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T(20 + log t)(R-h) 25 30 35 40 45 50 * x103 x103 x103 x103 x103 x103 Larson-Miller parameter = 12. 103 (R-h) 100,000 Larson-* Your answer is incorrect. For an 18-8 Mo stainless steel (Animated Figure 8.36), predict the time to rupture for a componen

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the stress as T(20+logt) graph. stress = 80 MPa Temperature T=973 K At sorpa T(20+ logtr) = 23.5 +10° 973 (20+logte) – 23.5 x

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