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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.

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Contents

Introduction
Mohr's Stress Circle
Pure Compression.
Principal Stresses Equal Tension And Compression.
A Two-dimensional Stress System.
Principal Strains In Three Dimensions.
Principal Stresses Determined From Principal Strains.
Analysis Of Strain.
Mohr's Strain Circle.
Volumetric Strain.
Strain Energy
Shear Strain Energy.
Page Comments

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 $\inline n$ with components $\inline n_1$ , $\inline n_2$ , $\inline n_3$ . 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 $\inline \theta$ with the Principle Planes.

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

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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 $\inline PM$ such that $\inline PL$ represents $\inline f_1$ and $\inline PM$ $\inline f_2$ . Note that the positive direction (Tension) is to the right.
On $\inline LM$ as a Diameter draw a Circle with centre $\inline O$
On this drawing $\inline f_2\;>\;f_1$ , but this is not a necessary condition.

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The radius $\inline OL$ represents the plane of $\inline f_1\;(BC)$
The radius $\inline OM$ represents the plane of $\inline f_2\;(AB)$
The Plane $\inline AC$ is obtained by rotating $\inline AB$ through $\inline \theta$ and if $\inline OM$ on the Stress Circle is rotated through $\inline 2\theta$ in the same direction, then the radius $\inline OR$ is obtained. This will be shown to represent the plane $\inline AC$ . (Note that $\inline OR$ could equally well be obtained by rotating $\inline OL$ clockwise through $\inline 180^{0} - 2\theta$ corresponding to rotating $\inline BC$ clockwise through $\inline 90^{0} - \theta$ )
Draw $\inline RN$ perpendicular to $\inline PM$

Then,

$PN = PO + ON$

$\therefore\;\;\;\;\;\;PN=\frac{1}{2}(f_1 + f_2) + \frac{1}{2}(f_2 - f_1)\:\cos2\theta$

$=f_1\left(\frac{1-\cos2\theta }{2} \right) + f_2\left(\frac{1 + \cos2\theta }{2} \right)$

$=f_1\sin^2\theta +f_2\cos^2\theta =f_\theta$

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

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

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

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

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.

$\inline f_\theta$ is a Tensile Stress in this case and $\inline s_\theta$ is considered positive if $\inline R$ is above $\inline PM$ . 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 $\inline AD$ , perpendicular to $\inline AC$ , are obtained from the radius $\inline OR$ ' which is at $\inline 180^{0}$ to $\inline OR$ .

i.e., $\inline f_{\theta }' = PN'$ and $\inline s_{\theta }' = R'N'$ , the latter being of the same magnitude as $\inline s_{\theta }$ but of the opposite type which tends to give an anticlockwise rotation to the dotted element.

The Maximum Shear Stress occurs when $\inline RN = OR$ (i.e. $\inline \theta = 45^{0}$ ) and is equal in magnitude to $\inline \displaystyle OR = \frac{1}{2}(f_2 - f_1)$
The maximum Value of $\inline \phi$ is obtained when $\inline PR$ 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 $\inline f$ is a Compressive Stress then the other Principle Stress is zero.

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

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

$\inline f_\theta = PN$ Compressive

$\inline s_\theta = RN$ Positive

And the Maximum Shear Stress $\inline q$ occurs when $\inline \displaystyle \theta = 45^{0}$ and is given by:

$q = OR = \frac{1}{2}f$

Principal Stresses Equal Tension And Compression.

Let $\inline \displaystyle \theta$ be the angle measured anticlockwise from the Plane of $\inline f$ Tensile.

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$\inline PM = f$ to the right.

$\inline PL = f$ to the left.

Hence $\inline O$ coincides with $\inline P$ .

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

$\inline \displaystyle s_\theta = RN$ . When $\inline \theta = 45^{0},\;\;s_\theta$ reaches its maximum value (Numerically equal to $\inline f$ ) 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 $\inline 4\;tons/sq.in.$ and $\inline 6\;tons/sq.in.$

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 $\inline \displaystyle \theta$ is inclined to the plane of $\inline 4\;tons/sq.in.$ compression. In the right hand diagram $\inline Pl = 6$ and $\inline PM = 4$ . The maximum angle $\inline \displaystyle \phi$ is found when $\inline PR$ is a tangent to the Stress Circle. $\inline OR = 1$ and $\inline PO = 5$ .

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From the right hand diagram:

$\phi = \sin^{-1}\;\frac{1}{5} = 11^{0}30'$

$f_r = PR = \sqrt{5^2 - 1^2} = 4.9\;tons\;in.^{-2}$

$2\theta = 90^{0} - \phi$

$\therefore\;\;\;\;\;\;\theta = 39^{0}15'$

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

$\inline \theta = 39^{0}15'$

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.

