How do you calculate the radial thermal expansion (or contraction) of a ring? For example, if you had a spool piece that was not constrained in any way, and you applied heat, how would you calculate the new ID and OD? Is it as simple as the original ID or OD times the thermal coefficient times the change in temperature? Logically, it seems like it would be this simple, but then I think there might be more to this than I realize, so I'd better double check. I can't find a similar example in any of my books so far, which seems odd.
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Thermal (radial) expansion of a ring 6
- Thread starter packdad
- Start date
I would take the circumference of material present and compute the new length based on difference in temperature. You are correct, use the thermal coefficient of expansion. Then dividing through by "pi" would give OD as in circumference of a circle is diameter times "pi".
In the real party world, I imagine some small loss of wall thickness. You would need to use Poisson's Ration for the material in question, knowing the final wall thickness. At any rate, volume of material is conserved so a guy could easily find the necessary circumference based on changes in the other two dimensions and Poisson's Ratio.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
In the real party world, I imagine some small loss of wall thickness. You would need to use Poisson's Ration for the material in question, knowing the final wall thickness. At any rate, volume of material is conserved so a guy could easily find the necessary circumference based on changes in the other two dimensions and Poisson's Ratio.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
Yes. It is that simple, which is probably why you cannot find it in any text books. For unconstrained uniform thermal expansion all linear dimensions want to increase in the same proportion, be they straight lines like radii or curved lines like circumferences or even lines across holes.
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3
- #4
There is no chance of the wall thickness increasing as the ring thermally expands, not a chance at all, Corus.
Material volume of the ring is conserved before and after thermal expansion. Cut the ring open and flatten out the circumference; apply tension. Does the openned ring not behave similar to a wire under tension? I can't see the wall thickness increasing!
Rather, you would see the wall experiencing "necking". As the circumference increases due to diametrical expansion the ring MUST decrease in both width and wall thickness. In other words:
eX = (Sx - vSy - vSz) / E
eY = (-vSx + Sy - vSz) / E
eZ = (-vSx - vSy + Sz) / E
which is the strain matrix for triaxial loading of the wall element in basis <x,y,z>. You can easily see the loss in dimensions of the latter two basis as strain increases in the first basis. The role of the negative sign explicitly denotes LOSS, not gain in dimensional value(s).
Clearly, given e=strain, S=stress, E=Youngs Modulus and v=Poisson Ratio then: e=a dT L for a=coefficient of thermal expansion, dT=temperature difference and L=length. Stress and strain are intimently linked via Hookes Law, S=Ee for e=(L-L')/L, L'=final length.
Therefore a guy can clearly see the influence of material properties and mass conservation of the ring under thermal flux. The gain in one dimension obviously implies loss in measurement of the latter two dimensions.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
Material volume of the ring is conserved before and after thermal expansion. Cut the ring open and flatten out the circumference; apply tension. Does the openned ring not behave similar to a wire under tension? I can't see the wall thickness increasing!
Rather, you would see the wall experiencing "necking". As the circumference increases due to diametrical expansion the ring MUST decrease in both width and wall thickness. In other words:
eX = (Sx - vSy - vSz) / E
eY = (-vSx + Sy - vSz) / E
eZ = (-vSx - vSy + Sz) / E
which is the strain matrix for triaxial loading of the wall element in basis <x,y,z>. You can easily see the loss in dimensions of the latter two basis as strain increases in the first basis. The role of the negative sign explicitly denotes LOSS, not gain in dimensional value(s).
Clearly, given e=strain, S=stress, E=Youngs Modulus and v=Poisson Ratio then: e=a dT L for a=coefficient of thermal expansion, dT=temperature difference and L=length. Stress and strain are intimently linked via Hookes Law, S=Ee for e=(L-L')/L, L'=final length.
Therefore a guy can clearly see the influence of material properties and mass conservation of the ring under thermal flux. The gain in one dimension obviously implies loss in measurement of the latter two dimensions.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
packdad
Years ago we ran into a similar problem. We wanted swell rates for composite materials immersed in water for long periods. There were cylinders. Water temperature variation was another factor.
We were unable to measure any significant difference between cylinders of the material v solid bar of the same material.
Years ago we ran into a similar problem. We wanted swell rates for composite materials immersed in water for long periods. There were cylinders. Water temperature variation was another factor.
We were unable to measure any significant difference between cylinders of the material v solid bar of the same material.
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1
- #7
Cockroach
please reconsider your position with fresh mind: your assumption of no change in volume is the faulty one.
An unconstrained body subject to a uniform rise in temperature will see all of its dimensions increased by the factor 1+[α][Δ]T and its volume by the factor 1+3[α][Δ]T, [α] being the coefficient of linear thermal expansion.
prex
Online tools for structural design
please reconsider your position with fresh mind: your assumption of no change in volume is the faulty one.
