Principal Stresses and Directions - Civil Engineering (CE) PDF Download

Principal stresses and directions 
At a given location in the body the magnitude of the normal and shear traction depend on the orientation of the plane. An isotropic body can fail along any plane and hence we require to find the maximum magnitude of these normal and shear traction and the planes for which these occur. As we shall see on planes where the maximum or minimum normal stresses occur the shear stresses are zero. However, in the plane on which maximum shear stress occurs, the normal stresses will exist.

Maximum and minimum normal traction 
In order to obtain the maximum and minimum values of σ_n, the normal traction, we have to find the orientation of the plane in which this occurs. This is done by maximizing (4.5) subject to the constraint that | n | = 1. This constrained optimization is done by what is called as the Lagrange-multiplier method. Towards this, we introduce the function

L(n, λ*) = n · σn − λ* [| n |² −1],                         (4.10) 

where λ* is the Lagrange multiplier and the condition | n |² − 1 = 0 characterizes the constraint condition. At locations where the extremal values of L occurs, the derivatives
 must vanish, i.e.,



To obtain these equations we have made use of the fact that Cauchy stress tensor is symmetric. Thus, we have to find three  and  ’s such that



(a = 1, 2, 3; no summation) which is nothing but the eigenvalue problem involving the tensor σ with the Lagrange multiplier being identified as the eigenvalue. Hence, the results of section 2.5 follows. In particular, for (4.13a) to have a non-trivial solution



where



the principal invariants of the stress σ. As stated in section 2.5, equation (4.14) has three real roots, since the Cauchy stress tensor is symmetric. These roots ( ) will henceforth be denoted by σ₁, σ₂ and σ₃ and are called principal stresses. The principal stresses include both the maximum and minimum normal stresses among all planes passing through a given x.
The corresponding three orthonormal eigenvectors , which are then characterized through the relation (4.13) are called the principal directions of σ. The planes for which these eigenvectors are normal are called principal planes. Further, these eigenvectors form a mutually orthogonal basis since the stress tensor σ is symmetric. This property of the stress tensor also allows us to represent σ in the spectral form



It is immediately apparent that the shear stresses in the principal planes are zero. Thus, principal planes can also be defined as those planes in which the shear stresses vanish. Consequently, σ_a’s are normal stresses.


Maximum and minimum shear traction 
Next, we are interested in finding the direction of the unit vector n at x that gives the maximum and minimum values of the shear traction, τ_p. This is important because in many metals the failure is due to sliding of planes resulting due to the shear traction exceeding a critical value along some plane. In the following we choose the eigenvectors {} of σ as the set of basis vectors. Then, according to the spectral decomposition (4.16), all the nondiagonal matrix components of the Cauchy stress vanish. Then, the traction vector t_(p) on an arbitrary plane with normal p could simply be written as



where  It then follows from (4.9) that



where σ_p is the normal traction on the plane whose normal is p.  
With the constraint condition | p |² = 1 we can eliminate p₃ from the equation (4.18). Then, since the principal stresses, σ₁, σ₂, σ₃ are known,  is a function of only p₁ and p₂. Therefore to obtain the extremal values of  we differentiate   with respect to p₁ and p₂ and equate it to zero¹ :



The above set of equations has three classes of solutions.

• Set - 1: p₁ = p₂ = 0 and hence p₃ = ±1, obtained from the condition that | p |² = 1.

• Set - 2: p₁ = 0, p₂ = ±1/  and hence p₃ = ±1/

• Set - 3: p₁ = ±1/, p₂ = 0 and hence p3 = ±1/

Instead of eliminating p₃ we could eliminate p₂ or p₁ initially and find the remaining unknowns by adopting a procedure similar to the above. Then, we shall find the extremal values of τ_p could also occur when

• Set - 4: p₁ = p₃ = 0 and hence p2 = ±1.

• Set - 5: p₂ = p₃ = 0 and hence p1 = ±1.

• Set - 6: p₃ = 0, p₁ = ±1/  and hence p₂ = ±1/ .

1Recognize that the extremal values of both τp and  occur at the same location.


Substituting these solutions in (4.18) we find the extremal values of τ_p. Thus, τ_p = 0 when    and



Consequently, the maximum magnitude of the shear traction denoted by τ_max is given by the largest of the three values of (4.21b) - (4.23b). Thus, we obtain

                        (4.24)

where σ_max and σ_min denote the maximum and minimum magnitudes of principal stresses, respectively. Recognize that the maximum shear stress acts on a plane that is shifted about an angle of ±45 degrees to the principal plane in which the maximum and minimum principal stresses act. In addition, we can show that the normal traction σ_p to the plane in which τ_max occurs has the value σ_p = (σ_max + σ_min)/2.