What is the largest principle strain for the strain element shown on the left?

The given strain components are,

       ε_x = 310×10^-6 = 310 μ
     ε_y = -140×10^-6 = -140 μ
     γ_xy = 260×10^-6 = 260 μ

where m represents micro-strains.

The principal strain equations could be used directly, but another way is to find the principal angles, and then use them to find the principal strains. The principle angles for this strain state are

     tan 2θ_p =  γ_xy / (ε_x - ε_y)
                = 260/(310 + 140) = 0.5778
     θ_p = (tan^-1 0.5778) / 2
     θ_p = 15.0°, 105.0°

Notice there are two possibilities. Both will give the same principal strains as will be shown below. For 15^o, the two normal strains are,

     ε_x' = [310 + (-140)] / 2 + [310 - (-140) ] / 2 cos30
              + 260/2 sin30
         = 85 + 194.9 + 65.0 = 344.9 μ

     ε_y' = [310 + (-140)] / 2 - [310 - (-140)] / 2 cos30
             - 260 / 2 sin30
         = 85 - 194.9 - 65 = -174.9 μ

The principal strains are,

     ε₁ = ε_x' = 344.9 μ @ 15°
     ε₂ = ε_y' = -174.9 μ @ 15°

     ε_x' = [310 + (-140)] / 2 + [310 - (-140)] / 2 cos210
              + 260/2 sin210
         = 85 - 194.9 - 65 = -174.9 μ

     ε_y' = [310 + (-140)] / 2 - [310 - (-140)] / 2 cos210
             - 260/2 sin210
         = 85 + 194.9 + 65.0 = 344.9 μ

The principal strains are,

     ε₁ = ε_y' = 344.9 μ @ 105°
     ε₂ = ε_x' = -174.9 μ @ 105°

As expected, the principal strain values are the same, the new coordinate, x' and y' are just reversed. But actually, the direction are the same sine this angle is 90 degree off from the previous angle.