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.

i would think klcio3

i. like it i'm gonna edit it tomorrow