Calculating Steel Rod Life with 8000 psi at 1000 °F

[Trevorman]

1. Problem statement

A steel rod supporting a stress of 8000 psi at 1000 ◦ F is not to exceed 5 % creep strain. Knowing that the steady-state creep rate can be expressed by an equation of the form

$\dot{\epsilon}_{s}^{C}=B|\sigma |^{n} exp \left(\frac{-Q}{kT} \right)$

• ε^⋅, strain rate
• B, Material variable
• σ, stress
• n, material variable
• k, boltzmanns constant
• Q, creep activation energy
• T, Temperature

where Q is the creep activation energy, determine the constants from the data for the steel in Figure P3.6 and estimate the life of the rod. ( R = F +460.)

http://imageshack.com/a/img913/4879/oCkvQ3.png [Broken]

Homework Equations

Strain
$\ln \dot{\epsilon} = \ln \frac{\epsilon}{t}$
strain = strain rate / time

The Attempt at a Solution

Reciprocal temperature at 1000 °F
T1 = 1000 + 460 = 1460 R
gives

1/T1 = 6.8493∗10−4

Linearizing the power law creep equation

$\ln \dot{\epsilon}= \ln B + n\ln \sigma - \frac{Q}{kT}$

The slope in the figure is calculated with
$\frac{k}{Q} = \frac{ \log_{10}(\frac{\dot{\epsilon_2}}{\dot{\epsilon_1}}) }{R^{-1}_2 -R^{-1}_1}$=-5 * 10^4

LineExpressed as (x1,y1),(x2,y2),(x3,y3)
I have extended the line to the y-axis to get a third value

L15000 = (10^-8 , 8.4), (10^-7 , 8.2), (0.1, 0.7)
L10000 = (10^-8 , 7.8), (10^-7 7.6), (0.1, 6.4)
L5000 = (10^-8 , 7.4), (10^-7, 7.2), (0.1, 0.6)

Now i have 2 unknown (B and n)

I use the linearized equation and trying to get the values by regression

$\ln \dot{\epsilon}+\frac{Q}{kT} = \ln B + n \ln \sigma$

I get
ln(B) = 126.7
n = -15.2

The time is calculated with (from the linearized equation and the relationship with strain rate and time)
$\ln t= \ln \epsilon + \frac{Q}{kT}-n*\ln \sigma - \ln B$

The result I get is
ln(t) = 7.34


This is however a invalid answer and I think the value of $\frac{Q}{kT}$ should be between 50 and 100. I have tried several times to correct this assignment but never been able to solve it.

Thank you very much!

 

[Chestermiller]

1. Problem statement

A steel rod supporting a stress of 8000 psi at 1000 ◦ F is not to exceed 5 % creep strain. Knowing that the steady-state creep rate can be expressed by an equation of the form

$\dot{\epsilon}_{s}^{C}=B|\sigma |^{n} exp \left(\frac{-Q}{kT} \right)$

• ε^⋅, strain rate
• B, Material variable
• σ, stress
• n, material variable
• k, boltzmanns constant
• Q, creep activation energy
• T, Temperature

where Q is the creep activation energy, determine the constants from the data for the steel in Figure P3.6 and estimate the life of the rod. ( R = F +460.)

http://imageshack.com/a/img913/4879/oCkvQ3.png [Broken]

Homework Equations

Strain
$\ln \dot{\epsilon} = \ln \frac{\epsilon}{t}$
strain = strain rate / time

The Attempt at a Solution

Reciprocal temperature at 1000 °F
T1 = 1000 + 460 = 1460 R
gives

1/T1 = 6.8493∗10−4

Linearizing the power law creep equation

$\ln \dot{\epsilon}= \ln B + n\ln \sigma - \frac{Q}{kT}$

The slope in the figure is calculated with
$\frac{k}{Q} = \frac{ \log_{10}(\frac{\dot{\epsilon_2}}{\dot{\epsilon_1}}) }{R^{-1}_2 -R^{-1}_1}$=-5 * 10^4

Why did you use the log to the base 10, if the equation calls for the natural log? Also, the left hand side should be Q/k.

Please show me the numbers you plugged into this equation, this time using the natural log.

