Interpretation of Stress Components

In order to interpret the components of σmn consider equilibrium of an

Interpretation of

Equilibrium Equations 4.3.1.1 Internal Equilibrium

We introduce the stresses ˜σmn as the force per unit area in the direction

of xnacting on the face with unit normal nm, which points in the positive

direction of the xm-axis. At this point we do not know whether these

stresses are equal to the generalized stresses σmn although we probably

have a strong suspicion that they are. And, indeed it is our intention to prove that ˜σmn= σmn. dx3 (˜σ32+ ˜σ32,3dx3) dx1dx2 x2 x1 x3 (˜σ33+ ˜σ33,3dx3) dx1dx2 dx1 dx2 (˜σ22+ ˜σ22,2dx2) dx3dx1 (˜σ23+ ˜σ23,2dx2) dx3dx1 (˜σ12+ ˜σ12,1dx1) dx2dx3 (˜σ21+ ˜σ21,2dx2) dx3dx1 (˜σ31+ ˜σ31,3dx3) dx1dx2 (˜σ13+ ˜σ13,1dx1) dx2dx3 (˜σ11+ ˜σ11,1dx1) dx2dx3

Fig. 4.2: Stress resultants on the positive faces of an infinites- imal parallelepiped.

The body forces and the stress resultants on the negative faces are not shown.

Let us establish equilibrium in the x1-direction first. In order to do so,

note that if the forces acting on the face with a normal in the negative x1-

direction are ˜σ1jdx2dx3, and that they must be (˜σ1j+ ˜σ1j,1dx1) dx2dx3at

the opposite face with analogous expressions for forces on the other faces. Thus, equilibrium in the x1-direction requires

Equilibrium in the x1-direction 0 = + (˜σ11+ ˜σ11,1dx1) dx2dx3− ˜σ11dx2dx3 + (˜σ21+ ˜σ21,2dx2) dx3dx1− ˜σ21dx3dx1 + (˜σ31+ ˜σ31,3dx3) dx1dx2− ˜σ31dx1dx2 + ¯q1dx1dx2dx3 (4.77)

where ¯q1is the component in the x1-direction of the body force.

Since dx1dx2dx3by implication does not vanish, we get

˜

σn1,n+ ¯q1= 0 (4.78)

which is the same as (4.75) with index m = 1 when the symmetry of σmn

in (4.75) is exploited. By analogy we may arrive at equations expressing equilibrium in the two other directions and realize that we recover (4.75). Now we know much about “how ˜σijlooks.” However, we still need to in-

vestigate moment equilibrium in order to confirm the symmetry property of ˜σmn, and we must also interpret the static boundary conditions (4.76) in

order to get the full interpretation of the stresses, see below.

As regards moment equilibrium we compute the moment about the cen- ter of the parallelepiped and commence by writing the moment about the x1-direction Moment equilibrium about the x1-direction 0 = − ˜σ32dx1dx21₂dx3− (˜σ32+ ˜σ32,3dx3) dx1dx2
1₂dx3 + ˜σ23dx3dx112dx2+ (˜σ23+ ˜σ23,2dx2) dx3dx1
12dx2 or 0 = − ˜σ32−1₂˜σ32,3dx3+ ˜σ23+1₂σ˜23,2dx2 (4.79)

Because ˜σmn and ˜σmn,k are finite while dxk is infinitely small, (4.79b)

implies ˜

σ32= ˜σ23 (4.80)

If we repeat the above procedure for the moment equilibrium about the two other directions we realize that we have proved the symmetry property of ˜σmn

Symmetry of ˜σmn ˜σmn= ˜σnm (4.81)

4.3.1.2 Static Boundary Conditions

In order to establish the static boundary conditions consider Fig. 4.3, which is spanned by dx1, dx2and dx3in the directions of the respective axes. The

total body force on the tetrahedron is d ¯P , and −dFj, j ∈ [1, 3] is the total

load acting on the face with the normal in the negative xj-direction. Finally,

dFn is the total load on the inclined face. Equilibrium of the tetrahedron

requires

Equilibrium of

tetrahedron 0= −dF1− dF2− dF3+ dFn+ d ¯P (4.82)

