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shear stress in a hollow circular beam

shear stress in a hollow circular beam

shear stress in a hollow circular beam

(OP)
Hello, everyone,

Is there anyone knows that the general formula to compute the shear stress at any point on the cross section of a hollow circular beam? Thanks.

RE: shear stress in a hollow circular beam

I don't think you can draw any conclusions about the stress without knowing something about the loading.  Torsional or axial or lateral-bending or something else?  If it is a lateral beam bending problem, where are the loads applied?

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RE: shear stress in a hollow circular beam

(OP)
electricpete,

Yes, it's beam bending, on a cross section with a shear force V.  

RE: shear stress in a hollow circular beam

You're not giving enough information. How is the beam loaded?  If the loading is uniformly distributed, that creates one shear characteristic...if the load is a single or multiple point loads, the shear will be different, therefore different approaches to the analysis would be necessary.

You first need to determine at what point along the beam the maximum shear occurs.  Then you apply the shear to that cross section to determine shear stress.

Hollow circular sections are not often used as beams.  What is your application?

RE: shear stress in a hollow circular beam

I get the maximum shear at y=0 (developed from Timoshenko formula for solid cylindrical beam)
4V*(R^3-r^3)/[3A*(R-r)(R^2+r^2)]
For r=0 (solid) this becomes
4V/3A

V vertical shear force
R,r outer and inner radii
A cross sectional area

RE: shear stress in a hollow circular beam

Unless the "hollow beam" creates a "narrow-walled-pipe" crushing, bending problem: The original poster has NOT stated his diameter, wall thickness, nor "how" the shear is being applied.

A 3/8" wire rope, for example, strung over a 36 inch dia 1/16" thick-walled piece of sheet metal duct holding up a 2000 pound load is NOT going to resist shear like a 4" dia sched 160 pipe being used as a 12 inch long beam lifting that same load.

 

RE: shear stress in a hollow circular beam

http://books.google.com/books?id=ZwHaHnNqLPwC&dq=TIMOSHENKO+%22THEORY+OF+ELASTICITY%22&printsec=frontcover&;source=bl&ots=VdBV5qEFQW&sig=gQLotwmQjJXglmAFB8Ug7YPQ89Y&hl=en&ei=O5IsSu6fOJ6ytAOv34SnCw&sa=X&oi=book_result&ct=result&resnum=4#PPA358,M1

Page 354 begins a discussion of bar bending.
Page 358 begins a discussion of circular bar bending.
Page 359 has the afore-mentioned 4/3 of P/A... assuming the shear stress is uniformly distributed....

It's a wee bit above my head.

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RE: shear stress in a hollow circular beam

Well, zekeman and electricpete, the formula for a solid circle is not directly applicable to a hollow one... (see also thread507-243730: Shear Deflection Constant for Thick-Walled Cylinders).
The general formula for shear stress in a section, assuming τxz is uniformly distributed along the horizontal diameter of the cross section, is
τxz=PSi/Ibi with meaning of symbols easily gathered from the elementary theory of beams.
For a hollow circle, neglecting the upper and lower portions, where a single segment is intercepted by a chord (and the shear stress is clearly not at its maximum), we have:
I=π(R4-r4)/4=A(R2+r2)/4
Si=(Ci3-ci3)/12
bi=Ci-ci
Ci and ci being the lengths of the segments intercepted by a chord on the outer and inner circles respectively.
Si/bi=(Ci2+Cici+ci2)/12
and at the centroid, where we know that the shear stress is maximum and τyz=0
τ=(4P/3A)(R2+Rr+r2)/(R2+r2)=2P/A
so substantially different from the value for a solid circle (if I'm not in error, of course)

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: shear stress in a hollow circular beam

Oops... in the last equation above, the second equal sign should be replaced by a ≅ symbol: the last value is an approximation for a thin hollow circle.

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: shear stress in a hollow circular beam

how about shear stress = V*Q/(I*t) ?
 

RE: shear stress in a hollow circular beam

Quote
"Well, zekeman and electricpete, the formula for a solid circle is not directly applicable to a hollow one... (see also thread507-243730: Shear Deflection Constant for Thick-Walled Cylinders: Shear Deflection Constant for Thick-Walled Cylinders)."

