MOHR'S CIRCLE FOR STRESS

MECHANICS - CASE STUDY
		Introduction
Stress State at the Bucket of a Tractor		To better understand the stress state for a new tractor, a structural analysis computer program was used to determine stresses in the tractor's front-end bucket. For the location shown, it was determined that the stresses are σx = 30 psi, σy = 10 psi, and τxy = 20 psi.
		Question
		Using Mohr's circle, what is 1) the principal direction and principal normal stresses, 2) maximum shear direction and the maximum shearing stress, and 3) the stresses at an angle of 15o?
		Approach
		Construct the basic Mohr's circle for the given stress state. Determine the principal direction from the Mohr's circle diagram, and then the principal stresses. Find the maximum shear stress direction from the Mohr's circle diagram. Rotate the stress element 15o using Mohr's circle

MECHANICS - THEORY
		Stress Rotation Equation
Rotated Stresses using Stress Rotation Equations		Previously, the stress transformation equations were developed to calculate the stress state at different orientations. These equations were       Plotting these equations show that every 180 degrees rotation, the stress state repeats. In 1882, Otto Mohr noticed that these relationships could be graphically represented with a circle. This was a tremendous help in the days of slide rulers when using complex equations, like the stress transformation equations, was time consuming.
		Mohr's Circle
		Mohr's circle is not actually a new derived formula, but just a new way to visualize the relationships between normal stresses and shear stresses as the rotation angle changes. To determine the actual equation for Mohr's circle, the stress transformation equations can be rearranged to give,       Each side of these equations can be squared and then added together to give Grouping similar terms and canceling other terms gives       Using the trigonometry identity, cos22θ + sin22θ = 1, gives
Basic Mohr's Circle for Stress		This is basically an equation of a circle. The circle equation can be better visualized if it is simplified to      where      This circle equation is plotted at the left using r and σave. One advantage of Mohr's circle is that the principal stresses, σ1, σ2 , and the maximum shear stress, τmax, are easily identified on the circle without further calculations.
		Rotating Stresses with Mohr's Circle
Stress Rotation with Mohr's Circle		In addition to identifying principal stress and maximum shear stress, Mohr's circle can be used to graphically rotate the stress state. This involves a number of steps. On the horizontal axis, plot the circle center at σavg = (σx + σy)/2. Plot either the point (σx , τxy) or (σy , -τxy). Note the sign change if plotting σy and the vertical axis, τxy, is positive downward. Draw a line from the center to the point plotted in step two (blue line in the diagram). This line should extend from one side of the circle to the other. Radius, r, can now be measured from the graph. The circle itself can be drawn since the center and one point on the circle is known (compass works well for this). The principal stresses and maximum shear stress can be identified on the graph. The line drawn in step 3 can be rotated twice the rotation angle, 2θ, in the counter-clockwise direction. It is important that the angle is twice the desired rotated angle. The new stress state is the intersection of the new line (green in the diagram) and the circle.
		The angle, 2θp, for the principal stresses is simply the half the angle from the blue line to the horizontal axis. Remember, Mohr's circle is just another way to visualize the stress state. It does not give additional information. Both the stress transformation equations and Mohr's circle will give the exactly same values

MECHANICS - CASE STUDY SOLUTION
Initial Local Stress Element		For a particular point in a structure, the local stresses were found to be 30 psi and 10 psi in the horizontal (x) and vertical (y) directions, respectively. The shear stress was found to be 20 psi. To help understand this stress state, a Mohr's circle will be constructed and used to find the 1) principal direction and principal stresses, 2) maximum shear stress direction and the maximum shear stress, and 3) the stress state if the element is rotated 15 degrees (counter-clockwise).
		Basic Mohr's Circle
Basic Mohr Circle for Stress		To construct a Mohr circle for a given stress state, first find the average normal stress, which will be the location of the circle's center. For this problem, it is      σavg = (σx + σy)/2 = (30 + 10)/2 = 20 psi This value is plotted on the graph at the left. Remember, the average normal stress, σavg, is always on the horizontal axis. Next, at least one point on the circle's edge is needed to define the circle's radius. There are two points that can be found immediately without any calculation, (σx, τxy) or (σy, -τxy). Note that both σx and σy are plotted on the same horizontal axis. This may seem strange at first, but it works well since they are both normal stresses. Generally, a line is drawn between the two points and the center. All three should be in a straight line. If they are not, this is an early indication something is wrong. The circle radius can be determined by using the right triangle with vertices (20, 0), (30, 0) and (30, 20).
		Principal Direction and Principal Stresses
Principal Stresses and Direction		The principal direction is where the normal stresses, σx and σy are at a maximum or minimum. This condition is represented by the intersection of the circle and the horizontal axis (shown in the diagram as a orange line). To get to this condition, the current stress state (blue line), needs to be rotated to the horizontal by an angle of 2θp in the counter-clockwise direction. Using geometry in the diagram, this gives      tan(2θp) = 20/(30 - 20) = 2      θp = 31.72o The actual principal stresses can be found using the circle center and the radius. They are,      σ1 = σavg + r = 20 + 22.36 = 42.36 psi      σ2 = σavg - r = 20 - 22.36 = -2.36 psi
		Maximum Shear Stress Direction and Maximum Shear Stress
Maximum Shear Stresses and Direction		The maximum shear stress occurs at the top or bottom of the circle. Thus, the current stress state, blue line, needs to be rotated clockwise (negative direction) by angle 2θτ. From the circle geometry, this angle is      tan(-2θτ-max) = (30 - 20)/20 = 0.5      θτ-max= 13.28o The maximum shear stress (or minimum) is simply the radius of the circle which was found previously to be      τmax = r = 22.36 psi
		Arbitrary Rotation
Stress State at an Arbitrary Angle		Mohr's circle can also be used to find a new stress state for an arbitrary rotation angle. The new stress state is identified by rotating the current stress state (blue line) by twice the angle, 2θ. The new stress state is shown on the diagram as a green line. Positive rotation angles are counter-clockwise. Using geometry, the stress points on the circle are      σx′ = σavg + r cos(2θp - 2θ)           = 20 + 22.36 cos(63.44 - 30) = 38.66 psi      σy′ = σavg - r cos(2θp - 2θ)           = 20 - 22.36 cos(63.44 - 30) = 1.34 psi      τx′y′ = r sin(2θp - 2θ)           = -22.36 sin(63.44 - 30) = 12.32 psi The initial stress state (blue line) and rotated stress state (green line) are also shown in the diagram as a small stress element to understand the orientation of the Mohr's circle stresses

MECHANICS - EXAMPLE
Mohr's Circle		Example
		Which Mohr circle is correct for the given stress state
		Solution
		The given stress components for the stress element are,        σx = 30 MPa      σy = -60 MPa      τxy = 50 MPa
Rotated Stress Element		Mohr's Circle center will be at the normal stress average,      σaverage = [30 + (-60)]/2 = -15 MPa which is plotted on the diagram at the left. Next, a point on the circle is needed. Two different points can be used,      Point 1: (σx , τxy) = (30, 50)      Point 2: (σy, -τxy) = (-60, -50) Remember, the shear stress is plotted positive in the downward direction. From the plotted center and points on the circle, the radius (shear maximum) can be determined.      R = τmax = [(30+15)2 + 502]1/2 = 67.3 MPa Principal stresses are      σ1 = -15 + 67.3 = 52.3 MPa      σ2 = -15 - 67.3 = -82.3 MPa The principle direction is       tan 2θ1 = 50/(15+30)       2θ1 = 48.0o The correct diagram is 1)