Strain Gages on Hydraulic Chair Piston

A hydraulic piston is used to move a dental chair up and down. To help assist in designing the piston, three strain gages have been attached directly to the piston as shown in the diagram on the left. When the chair is raised, the strain gages give strains as

     ε_a = 80×10^-6 m/m
     ε_b = 60×10^-6 m/m
     ε_c = 20×10^-6 m/m

(1) Determine the principal strains and the principal strain directions for the given set of strains.

(2) Compute the strain in a direction -60° (clockwise) with the x axis.

The first step is to transform the strains to the standard x-y coordinate directions. This can be done by rotating the three unknown strains,  ε_x, ε_y and γ_xy, into the three known stains ε_a, ε_b and ε_c. Using the equations for a 45° strain rosette, these strains are,

     
Solving for ε_x, ε_y and γ_xy gives,

     ε_x = ε_a    ε_y = ε_c    γ_xy = 2ε_b - (ε_a + ε_c)

Substituting the actual strains into the equations gives,

     ε_x = 80×10^-6 m/m
     ε_y = 20×10^-6 m/m
     γ_xy = 20×10^-6 rad 

Mohr's Circle for Principal Strains

To determine the principal strains, the strains are plotted on a Mohr's circle as shown in the diagram on the left. The coordinates of point A are ε_x = 80 and γ_xy/2 = 10 and the coordinates of B are ε_y = 20 and -γ_xy/2 = -10. The x axis is represented by the radius CA, and the y axis by the radius CB.

Radius of the circle is

     R = ((80 - 50)² + (10)²)^1/2 = 31.62

The principal strains are,

     ε₁ = (50 + 31.62)×10^-6
         = 81.62 ×10^-6 m/m

     ε₂ = (50 - 31.62)×10^-6
         = 18.38 ×10^-6 m/m

The angle between the maximum strain axis and the x axis is,

     θ = (1/2) (∠ACD) = (1/2) tan^-1(10/30) = 9.217 °

To compute the strains in a new direction, Mohr's circle can again be used. For strains in a direction 60° clockwise with the x axis, rotate the x axis two times 60° (i.e., 120°) as shown with line FG. The strain at that direction are,

      ε_x^* = {50 - 31.62 cos(180 - 120 - 18.43)}×10^-6
           = (50 - 23.66) ×10^-6 m/m
           = 26.34 ×10^-6 m/m

To compute the strain in a direction 90° with this line, it can be rotated two times 90° or 180°, which is diametrically opposite to it.

      ε_y^* = {50 + 31.62 cos(180 - 120 - 18.43)}×10^-6
              = (50 + 23.66) ×10^-6 m/m
          = 73.66 ×10^-6 m/m

Corresponding shear strain is,

     γ_xy^* = 2 {31.62 sin(180 - 120 - 18.43)}×10^-6
           = 41.96×10^-6 rad

The new strains are represented as point F and G on the diagram at the left.