shear stress in a hollow circular beam

[feaeng1]

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.

 

[electricpete]

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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[feaeng1]

electricpete,

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

 

[Ron]

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?

 

[zekeman]

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

 

[racookpe1978]

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.

 

[electricpete]

http://books.google.com/books?id=Zw...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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[prex]

Well, zekeman and electricpete, the formula for a solid circle is not directly applicable to a hollow one... (see also thread507-243730).
The general formula for shear stress in a section, assuming [τ]_xz is uniformly distributed along the horizontal diameter of the cross section, is
[τ]_xz=PS_i/Ib_i 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
S_i=(C_i³-c_i³)/12
b_i=C_i-c_i
C_i and c_i being the lengths of the segments intercepted by a chord on the outer and inner circles respectively.
S_i/b_i=(C_i²+C_ic_i+c_i²)/12
and at the centroid, where we know that the shear stress is maximum and [τ]_yz=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

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[prex]

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

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[rb1957]

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

 

[zekeman]

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)."

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.

 

[zekeman]

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

 

[prex]

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 R[≅]r (thin hollow circle, presumably what the OP was looking for) resolves into [τ]=2P/A

prex

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[rb1957]

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)

 

[feaeng1]

Wow, This topic generates so many response. Thanks all.

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

 

[zekeman]

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.

 

[zekeman]

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))]

 

[prex]

My results:
for y>=r:
[&tau;]=(P/3A)(R²-y²)/(R²+r²)
for y<r and taking x²=y²/(R²+r²):
[&tau;]=(P/3A)(1-2x²+[&radic;](x⁴-x²+R²r²/(R²+r²)²))

prex

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[prex]

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

prex

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[rb1957]

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

zeke has an extra sqrt

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