Bending force calculators

[m_c]

Does anybody have any suggestions for a basic bending force / deflection calculator?
My googlefu is failing tonight.

I'm working on an idea, and just want a rough estimate for how much force a bit 25mm square mild steel bar could support without permanently bending, when supported 200mm apart.

 

[Maker]

Look up the Euler-Bernoulli beam equation for your particular load and support situation. The maximum force will depend on your allowable deflection.

 

[atomant48]

There’s a calculator on sunsetpatios.com.au

random link but ive used it a fair bit and it seems good. Simple answer to your question though, is a fair bit i would say! Depends entirely on how the load is being applied of course, but 200mm is not much of a span.

edit just took a loook… point load at 500kg negligible (1/10 mm) deflection. You’d have to check the shear stress though.

 

[mtt.tr]

Beam deflection app on Android it's pretty damn good.

 

[m_c]

Look up the Euler-Bernoulli beam equation for your particular load and support situation. The maximum force will depend on your allowable deflection.

I'm aware of the full equation requirements, but was just looking for some ball park figures without going into the full details.

There’s a calculator on sunsetpatios.com.au

random link but ive used it a fair bit and it seems good. Simple answer to your question though, is a fair bit i would say! Depends entirely on how the load is being applied of course, but 200mm is not much of a span.

edit just took a loook… point load at 500kg negligible (1/10 mm) deflection. You’d have to check the shear stress though.

Just tried that calculator, and it's given me some figures to work with, although I get 0.09mm deflection with a 3000kg load.

Beam deflection app on Android it's pretty damn good.

I'll check that out.

 

[atomant48]

Just tried that calculator, and it's given me some figures to work with, although I get 0.09mm deflection with a 3000kg load.

Point load vs continuous load, i suspect.

 

[mtt.tr]

I'm aware of the full equation requirements, but was just looking for some ball park figures without going into the full details.

Just tried that calculator, and it's given me some figures to work with, although I get 0.09mm deflection with a 3000kg load.

I'll check that out.

Blue ketchup is the development company

 

[Agroshield]

I'm working on an idea, and just want a rough estimate for how much force a bit 25mm square mild steel bar could support without permanently bending

It might be difficult to find what you are asking directly from a calculator. You would have to take the numbers they give you and do one last (not very complicated) number crunch on them.

When you apply a load to the beam you have defined, most of the calculators will tell you the stress in the beam and the deflection that results.

Your question is asking what load will produce a stress in the structure equal to 100% of the elastic limit.

Maybe some of the calculators will tell you what percentage of the elastic limit stress for the material the calculated stress represents but you might have to look up the relevant elastic limit stress for the material and do the calculation yourself (random link here says 8 x 10^8 N/m^2 https://byjus.com/question-answer/the-elastic-limit-of-steel-is-8-times-10-8-nm-2-and-its-young/ ).

It is possible but it is an unusual thing to want to calculate: for 95% of beams, the limiting criteria is deflection (normally limited to span/360).

 

[m_c]

It might be difficult to find what you are asking directly from a calculator. You would have to take the numbers they give you and do one last (not very complicated) number crunch on them.

When you apply a load to the beam you have defined, most of the calculators will tell you the stress in the beam and the deflection that results.

Your question is asking what load will produce a stress in the structure equal to 100% of the elastic limit.

Maybe some of the calculators will tell you what percentage of the elastic limit stress for the material the calculated stress represents but you might have to look up the relevant elastic limit stress for the material and do the calculation yourself (random link here says 8 x 10^8 N/m^2 https://byjus.com/question-answer/the-elastic-limit-of-steel-is-8-times-10-8-nm-2-and-its-young/ ).

It is possible but it is an unusual thing to want to calculate: for 95% of beams, the limiting criteria is deflection (normally limited to span/360).

I'm aware of how the numbers should be calculated, but this is for something where I just want to make sure 25mm bar would be good enough (I happen to have a length in the workshop!).
It involves some machining, so I just wanted to make sure it will be strong enough before I start, rather than waste time and have it bend on first use.

Provided it can take a couple tonne of pointed load on an unconstrained beam, it'll work for what I need, as the actual load is going to be spread out, and I suspect if it gets over a couple tonne of load, the support isn't going to be my main problem.

 

[Misterg]

random link here says 8 x 10^8 N/m^2

That's a bit(!) optimistic for mild steel - I think that the relevant property would be the yield stress which might be as low as ~250 MPa ( or N/mm^2) for "mild steel".

If we're talking about a solid, square bar, rather than an 'I' beam or a hollow section it'll probably fail in bending (for more complicated shapes, one would also need to consider other modes of failure).

The bending stress (s) in a beam = (M x y) / I ......(1)

M = bending moment, y = distance from the neutral axis of the beam., I = moment of inertia of the beam cross section.

For a square section beam, I= (t^4)/12 For the beam in question I= (25^4)/12 = 32,552 mm^4
Maximum bending stress occurs at the outer face of the beam, so y = 12.5mm in this case.
If we want to find out when the maximum bending stress reaches the yield stress of the material, then s = 250 N/mm^2 (for the sake of argument)

Rearranging (1):

M = (I x s)/y = (32552mm^4 x 250N/mm^2) / 12.5mm = 65,1041 Nmm (651 Nm)

See here for the bending moment calculation for various loading scenarios (scroll down to the pictures at the bottom)..

From that page, for a beam on point supports separated by distance (L), with a central point load (F):

Maximum bending moment (M) = (F x L)/4 .....(2) (NB: different loading or support scenarios will have a different formula here!)

Rearranging (2):

F = (M x 4)/L = (651041 x 4) / 200 = 13,020N (13 kN) or roughly 1300 Kg


Latest reply arrived after I'd typed this -

My answer: Not looking good for a couple of tonnes as a central point load. (IMHO , FWIW)

(Not a Structural Engineer and make a lot of mistakes with maths.)

 

[carbonarc]

I'm aware of how the numbers should be calculated, but this is for something where I just want to make sure 25mm bar would be good enough (I happen to have a length in the workshop!).
It involves some machining, so I just wanted to make sure it will be strong enough before I start, rather than waste time and have it bend on first use.

Provided it can take a couple tonne of pointed load on an unconstrained beam, it'll work for what I need, as the actual load is going to be spread out, and I suspect if it gets over a couple tonne of load, the support isn't going to be my main problem.

Hi m_c:. What sort of reduction in cross section will the machining you mention cause?

 

[slim_boy_fat]

I'm working on an idea....

Just out of interest, what's the application?

 

[m_c]

It's for holding a shaft in a press for assembly.
Think open frame vice type setup, as the shaft needs to pass through it for clamping.

The machined parts need to sit on top of the square bar, but I'm probably going to double up the bar, as it'll make the design a bit simpler to weld the major parts together.