The tool SM-07 computes stress or strain components of elastic bodies in two and three dimensions by Hooke’s law. The tool also determines principle stress/strain and maximum in-plane shear stress/strain for two-dimensional problems.

In addition, a tutorial on how to use the tool in computation is provided and a subject review on fundamental theories and useful formulas is presented.

## Tutorial

Example: A plane stress problem

The input in the tool is

After clicking "Run", the results will be shown as follows in another page.

Stresses:

σx = -50.0

σy = 20.0

τxy = -15.0

Young's Modulus (E) = 0.2

Poisson's Ratio (ν) = 0.08

Strains:

εx = -2.5800E2

εy = 1.2000E2

γxy = -1.6200E2

Out-of-Plane Strain Components:

εz = 1.2000E1

γyz = 0

γxz = 0

------------------------

Principal stresses:

σ1 = 2.3079E1 with angle θp1 = -78.40070474 degree

σ2 = -5.3079E1 with angle θp2 = 11.59929526 degree

Maximum in-plane shear stress:

τmax = 3.8079E1 with angle θt = -33.40070474 degree and 56.59929526 degree

On the plane of τmax, average normal stress σavg = (σxy)/2 = -1.5000E1

## Subject Review

Hooke’s law in stress analysis of elastic bodies is a relationship between force and deformation or one between stress and strain.

### Hooke’s Law in Three Dimensions

Hooke’s law for isotropic linearly elastic materials in three dimensions is described by

where

Shear modulus is related to Young’s modulus and Poisson’s ratio by

See Fig. 1 for the sign convention and orientation of stress components. The sign convention and orientation for strain components is similarly defined. The unit of stress components is force per unit area; see Table 1 for commonly used units and conversion. Strain components are dimensionless quantities, whose units are such as micrometers per meter (mm/m), inches per inch (in/in), and μ (10-6).

Figure 1. Stress components in three dimensions

Table 1.Units of Stress

Hooke’s law can also be presented in the following equivalent form

which is obtained by converting Eq. (1) to express stress components in terms of strain components.

### Hooke’s Law for Plane Stress Problems

In a plane stress problem defined in the x-y plane, stress components σz, τxz, τyz are negligible. This reduces Hooke’s law in three dimensions (Eq. (1)) to

The stress components on an element in the x-y plane are shown in Fig. 2. Also by Eq. (1), the out-of-plane strain components become

Figure 2. Stress components in the x-y plane

The principal stresses (maximum and minimum normal stresses) and the orientation θ of the planes of the element with respect to the x-axis (see Fig. 3) are determined by

where the second equation has two roots θp1 and θp2, which are 90apart. Note that no shear stress acts on the planes of principal stresses, as shown in Fig. 3.

Figure 3. Principal stress

The maximum shear stress and the orientation of the planes of the element with respect to the x-axis (see Fig. 4) are given by

where the second equation has two roots, θs1 and θs2, which are 90apart. The orientation θp of the planes of principal stresses and the orientation θs of the planes of maximum shear stress are 45apart. On each of the planes of maximum stress, there acts a normal stress (average normal stress), as shown in Fig. 4, which is given by

Figure 4. Maximum shear stress and average normal stress

### Hooke’s Law for Plane Strain Problems

In a plane strain problem defined in the x-y plane, strain components εz, εxz, εyz are negligible. In this case, Hooke’s law in three dimensions (Eq. (2)) is reduced

The out-of-plane stress components by Eq. (2) are given by

Mathematically, Hooke’s law for a plane stress problem (Eq. (3)) and Hooke’s law for a plane strain problem (Eq. (8)) are convertible to each other. Nevertheless, these problems have different physical meanings, and have different out-of-plane stress and strain components.

Similar to the plane stress problem, the principal strains (maximum and minimum normal strains) and the orientation of planes of principal strains are given by

where the second equation has two roots θp1 and θp2, which are 90apart. No shear strain acts on the planes of the principal strains.

The maximum in-plane shear strain and the associate orientation are determined by

where the second equation has two roots, θs1 and θs2, which are 90apart. The θp and θs are 45apart. On each of the planes of maximum in-plane shear stress, there acts a normal strain (average normal strain) that is given by

Figures 3 and 4 can be used to show principal strains, maximum shear strain and average normal strain if the stress quantities therein are replaced by corresponding strain quantities.

### References

1. Gere, J. M., and Stephen P. Timoshenko, S. P., 1997, Mechanics of Materials, 4th edition, PWS Publishing Co., Boston.

2. Popov, E. P., 1998, Engineering Mechanics of Solids, 2nd ed., Prentice Hall, Upper Saddle River, New Jersey.

3. Riley, W.F., Sturges, L. D., and Morris, D. H., 1999, Mechanics of Materials, 5th ed., John Wiley and Sons, Inc., New York.

4. Shames, I. H., and Pitarresi, J. M., 2000, Introduction to Solid Mechanics, 3rd ed., Prentice Hall, Upper Saddle River, New Jersey.

5. Bedford, A., and Liechti, K. M., 2000, Mechanics of Materials, Prentice Hall, Upper Saddle River, New Jersey.

6. Yang, B., 2005, Stress, Strain, and Structural Dynamics: An Interactive Handbook of Formulas, Solutions, and MATLAB Toolboxes, Elsevier Science, Boston.