Last time we introduced a displacement u_i (i = 1, 2, 3 or x, y, z) from an initial position as a result of some deformation. The strain components were defined in terms of the displacements due to infinitesimal deformations. We assumed that higher order terms beyond first-order may be neglected.

14. General Formulation of Elastic Strain and Stress [11]

The components of the strain tensor are written out in Cartesian coordinates as

(13-1)

Note that the shear strains are considered symmetric:

(13-2)

e_zy = e_yz
e_xz = e_zx
e_yx = e_xy

We can also write the strain tensor components in cylindrical and spherical coordinates. Figure 77 shows the displacement in the cylindrical coordinate system.

 

Figure 77. Deformation in Cylindrical Coordinates.

 

 

(13-3)

 

Figure 78 shows the displacement in the spherical coordinate system.

 

Figure 78. Deformation in Spherical Coordinates.

 

(13-4)

 

The stress at a point P within an elastic solid is examined as follows. Consider Figure 79 in which the solid is divided into two regions, (1) below P and (2) above P, a surface S. A surface element DS surrounds the point P and has a unit normal vector n. The forces exerted by the material in region 2 on the material in region 1 can be reduced to single force DF and a torque DG as shown. The torque is also called a moment in many texts.

 

Figure 79. A surface is drawn within an elastic solid, dividing it into two regions. Point P is within a small surface element DS. The unit normal vector is drawn from region 1 to region 2.

Let's assume that the following is correct:
 

(13-5)

where F_n is the stress vector across the surface at P. The general form of F_n is

 

F_n = F_nxe₁ + F_nye₂ + F_nze₃.

 

The forces acting on a volume element DV of the elastic material are separated into surface and bulk forces. An example of a bulk force is that due to gravity. We can assume that the forces acting on a small volume element surrounding P are equivalent to a force DR acting at P and a torque (or moment) DM. In this case, we have

(13-6)

Here F is a bulk force per unit volume that is a continuous function of (x, y, z). Let's have a look at a volume surrounding the point P in the shape of a rectangular parallelepiped with sides 2a, 2b, and 2c as shown in Figure 80.

 

Figure 80. The stresses labeled t_xy and t_yx produce a torque about the z axis.

The moment of the volume in the z direction is produced by the bulk forces acting on it. Since the element remains in equilibrium, the sum of the torques must be zero:
 

(13-7)

We note that the volume of the parallelepiped is V = 8abc, so Equation 13-7 becomes,
 

(13-8)

As the DV goes to zero, DM goes to zero (based on Equation 13-6). This leads to the symmetry,

t_yx = t_xy

 

Similarly, one can show that the other stresses would also be symmetrical:
t_yz = t_zy
t_xz = t_zx

 

The tetrahedron shown in Figure 81 has three sides parallel to the coordinate planes. Let's consider forces acting in the y direction. If DS is the area ABC and the distance of P to the ABC plane is h, then we can write

area PBC = lDS
area PCA = mDS
area PAB = nDS

 

Figure 81. A tetrahedral volume element PABC is shown. The unit vector n is normal to the plane ABC.

Also the volume of the tetrahedron is DV = (1/3)hDS. If the tetrahedron is in equilibrium, then

(13-9)

If h and DS are small, then the surface stresses can be approximated by their values at P. Dividing, Equation (13-9) by DS and allowing DV to go to zero gives

(13-10)

This leads to the three equations for the stresses:

(13-11)

The normal stresses, t_xx, t_yy, and t_zz, are usually written s_x, s_y, and s_z, respectively. The nine components of the stress indicated on the right hand side of Equation 13-11 make up the stress tensor which is of rank 2.

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