Chet

 

[Trevorman]

Why did you use the log to the base 10, if the equation calls for the natural log? Also, the left hand side should be Q/k.

Please show me the numbers you plugged into this equation, this time using the natural log.

Chet

Hi, the picture I attached is where i get the equation k/Q from. Q/k is the slope of the line, and since the graph is a semilog graph i use log10(eps2) - log10(eps1) = log10(eps2/eps1). The Y-values is the $R_1^-1$ and $R_2^-1$.

The numbers are:
$\epsilon_2$ = 10^-7
$\epsilon_1$ = 10^-8
$R_1^{-1}$ = 8.4*10^-4
$R_2^{-1}$. 8.2*10^-4
(the scale seems to be 10^-4 in the figure)

The slope (dY/dX) in the figure
$\frac{Q}{k} = \frac{8.4*10^{-4} -8.2*10^{-4}}{\ln(\frac{10^{-8}}{10^{-7}}) }= -8.6858*10^{-6}$

 

[Chestermiller]

Hi, the picture I attached is where i get the equation k/Q from. Q/k is the slope of the line, and since the graph is a semilog graph i use log10(eps2) - log10(eps1) = log10(eps2/eps1). The Y-values is the $R_1^-1$ and $R_2^-1$.

The numbers are:
$\epsilon_2$ = 10^-7
$\epsilon_1$ = 10^-8
$R_1^{-1}$ = 8.4*10^-4
$R_2^{-1}$. 8.2*10^-4
(the scale seems to be 10^-4 in the figure)

The slope (dY/dX) in the figure
$\frac{Q}{k} = \frac{8.4*10^{-4} -8.2*10^{-4}}{\ln(\frac{10^{-8}}{10^{-7}}) }= -8.6858*10^{-6}$

This is not correct. The equation calls for the use of natural logs. I don't care that the graph is shown in common logs, since this has nothing to do with the calculation.

Here's my calculation for the 5000 psi line:

$ε = 10^{-1}$ at (1/T) = $6.0\times 10^{-4}$

$ε = 10^{-7}$ at (1/T) = $7.2\times 10^{-4}$
$$-\frac{Q}{k}=\frac{Δlnε}{Δ(1/T)}=\frac{\ln{(10^{-6})}}{1.2\times 10^{-4}}$$
This gives:

$\frac{Q}{k}=+95900$ degrees R

Chet

 

[Trevorman]

This is not correct. The equation calls for the use of natural logs. I don't care that the graph is shown in common logs, since this has nothing to do with the calculation.

Here's my calculation for the 5000 psi line:

$ε = 10^{-1}$ at (1/T) = $6.0\times 10^{-4}$

$ε = 10^{-7}$ at (1/T) = $7.2\times 10^{-4}$
$$-\frac{Q}{k}=\frac{Δlnε}{Δ(1/T)}=\frac{\ln{(10^{-6})}}{1.2\times 10^{-4}}$$
This gives:

$\frac{Q}{k}=+95900$ degrees R

Chet

Thank you very much, the equation is correct but I think you calculated with $\ln{10^{-5}} instead of$ $\ln{10^{-6}}$
and

$\frac{Q}{k} =+115129.25$


That seems reasonable, however when I do the regression the n-value is the same for all but can I assume that the lnB value have a linear relationship with the stress?

http://imageshack.com/a/img538/9776/DB2Hir.jpg [Broken]

As you see there will be three lnB values for each stress case. Should i interpolate the lnB for 8000psi?

 

[Chestermiller]

Thank you very much, the equation is correct but I think you calculated with $\ln{10^{-5}} instead of$ $\ln{10^{-6}}$
and

$\frac{Q}{k} =+115129.25$

Yes. You're right. Thanks for catching that error.

That seems reasonable, however when I do the regression the n-value is the same for all but can I assume that the lnB value have a linear relationship with the stress?

As you see there will be three lnB values for each stress case. Should i interpolate the lnB for 8000psi?

This can't be. There is only one value of B. The people who dreamed up this problem chose values for B, n, and Q/k in advance, and then plotted up the graph. This is not experimental data.

There are two ways of getting n, independently of B.

