Evaluate the principal stresses and principal planes:
The state of stress at a point in a loaded solid is prescribed on two faces of an element whose shape is a triangular prism as display in Figure. Evaluate the principal stresses and principal planes.
Figure
Solution
Here, we have to first obtain the value of x and subsequent calculations will be a standard set.
Given σ_{y} = 48 MPa
τ_{yx} = 36 MPa ∴ τ_{xy} = - 36 MPa
σ30^{o} = 60 MPa (aspect angle of plane BC is + 30^{o})
σ30° = (σ_{x} +σ_{y}/2) +(σ_{x} -σ_{y}/2) cos (2× 30°) +τ_{xy} sin (2 × 30°)
60 = (σ_{x} + 48/2) + (σ_{x} - 48/2) cos 60^{o} + (- 36) sin 60^{o}
60 = σ_{x}/2 + 24 +σ_{x}/2 cos 60° -24 cos 60° -36 sin 60°
i.e. 60 = σ_{x}/2 (1+ cos 60°) +24 -24 cos 60° -36 sin 60°
∴ σ_{x} =2/1+0.5 [60-24-(-24×0.5)-(-36)× 0.866]
= 2/1.5[79.177] = 105.57 MPa
∴ σ_{1} = 76.835 + 46.093 =122.928 MPa
σ_{2} = 76.835 - 46.093 =30.742 MPa
Let the aspect angles of the principal planes be Φ
tan 2Φ = 2τ_{xy}/σ_{x} - σ_{y} = 2 × (- 36)/105.57 - 48 = -51.35° or 128.65°
Φ = - 25.675° or 64.325°
Substituting Φ = - 25.675^{o} (or 2Φ = - 51.35^{o}) in expression for σn
σn = (105.57 + 48/2 ) + (105.57 - 48/2) cos (- 51.35^{o} ) - 36 sin (- 51.35^{o} )
= 122.92 MPa
∴ Φ_{1} = - 25.675^{o }and Φ_{2} = 64.325^{o}