Transcript

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P4 Stress and Strain Dr. A.B. ZavatskyHT08

Lecture 6

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

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Stress Transformation Equations

σx σx

σy

τxy

τyx

τxy

τyx

xy

x

y

σx1

θ

y1

x1

σx1

σy1

σy1

τx y1 1 τy x1 1

τx y1 1

τy x1 1

( )θτθ

σστ

θτθσσσσ

σ

2cos2sin2

2sin2cos22

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1

xyyx

yx

xyyxyx

x

+−

−=

+−

++

=

If we vary θ from 0° to 360°, we will get all possible values of σx1 and τx1y1for a given stress state. It would be useful to represent σx1 and τx1y1 as functions of θ in graphical form.

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

( )θτθ

σστ

θτθσσσσ

σ

2cos2sin2

2sin2cos22

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1

xyyx

yx

xyyxyx

x

+−

−=

+−

=+

−

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

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211

2

1 22 xyyx

yxyx

x τσσ

τσσ

σ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=+⎟⎟

⎠

⎞⎜⎜⎝

⎛ +−

Define σavg and R

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22 xyyxyx

avg R τσσσσ

σ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

+=

4

Substitue for σavg and R to get

( ) 2211

21 Ryxavgx =+− τσσ

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

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

The construction of Mohr’s 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 Mohr’s Circle

x

y

σx1

θ

y1

x1

σx1

σy1

σy1

τx y1 1 τy x1 1

τx y1 1

τy x1 1

( ) 2211

21 Ryxavgx =+− τσσ

σx1

τx1y1

R

2θσavg

Notice that shear stress is plotted as positive downward.

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

Notice that although 2θ appears in Mohr’s circle, θ appears on the stress element.

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Procedure for Constructing Mohr’s Circle

1. Draw a set of coordinate axes with σx1 as abscissa (positive to the right) and τx1y1 as ordinate (positive downward).

2. Locate the centre of the circle c at the point having coordinates σx1 = σavg and τx1y1 = 0.

3. Locate point A, representing the stress conditions on the x face of the element by plotting its coordinates σx1 = σx and τx1y1 = τxy. Note that point A on the circle corresponds to θ = 0°.

4. Locate point B, representing the stress conditions on the y face of the element by plotting its coordinates σx1 = σy and τx1y1 = −τxy. Note that point B on the circle corresponds to θ = 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 Mohr’s circle through points Aand B. This circle has radius R.

(based on Gere)

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σx σx

σy

τxy

τyx

τxy

τyx

xy

σx1

τx1y1

σavg

cR

A (θ=0)

σx

τxy

AB (θ=90)

σy

-τxy

B

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Stresses on an Inclined Element 1. On Mohr’s circle, measure an angle 2θ counterclockwise from

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

2. The angle 2θ locates the point D on the circle, which has coordinates σx1 and τx1y1. 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 2θ + 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 θ, the point on Mohr’s circle corresponding to the x1 face moves counterclockwise through an angle 2θ.

(based on Gere)

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σx σx

σy

τxy

τyx

τxy

τyx

xy

A (θ=0)

σx1

τx1y1

c

R

B (θ=90)A

B

x

y

σx1

θ

y1

x1

σx1

σy1

σy1

τx y1 1 τy x1 1

τx y1 1

τy x1 1

2θ

D

D (θ)

σx1

τx1y1

σy1

-τx1y1

2θ+180E (θ+90)

E

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Principal Stresses

σx σx

σy

τxy

τyx

τxy

τyx

xy

A (θ=0)

σx1

τx1y1

c

R

B (θ=90)A

B

x

y

σ1

θp1

σ1

σ2

σ2

θp2

P1

2θp1

P2

2θp2

σ1σ2

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Maximum Shear Stress

σx σx

σy

τxy

τyx

τxy

τyx

xy

A (θ=0)

σx1

τx1y1

c

R

B (θ=90)

A

B

2θs

σs

τmax

τmin

x

y

σs

θs

σs

σs

σs

τmaxτmax

τmax

τmax

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 Mohr’s 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

σ

τ

152

50802

−=+−

=+

== yxavgc

σσσ

c

A (θ=0)

