Beam bending - calculating effects of shear deformation 4

[electricpete]

If we are given a set of loadings w(x) on a beam, we can integrate 4 times (with a few constants) to get V(x), M(x), Slope(x), Yb(x).

This is the displacement due to bending... dependent upon E, and assumes G infinite (no shear deflection). The error is small for anything other than very short fat beam.

If we now want to calculate the deflection due to shear alone, couldn't we analyse a beam with infinite E and finite G. It seems that deflection would be:
Ys(x) = Integral [Gamma dx]
where Gamma = shear strain = Tau/G = V*A/G
Ys(x) = Integral[V(x) * A dx] / G
I think this requires an assumption that shear stress is uniformly distributed... seems reasonable.

Then we could find Y(x) = Yb(x) + Ys(x).
Yb = displacement from typical bending cal
Ys = displacement from shear calc

Would this be exactly correct? Or close to correct? Or totally wrong?
Thanks in advance.


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[desertfox]

Hi elecricpete

I have never really had to calculate shear deflection on a beam, but according to my mechanics book I am looking at for a simply supported beam with a udl then:-

d[²][ν]s/dx[²]= -w/GxA*y

note:-
A*y = shear area associated with shear parallel to y axis,
for a rectangle A*y= 5/6*A (A being the area)

w = load intensity

vs= deflection

Not sure whether your loadings a udl or point load or whether your beam is a solid or hollow section, a hollow section will make a difference if its a thin walled tube in relation to the shear area.
Otherwise I agree with your thoughts.

desertfox

 

[desertfox]

Hi electricpete

I found this article on shear deflection of beams although it deals with beams made from wood, the formula and approach seem applicable to any beam, either way its quite interesting and a very old paper.

http://www.fpl.fs.fed.us/documnts/fplmisc/rpt1309.pdf

desertfox

 

[flash3780]

I seem to remember that famous Irishman, Tim O'Shenko, came up with a formulation for beam bending that takes into account shear deformation. Check the following link:

http://en.wikipedia.org/wiki/Timoshenko_beam_theory

 

[electricpete]

Yes, I am well familiar with Mr O'Shenko (LOL) and the fact that his beam model accounts for shear deformation (and rotary inertia in the case of dynamic analysis).

I have never got to the point of using that model for analysis but looking at the differential equation I believe it is correct that the displacement solution can be separated into one portion associated with bending and one portion associated with shear (which is good... allows us to compare the familiar bending one to the shear one).

The part I am not quite figuring out is this mysterious shear factor Kappa in the wikipedia link.

I think it is the 5/6 in desertfox's result fo rrectangular beam.

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[desertfox]

hi

Yes thats correct e-pete from that link I found this:-

?, called the Timoshenko shear coefficient, depends on the geometry. Normally, ? = 5 / 6 for a rectangular section

desertfox

 

[flash3780]

I have to admit, I shamelessly stole that joke from another post.

Nonetheless, kappa is always a troublemaker. There's a 2001 article in the ASME Journal of Applied Mechanics by J. R. Hutchinson that might be worth looking into called "Shear Coefficients for Timoshenko Beam Theory". I'm having trouble logging in at the moment, but you can get it on the ASME Digital library. Otherwise, might be worth a trip to your local engineering college library.

Quite honestly, I've never found a situation where the beam theory that the other Irishman, Bernie Noulli, came up with wasn't sufficient.

 

[electricpete]

What is that Kappa factor for a round geometry such as a shaft?

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[TERIO]

There is some good discussion in Robert Jones, "Buckling of Bars, Plates, and Shells" in sections 2.7.2 and 2.7.3.

http://books.google.com/books?id=UzVBr8b_jS8C&lpg=PP1&client=firefox-a&pg=PA145#v=onepage&q=&f=false

Note he uses a value [α]=1/[κ]. The [κ] value makes the strain energy of a uniform shear stress on the beam cross-section equal to the strain energy of the actual shear stress distribution.

 

[flash3780]

For a circular cross section, kappa = 9/10.
(Ref. Rao, "Mechanical Vibrations")

 

[prex]

electricpete, I'm afraid that your statement is not true.
That wiki article incorrectly states IMHO that the two equations may be decoupled in the static case. In fact, if you look at the second equation, it contains two unknown functions, w and [φ], so shear cannot be solved alone without solving for bending first.
In other words shear cannot be solved without bending, or even, more generally, there is no shear in a beam with no bending.
In the first site below, under Beams -> Bending+shear you can see numerical solutions for shear separated from bending: I would be interested to know if you find something odd in those.

prex

http://www.xcalcs.com

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http://www.megamag.it

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http://www.levitans.com

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[electricpete]

Prex,

I focused on this equation:
E*I *d^4(w)/dx^4 = q(x) – [E*I* d^2(q)/dx^2]/[Kappa*A*G]

Displacement shows up only on left side. The right side doesn’t have displacement or any of its derivatives, so can be viewed as sum of two forcing functions: q(x) and – [E*I* d^2(q)/dx^2]/[Kappa*A*G]. So it suggests immediately that the complete solution can be solved by superposition considering each of the forcing functions alone.

The first term q(x) on on rhs is the familiar applied force for normal beam bending calc.

