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

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

23287/Compound-SnS-p2-0017.png
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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 PM such that PL represents f_1 and PM f_2. Note that the positive direction (Tension) is to the right.
  • On LM as a Diameter draw a Circle with centre O
  • On this drawing f_2\;>\;f_1, but this is not a necessary condition.
23287/Compound-SnS-p2-0018.png
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  • The radius OL represents the plane of f_1\;(BC)
  • The radius OM represents the plane of f_2\;(AB)
  • The Plane AC is obtained by rotating AB through \theta and if OM on the Stress Circle is rotated through 2\theta in the same direction, then the radius OR is obtained. This will be shown to represent the plane AC. (Note that OR could equally well be obtained by rotating OL clockwise through 180^{0} - 2\theta corresponding to rotating BC clockwise through 90^{0} - \theta)
  • Draw RN perpendicular to 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

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

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

Also, the Resultant Stress is given by: 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: \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.

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

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

23287/Compound-SnS-p2-0019.png
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If \displaystyle \theta is the angle measured from the Plane of zero Stress then in the above diagram PL = f numerically. It will be measured to the left for Compression. PM = 0

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

f_\theta  = PN Compressive

s_\theta  = RN Positive

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

Principal Stresses Equal Tension And Compression.

Let \displaystyle \theta be the angle measured anticlockwise from the Plane of f Tensile.
23287/Compound-SnS-p2-0020.png
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PM = f to the right.

PL = f to the left.

Hence O coincides with P.

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

\displaystyle s_\theta  = RN. When \theta  = 45^{0},\;\;s_\theta reaches its maximum value (Numerically equal to 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 4\;tons/sq.in. and 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 \displaystyle \theta is inclined to the plane of 4\;tons/sq.in. compression. In the right hand diagram Pl = 6 and PM = 4. The maximum angle \displaystyle \phi is found when PR is a tangent to the Stress Circle. OR = 1 and PO = 5.

13108/img__ss_2_0021.jpg
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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
  • \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.

23287/Compound-SnS-p2-0023.png
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Consider the Strains produced by each Stress separately:

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

\displaystyle f_2 will cause:
  • Strain \displaystyle \frac{f_2}{E} in the direction of \displaystyle f_2
  • Strain \displaystyle -\:\frac {f_2}{m\;E} in the direction of \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 \displaystyle f_1
\;\;\;\;\;\;\;\;\;\;e_1 = \frac{f_1}{E} - \frac{f_2}{m\;E}
  • Strain in the direction of \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 f_1,\;\;f_2, and f_3 are given by :

23287/Compound-SnS-p2-0025.png
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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 \;\;\;f_3 = 0, Then \;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 6\;tons/sq.in.
Workings
Let the Stresses be f_1,\;f_2, and f_3 and the corresponding Strains 3k, 4k, and 5k.

Then :

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)

And

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 6\;tons/sq.in.

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

Principal Stresses Determined From Principal Strains.

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

Subtracting Equation (6) from (5)

From (5) and (7)

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 \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 f_1 and f_2 And,

Analysis Of Strain.

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

23287/Compound-SnS-p2-0026.png
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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 OP, of length r , is the diagonal of a rectangle which under the given Strains distorts into the dotted parallelogram. P moves to P'. It must be remembered that the actual Strains are small.

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 \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

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

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

In order to evaluate \displaystyle e_x,\;\;e_y and \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 \phi  = 0 and e_x = e_1, 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 \displaystyle 550\times 10^{-6}\;,\; - 100\times 10^{-6} and 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.

E = 30\times10^6\;lb.in^{-2} and \displaystyle\frac{1}{m} = 0.3
Workings
13108/img__ss_2_0027.jpg
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Taking the X-axis in the direction of the \displaystyle 550\times 10^{-6} Strain, \displaystyle e_x,\;\;e_y and \phi are determined from equation (11) with \displaystyle \theta  = 0,\;\;60^{0} and 120^{0} from the three measured Strains. Hence,

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}}

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}}

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), 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 \displaystyle e_1 and e_2 (To the 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, \;\;\;\;\;\theta \;= - 11.2^{0} or 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 \;\;\;\;\;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.

23287/Compound-SnS-p2-0028.png
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The diagram shows the relationship between \displaystyle e_x,\;e_y,\;\phi and \theta and the Principal Strains \displaystyle e_1 and e_2 as given by equations (12) and (14).

Note that \displaystyle PO = \frac{1}{2}(e_x + e_y) and 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 \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 P and at distances on either side proportional to the given Strains. From R on the central line (i.e. \displaystyle e_{120} in this case),lines are set off at \displaystyle 60^{0} and 120^{0} to the vertical, to cut the corresponding Strain Verticals in Q and S. The Strain Circle then passes through QRS and the Principal Strains are :

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

The radius OS gives the Strain condition in the X direction and the angle SOM = 22^{0}. The direction of \displaystyle e_1 is then at \displaystyle \frac{1}{2}\times 22 = 11^{0} clockwise from the X-axis and \displaystyle e_2 is at right angles to \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 x, y, z is under the action of three principal Stresses \displaystyle f_1,\;f_2, and f_3.

23287/Compound-SnS-p2-101-2.png
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Then if \displaystyle e_1,\;e_2, and e_3 are the corresponding linear Strains, the dimensions become : x+e_1\;x,\;\;y+e_2\;y, and 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: \;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 = \displaystyle\frac{(f_1 + f_2 + f_3)(1 - \displaystyle\frac{2}{m})}{E}

Strain Energy

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

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

= \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 \displaystyle f_3 = 0

Example:

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

Calculate the volumetric Strain and the resilience.

E = 6000\;tons/sq.in. and \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 = \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 1.5\times10^{-4}
  • The resilience is 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 \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 \displaystyle f_1,\;f_2, and 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 \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 \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 E and C will be discussed in "Elastic Constants".