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Stress and Strain Dr. A.B. Zavatsky

Mohrs Circle for Plane Stress Transformation equations for plane stress. Procedure for constructing Mohrs circle. Stresses on an inclined element. Principal stresses and maximum shear stresses. Introduction to the stress tensor.

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Stress Transformation Equations

x x

y

xy

yx

xy

yx

xy

x

y

x1

y1

x1

x1

y1

y1

x y1 1y x1 1

x y1 1

y x1 1

2cos2sin2

2sin2cos22

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1

xyyx

yx

xyyxyx

x

If we vary from 0 to 360, we will get all possible values of x1 and x1y1for a given stress state. It would be useful to represent x1 and x1y1 as functions of in graphical form.

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To do this, we must re-write the transformation equations.

2cos2sin2

2sin2cos22

11

1

xyyx

yx

xyyxyx

x

Eliminate by squaring both sides of each equation and adding the two equations together.

22

211

2

1 22 xyyx

yxyx

x

Define avg and R

22

22 xyyxyx

avg R

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Substitue for avg and R to get

2211

21 Ryxavgx

which is the equation for a circle with centre ( avg,0) and radius R.

This circle is usually referred to as Mohrs circle, after the German civil engineer Otto Mohr (1835-1918). He developed the graphical technique for drawing the circle in 1882.

The construction of Mohrs circle is one of the few graphical techniques still used in engineering. It provides a simple and clear picture of an otherwise complicated analysis.

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Sign Convention for Mohrs Circle

x

y

x1

y1

x1

x1

y1

y1

x y1 1y x1 1

x y1 1

y x1 1

2211

21 Ryxavgx

x1

x1y1

R

2avg

Notice that shear stress is plotted as positive downward.

The reason for doing this is that 2 is then positive counterclockwise, which agrees with the direction of 2 used in the derivation of the tranformation equations and the direction of on the stress element.

Notice that although 2 appears in Mohrs circle, appears on the stress element.

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Procedure for Constructing Mohrs Circle

1. Draw a set of coordinate axes with x1 as abscissa (positive to the right) and x1y1 as ordinate (positive downward).

2. Locate the centre of the circle c at the point having coordinates x1 avg and x1y1 0.

3. Locate point A, representing the stress conditions on the x face of the element by plotting its coordinates x1 x and x1y1 xy. Note that point A on the circle corresponds to = 0.

4. Locate point B, representing the stress conditions on the y face of the element by plotting its coordinates x1 y and x1y1 xy. Note that point B on the circle corresponds to = 90.

5. Draw a line from point A to point B, a diameter of the circle passing through point c. Points A and B (representing stresses on planes at 90 to each other) are at opposite ends of the diameter (and therefore 180 apart on the circle).

6. Using point c as the centre, draw Mohrs circle through points Aand B. This circle has radius R.

(based on Gere)

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x x

y

xy

yx

xy

yx

xy

x1

x1y1

avg

cR

A ( =0)

x

xy

AB ( =90)

y

- xy

B

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Stresses on an Inclined Element 1. On Mohrs circle, measure an angle 2 counterclockwise from

radius cA, because point A corresponds to = 0 and hence is the reference point from which angles are measured.

2. The angle 2 locates the point D on the circle, which has coordinates x1 and x1y1. Point D represents the stresses on the x1 face of the inclined element.

3. Point E, which is diametrically opposite point D on the circle, is located at an angle 2 + 180 from cA (and 180 from cD). Thus point E gives the stress on the y1 face of the inclined element.

4. So, as we rotate the x1y1 axes counterclockwise by an angle , the point on Mohrs circle corresponding to the x1 face moves counterclockwise through an angle 2 .

(based on Gere)

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x x

y

xy

yx

xy

yx

xy

A ( =0)

x1

x1y1

c

R

B ( =90)A

B

x

y

x1

y1

x1

x1

y1

y1

x y1 1y x1 1

x y1 1

y x1 1

2

D

D ( )

x1

x1y1

y1

- x1y1

2 +180E ( +90)

E

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Principal Stresses

x x

y

xy

yx

xy

yx

xy

A ( =0)

x1

x1y1

c

R

B ( =90)A

B

x

y

1

p

1

2

2

p2

P1

2 p1

P2

2 p2

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Maximum Shear Stress

x x

y

xy

yx

xy

yx

xy

A ( =0)

x1

x1y1

c

R

B ( =90)

A

B

2 s

s

max

min

x

y

s

s

s

s

s

maxmax

max

max

Note carefully the directions of the

shear forces.

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Example: The state of plane stress at a point is represented by the stress element below. Draw the Mohrs circle, determine the principal stresses and the maximum shear stresses, and draw the corresponding stress elements.

