4 Force and Moment Equilibrium, Stress Tensor Symmetry

FIGURE 3.7

ˆ

Material volume showing surface traction vector ti(n) on infinitesimal area element dS, and body

force vector bi acting on infinitesimal volume element dV at xi.



so that Eq 3.4-1 becomes



∫ (σ

V



ji,j



)



+ ρ bi dV = 0



(3.4-2)



This equation must be valid for arbitrary V (every portion of the body is in

equilibrium), which requires the integrand itself to vanish, and we obtain

the so-called local equilibrium equations



σ ji,j + ρ bi = 0



(3.4-3)



In addition to the balance of forces expressed by Eq 3.4-1, equilibrium

requires that the summation of moments with respect to an arbitrary point

must also be zero. Recall that the moment of a force about a point is defined

by the cross product of its position vector with the force. Therefore, taking

the origin of coordinates as the center for moments, and noting that xi is the

position vector for the typical elements of surface and volume (Figure 3.7),

we express the balance of moments for the body as a whole by



∫ε

S



(

x j tkn ) dS +

ˆ



ijk



∫ε

V



ijk



x j ρ bk dV = 0



(3.4-4)



(ˆ

As before, using the identity tkn ) = σ qk nq and Gauss’s divergence theorem,

we obtain



∫ ε ( x σ )

V



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ijk



j



qk , q



+ x j ρ bk  dV = 0







or



∫ ε [x

V



ijk



)]



(



σ qk + x j σ qk,q + ρ bk dV = 0



j,q



But xj,q = δjq and by Eq 3.4-3, σ qk,q + ρ bk = 0 , so that the latter equation immediately above reduces to



∫ε

V



ijk



σ jk dV = 0



Again, since volume V is arbitrary, the integrand here must vanish, or



ε ijkσ jk = 0



(3.4-5)



By a direct expansion of the left-hand side of this equation, we obtain for

the free index i = 1 (omitting zero terms), ε123σ23 + ε132σ32 = 0, or σ23 – σ32 = 0

implying that σ23 = σ32. In the same way for i = 2 and i = 3 we find that

σ13 = σ31 and σ12 = σ21, respectively, so that in general



σ jk = σ kj



(3.4-6)



Thus, we conclude from the balance of moments for a body in which concentrated body moments are absent that the stress tensor is symmetric, and

Eq 3.4-3 may now be written in the form



σ ij , j + ρ bi = 0



or



١ ⋅ ␴ + ρb = 0



(3.4-7)



Also, because of this symmetry of the stress tensor, Eq 3.3-8 may now be

expressed in the slightly altered form

ti( n) = σ ij nj

ˆ



or



ˆ

t(n ) = ␴ ⋅ n

ˆ



(3.4-8)



In the matrix form of Eq 3.4-8 the vectors ti( n ) and nj are represented by

column matrices.

ˆ



3.5



Stress Transformation Laws



Let the state of stress at point P be given with respect to Cartesian axes

Px1x2x3 shown in Figure 3.8 by the stress tensor ␴ having components σij.

We introduce a second set of axes Px1x2 x3 , which is obtained from Px1x2x3

′ ′ ′

by a rotation of axes so that the transformation matrix [aij] relating the two

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FIGURE 3.8

Rectangular coordinate axes Px′1 x′2 x′3 relative to Px1x2x3 at point P.



is a proper orthogonal matrix. Because ␴ is a second-order Cartesian tensor,

its components σij in the primed system are expressed in terms of the

unprimed components by Eq 2.5-13 as



σ ij = aiqσ qma jm

′



or



␴′ = A␴A T



(3.5-1)



The matrix formulation of Eq 3.5-1 is very convenient for computing the primed

components of stress as demonstrated by the two following examples.



Example 3.5-1

Let the stress components (in MPa) at point P with respect to axes Px1x2x3

be expressed by the matrix



[σ ]

ij



1



= 3

2





3

1

0



2



0

−2





′ ′ ′

and let the primed axes Px1 x2 x3 be obtained by a 45° counterclockwise

′

rotation about the x3 axis. Determine the stress components σ ij .



