APPLICATIONS/ MOHR’S CIRCLE

Special Forms of the Stress Tensor Uniaxial stress is given by: Fig. 1. Examples of special state of stress: (a) Uniaxial; (b) biaxial; (c) hydrostatic; (d) pure shear

Biaxial Stress:

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

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.

Applying the transformation law, we have: old new l11 l12 l21 l22 l11 = ; l12 = ; l22 = ; l21 = - From Eq. 4-14 s 11 = 2l11l12s12 = s12 s 22 = 2l21l22s12 = -s12 Hence, we have =

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

Important Stresses in Plasticity It is often useful to decompose ij into two components: (4-33) Deviatoric, or Deviator, or Shear stress Tensor Spherical, or Hydrostatic, or Dilatational stress Tensor [Responsible for Failure] [Does Not cause Deformation]

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. (4-34)

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.

2-D TRANSFORMATION Recall the three equations obtained for the 2-D transformation of stress: (4-35a) (4-35b) (4-35c)

Two angles n separated by 90o satisfies Eq. 3-36 ***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 (4-36)

Note that the shear stress at the n orientation is found to be zero. 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 (4-37)

The corresponding shear stress from Eq. 4-35c is 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. (4-38) (4-39)

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 (4-40)

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.

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

Solution (B) 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.

The Development is as follows: Mohr’s Circle 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 (4-41)

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: (4-42) (4-43) (4-44)

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

Problem 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

Solution 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.

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:

The angle with the  - axis is A counter-clockwise rotation of the diameter of the circle by this 2n 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

Mohr’s circle corresponding to the state of stress , MPa 60 (60, 40.3) (25, 20) (100.3, 0) , MPa (19.7, 0) 2n (95, -20) (60, -40.3) Mohr’s circle corresponding to the state of stress