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Transformation of Plane Stress

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The most general state of stress at a given point Q may be represented by six components. Three of these components, σ_x, σ_y, and σ_z, define normal stresses exerted on the faces of a small cubic element centered at Q and of the same orientation as the coordinate axes, and the other three, τ_xy, τ_yz, and τ_zx, are the components of the shearing stresses on the same element. τ_yx = τ_xy, τ_zy = τ_zy, and τ_xz = τ_zx. The same state of stress will be represented by a different set of components if the coordinate axes are rotated.

For the sake of simplicity, this three dimensional viewpoint is reduced to a two dimensional examination. Therefore, two faces of the cubic element are free of stress. The element is oriented such that these two faces are perpendicular to the z-axis. This means that σ_z = τ_zx = τ_zy = 0, and the only remaining stress components are σ_x, σ_y, and τ_xy. This kind of situation occurs in a thin plate subject to forces acting in the midplane of the plate. It also occurs on the free surface of a structural element or machine componenet, such as at any point of the surface of that element or componenet that is not subjected to an external force.

Assume that a state of plane stress exists at point Q with σ_z = τ_zx = τ_zy = 0, and that it is defined by the stress components σ_x, σ_y, and τ_xy associated with the element shown in the following figure.

The components of interest are σ_x', σ_y', and τ_x'y' associated with the element after it has been rotated through an angle θ about the z-axis, and to express thesse componenets in terms of σ_x, σ_y, τ_xy, and θ.

In order to determine the normal stress σ_x' and the shearing stress τ_x'y' exerted on the face perpendicular to the x'-axis, we consider a prismatic element with faces respectively perpendicular to the x, y, and x' axes. If the area of the oblique face is denoted by ΔA, the areas of the vertical and horizontal faces are respectively equal to ΔA cosθ and ΔA sinθ. It follows that the forces exerted on the three faces are as shown below.

Using components along the x' and y' axes, the equilibrium equations follow:

ΣF_x' = 0:    σ_x' ΔA – σ_x(ΔA cos θ) cos θ – τ_xy(ΔA cos θ) sin θ – σ_y(ΔA sin θ) sin θ – τ_xy(ΔA sin θ) cos θ = 0

ΣF_x' = 0:    τ_x'y' ΔA + σ_x(ΔA cos θ) sin θ – τ_xy(ΔA cos θ) cos θ – σ_y(ΔA sin θ) cos θ + τ_xy(ΔA sin θ) sin θ = 0

Solving the first equation for σ_x' and the second for τ_x'y' :

(Eq1)    σ_x' = σ_x cos² θ + σ_y sin² θ + 2τ_xy sin θ cos θ

(Eq2)    τ_x'y' = –(σ_x – σ_y) sin θ cos θ + τ_xy(cos² θ – sin² θ)

Recalling the trigonometric relations:

sin 2θ = 2 sin θ cos θ	cos 2θ = cos2 θ – sin2 θ
cos2 θ = 1 + cos 2θ 2	sin2 θ = 1 – cos 2θ 2

then Eq1 is rewritten:

σx' = σx

1 + cos 2θ 2

+ σy

1 – cos 2θ 2

+ τxy sin 2θ

or

(Eq3)

σx' = σx + σy 2 + σx – σy 2 cos 2θ + τxy sin 2θ

Using the trigonometric relations from the first row, Eq2 is rewritten:

(Eq4)

τx'y' = – σx – σy 2 sin 2θ + τxy cos 2θ

The expression for the normal stress σ_y' is obtained by replacing θ in Eq3 by the angle θ + 90° that the y' axis forms with the x axis. Since cos (2θ + 180°) = –cos 2θ and sin (2θ + 180°) = –sin 2θ, the following results:

(Eq5)

σy' = σx + σy 2 – σx – σy 2 cos 2θ – τxy sin 2θ

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