Plane Stress and Strain State
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Plane stress and strain state
Generalised HOOKE's Law
Assuming isotropic material properties in all spatial and axial directions, the generalised HOOKE's law (Eq. 1) is as follows:
| (1) |
It can be seen that the strain in the respective axial direction is primarily caused by the stress in the same axis, but an additional strain component from the other stress directions is added, depending on Poisson's ratio ν and the modulus of elasticity E. In the case of uniaxial stress in a tensile or compression test, in addition to the longitudinal strain εx, there is also transverse strain εy and εz, i.e. in the width and thickness directions of the test specimen, if a plane stress state is present. This is usually the case when a sufficiently slender test specimen is used, which allows the transverse strain to be measured in the width or thickness direction (Fig. 1).
| Fig. 1: | Deformation of the test specimen under load in a plane stress state |
In addition to the normal stress σx (Eq. 2), loading causes longitudinal strain εx or εL (Eq. 3) and transverse strain in the width and thickness directions (Eq. 4).
| (2) |
| (3) |
| (4) |
In the case of isotropy and a homogeneous material state, the strain in the width direction εy and in the thickness direction εz (Eq. 5) are identical, with the absolute value being specified for both transverse strains. The stresses σy and σz are equal to zero in the plane stress state.
| (5) |
Determination of Poisson's ratio
The Poisson's ratio, also known as the transverse contraction number, is then calculated according to Eq. (6):
| (6) |
If, instead of the slim test specimens, thicker and/or wider prismatic test specimens are used as shown in Fig. 2, no measurable strain signal will be obtained in either the thickness or width direction, as the test specimen is now in a plane strain state.
| Fig. 2: | Deformation of the test specimen under load in a plane state of strain |
The resulting longitudinal strain εx should be identical, which is why an increased normal stress σx is required due to the larger cross-section A0. To explain this situation, we can assume an identical geometry as in Fig. 1, whereby a possible change in the width of the test specimen is prevented by lateral abutments (Fig. 3). If these supports could also be equipped with a large number of load cells (see: electro-mechanical force transducer and piezoelectric force transducer), the forces in the width and thickness directions of the test specimen could be measured. Under load, this would result in an identical longitudinal strain εx according to Eq. (3) and a higher normal stress σx. In this case, Poisson's ratio ν can no longer be calculated because the strains εy and εz are equal to zero. Since forces in the y and z directions would be measured in this simulated state, stresses also arise in these directions, the magnitude of which can be calculated according to Eq. (1) [1].
| Fig. 3: | Simulation of the slender test specimen in a plane strain state |
See also
- Uniaxial stress state
- Energy elasticity
- Crack model according to IRWIN and Mc CLINTOCK
- Plastic zone
- Plastic hinge model
References
| [1] | Bierögel, C.: Quasi-Static Test Methods. In: Grellmann, W., Seidler, S. (Eds.): Polymer Testing. Carl Hanser, Munich (2022) 3rd Edition, pp. 101–143 (ISBN 978-1-56990-806-8; E-Book: ISBN 978-1-56990-807-3; see AMK-Library under A 23) |



