Answer:
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.
