Simple Cantilever Beam Equation 3

[dfowler]

I am a mechanical engineer who seems to be a bit out of practice with my old statics and physics equations. The company I work for builds garage doors and we have an aluminum extruded strip that runs the length of the bottom of each garage door. This strip has a trough that allows us to attach our astragal which is a vinyl flap that helps seal out light and moisture from under the door. We want to rollform a shape out of steel to replace the extruded aluminum astragal retainer. The side profile of the part would be roughly a 'Z' shape with the bottom holding the vinyl astragal and the top protruding and acting as a step to help close the garage door when done manually.
I am wondering if a .034" thick 45ksi steel shape that protrudes 1.25" would permanently deflect when a person weighing 200 lbs steps on it. I think this is a fairly basic cantilever beam calculation but unfortunately I have forgotten most of that. This is a cantilever beam that protrudes 1.25" long and is .034" thick with a width of 6 to 20 feet. Any help would be greatly appreciated. Thank you.

 

[miecz]

Yes, it will permanently deflect. Try 3/16" material.

 

[IFRs]

Yes it will bend. For a beam with one end fixed and one end free, a load on the end produces a bending moment of P * L or in this case 250 inch pounds. The deflection equation is M * L^3 / (2 * E * I ). E for steel is 30E6, for aluminum is 10.2E6. L could be the length of a shoe - 10" or so or 3" if just the toe. I would be then be 3 * (0.034)^2 / 12. The stress is equal to M * C / I, where C is 0.034 / 2. I have ot worked out the numbers but I will check yours when you post them!!

 

[dfowler]

Thanks for the replies guys.

Here is the equation that I have used for deflection:
defl = AppliedLoad*LengthofStep^3)/(3*ElasticModulus*MomentofInertia)
= P * L^3 / (3 * E * I)

And here is the equation that I have used for stress:
stress = AppliedLoad*LengthofStep*1/2MetalThickness/MomentofInertia
= P * L * C / I

The results seem to be reasonable so I think the calculation formulas are correct. If anyone has the chance to check them however that would be great. Thanks again for the help.

 

[miecz]

Yes, you have the correct formulas. Take care with your units. You'll want to limit your bending stress to around 75% of your yield stress. The difficult part here is in the calculation of the moment of inertia. Specifically, the effective width, "b", in the formula I=bt3. For this case, I would use the width of a shoe plus twice the length of the cantilever, or about 6 inches.

 

[miecz]

Oops, I=bt3/12.

 

[IFRs]

Can you perform a test? Will there be corrosion? I like to use a factor of safety of 2 on the yield stress. In this application, will people jump or bounce on the lip?

 

[dfowler]

Yes, we are making up a few samples to test before we go ahead with this idea. It looks like it will hold but the numbers are too close to failure to assume the math is good enough. We will use a galvanized material as well since the bottom of a garage door is subjected to quite a bit of moisture. I hope people do not want to jump on this piece. Garage door tosion springs are ideally wound so that there is no force required to lower the door to the ground, only the friction in the system needs to be defeated. This would mean that maybe 10lbf could reasonably be used to close a garage door all the way to the ground. Someone could momentarily lean into the door to put a significant portion of their body weight onto the bottom door step but this would cause the door to crash downward. I think a reasonable person would either not do this or would quickly remove that force.