roarks formula confusion 5

[greycloud]

Greatings everyone

I'm trying to get how roark in his book "formulas for stress and strain" got the stress equation for bending stresses in rectangular plates.

I noticed that the equation is composed of the moment formula divided by just T^2 so how did he arrive at this result.

I hope you can help me resolve this mystery.

Thanks in advance

 

[prex]

Well, β in the above discussions depends indeed on the aspect ratio, but the case with b/a=∞ has been taken for direct comparison with the beam equation.
And if you look at Timoshenko (you should, as you cite it so often) you'll see that the factor β for the center moment in a simply supported rectangular plate equals 0.125 (b/a=∞), and you see that this is 0.75/6 or 1/8!
Also, if you compare β values in table 8 page 120 of Timoshenko with β values by Roark, you'll see that they exactly differ by a factor of 6.
So what are we discussing about? β is not a physical quantity, it is just a numerical coefficient, and any author can use its own definition of it without being in contrast with any theory.

prex
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[IDS]

beta is a function of the aspect ratio but from your derivation it doesn't seem to be affected by the aspect ratio.

I hope the additional details given by prex above will clear this up. I just looked at the beta factor for a very long plate (where the bending moment/m width is the same as for a beam 1 metre wide), because this is the easiest to check, but you can see in Roark that beta varies depending on the configuration of the plate. Anyway, it's good to get confirmation that Roark's values are indeed equal to Timoshenko's multiplied by 6.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[greycloud]

well thank you indeed for the clarification prex u are correct indeed and thanks Doug for your help as well and to everyone else now it makes since.

since we are on this does anyone know what is the moment formula for a rectangular plate subjected to a uniform moment at middle of longer side.

 

[rb1957]

"uniform moment at middle of longer side" ? maybe a point (discrete) moment at the middle of the long side ??

how is the plate supported ? corners only ? pinned along all edges ?? fixed ??

Quando Omni Flunkus Moritati

 

[greycloud]

it is the same case as moments around edges but instead applied in middle of the plate.

the plate is clamped from all sides

 

[BAretired]

Also, if you compare β values in table 8 page 120 of Timoshenko with β values by Roark, you'll see that they exactly differ by a factor of 6.

They should not differ by a factor of exactly 6 because, for a plate, the section modulus includes a factor of (1 - ν²) where ν is Poisson's Ratio. When ν is 0.3 in the case of steel, S = bt²/5.46.

BA

 

[rb1957]

if the plate has fixed edges, and the moment is applied on the edge ... then the plate will see nothing !?

Quando Omni Flunkus Moritati

 

[BAretired]

rb, I'm guessing that the moment is applied at midspan in both directions but it isn't clear. In any case, it belongs in a separate thread.

BA

 

[greycloud]

BA.: stress equations for plates in timoshinko's use 6 always. the usage of the factor u are talking about was included in buckling strength as I remember.

rb:Regarding the bending case as BA said the moment is applied midspan of the plate along the shorter length with all edges clamped.it needs a separate thread indeed.

 

[BAretired]

BA.: stress equations for plates in timoshinko's use 6 always. the usage of the factor u are talking about was included in buckling strength as I remember.

On p. 127 of "Plates and Shells" which you quoted earlier, Timoshenko uses the factors β and β1 to calculate Mx and My for Hydrostatic Loading on a simply supported rectangular plate. He does not provide stress equations.

If Roark (I don't have his book) has interpreted the stress in the plate from Timoshenko's work, he should not be using a section modulus of bt²/6. He should be using the plate section modulus of bt²/6(1 - ν²).

BA

 

[prex]

BA, the numerical factor we were speaking about is exactly 6.
The factor (1-ν²) is included in the values for β, as you can see from the fact that Timoshenko (p.120 as well as p.127) specifies ν=0.3 in the tables for β. Those tables serve to calculate plate moments, and the stress is simply 6M/t² (M=moment per unit length).

prex
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[IDS]

I was going to say what prex just said.

I would add that my earlier statement that the bending moment per unit width in a very long slab is equal to that in a beam under the same (unit width) load was overly simplistic. Nonetheless, that's the way it works out.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[BAretired]

prex and IDS,

Yes, you are both right. Sorry, greycloud for rocking the boat.

BA

 

[greycloud]

no problem man

 

[IDS]

Boat-rocking is good!

It prompted me to do a bit of research on the theory, confirm that the different looking formulas in Wikipedia, Pilkey and a couple of other sources all gave the same answer, which was the same as the Roarke values, and compare that with some quick FE analysis.

One thing I did discover is that although the formulas converge quite quickly for plates with an aspect ratio of up to about 5:1, for very long plates they converge quite slowly. I couldn't work out why the moment was increasing above wl^2/8 as the aspect ratio increased; it turned out I was just doing too few iterations.

Amazing that all this was worked out in the early 1800s.


Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[greycloud]

good for you man, you mentioned a couple of sources other than pelky and roark so can u give me their names please

 

[IDS]

Just a couple of links found on Google:

http://homepages.engineering.auckla.../06_PlateTheory_05_RectangularCoordinates.pdf

http://imechanica.org/files/theory of plates.pdf

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[greycloud]

thanks

 

[BAretired]

I still have a problem understanding how the stress in a plate can be identical to the stress in a rectangular beam when the moment per unit width is identical.

In the case of a beam, the top fibers compress and are free to expand sideways due to Poisson's Ratio. The bottom fibers stretch and are free to shrink sideways due to Poisson's Ratio. Under load, the beam's cross section is not exactly rectangular, but trapezoidal.

In the case of a plate, the top fibers are compressed but are not free to expand sideways. The bottom fibers stretch but are not free to shrink sideways. Under load, the cross section remains rectangular. If the applied moment is Mx, then My is required to prevent each little section of plate from becoming a trapezoid. This requires strain energy in the Y direction.

My current quandary is why that does not decrease the stress in the X direction. A uniformly loaded simply supported plate of span L and infinite width has a maximum moment of wL²/8. Does it have precisely the same stress as a square beam of the same depth?

For me, it is not intuitive.









BA

 

[IDS]

For me, it is not intuitive.

It wasn't for me either, which is why I wasted* half a day looking into it.

For me the non-intuitive part was that the longitudinal curvature at the mid-span of a very long slab must be close to zero, so how can there be a longitudinal moment?

I re-intuited it by imagining a very long slab (if we build it round the Equator it will be effectively infinite) simply supported on the two long sides. If we now cut a unit length slice of the slab with frictionless cuts of zero width there will be no change in the state of the cut slice, because there is zero shear transfer across any transverse section, and the slice is still restrained longitudinally.

If we now increase the width of the cuts to allow the two cut faces of the slab to rotate; the base of the slab will expand in the longitudinal direction, and the top will contract. This will tend to reduce the transverse stresses, due to the Poisson's Ratio effect, so the slab will deflect downwards to maintain moment equilibrium. The transverse bending moment at mid-span will always be wL^2/8, and the maximum transverse stresses will always be wL^2/8/(d^2/6), to maintain moment equilibrium, but the deflection will increase (to 5wL^4/384EI), depending on the Poisson's Ratio. For a Poisson's Ratio of zero the longitudinal moment would have been zero before the cut, so the deflection would already have been 5wL^4/384EI; for any positive Poisson's Ratio the deflection (before the cut) would have been less.

The Poisson's Ratio effect provides an effective prestress in the longitudinal direction. This does not reduce the transverse bending moment (which is controlled by the applied loads), and hence does not change the transverse stresses, but it does reduce the vertical deflection.


* Not really wasted.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/