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# Compound Stress and Strain Part 2

A description of Mohr's Stress and Strain Circles, two and three dimensional Stress and Strain systems, and Strain Energy.

## Introduction

This is the second part in our discussion on the topic of Compound Stress and Strain. In this section we analyse the state of Stress at a point with a graphical representation using Mohr's Circle. In the latter half, we also look at how Mohr's circle can be adapted to represent direct or linear strain, and shear strain.

See Also the section on Compound Stress and Strain Part 1 .

## Mohr's Stress Circle

Mohr's circle is a two-dimensional graphical representation of the state of stress at a point. The abscisa and ordinate of each point on the circle are the normal stress and shear stress components respectively, acting on a particular cut plane with a unit vector $\inline&space;n$ with components $\inline&space;n_1$, $\inline&space;n_2$, $\inline&space;n_3$. In other words, the circumference of the circle is the locus of points that represent the state of stress on individual planes at all their orientations.

Mohr's Stress Circle allows the Stress on any plane which makes an angle $\inline&space;\theta$ with the Principle Planes.

In the figure $\inline&space;f_1$ and $\inline&space;f_2$ are the Principle Stress on the Principle Planes $\inline&space;BC$ and $\inline&space;AB$.

A diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circle. The diameters are the longest chords of the circle.

To draw the circle:
• Draw a line $\inline&space;PM$ such that $\inline&space;PL$ represents $\inline&space;f_1$ and $\inline&space;PM$ $\inline&space;f_2$. Note that the positive direction (Tension) is to the right.
• On $\inline&space;LM$ as a Diameter draw a Circle with centre $\inline&space;O$
• On this drawing $\inline&space;f_2\;>\;f_1$, but this is not a necessary condition.
• The radius $\inline&space;OL$ represents the plane of $\inline&space;f_1\;(BC)$
• The radius $\inline&space;OM$ represents the plane of $\inline&space;f_2\;(AB)$
• The Plane $\inline&space;AC$ is obtained by rotating $\inline&space;AB$ through $\inline&space;\theta$ and if $\inline&space;OM$ on the Stress Circle is rotated through $\inline&space;2\theta$ in the same direction, then the radius $\inline&space;OR$ is obtained. This will be shown to represent the plane $\inline&space;AC$. (Note that $\inline&space;OR$ could equally well be obtained by rotating $\inline&space;OL$ clockwise through $\inline&space;180^{0}&space;-&space;2\theta$ corresponding to rotating $\inline&space;BC$ clockwise through $\inline&space;90^{0}&space;-&space;\theta$)
• Draw $\inline&space;RN$ perpendicular to $\inline&space;PM$

Then,
$PN&space;=&space;PO&space;+&space;ON$
$\therefore\;\;\;\;\;\;PN=\frac{1}{2}(f_1&space;+&space;f_2)&space;+&space;\frac{1}{2}(f_2&space;-&space;f_1)\:\cos2\theta$
$=f_1\left(\frac{1-\cos2\theta&space;}{2}&space;\right)&space;+&space;f_2\left(\frac{1&space;+&space;\cos2\theta&space;}{2}&space;\right)$
$=f_1\sin^2\theta&space;+f_2\cos^2\theta&space;=f_\theta$

$\inline&space;f_\theta$ is the normal Stress Component on $\inline&space;AC$ (See Part 1 of Compound Stress and Strain).

And $\inline&space;RN&space;=&space;\frac{1}{2}(f_2&space;-&space;f_1)\sin2\theta&space;&space;=&space;s_\theta$ where $\inline&space;s_\theta$ is the Shear Stress Component on $\inline&space;AC$

Also, the Resultant Stress is given by: $\inline&space;f_r&space;=&space;\sqrt{f_\theta&space;^2&space;+&space;s_\theta&space;^2}&space;=&space;PR$

The inclination of the resultant Stress to the Normal of the plane is given by: $\inline&space;\phi&space;&space;=&space;\angle\;&space;RPN$

A shear stress is defined as the component of stress coplanar with a material cross section. Shear stress arises from a force vector perpendicular to the surface normal vector of the cross section.

$\inline&space;f_\theta$ is a Tensile Stress in this case and $\inline&space;s_\theta$ is considered positive if $\inline&space;R$ is above $\inline&space;PM$. A positive Shear Force is one which tends to give a clockwise rotation to a rectangular element (Shown dotted in the first Diagram).

