY1/dY2=5
F1dY1=F2dY2
F2/F!=5
Mechanical advantage 1/5

 

[Compositepro]

Assuming this is a statics problem, the vertical force at each of the three center pivots of the scissors must be equal. The top lever arm has a 3:1 ratio. The direction of the force arrows in the drawing is somewhat confusing. They are not in balance for a statics problem.

 

[JLKDS]

Hi guys, I do apologies for the confusion in the diagram, the force at A was meant to be in the opposite direction as this is the counteracting force that will balance the mechanism making it static. I realise the issue regarding mechanical adavantage or should I say disadvantage with making a big movement from a small input movement in that is always going to require a greater force at the input. That is a the movement I require but I didn't calculate the ratio was that high as 1:5 so thanks for cleaning that up.

Cheers

 

[TheTick]

In my last job, I analyzed dozens upon dozens of mechanical lift mechanisms. I'm pretty adept at finding mathematical solutions, but I needed a way to keep up with all of the various "inventions", and allow the designers to analyze changes to their own designs.

The solution was simple: find the net mechanical advantage. The designer would sketch the framework in SolidWorks. Then, a macro in a spreadsheet would take the mechanism through various positions, "wiggle" it, and measure and graph the resulting mechanical advantage at each position.

Usually, the input was the length of a line representing a linear actuator, and the output was the net height of the system center of gravity.

The results of this method were great. Reality met calculations, the way they ought.

 

[jlnsol]

Tick,
Thanks for sharing this analyzing method. Very nice. "a macro in a spreadsheet"... A question I have though: Did the macro take the mechanism sketch automatically through various positions? Or was it necessary for the designer to start up the macro again and again per each sketch step? Did you make a table in the spreadsheet first for the various positions that you want to know and then let the macro step through the table?

 

[chicopee]

This problem should be of the type whereby the virtual work method is used. Virtual work is covered under Statics and Dynamics, college college course offered to ME's and CE's.