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Thread: Scissor lift strength

  1. #1
    Associate Engineer
    Join Date
    Oct 2012
    Location
    Ukraine
    Posts
    6

    Lightbulb 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?

  2. #2
    Associate Engineer
    Join Date
    Oct 2012
    Location
    Ukraine
    Posts
    6
    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?

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