Biaxial tension

There are four commonly occurring states of stress, shown in Fig. 3.2. The simplest is that of simple tension or compression (as in a tension member loaded by pin joints at its ends or in a pillar supporting a structure in compression). The stress is, of course, the force divided by the section area of the member or pillar. The second common state of stress is that of biaxial tension. If a spherical shell (like a balloon) contains an internal pressure, then the skin of the shell is loaded in two directions, not one, as shown in Fig. 3.2. This state of stress is called biaxial tension (unequal biaxial tension is obviously the state in which the two tensile stresses are unequal). The third common state of stress is that of hydrostatic pressure. This occurs deep in the earth s crust, or deep in the ocean, when a solid is subjected to equal compression on all sides. There is a convention that stresses are positive when they pull, as we have drawn them in earlier figures. Pressure,... 

Determination of the fourth-rank tensor term F. 2 remains. Basically, F.,2 cannot be found from any uniaxial test in the principal material directions. Instead, a biaxial test must be used. This fact should not be surprising because F-,2 is the coefficient of the product of a. and 02 in the failure criterion. Equation (2.140). Thus, for example, we can impose a state of biaxial tension described by a, = C2 = c and all other stresses are zero. Accordingly, from Equation (2.140),... 

Various tests may be used to determine the survivability of unexposed polymeric GMs. Puncture tests are frequently used to estimate the survivability of FMLs in the held. During a puncture test, a 5/16 steel rod with rounded edges is pushed down through the membrane. A very hexible membrane that has a high strain capacity under biaxial tension may allow that rod to penetrate almost to... 

Since more trenches are being used in double FML landfills, the impact of waste settlement along such trenches should be considered. Figure 26.17 is a simple evaluation of the impact of waste settlement along trenches on the FML. Settlements along trenches will cause strain in the membrane, even if the trench is a very minor ditch. Knowing that when biaxial tension is applied to HDPE, the material fails at a 16-17% strain, it is possible that the membrane will fail at a moderate settlement ratio. 

Sharma (90) describes an apparatus which uses a tubular specimen which may be strained longitudinally with internal pressurization. Loading rate can be varied, and strains are measured by clip gages. This device permits characterization of propellant-like materials in various biaxial tension-tension and biaxial tension-compression stress fields. A schematic of the apparatus is shown in Figure 24. 

Uniaxial tensile criteria can lead to gross inaccuracies when applied to situations where combined stresses lead to failure in multiaxial stress fields. Often one assumes that combined stresses have no influence and that the maximum principal stress governs the failure behavior. An improved approach applied to biaxial tension conditions relies upon a pragmatic biaxial correction factor which is applied to uniaxial data,... 

Sharma (90) has examined the fracture behavior of aluminum-filled elastomers using the biaxial hollow cylinder test mentioned earlier (Figure 26). Biaxial tension and tension-compression tests showed considerable stress-induced anisotropy, and comparison of fracture data with various failure theories showed no generally applicable criterion at the strain rates and stress ratios studied. Sharma and Lim (91) conducted fracture studies of an unfilled binder material for five uniaxial and biaxial stress fields at four values of stress rate. Fracture behavior was characterized by a failure envelope obtained by plotting the octahedral shear stress against octahedral shear strain at fracture. This material exhibited neo-Hookean behavior in uniaxial tension, but it is highly unlikely that such behavior would carry over into filled systems. 

The stress state variable >fr is depicted in Figure 2 and has the following important special values = 45° represents equal biaxial tension 0° and 90° uniaxial tension —90° and 180° uniaxial compression —45° and 135° pure shear and —135° equal biaxial compression. All biaxial stress states can be represented by a range of values from 45° to —135°. 

It is emphasized that defines the state of stress and should not be confused with the orientation angle p. For example, — 45° and p = 45° represents equal biaxial tension acting on a sheet whose ellipse is at 45° to one of the chosen external principal stress axes. This is not equivalent to — 0 and ft = 0, which represents uniaxial tension parallel to the elliptic major axis (which is equivalent to = 90° and p — 90°). Because of symmetry considerations, all possible combinations of elliptic... 

