Calculations: Creep of Metals and the Life of a Rod

In summary: Q/kT## as a function of ln(Sigma):From the second document:## y = ax+b \rightarrow y = 15.08x + \frac {Q} {k} ####y = 20.01x + \frac {Q} {k}##From where we get Q/k = 0.7273 and 0.7.Thus, 0.7273 = 16.559 + 0.0604/T -> T = 810.59.For the other equation we get 0.7 = 16.559 + 0.0604/T -> T = 819.44.Therefore, the estimated life of the rod
  • #1
Nadia
4
2
TL;DR Summary
Exercise about estimating the life of a rod given a graph and an equation to work with. (Ex. 3.6 from Inelastic Deformation of Metals: Models, Mechanical Properties, and Metallurgy from D. C. Stouffer and L. Thomas Dame)
I'm sorry I'm uploading lots of images because I don't know how to write equations here.

PROBLEM DESCRIPTION

I have to solve this problem:

1673261486756.png

Figure P3.6:
1673261515670.png

MY SOLUTION

I did this:

1673261607364.png


THE CORRECTION

I got this as a correction:

- Don't use the middle line in figure P3.6 in my calculations.
- Start finding Q/k but for the top and bottom line, if they are not identical, deal with it.
- You will now have the temperature dependance.
- Plot ln(eps-dot)+Q/kT as a function of ln(Sigma)
- You get two points from which you can calculate n and B

SOLUTION AFTER THE CORRECTION

1673261884009.png


I've now done this part but I think it's wrong and I also don't know how to keep going with the plot.
 
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  • #3
FEAnalyst said:
It's hard to read those hand notes. It would certainly help if you rewrote the equations using LaTeX: https://www.physicsforums.com/help/latexhelp/
Okay! Thanks, I didn't see this. I'm gonna give it a try
 
  • #5
In this thread, the problem is presented using images:

https://www.physicsforums.com/threads/creep-of-metals.1048878/

PROBLEM DESCRIPTION

I have to solve this problem:

3.6 A steel rod supporting a stress of 8000 psi at 1000ºF is not to exceed 5% creep strain. Knowing that the steady-state creep rate can be expressed by an equation of the form
$$ε = B|∂|^n exp (-Q/kT)$$,

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. (ºR = ºF - 460).

Figure P3.6:
1673261515670-png.png


MY SOLUTION
##y = ax+b##, ##a=\frac {\Delta y} {\Delta x} = \frac {y_2-y_1} {ln(\frac {x_2} {x_1})}##
$$ε = B|∂|^n exp (-Q/kT),$$

##1/T=m ln ε + n##

##lnε=lnB+nln|∂|-Q/(kT) \rightarrow lnε -b = \frac {-Q} {k} · \frac {1} {T}##
## \rightarrow \frac {-k} {Q} · \{ lnε - lnb \} = \frac {1} {T}##
## \rightarrow \frac {k} {Q} lnb - \frac {k} {Q} lnε = \frac {1} {T}##
## \rightarrow a = \frac {k} {Q} = \frac {\frac {1} {R_2} - \frac {1} {R_1}} {ln (\frac {ε_2} {ε_1}) }##

Using points from the graph:

##a = \frac {k} {Q} = \frac {\frac {1} {6.5*10^{-4}} - \frac {1} {6.85*10^{-4}}} {ln (\frac {0.006} {0.0008}) }=3.901*10^{-7}##

##\rightarrow ln0.0008 · lnB+n ln 8000 - \frac {1} {3.901*10^{-7}} · \frac {1} {811} ##

##\rightarrow ln0.0008 + \frac {1} {3.901*10^{-7}} · \frac {1} {811} -n ln8000 = lnB##

##ln0.0003=lnB + n ln5000 - \frac {1} {3.901*10^{-7}} · \frac {1} {T_2}##,

where ##\frac {1} {R} = 6.5*10^{-4} \rightarrow F = 1078,46 \rightarrow T_2 = 855K##

