Understanding Mohr's Circle Formula: Angle Placement Outside the Triangle

[chetzread]

Homework Statement

this is actually mohr's circle formula, forget about the theory,let's focus on the mathematics part. I couldn't understand why the tan( 2 θs1) = -(σx -σy) / 2τxy ?
2θs1 is outside the triangle

Homework Equations

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?

 

[Simon Bridge]

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?

Lessee ... looking at the triangle on the right, basic trig says that $\tan(180-2\theta_{S1}) = O/A$ where $O=\tau_{xy}$ and $A=\frac{1}{2}(\sigma_x-\sigma_y)$ ...

 

[chetzread]

Lessee ... looking at the triangle on the right, basic trig says that $\tan(180-2\theta_{S1}) = O/A$ where $O=\tau_{xy}$ and $A=\frac{1}{2}(\sigma_x-\sigma_y)$ ...

which triangle do you mean? , where is it?
tan [ theta(S1) ] is (σx- σy) / 2 τxy , where $A=\tau_{xy}$ and $O=\frac{1}{2}(\sigma_x-\sigma_y)$

 

[haruspex]

which triangle do you mean? , where is it?
tan [ theta(S1) ] is (σx- σy) / 2 τxy , where $A=\tau_{xy}$ and $O=\frac{1}{2}(\sigma_x-\sigma_y)$

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: $y=r\sin(\theta)$, $x=r\cos(\theta)$.

 

[chetzread]

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: $y=r\sin(\theta)$, $x=r\cos(\theta)$.

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?

 

[haruspex]

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?

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.

 

[chetzread]

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.

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?

 

[haruspex]

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?

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?

 

[chetzread]

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?

for the triangles that i drew earlier, it's measured from tau (vertical ) axis , right ? why you said it is measured from sigma axis ?

 

[haruspex]

for the triangles that i drew earlier, it's measured from tau (vertical ) axis , right ? why you said it is measured from sigma axis ?

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.

 

[chetzread]

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.

Why? The angle that I draw start from positive y(tau) axis, right?

 

[haruspex]

Why? The angle that I draw start from positive y(tau) axis, right?

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.

 

[chetzread]

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.

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 ?

 

[haruspex]

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 ?

If you now are measuring the angles from the tau axis, yes, except that the tangent will now be positive.

 

[chetzread]

If you now are measuring the angles from the tau axis, yes, except that the tangent will now be positive.

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 ?

 

[haruspex]

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 ?

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.

 

[chetzread]

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.

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 ...

 

[haruspex]

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 ...

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.)

 

[chetzread]

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.)

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 can't measure it directly from negative vertical axis ?

 

[haruspex]

refer to the green colour

Green? Where? Did you mean to attach an image in post #13?

180 - α = 2θs1

Ok, if you switch to measuring clockwise that will switch the sign.

why we can't measure it directly from negative vertical axis ?

Doesn't help. tan(θ+π)=tan(θ).

 

[chetzread]

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).

here it is . sorry

 

[chetzread]

Doesn't help. tan(θ+π)=tan(θ)

why it's not tan(θ-π) ?

 

[haruspex]

why it's not tan(θ-π) ?

tan(θ+π) is the same as tan(θ-π).

 

[chetzread]

tan(θ+π) is the same as tan(θ-π).

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 )
Is my idea correct ?

 

[haruspex]

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 )
Is my idea correct ?

If you measure the angles clockwise from the vertical axis, as you have with your green arrows, you will get the tan to come out the way you want.

 

[chetzread]

If you measure the angles clockwise from the vertical axis, as you have with your green arrows, you will get the tan to come out the way you want.

so , where should be the 2θs1 located ? i have another notes here , where the 2θs1 is measured from the positive axis ... and it's less than 90 degree , whereas in the case in post#21 , it's more than 90 degree.

 

[haruspex]

so , where should be the 2θs1 located ? i have another notes here , where the 2θs1 is measured from the positive axis ... and it's less than 90 degree , whereas in the case in post#21 , it's more than 90 degree.

It is unclear how the angles are defined, exactly. The diagram merely shows two angles in a geometric sense, i.e. as the angles between two lines. One is between 0 and 90, while the other is between 180 and 270. In each case, that means the tangent should be positive.
But if the angles are defined as being measured clockwise from the vertical then they are between 0 and -90 and between -180 and -270. In this view the tangents would be negative.
I suspect that, for the purposes of the text, it will not matter which view is taken as long as you are consistent.