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Consider the Strains produced by each Stress separately:

$\inline \displaystyle f_1$ will cause:

Strain $\inline \displaystyle \frac{f_1}{E}$ in the direction of $\inline \displaystyle f_1$
Strain $\inline \displaystyle - \frac{f_1}{m\;E}$ in the direction of $\inline \displaystyle f_2$

$\inline \displaystyle f_2$ will cause:

Strain $\inline \displaystyle \frac{f_2}{E}$ in the direction of $\inline \displaystyle f_2$
Strain $\inline \displaystyle -\:\frac {f_2}{m\;E}$ in the direction of $\inline \displaystyle f_2$

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 $\inline \displaystyle f_1$

$\;\;\;\;\;\;\;\;\;\;e_1 = \frac{f_1}{E} - \frac{f_2}{m\;E}$

Strain in the direction of $\inline \displaystyle f_2$

$\;\;\;\;\;\;\;\;\;\;e_2 = \frac{f_2}{E} - \frac{f_1}{m\;E}$

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 $\inline f_1,\;\;f_2$ , and $\inline f_3$ are given by :

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$e_1 = \frac{f_1}{E} - \frac{f_2}{m\;E} - \frac{f_3}{m\;E}$
(2)
$e_2 = \frac{f_2}{E} - \frac{f_3}{m\;E} - \frac{f_1}{m\;E}$
(3)
$e_3 = \frac{f_3}{E} - \frac{f_1}{m\;E} - \frac{f_2}{m\;E}$
(4)

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 $\inline \;\;\;f_3 = 0$ , Then $\inline \;e_3\;= - \displaystyle\frac{f_1}{m\;E} - \displaystyle\frac{f_2}{m\;E}$ ).

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 $\inline 6\;tons/sq.in.$

Workings

Let the Stresses be $\inline f_1,\;f_2$ , and $\inline f_3$ and the corresponding Strains $\inline 3k$ , $\inline 4k$ , and $\inline 5k$ .

Then :

$3\;k\;E = f_1 - 0.286\;(f_2 + f_3)$

(1)

$4\;k\;E = f_2 - 0.286\;(f_3 + f_1)$

(2)

$5\;k\;E = f_3 - 0.286\;(f_1 + f_2)$

(3)

Subtracting Equation (1) from (3)

$f_3 - f_1 - 0.286\;(f_1 - f_3) = 2\;k\;E$

$\therefore\;\;\;\;\;\;f_3 - f_1 = \frac{2\;k\;E}{1.286}$

Re-writing equations (3) and (2)

$\frac{f_3}{0.286} - f_1 - f_2 = \frac{5\;k\;E}{0.286}$

(4)

And

$\;\;\;\;\;f_2 - 0.286\;f_3 - 0.286\;f_1 = 4\;k\;E$

(5)

From Equations (4) and (5)

$1.924\;f_3 = 19.5\;k\;E$

From this and the other equations above :

$f_3 = 10.14\;k\;E$

$f_1 = 8.58\;k\;E$

$f_2 = 9.34\;k\;E$

Hence the Ratios of the stresses are:

$f_1:f_2:f_3 = 0.847:0.921:1$

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

Then $\inline f_1=5.08\;tons\;in.^{-2}$ , $\inline f_2=5.53\;tons\;in.^{-2}$ and $\inline f_3=6\;tons\;in.^{-2}$

Solution

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

Principal Stresses Determined From Principal Strains.

Rearranging equations (2),(3) and (4) as:

$E\;e_1 = f_1\;-\frac{f_2}{m} - \frac{f_3}{m}$

(5)

$E\;e_2 = f_2\;-\frac{f_3}{m} - \frac{f_1}{m}$

(6)

$E\;e_3 = f_3\;-\frac{f_1}{m} - \frac{f_2}{m}$

(7)

Subtracting Equation (6) from (5)

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

(8)

From (5) and (7)

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

(9)

Subtracting (8) from (9)

$E[(m - 1)e_1 + e_2 + e_3] = f_1\left(m - 1 - \frac{2}{m} \right)$

$= f_1(m + 1)\left(\frac{m - 2}{m} \right)$

$\therefore\;\;\;\;\;\;f_1 = \frac{E\;m[(m - 1)\;e_1 + e_2 + e_3]}{(m + 1)(m - 2)}$

Similarly,

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

And,

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

(b) A Two Dimensional Stress System where $\inline \displaystyle f_3 = 0$

$E\;e_1 = f_1 - \frac{f_2}{m}$

$E\;e_2 = f_2 - \frac{f_1}{m}$

Solving these two equations for $\inline f_1$ and $\inline f_2$

$f_1 = \frac{E\;m}{m^2 - 1}\left(m\;e_1 + e_2 \right)$

(10)

And,

$f_2 = \frac{E\;m}{m^2 - 1}\left(e_1 + m\;e_2 \right)$

(11)

Analysis Of Strain.

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

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

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

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

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

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

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

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

(12)

The Principal Strains $\inline \displaystyle e_1$ and $\inline e_2$ are the maximum and Minimum values of Strain. These occur at values of $\inline \displaystyle \theta$ obtained by equating $\inline \displaystyle \frac{de_\theta }{d\theta }$ to Zero,i.e.,

$\tan\;2\theta = \phi\; (e_x - e_y)$

(13)

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

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

(14)

In order to evaluate $\inline \displaystyle e_x,\;\;e_y$ and $\inline \phi$ (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 $\inline \phi = 0$ and $\inline e_x = e_1$ , $\inline e_y = e_2$ ).