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

152

50802

yxavgc

c

A ( =0)

A

B ( =90)B

6.692565

251550

22

22

R

R

R

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MPa6.84MPa6.54

6.6915

2

1

2,1

2,1 Rc

maxMPa15

MPa6.69

s

maxc

R

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84.6 MPa

84.6 MPa

54.6 MPa

x

y

54.6 MPa

10.5o

100.5o

c

A ( =0)

B ( =90)R

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

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2 2

2 1

5.105.100

2011800.2120.212

3846.01580

252tan

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1

2

2

2

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c

A ( =0)

B ( =90)R

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

2 2

2

15 MPa

15 MPa

15 MPa

x

y

15 MPa

69.6 MPa

-34.5o55.5o

2 smax

5.550.111900.212

0.212

max

max

2

s

s

2 smin

taking sign convention into account

5.340.69)0.2190(2

0.212

min

min

2

s

s

max

min

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80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPaA ( =0)

B ( =90)25.8 MPa

25.8 MPa

4.15 MPa

x

y

4.15 MPa68.8 MPa

x1

y1

-30o

Example: The state of plane stress at a point is represented by the stress element below. Find the stresses on an element inclined at 30 clockwiseand draw the corresponding stress elements.

-60

-60+180

C ( = -30)

CD ( = -30+90)

D

2 2

x1 = c R cos(2 2+60)y1 = c + R cos(2 2+60)

x1y1= -R sin (2 2+60)x1 = -26

y1 = -4x1y1= -69

2

= -302 = -60

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A ( =0)

B ( =90)

Principal Stresses 1 = 54.6 MPa, 2 = -84.6 MPaBut we have forgotten about the third principal stress!

Since the element is in plane stress ( z = 0), the third principal stress is zero.

1 = 54.6 MPa2 = 0 MPa3 = -84.6 MPa

123This means three Mohrs circles can be drawn, each based on two principal stresses:

1 and 3

1 and 2

2 and 3

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1

3

1

3

123

1

3

1

3

113

3

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x x

y

xy

yx

xy

yx

xy

The stress element shown is in plane stress.What is the maximum shear stress?

A

B

x1

x1y1

A

B

3 12

22131

)3,1max(

221

)2,1max(

22232

)3,2max(

overall maximum

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zzzyzx

yzyyyx

xzxyxx

y

z

xxx xx

yy

yy

zz

xy

yx

xzzx

zy

yz

Introduction to the Stress Tensor

Normal stresses on the diagonalShear stresses off diagaonal

xy = yx, xz = zx, yz = zy

The normal and shear stresses on a stress element in 3D can be assembled into a 3x3 matrix known as the stress tensor.

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From our analyses so far, we know that for a given stress system, it is possible to find a set of three principal stresses. We also know that if the principal stresses are acting, the shear stresses must be zero. In terms of the stress tensor,

zzzyzx

yzyyyx

xzxyxx

3

21

000000

In mathematical terms, this is the process of matrix diagonaliza-tion in which the eigenvalues of the original matrix are just the principal stresses.

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80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

Example: The state of plane stress at a point is represented by the stress element below. Find the principal stresses.

50252580

yyxxyxM

We must find the eigenvalues of this matrix.

Remember the general idea of eigenvalues. We are looking for values of such that:Ar = r where r is a vector, and A is a matrix.Ar r = 0 or (A I) r = 0 where I is the identity matrix.

For this equation to be true, either r = 0 or det (A I) = 0.Solving the latter equation (the characteristic equation) gives us the eigenvalues 1 and 2.

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6.54,6.840462530

0)25)(25()50)(80(

05025

2580det

2So, the principal stresses are 84.6 MPa and 54.6 MPa, as before.

Knowing the eigenvalues, we can find the eigenvectors. These can be used to find the angles at which the principal stresses act. To find the eigenvectors, we substitute the eigenvalues into the equation (A I ) r= 0 one at a time and solve for r.

00

6.545025256.5480

00

50252580

yx

yx

1186.0

186.000

64.425256.134

yxyx

is one eigenvector.

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00

)6.84(502525)6.84(80

00

50252580

yx

yx

1388.5

388.500

6.13425256.4

yxyx

is the other eigenvector.

Before finding the angles at which the principal stresses act, we can check to see if the eigenvectors are correct.

CMCD

MCD

1

50252580

11388.5186.0

6.84006.54

033.0179.0967.0179.0

factors-co ofmatrix wheredet

1

1

1

C

AAC

C T

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6.84006.54

11388.5186.0

50252580

033.0179.0967.0179.0

D

To find the angles, we must calculate the unit eigenvectors:

183.0938.0

1388.5

983.0183.0

1186.0

And then assemble them into a rotation matrix R so that det R = +1.

1)183.0)(183.0()983.0)(983.0(det983.0183.0183.0983.0

RR

RMRDR Tcossinsincos

The rotation matrix has the form

So = 10.5, as we found earlier for one of the principal angles.

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6.54006.84

983.0183.0183.0983.0

50252580

983.0183.0183.0983.0

D

D

RMRD T

Using the rotation angle of 10.5, the matrix M (representing the original stress state of the element) can be transformed to matrix D (representing the principal stress state).

84.6 MPa

84.6 MPa

54.6 MPa

x

y

54.6 MPa

10.5o

100.5o

So, the transformation equations, Mohrs circle, and eigenvectors all give the same result for the principal stress element.

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15.48.688.688.25

866.05.05.0866.0

50252580

866.05.05.0866.0

yyxxyx

T

M

M

RMRM

866.05.05.0866.0

)30cos()30sin()30sin()30cos(

R

25.8 MPa

25.8 MPa

4.15 MPa

x

y

4.15 MPa68.8 MPa

x1

y1

-30o

Finally, we can use the rotation matrix approach to find the stresses on an inclined element with = -30.

Again, the transformation equations, Mohrs circle, and the stress tensor approach all give the same result.