##ln0.0003=lnB + n ln5000 - \frac {1} {3.901*10^{-7}} · \frac {1} {855}##

##ln0.0003+\frac {1} {3.901*10^{-7}} · \frac {1} {855}-n ln5000=lnB##

##ln0.0008+\frac {1} {3.901*10^{-7} · 811} -n ln8000=ln0.0003+\frac {1} {3.901*10^{-7}} · \frac {1} {855}-n ln5000##

##3153.71-n ln8000 = 2990.07-n ln5000##

##163.64 = n ln8000 - n ln5000= n (ln8000-ln5000) = n 0.47##

##n = \frac {163.64} {0.47} = 348.16##

##-> ln0.0003 + \frac {1} {3.901*10^{-7} · 855} - 348.16 · ln5000 = lnB \rightarrow 24,6598 = lnB##

##e^{24.6598} = e^{lnB} = B \rightarrow B = 5.124 * 10^{10}##

From the relationship between strain rate and time:
##ln t = lnε + \frac {Q} {kT} - n ln∂ - lnB = 0.064##

##\rightarrow t=e^{0.064} = 1.067##

THE CORRECTION

I got this as a correction:

- Don't use the middle line in figure P3.6 in my calculations.
- Start finding Q/k but for the top and bottom line, if they are not identical, deal with it.
- You will now have the temperature dependance.
- Plot ln(eps-dot)+Q/kT as a function of ln(Sigma)
- You get two points from which you can calculate n and B

SOLUTION AFTER THE CORRECTION
From the first document:

## a_T = \frac {k} {Q} = \frac {\frac {1} {R_2} - \frac {1} {R_1}} {ln (\frac {ε_2} {ε_1})} = \frac {\frac {1} {7.2*10^{-4}} - \frac {1} {8*10^{-4}}} {ln (\frac {0.01} {1*10^{-6}}) } = 15.08 ##

##a_B = \frac {k} {Q} = \frac {\frac {1} {R_2} - \frac {1} {R_1}} {ln (\frac {ε_2} {ε_1})} = \frac {\frac {1} {6.2*10^{-4}} - \frac {1} {7*10^{-4}}} {ln (\frac {0.01} {1*10^{-6}}) } = 20.01##

They are not identical, so we interpolate:

##\frac {15000-5000} {8000-5000} = \frac {20.01-15.08} {x-15.08}##

From where ##x = 16.559##
##\frac {k} {Q} = 16.559## temperature dependance -> ##ln(ε) + \frac {Q} {kT} -> ln(ε) + 0.0604/T##
 
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FAQ: Calculations: Creep of Metals and the Life of a Rod

What is creep in metals?

Creep in metals is the tendency of a solid material to slowly move or deform permanently under the influence of mechanical stresses. It occurs over long periods, especially at high temperatures, and is a critical factor in the life expectancy of metal components under stress.

How is the life of a rod determined under creep conditions?

The life of a rod under creep conditions is determined by calculating the time it takes for the rod to reach a specific amount of deformation or to fail under a constant load at a given temperature. This involves using creep rate equations and material constants derived from empirical data.

What factors influence the creep rate of metals?

The creep rate of metals is influenced by several factors, including the applied stress, temperature, material properties, grain size, and the presence of any impurities or alloying elements. Higher temperatures and stresses generally increase the creep rate.

What are the common methods for calculating creep deformation?

Common methods for calculating creep deformation include empirical models like the Norton-Bailey law, which relates the creep rate to stress and temperature, and more complex constitutive models that account for different stages of creep (primary, secondary, and tertiary). These calculations often require experimental data to determine material-specific constants.

How can the creep life of a rod be extended?

The creep life of a rod can be extended by reducing the operating temperature and stress, using materials with higher creep resistance, implementing better cooling systems, and designing components with thicker cross-sections to distribute stress more evenly. Regular inspections and maintenance can also help detect early signs of creep deformation.

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