Ratio of applied stress, Mechanical Engineering

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Ratio of applied stress:

What should be the ratio of applied stress, s, to yield stress sY in a large thin plate so that plastic spread along lint of crack is equal to  1 /10 crack length?

Solution

For large plate the crack length is 2a,

1542_Ratio of applied stress.png

Along crack length, q = 0

2146_Ratio of applied stress1.png

sxx and sxy vanish along crack in x-direction. For plastic deformation syy = sY = yield stress.

The plastic spread is to be upto a distance of r = 2a/10  = a/5

1185_Ratio of applied stress2.png

i.e. a stress which is only 63.25% of yield stress will be able to cause a plastic deformation equal to  1 / 10 of crack length.


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