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Consider a metal single crystal oriented such that the normal to the slip plane and slip direction are at angles of 60° and 35°, respectively, with the tensile axis. If the critical resolved shear stress is 8 MPa, will the applied stress of 12 MPa cause the single crystal to yield? If not, what stress will be necessary?

Respuesta :

opudodennis

Answer:

19.5324 MPa

Explanation:

Information provided

Angle between the normal to the slip plane with tensile axis, [tex]\alpha=60^{o} [/tex]

Angle by slip direction with tensile axis, [tex]\beta=35^{o}[/tex]

Critical resolved shear stress, [tex]\tau_{c}=8 MPa[/tex]

Applied stress [tex]\sigma=12 MPa[/tex]

Shear stress at slip plane

[tex]\tau=\sigma cos\alpha cos\beta[/tex]

[tex]\tau=12cos60^{o}cos35^{o}=4.915 MPa[/tex]

[tex]\tau <\tau_{c}[/tex] hence crystal won’t yield

Applied stress, [tex]\sigma[/tex] for crystal to yield is given by

[tex]\sigma=\frac {\tau_{c}}{cos\alpha cos\beta}[/tex]

[tex]\sigma=\frac {8}{cos60cos35}=19.53239342 MPa[/tex]

[tex]\sigma=19.5324 MPa [/tex]

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