For an 18-8 Mo stainless steel (Animated Figure 8.36), predict the time to rupture for a component that is subjected to a stress of 80 MPa(11.600 psi) at 700°C (973 K). hours eTextbook and Media Attempts: 0 of 3 used Submit Answer Save for Later

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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For an 18-8 Mo stainless steel (Animated Figure 8.36), predict the time to rupture for a component that is subjected to a stress of 80
MPa(11,600 psi) at 700°C (973 K).
hours
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Transcribed Image Text:For an 18-8 Mo stainless steel (Animated Figure 8.36), predict the time to rupture for a component that is subjected to a stress of 80 MPa(11,600 psi) at 700°C (973 K). hours eTextbook and Media Save for Later Attempts: 0 of 3 used Submit Anwer
Stress (MPa)
Stress (psi)
T(20 + log t,)(R-h)
Larson-Miller parameter =
42.42 10 (R-h)
25
30
35
x103 x103
40
45
50
x103
x103 x103 x103
Larson-Miller parameter =
100,000 23.24 - 10 (K-h)
Stress = 80 MPA
Stress = 11.6 10 psi
%3D
100
10,000
10
1000
24
x103
20
12
x103
16
x103
28
x103
x103
T(20 + log t,)(K-h)
Transcribed Image Text:Stress (MPa) Stress (psi) T(20 + log t,)(R-h) Larson-Miller parameter = 42.42 10 (R-h) 25 30 35 x103 x103 40 45 50 x103 x103 x103 x103 Larson-Miller parameter = 100,000 23.24 - 10 (K-h) Stress = 80 MPA Stress = 11.6 10 psi %3D 100 10,000 10 1000 24 x103 20 12 x103 16 x103 28 x103 x103 T(20 + log t,)(K-h)
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