In this article we will discuss about how to analyze stress in metal forming processes.
Stress State at a Point:
Let us first examine what comprises a complete state of stress at a point. State of stress at a point comprises the stress vectors or components of stress vectors acting on three mutually perpendicular planes passing through the point. With the knowledge of these components of stress state we are able to determine normal and shear stresses on any plane passing through the point.
Figure 3.1 shows the stress vectors Sa, Sb and Sc acting on three faces of a cubical element at a point P. Here the following vectors are denoted by bold letters. When the sides of the cubical element are reduced to zero the three faces of the cube would pass through the point.
The stress vectors acting on the faces of the cubical element may be decomposed into stress components acting in the directions of chosen co-ordinate axes. Thus the stress vector Sa acting on the face ABCD is decomposed into three components, one along the normal to the face, while other two are parallel to the face but are directed along the other two co-ordinate axes. The three components are denoted as σxx, σxy and σxz.
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Note that the first subscript in all the three components is x, because these three components are acting on a plane which is perpendicular to x-axis. The first subscript denotes the orientation of the normal to the plane on which the stress component is acting.
The second subscript gives the direction of stress component itself. Thus σxy is acting on the plane normal to x-axis and is directed along y-axis. Similarly σxz is acting on the same plane but is directed along z-axis. The stress components on other two faces are also designated similarly.
Thus we see that with every component of stress two directions are associated. From the above it is clear that stress state comprises nine components as shown in Fig. 3.1. However the three shears components σxy, σxz and σvz are respectively equal to the three components σyx, σzx and σzv. Thus we need to determine only six components of stress, i.e. σxx, σxy, σyy, σxz, σyz and σzz to know the complete stress state at a point.
In the following, in view of convenience, we may take the axes as x1, x2 and x3 instead of x, y and z, in that case, the nine components of stress state are denoted as σ11, σ12, σ13, σ21, σ22, σ23, σ31, σ32 and σ33.
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These components are in fact the components of a second order symmetric stress tensor. The second order means there are two directions associated with each component. It is symmetric because σ12 = σ21, σ13 = σ31 and σ23 = σ32. Therefore, we may denote the stress state at the point P by components of a stress tensor T as given below-
Stresses on an Inclined Plane:
Here we determine stresses (normal and shear) on any plane passing through a point at which the stress state is known. Consider a plane ABC inclined to the three Cartesian co-ordinate axes x1, x2 and x3 (Fig. 3.2) and intersecting them at points A, B and C respectively. Thus ABCP forms a trapezoid.
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The stress components acting on the three perpendicular faces of this trapezoid, which lie in co-ordinate planes are the components of stress tensor or stress state at the point P and are completely known. We have to find the stress vector S acting on the inclined plane ABC. Let S1, S2 and S3 be the magnitudes of the three components of the vector S in the directions of co-ordinate axes. Let ‘As‘ be the area of the triangular face ABC, and A1, A2 and A3 be areas of the triangular faces BPC, CPA and APB respectively.
The trapezoid ABCP would be in equilibrium if the resultant forces acting along the three axes are zero.
Therefore, we can write the following equations:
In the above equation and in the following a repeated subscript in a term means that there is summation of terms over the values of the subscript. The RHS of Eqn. (3.4) is a sum of three terms formed for j = 1, j = 2, and j = 3.
The stress normal to the plane ABC (σn) is obtained by projecting the three components S1, S2 and S3 on to the normal. Thus σn can be obtained by dot product of vector S with the unit normal vector n as given below-
If we go on changing the orientation of the inclined plane we shall come across a set of three perpendicular planes on which the shear stresses are zero. These planes are called principal planes and the normal stresses acting on them are called principal stresses.
Principal Stresses:
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When the total stress vector on a plane is directed along the normal to the plane, the stress vector is called principal stress and the plane is called principal plane. There is no shear stress component on a principal plane.
For a given stress state there is only one set of three principal planes and only one set of three principal stresses acting on them. Now again consider (Fig. 3.2). For some orientations of the inclined plane the shear stress component will become zero and the total stress vector on these planes would be directed in the direction of their normal. Therefore, in such cases, the stress vector will be a scalar multiple of unit normal vector as given by the following equation.
If we take n1, n2 and n3 as variables, the above three Eqns. (3.12 a-c) are linear homogeneous equations in n1, n2 and n3. For a non-trivial solution of these equations, the determinant of coefficients of the variables in the above three equations must be zero. Therefore,
The Eqn. (3.13) is a cubic equation in σ and the three roots of this equation are the three principal stresses. The above equation, after expanding the determinant and multiplying by -1 may be written as follows.
Since there is only one set of three principal stresses, therefore, the Eqn. (3.14) must remain same irrespective of the changes in co-ordinate system, because, it has to have the same three roots. With the change of co-ordinate system the different stress components change, however, the coefficients I, II and III do not change, hence, these coefficients are called invariants of stress tensor.
This important quality of invariability of these coefficients is helpful in writing the constitutional equations of materials behavior. The first invariant denotes that the sum of three normal stresses acting on three mutually perpendicular planes is invariant for a stress state. This sum is also equal to the sum of three principal stresses.
Let us denote the principal stresses by σ1, σ2 and σ3. These are illustrated in Fig. 3.3. This stress state is equivalent to the one shown in Fig. 3.1 from which principal stresses are derived. However, the directions of co-ordinate axes in the two cases are, of course, different.
Planes of Maximum Shear Stresses:
If we change the orientation of the inclined plane (Fig. 3.2), we shall come across a set of six planes on which the shear stresses are the maximum but normal stresses are not zero. These are called planes of maximum shear stress. The direction cosines of these planes may be determined as given below.
The shear stress on an inclined plane is given by following Eqn.-
Figure 3.4 illustrates all the six planes of maximum shear stresses. As you will notice these planes are inclined at 45 degrees to the principal planes.
Equations of Motion:
A deformable body subjected to external forces, sets up internal resisting stresses due to movement of its atoms. These stresses vary from point to point inside the body. The equilibrium at any point of the body is established by Newton’s second law of motion.
Let us consider a cubical element with dimensions dx1, dx2, and dx3 at a point P inside the stressed body (Fig. 3.10). The figure shows the nine components of stress tensor acting on the back faces of the cube as well as the nine components along with their variations over the dimensions dx1, dx2, and dx3 on the front faces of the cube.
For equilibrium, the resultant forces along each of the three axes must be zero. Let F1, F2 and F3 be the components of body force, acting along x1, x2 and x3 respectively. By equating the forces along x1 direction we get,
where ρ is the density of the material. The R.H.S. of the above equation consists of acceleration force along x1– axis. By neglecting the body force component F1, the above equation may be simplified to the following.
These equations are called equations of equilibrium.
The above three equation involve nine components of stresses out of which six are independent, because σ12 = σ21, σ23 = σ32 and σ13 = σ31. For the solution of the six stress components we need six equations. Here we have got only three equations.
The other three equations are obtained from equations of compatibility which are given below. The compatibility equations are in strains which are converted into stresses with the help of stress- strain relations. Thus we get six equations which along with boundary conditions make a complete system of equations for the solution of the six stress components.
Equivalent Stress:
Equivalent stress and equivalent strains (also called effective or generalized stress and strain) are often used in theory of plasticity. They are also called representative stress and representative strain. The equivalent stress σ̅ is defined as-
In case of uni-axial tension or compression, i.e. σ1 ≠ 0 while σ2 = σ3 = 0. Hence the equivalent stress becomes,
σ̅ = σ1