Nobody said it was; if you read my post carefully, I gave the solution which is yours , absent the -P/A which is puzzling to me. since I used Timoshenko's basic approach to get the hollow cylinder
I pointed  that the  solution I presented degenerated to the solid one (r=0) which is well known.
Please explain why yours does not.

 

RE: shear stress in a hollow circular beam

Corresction,
Prex,
I meant your -2P/A .

RE: shear stress in a hollow circular beam

Sorry zekeman, I wasn't that careful: your solution is exactly the same as mine, and (of course) both correctly default to τ=4P/3A for r=0.
What I added as an important point, is that this solution for Rr (thin hollow circle, presumably what the OP was looking for) resolves into τ=2P/A

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: shear stress in a hollow circular beam

now the OP does ask for "any" point, so the general expresion VQ/(It) is a better answer.  the math seems to get messy, 'cause it looks hard to relate ci to Ci (Ci = Rsin(theta), ci = sqrt(r^2-(Rcos(theta))^2)

RE: shear stress in a hollow circular beam

(OP)
Wow, This topic generates so many response. Thanks all.

Yes, for the maximum shear stress: it's correct, τ=(4P/3A)(R2+Rr+r2)/(R2+r2), for a hollow circular cross section and τ=2P/A for thin wall.

How about shear stress at any point? Do I have to use VQ/(It) as rb1957 said? Is there any formular just for a hollow circular cross section?

Thanks again to you all.  

RE: shear stress in a hollow circular beam

Shear stress
y1=distance to shear point from neutral axis
For y1>r:
4V*R*(R^2-y1^2)^1/2)/[3*A*(R^2-r^2)]

Fory1<r:
8*V/[(R^2-y1^2)^3/2-(r^2-y1^2)^3/2]/[3*A*(R^2-r^2)]
r,R inner and outer radii
Wouldn't bet the farm on it being error free.
 

RE: shear stress in a hollow circular beam

Correction
fory1>r:
4V*R*(R^2-y1^2)^1/2)/[3*A*(R^2+r^2)]

Fory1<r:
8*V/[(R^2-y1^2)^3/2-(r^2-y1^2)^3/2]/[6*A*(R^2+r^2)*((R^2-y1^2)^1/2-
(r^2-y1^2)^1/2))]

RE: shear stress in a hollow circular beam

My results:
for y>=r:
τ=(P/3A)(R2-y2)/(R2+r2)
for y<r and taking x2=y2/(R2+r2):
τ=(P/3A)(1-2x2+√(x4-x2+R2r2/(R2+r2)2))

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: shear stress in a hollow circular beam

A factor of 4 is missing above. Again:
for y>=r:
τ=(4P/3A)(R2-y2)/(R2+r2)
for y<r and taking x2=y2/(R2+r2):
τ=(4P/3A)(1-2x2+√(x4-x2+R2r2/(R2+r2)2))
and for a thin tube (any y):
τ=(2P/A)(1-y2/R2)

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads

RE: shear stress in a hollow circular beam

prex is at least dimensionally correct (for y>r)

zeke has an extra sqrt

other than that i think prex is right (again!)

RE: shear stress in a hollow circular beam

Quote from rb
"other than that i think prex is right (again!) "

I didn't know that you were the arbiter of last resort.
If so, I would think you could read  dimensionality correctly and not make a case for correctness based on dimensionality alone.
I don't think Prex needs fans like you.

RE: shear stress in a hollow circular beam

actually zeke, i ran a bunch of numbers thru excel, things seemed to line up, including the somewhat surprising simplification for a thin ring (on 1st glance the bracketed term doesn't look as though it's going to converge onto the thin ring  solution).

for y>r, the difference between your post and prex's is and extra sqrt ... whihc i took (incorrectly) to mess with the units.  on closer inspection it doesn't, so i humbly apologise for taking your name in vain.  however, R*sqrt(R^2-y^2) is quite different to (R^2-y^2) which is correct for a solid disc (which the tube is for y>r)

and since when is my opinion "the arbitor of last resort" ... my opinion is mine, to be agreed with, argued with, ignored, spurned, as you (whoever) wills.

have a nice day.

