Question 4.3.1: If the state of stress for an element is shown in Figure 4.2...
If the state of stress for an element is shown in Figure 4.23a, we may also express the state of stress on a wedge of angle α = 22.5 ̊. Because this wedge (ABC) is part of the original element, the stresses on faces AC and BC are known. They appear again on the wedge in Figure 4.19b. The unknown normal and shear stresses acting on face AB, σα and τα are what we want to find.
Face AB has area dA, in m² . The area corresponding to line AC is dA × cos α = 0.924dA; the area corresponding to line BC is dA × sin α = 0.383dA.
Next we can obtain the forces on the faces, F_i in Figure 4.23c, by multiplying the stresses by their respective areas. All F_i are in MN.
F_1 = 3 MPa × 0.924dA = 2.78dA.
F_2 = 2 MPa × 0.924dA = 1.85dA.
F_3 = 2 MPa × 0.383dA = 0.766dA.
F_4 = 1 MPa × 0.383dA = 0.383dA.
To keep the wedge in equilibrium, the unknown forces due to unknown normal and shear stresses must balance these forces:
\sum{F_N}=0 N = F_1 \cos α – F_2 \sin α – F_3 \cos α + F_4 \sin α = 1.29dA .
\sum{F_S}=0 S = F_1 \sin α + F_2 \cos α – F_3 \sin α – F_4 \cos α = 2.12dA .
Since forces N and S act on the plane defined by AB, whose area is dA, we divide these values by dA to find the stresses. Thus, σ_α = 1.29 MPa and τ_α = 2.12 MPa, in the directions shown in Figure 4.23b.
Conceptually, this is all we are doing in this section. This approach is the starting point for all of the seemingly more sophisticated analyses to follow.
We want to generalize the approach of the example to any initial element and to any inclined wedge. This is illustrated in Figure 4.24. Again, we want to determine the transformed stresses (in the “prime” directions, as in Figure 4.22); again, we apply the equations of equilibrium to our wedge.
Equilibrium in the x′ and y′ directions requires (check these results as an exercise)
\sigma _{x^′}=\frac{\sigma _x +\sigma _y}{2} + \frac{\sigma _x -\sigma _y}{2} \cos2 \theta + \tau_{xy} \sin 2 \theta , (4.40)
\tau _{x^′y^′}= \frac{\sigma _x -\sigma _y}{2} \sin 2 \theta + \tau_{xy} \cos 2 \theta . (4.41)
These are the general expressions for the normal and shear stress on any plane located by the angle θ . Clearly, we must know the state of stress in the initial (x, y, z) orientation to find these transformed stresses. The quantities σ_x , σ_y , and τ_{xy} are initially known.
To find the normal stress on the face perpendicular to the wedge face dA (i.e., σ_{y^′} ) we replace θ by θ + 90 ̊ in the equation for σ_{x^′} and obtain
\sigma _{y^′}=\frac{\sigma _x +\sigma _y}{2} – \frac{\sigma _x -\sigma _y}{2} \cos2 \theta – \tau_{xy} \sin 2 \theta . (4.42)
If we add this to the equation for σ_{x^′} we see that σ_{x^′} + σ_{y^′} = σ_x + σ_y , meaning that the sum of the normal stresses remains invariant, regardless of orientation.
In plane strain problems where ε_z = γ_{zx} = γ_{zy} = 0 , a normal stress σ_z can develop.
This stress is given as σ_z = ν(σ_x + σ_y ) , where ν is Poisson’s ratio. However, the forces resulting from this stress do not enter into the relevant equilibrium equations used to derive stress transformation relations. These equations for σ_{x^′} and σ_{y^′} are applicable for plane stress and plane strain.
There are two angles by which our axes can be rotated to achieve the limiting stress cases of extreme normal and extreme shear stress.
