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**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!

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