| Principle Stresses and Maximum Shear Stress |
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, τxy, τyz, and τzx, are the components of the shearing stresses on the same element. τyx = τxy, τzy = τzy, and τxz = τzx. 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 = τzx = τzy = 0, and the only remaining stress components are σ;x, σ;y, and τxy. 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 = τzx = τzy = 0, and that it is defined by the stress components σ;x, σ;y, and τxy associated with the element shown in the following figure.
The components of interest are σ;x', σ;y', and τx'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, τxy, and θ;.
In order to determine the normal stress σ;x' and the shearing stress τx'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:
σ;Fx' = 0: σ;x' ΔA – σ;x(ΔA cos θ;) cos θ; – τxy(ΔA cos θ;) sin θ; – σ;y(ΔA sin θ;) sin θ; – τxy(ΔA sin θ;) cos θ; = 0
σ;Fx' = 0: τx'y' ΔA + σ;x(ΔA cos θ;) sin θ; – τxy(ΔA cos θ;) cos θ; – σ;y(ΔA sin θ;) cos θ; + τxy(ΔA sin θ;) sin θ; = 0
Solving the first equation for σ;x' and the second for τx'y' :
(Eq1) σ;x' = σ;x cos2 θ; + σ;y sin2 θ; + 2τxy sin θ; cos θ;
(Eq2) τx'y' = –(σ;x – σ;y) sin θ; cos θ; + τxy(cos2 θ; – sin2 θ;)
Recalling the trigonometric relations:
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sin 2θ; = 2 sin θ; cos θ;
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cos 2θ; = cos2 θ; – sin2 θ;
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then Eq1 is rewritten:
| σ;x' = σ;x | | + σ;y | | + τxy sin 2θ; |
or
| (Eq3) | | σ;x' = | | + | | cos 2θ; + τxy sin 2θ; |
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Using the trigonometric relations from the first row, Eq2 is rewritten:
| (Eq4) | | τx'y' = – | | sin 2θ; + τxy cos 2θ; |
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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) | | σ;y' = | | – | | cos 2θ; – τxy sin 2θ; |
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Eq3 and Eq4 from mechanics of materials lesson 'Transformation of Plane Stress' are shown below as Eq1 and Eq2 below:
| (Eq1) | | σ;x' = | | + | | cos 2θ; + τxy sin 2θ; |
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| (Eq2) | | τx'y' = – | | sin 2θ; + τxy cos 2θ; |
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The equations above can be used as the parametric equations of a circle. This means that, if a rectangular set of axes is chosen and a point M of abscissa σx' and ordinate τx'y' is plotted for any given value of the parameter θ;, all the points thus obtained will lie on a circle. To establish this property θ is eliminated from Eq1 and Eq2. This is done by first transposing (σ;x)/2 in Eq1 and squaring both members of the equation, then squaring both members of Eq2, and finally adding member to member the two equations obtained in this fashion:
| σ;x' = | | + | | cos 2θ; + τxy sin 2θ; |