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 30^o (counter-clockwise).

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

     ε_avg = (ε_x + ε_y)/2 = (-800 μ + 400 μ)/2 = -200 μ

This value is plotted on the graph at the left. Remember, the average normal strain, ε_avg, 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, (ε_x, γ_xy/2) or (ε_y, -γ_xy/2). Note that both ε_x and ε_y 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).

The principal direction is were the normal strains, ε_x and ε_y 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 2θ_p in the clockwise direction (negative direction). Using geometry in the diagram, this gives

     tan(2θ_p) = 200 μ / [-800 μ - (-200 μ)] = -1/3

     θ_p = -9.22^o

The actual principal strains can be found using the circle center and the radius. They are,

     ε₁ = ε_avg + r = -200 μ + 632.5 μ = 432.5 μ

     ε₂ = ε_avg - 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.

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 2θ_γ. From the circle geometry, this angle is

     tan(2θ_γ-max) = (800 μ - 200 μ) / 200 μ = 3

     θ_γ-max= 35.78^o

The maximum shear strain (or minimum) is simply twice the radius of the circle, which was found previously to be

     γ_max / 2 = r = 632.5 μ

     γ_max = 1,265 μ

Strain State at 30^o 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

     ε_x′ = ε_avg - r sin(90 - 2θ_p - 2θ)

        = -200 μ - 632.5 μ sin(90 - 18.43 - 60) = -326.9 μ

     ε_y′ = ε_avg + r sin(90 - 2θ_p - 2θ)

        = -200 μ + 632.5 μ sin(90 - 18.43 - 60) = -73.14 μ

     γ_x′y′ / 2 = r cos(90 - 2θ_p - 2θ)

        = 632.5 μ cos(90 - 18.43 - 60) = 619.6 μ

     γ_x′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.