Calculate stress to cause failure - creep parameters, Mechanical Engineering

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 t1 = 500 hours, T1 = 1373 K at σ1 = 34 MPa

m1  = T1  (ln t1  - C ) = 1373 (ln 500 + 85.75)

= 1373 (6.2 + 85.75) = 12.62 × 104

From same equation with t2 = 3 hours, T2 = 1308 K at σ2 = 136 MPa

m2  = 1308 (ln 3 + 85.75) = 1308 (1.1 + 85.75)

    = 11.36 × 104

Using values of σ1, m1 and σ2, m2 in Eq. (25)

ln 34 = k1 × 12.62 × 104 + c1  = 3.526

ln 136 = k1 × 11.36 × 104 + c1  = 4.913

∴          k1 =      3.526 - 4.913/104 (12.62 - 11.36)

                          = (1.387 /1.26)× 10- 4 = - 1.1 × 10- 4

∴ c1  = 4.913 + 1.1 × 11.36 = + 17.409

∴ ln σ= - 1.1 × 10- 4 m + 17.409 -------------(i)

m3  = 1173 (ln 30, 000 + 85.75)

       = 1173 (10.31 + 85.75)

              = 11.27 × 104

Use the value of m3 in (i)

ln σ3  = - 1.1 × 10 - 4  × 11.27 × 104 + 17.409

= - 12.397 + 17.409 = 5.019

∴          σ3  = 150.2 MPa

which means that a stress of 150.2 MPa 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.

Posted Date: 1/23/2013 2:27:11 AM | Location : United States







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