The state of plane strain on an element is:
εx=-300 x10⁻⁶, εy=0, and γxy=150 x 10⁻⁶.
Determine the equivalent state of strain which represents (a) theprincipal strains, and (b) the maximum in-plane shear strainand the associated average normal strain. Specify theorientation of the corresponding elements for these states of strain with respect to the original element.

a. ε₁=-0.000317

   ε₂=0.000017

θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

θ= -13.28° and  76.72°

To determine the direction of ε₁ and ε₂

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain

=3.335 *10^-4

ε(avg) =150 *10^-6

orientation of γmax

θ = 31.71 or -58.29

To determine the direction of γmax

= 1.67 *10^-4

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