Solution

For a positive rotation θ about x3 as shown by the sketch, the transformation

matrix [aij] has the general form

 cos θ



aij = − sin θ



0





[ ]

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sin θ

cos θ

0



0



0

1





FIGURE E3.5-1

Rotation of axes x1 and x2 by 45° about x3 axis.



Thus, from Eq 3.5-1 expressed in matrix form, a 45° rotation of axes requires

 1/ 2



σ ij = −1 / 2

′



0





[ ]



 4



= 0

 2





0

−2

− 2



0  1



0 3

1 2





1/ 2

1/ 2

0



3

1

0



2 1 / 2



0 1 / 2

−2 

0





−1 / 2

1/ 2

0



0



0

1





2 



− 2

−2 





Example 3.5-2

Assume the stress tensor ␴ (in ksi) at P with respect to axes Px1x2x3 is

represented by the matrix



[σ ]

ij



 18



= 0

−12





0

6

0



−12



0

24





If the x1 axis makes equal angles with the three unprimed axes, and the x2

′

′

axis lies in the plane of x1x3 , as shown by the sketch, determine the primed

′

components of ␴ assuming Px1 x2 x3 is a right-handed system.

′ ′ ′



Solution

We must first determine the transformation matrix [aij]. Let β be the common

angle which x1 makes with the unprimed axes, as shown by the sketch. Then

′

a11 = a12 = a13 = cosβ and from the orthogonality condition Eq 2.5-4 with i = j =

1, cosβ = 1 3 . Next, let φ be the angle between x2 and x3. Then a23 = cosφ =

′

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FIGURE E3.5-2

Rotated axes x′1 and x′2 with respect to Ox1x2x3.



sinβ = 2 6 . As seen from the obvious symmetry of the axes arrangement,

x2 makes equal angles with x1 and x2, which means that a21 = a22. Thus, again

′

from Eq 2.5-4, with i = 1, j = 2, we have a21 = a22 = − 1 6 (the minus sign is

required because of the positive sign chosen for a23). For the primed axes to

ˆ3 ˆ′ ˆ2

be a right-handed system we require e ′ = e1 × e ′ , with the result that a31 =

1 2 , a32 = − 1 2 , and a33 = 0. Finally, from Eq 3.5-1,

 1/ 3



σ ij = −1 / 6

′

 1/ 2





1/ 3

−1 / 6

−1 / 2



[ ]



 8



= 2 2

 0





3.6



2 2

28

−6 3



1 / 3   18



2/ 6  0

0 −12





0

6

0



−12 1 / 3



0 1 / 3

24 1 / 3





−1 / 6

−1 / 6

2/ 6



1/ 2 



−1 / 2 

0









−6 3

12 



0



Principal Stresses, Principal Stress Directions



Let us turn our attention once more to the state of stress at point P and

assume it is given with respect to axes Px1x2x3 by the stress tensor σij. As we

saw in Example 3.3-1, for each plane element of area ∆S at P having an

ˆ

outward normal ni, a stress vector ti( n ) is defined by Eq 3.4-8. In addition,

as indicated by Figure 3.9A, this stress vector is not generally in the direction

of ni. However, for certain special directions at P, the stress vector does indeed

act in the direction of ni and may therefore be expressed as a scalar multiple

of that normal. Thus, as shown in Figure 3.9B, for such directions

ti( n ) = σ ni

ˆ



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(3.6-1)



FIGURE 3.9A

Traction vector at point P for an arbitrary plane whose normal is ni.



FIGURE 3.9B

Traction vector at point P for a principal plane whose normal is ni*.



where σ is the scalar multiple of ni. Directions designated by ni for which

Eq 3.6-1 is valid are called principal stress directions, and the scalar σ is called

a principal stress value of σij . Also, the plane at P perpendicular to ni is referred

to as a principal stress plane. We see from Figure 3.9B that because of the

ˆ

perpendicularity of t (n) to the principal planes, there are no shear stresses

acting in these planes.

The determination of principal stress values and principal stress directions

follows precisely the same procedure developed in Section 2.6 for determining principal values and principal directions of any symmetric second-order

tensor. In properly formulating the eigenvalue problem for the stress tensor

ˆ

we use the identity ti( n ) = σ ji n j and the substitution property of the Kronecker

delta to rewrite Eq 3.6-1 as



(σ

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ji



)



− δ ijσ n j = 0



(3.6-2)