00:01
Let's discuss this question.
00:01
So here we need to draw the most circle of stress for the given state of stress at points and determine the following given.
00:09
So now for example, if this is the diagram, here we can say this will be the direction.
00:32
So here we have 50 mp a and this is f x that is equal to 125 mp a and f x and f y will be equal to my minus 75 mpa and the eq is equal to 50 mpa.
00:53
Now solving for a that is principal stresses rp slash rp square is equal to sigma x plus sigma epa sigma p slash sigma p square dash is equal to sigma square x plus sigma divided by 2 plus minus half square root of sigma x minus sigma x squared x square plus 4 t x y minus sigma x square xy so here we can say this will be equal to 125 minus 75 divided by 2 plus minus half under root of minus 75 minus 125 square plus 4 multiplied by 50 square.
02:03
So this is equal to 50 divided by 2 plus minus half square root of 200 square plus 4 multiplied by 2 500.
02:15
So here we can say this is equal to sigma p even that is equal to 136 .80 and sigma p 2 that is equal to minus 86 .80 and sigma p 2 that is equal to minus 86.
02:26
Point 80m p a now therefore here we can write down the values of both that is sigma p1 136 .80 mp a and sigma p2 is 86 minus 86m p a now further here we can say solving for b maximum in plain shear stress so, t max is equal to sigma p1 minus sigma p2 divided by 2 and this is equal to 136 .80 plus 86 .80 divided by 2.
03:22
So this will be equal to 118 .1 .11 .8.
03:34
So this will be.
03:38
Now we need to draw the max circles representation...