angle outside triangle

Mar 2014
909
2
malaysia
[FONT=&quot]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 ?[/FONT]
[FONT=&quot]2θs1 is outside the triangle[/FONT]
[FONT=&quot]3153.jpg[/FONT]
[FONT=&quot]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?[/FONT]
 

chiro

MHF Helper
Sep 2012
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Australia
Hey xl5899.

Do you know the relationship between tan(theta) and the gradient (i.e. (y2-y1)/(x2-x1))?
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
It's not that - it's in terms of the gradient.

In a normal line (that's embedded in two-dimensional space) you have y - y0 = m(x-x0) where m = tan(theta) = (y2-y1)/(x2-x1).

The rate of change that is m in terms of tan(theta) is what I'm discussing above.
 
Mar 2014
909
2
malaysia
It's not that - it's in terms of the gradient.

In a normal line (that's embedded in two-dimensional space) you have y - y0 = m(x-x0) where m = tan(theta) = (y2-y1)/(x2-x1).

The rate of change that is m in terms of tan(theta) is what I'm discussing above.
But , the problem is the angle is outside the triangle....
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
We are referring to the lines made by the triangle that form the gradient.

You can apply them to any triangle.
 
Mar 2014
909
2
malaysia
We are referring to the lines made by the triangle that form the gradient.

You can apply them to any triangle.
can you explain further?I'm still blurred...
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Let m = (y2-y1)/(x2-x1) and you have points on the line (x1,y1) and (x2,y2).

In the triangle you have two points - one in the top left hand side and one in the bottom right.

For the top left you have (x2,y2) = (txy, [sigma_x - sigma_y]/2) and (x1,y1) = (0,0)

Since m = tan(theta) you have

m
= tan(theta)
= [[sigma_x - sigma_y]/2] / txy
= [sigma_x - sigma_y] / {2*txy}

which is what the result is.

You can do the other triangle by using the bottom right vertex for (x2,y2) and get the result that way.
 
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Mar 2014
909
2
malaysia
Let m = (y2-y1)/(x2-x1) and you have points on the line (x1,y1) and (x2,y2).

In the triangle you have two points - one in the top left hand side and one in the bottom right.

For the top left you have (x2,y2) = (txy, [sigma_x - sigma_y]/2) and (x1,y1) = (0,0)

Since m = tan(theta) you have

m
= tan(theta)
= [[sigma_x - sigma_y]/2] / txy
= [sigma_x - sigma_y] / {2*txy}

which is what the result is.

You can do the other triangle by using the bottom right vertex for (x2,y2) and get the result that way.
it should be (x2,y2) = ([sigma_x - sigma_y]/2 , txy) , the y-axis(vertical axis) represent txy , right ?
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Yes - I'm sorry about that you are correct.
 
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