Mechanical Behaviour of Materials · Creep · 2 marks · Numerical answer
Steady creep rate rises by a factor 20 when temperature rises from 890 to 980 K. At what temperature is the rate five times its 890 K value? Give nearest integer K.
At fixed stress, use $\dot\epsilon=Ae^{-Q/(RT)}$. Taking rate ratios eliminates A. Thus $\ln5/\ln20=(1/890-1/T)/(1/890-1/980)$. Solving gives T ≈ 936 K. This assumes the same activation energy and creep mechanism over the temperature range.
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