Tungsten is being used at half its melting point (Tm≈3,400◦C) and astress level of 160 MPa. An engineer suggests increasing the

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Tungsten is being used at half its melting point (Tm≈3,400◦C) and astress level of 160 MPa. An engineer suggests increasing the

grain size by afactor of 4 as an effective means of reducing the creep rate.(a)Do you agree with the engineer? Why? What if the stress level were equalto 1.6 MPa?(b)What is the predicted increase in length of the specimen after 10,000hours if the initial length is 10 cm?

Engineering

1 answer:

Paladinen [302] · Answer 1 · 2022-02-28T16:30:57+03:00

Answer:

Explanation:

The missing diagram is attached in the image below which shows the deformation map of the Tungsten.

Given that:

Stress level $\sigma = 160 MPa$

T = 0.5 Tm

$\implies \dfrac{T}{Tm} = 0.5$

G = 160 GPa

$\implies \dfrac{\sigma}{G} = 10^{-3}$

a)

The regulating creep mechanism is dislocation driven, as we can see from the deformation mechanism.

The engineer's recommendation would not be approved because increasing grain size results in a decrease in the grain-boundary count, preferring dislocation motion. The existence of grain borders is a hindrance to dislocation motion, as the dislocation principle explicitly states. To stop the motion, we'll need a substance with finer grains, which would result in more grain borders, or a material with higher pressure. In the case of Nabarro creep, which is diffusion-driven, an engineer's recommendation would be useful.

b)

If stress level reduced to $\sigma = 1.6 MPa$

$\implies \dfrac{\sigma }{G} = 10^{-5}$

Cable creep is now the controlling creep mode, which entails tension-driven atom diffusion along grain borders to elongate grain along the stress axis, a process known as grain-boundary diffusion. Cable creep is more common in fine-grained materials. As a result, the engineer's advice would succeed in this case. The affinity for cable creep is reduced when the grain size is increased.

c)

From the map of creep mechanism for $\dfrac{\sigma}{G} = 10^{-3} \ and \ \dfrac{T}{Tm} = 0.5$