Calculate How Much Load A Flat Carbon Steel Bar Can Hold

[rc0213]

Can someone remind me how to calculate a load a flat bar can hold?

The material is steel. The dimensions are 1/4" X 3" X 6".

Thanks,

 

[BrianE22]

Quite a bit if it is sitting on a flat thick steel plate.

 

[IRstuff]

That's something that you can find in Roark's

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[BridgeSmith]

P*L/4 = M = f*S --> P = 4*f*S/L

P is the concentrated load
L is is the distance between support points
f is the strength of the material (yield or ultimate, depending on how much deflection is allowed)
S is the section modulus = b*t^2/6

Rod Smith, P.E., The artist formerly known as HotRod10

 

[GregLocock]

Well yes, but where's the teaching moment in that?

I always go back to the only useful structural equation (I exaggerate)

In the linear range M/I=s/y=E/R

With that little thing you can build models of tapered springs, beams with any load distribution, any number of supports, and composites.

In the rather banal case you have selected, at the centre of the span M=PL/4, I is 1/12bt^3, and y is t/2






Cheers

Greg Locock


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[BrianE22]

Beam deflections should be covered in your strength of mechanics text books (or something similarly worded). If you're an internet fan and you just want the answer:

https://www.engineersedge.com/calculators.htm

You have to pick out the supports and loading case.

 

[BAretired]

The question is not clear. A 1/4" x 3" x 6" flat bar can resist a tension of 3Fy/4 parallel to the 6" dimension and 3Fy/2 parallel to the 3" dimension.

BA

 

[drawoh]

BridgeSmith,

You are assuming the bar is loaded as a cantilever.

--
JHG

 

[BAretired]

BridgeSmith,

You are assuming the bar is loaded as a cantilever.

No, BridgeSmith is assuming the bar is loaded as a simple span with concentrated load at midspan.

BA

 

[cole44]

P*L/4 = M = f*S --> P = 4*f*S/L

P is the concentrated load
L is is the distance between support points
f is the strength of the material (yield or ultimate, depending on how much deflection is allowed)
S is the section modulus = b*t^2/6

Hope this helps!

 

[BAretired]

P*L/4 = M = f*S --> P = 4*f*S/L

P is the concentrated load
L is is the distance between support points
f is the strength of the material (yield or ultimate, depending on how much deflection is allowed)
S is the section modulus = b*t^2/6

This assumes that the OP is referring to a concentrated load at midspan of a simple span, which is not clear from his post.

If P is the applied concentrated load, then f is the allowable stress, about 0.6Fy.
If P is the factored concentrated load, then f = φF_y where φ = 0.9, and S = plastic modulus = b*t^2/4.
In either case, deflection should be checked for acceptability.

F_ultimate should never be used for this calculation.

BA

 

[BridgeSmith]

Thanks, cole44 for reposting what I wrote, but did you have a comment? Btw, it's common etiquette when you quote someone to clarify that it's a quote from someone else, and not your own.

This assumes that the OP is referring to a concentrated load at midspan of a simple span, which is not clear from his post.

Agreed. I did assume the bar was in bending about its 3" width dimension, simply supported at the ends of the 6" span. There are numerous loading conditions possible. Maybe I should have followed my first inclination and not bothered to respond to this thread. It's an ill-defined question that even a second year engineering student should have no problem with.

S = plastic modulus = b*t^2/4.

If ultimate tensile strength of the material is to be used, I suppose it would be correct to use the plastic section modulus to calculate capacity. As I alluded to in my previous post, the correct stress limit and section properties to use would depend on what constitutes failure - deflection beyond a certain value, yielding, or rupture/fracture.

Rod Smith, P.E., The artist formerly known as HotRod10