• The Stresses on the plane $\inline&space;AD$, perpendicular to $\inline&space;AC$, are obtained from the radius $\inline&space;OR$' which is at $\inline&space;180^{0}$ to $\inline&space;OR$.

i.e., $\inline&space;f_{\theta&space;}'&space;=&space;PN'$ and $\inline&space;s_{\theta&space;}'&space;=&space;R'N'$, the latter being of the same magnitude as $\inline&space;s_{\theta&space;}$ but of the opposite type which tends to give an anticlockwise rotation to the dotted element.
• The Maximum Shear Stress occurs when $\inline&space;RN&space;=&space;OR$ (i.e. $\inline&space;\theta&space;&space;=&space;45^{0}$) and is equal in magnitude to $\inline&space;\displaystyle&space;OR&space;=&space;\frac{1}{2}(f_2&space;-&space;f_1)$
• The maximum Value of $\inline&space;\phi$ is obtained when $\inline&space;PR$ is a tangent to the Stress Circle.

Two particular cases which were considered analytically in Part 1 are now dealt with using this method.

## Pure Compression.

If $\inline&space;f$ is a Compressive Stress then the other Principle Stress is zero.

If $\inline&space;\displaystyle&space;\theta$ is the angle measured from the Plane of zero Stress then in the above diagram $\inline&space;PL&space;=&space;f$ numerically. It will be measured to the left for Compression. $\inline&space;PM&space;=&space;0$

Hence: $\inline&space;OR&space;=&space;\displaystyle\frac{1}{2}f$

$\inline&space;f_\theta&space;&space;=&space;PN$ Compressive

$\inline&space;s_\theta&space;&space;=&space;RN$ Positive

And the Maximum Shear Stress $\inline&space;q$ occurs when $\inline&space;\displaystyle&space;\theta&space;&space;=&space;45^{0}$ and is given by:
$q&space;=&space;OR&space;=&space;\frac{1}{2}f$

## Principal Stresses Equal Tension And Compression.

Let $\inline&space;\displaystyle&space;\theta$ be the angle measured anticlockwise from the Plane of $\inline&space;f$ Tensile. $\inline&space;PM&space;=&space;f$ to the right.

$\inline&space;PL&space;=&space;f$ to the left.

Hence $\inline&space;O$ coincides with $\inline&space;P$.

$\inline&space;\displaystyle&space;f_\theta&space;&space;=&space;PN$ and is Tensile for $\inline&space;\displaystyle&space;\theta$ between $\inline&space;\displaystyle&space;\pm&space;45^{0}$ and Compressive for $\inline&space;\displaystyle&space;\theta$ between $\inline&space;\displaystyle&space;45^{0}$ and $\inline&space;135^{0}$.

$\inline&space;\displaystyle&space;s_\theta&space;&space;=&space;RN$. When $\inline&space;\theta&space;&space;=&space;45^{0},\;\;s_\theta$ reaches its maximum value (Numerically equal to $\inline&space;f$) on those Planes where the Normal Stress is Zero (i.e. Pure Shear).

Example:
[imperial]
##### Example - Example 1
Problem
A piece of material is subjected to two compressive Stresses at right angles, their values being $\inline&space;4\;tons/sq.in.$ and $\inline&space;6\;tons/sq.in.$

Find the position of the Plane across which the resultant Stress is most inclined to the Normal and determine the value of this resultant Stress.
Workings
In the left hand diagram the angle $\inline&space;\displaystyle&space;\theta$ is inclined to the plane of $\inline&space;4\;tons/sq.in.$ compression. In the right hand diagram $\inline&space;Pl&space;=&space;6$ and $\inline&space;PM&space;=&space;4$. The maximum angle $\inline&space;\displaystyle&space;\phi$ is found when $\inline&space;PR$ is a tangent to the Stress Circle. $\inline&space;OR&space;=&space;1$ and $\inline&space;PO&space;=&space;5$.

From the right hand diagram:
$\phi&space;&space;=&space;\sin^{-1}\;\frac{1}{5}&space;=&space;11^{0}30'$
$f_r&space;=&space;PR&space;=&space;\sqrt{5^2&space;-&space;1^2}&space;=&space;4.9\;tons\;in.^{-2}$
$2\theta&space;&space;=&space;90^{0}&space;-&space;\phi$
$\therefore\;\;\;\;\;\;\theta&space;&space;=&space;39^{0}15'$
which gives the position of the plane required.

It is also possible to use Mohr's Stress Circle in the reverse sense to find the magnitude and direction of the Principal Stresses in a given Stress system. An example of this is shown below.
Solution
• $\inline&space;\theta&space;&space;=&space;39^{0}15'$

## A Two-dimensional Stress System.

It has been shown that every system can be reduced to the action of pure normal Stresses on the Principal Planes.