Finally,if the Strains are caused by Stresses in two dimensions only, then the Principal Stresses can be determined by equations (10) and (11).

Example:

[imperial]

Example - Example 5

Problem

The measured Strains in three directions inclined at 60 degrees to one another are $\inline \displaystyle 550\times 10^{-6}\;,\; - 100\times 10^{-6}$ and $\inline 150\times 10^{-6}.$ .

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.

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

Workings

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

$e_0 = 550\times10^{-6} = \frac{1}{2}(e_x + e_y) = e_x$

(6)

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

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

(7)

And,

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

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

(8)

Adding equations (7) and (8)

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

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

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

From equations (7) and (8)

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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The diagram shows the relationship between $\inline \displaystyle e_x,\;e_y,\;\phi$ and $\inline \theta$ and the Principal Strains $\inline \displaystyle e_1$ and $\inline e_2$ as given by equations (12) and (14).

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

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

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

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

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 $\inline x$ , $\inline y$ , $\inline z$ is under the action of three principal Stresses $\inline \displaystyle f_1,\;f_2$ , and $\inline f_3$ .

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

The Volumetric Strain = Increase in volume / Original volume

$= \frac{x(1+e_1)\times y(1+e_2)\times z(1+e_3)-x\;y\;z}{x\;y\;z}$

$(1 + e_1)(1 + e_2)(1 + e_3) - 1$

$= 1 + e_1 + e_2 + e_3 + e_1e_2 + e_2e_3 + e_3e_1 + e_1e_2e_3 - 1$

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

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 = $\inline \displaystyle\frac{(f_1 + f_2 + f_3)(1 - \displaystyle\frac{2}{m})}{E}$

Strain Energy

The Strain Energy $\inline U$ 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 $\inline \displaystyle f_1,\;f_2$ , and $\inline f_3$ . If the corresponding Strains are $\inline \displaystyle e_1,\;e_2$ , and $\inline e_3$ then, since the Stresses are applied gradually from zero, the Total work done = $\inline \displaystyle = \sum{\frac{1}{2} f\;e}$ .

Using equations (2),(3),(4)

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

(15)

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

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

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

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

(16)

Example:

[imperial]

Example - Example 6

Problem

The Principal Stresses at a point in an elastic material are $\inline 6\;tons/sq.in.$ tensile $\inline 2\;tons/sq.in.$ tensile $\inline 5\;tons/sq.in.$ compressive.

Calculate the volumetric Strain and the resilience.

$\inline E = 6000\;tons/sq.in.$ and $\inline \displaystyle\frac{1}{m} = 0.35.$

Workings

Using equation (14), Volumetric Strain,

$= (f_1 + f_2 + f_3)\left(\frac{1 - 2/m}{E} \right)= (6 + 2 - 5)\left(\frac{1\;-0.7}{6000} \right)\; = 1.5\times10^{-4}$

Resilience = $\inline \displaystyle\frac{1}{2\times 6000}\times [6^2+2^2+(-5)^2\;-2\times 0.35(6\times 2 - 2\times 5 - 5\times 6)]$

$= \frac{1}{12,000}\times (65 + 0.7\times 28)\;in.tons\;\; in^{-3}$

$= \frac{84.6\times2240}{12,000}\;\;in.lb.\;\;in^{-3} = 15.8\;in.lb.\;\;in^{-3}$

Solution

The volumetric Strain is $\inline 1.5\times10^{-4}$
The resilience is $\inline 15.8\;in.lb.\;\;in^{-3}$

Shear Strain Energy.

Writing,

$f_1 = \frac{1}{3}(f_1 + f_2 + f_3) + \frac{1}{3}(f_1 - f_2) + \frac{1}{3}(f_1 - f_3)$

$f_2 = \frac{1}{3}(f_1 + f_2 + f_3) + \frac{1}{3}(f_2 - f_1) + \frac{1}{3}(f_2 - f_3)$

$f_3 = \frac{1}{3}(f_1 + f_2 + f_3) + \frac{1}{3}(f_3 - f_1) + \frac{1}{3}(f_3 - f_2)$

Then under the action of the mean stress $\inline \displaystyle \frac{1}{3}(f_1 + f_2 + f_3)$ 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 (16) and may be called volumetric Strain Energy

$=\left(\frac{3}{2E} \right)\left(\frac{(f_1+f_2+f_3)}{3} \right)^2\times \left(1-\frac{2}{m} \right)$

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

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

Thus,

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

$-\left\{(f_1+f_2+f_3)^2 \times \frac{1 - \displaystyle\frac{2}{m}}{6E} \right\}$

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

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

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

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 $\inline \displaystyle \pm \;s,\;0$ and by substitution:

Shear Straing Energy

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

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