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APPLICATIONS/MOHR’S CIRCLE

2

• Uniaxial stress is given by:

000

000

00

Special Forms of the Stress Tensor

Fig. 1. Examples of special state of

stress: (a) Uniaxial; (b) biaxial; (c)

hydrostatic; (d) pure shear

3

• Biaxial Stress:

0 0 0

00

00

2

1

4

• Hydrostatic Pressure:

• It is a occurs in liquids; it is a special case of triaxial stress, when the three principal stresses are equal.

p

p

p

00

00

00

5

• Pure Stress:

• It is a special case of biaxial stress, as will be seen by performing a 45o rotation.

• Students are required to show that the above statement is correct.

000

00

00

6

• Applying the transformation law, we have:

old

new l11 l12

l21 l22

l11 = ; l12 = ; l22 = ;

l21 = -

From Eq. 4-14

11 = 2l11l1212 = 12

22 = 2l21l2212 = -12

Hence, we have=

22

22

22

22

0

0

0

0

7

• Fig. 2. A state of pure shear stress and strain in an isotropic material.

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Important Stresses in Plasticity

• It is often useful to decompose ij into two components:

'''ijijijij

Deviatoric, or Deviator,or Shear stress Tensor

Spherical, or Hydrostatic,or Dilatational stress Tensor

)(

)(

)(

111212

231112

131211

m

m

m

m

m

m

00

00

00

(4-33)

[Responsible for Failure] [Does Not cause Deformation]

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• It can be found that m has to be equal to:

• The deviatoric (or shear) components of stress are responsible for failure (under shear) or distortion.

• The hydrostatic (or spherical) stress produces volume change and does not cause any plastic deformation. This explains why shrimp can live in great ocean depths without problems.

• The above stress components are often used in plasticity.

331332211 I

m (4-34)

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• Consider a shrimp under two conditions:(a) depth of 100 m, with hydrostatic pressure of 1.0 MPa

acting on it, and

(b) squeezed between our fingers, with an applied stress (uniaxial) of about 0.5 MPa.

– While the Hydrostatic pressure will not bother the shrimp, it will certainly crush under the applied tensile stress.

– The difference between the two cases is the deviatoric stress, which is 0 and 0.25 MPa for cases (a) and (b) respectively.

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2-D TRANSFORMATION

• Recall the three equations obtained for the 2-D transformation of stress:

2sin2cos22

' xyyxyx

x

2sin2cos22

' xyyxyx

y

2cos2sin2

'' xyyx

yx

(4-35a)

(4-35c)

(4-35b)

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• ***As in all Transformation cases, Eqs. 4-35 give the variation of and with direction in the material***.

• The direction being specified by the angle relative to the originally chosen x-y coordinate system.

• Taking the derivative d/d of Eq. 4-35(a) and equating the result to zero gives the coordinate axes rotations for the maximum and minimum values of .

• Two angles n separated by 90o satisfies Eq. 3-36

yx

xyn

22tan

(4-36)

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• The corresponding maximum and minimum normal stresses from Eq. 4-35, called the principal normal stresses can be obtained by substituting 4-36 into 4-35, and is given as:

• Note that the shear stress at the n orientation is found to be zero.

• Conversely, if the shear stress is zero, then the normal stresses are the principal normal stresses

2

2

21 22, xy

yxyx

(4-37)

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• Similarly, Eq. 4-35c and d/d = 0 gives the coordinate axes rotation for the maximum shear stress.

• The corresponding shear stress from Eq. 4-35c is

• This is the maximum shear stress in the x-y plane and is called the principal shear stress.

xy

yxs

22tan

(4-38)

2

2

max 2 xyyx

(4-39)

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max can be expressed in terms of the principal normal stresses 1 and 2 by substituting Eq. 4-39 into 4-37 to obtain

• The absolute value is necessary for max due to the two roots in Eq. 4-39

221

max

(4-40)

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Example

• At a point of interest on the free surface of an engineering component, the stresses with respect to a convenient coordinate system in the plane of the surface are x = 95, y = 25, xy = 20 MPa. Determine the principal stresses and the orientations of the principal planes.

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Solution (A)• Substitution of the given values into Eq. 4-36 gives the

angle to the coordinate axes of the principal normal stresses.

n = 14.9o (CCW)

Substituting into Eq. 4-37 gives the principal normal stresses

1, 2 = 60 40.3 = 100.3, 19.7 MPa

742

2tan

yx

xyn

22

21 )2

(2

, xyyxyx

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• Alternatively, a more rigorous procedure is to use = n = 14.9 in Eq. 4-35a, which gives

= x’ = 1 = 100.3 MPa

• Use of = n + 90o = 104.9o in Eq. 4-35a then gives the normal stress in the other orthogonal direction.

= y’ = 2 = 19.7 MPa

The zero value of at = n can also be verified by using Eq. 4-35c.

Solution (B)

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• A convenient graphical representation of the transformation equations for plane stress.

• On versus coordinates, these equations can be shown to represent a circle, called Mohr’s circle.

• The Development is as follows:– Isolate all terms of Eq. 4-35a containing on one side.

– Then square both sides of 4-35a and 4-35c and sum.

– Invoke simple trigonometric identities to eliminate , and we obtain

Mohr’s Circle

22

22

22 xyyxyx

(4-41)

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• This equation is of the form:

which is the equation of a circle on a plot of versus with center at the coordinates (a, b) and radius r, where

Comparison with Eq. 4-37 and 4-39, revealed that the maximum and minimum normal stresses can be written as:

222 rba (4-42)

2

2

2,0,

2 xyyxyx rba

(4-43)

max21 2,

yxra (4-44)

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• Confusion concerning the signs in Mohr’s circle can be resolved as follws:– For shear, the portion that causes clockwise rotation is

considered positive, and the portion that causes counter-clockwise rotation is considered negative

– For normal stresses, tension is positive, and compression is negative.

• Graduate students are required to read up the 3-D Mohr’s circle.

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• Repeat the previous problem using Mohr’s circle method. Recall that the original state of stress is

x = 95, y = 25, and xy = 20 MPa

Problem

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• The circle is obtained by plotting two points that lie at opposite ends of a diameter

– A negative sign is applied to xy for the point associated with x because the shear arrows on the same planes as x tend to cause a counter-clockwise rotation.

– Similarly, a positive sign is used for xy when associated with y, due to the clockwise rotation.

Solution

MPaxyx 20,95,,

MPaxyx 20,25,,

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• Recall Eq. 4-43

• The center of the circle is (a, b), which lies on the axis:

• The hypotenuse of the triangle/ radius of the circle is also the principal shear stress, which is given as:

MPayxctr 0,600,

2, .

2

2

2,0,

2 xyyxyx rba

MPar xyyx 3.402035

2222

2

max

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• The angle with the - axis is

• A counter-clockwise rotation of the diameter of the circle by this 2n gives the horizontal diameter that corresponds to the principal normal stress.

• Their values are obtained from Eq. 4-44:

1, 2 = 100.3 and 19.7 MPa

)(74.292

3520

2tan

CCWon

n

max21 2,

yxra

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(95, -20)

(60, 40.3)

(19.7, 0)(100.3, 0)

(25, 20)

60, MPa

, MPa2n

Mohr’s circle corresponding to the state of stress (60, -40.3)