Consider the Strains produced by each Stress separately:

$\inline&space;\displaystyle&space;f_1$ will cause:
• Strain $\inline&space;\displaystyle&space;\frac{f_1}{E}$ in the direction of $\inline&space;\displaystyle&space;f_1$
• Strain $\inline&space;\displaystyle&space;-&space;\frac{f_1}{m\;E}$ in the direction of $\inline&space;\displaystyle&space;f_2$

$\inline&space;\displaystyle&space;f_2$ will cause:
• Strain $\inline&space;\displaystyle&space;\frac{f_2}{E}$ in the direction of $\inline&space;\displaystyle&space;f_2$
• Strain $\inline&space;\displaystyle&space;-\:\frac&space;{f_2}{m\;E}$ in the direction of $\inline&space;\displaystyle&space;f_2$

Since the Strains are all small, the resultant strains are given by the algebraic sum of those due to each Stress separately, i.e.,
• Strain in the direction of $\inline&space;\displaystyle&space;f_1$
$\;\;\;\;\;\;\;\;\;\;e_1&space;=&space;\frac{f_1}{E}&space;-&space;\frac{f_2}{m\;E}$
• Strain in the direction of $\inline&space;\displaystyle&space;f_2$
$\;\;\;\;\;\;\;\;\;\;e_2&space;=&space;\frac{f_2}{E}&space;-&space;\frac{f_1}{m\;E}$

The normal conventions apply and Tensile Stress is positive and Compressive Stress negative. A positive Stress represents an increase in dimensions in that direction.

## Principal Strains In Three Dimensions.

Using a similar argument to that used in the previous paragraph t can be shown that the Principal Strains in the direction $\inline&space;f_1,\;\;f_2$, and $\inline&space;f_3$ are given by :

• $e_1&space;=&space;\frac{f_1}{E}&space;-&space;\frac{f_2}{m\;E}&space;-&space;\frac{f_3}{m\;E}$
• $e_2&space;=&space;\frac{f_2}{E}&space;-&space;\frac{f_3}{m\;E}&space;-&space;\frac{f_1}{m\;E}$
• $e_3&space;=&space;\frac{f_3}{E}&space;-&space;\frac{f_1}{m\;E}&space;-&space;\frac{f_2}{m\;E}$

It must be remembered that Stress and Strain in any given direction are not proportional when Stress exists in more than one dimension.
Strain can exist without Stress in the same direction (e.g. If $\inline&space;\;\;\;f_3&space;=&space;0$, Then $\inline&space;\;e_3\;=&space;-&space;\displaystyle\frac{f_1}{m\;E}&space;-&space;\displaystyle\frac{f_2}{m\;E}$ ).

Example:
[imperial]
##### Example - Example 4
Problem
A piece of material is subjected to three perpendicular Tensile Stresses and the Strains in the three directions are in the ratio of 3:4:5.

If Poisson's Ratio is 0.286 find the ratio of the Stresses and their value if the greatest is $\inline&space;6\;tons/sq.in.$
Workings
Let the Stresses be $\inline&space;f_1,\;f_2$, and $\inline&space;f_3$ and the corresponding Strains $\inline&space;3k$, $\inline&space;4k$, and $\inline&space;5k$.

Then :

$3\;k\;E&space;=&space;f_1&space;-&space;0.286\;(f_2&space;+&space;f_3)$
$4\;k\;E&space;=&space;f_2&space;-&space;0.286\;(f_3&space;+&space;f_1)$
$5\;k\;E&space;=&space;f_3&space;-&space;0.286\;(f_1&space;+&space;f_2)$

Subtracting Equation (1) from (3)

$f_3&space;-&space;f_1&space;-&space;0.286\;(f_1&space;-&space;f_3)&space;=&space;2\;k\;E$
$\therefore\;\;\;\;\;\;f_3&space;-&space;f_1&space;=&space;\frac{2\;k\;E}{1.286}$

Re-writing equations (3) and (2)

$\frac{f_3}{0.286}&space;-&space;f_1&space;-&space;f_2&space;=&space;\frac{5\;k\;E}{0.286}$
And
$\;\;\;\;\;f_2&space;-&space;0.286\;f_3&space;-&space;0.286\;f_1&space;=&space;4\;k\;E$

From Equations (4) and (5)
$1.924\;f_3&space;=&space;19.5\;k\;E$

From this and the other equations above :
$f_3&space;=&space;10.14\;k\;E$
$f_1&space;=&space;8.58\;k\;E$
$f_2&space;=&space;9.34\;k\;E$

