1. ## Scissor lift strength

I want to specify one point.
When I calculate Force Required for Equilibrium at Load Rx, and it's value is 70000 N. And as shown here [url]http://www.engineersedge.com/mechanics_machines/scissor-lift.htm[/url] on "Free Body Diagrams:" length of one scissor member 1,8 meter. For calculating longitudinal moment of inertia (Wx) of my beam cross section I need to use formula [B]bending moment[/B]= (Rx*length of one scissor member)/4
then [B]Stress[/B]=[B]bending moment[/B]/Wx and finally
Wx=[B]bending moment[/B]/[B]Stress[/B]
am I right?   Reply With Quote

2. So, nobody knows?
I think this is too much. For example, if we have, for example a W=payload waight= 4000 N; Wf=frame weight=4000 N; L=length of the scissor arm=1,8 m; Theta=angle between the scissor arm and the horizontal=5 degree; so:
Force Required for Equilibrium at Load Rx= (W+Wf/2)/Tan theta = 4+2/0,087 = 69 kN (!!!)
Bending moment = Rx*(L/2)/4 = 69*0,9/4 = 15,5 kN*m
Now this moment require rectangular tube about 140x120x6 This is too much (((
may be mistake somewhere?  Reply With Quote

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•