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2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 1

Stress Transformation, Principal Stresses and Mohr’s Circle (Chapter 5)

Consider Conservation of Linear Momentum and assume there is no mass flux and that body forces are negligible (only the traction terms remain). The result is static equilibrium of stresses on a differential volume:

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 2Consider a stress state where the only non-zero stresses applied to the volume occur in a 2-D plane oriented along the coordinate axes. This is called plane stress. For example, we can have plane stress in the x-y plane, the x-z plane, or the y-z plane.

For example, plane stress in the x-y plane is shown below:

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 3Assume plane stress (stresses in x-y plane only) so that the traction (stress) tensor becomes:

We now consider the question of resolving the given stress components in the x-y directions into stresses oriented in a different direction. For example, the resultant stresses on a plane which has a normal which makes an angle with the x-axis . A further question is whether there is some plane where the stresses are a maximum or are zero.

Consider the following example:

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 4Consider a column loaded by a compressive pressure (traction) as shown below in left figure. Draw two different free-bodys as shown:

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 5For free-body 1, where we cut the structure normal to the y axis, we obtain only the normal stress on the cutting plane. However, in fb#2, both normal and shear stresses exist.

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 6In the previous column example, we see that for a column loaded only in compression, if one takes a cutting plane at some angle, then shear stresses must exist on the cut plane. Consider a more general problem.

Suppose that we want to determine the internal stresses at some point “O” in the body. We could use an x-y coordinate system as shown in the middle figure, but we could also use and x’-y’ coordinate system as shown in the right figure. Will we get different results for . DEFINITELY, YES!

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 7Consider a solid body such as that shown below. Suppose that we start with the state of stress defined in x-y coordinates.

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 8We wish to determine the state of stress in the x'-y' coordinate system. We pass a cutting plane through point "O" which a unit normal vector as shown below to obtain:

A 2-D picture of the stress-state may be easier to work with:


Note that the unit vector as well as the x'-axis makes an angle with the x-axis (measured CCW from the x-axis). From our work on tractions in chapter 4, we have:


= traction vector

= Cauchy stresses


From Cauchy's formula, we have or

The expression is a general result in 3-D which gives the projection of Cauchy stress tensor onto a plane whose unit normal is given by .

The traction vector on the inclined face, as given above, is written terms of it's x and y components. It is much more


informative and useful to write the traction vector in terms a normal component and a parallel (shear) component as shown below. Or, because the unit normal and x'-axis are in the same direction, we would actually be determining the stresses

and in the x'-y' coordinate system as shown below:


The normal component (also called ) is first obtained from the dot product of the unit normal and the traction vector:

or as a vector

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 14and the shear (parallel) component is obtained from vector addition

Now, lets carry out these vector operations to obtain and . The unit normal vector in 2-D (x-y) is given by

Cauchy's formula in vector notation

Normal component of traction


Use the double angle trig identities to rewrite above equation


With the double angle trig identities, becomes

Shear component of traction

The shear component, , can be obtained from vector algebra, ie, or

or Evaluating the above and using the double angle formulas, we obtain for the shear component:


An alternate procedure to obtain the shear component:

Define the unit normal in the direction of y' to be :

The shear component is the component of in the direction of so that


These last two results allow us to transform the stresses from an x-y coordinate system to an x'-y' coordinate system and are called the stress transformation equations.

We could obviously use either the notation or , and or . We will choose to use and . After squaring

both sides of the and equations, adding the results to obtain one equation and using trig identities, we obtain the following

The above is similar to the equation of a circle of radius r located at x=a and y=b, ie,

(x-a)2 + (y-b)2 = r2


Thus, the relation defining the normal and shear components (in terms of the x and y stresses) can be drawn as a circle if we choose the following:

This leads to graphical representation of the stress transformation equations known as Mohr's Circle.


Important note on sign convention for shear stress used in constructing Mohr’s Circle.

When defining the Cauchy stress, a positive shear stress on the positive x face was in the positive y coordinate direction.

Because the calculation for the shear component above, ie,

which involves a square root, there is an

uncertainty in the + direction of the shear component. In order for the Mohr’s circle graphical representation to be used properly, we must adopt a sign convention:

Shear stresses on opposite faces that form a clockwise couple about the center are positive on the Mohr’s circle .


A positive (in the stress tensor) is plotted on Mohr's circle as negative for the x-face and positive for y-face.


Mohr’s Circle graphically represents the Cauchy formula for . It allows one to graphically determine the normal and shear stress on any plane relative to the x-y axes.

Its most important use is to determine the principal stresses (the maximum and minimum values of the normal stress, , where there is no shear stress), the maximum shear stress,

, and the orientation of the planes on which these occur.

The principal stress is the normal stress that occurs on a plane where no shear stress exists.


Construction of Mohr’s Circle (method 1)

1. Locate center on axis at .

2. Draw circle with radius .

3. Locate the two points on circle with values of the stress components on the x-face and y-face. These two points lie on a diameter line of circle which passes through the center.

4. Determine max/min values of and and their plane orientation (angle from x or y face). Remember: angles on Mohr’s circle are twice the real world.


Construction of Mohr’s Circle (method 2)

1. Plot the values of normal and shear stress from the x-face ( and ) on the and axes. Observe Mohr’s circle assumption

on positive shear (shear is positive if moment due to shear is CW).

