Lecture 6 Mohr's Circle for Plane Stress

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1 P4 Stress and Strain Dr. A.B. Zavatsky HT08 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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Transcript of Lecture 6 Mohr's Circle for Plane Stress

  • 1

    P4 Stress and Strain Dr. A.B. ZavatskyHT08

    Lecture 6

    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

    xyyx

    xy

    x

    y

    x1

    y1

    x1

    x1

    y1

    y1

    x y1 1 y x1 1

    x y1 1 y x1 1

    ( )

    2cos2sin2

    2sin2cos22

    11

    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

    ( ) 221121 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 1 y x1 1

    x y1 1 y x1 1

    ( ) 221121 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

    xyyx

    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

    xyyx

    xy

    A (=0)

    x1

    x1y1

    c

    R

    B (=90)A

    B

    x

    y

    x1

    y1

    x1

    x1

    y1

    y1

    x y1 1 y 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

    xyyx

    xy

    A (=0)

    x1

    x1y1

    c

    R

    B (=90)A

    B

    x

    y

    1 p1

    1

    2

    2

    p2P1

    2p1

    P2

    2p2

    12

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    Maximum Shear Stress

    x x

    y

    xy

    yx

    xyyx

    xy

    A (=0)

    x1

    x1y1

    c

    R

    B (=90)

    A

    B

    2s

    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

    12

    MPa6.84MPa6.54

    6.6915

    2

    1

    2,1

    2,1

    ==

    =

    =

    Rc

    maxMPa15

    MPa6.69

    s

    max====

    cR

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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

    12

    22

    21

    ==

    =+==

    =

    =

    5.105.100

    2011800.2120.212

    3846.01580

    252tan

    21

    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

    22

    2

    15 MPa

    15 MPa

    15 MPa

    x

    y

    15 MPa

    69.6 MPa

    -34.5o55.5o

    2smax

    =

    =+==

    5.550.111900.212

    0.212

    max

    max

    2

    s

    s

    2smin

    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

    22

    x1 = c R cos(22+60)y1 = c + R cos(22+60)

    x1y1= -R sin (22+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

    xyyx

    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

    32

    1

    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

    2

    ==+

    =

    =

    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 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 T=

    =

    cossinsincos

    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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    =

    =

    =

    =

    11

    15.48.688.688.25

    866.05.05.0866.0

    50252580

    866.05.05.0866.0

    yyxxyx

    T

    M

    M

    RMRM

    =

    =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 = -30.

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