The intensity of the force vector −dF1is ˜t1etc., the intensity of dFnis

eτ, and the intensity of d ¯Pis ¯q, and thus (4.82) requires

0= −˜t112dx2dx3− ˜t212dx3dx1− ˜t312dx1dx2+eτdA + ¯qdV (4.83)

Equilibrium Equations

volume of the tetrahedron. Recall that ˜σijare the force intensities resolved

in terms of the base vectors im, m ∈ [1, 3]

˜t_i= ˜σijij (4.84)

Further, resolveeτ in terms of the base vectors ij, j ∈ [1, 3] to get

eτ = ˜τjij (4.85)

and get

0= −˜σ1jij1₂dx2dx3− ˜σ2jij1₂dx3dx1− ˜σ3jij1₂dx1dx2

+˜τjijdA + ¯qdV

(4.86) Obviously, the last term is of order 3 in the differentials, while the others are of order 2, and therefore it may be omitted in the following.

i2 _x 2 i1 d ¯P −dF1 dFn x1 −dF3 −dF2 i3 x3

Fig. 4.3: Stresses and loads on an infinitesimal tetrahedron. The areas of the faces that are parallel with the coordinate planes may be computed as

dA1=12dx2dx3= n · i1dA

dA2=12dx3dx1= n · i2dA

dA3=1₂dx1dx2= n · i3dA

(4.87)

and thus (4.86) may yield

0_{= −˜σ}1jijdA (n · i1) − ˜σ2jijdA (n · i2) − ˜σ3jijdA (n · i3)

+˜τjijdA

(4.88)

Resolve n is resolved in terms of ik

n= nkik ⇒ n · im= nkik· im= nkδkm= nm (4.89)

and (4.88) becomes

0= −˜σijijni+ ˜τjij (4.90)

Since this must hold for all ij, which are linearly independent, we get

the final result

Projection of

stress τj= ˜σijni (4.91)

This expression, which holds for all inclined surfaces—not only on the static boundary ST—is similar to (4.76).

Now that we have seen that the equilibrium equations derived from the principle of virtual displacements and by direct formulation of equilibrium are identical, we may finally interpret σmnas the stress components ˜σmn,

σmn= ˜σmn

which are the forces per unit area in the direction of xnacting on the face

with unit normal nm.

Actually, we might just as well have acknowledged the fact that in the infinitesimal theory |ui,j| ≪ 1 and therefore (2.59)

tik,i+ (tijuk,j),i+ ¯qk= 0 (4.92)

would become

tik,i+ ¯qk= 0 (4.93)

which, except for notational differences, is the same as (4.75). Since the in- finitesimal theory does not distinguish between undeformed and deformed configuration4.8_{the components of the stress σ}

ijmust therefore be the force

per unit undeformed area resolved in terms of the undeformed base vec- tors ij, just as the Piola-Kirchhoff stresses are the forces per unit undeformed

area resolved in terms of the deformed base vectors gj, see Section 2.3.1.

If we follow this line of reasoning, we may immediately write the equiv- alent of (2.71)

τm= (δmk+ um,k) n0jtjk (4.94) 4.8_{Strictly speaking, this is not completely true in that the strains and the displace-} ments are expressions of the difference between undeformed and deformed configurations. For example, while 1 + ε11≈ 1 is true because the theory assumes that |ε11| is small, it does not necessarily imply that (1 + ε11) − 1 ≈ 0. In most other instances it is neither possible nor desirable to try to differentiate between the two configurations when we em- ploy the infinitesimal theory. This is a fundamental difficulty inherent in the infinitesimal theory, as mentioned in Chapter 1, and the most important reason why I chose to begin with a theory that does not have such paradoxes built in.

Equilibrium Equations for the infinitesimal theory, namely

τm= njσjm (4.95)

which gives the relation between the stress tensor σjm and the stress on

a plane with the unit normal nj. On the static boundary ST0, where the

boundary stress resultant ¯τmis given, (4.95) becomes

Static boundary conditions

¯

τm= njσjm (4.96)

Thus, the static boundary condition (4.76) may be interpreted in a straightforward manner through (4.96).