Hence the Ratios of the stresses are:
$f_1:f_2:f_3&space;=&space;0.847:0.921:1$

If the greatest Stress is $\inline&space;6\;tons/sq.in.$

Then $\inline&space;f_1=5.08\;tons\;in.^{-2}$, $\inline&space;f_2=5.53\;tons\;in.^{-2}$ and $\inline&space;f_3=6\;tons\;in.^{-2}$
Solution
• The ratio of the Stresses are $\inline&space;f_1:f_2:f_3&space;=&space;0.847:0.921:1$
• $\inline&space;f_1=5.08\;tons\;in.^{-2}$, $\inline&space;f_2=5.53\;tons\;in.^{-2}$ and $\inline&space;f_3=6\;tons\;in.^{-2}$

## Principal Stresses Determined From Principal Strains.

Rearranging equations (1),(2) and (3) as:

$E\;e_1&space;=&space;f_1\;-\frac{f_2}{m}&space;-&space;\frac{f_3}{m}$
$E\;e_2&space;=&space;f_2\;-\frac{f_3}{m}&space;-&space;\frac{f_1}{m}$
$E\;e_3&space;=&space;f_3\;-\frac{f_1}{m}&space;-&space;\frac{f_2}{m}$

Subtracting Equation (5) from (4)

$E(e_1&space;-&space;e_2)&space;=&space;(f_1&space;-&space;f_2)\left(1&space;+&space;\frac{1}{m}&space;\right)$

From (4) and (6)

$E(m\;e_1&space;+&space;e_3)&space;=&space;f_1\left(1&space;-&space;\frac{1}{m}&space;\right)&space;-&space;f_2\left(1&space;+&space;\frac{1}{m}&space;\right)$

Subtracting (7) from (8)
$E[(m&space;-&space;1)e_1&space;+&space;e_2&space;+&space;e_3]&space;=&space;f_1\left(m&space;-&space;1&space;-&space;\frac{2}{m}&space;\right)$
$=&space;f_1(m&space;+&space;1)\left(\frac{m&space;-&space;2}{m}&space;\right)$
$\therefore\;\;\;\;\;\;f_1&space;=&space;\frac{E\;m[(m&space;-&space;1)\;e_1&space;+&space;e_2&space;+&space;e_3]}{(m&space;+&space;1)(m&space;-&space;2)}$

Similarly,
$f_2&space;=&space;\frac{E\;m[e_1&space;+&space;(m&space;-&space;1)e_2&space;+&space;e_3]}{(m&space;+&space;1)(m&space;-&space;2)}$

And,
$f_3&space;=&space;\frac{E\;m[e_1&space;+&space;e_2&space;+&space;(m&space;-&space;1)e_3]}{(m&space;+&space;1)(m&space;-&space;2)}$

(b) A Two Dimensional Stress System where $\inline&space;\displaystyle&space;f_3&space;=&space;0$
$E\;e_1&space;=&space;f_1&space;-&space;\frac{f_2}{m}$
$E\;e_2&space;=&space;f_2&space;-&space;\frac{f_1}{m}$

Solving these two equations for $\inline&space;f_1$ and $\inline&space;f_2$
$f_1&space;=&space;\frac{E\;m}{m^2&space;-&space;1}\left(m\;e_1&space;+&space;e_2&space;\right)$
And,
$f_2&space;=&space;\frac{E\;m}{m^2&space;-&space;1}\left(e_1&space;+&space;m\;e_2&space;\right)$

## Analysis Of Strain.

If $\inline&space;e_x\;,\;e_y$ and $\inline&space;\phi$ are the linear and Shear Strains in the plane $\inline&space;XOY$, then we require an expression for $\inline&space;\theta$, the linear Strain in a direction inclined at an angle $\inline&space;\theta$ to $\inline&space;OX$ in terms of $\inline&space;e_y\;,\;\phi$ and $\inline&space;\theta$.

Shear strain refers to a deformation of a solid body in which a plane in the body is displaced parallel to itself relative to parallel planes in the body; quantitatively, it is the displacement of any plane relative to a second plane, divided by the perpendicular distance between planes.

In the diagram the line $\inline&space;OP$, of length $\inline&space;r$ , is the diagonal of a rectangle which under the given Strains distorts into the dotted parallelogram. $\inline&space;P$ moves to $\inline&space;P'$. It must be remembered that the actual Strains are small.

$\inline&space;PP'&space;=&space;PQ\;\cos\theta&space;&space;+&space;QR\;\sin\theta&space;&space;+&space;RP'\;\cos\theta$ (Approx.)

