4. The state of stress at a point is = 140 MPa, = -220 MPa, = 160 MPa, τxy = -80 MPa, τyz =

Step 1: The principal stresses are the eigenvalues of the stress tensor. The stress tensor is gi

4. The state of stress at a point is = 140 MPa, = -220 ...

4. The state of stress at a point is $\sigma_x = 140 \text{ MPa}$, $\sigma_y = -220 \text{ MPa}$, $\sigma_z = 160 \text{ MPa}$, $\tau_{xy} = -80 \text{ MPa}$, $\tau_{yz} = 100 \text{ MPa}$, $\tau_{xz} = -120 \text{ MPa}$. [Answers: 301.72 MPa, 29.83 MPa, -251.55 MPa, max shear = 276.6 MPa] a. Calculate the principal stresses. b. Draw the 3D Mohr's circle for the state of stress. c. Calculate the maximum shear stress.

Transcript

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...

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