2. Plot the values of normal and shear stress from the y-face ( and ) on the and axes.

3. The above two points form the diameter of Mohr’s circle whose center is located where the diameter line intersects the axis.

4. Determine max/min values of and and their plane orientation (angle from x or y face). Remember: angles on Mohr’s circle are twice the real world.


Some reminders:1. For determining and from the defining equations, the

angle between the normal to any plane and the x-axis is defined positive in the CCW direction from the x-axis.

2. When plotting shear stresses on Mohr’s circle, a shear stress is considered positive if it produces a CW moment.

3. All angles on Mohr’s circle are twice the real world [the x- and y-face stresses are 90 in the x-y coordinate system, but 180 apart on Mohr’s circle (opposite ends of circle diameter)].

4. The planes of principal stress and maximum shear stress are 90 apart on Mohr’s circle and thus 45 apart in the real world!!!!

5. The angle defining the planes where the maximum values of and occur can also be obtained by calculus. If


Then taking the derivative with respect and setting equal to zero gives the max/min values:

The above can be solved for two roots in the range: -90+90 which define the normal for the planes of max/min normal stress with respect to the x-axis. (see the MAPLE example)

The max value of shear stress can be obtained in a similar manner. The defining plane of max shear is always 45 from plane of max/min normal stress.

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 29Mohr's Circle Example. Consider the plane stress state given by the stress tensor:

1. Determine center of circle, point A.

2. Determine point B, x-face =(50,-10)


3. Determine point C, y-face =(20,+10)

4. Draw Mohr’s circle (plot x face first with :

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 315. Compute principal stresses and max shear stress.

Note that we have two principal stresses: and . These are located 180 on Mohr's circle, but 90 in the real world. In relation to the stress transformation equations, is the normal stress in the x'-axis direction (which is oriented at an angle to


the x-axis) and is the normal stress in the y'-axis direction. The planes of maximum shear occur 45 (real world) from the planes of principal stress.Plot the principal stresses and max shear stresses on Mohr’s Circle:


6. Identify orientation of planes for principal stresses relative to x-face plane.


The plane for the second principal stress is CCW from x-face (real world). Note: In the above calculation, a formula was written down and used for . This is discouraged. It is far simpler (and usually less mistakes) to look at Mohr’s circle and apply trigonometry to calculate the angles.

7. Identify orientation of planes for maximum shear stress (bottom of circle) relative to x-face plane.


OR, plane of max shear stress (bottom of circle) is 45 CW in the real world from the principal stress plane with .Note:

Note that the plane of max shear stress will generally have non-zero normal stress! In this case, the plane of max shear stress has a normal stress .

8. Draw three free bodies: with the x-y stresses, with the principal stresses, and with the max shear stresses.

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 37Note the following: To obtain the orientation for principal stresses, the x-face has

been rotated 16.85 deg CCW in the real world (33.7 deg on Mohr’s circle).

To obtain the orientation for maximum shear stresses, the x-face has been rotated 28.15 deg CW in the real world (56.3 deg on Mohr’s circle), OR, equivalently a rotation of 45 deg CW in the real world from the principal stress plane.


Generalized Plane Stress

In the previous development of the Mohr’s Circle, we considered only the stresses in the x-y plane that resulted in two principal stresses and (remember, that principal stresses occur on planes where there is no shear stress). We can also use Mohr’s Circle for the case when the third plane (for example, the z plane) contains only a normal stress . The stress tensor is


Since there are no shear stresses on the z plane, is a principal stress, i.e., .

We note that in the x-y plane, Mohr’s circle is defined by a circle with coordinates of ( , 0) and ( , 0). Thus, the circle is (or can be) defined by the principal stresses. It follows that the Mohr’s circle for the x-z plane would also be formed the principal stresses in the x-z plane, in this case

and . Similarly, Mohr’s circle in the y-z plane is defined by and . We can draw all three of these circles on one diagram as shown below.

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 41Mohr's Circle Generalized Plane Stress Example. Consider the plane stress state given by the stress tensor:

Steps 1-7 will be identical to the previous example for the x-y plane (since both examples have exactly the same stress components in the x-y plane). Hence, for the x-y plane we obtain Mohr’s circle:


To complete the solution we note that there are no shear stresses in the z plane and hence is automatically a principal stress. Thus, ksi. Adding the third principal stress to Mohr’s circle, we obtain:


Hence, the principal stresses are . The maximum shear stress is the radius of the largest Mohr’s


circle. Thus, when generalized

plane stress is considered.

2001, W. E. Haisler Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle 45Some practice problems for Plane and Generalized Plane Stress:

1. (plane stress in x-y plane)

2. (plane stress in y-z plane)

3.4.5.6.7. (hydrostatic

compressive stress)

8. (generalized plane stress)

9. (generalized plane stress)

10.


Note on Three-Dimensional Stress Analysis

For a general stress tensor (3-D), it can be shown that the principal stresses are defined by the following eigenvalue problem:

or

The above yields a cubic equation in which can be solved for the three principal stress components. (See Sec. 2.2.7 of Introduction to Aerospace Structural Analysis by Allen and Haisler).

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