$=&space;(r\cos\theta&space;\times&space;e_x)\cos\theta&space;&space;+&space;(r\sin\theta&space;\times&space;e_y)\sin\theta&space;&space;+&space;(r\sin\theta&space;\times&space;\phi&space;)\cos\theta$

$=&space;r\times&space;e_x\times&space;\cos^2\theta&space;&space;+&space;r\times&space;e_y\times&space;\sin^2\theta&space;&space;+&space;r\;\phi&space;\;\sin\theta&space;\times&space;\cos\theta$

But by definition $\inline&space;\displaystyle&space;e_\theta&space;&space;=&space;\frac{PP'}{r}$

$=\frac{1}{2}\;e_x(1&space;+&space;\cos2\theta&space;)+\frac{1}{2}\;e_y(1&space;-&space;\cos2\theta&space;)+\frac{1}{2}\;\phi\;&space;\sin2\theta$

$=&space;\frac{1}{2}(e_x+e_y)+\frac{1}{2}(e_x-e_y)\;\cos2\theta&space;+\frac{1}{2}\;\phi&space;\;\sin2\theta$
The Principal Strains $\inline&space;\displaystyle&space;e_1$ and $\inline&space;e_2$ are the maximum and Minimum values of Strain. These occur at values of $\inline&space;\displaystyle&space;\theta$ obtained by equating $\inline&space;\displaystyle&space;\frac{de_\theta&space;}{d\theta&space;}$ to Zero,i.e.,
$\tan\;2\theta&space;&space;=&space;\phi\;&space;(e_x&space;-&space;e_y)$

Then as for the Principal Stresses $\inline&space;\displaystyle&space;e_1$ and $\inline&space;e_2$ are given by :

$\frac{1}{2}(e_x&space;+&space;e_y)\;\pm&space;\frac{1}{2}\sqrt{(e_x&space;-&space;e_y)^2&space;-&space;\phi&space;^2}$

In order to evaluate $\inline&space;\displaystyle&space;e_x,\;\;e_y$ and $\inline&space;\phi$ (and hence the Principal Strains) it is necessary to know the linear Strains in any three directions at a particular point.

(Note: If the principal direction are known then only two Strains are required,since $\inline&space;\phi&space;&space;=&space;0$ and $\inline&space;e_x&space;=&space;e_1$, $\inline&space;e_y&space;=&space;e_2$).

Finally,if the Strains are caused by Stresses in two dimensions only, then the Principal Stresses can be determined by equations (9) and (10).

Example:
[imperial]
##### Example - Example 5
Problem
The measured Strains in three directions inclined at 60 degrees to one another are $\inline&space;\displaystyle&space;550\times&space;10^{-6}\;,\;&space;-&space;100\times&space;10^{-6}$ and $\inline&space;150\times&space;10^{-6}.$.

Calculate the magnitude and direction of the Principal Strains in this plane. If there is no Stress perpendicular to the given Plane, determine the Principal Stresses at the point.

$\inline&space;E&space;=&space;30\times10^6\;lb.in^{-2}$ and $\inline&space;\displaystyle\frac{1}{m}&space;=&space;0.3$
Workings

Taking the $\inline&space;X$-axis in the direction of the $\inline&space;\displaystyle&space;550\times&space;10^{-6}$ Strain, $\inline&space;\displaystyle&space;e_x,\;\;e_y$ and $\inline&space;\phi$ are determined from equation (11) with $\inline&space;\displaystyle&space;\theta&space;&space;=&space;0,\;\;60^{0}$ and $\inline&space;120^{0}$ from the three measured Strains. Hence,
$e_0&space;=&space;550\times10^{-6}&space;=&space;\frac{1}{2}(e_x&space;+&space;e_y)&space;=&space;e_x$

$e_{60}\;=&space;-&space;100\times&space;10^{-6}&space;=&space;\frac{1}{2}(e_x&space;+&space;e_y)&space;-&space;\frac{1}{4}(e_x&space;-&space;e_y)&space;+&space;\frac{1}{2}\phi&space;\;\sqrt{\frac{3}{2}}$

$=&space;\frac{1}{4}(e_x&space;+&space;3\;e_y)&space;-&space;\frac{1}{4}\;\phi&space;\;\sqrt{3}$

And,
$e_{120}&space;=&space;150\times&space;10^{-6}&space;=&space;\frac{1}{2}\;(e_x&space;+&space;e_y)&space;-&space;\frac{1}{4}\;(e_x&space;-&space;e_y)&space;+&space;\frac{1}{2}\;\phi&space;\;\sqrt{\frac{3}{2}}$

$=&space;\frac{1}{4}\;(e_x&space;+&space;3e_y)&space;-&space;\frac{1}{2}\;\sqrt{3}$

$\frac{1}{2}\;(e_x&space;+&space;3e_y)&space;=&space;50\times&space;10^{-6}$

$\therefore\;\;\;\;\;\;e_y&space;=&space;\frac{1}{3}(100&space;-&space;e_x)\;10^{-6}$

