Which Angle Beam Formulas Are Needed for Building a Motor Home Bed Frame?

[xexorz]

I need assistance identifying the required formulae to choose materials in building a span.

The actual application is a bed frame of all things - but it can only be end-supported (simple end support).

The entire bed will move up and down on rails (this is in a motor home - which is why weight is critical - I need it to be light enough to move up and down and low enough mass not to rip free from its housing in a crash). I mention this to answer the obvious "Why not center support" and "Why not steel / wood etc" questions which is exactly what I'd ask in the same position.

The current implementation of this design uses steel angle L (totally over-sized and very heavy). In the event of a crash I'm quite certain it will become a deadly projectile as it rips free from the rails. To boot, it is very difficult to push up into the locked position overhead. So I've decided to use aluminum - but am not sure how much I need.

The span is 80 inch and will require 2 parallel angle-L beams made of aluminum. Wooden decking will lay across the flats of the two L's (plywood which will be pop-riveted into the flats of the aluminum).

The aluminum I have access to is 6061-T6

Ultimate Tensile Strength, psi 45,000
Yield Strength, psi 40,000
Brinell Hardness 95
Rockwell Hardness B60I can get some standard denominations of L angle

1x1x1/4
2x2x1/4
3x3x1/4

Can get other sizes but instead of solving for a specific set of sizes I'd like to test each on a spreadsheet when finished.

The static load is 500lbs maximum (My wife and I with a good amount of margin) - I am not sure what to use for a dynamic load (any thoughts?)

I have never done any real mechanical engineering so I'm not sure what equations I'm looking for. I guess I would be interested in seeing deflection but have no idea how to determine where failure should happen.

Simplest way to design it may be center loading (worst case?)

I'd like to build this with economy of both materials and weight but with sufficient margin that it doesn't collapse - contradictions in terms but I think you know what I mean :)

If possible, please direct me to the equations I need. My math is fairly good up to some calculus but I'm not familiar with the engineering symbols and constants - please let me know what each element in the equation is (or direct me to a good resource with the same).

I'm eager to learn and really appreciate the help!

Be well,

George

p.s. will post images / video of finished project! Current system is being disassembled soon after I write this!

 

[edgepflow]

The loading is somewhere between distributed and point load, so I would choose point loading. For a beam simply supported and point loaded in the middle, the maximum stress and deflection is:

stress_max = W L / 4 Z

deflection_max = W L^3 / (48 * E * I)

where,

W = weight load
L = length of beam
Z = section modulus
E = modulus of elasticity
I = moment of inertia

For your weight load to account for "dynamic loading" use an impact factor FI = 1.5 to account for extra force when someone climbs on the bed.

So use W = 500 lbf X FI = 500 X 1.5 = 750 lbf.

For the section modulus and moment of inertia, there are formulas for L shape but they are a little unpleasant. Do an internet search and you can find tables.

A good structural rule-of-thumb is to limit deflection to 1 inch for every 240 inches of beam length. Thus, for an 7 ft long bed, the max deflection should be under 0.35 inch.

Finally, limit your calculated stress to 30 - 50% of the yield strength.

 

[xexorz]

Thanks much edgepflow!

I can't seem to find a chart with the moment of inertia but I did find an equation for it and put it into a PHP program - could you check this against a known L section to see if it jives or if I made an error? If not, can you point me out to a list of L moment of inertia?

http://infinitycyberworks.com/maths/rect_l_beam.php"

Will work out the the section modulus next - unless you have a resource for this as well?

Thanks again for the pearls!

-George

 

[nvn]

xexorz: When I tested it, it appears the program sometimes gave wrong answers for moment of inertia. Therefore, it currently seems there might be mistakes in the program.

 

[xexorz]

Thank you nvn!

Would you please provide me with input variables and expected results (the one(s) that are producing bad results) so I can debug?

Thank you!

George

 

[edgepflow]

xexorz,

This link has the formulas and also tables for steel L shapes (not sure how close alum L shapes are to steel).

http://www.efunda.com/designstandards/beams/BasicLBeam.cfm

 

[nvn]

xexorz: Per your request in post 5, here are some input values and expected results; t = 3 mm, a = 30 mm, b = 50 mm, cx = 6.759 740 mm, cy = 16.759 740 mm, Ixc = 60 357.67 mm^4, Iyc = 16 867.67 mm^4.

Change your "I" parameter to J = 0.3333(a + b - t)(t^3).

Zx = Ixc/y; Zy = Iyc/z.

 

[xexorz]

Thanks for the corrections - I've updated the code and found the problem with cy (I hope)

Do the maths look right now?

I still haven't figured out which beam I need :)

Thanks again to all for the help - I really do appreciate your time.

 

[nvn]

xexorz: y and lowercase z are now correct. Iyc is incorrect. Notice, it does not match post 7. Also, change the name "I" to "J," and list some decimal places after it. Also, uppercase Z is incorrect.

 

[xexorz]

Ok I think I've got it now! Now, to translate the results:

I'm using input:

t = 0.25 inches
a = 3 inches
b = 3 inches
W = 750 pounds
L = 80 inches
E = 10000000 (Aluminum? I hope)

Results for deflection: 26.71 Inches?? This must mean it failed I assume?

 

[nvn]

xexorz: You have an L76.2 x 76.2 x 6.35 mm beam, of length L = 2032 mm. I would say tensile yield strength might be closer to Sty = 262 MPa, and tensile ultimate strength closer to Stu = 283 MPa. E = 68.95 GPa, and G = 25.92 GPa. One can conservatively analyze this as a point load, because the plywood will tend to distribute the loading somewhat. I think you can use a weight of P = 2000 N; and I currently think you could get away with an impact factor of FI = 1.30.

An L76.2 x 76.2 x 6.35 mm appears to be overstressed. It currently looks like an L76.2 x 76.2 x 7.938 mm, or an L80 x 80 x 8 mm, made of Al 6061-T6, would probably be adequate. The midspan deflection, during peak dynamic loading, would then be 10.5 mm; and it appears the midspan torsional tilt angle, during peak dynamic loading, would not exceed 6.0 deg.