RE: shear stress in a hollow circular beam

My answer for y>r was in fact the maximum shear stress at the curvature  points ( from Timoshenko) but the midrange should be as Prex's.
The only test I have for my y1<r solution (with some factoring and typos) would be to check them at y1=r, in which I confirm the equality of my corrected 2 solutions, but can't guarantee it.
Coreected solutions:
For y1>r:
4V*(R^2-y1^2))/[3*A*(R^2+r^2)]

For y1<r:
4*V*[(R^2-y1^2)^3/2-(r^2-y1^2)^3/2]/[3*A*(R^2+r^2)*((R^2-y1^2)^1/2-
(r^2-y1^2)^1/2))]
 
 

RE: shear stress in a hollow circular beam

(OP)
So it's solved. Assume that the shear stress is parallel to y axis and uniformly distributed along chord.
 

RE: shear stress in a hollow circular beam

Not exactly. The τ calculated above is in fact a τxz and is indeed assumed constant over a chord in the approximation used.
You must then add a τyz that, at each end of the chord, is such that the total τ is tangent to the boundary, and may be assumed to vary linearly along the chord.

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

RE: shear stress in a hollow circular beam

yeah, sure there is a distribution of shear stress across the wall, but for all practical purposes these equations give the shear stress acting on the wall of the tube.  if you integrate shear stress over area and then do the vector component you've get the applied shear force. particularly if you're using the thin wall equation.  

RE: shear stress in a hollow circular beam

just a thought, which i haven't thought enough about, ... what is the implication of the varying shear stress on the normal stress distribution on the section ?

the normal stress distribution on a section in bending is linear.  doesn't this imply a constant shear stress ?

the noraml stress at different sections along the beam increase linearly; doesn't this imply constant shear stress ??

aren't these two shears the same ???

i'll have to look into the derivation of flexural shear (VQ/It).   

RE: shear stress in a hollow circular beam

(OP)
I believe the shear stress is the function of y^2 while the bending stress is the function of y.

RE: shear stress in a hollow circular beam

there's something to think on ... is shear stress proportional to the derivative of normal stress or is normal stress proportional to the derivative of shear stress (as feajob suggests above)?  I pretty much automatically thought that shear stress was related to the derivative of normal stress, that a change in normal stress is caused by shear stresses ??

RE: shear stress in a hollow circular beam

Neither of the two, I suppose.
What we can derive from the equations of equilibrium:
-from one section of the beam to another one, assuming no distributed or concentrated loads act between the two on the beam: V=dM/dz, and of course, in the absence of external forces, M varies linearly (or is constant) and V is constant (or zero). Here only the global characteristics V and M are meaningful, not the stresses
-from the equilibrium of a section of the beam cut between two sections and a plane parallel to the axis z and orthogonal to the shearing load, we get VQ/It (or PSi/Ibi as I wrote it above), so that this relationship is a direct consequence of the normal stress due to bending varying linearly over section depth.
Two (possibly) interesting notations for feaeng1:
-the usual approximation of considering a constant τyz over a chord leads inevitably to an infinite shear stress at the points on the inside diameter in line with the load; this perhaps explains why this case is normally not treated in the books ponder, but is generally not a problem, as no one is interested to the shear stress elsewhere than at the neutral axis (so feaeng1, could you tell us why are you so interested by the distribution of shear over the section?)
-the distribution of shear stress for a thin tube is the same as for a rectangle having the same section depth and a width a bit over the double of the thickness (in rethinking to this very profound discovery, can't really see what usage we could do of it...blush)

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

RE: shear stress in a hollow circular beam

(OP)
Acturally we are trying to find out where the maximum von Misses for a combinded load case. For the pure shear, the max von Misses occurs at the neutral axis. For the pure bending, the max von Misses is located on the outter surface. We are wondering that if the max von Misses could be in somewhere in between for a combined load case.

RE: shear stress in a hollow circular beam

really every practical case is a combination of shear and moment ... ok, you can create some structures waht have pure bending, i have touble thinking of a beam (outside of a textbook) in pure shear.

what you might look at varying the moment stress (consider a cantilever with different sections at different distances from the load).  you know what the shear stresses are, they don't change.  once the peak bending stress > peak shear stress, the critical stress is probably the extreme fiber; but what if the peak bending stress is 1/2 the peak shear stress.  i'd play with the von mises formulae, with noraml stress = k*shear stress.  it makes sense to me that the critical location is at the neutral axis for pure shear, and as soon as you add moment stresses the critical location moves towards the extreme fiber.

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