Using equation (6), $\inline&space;e_y\;=&space;-&space;150\times&space;10^{-6}$

From equations (7) and (8)

$\frac{1}{4}\phi&space;\;\sqrt{3}&space;=&space;[&space;-&space;100&space;-&space;\frac{1}{4}(550&space;-&space;450)]\;10^{-6}$

The Direction of the Principal Strains $\inline&space;\displaystyle&space;e_1$ and $\inline&space;e_2$ (To the $\inline&space;X$-axis) are found using equation (11)

$\tan\;2\theta&space;&space;=&space;\frac{\phi&space;}{e_x&space;-&space;e_y}\;=&space;-&space;\frac{500}{700\times\sqrt{3}}\;=&space;-&space;0.4125$

$\therefore\;\;\;\;\;\;2\theta&space;\;=&space;-&space;22.4^{0}\;\;\;or\;\;\;180^{0}&space;-&space;22.4^{0}$

Therefore, $\inline&space;\;\;\;\;\;\theta&space;\;=&space;-&space;11.2^{0}$ or $\inline&space;78.8^{0}$

The Principal Strains are found using equation (13) and are:

$\frac{1}{2}(e_x&space;-&space;e_y)\pm&space;\frac{1}{2}\sqrt{(e_x-e_y)^2+\phi&space;^2}=200\times&space;10^{-6}\pm&space;\frac{1}{2}\sqrt{\left[700^2&space;+&space;\frac{500^2}{3}&space;\right]}\;\times&space;10^{-6}$

$=&space;(200\;\pm&space;\;379)\times&space;10^{-6}$

$\therefore\;\;\;\;\;e_1&space;=&space;(200&space;+&space;379)\times&space;10^{-6}&space;=&space;579\times&space;10^{-6}$

And $\inline&space;\;\;\;\;\;e_1&space;=&space;(200&space;-&space;379)\times&space;10^{-6}\;=&space;-&space;179\times&space;10^{-6}$

For a two dimensional Stress system and using equations (9) and (10)

$f_1&space;=&space;\frac{30}{0.3(1/0.3^2&space;-&space;1)}\left(\frac{579}{0.3}&space;-&space;179&space;\right)&space;=&space;17,300\;lb.in.^{-2}$

$f_2&space;=&space;\frac{30}{0.3(1/0.3^2&space;-&space;1)}\left(579&space;-&space;\frac{179}{0.3}&space;\right)\;=&space;-&space;180\;lb.in.^{-2}$

## Mohr's Strain Circle.

It is now apparent that Mohr's Circle can also be used to represent Strains. The horizontal axis represents linear Strain and the vertical axis half the Shear Strain.

The diagram shows the relationship between $\inline&space;\displaystyle&space;e_x,\;e_y,\;\phi$ and $\inline&space;\theta$ and the Principal Strains $\inline&space;\displaystyle&space;e_1$ and $\inline&space;e_2$ as given by equations (11) and (13).

Note that $\inline&space;\displaystyle&space;PO&space;=&space;\frac{1}{2}(e_x&space;+&space;e_y)$ and $\inline&space;OR&space;=&space;\displaystyle\frac{1}{2}\sqrt{(e_x&space;-&space;e_y)^2&space;+&space;\phi&space;^2}$

The Strain Circle can be constructed if the linear Strains in three directions at a point and in the same plane are known. The problem of the last exercise will now be solved using this method.

The given Strains are $\inline&space;\displaystyle&space;e_0\;,\;\;e_{60}\;,\;\;e_{120}$. The Construction of the circle is similar to the Stress Circle. Vertical lines are drawn in relative positions to a datum through $\inline&space;P$ and at distances on either side proportional to the given Strains. From $\inline&space;R$ on the central line (i.e. $\inline&space;\displaystyle&space;e_{120}$ in this case),lines are set off at $\inline&space;\displaystyle&space;60^{0}$ and $\inline&space;120^{0}$ to the vertical, to cut the corresponding Strain Verticals in $\inline&space;Q$ and $\inline&space;S$. The Strain Circle then passes through $\inline&space;QRS$ and the Principal Strains are :

$\inline&space;e_1&space;=&space;PM&space;=&space;580\times&space;10^{-6}$ And $\inline&space;\;\;\;\;\;e_2&space;=&space;PL\;=&space;-&space;180\times&space;10^{-6}$

The radius $\inline&space;OS$ gives the Strain condition in the $\inline&space;X$ direction and the angle $\inline&space;SOM&space;=&space;22^{0}$. The direction of $\inline&space;\displaystyle&space;e_1$ is then at $\inline&space;\displaystyle&space;\frac{1}{2}\times&space;22&space;=&space;11^{0}$ clockwise from the $\inline&space;X$-axis and $\inline&space;\displaystyle&space;e_2$ is at right angles to $\inline&space;\displaystyle&space;e_1$.