An unconstrained body subject to a uniform rise in temperature will see all of its dimensions increased by the factor 1+[α][Δ]T and its volume by the factor 1+3[α][Δ]T, [α] being the coefficient of linear thermal expansion.
prex
Online tools for structural design
strokersix
Mechanical
I also believe corus is correct. This problem does not involve applied mechanical loading.
Thanks for the input, but no chance.
So we're talking a cube of wall material of volume V=x^3 and after application of heat, V'=[x(1+a dt)]^3? What's the catch? Are we creating mass to keep material density constant?
This does not correlate to my personal experiences! On the other hand, I guess miracles, and I am a practicing Catholic, are beyond the realm of science.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
So we're talking a cube of wall material of volume V=x^3 and after application of heat, V'=[x(1+a dt)]^3? What's the catch? Are we creating mass to keep material density constant?
This does not correlate to my personal experiences! On the other hand, I guess miracles, and I am a practicing Catholic, are beyond the realm of science.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
I guess Prex and CRG are saying the same thing. Since mass is conserved in the heating process, density decreases since volume increases, i.e. each dimension by "a dT L".
To me, and obviously it is just me, something here just doesn't seem right. If a mild steel circular ring of 3.0 ID is 0.5 inches wide and made of flat bar 0.125 inches thick, then heating it from room temperature 500C does the following:
1) the volume of the ring changes from V to V',
2) therefore the density is less, heated than cooled.
Can anyone suggest final dimensions for the circular ring? What is the hoop stress as the result of heating?
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
To me, and obviously it is just me, something here just doesn't seem right. If a mild steel circular ring of 3.0 ID is 0.5 inches wide and made of flat bar 0.125 inches thick, then heating it from room temperature 500C does the following:
1) the volume of the ring changes from V to V',
2) therefore the density is less, heated than cooled.
Can anyone suggest final dimensions for the circular ring? What is the hoop stress as the result of heating?
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada
The hoop stress (as any other stress) is zero.
Final dimensions are roughly as follows:
-ID increases by about half a mm
-width increases by less than one tenth of a mm
-thickness increases by about 2/100 of a mm
Of course we are also assuming that no significant residual stress is present in the ring after forming (and possibly welding).
prex
Online tools for structural design
Final dimensions are roughly as follows:
-ID increases by about half a mm
-width increases by less than one tenth of a mm
-thickness increases by about 2/100 of a mm
Of course we are also assuming that no significant residual stress is present in the ring after forming (and possibly welding).
prex
Online tools for structural design
Consider that a solid round bar is a cylinder with zero ID.
It is reasonable to expect its radius will increase/decrease proportional to temperature.
Now, visualise that round bar as a cylinder; the OD is the same, but now with ID > zero.
It is again reasonable to expect no change, meaning that the remaining radius, now the wall thickness, will expand proportionally exactly as before, and that is consistent with what we measured.
It is reasonable to expect its radius will increase/decrease proportional to temperature.
Now, visualise that round bar as a cylinder; the OD is the same, but now with ID > zero.
It is again reasonable to expect no change, meaning that the remaining radius, now the wall thickness, will expand proportionally exactly as before, and that is consistent with what we measured.
I believe that Cockroach is thinking of mechanical loads as opposed to thermal loads where objects expand/change volume etc. but there is no stress (unless restrained from expanding). The expressions for strain above exclude the thermal strain alpha x T, acting in all directions. Rewriting the expressions above would show that the stress is zero everywhere when you heat something up.
corus
corus
diamondjim
Mechanical
If the id and od expand at the same rate,
it does seem to follow that the wall thickness
would increase. Do you use the centroid to
figure the expansion? Also do you assume the
center is at the same temperature and the material
has reached a steady state at that temperature?
it does seem to follow that the wall thickness
would increase. Do you use the centroid to
figure the expansion? Also do you assume the
center is at the same temperature and the material
has reached a steady state at that temperature?
All dimensions increase so there's no need to take the centroid to calculate the expansion. If you don't have uniform temperatures then you'll have differential thermal expansion within the structure and hence stress, regardless of whether or not it's steady state.
corus
corus
EnglishMuffin
Mechanical
This question has been answered before. See thread404-61378
Equal degree of confusion on both threads!
Equal degree of confusion on both threads!
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2
- #19
To the credit of an old physics professor I had in college:
Think of unconstrained thermal expansion as photographic enlargement. Follows the same principle. The scale increase of the object is (1+alpha*dT). So the new dimension is [original dimension*(1+alpha*dT)].
Think of unconstrained thermal expansion as photographic enlargement. Follows the same principle. The scale increase of the object is (1+alpha*dT). So the new dimension is [original dimension*(1+alpha*dT)].
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