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shear stress in a hollow circular beam
- Thread starter feaeng1
- Start date
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.
"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.
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.
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.
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))]
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))]
- Thread starter
- #24
Not exactly. The [τ] calculated above is in fact a [τ]xz and is indeed assumed constant over a chord in the approximation used.
You must then add a [τ]yz 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
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You must then add a [τ]yz 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
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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.
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).
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).
- Thread starter
- #28
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 ??
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 [τ]yz 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]](/data/assets/smilies/ponder.gif "[ponder] [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]](/data/assets/smilies/blush.gif "[blush] [blush]")
)
prex
: Online engineering calculations
: Magnetic brakes and launchers for fun rides
: Air bearing pads
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 [τ]yz 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
, 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...
)prex
: Online engineering calculations
: Magnetic brakes and launchers for fun rides
: Air bearing pads
- Thread starter
- #31
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.
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.
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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