The Principal Stresses can best be obtained from the Principal Strains by using the same calculations as were used in the last Example.

## Volumetric Strain.

A rectangular solid of sides $\inline&space;x$, $\inline&space;y$, $\inline&space;z$ is under the action of three principal Stresses $\inline&space;\displaystyle&space;f_1,\;f_2$, and $\inline&space;f_3$.

Then if $\inline&space;\displaystyle&space;e_1,\;e_2$, and $\inline&space;e_3$ are the corresponding linear Strains, the dimensions become : $\inline&space;x+e_1\;x,\;\;y+e_2\;y$, and $\inline&space;z&space;+&space;e_3\;z$

The Volumetric Strain = Increase in volume / Original volume

$=&space;\frac{x(1+e_1)\times&space;y(1+e_2)\times&space;z(1+e_3)-x\;y\;z}{x\;y\;z}$
$(1&space;+&space;e_1)(1&space;+&space;e_2)(1&space;+&space;e_3)&space;-&space;1$
$=&space;1&space;+&space;e_1&space;+&space;e_2&space;+&space;e_3&space;+&space;e_1e_2&space;+&space;e_2e_3&space;+&space;e_3e_1&space;+&space;e_1e_2e_3&space;-&space;1$

Since the actual Strains are small this may be written as equal to: $\inline&space;\;e_1&space;+&space;e_2&space;+&space;e_3$

Thus it can be stated that the Volumetric Strain is the Algebraic sum of the three Principal Strains.

The Volumetric Strain can also be found using the Principal Stresses in which case:

Volumetric Strain = $\inline&space;\displaystyle\frac{(f_1&space;+&space;f_2&space;+&space;f_3)(1&space;-&space;\displaystyle\frac{2}{m})}{E}$

## Strain Energy

The Strain Energy $\inline&space;U$ is the Work done by the Stresses in Straining material.It is sufficiently general to consider a unit cube acted upon by the three Principal Stresses $\inline&space;\displaystyle&space;f_1,\;f_2$, and $\inline&space;f_3$ . If the corresponding Strains are $\inline&space;\displaystyle&space;e_1,\;e_2$, and $\inline&space;e_3$ then, since the Stresses are applied gradually from zero, the Total work done = $\inline&space;\displaystyle&space;=&space;\sum{\frac{1}{2}&space;f\;e}$.

Using equations (1),(2),(3)

$U&space;=&space;\frac{1}{2}f_1e_1&space;+&space;\frac{1}{2}f_1e_1&space;+&space;\frac{1}{2}f_3e_3$

$=&space;\left(\frac{1}{2E}&space;\right)\left\{f_1\left(f_1\;-\frac{f_2}{m}&space;-&space;\frac{f_3}{m}&space;\right)&space;+&space;f_2\left(f_2&space;-&space;\frac{f_3}{m}&space;-&space;\frac{f_1}{m}\right)&space;+&space;f_3\left(f_3&space;-&space;\frac{f_1}{m}&space;-&space;\frac{f_2}{m}&space;\right)&space;\right\}$

$=&space;\left(\frac{1}{2E}&space;\right)\left\{f_1^2&space;+&space;f_2^2&space;+&space;f_3^2&space;-&space;\left(\frac{2}{m}&space;\right)(f_1f_2&space;+&space;f_2f_3&space;+&space;f_3f_1)&space;\right\}$

For a two dimensional Strain system $\inline&space;\displaystyle&space;f_3&space;=&space;0$

$\therefore\;\;\;\;\;\;U&space;=&space;\left(\frac{1}{2E}&space;\right)\left\{f_1^2&space;+&space;f_2^2&space;-&space;\left(\frac{2}{m}&space;\right)f_1f_2&space;\right\}\;\;\;\;\;\text{per&space;unit&space;volume}$

Example:
[imperial]
##### Example - Example 6
Problem
The Principal Stresses at a point in an elastic material are $\inline&space;6\;tons/sq.in.$ tensile $\inline&space;2\;tons/sq.in.$ tensile $\inline&space;5\;tons/sq.in.$ compressive.

Calculate the volumetric Strain and the resilience.

