For each of the stress state given determine the principal normal stresses and the maximum shear stress. Note that all units are in MPa.

(a.)  xx = - 100,  yy = - 100

(b.)  xx = 50,  yy = 50,  zz = 50

(c.)  xx = - 100

(d.)  xx = 50,  yy = 50

(e.)  xx = - 100,  yy = - 100,  zz = - 100

(f.)  xx = 100

(g.)  xy = 100

a) σ₁₂ = - 100 MPa

τmax = 0 MPa

We have a bi-axial compression.

b) We have a hydrostatic tension.

σx = σy = σz = 50

c) σ₁ = - 100 MPa

σ₂ = 0 MPa

τmax = - 50 MPa

We have an uni-axial compression.

d) σ₁₂ = 50 MPa

τmax = 0 MPa

We have a bi-axial tension.

e) We have a hydrostatic compression.

σx = σy = σz = - 100 MPa

f) σ₁ = 100 MPa

σ₂ = 0 MPa

τmax = 50 MPa

We have an uni-axial tension.

Explanation:

a) Given σx = σy = - 100 MPa; τxy = 0

The principal normal stress can be obtained as follows

σ₁₂ = ((σx+σy) / 2) + - √ (((σx-σy) / 2) ²+τxy²)

⇒ σ₁₂ = ((-100-100) / 2) + - √ (((-100 - (-100)) / 2) ² + (0) ²) = - 100 MPa

The maximum shear stress can be obtained as follows

τmax = √ (((σx-σy) / 2) ²+τxy²)

⇒ τmax = √ (((-100 - (-100)) / 2) ² + (0) ²) = 0 MPa

We have a bi-axial compression.

b) We have a hydrostatic tension.

σx = σy = σz = 50

c) Given σx = - 100 MPa; σy = τxy = 0

The principal normal stress can be obtained as follows

σ₁₂ = ((σx+σy) / 2) + - √ (((σx-σy) / 2) ²+τxy²)

⇒ σ₁₂ = ((-100+0) / 2) + - √ (((-100-0) / 2) ² + (0) ²) = - 50 + - (-50) = MPa

σ₁ = - 100 MPa

σ₂ = 0 MPa

The maximum shear stress can be obtained as follows

τmax = √ (((σx-σy) / 2) ²+τxy²)

⇒ τmax = √ (((-100-0) / 2) ² + (0) ²) = - 50 MPa

We have an uni-axial compression.

d) Given σx = σy = 50 MPa

The principal normal stress can be obtained as follows

σ₁₂ = ((σx+σy) / 2) + - √ (((σx-σy) / 2) ²+τxy²)

⇒ σ₁₂ = ((50+50) / 2) + - √ (((50-50) / 2) ² + (0) ²) = 50 MPa

The maximum shear stress can be obtained as follows

τmax = √ (((σx-σy) / 2) ²+τxy²)

⇒ τmax = √ (((50-50) / 2) ² + (0) ²) = 0 MPa

We have a bi-axial tension.

e) We have a hydrostatic compression.

σx = σy = σz = - 100 MPa

f) Given σx = 100 MPa; σy = τxy = 0

The principal normal stress can be obtained as follows

σ₁₂ = ((σx+σy) / 2) + - √ (((σx-σy) / 2) ²+τxy²)

⇒ σ₁₂ = ((100+0) / 2) + - √ (((100-0) / 2) ² + (0) ²) = 50 + - (50) = MPa

σ₁ = 100 MPa

σ₂ = 0 MPa

The maximum shear stress can be obtained as follows

τmax = √ (((σx-σy) / 2) ²+τxy²)

⇒ τmax = √ (((100-0) / 2) ² + (0) ²) = 50 MPa

We have an uni-axial tension.