| Introduction | ||
 Strain State at a Single Location on an Oil Tanker | To better understand the strain state for an old oil tanker, strain gages where attached to the surface of the tanker at the location shown. With the help of the strain gage results, it was determined that the strains at that location were εx = -800 μ, εy = 400 μ, and γxy = 400 μ. | |
| Question | ||
| Using Mohr's circle, what is 1) the principal direction and principal normal strains, 2) maximum strain direction and the maximum shearing strain, and 3) the strains at an angle of 30o? | ||
| Approach | ||
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| Strain Rotation Equation | |||||
 Rotated Strains using Strain Rotation Equations | Previously, the Strain transformation equations were developed to calculate the strain state at different orientations. These equations were  Plotting these equations show that every 180 degrees rotation, the strain 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 strain 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 strains and shear strains as the rotation angle changes. To determine the actual equation for Mohr's circle, the strain transformation equations can be rearranged to give,  Each side of these equations can be squared and then added together to give  Grouping like terms and canceling other terms gives  Using the trigonometry identity, cos22θ + sin22θ = 1, gives  | ||||
 Basic Mohr's Circle for Strain | This is basically an equation of a circle. The circle equation can be better visualized if it is simplified to
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| Rotating Strains with Mohr's Circle | |||||
 Strain Rotation with Mohr's Circle | In addition to identifying principal strain and maximum shear strain, Mohr's circle can be used to graphically rotate the strain state. This involves a number of steps.
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| The angle, 2θp, for the principal strains is simply half the angle from the blue line to the horizontal axis. Remember, Mohr's circle is just another way to visualize the strain state. It does not give additional information. Both the strain transformation equations and Mohr's circle will give the exactly same values | |||||
 Initial Local Strain Element | For a particular point on the oil tank, the local strains where found to be -800 μ and 400 μ in the horizontal (x) and vertical (y) directions, respectively. The shear strain was found to be 400 μ. Recall, strain is unit-less and the symbol μ represents 10-6. To help understand this strain state, a Mohr's circle will be constructed and used to find the 1) principal direction and principal strains, 2) maximum shear strain direction and the maximum shear strain, and 3) the strain state if the element is rotated 30o (counter-clockwise). | |
| Basic Mohr's Circle | ||
 Basic Mohr Circle for Strain (all strains are micro-strains, μ) | To construct a Mohr circle for a given strain state, first find the average normal strain, which will be the location of the circle's center. For this problem, it is εavg = (εx + εy)/2 = (-800 μ + 400 μ)/2 = -200 μ This value is plotted on the graph at the left. Remember, the average normal strain, εavg, 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, (εx, γxy/2) or (εy, -γxy/2). Note that both εx and εy are plotted on the same horizontal axis. This may seem strange at first, but it works well since they are both normal strains. 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 a right triangle with vertices (-200, 0), (400, 0) and (400, -200).  | |
| Principal Direction and Principal Strains | ||
 Principal Strains and Direction (all strains are micro-strains, μ) | The principal direction is were the normal strains, εx and εy 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 strain state (blue line), needs to be rotated to the horizontal by an angle of 2θp in the clockwise direction (negative direction). Using geometry in the diagram, this gives tan(2θp) = 200 μ / [-800 μ - (-200 μ)] = -1/3 θp = -9.22o The actual principal strains can be found using the circle center and the radius. They are, ε1 = εavg + r = -200 μ + 632.5 μ = 432.5 μ ε2 = εavg - r = -200 μ - 632.5 μ = -832.5 μ It is interesting to see that the magnitude of ε2 is actually larger than the magnitude of ε1. | |
| Maximum Shear Strain Direction and Maximum Shear Strain | ||
 Maximum Shear Strains and Direction (all strains are micro-strains, μ) | The maximum shear strain occurs at the top or bottom of the circle. Thus, the current strain state, blue line, needs to be rotated counter-clockwise (positive direction) by angle 2θγ. From the circle geometry, this angle is tan(2θγ-max) = (800 μ - 200 μ) / 200 μ = 3 θγ-max= 35.78o The maximum shear strain (or minimum) is simply twice the radius of the circle, which was found previously to be γmax / 2 = r = 632.5 μ γmax = 1,265 μ | |
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| Arbitrary Rotation | ||
 Strain State at 30o Angle (all strains are micro-strains, μ) | Mohr's circle can also be used to find a new strain state for an arbitrary rotation angle. The new strain state is identified by rotating the current strain state (blue line) by twice the angle, 2θ. The new strain state is shown on the diagram as a green line. Positive rotation angles are counter-clockwise. Using geometry, the strain points on the circle are εx′ = εavg - r sin(90 - 2θp - 2θ) = -200 μ - 632.5 μ sin(90 - 18.43 - 60) = -326.9 μ εy′ = εavg + r sin(90 - 2θp - 2θ) = -200 μ + 632.5 μ sin(90 - 18.43 - 60) = -73.14 μ γx′y′ / 2 = r cos(90 - 2θp - 2θ) = 632.5 μ cos(90 - 18.43 - 60) = 619.6 μ γx′y′ = 1,239 μ The initial strain state (blue line) and rotated strain state (green line) are also shown in the diagram as small strain element to understand the orientation of the Mohr's circle strains | |
 Mohr's Circle - Strain | Example | |
| Determine what diagram is the correct Mohr's circle for the given strain state. | ||
| Solution | ||
 Mohr's Circle - Strain | The first step in constructing a Mohr's Circle is to locate its center along the normal strain axis (x-direction). center = (εx + εy)/2 = (100 + 400)/2 = 250μ Next, a point on the circle can be plotted, (εx, τxy/2) = (400μ, 100μ) or (εy, -τxy/2) = (100μ, -100μ) These points are shown on the diagram on the left. Now the radius can be determined, R2 = 1002 + (400 - 250)2 R = 180.3μ = 180.3 × 10-6 The current stress state is represented by a line at an angle of θ = tan-1(100/15) = 33.69o Diagram 3 is the correct representative of Mohr's circle for the given strain state | |
