shear deflection constant for thick-walled cylinders

huangyhg

shear deflection constant for thick-walled cylinders
hi.  i am trying to determine an equation for the shear deflection constant of a cantilevered cylinder, as a function of the wall thickness.
i can find the thin-walled factor in roark, where the value is equal to 2, but in researching the paper referenced by roark, "deflection of beams with special reference to shear deformations" dept. of agriculture 1924, i can't figure out how to derive this for a cylinder.  
does anyone have another reference?  any thoughts on what to do for shear deflection of a thick-walled tube?  i am trying to figure this out in order to come up with the correct rigidity of a large circular shear wall in my building.  
thanks for your help.
also, fyi, this was discussed in
see if you can find the reference below:
cowper, g. r., "the shear coefficient in timoshenko's beam theory." journal of applied mechanics, vol. 33, no. 2, june 1966 (transactions of the american society of mechanical engineers, vol. 88, series e), pp. 335-340.
good luck.  
the values of ffff"> as tabulated by roark must be interpreted as follows.
1)the value of ffff">=10/9=1.111 for the solid circle corresponds to an approximation where only the component τfff">xy is considered, and the component τfff">xz of the shear stress is neglected.
2)if the above approximation is removed, but still the common approximation, of considering τfff">xy constant over the chords perpendicular to the load, is maintained, then the value becomes ffff">=32/27=1.185
3)if the full solution for elastic stresses is obtained, now ffff">=1.175
4)from a numerical integration (an analytical integration can be done, but the resulting expressions are really cumbersome) i've determined that, for a thin hollow circle, ffff">=1.5 in the approximation under 1 above
5)from the same numerical approach, but considering the two components as in 2 above, and again for a thin hollow circle, i get the figure quoted by roark: ffff">=2
6)the approximation as in 5 above has the disadvantage of leading to an infinite shear stress at the inside diameter on the load line of action, so it's not necessarily better than the other one, at least for thin cylinders. however, as the calculated deflection is bigger, it could be considered as safer.
7)from the numerical approach as in 2 and 5, i've determined the following curve fit, that you can use for your purpose; the error with respect to the (almost) exact numerical solution is everywhere better than 1%:
ffff">=-1.9235xfff">3+2.6215xfff">2+0.1109xfff">+1.1773
where xfff">=rfff">i/rfff">o
as you can see, the picture is far from being crystal clear. good luck!
prex
thank you for the very thorough answer!  this is very helpful.
i would appreciate it if you could give me a little more background on how you came up with this equation - i would have to derive it myself if i want to use it on the project.  could you please tell me the equation that you integrated to arrive at the equation that you gave me (or the general approach) -  i myself am having trouble understanding what the first moment of area term is for this case, from the 1970 follow-up dept. of agriculture paper on the topic.  
or any kind of attachment with the derivation, if you've alreay completed it, would be awesome.  
thank you so much for your help!
you are getting me troubles! this is not really simple to put down and to explain: my basic book for structural mechanics is a gigantic performance of belluzzi in 4 volumes, but it's only available in italian, so i guess it wouldn't be of much help to you (and it gives anyway only the starting point!).
let me try to give at least the basics.
when the component τfff">xz is neglected, the expression is
ffff">=(1/aρfff">4)∫(sfff">i2/bfff">i)dy
where:
afff">=section area
ρfff">=radius of gyration
bfff">i=length of a chord intersected by section
sfff">i=static moment of one of the parts divided by bfff">i with respect to neutral axis.
to calculate the integral it is necessary to split the integral into two parts: for y going from 0 to rfff">i (where the chord is split into two equal segments) and for y from rfff">i to rfff">o (the chord is a single segment)(the result is then of course multiplied by 2).
the second part is easy:
sfff">i=bfff">i3/12
bfff">i=2rfff">ocosφ
dy=rfff">ocosφfff">dφ
where φfff"> is the angle from the horizontal to the chord under consideration at the outer diameter.
let's call:
xfff">=rfff">i/rfff">o,
yfff">=√(1-xfff">2)
φfff">=sin-1x
for a full circle the integral, already divided by aρfff">4, would give
(32/9π)[5φfff">/16]π/2-π/2=10/9 as expected.
for the hollow circle the second part of the integral (excluding the factor 1/aρfff">4) gives:
(2rfff">o6/9)(5(π/2-φfff">)/16-15xyfff">/32-3xyfff">(1-2xfff">2)/16+xyfff">(3-4xfff">2)(4xfff">2-1)/96)
as you see, if this is the easy part you can imagine what is left!
i think i'll stop here, it is really a too long route to explain everything. at least you had a picture of the problem and you'll be able to decide whether you can go on by yourself.
what i can suggest you more is that if you draw a straight line between the (known) extreme points ffff">=1.185 and ffff">=2 you'll obtain a value that's closer than 6% to the theoretical one. if you consider that the value ffff">=2 for a thin cylinder will be in fact somewhere between 1.5 and 2 (see my first post), the method should be fully acceptable.
i assume also, of course, that you are aware that a pure shear deformation assumes that the shape of the section is not changed by a local bending under the action of the load: a local deformation in bending will in almost any case be much higher than pure shear.
prex
prex is correct, the solution involves the use of energy method to solve a set of differential equations. go find the reference provided above, you should be able to get re-prints from asce.