Steadystate creep data taken for an iron at a stress of 132

Steady-state creep data taken for an iron at a stress of 132 MPa are given here; If it is know that the volume of the stress exponent for this alloy is 8.6, compute the steady-state creep rate at 1380 k and stress level of 83 MPa

Solution

The equation for steady state creep= e=A *(Stress^n)* exp(-(Q/RT)) ......equation 1

A= Constant

n= Stress exponent= 8.6

Q= Activation energy

T= temperatture

Taking ln on both sides of above equation

ln e=ln A+ n ln (stress) - (Q/RT)

Putting the given data

ln( 6.6* 10^(-4))= ln A+ n ln ( stress) - (Q/(R*1080)) ......equation 2

ln(8.6*10^(-2)= ln A+ n ln ( stress) - (Q/(R*1210)) ......equation 3

As n and stress are constanst for above two equations, subtracting equation 3 from 2 we get

Q/R= - 48954.56

Putting value of Q/R in equation 2 , we get the value of A

A=7.889* (10^(-42))

Now for the solution using equation no1

e=A *(Stress^n)* exp(-(Q/RT))

Data given

stress=83 Mpa

T= 1300 K

putting the values in equation 1 ,

we get e= 5.794*(10^(-9)) .......Answer