To find the response of that 2nd rhs term alone, ignore the 1st term:
E*I *d^4(w)/dx^4 = – [E*I* d^2(q)/dx^2]/[Kappa*A*G]

Divide by E*I
d^4(w)/dx^4 = – [ d^2(q)/dx^2]/[Kappa*A*G]

Integrate three times
d(w)/dx = – [ Integral q dx ]/[Kappa*A*G]

Substitute V = Integral q dx
d(w)/dx = – [ V ]/[Kappa*A*G]

Integrate one more time:
w(x) = - Integral (V)dx / [Kappa *A *G]

That looks pretty close to what I started out with in my original post (allowing notation w instead of y for displacement),

If we now want to calculate the deflection due to shear alone, couldn't we analyse a beam with infinite E and finite G. It seems that deflection would be:
Ys(x) = Integral [Gamma dx]
where Gamma = shear strain = Tau/G = V*A/G
Ys(x) = Integral[V(x) * A dx] / G[/quote electricpete]
Except:
1 – new factor of Kappa
2 – notation w instead of y for displacement.
3 – my original post had an error... obviously Tau/G = V*A/(G) was wrong and should've been Tau/G = V/(A*G) so the final expression should’ve had A in the denominator.
4 – Negative sign. I suspect that is based on some convention of signs that I have overlooked.

Maybe it’s too simplistic, but it seems to make sense to me.

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[prex]

What you state doesn't seem incorrect to me, but again this is not necessarily sufficient to solve a shear problem completely separate from a bending problem, as you are trying to do.
This should be because of the boundary conditions: in that equation on w alone you can only impose boundary conditions on w, but you need to constrain also the rotations (that are independent of the displacements) to solve a general shear problem.
The bending+shear problems in xcalcs, for example here, are solved with an energy minimization approach. At first I tried to do as you propose, minimizing the energy associated to shear separate from that of bending, but this didn't work, the solution was incorrect for statically determinate conditions, and impossible to find for indeterminate ones. The solutions you can see there are obtained by summing up the energies of bending and shear.
But we all know that the inverse is true: we can of course solve a problem of pure bending without shear!
This situation must have something in common with the fact that shear stresses alone cannot exist in an elastic material: if you have a shear stress in a certain direction, than you have other specific directions where those shear stresses disappear and only direct (principal) stresses exist (Mohr's circle).

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[electricpete]

Good point - I agree – each of the solutions must satisfy the boundary conditions: the bending solution and the shear solution. Can do it for some problems, but not for others.

Here’s one where it works: If we have a simply supported beam on two supports with applied loads between the bearings, then the shear solution will satisfy the boundary conditions. (If we adjust the single integration constant to make displacement at first bearing zero, the second will also be zero... I know this because the equation for bending shear is the same form as the equation for moment... both depend on integral of shear ... and moment comes is zero at both bearings).

But if that same simply supported beam on two supports has load applied to an overhung portion, then the shear solution does not satisfy the boundary condition (the bearing adjacent to the overhung load has non-zero shear so would have non-zero displacement in the shear solution of displacement).

Likewise if we have built in boundary conditions, the shear solution won’t work because it doesn’t have zero slope at that boundary.


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[electricpete]

Correction:
"...the bearing adjacent to the overhung load has non-zero shear so would have non-zero displacement in the shear solution..."
should've been:
"...the bearing adjacent to the overhung load has non-zero moment so would have non-zero displacement in the shear solution..."

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[electricpete]

Sorry: Here are two corrections in bold

Good point - I agree – each of the solutions must satisfy the boundary conditions: the bending solution and the shear solution. Can do it for some problems, but not for others.

Here’s one where it works: If we have a simply supported beam on two supports with applied loads between the bearings, then the shear solution will satisfy the boundary conditions. (If we adjust the single integration constant to make displacement at first bearing zero, the second will also be zero... I know this because the equation for displacement from bending shear is the same form as the equation for moment... both depend on integral of shear ... and moment comes is zero at both bearings).

But if that same simply supported beam on two supports has load applied to an overhung portion, then the moment solution does not satisfy the boundary condition (the bearing adjacent to the overhung load has non-zero shear so would have non-zero displacement in the shear solution of displacement).

Likewise if we have built in boundary conditions, the shear solution won’t work because it doesn’t have zero slope at that boundary.

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[electricpete]

Note - I didnt’ consider the moment=0 part of the simply-supported boundary condition in my discussion above.

It doesn’t really seem applicable that the shear solution would have to satisfy that... i.e. I don’t think we have to require 2nd derivative of displacement-from-shear solution to be 0 at simply supported boundary. But I’m not positive.


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[TERIO]

I'm not sure you have all the boundary conditions right here. For instance for a built-in end when considering shear deformations I think it is typical to consider the boundary conditions as

dY/dx=dYs/dx=V/(G*As)

and

dYb/dx=0

Just the slope due to bending is taken as zero, not the entire slope (see e.g. J.S. Przemieniecki, "Theory of Matrix Structural Analysis", pgs. 74&75).

 

[electricpete]

Thanks. I dont' have that reference. But it seems to make sense that the “slope boundary condition” of a fixed end may not apply to the shear displacement problem. The physical situation is that the support provides equal/opposite moment to the moment of the beam where it enters the support. The moment is important to the bending problem, but the momoent is not important to the shear problem (at least under assumptions about shear stress distribution).

That would mean that the superposition of bending and shear displacements is applicable to a wider range than I described. I’m not sure whether Prex would agree or not.

If I get some spare time on my hands, I will compare this to Castagliano’s strain energy approach to confirm check where it gives the same results (don’t anyone hold your breath). I am reasonably sure it cannot give correct results for the load which was overhung off of two simple supports... for reasons discussed above (cannot simultaneously satisfy the displacement boundary conditions at both supports).


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[desertfox]

hi elecricpete

The way I see it is that, yes you need to have bending on the beam to get shear stresses, however I believe once you meet that criteria you can then calculate the deformations seperately.

regards

desertfox