# Stress in a ring that is mounted on a solid shaft

Hello!

I want to calculate the stress in a ring that is mounted on a solid shaft due to the difference in temperature.

The temperature in the shaft is always higher than the temperature in the ring. The geometry is known, the shaft has a certain diameter and the outer diameter is also know.E-modulus and the thermal coefficient alpha is also known.

I just want to do a simple conservative estimate on the stress in the ring.

Thankful for any help with formulas etc.

Daniel

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Q_Goest
Homework Helper
Gold Member
To be conservative, assume the shaft is 'infinitely rigid' meaning the OD of the shaft doesn't change.

Calculate the OD of shaft and ID of ring at operating temperature.

Given the ID of the ring is smaller than the OD of the shaft, assume the ring stretches uniformly (ie: like a rubber band) on the shaft.

Use the basic equation:
S = E e
where
S = Stress
E = Modulus of elasticity for the ring
e = strain = (OD of the shaft - ID of ring) / (ID of ring)

How do I calculate the OD of the shaft and ID of the ring, does ID mean Inital Deflection? OD ?. Im not English or used to abbreviations.

Thanks again

Inner / Outer Diameter

What if the OD of the shaft is equal to the ID of the ring..?

If for example the temp. in the ring is 40 celcius and the shaft has a temp. of 60 celcius.
How do I use theese temperatures in my calculations?

FredGarvin
Technically, the point at which the two diameters are equal is not solveable. Assume that they will not be equal and make a judgement call on what the real diameters will be.

Coefficient of thermal expansion:
$$\frac{\Delta L}{L} = \alpha \Delta T$$

Thanks.

What if I make it a little bit more complicated than before.
Does the length of the shaft affect the stress in any matter?
Does poissons ratio of the shaft and ring affect the stress ?

Can I use superposition on the stresses?
S=E(ring)*e(ring)+E(shaft)*e(shaft) ? Or is that wrong.

The ring is of cemented carbide and the shaft is made of steel.

I have the material data and geometrical data
E(ring)=540 [GPa]
alpha(ring)=5.6 [1/Celcius]
poissons(ring) = 0.24
L=58 mm
ID=87 mm
OD=142 mm
E(shaft)= 205 [GPa]
alpha(shaft)= 11.5 [1/Celcius]
poissons(shaft)=0.30

Difference in temperature is 50 Celcius, the shaft is warmer than the ring.

I have attached a picture of my problem.

Im thankful for any help and the previous help.

#### Attachments

• temp.jpg
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nvn
Homework Helper
ladil123: I assume you meant to say alphas = 11.5e-6/C and alphar = 5.6e-6/C; please correct me if I am wrong, because I did not look up these values. I currently assume the shaft outside diameter (OD) is 87 mm; please correct me if I am misinterpreting.

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Thanks.

Yes, you are right regarding the alpha constant.

Thanks again

What if the same strain doesnt occur in the shaft.
Lets say that the strain in the shaft is twice as big as the strain in the ring, what would happen then ?

nvn
Homework Helper
ladil123: Here is how you compute the stress on an interference fit, press fit, or shrink fit for a shaft and collar.

(1) Description of parameters.
d1 = shaft inside diameter at reference temperature T1o (zero for solid shaft).
d2 = shaft outside diameter at reference temperature T1o.
d3 = collar inside diameter at reference temperature T2o.
d4 = collar outside diameter at reference temperature T2o.
d = resultant contact interface diameter.
E1 = shaft modulus of elasticity.
E2 = collar modulus of elasticity.
nu1 = shaft Poisson ratio.
nu2 = collar Poisson ratio.
alpha1 = shaft coefficient of thermal expansion (CTE).
alpha2 = collar CTE.
T1 = shaft temperature.
T2 = collar temperature.
T1o = reference temperature at which shaft dimensions d1 and d2 are measured.
T2o = reference temperature at which collar dimensions d3 and d4 are measured.
sd = surface roughness deformation = 1.2(Ra1 + Ra2).
Ra1 = shaft outside diameter surface roughness, Ra value.
Ra2 = collar inside diameter surface roughness, Ra value.
pc = contact pressure between shaft and collar.
sigma_t1 = hoop stress at shaft inside diameter.
sigma_t2 = hoop stress at shaft outside diameter.
sigma_t3 = hoop stress at collar inside diameter.

