Calculate stress to cause failure - Creep Parameters:
From creep tests on 713 C alloy the constant C in Larson-Miller parameter is determined as - 85.75. In rupture test a specimen of this material fails after 500 hours at 34 MPa and 1373 K while another specimen at a stress level 136 MPa and at temperature of 1308 K fails after 3 hours. Calculate stress to cause failure after 30,000 hours at 1173 K.
Solution
From Eq. (23) with t_{1} = 500 hours, T_{1} = 1373 K at σ_{1} = 34 MPa
m_{1} = T_{1} (ln t_{1} - C ) = 1373 (ln 500 + 85.75)
= 1373 (6.2 + 85.75) = 12.62 × 10^{4}
From same equation with t_{2} = 3 hours, T_{2} = 1308 K at σ_{2} = 136 MPa
m_{2} = 1308 (ln 3 + 85.75) = 1308 (1.1 + 85.75)
= 11.36 × 10^{4}
Using values of σ_{1}, m_{1} and σ_{2}, m_{2} in Eq. (25)
ln 34 = k_{1} × 12.62 × 10^{4} + c_{1} = 3.526
ln 136 = k_{1} × 11.36 × 10^{4} + c_{1} = 4.913
∴ k_{1} = 3.526 - 4.913/10^{4} (12.62 - 11.36)
= (1.387 /1.26)× 10^{- 4 }= - 1.1 × 10^{- 4}
∴ c_{1} = 4.913 + 1.1 × 11.36 = + 17.409
∴ ln σ= - 1.1 × 10^{- 4} m + 17.409 -------------(i)
m_{3} = 1173 (ln 30, 000 + 85.75)
= 1173 (10.31 + 85.75)
= 11.27 × 10^{4}
Use the value of m_{3} in (i)
ln σ_{3} = - 1.1 × 10 ^{- 4} × 11.27 × 10^{4} + 17.409
= - 12.397 + 17.409 = 5.019
∴ σ_{3} = 150.2 MPa
which means that a stress of 150.2 MP_{a} will cause specimen to rupture at 1173 K after 30,000 hours.
Note that performing test for 30,000 hours is highly time consuming and this result has been obtained by performing two tests which required 500 + 3 = 503 hours.