The obtained stress which is acting from every direction is at 45 degree.
The transformation equations will be used to calculate the stresses acting on an inclined element.
(σ x+σ y )/2 =75 MPa
(σ x-σ y )/2 =35 MPa
Tor xy = 28 MPa
sin 2θ = sin 90° = 1
cos 2θ = cos 90° = 0
Substituting these values into Eqs
σ x1 =(σ x+σ y )/2+ (σ x-σ y )/2 cos2θ+ τxy sin2θ
= 130 MPa
τ x 1 y 1 = (σ x-σ y )/sin2θ + τxy cos2θ
σ y 1= (σ x+σ y )/2- (σ x-σ y )/cos2θ+ τxy sin2θ
= 47Mpa
Elements of stress
We can easily obtain the stresses acting on all sides of an element oriented at = 45° from these results.
The arrows indicate the true directions of the stresses. Take special note of the shear stress directions, which all have the same magnitude. Also, keep in mind that the sum of the normal stresses remains constant at 150 MPa.
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