- 1.1 Introduction
- 1.2 Scope of Treatment
- 1.3 Analysis and Design
- 1.4 Conditions of Equilibrium
- 1.5 Definition and Components of Stress
- 1.6 Internal Force-Resultant and Stress Relations
- 1.7 Stresses on Inclined Sections
- 1.8 Variation of Stress Within a Body
- 1.9 Plane-Stress Transformation
- 1.10 Principal Stresses and Maximum in-plane Shear Stress
- 1.11 Mohr's Circle for Two-Dimensional Stress
- 1.12 Three-Dimensional Stress Transformation
- 1.13 Principal Stresses in Three Dimensions
- 1.14 Normal and Shear Stresses on an Oblique Plane
- 1.15 Mohr's Circles in Three Dimensions
- 1.16 Boundary Conditions in Terms of Surface Forces
- 1.17 Indicial Notation
1.10 Principal Stresses and Maximum in-plane Shear Stress
The transformation equations for two-dimensional stress indicate that the normal stress s x' and shearing stress t x'y' vary continuously as the axes are rotated through the angle q. To ascertain the orientation of x'y' corresponding to maximum or minimum s x' , the necessary condition ds x' /dq = 0 is applied to Eq. (1.18a). In so doing, we have
Inasmuch as tan 2q = tan(p + 2q), two directions, mutually perpendicular, are found to satisfy Eq. (1.19). These are the principal directions, along which the principal or maximum and minimum normal stresses act. Two values of q p , corresponding to the s 1 and s 2 planes, are represented by and , respectively.
When Eq. (1.18b) is compared with Eq. (a), it becomes clear that t x'y' = 0 on a principal plane. A principal plane is thus a plane of zero shear. The principal stresses are determined by substituting Eq. (1.19) into Eq. (1.18a):
Note that the algebraically larger stress given here is the maximum principal stress, denoted by s 1. The minimum principal stress is represented by s 2. It is necessary to substitute one of the values q p into Eq. (1.18a) to determine which of the two corresponds to s 1.
Similarly, employing the preceding approach and Eq. (1.18b), we determine the planes of maximum shearing stress. Thus, setting dt x'y' /dq = 0, we now have (s x – s y )cos 2q + 2t xy sin 2q = 0 or
The foregoing expression defines two values of q s that are 90° apart. These directions may again be denoted by attaching a prime or a double prime notation to q s . Comparing Eqs. (1.19) and (1.21), we also observe that the planes of maximum shearing stress are inclined at 45° with respect to the planes of principal stress. Now, from Eqs. (1.21) and (1.18b), we obtain the extreme values of shearing stress as follows:
Here the largest shearing stress, regardless of sign, is referred to as the maximum shearing stress, designated t max. Normal stresses acting on the planes of maximum shearing stress can be determined by substituting the values of 2q s from Eq. (1.21) into Eqs. (1.18a) and (1.18c):
The results are illustrated in Fig. 1.14. Note that the diagonal of a stress element toward which the shearing stresses act is called the shear diagonal. The shear diagonal of the element on which the maximum shearing stresses act lies in the direction of the algebraically larger principal stress as shown in the figure. This assists in predicting the proper direction of the maximum shearing stress.
Figure 1.14 Planes of principal and maximum shearing stresses.