(2) Apply temperatures T1 and T2.
d1 = d1*[1 + alpha1*(T1 - T1o)]
d2 = d2*[1 + alpha1*(T1 - T1o)]
d3 = d3*[1 + alpha2*(T2 - T2o)]
d4 = d4*[1 + alpha2*(T2 - T2o)]

(3) Initially estimate d as follows.
d = 0.5(d2 + d3)

(4) Compute sd if surface roughnesses are known; or you can set sd = 0 if you want to neglect yielding of surface roughness asperities.
sd = 1.2(Ra1 + Ra2)

(5) Compute contact interface pressure.
pc = [(d2 - d3 - sd)/d]/{[(d^2 + d1^2)/(E1*(d^2 - d1^2))] + [(d4^2 + d^2)/(E2*(d4^2 - d^2))] - (nu1/E1) + (nu2/E2)}

(6) Refine d, and repeat step 5 one time. You can skip this step if the shaft or collar is not relatively flexible compared to the other member.
d = d2*[1 - (pc/E1)(1 - nu1)]

(7) Compute shaft and collar hoop stress.
sigma_t1 = pc*2(d^2)/(d1^2 - d^2)
sigma_t2 = pc*(d1^2 + d^2)/(d1^2 - d^2)
sigma_t3 = pc*(d4^2 + d^2)/(d4^2 - d^2)

Hoop stress at the collar outside diameter is not given, above, because it is not maximum. Radial stress at the shaft outside diameter and collar inside diameter is sigma_r2 = sigma_r3 = -pc. Radial stress at the shaft inside diameter (or center) and collar outside diameter is sigma_r1 = sigma_r4 = 0. Combine the radial and hoop stress using an appropriate failure theory. For ductile materials, use von Mises theory. The foregoing analysis is applicable only if the combined stress is below the material yield strength. Also, the foregoing analysis and formulas are applicable only if pc in item 5 is positive.

For example, for your given problem in post 7, if T1o = T2o = 20 C, T2 = 30 C, T1 = 80 C, and sd = 1.2(0.7 micrometer + 1.0 micrometer) = 2.04 micrometer, we have the following.

d1 = 0 mm, d2 = 87 mm, E1 = 205 GPa, nu1 = 0.30, alpha1 = 11.5e-6/C
d3 = 87 mm, d4 = 142 mm, E2 = 540 GPa, nu2 = 0.24, alpha2 = 5.6e-6/C
d1 = (0.000 m)[1 + (11.5e-6/C)(80 C - 20 C)] = 0 mm
d2 = (0.087 m)[1 + (11.5e-6/C)(80 C - 20 C)] = 87.060 030 mm
d3 = (0.087 m)[1 + (5.6e-6/C)(30 C - 20 C)] = 87.004 872 mm
d4 = (0.142 m)[1 + (5.6e-6/C)(30 C - 20 C)] = 142.007 952 mm
d = 0.5(d2 + d3) = 87.032 451 mm

(5) Compute contact interface pressure.
pc = [(0.087 060 030 - 0.087 004 872 - 2.04e-6)/0.087 032 451]/{[(0.087 032 451^2 + 0^2)/((205e9 Pa)*(0.087 032 451^2 - 0^2))] + [(0.142 007 952^2 + 0.087 032 451^2)/((540e9 Pa)*(0.142 007 952^2 - 0.087 032 451^2))] - (0.30/(205e9 Pa)) + (0.24/(540e9 Pa))} = 76.8773 MPa

(6) Refine d, and repeat step 5 one time. (This step is not expressly required in your particular problem, but I will perform it anyway.)
d = 0.087 060 030{1 - [(76 877 259 Pa)/(205e9 Pa)](1 - 0.30)} = 87.037 176 mm
pc = 76.8693 MPa

(7) Compute shaft and collar hoop stress.
sigma_t1 = (76 869 334 Pa)*2*0.087 037 176^2/(0^2 - 0.087 037 176^2) = -153.7 MPa
sigma_t2 = (76 869 334 Pa)(0^2 + 0.087 037 176^2)/(0^2 - 0.087 037 176^2) = -76.88 MPa
sigma_t3 = (76 869 334 Pa)(0.142 007 952^2 + 0.087 037 176^2)/(0.142 007 952^2 - 0.087 037 176^2) = 169.4 MPa

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FredGarvin
Nice post NVN. Thanks for taking the time to type all of that out.

yes, i´m really glad for the effort!
Thanks again

Does this come from thick walled tube theory ? What are the starting equations.? I tried to do it with thick walled tube theory, but I failed. Maybe because I did it wrong...

nvn
Homework Helper
Yes, it partially comes from thick-walled cylinder theory. In the following, I am using the same numbering notation as previously defined in post 11.

(1) You know delta = diametral deflection = d2 - d3 = delta3 - delta2.
(2) You know the resultant diameter is approximately d = 0.5(d2 + d3).
(3) You know hoop strain eps2 = delta2/d, and eps3 = delta3/d.
(4) But from Hooke's law for a two-dimensional stress state (hoop and radial stress), eps2 = (sigma_t2 - nu1*sigma_r2)/E1, and eps3 = (sigma_t3 - nu2*sigma_r3)/E2.
(5) From thick-walled cylinder theory, you know sigma_t2 and sigma_t3, given in item 7 of post 11. And you know sigma_r2 = sigma_r3 = -pc.
(6) Substitute equations 3 and 5, above, into eqs. 4, then solve for delta2 and delta3. Substitute this result into eq. 1, and solve for pc.

How do I get the radial stress? as a function of the distance from center
And the hoop stress at the rings outer radius?

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nvn