In many real situations, some of the components of the stress tensor (Eqn. 4-1) are zero. E.g., Tensile test • For most stress states there is one set of coordinate axes (1, 2, 3) along which the shear stresses vanish. The normal stresses, 1, 2, and 3 along these axes are principal stresses.
Stress Acting in a General Direction • p ds = force acting on a surface element ds • Area ds is defined by the unit vector (normal to it) that passes through R • In order to know how p ds changes with the orientation of the area, we have to consider a system of orthogonal axes and the summation of forces Figure 4.3 Force p ds acting on surface element ds.
Figure 4.4 Same situation as Fig. 4-3 referred to a system of orthogonal axes. • The summation of the forces along the axes are: (4-17)
Where lm1, lm 2, and lm3 are the direction cosines between the normal to the oblique plane and the x1, x2, x3 axes. • In indicial notation, Eqn. 4-17 can be written as: • This equation defines ij as a tensor, because it relates to vectors p and according to the relationship for tensor (4-18)
Determination of Principal Stresses • The shear stresses acting on the faces of a cube referred to its principal axes are zero. • This means that the total stress is equal to the normal stress. • The total stress is: • The normal stress is: • If pi and N coincide then (4-18) (4-19) (4-20)
(4-21) • Applying equations 4-18 and 4-20 to p1, we have OR • Similarly, for p2 and p3, (4-22) (4-23) (4-24)
The solution of the system of Eqs. 4-22, 4-23 and 4-24 is given by the following determinant • Solution of the determinant results in a cubic equation in , (4-25) (4-26)
The three roots of Eq. 4-26 are the principal stresses 1, 2 and3, of which 1> 2 >3 • To determine the direction of the principal stresses with respect to the original x1, x2 and x3 axes, • we substitute 1, 2 and3 successively back into Eqs. (4-22), (4-23) and (4-24). • solve the resulting equations simultaneously for the direction cosines • and use:
Notes: • The convention (notation) in your text book is different. It uses respectively • There are three combinations of stress components in Eq. 4-26 that make up the coefficient of the cubic equation, and these are: (4-27) (4-28) (4-29)
Notes (cont): • The coefficients I1,I2 andI3 are independent of the coordinate system, and are therefore calledinvariants. • This means that the principal stresses for a given stress state are unique. Example: The first invariant I1 states that the sum of the normal stresses for any orientation in the coordinate system is equal to the sum of the normal stresses for any other orientation.
Notes (cont): • For any stress state that includes all shear components as in Eq. 4-1, a determination of the three principal stresses can be made only by finding the three roots. • The invariants is important in the development of the criteria that predict the onset of yielding.
The invariants of the stress tensor may be determined readily from the matrix of its components. Since s12=s21, etc., the stress tensor is a symmetric tensor. • The first invariant is the trace of the matrix, i.e. , the sum of the main diagonal terms. I1 = s11 + s22+ s33
The second invariant is the sum of the principal minors. • Thus taking each of the principal (main diagonal) terms in order and suppressing that row and column we have
Finally, the third invariant is the determinant of the entire • matrix of the components of the stress tensor. • The cubic equation can be expressed in terms of the stress • invariants. (4-30)
Since the principal normal stresses are roots of an equation involving the stress invariants as coefficients, their values are also invariant, that is, not dependent on the choice of the original coordinate system. • It is common practice to assign the subscripts 1, 2, and 3 in order to the maximum, intermediate, and minimum values.
Examples: • (1) Consider a stress state where 11 = 10, 22 = 5, 12 = 3 (all in ksi) and 33 = 31 = 32 = 0 Find the principal stresses.
Solution Using Eqs. 4-27 to 4-29, we obtain I1 = 15, I2 = 41 and I3 = 0 Substitute values into Eq. 4-30, and we have The roots of this quadratic give the two principal stresses in the x-y plane. They are:
Examples: • (2) Repeat example 1, where all the stresses are the same except that 33 = 8 instead of zero.
Solution Using Eqs. 4-27 to 4-29, we obtain I1 = 23, I2 = 161 and I3 = 328 Substitute values into Eq. 4-30, and we have The three roots are
Principal Shear Stresses • Principal shear stresses, 1, 2 and 3 are define in analogy with the principal stresses. • In order to understand how to derive the values, students are advised to see pages 29 and 30 of the text. • The principal shear stresses occur along the direction that bisect any two of the three principal axes
Principal Shear Stresses (cont.) • The numeric values of the Principal shear stresses are: Since 1 > 2 > 3, 2 is the maximum shear stress i.e. • In materials that fail by shear (as most metals do) the orientation of the maximum shear is very important. (4-31) (4-32)