1. Hold (1/T) constant and evaluate n = (Δlnε/Δlnσ)

2. Hold ε constant, and evaluate $n = \frac{Q}{k}\frac{Δ(1/T)}{Δ\ln σ}$

Both methods should give the same answer. Try it both ways using the 5000 psi and the 15000 psi lines, and confirm that you get the same answer either way.

Then, to get lnB, choose one point at the middle of the 10000 psi line, and plug in everything else.

Chet

 

[Trevorman]

Yes. You're right. Thanks for catching that error.

This can't be. There is only one value of B. The people who dreamed up this problem chose values for B, n, and Q/k in advance, and then plotted up the graph. This is not experimental data.

There are two ways of getting n, independently of B.

1. Hold (1/T) constant and evaluate n = (Δlnε/Δlnσ)

2. Hold ε constant, and evaluate $n = \frac{Q}{k}\frac{Δ(1/T)}{Δ\ln σ}$

Both methods should give the same answer. Try it both ways using the 5000 psi and the 15000 psi lines, and confirm that you get the same answer either way.

Then, to get lnB, choose one point at the middle of the 10000 psi line, and plug in everything else.

Chet

Hi thank you so much for helping me.

Using the formulas you specified gives those results


This is one of the equations you specified, but i get the same result with the other one. Do you have any idea of how I will proceed with this?

$n = \frac{Q}{k}\frac{(8.2-7.6)×10^{-4}}{\log{\frac{15000}{10000}}} = 17.04$
$n = \frac{Q}{k}\frac{(8.2-7.2)×10^{-4}}{\log{\frac{15000}{5000}}} = 10.47$

Thank you!

 

[Chestermiller]

Hi thank you so much for helping me.

Using the formulas you specified gives those results


This is one of the equations you specified, but i get the same result with the other one. Do you have any idea of how I will proceed with this?

$n = \frac{Q}{k}\frac{(8.2-7.6)×10^{-4}}{\log{\frac{15000}{10000}}} = 17.04$
$n = \frac{Q}{k}\frac{(8.2-7.2)×10^{-4}}{\log{\frac{15000}{5000}}} = 10.47$

Thank you!

I can see that there's a problem. The stress increases by a factor of 2 between 5000 and 10000, and by a factor of 1.5 between 10000 and 15000. Yet the 10000 line is closer to the 5000 line than to the 15000 line. If your equation were an exact match to the data, the 10000 line would be closer to the 15000 line. So the equation is not going to be a perfect fit, and you need to reach a judgement call on what to do. I would use the value of n determined over the full range from 5000 to 15000. This would be n = 10.5. This won't result in a perfect fit at a stress of 10000, but, oh well, it's about the best your going to be able to do with the equation you are using. I would then continue by neglecting the 10000 line all together. Your final fit will be a perfect fit to the 5000 line and the 15000 line, and only an approximate fit at 10000.

So, using n= 10.5, pick any convenient point on either the 5000 line or the 15000 line, and solve for ln B. Then calculate ε at the required temperature and stress. See where this comes out on the graph, and see if its location makes sense.

Chet

 

[Trevorman]

I can see that there's a problem. The stress increases by a factor of 2 between 5000 and 10000, and by a factor of 1.5 between 10000 and 15000. Yet the 10000 line is closer to the 5000 line than to the 15000 line. If your equation were an exact match to the data, the 10000 line would be closer to the 15000 line. So the equation is not going to be a perfect fit, and you need to reach a judgement call on what to do. I would use the value of n determined over the full range from 5000 to 15000. This would be n = 10.5. This won't result in a perfect fit at a stress of 10000, but, oh well, it's about the best your going to be able to do with the equation you are using. I would then continue by neglecting the 10000 line all together. Your final fit will be a perfect fit to the 5000 line and the 15000 line, and only an approximate fit at 10000.

So, using n= 10.5, pick any convenient point on either the 5000 line or the 15000 line, and solve for ln B. Then calculate ε at the required temperature and stress. See where this comes out on the graph, and see if its location makes sense.

Chet

Thank you Chet I made the approximation and it's good enough I think.

You saved my last hair that I have left on my head!

 

[Chestermiller]

Thank you Chet I made the approximation and it's good enough I think.

You saved my last hair that I have left on my head!

LOL. My pleasure.

Chet

 

[manjit]

Can you post full solution for this problem, I am getting very hard time