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Angle outside triangle

  1. Jul 23, 2016 #1
    1. The problem statement, all variables and given/known data
    this is actually mohr's circle formula, forget about the theory,let's focus on the mathematics part. I couldnt understand why the tan( 2 θs1) = -(σx -σy) / 2τxy ?
    2θs1 is outside the triangle
    2. Relevant equations


    3. The attempt at a solution
    theorically, tan(180-α )= -tan( α ) , so tan( 2 θs1) = -(σx -σy) / 2τxy , the vertical staright line of triangle should represent -(σx -σy) , whereas horizontal axis represent 2τxy , am i right?
     

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  3. Jul 23, 2016 #2

    Simon Bridge

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    Lessee ... looking at the triangle on the right, basic trig says that ##\tan(180-2\theta_{S1}) = O/A## where ##O=\tau_{xy}## and ##A=\frac{1}{2}(\sigma_x-\sigma_y)## ...
     
  4. Jul 23, 2016 #3
    which triangle do you mean? , where is it?
    tan [ theta(S1) ] is (σx- σy) / 2 τxy , where ##A=\tau_{xy}## and ##O=\frac{1}{2}(\sigma_x-\sigma_y)##
     
  5. Jul 23, 2016 #4

    haruspex

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    Simon means the grey triangle on the right, and I agree with his equation.
    The tangent will be negative, and it will be vertical displacement divided by horizontal displacement, so ##\frac{2\tau}{\Delta \sigma}##.
    You can check this by thinking about the coordinates of the top corner of the left-hand grey triangle in polar: ##y=r\sin(\theta)##, ##x=r\cos(\theta)##.
     
  6. Jul 23, 2016 #5
    so , the diagram is wrong ?
    because it's tan [ theta(S1) ] is (σx- σy) / 2 τxy , where ##A=\tau_{xy}## and ##O=\frac{1}{2}(\sigma_x-\sigma_y)##
    so , the triangle should be drawn like this rather the grey triangles given by the author , am i right?
     

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  7. Jul 23, 2016 #6

    haruspex

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    Merely considering a different triangle is not going to change the relationship.
    To get the tangent to be sigma/tau you need theta to be measured from a tau axis, not from a sigma axis.
     
  8. Jul 23, 2016 #7
    i have redrawn the triangle in my previous post using red colour, and the angle is measured from the tau axis, so is my idea correct? the triangles shown by author are wrong?
     
  9. Jul 23, 2016 #8

    haruspex

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    The image you posted with the red triangles still shows the angles as measured from the sigma axis.
    Did you intend that the angles are now the acute angles within the red triangles?
     
  10. Jul 23, 2016 #9
    for the triangles that i drew earlier, it's measured from tau (vertical ) axis , right ? why you said it is measured from sigma axis ?
     
  11. Jul 23, 2016 #10

    haruspex

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    Your diagram still shows the angles as defined by two black curved arrows starting at the positive sigma axis. I see no other indication in the diagram of how the angles are defined.
     
  12. Jul 24, 2016 #11
    Why? The angle that I draw start from positive y(tau) axis, right?
     
  13. Jul 24, 2016 #12

    haruspex

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    The only difference I can see between your original diagram and the one in post #5 is that two triangles have been coloured in red. Both diagrams show the theta angles by means of black curved arrows starting at the positive sigma (x) axis.
     
  14. Jul 24, 2016 #13
    sorry , i forgot to label the angle in the new triangle that i drew , here's what i mean ...
    According to the notes , tan( 2 θs1) = -(σx -σy) / 2τxy , are they correct now ?
     
  15. Jul 24, 2016 #14

    haruspex

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    If you now are measuring the angles from the tau axis, yes, except that the tangent will now be positive.
     
  16. Jul 24, 2016 #15
    why will tan(2 θs1) will be negative ?
    as we can see from the figure , -(σx - σy) is negative value . 2τxy is positive , so tan(2 θs1) will be negative value , right ?
     
  17. Jul 24, 2016 #16

    haruspex

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    If you measure the angle anticlockwise from the vertical axis it will be between 0 and 90 degrees or between 180 and 270 degrees. Tan is positive through both of those ranges.
     
  18. Jul 24, 2016 #17
    how to make tan (2 θs1) and tan (2 θs2 ) positive ?
    From the figure , we can see that tan( 2 θs1) and tan (2 θs2 ) = - (σx -σy) / 2τxy , both are negative ....
     
  19. Jul 24, 2016 #18

    haruspex

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    The diagram labels the distance from the vertical axis to the left hand dashed line as - (σx -σy), but it is not. As a distance it is + (σx -σy). The x coordinate of that dashed line is - (σx -σy), but that is another matter.
    With standard Cartesian and polar coordinates, it is true that a point with coordinates (x, y) is at (r, θ), where y/x=tan(θ). But since we have had to switch theta to be measured from the positive vertical axis, we now have instead that x/y=-tan(θ).
    (This is connected with the fact that the product of the gradients of two lines at right angles is -1.)
     
  20. Jul 24, 2016 #19
    well , refer to the green colour , igonre the black colour label , so 2θs1 is measured in this way ? so , tan2θs1 = - tan(α) = - (-(σx - σy)) / 2τxy = (σx - σy)) / 2τxy
    P/s : I let 180 - α = 2θs1 , so tan (180 - α ) = -tan( α )
    Am i right ?
    why we need to switch theta to be measured from the positive vertical axis ? why we cant measure it directly from negative vertical axis ?
     
  21. Jul 24, 2016 #20

    haruspex

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    Green? Where? Did you mean to attach an image in post #13?
    Ok, if you switch to measuring clockwise that will switch the sign.
    Doesn't help. tan(θ+π)=tan(θ).
     
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