$\inline&space;E&space;=&space;6000\;tons/sq.in.$ and $\inline&space;\displaystyle\frac{1}{m}&space;=&space;0.35.$
Workings
Using equation (14), Volumetric Strain,
$=&space;(f_1&space;+&space;f_2&space;+&space;f_3)\left(\frac{1&space;-&space;2/m}{E}&space;\right)=&space;(6&space;+&space;2&space;-&space;5)\left(\frac{1\;-0.7}{6000}&space;\right)\;&space;=&space;1.5\times10^{-4}$

Resilience = $\inline&space;\displaystyle\frac{1}{2\times&space;6000}\times&space;[6^2+2^2+(-5)^2\;-2\times&space;0.35(6\times&space;2&space;-&space;2\times&space;5&space;-&space;5\times&space;6)]$
$=&space;\frac{1}{12,000}\times&space;(65&space;+&space;0.7\times&space;28)\;in.tons\;\;&space;in^{-3}$
$=&space;\frac{84.6\times2240}{12,000}\;\;in.lb.\;\;in^{-3}&space;=&space;15.8\;in.lb.\;\;in^{-3}$
Solution
• The volumetric Strain is $\inline&space;1.5\times10^{-4}$
• The resilience is $\inline&space;15.8\;in.lb.\;\;in^{-3}$

## Shear Strain Energy.

Writing,
$f_1&space;=&space;\frac{1}{3}(f_1&space;+&space;f_2&space;+&space;f_3)&space;+&space;\frac{1}{3}(f_1&space;-&space;f_2)&space;+&space;\frac{1}{3}(f_1&space;-&space;f_3)$
$f_2&space;=&space;\frac{1}{3}(f_1&space;+&space;f_2&space;+&space;f_3)&space;+&space;\frac{1}{3}(f_2&space;-&space;f_1)&space;+&space;\frac{1}{3}(f_2&space;-&space;f_3)$
$f_3&space;=&space;\frac{1}{3}(f_1&space;+&space;f_2&space;+&space;f_3)&space;+&space;\frac{1}{3}(f_3&space;-&space;f_1)&space;+&space;\frac{1}{3}(f_3&space;-&space;f_2)$

Then under the action of the mean stress $\inline&space;\displaystyle&space;\frac{1}{3}(f_1&space;+&space;f_2&space;+&space;f_3)$ there will be volumetric Strain with no distortion of shape (i.e.no shear Stress anywhere).

The Strain energy under this mean Stress acting in each direction can be derived from equation (15) and may be called volumetric Strain Energy
$=\left(\frac{3}{2E}&space;\right)\left(\frac{(f_1+f_2+f_3)}{3}&space;\right)^2\times&space;\left(1-\frac{2}{m}&space;\right)$

The other terms in the arrangement of $\inline&space;\displaystyle&space;f_1,\;f_2$, and $\inline&space;f_3$ are proportional to the maximum Shear Stress values in the three planes and will cause a distortion of the shape.

Shear strain Energy $\inline&space;\displaystyle&space;\mathbf{U_s}$ is defined as the Total Strain Energy and the Volumetric Strain Energy.

Thus,
$U_s=\left(\frac{1}{2E}&space;\right)\left\{\left(f_1^2+f_2^2&space;+&space;f_3^2&space;\right)&space;-&space;\left(\frac{2}{m}&space;\right)(f_1f_2&space;+&space;f_2f_3+f_3f_1)&space;\right\}\;&space;-$
$-\left\{(f_1+f_2+f_3)^2&space;\times&space;\frac{1&space;-&space;\displaystyle\frac{2}{m}}{6E}&space;\right\}$

$=\left(\frac{1}{6E}&space;\right)\left\{(f_1^2&space;+&space;f_2^2&space;+&space;f_3^2)(3&space;-&space;1\;+\frac{2}{m})-(f_1f_2&space;+&space;f_2f_3&space;+&space;f_3f_1)(\frac{6}{m}&space;+&space;2&space;-&space;\frac{4}{m}&space;)\right\}$

$=&space;\left(\frac{1&space;+&space;\displaystyle\frac{1}{m}}{6E}&space;\right)\left\{(f_1^2&space;+&space;f_2^2&space;+&space;f_3^2)-2\;(f_1f_2&space;+&space;f_2f_3\;+f_3f_1\right\}$

$=&space;\left(\frac{1}{12C}&space;\right)\left\{(f_1&space;-&space;f_2)^2&space;+&space;(f_2&space;-&space;f_3)^2&space;+&space;f_3&space;-&space;f_1)^2&space;\right\}$

The quantities in the brackets are each twice the maximum Shear Stress in their respective planes.

In a pure Shear Stress system the Principal Stresses are $\inline&space;\displaystyle&space;\pm&space;\;s,\;0$ and by substitution:

Shear Straing Energy
$=\left(\frac{1}{12\;C}&space;\right)[(2s)^2+(-s)^2+(-&space;s)^2]=\frac{s^2}{2\;C}$

Note: The relationship between $\inline&space;E$ and $\inline&space;C$ will be discussed in "Elastic Constants".