Determine the direction of slip line at the workpiece tool interface in the condition of coulomb friction when the value of COF is 0.2 and the normal stress is 100 N/mm^2?. The value of the shear yield strength is 84 N/mm^2?. Also explain the direction of slip line theory with coulomb friction

Determine the direction of slip line at the workpiece tool interface in the condition of coulomb friction when the value of COF is 0.2 and the normal stress is 100 N/mm^2?. The value of the shear yield strength is 84 N/mm^2?. Also explain the direction of slip line theory with coulomb friction

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Determine the direction of slip line at the workpiece tool interface in the condition of coulomb
friction when the value of COF is 0.2 and the normal stress is 100 N/mm^2?. The value of the
shear yield strength is 84 N/mm^2?. Also explain the direction of slip line theory with coulomb
friction.

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I am only interested in part 6-27(a). I see that Bartleby has answered this question before but I have a question regarding the yielding factor of safety. Why is the yielding factor of safety calculated as ( n=Sy/sigma_max), but not as ( n=Sy/Kt*sigma_max )? Since there is a stress concentration factor and the sigma_max is technically considered the nominal stress. My understanding is that when a ductile material is subjected to a static load the effects of stress concentration near the discontinuity are reduced to local yielding and for this reason not considered in the yielding factor of safety? Because under the static load, when the stress near the discontinuity reaches the yield point, local plastic deformation takes place and increases the yield strength hence sustaining higher loads. Here the the ductile material is subject to a dynamic load, Would the yielding factor of safety be the same reasoning as for static load?
I am only interested in part 6-27(a). I see that Bartleby has answered this question before but I have a question regarding the yielding factor of safety. Why is the yielding factor of safety calculated as ( n=Sy/sigma_max), but not as ( n=Sy/Kt*sigma_max )? Since there is a stress concentration factor and the sigma_max is technically considered the nominal stress.
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