Ramp Question

[kiskaw]

Ok guys here is a question for you
I have purchased a truck with a flatbed car hauler on it The problem is one ramp needs to be repaired.
I would just like to build new ones the old ramps are made from 1-1/2 x 1/4 angle about 10 foot long kind of a ladder
I want to make the sides out of 1-1/2x1-1/2x3/16 square tubing with the 1-1/2x1/4 cross bars
The question is how does the strength of the tube compare to the strength of the angle?

I have read other threads that say it is just as easy to buy them as build them but I usually find that when I build it myself it will take a lot more to tear it up

NO SUCH THING AS OVER KILL

[steve45]

The 1-1/2" x 3/16" square tube will resist bending more than the 1-1/2" x 1/4" angle.

Will your crossbars be angle?

[tapwelder]

What is the problem with the damaged ramp? Most ramps I have seen are angle. So rather than almost double the weight of the ramp, I would go with a larger dimension angle-- if bending is a concern.

[kiskaw]

The 1-1/2" x 3/16" square tube will resist bending more than the 1-1/2" x 1/4" angle.

Will your crossbars be angle?

The cross bars would be angle the same size as the original ramps are

tapwelder
The problem with the original ramp is the guy that had the truck put way to much load on it they are designed to haul cars and he put a tractor on it
I however would like to have extra strength so if I get ahold of a 3/4 ton 4wd pickup with a diesel I don't have to worry

The extra weight of the tube isn't really an issue it figures out to be 84.32LBS. versus 70.20 LBS.
If this will increase my strength substantially then I am good with that

[Coalsmoke]

I would make the ramp beams 2"x1(or 1.5") x 3/16" to increase your carrying capacity over the angle. Some diesels like the Dodge with the Cummins can get heavy. Keep in mind that most square tubing conforms to a 55w steel spec, while angle conforms to a A36 or 44 spec. IOW, the square tube is stronger, not just by design of it being square (or rectangular) but also because of its tensile and yield strength properties.

[awright]

The strength of a structural shape in bending is proportional to the Area Moment of Inertia of the section around the axis normal to the bending moment, I. The calculated values below were calculated with a friend's calculator with which I am not familiar, so there is the possibility of error. You may find a website that allow you to enter dimensions and automatically calculate I.

I calculate the moment of inertia of the 1-1/2" x 3/16" square tube to be 0.28828 in.^4. The formula is:

I = (a^4 - b^4)/12, where a = outside dimension of square tube and b = inside dimension of square tube. Units are whatever unit of measure you use to the fourth power.

Now it gets complicated. The formula for an angle is much more complex due to the absence of symmetry around the neutral axis of the angle as a beam.

For an angle of equal leg length a and thickness t, the distance of the neutral axis (axis y-y) from the extreme fiber (end of the vertical leg) is y.

y = a - [(a^2 + at - t^2)/2(2a-t) inches (assuming you use inches as your units).

The area moment of inertia around the y-y axis is:

I = [t(y^3) + a(a-y)^3 - (a-t)(a-y-t)^3]/3 inches^4.

My calculation of I for a 1.5" x 1/4" equal angle is 0.21578 in.^4, but you better check that.

If my numbers are correct, the box tube is about 33% stronger in bending than the angle. But run your own calculations.

Have fun.

awright

[steve45]

I didn't bother to calculate the polar moment of inertia for these cross-sections. I just looked at the thickness of the vertical sections and saw that the tube had significantly more area. I considered that the horizontal sections have virtually no strength to resist bending in the vertical direction.

[kiskaw]

Thanks for all of your help
I am thinking that a 2" will fit into the pocket where the ramps go so I may move up to a 2x2x1/4 tube that will surely be strong enough for anything I will haul

Thanks again

[awright]

kiskaw, that 2" x 1/4" tube will be WAY stronger than the original angles. You can use my formula to calculate just how much stronger, if you care. (Its not a bad idea to know.)

Actually, steve45, the MAJOR contributors to strength of a beam are the "horizontal" elements at the top and bottom of the beam, not the vertical web(s). In fact, the most frequently used structural shapes for beams, "I" beams or WF beams of various configurations, are designed to put most of the material at the extreme top and bottom of the beam. The vertical web is there not primarily as a direct contributor to the bending stiffness of the beam, but as the element that holds the top and bottom horizontal elements contributing most to the stiffness in position without allowing them to buckle or to shear relative to each other.

Calculation of the moment of inertia of the combination of the top and bottom elements alone, separated by an imaginary web of zero thickness, followed by calculation of the moment of inertia of the vertical web alone, with no top and bottom flanges, will show the relative contribution of each of the element types. You can do the same calculation for the horizontal and vertical elements of a box beam with similar results, albeit not as extreme because there is proportionally more meat in the vertical sides of a box beam than in the web of an "I" beam.

Incidentally, the "polar" moment of inertia is a different (but closely related) beast than the area moment of inertia for beam calculations and is used to calculate resistance of a shaft to torque or for rotary acceleration or resonance calculations, not bending calculations.

I have to disagree with you that intuition is an adequate substitute for calculations based on established principles of strength of materials (the same ones used to build bridges and buildings), even for simple projects. People get killed that way. Even though your conclusion about the relative strength of the box beam and the angle was correct, your intuition about why was wrong.

The calculations are really not that difficult, and if only I could find my slide rule and my Popov, "Strength of Materials" text from 45 years ago, I'd prove it to you!

Have fun and be safe.

awright

P.S.

steve45, rereading the phrasing of your comment, I think I see where your intuition threw you off. Is it possible that you thought of the top and bottom elements of the box beam as two thin flat bars that would obviously sag seriously under their own weight? They would if isolated from the box, but that's not how a beam works. The top element of a box beam is in compression and the bottom element is in tension. If they can be restrained in their relative positions and spacing, they contribute most to the strength of the beam because virtually all of the imaginary "fibers" in them are at the greatest distance from the neutral axis (the mid line of a symmetrical shape). The vertical sides perform that task of holding the top and bottom in position and not allowing them to shear relative to each other. (If they are allowed to shear relative to each other, you are back to the sagging flat plates.) Because the "fibers" of the vertical sides vary in distance from the neutral axis from zero to the maximum distance, many of those "fibers" don't contribute much to the bending strength of the beam, since, for a uniform material, the contribution of each "fiber" is proportional to the third power of its distance from the neutral axis.

Hope this clarifys the issue a bit. Actually, any of you involved in fabrication involving selection of structural component sizes would find a few hours studying a textbook on strength of materials fascinating and informative. I did, even though I have almost never used the information in my 45 year career in noise and vibration engineering. And don't be intimidated by the math. It's not that complicated for simple structural shapes and is tabulated in many handbooks.