Consider nickel with the same grain size and processing as the material of Fig. 15.16. This material is subjected to a tensile stress of 6 MPa at a temperature of 900°C. What is the approximate strain rate, and what creep mechanism is dominant?

Question

Accepted Answer

To enter the map, we need τ and G from Eqs. 15.12

[latex]\sigma=\sqrt{3} au, \quad \dot{\varepsilon}=\dot{\gamma} / \sqrt{3}[/latex] (15.12)

and 15.11,

[latex]G=G_{300}-h(T-300)[/latex] (15.11)

which give the normalized shear stress τ/G. We have

[latex] au=\frac{\sigma}{\sqrt{3}}=\frac{6 MPa }{\sqrt{3}}=3.464 [/latex] MPa

[latex]G=G_{300}-h(T-300)=78,900-29.3(1173-300)=53,320 [/latex] MPa

[latex] au / G=6.50 imes 10^{-5}[/latex]

where T = 900 + 273 = 1173 K is substituted for obtaining G.

For the temperature, either use the upper scale and enter the graph directly with T = 900°C, or enter the lower scale with [latex]T/T_{m} = 1173/1726 = 0.680[/latex]. To enter the τ/G scale, it may be useful to calculate log (τ/G) = −4.19, allowing linear scaling between [latex]τ/G = 10^{−4} ext{ and } 10^{−5}[/latex] (that is, between log (τ/G) = −4 and −5). Entering the map with these values and interpolating between curves gives a strain rate of approximately

[latex]\dot{\gamma} \approx 10^{-6.7}=2.0 imes 10^{-7} s ^{-1}, \quad \dot{\varepsilon}=\dot{\gamma} / \sqrt{3}=1.2 imes 10^{-7} s ^{-1}[/latex]

This point falls within the region of dominance of the high-temperature type of power law creep.

Comment The value calculated actually represents quite a high strain rate for any practical application. For example, in one day, the accumulated strain would be

[latex]\varepsilon=\dot{\varepsilon} t=\left(1.2 imes 10^{-7} s ^{-1} ight)(3600 s / h )(24 h )=0.010=1.0 \%[/latex]

Question 15.1: Consider nickel with the same grain size and processing as t...

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