Statics - calculation of force on a wall mouting

Aug 19, 2006 3 Replies
K

Hi



Many years since I had some classes in physics regarding forces and torque.



I have a sun-shield I have to mount on a wall, but I would like to be able to calculate the forces on the mount on the wall



See drawing:



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The left drawing is the mount. One metalplate is mounted sligtly tilted



on the wall using two nails. A metal rod is welded in a right angle to this one - and in the end of this rod a weight exerts a force F on the metal rod.



The draing on the right is my free-force diagram. The force, F, is converted to a torque Ma, that is in the middle of the mounting plate L2. The rod has length L1



Torque Ma = F*Cos(alpha) * L1.



My problem is to calculate exactly the force on the upper nail. The lower nail (point C) only supports the plate and the mount rotates around this hinge.



Is the force on the upper nail (B) equal to:



Mb = Fb * L2 => Fb = Mb / L2?



Is that correct that even though the rod is welded at the middle of the



mount, the torque is reffered to point C (so that a torque on a non-ridig object is translated to the end of the object?)



Hope you can help me :-)



Thanks



Klaus



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B

you've got it worked out

the force on the nail is just the sun shade weight modified by the ratio of the sun shade are to the nail spacing

if the nails are 8" apart & the sun shade arm is 32"....then the force on the nail is 4x higher than sun shade weight approximately since the reaction couple (moment) is really the edge of the moutning bracket to the upper nail.

Basically you've got an applied moment (couple) & a reaction couple (moment)

BTW I would recommend against nails......not great in withdrawl.....use screws

be conservative in your design & calcs....how large / heavy is the sun shade? how long is the arm? what are the consequences of failure?

cheers Bob

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C

Review alt.engineering group for response.

CID...

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D

On 2006-08-19 in

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Kragelund wrote (paraphrasing):

Klaus: Yes, that's correct. Forces and moments can be referenced from anywhere, even if the body is nonrigid. Assuming the bending stiffness of your metal plate is relatively high, your choice to sum moments about point C is a good one. Therefore, assuming the self weight of your rod is negligible compared to the applied load F, then summing moments about the lower nail (point C), and solving for the tensile force (P) on the upper nail, we obtain P = (L1/L2)*F*cos(alpha) -

0.5*F*sin(alpha). And the shear force on the nail is V = 0.5*F*cos(alpha).

(You inadvertently omitted the component of F parallel to the rod, in your free-body diagram and in your summation of moment about point C.)

The nail tensile stress is sigma = P/A. Obviously, ensure FS*sigma < Sty, where FS = factor of safety, and Sty = fastener tensile yield strength. (Sty = 250 MPa for mild steel.) There's no need to check the nail shear stress, tau, if L1 > L2, because you automatically know it doesn't govern (since shear yield strength Ssy = 0.577*Sty for ductile metals). There's also no need to check the nail combined stress if L1 >

4*L2, since the nail combined stress doesn't significantly increase beyond tensile stress, sigma, for large values of L1.
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