# Advanced Mechanics of Materials and Applied Elasticity: Analysis of Stress

This chapter is from the book

## 1.7 Stresses on Inclined Sections

The stresses in bars, shafts, beams, and other structural members can be obtained by using the basic formulas, such as those listed in Table 1.1. The values found by these equations are for stresses that occur on cross sections of the members. Recall that all of the formulas for stress are limited to isotropic, homogeneous, and elastic materials that behave linearly. This section deals with the states of stress at points located on inclined sections or planes under axial loading. As before, we use stress elements to represent the state of stress at a point in a member. However, we now wish to find normal and shear stresses acting on the sides of an element in any direction.

The directional nature of more general states of stress and finding maximum and minimum values of stress are discussed in Sections 1.10 and 1.13. Usually, the failure of a member may be brought about by a certain magnitude of stress in a certain direction. For proper design, it is necessary to determine where and in what direction the largest stress occurs. The equations derived and the graphical technique introduced here and in the sections to follow are helpful in analyzing the stress at a point under various types of loading. Note that the transformation equations for stress are developed on the basis of equilibrium conditions only and do not depend on material properties or on the geometry of deformation.

We now consider the stresses on an inclined plane a–a of the bar in uniaxial tension shown in Fig. 1.6a, where the normal x' to the plane forms an angle q with the axial direction. On an isolated part of the bar to the left of section a–a, the resultant P may be resolved into two components: the normal force Px' = P cos q and the shear force Py' = –P sin q, as indicated in Fig. 1.6b. Thus, the normal and shearing stresses, uniformly distributed over the area Ax' = A/cos q of the inclined plane (Fig. 1.6c), are given by

Equation 1.11a

Equation 1.11b

where s x = P/A. The negative sign in Eq. (1.11b) agrees with the sign convention for shearing stresses described in Section 1.5. The foregoing process of determining the stress in proceeding from one set of coordinate axes to another is called stress transformation.

Equations (1.11) indicate how the stresses vary as the inclined plane is cut at various angles. As expected, s x' is a maximum (s max) when q is 0° or 180°, and t x'y' is maximum (t max) when q is 45° or 135°. Also, . The maximum stresses are thus

Equation 1.12

Observe that the normal stress is either maximum or a minimum on planes for which the shearing stress is zero.

Figure 1.7 shows the manner in which the stresses vary as the section is cut at angles varying from q = 0° to 180°. Clearly, when q > 90°, the sign of t x'y' in Eq. (1.11b) changes; the shearing stress changes sense. However, the magnitude of the shearing stress for any angle q determined from Eq. (1.11b) is equal to that for q + 90°. This agrees with the general conclusion reached in the preceding section: shearing stresses on mutually perpendicular planes must be equal.

We note that Eqs. (1.11) can also be used for uniaxial compression by assigning to P a negative value. The sense of each stress direction is then reversed in Fig. 1.6c.

#### Example 1.1. State of Stress in a Tensile Bar

Compute the stresses on the inclined plane with q = 35° for a prismatic bar of a cross-sectional area 800 mm2, subjected to a tensile load of 60 kN (Fig. 1.6a). Then determine the state of stress for q = 35° by calculating the stresses on an adjoining face of a stress element. Sketch the stress configuration.

Solution

The normal stress on a cross section is

Introducing this value in Eqs. (1.11) and using q = 35°, we have

 s x' = s x cos2 q = 75 (cos 35°)2 = 50.33 MPa t x'y' = –s x sin q cos q = –75(sin 35°)(cos 35°) = –35.24 MPa

The normal and shearing stresses acting on the adjoining y' face are, respectively, 24.67 MPa and 35.24 MPa, as calculated from Eqs. (1.11) by substituting the angle q + 90° = 125°. The values of s x' and t x'y' are the same on opposite sides of the element. On the basis of the established sign convention for stress, the required sketch is shown in Fig. 1.8.

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