A

B (θ=90)B

( )( ) ( )

6.692565

251550

22

22

=+=

+−−=

R

R

R

σ1σ2

MPa6.84MPa6.54

6.6915

2

1

2,1

2,1

−==

±−=

±=

σσ

σ

σ Rc

τmaxMPa15

MPa6.69

s

max−====

cR

στ

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84.6 MPa

84.6 MPa

54.6 MPa

x

y

54.6 MPa

10.5o

100.5o

σ

τ

c

A (θ=0)

B (θ=90)R

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

σ1σ2

2θ2

2θ1

°=°=

°=°+=°=

=−

=

5.105.100

2011800.2120.212

3846.01580

252tan

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1

2

2

θθ

θθ

θ

2θ

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σ

τ

c

A (θ=0)

B (θ=90)R

80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPa

2θ2

2θ

15 MPa

15 MPa

15 MPa

x

y

15 MPa

69.6 MPa

-34.5o

55.5o

2θsmax

°=

°=°+=°=

5.550.111900.212

0.212

max

max

2

s

sθ

θθ

2θsmin

taking sign convention into account

°−=

°−=−−=°=

5.340.69)0.2190(2

0.212

min

min

2

s

sθ

θθ

τmax

τmin

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80 MPa 80 MPa

50 MPa

xy

50 MPa

25 MPaA (θ=0)

σ

τ

B (θ=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 (θ = -30°)

CD (θ = -30+90°)

D

2θ2

σx1 = c – R cos(2θ2+60)σy1 = c + R cos(2θ2+60)

τx1y1= -R sin (2θ2+60)σx1 = -26

σy1 = -4τx1y1= -69

2θ

θ = -30°2θ = -60°

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σ

τ

A (θ=0)

B (θ=90)

Principal Stresses σ1 = 54.6 MPa, σ2 = -84.6 MPaBut we have forgotten about the third principal stress!

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

σ1 = 54.6 MPaσ2 = 0 MPaσ3 = -84.6 MPa

σ1σ2σ3This means three Mohr’s circles can be drawn, each based on two principal stresses:

σ1 and σ3

σ1 and σ2

σ2 and σ3

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σ1

σ3

σ1

σ3

σ

τ

σ1σ2σ3

σ1

σ3

σ1

σ3

σ1σ1σ3

σ3

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σx σx

σy

τxy

τyx

τxy

τyx

xy

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

A

B

σx1

τx1y1

A

B

σ3 σ1σ2

22131

)3,1max(σσστ =

−=

221

)2,1max(σστ −

=

22232

)3,2max(σσστ =

−=

overall maximum

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⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

zzzyzx

yzyyyx

xzxyxx

στττστττσ

y

z

xσxx σxx

σyy

σyy

σzz

τxy

τyx

τxzτzx

τzy

τyz

Introduction to the Stress Tensor

Normal stresses on the diagonalShear stresses off diagaonal

τxy = τyx, τxz = τzx, τyz = τzy

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

στττστττσ

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

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1

000000

σσ

σ

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

σττσ

We must find the eigenvalues of this matrix.

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

For this equation to be true, either r = 0 or det (A – λI) = 0.Solving the latter equation (the “characteristic equation”) gives us the eigenvalues λ1 and λ2.

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6.54,6.840462530

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

05025

2580det

2

−==−+

=−−−−−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−−−

λλλ

λλλ

λ

So, 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 – λI ) r= 0 one at a time and solve for r.

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−−−

00

6.545025256.5480

00

50252580

yx

yx

λλ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

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

λλ

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

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 T=′⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

θθθθ

cossinsincos

The rotation matrix has the form

So θ = 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, Mohr’s circle, and eigenvectors all give the same result for the principal stress element.

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⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−−−

=′

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−−⎟⎟⎠

⎞⎜⎜⎝

⎛ −=′

=′

11

15.48.688.688.25

866.05.05.0866.0

50252580

866.05.05.0866.0

yyxxyx

T

M

M

RMRM

σττσ

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛°−°−°−−°−

=866.05.0

5.0866.0)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 θ = -30°.

Again, the transformation equations, Mohr’s circle, and the stress tensor approach all give the same result.

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