Steven Dutch, Natural and Applied Sciences,
University of Wisconsin - Green Bay

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What is the formula for the radius r of the ellipse at left?
The familiar Cartesian coordinate formula is shown below the ellipse. We begin by noting that: x = r cos A y = r sin A and substituting those formulas into the Cartesian equation. |

When we do that substitution, we get:

r^{2}cos^{2}A/a^{2} + r^{2}sin^{2}A/b^{2} =
1, from which we can isolate r to get:

cos^{2}A/a^{2} + sin^{2}A/b^{2} =
1/r^{2}

We can eliminate the squared trigonometric terms by recalling that:

(cos^{2}A - sin^{2}A) = (2cos^{2}A - 1) = (1 - 2sin^{2}A)
= cos2A and therefore:

- cos
^{2}A = (1 + cos2A)/2 - sin
^{2}A = (1 - cos2A)/2

Plugging these values back into the formula for 1/r^{2}, we get:

(1/a^{2})(1 + cos2A)/2 + (1/b^{2})(1 - cos2A)/2 = 1/r^{2}

We can now isolate the cos2A terms to get:

(1/b^{2} + 1/a^{2})/2 - (1/b^{2} - 1/a^{2})(cos2A)/2
= 1/r^{2}

This looks surprisingly like our Mohr Circle derivation for stress, except we derived that strictly from vector analysis without thinking of ellipses at all (It's no accident). The only slightly puzzling feature is that we've reversed the a and b terms and there is a minus sign in front of the trigonometric term.

The reason for the changes is clear at left. In this Mohr
Circle space, all radii are mapped as their inverse squares (called quadratic
extensions). Since a is bigger, its inverse square is smaller. All the
other rules of Mohr Circles still apply.
Now, what does h (red) represent? The radius of the Mohr Circle is (1/b |

We might suspect (rightly, it turns out) that h has something to do with shear, as the corresponding coordinate does on the Mohr Circle for stress. Let's begin with a circle and deform it into our ellipse as shown below:

On the left we have a circle with a tangent line P'Q'. On the circle the radius is perpendicular to the tangent. On the right the diagram has been flattened by a factor b/a. All the triangles on the left are similar, but not on the right. Also the radius is no longer perpendicular to the tangent. The angle C, the change in angle due to flattening, is a measure of shear. Dimensions in the horizontal direction are unchanged. We have to do a tedious bit of trig manipulation but it's perfectly straightforward.

Tan B' = OQ'/OP and Tan B = OQ/OP = (b/a)OQ'/OP = (b/a) Tan B'

Tan A = y/x = (b/a)y'/x = (b/a) Tan A'

Tan C = Cot(A+B) = (1 - Tan A Tan B)/(Tan A + Tan B)

Now:

- Tan B = (b/a) Tan B',
- Tan B' = Cot A' = 1/Tan A', and
- Tan A' = (a/b)Tan A. Therefore:
- Tan B = (b
^{2}/a^{2})/ Tan A. Thus,

Tan C =

- (1 - Tan A Tan B)/(Tan A + Tan B) =
- (1 - (b
^{2}/a^{2})Tan A/Tan A)/(Tan A + (b^{2}/a^{2})/ Tan A) = - (1 - (b
^{2}/a^{2})/(Tan A + (b^{2}/a^{2})/ Tan A) = - (a
^{2}- b^{2})/(a^{2}Tan A + b^{2}/Tan A) = - (a
^{2}- b^{2})Tan A/(a^{2}Tan^{2}A + b^{2}) = - (a
^{2}- b^{2})(Sin A/Cos A)/(a^{2}Sin^{2}A/Cos^{2}A + b^{2}) = - (a
^{2}- b^{2})(Sin ACos A)/(a^{2}Sin^{2}A + b^{2}Cos^{2}A)

We recall that cos^{2}A/a^{2} + sin^{2}A/b^{2} =
1/r^{2}. The bottom term in the formula is thus equal to a^{2}b^{2}/r^{2}.
So we can now write:

- Tan C = (a
^{2}- b^{2})(Sin ACos A)/(a^{2}b^{2}/r^{2}) - = r
^{2}(a^{2}- b^{2})/a^{2}b^{2})(Sin ACos A) - = r
^{2}(1/b^{2}- 1/a^{2})(Sin ACos A) Now Sin 2A = 2 Sin A Cos A, so we have: - = (r
^{2}(1/b^{2}- 1/a^{2})/2)Sin 2A - = r
^{2}h

So: h = Tan C/r.
^{2} = (1/r^{2})Tan CWriting it this way allows us to see that the geometrical relationship at left must hold. |

The equation of a hyperbola is similar to that of an ellipse, differing only in one sign:

**x ^{2}/a^{2} - y^{2}/b^{2} = 1**

Note a couple of significant facts:

- When x = 0, y
^{2}= -b^{2}. In other words,**there is no y-intercept**. - When y = 0, x
^{2}= a^{2}. The x-intercepts are x = a and x = -a. - When x and y approach infinity, y/x approaches b/a. In other words, the
hyperbola approaches the lines y = xb/a and y = -xb/a. These lines are
called
*asymptotes.*

We can proceed exactly as we did for the ellipse, and get the polar equation

r^{2}cos^{2}A/a^{2} - r^{2}sin^{2}A/b^{2} =
1, from which we can obtain

(1/a^{2} - 1/b^{2})/2 + (1/b^{2} + 1/a^{2})(cos2A)/2
= 1/r^{2}

This formula leads to the Mohr Circle shown here.
Where the Mohr Circle crosses the vertical axis, 1/r |

The Mohr Circle above is a little different from the ellipse only because we chose the ellipse orientation so the long axis was always horizontal and labelled a. For both the ellipse and hyperbola, the ratio b/a affects the relative proportions in the x- and y- directions. It doesn't make sense to speak of a "long" or "short" axis of a hyperbola, and the ratio b/a can have any value. The larger b/a is, the more open the hyperbolas.

If the Mohr circle passes through the origin, the conic section is on the
borderline between an ellipse and a hyperbola, a parabola. However, it can't be
drawn. A Mohr Circle through the origin means 1/b^{2} = 0 or b infinite.
The equation for an ellipse reduces from x^{2}/a^{2} + y^{2}/b^{2} = 1
to x^{2}/a^{2} = 1 or x = +/-a. The equation for a
hyperbola reduces from x^{2}/a^{2} - y^{2}/b^{2} = 1
to x^{2}/a^{2} = 1 and again to x = +/-a. In the case of
the ellipse, we imagine we are in the middle of an ellipse with an infinitely
long b axis.

A circle deformed uniformly becomes an ellipse, so it's easy to see the connection between strain and ellipses. But why do Mohr Circles work for stress? We said nothing about ellipses in deriving the Mohr Circle for stress. Because there's a connection between stress and ellipses, too. In fact, all tensor quantities (of which stress and strain are two examples) can be described in terms of conic sections.

If we have two principal stresses, S1 and S2, we can draw an ellipse with
axes 1/S1^{2} and 1/S2^{2}, and draw the ellipse x^{2}/S1^{2} +
y^{2}/S2^{2} = 1. For any given plane, we plot the pole to
the plane. The normal stress is the inverse square of the radius of the ellipse
(1/r^{2}) along the pole direction. The shear stress is TanC/r^{2}.
This ellipse is called the *stress ellipse*. For
situations where one stress is tensional and one compressional, the Mohr circle
corresponds to a hyperbola rather than an ellipse, so the proper name for stress
ellipses or hyperbolas is *stress conic.*

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*Created 29 November 2000, Last Update 1 December 2000*

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