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1.14 Normal and Shear Stresses on an Oblique Plane

A cubic element subjected to principal stresses s 1, s 2, and s 3 acting on mutually perpendicular principal planes is called in a state of triaxial stress (Fig. 1.22a). In the figure, the x, y, and z axes are parallel to the principal axes. Clearly, this stress condition is not the general case of three-dimensional stress, which was taken up in the last two sections. It is sometimes required to determine the shearing and normal stresses acting on an arbitrary oblique plane of a tetrahedron, as in Fig. 1.22b, given the principal stresses or triaxial stresses acting on perpendicular planes. In the figure, the x, y, and z axes are parallel to the principal axes. Denoting the direction cosines of plane ABC by l, m, and n, Eqs. (1.26) with s x = s 1, t xy = t xz = 0, and so on, reduce to

Equation a


Figure 1.22

Figure 1.22 Elements in triaxial stress.

Referring to Fig. 1.22a and definitions (a), the stress resultant p is related to the principal stresses and the stress components on the oblique plane by the expression

Equation 1.36


The normal stress s on this plane, from Eq. (1.28a), is found as

Equation 1.37


Substitution of this expression into Eq. (1.36) leads to

Equation 1.38a



Equation 1.38b


Expanding and using the expressions 1 – l 2 = m 2 + n 2, 1 – n 2 = l 2 + m 2, and so on, the following result is obtained for the shearing stress t on the oblique plane:

Equation 1.39


This clearly indicates that if the principal stresses are all equal, the shear stress vanishes, regardless of the choices of the direction cosines.

For situations in which shear as well as normal stresses act on perpendicular planes (Fig. 1.22b), we have px , py , and pz defined by Eqs. (1.26). Then, Eq. (1.37) becomes

Equation 1.40



Equation 1.41


where s is given by Eq. (1.40). Formulas (1.37) through (1.41) represent the simplified transformation expressions for the three-dimensional stress.

It is interesting to note that substitution of the direction cosines from Eqs. (a) into Eq. (1.25) leads to

Equation 1.42


which is a stress ellipsoid having its three semiaxes as the principal stresses (Fig. 1.23). This geometrical interpretation helps to explain the earlier conclusion that the principal stresses are the extreme values of the normal stress. In the event that s 1 = s 2 = s 3, a state of hydrostatic stress exists, and the stress ellipsoid becomes a sphere. In this case, note again that any three mutually perpendicular axes can be taken as the principal axes.

Figure 1.23

Figure 1.23 Stress ellipsoid.

Octahedral Stresses

The stresses acting on an octahedral plane is represented by face ABC in Fig. 1.22b with QA = QB = QC. The normal to this oblique face thus has equal direction cosines relative to the principal axes. Since l 2 + m 2 + n 2 = 1, we have

Equation b


Plane ABC is clearly one of eight such faces of a regular octahedron (Fig. 1.24). Equations (1.39) and (b) are now applied to provide an expression for the octahedral shearing stress, which may be rearranged to the form

Equation 1.43


Figure 1.24

Figure 1.24 Stresses on an octahedron.

Through the use of Eqs. (1.37) and (b), we obtain the octahedral normal stress:

Equation 1.44


The normal stress acting on an octahedral plane is thus the average of the principal stresses, the mean stress. The orientations of s oct and t oct are indicated in Fig. 1.24. That the normal and shear stresses are the same for the eight planes is a powerful tool for failure analysis of ductile materials (see Sec. 4.8). Another useful form of Eq. (1.43) is developed in Section 2.15.

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