Traction on an Arbitrary Plane Cauchys Stress Tensor

7 Let us now consider the problem of determining the traction acting on the surface of an oblique plane (characterized by its normal n) in terms of the known tractions normal to the three principal axis, 11, t2 and t3. This will be done through the so-called Cauchy's tetrahedron shown in Fig. 2.3.


3 /




Figure 2.3: Cauchy's Tetrahedron

8 The components of the unit vector n are the direction cosines of its direction:

n1 = cos( Z AON); n2 = cos(Z BON); n3 = cos(Z CON); (2.3)

The altitude ON, of length h is a leg of the three right triangles ANO, BNO and CNO with hy-pothenuses OA, OB and OC. Hence h = OAn1 = OBn2 = OCn3 (2.4)

9 The volume of the tetrahedron is one third the base times the altitude

which when combined with the preceding equation yields

10 In Fig. 2.3 are also shown the average values of the body force and of the surface tractions (thus the asterix). The negative sign appears because t* denotes the average traction on a surface whose outward normal points in the negative x, direction. We seek to determine t* .

11 We invoke the momentum principle of a collection of particles (more about it later on) which is postulated to apply to our idealized continuous medium. This principle states that the vector sum of all external forces acting on the free body is equal to the rate of change of the total momentum1. The total momentum is / vdm. By the mean-value theorem of the integral calculus, this is equal

Am to v* Am where v* is average value of the velocity. Since we are considering the momentum of a given

collection of particles, Am does not change with time and Am dj- = p* AV where p* is the average density. Hence, the momentum principle yields dv*

t*nAS + p*b*AV - t*AS1 - t* AS2 - t*AS3 = p*AV— Substituting for AV, ASj from above, dividing throughout by AS and rearanging we obtain tn + 1 hp*b* = t*ni + t*n + t*3n3 + 3 hp*

We observe that we dropped the asterix as the length of the vectors approached zero.

12 It is important to note that this result was obtained without any assumption of equilibrium and that it applies as well in fluid dynamics as in solid mechanics.

13 This equation is a vector equation, and the corresponding algebraic equations for the components of tn are


= CTiini + CT21«2 +


= ^12«1 + ^22«2 + ct32«3


= + ^23«2 + a33n3

Indicial notation


= ajinj

dyadic notation


= n-CT = aT •n

14 We have thus established that the nine components aij are components of the second order tensor, Cauchy's stress tensor.

15 Note that this stress tensor is really defined in the deformed space (Eulerian), and this issue will be revisited in Sect. 4.6.

Example 2-1: Stress Vectors

if the stress tensor at point P is given by


-5 0 "

( t1

a =


3 1

= S t2


1 2

We seek to determine the traction (or stress vector) t passing through P and parallel to the plane ABC where A(4,0, 0), B(0, 2,0) and C(0, 0, 6). Solution:

The vector normal to the plane can be found by taking the cross products of vectors AB and AC:

1 This is really Newton's second law F = ma = md

The unit normal of N is given by Hence the stress vector (traction) will be




I 3 6 2 I L 7 7 7 J







0 0

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