The effect of stress state on the maximum algebraic stress concentration for various orientations of the ellipse (ft) is shown in Figure 4 for an ellipticity R = 0.1. The uniaxial tension condition = 0) generates the highest stress concentration. Equal biaxial tension ( = 45°) generates the same stress concentration for any ellipse orientation p. This is because the stress state is planar isotropic, and any pair of orthogonal axes can be chosen as the external principal axes. Consequently the orientation p is without effect. 

Maximum absolute stress concentration factors are shown in Figure 5, again for an ellipticity R = 0.1. As before, all orientations p give the same stress concentration for equal biaxial tension ( = 45°). In addition equal biaxial compression ( = —135°) gives the same stress concentration since absolute stress concentration is being considered. The shear stress concentration is given by Q /2. The curves shown in Figure 5 are symmetrical about the pure shear condition ( = —45°). 

Assuming planar symmetry, for a two dimensionally constrained film the shrinkage stress would approach 16 MPa in equal biaxial tension. 

For macroscopically isotropic polymers, the Tresca and von Mises yield criteria take very simple analytical forms when expressed in terms of the principal stresses cji, form surfaces in the principal stress space. The shear yield surface for the pressure-dependent von Mises criterion [Eqs (14.10) and (14.12)] is a tapering cylinder centered on the applied pressure increases. The shear yield surface of the pressure-dependent Tresca criterion [Eqs (14.8) and (14.12)] is a hexagonal pyramid. To determine which of the two criteria is the most appropriate for a particular polymer it is necessary to determine the yield behavior of the polymer under different states of stress. This is done by working in plane stress (ct3 = 0) and obtaining yield stresses for simple uniaxial tension and compression, pure shear (di = —CT2), and biaxial tension (cti, 0-2 > 0). Figure 14.9 shows the experimental results for glassy polystyrene (13), where the... 

Figure 14.9 Section of the yield surface in the plane 03 = 0 choosing the Tresca criterion (hexagonal envelope) and von Mises criterion (elliptical envelope) for polystyrene. The points correspond to experiments performed under pure shear (gi = -CJ2), biaxial tension (oj, 03 > 0), and uniaxial tension and compression. (From Ref. 13.)...

Sternstein and Ongchin (28) considered that if cavitation occurs in crazes the criterion for crazing initiation should include the dilative stress component. They proposed the criterion to fit the experimental data for surface craze initiation in PMMA when the polymer is subjected to biaxial tension. The segmental mobility of the polymer will increase due to dilative stresses, thus provoking cavitation and the orientation of molecular segments along the maximum stress direction. 

Consider, therefore, an infinitely large plate of elastic material of thickness B, containing a through-thickness crack of length 2a, and subjected to uniform biaxial tension (a) at infinity as shown in Fig. 2.5. Let U = potential energy of the system, Uo = potential energy of the system before introducing the crack, Ua = decrease... 

Figure 2.5. An infinitely large plate of elastic material containing a through-thickness central crack of length 2a and subjected to uniform biaxial tension a.

The functions fi(y) and fzix) depend on the boundary conditions and cannot be chosen arbitrarily. It will be shown later, for example, that for a cracked plate under remote biaxial tension, the boundary condition on shearing stresses (r y = 0 at infinity) requires that the sum of the last two terms be zero. As such, the derivatives must be equal to a constant of opposite sign, which leads to fi(y) = Ay and fiix) = -Ax. Here, the contribution of/i(y) and/2(x) to the u and v displacements represents rigid body rotation and is not considered further in the stress analysis of cracked bodies. It should be noted that the constant A cannot be arbitrarily neglected. 

Central Crack in an Infinite Plate under Biaxial Tension (Griffith Problem)... 

The case of an infinitely large thin plate, containing a central through-thickness crack of length 2a, subjected to remote, uniform biaxial tension is considered (Fig. 3.3). The boundary conditions are as follows ... 

Figure 3.3. A central through-thickness crack in an infinitely large plate snbjected to remote, nniform biaxial tension.

In other words, fi(y) = -Ay and f (x) = Ax, which correspond to rigid body rotation about the z-axis. The sign is chosen to be consistent with a counter-clockwise rotation. It is clear that the constant A could not be arbitrarily neglected it is zero only for the case of equal biaxial tension. 

The equilibrium small-strain elastic behavior of an "incompressible" rubbery network polymer can be specified by a single number—either the shear modulus or the Young s modulus (which for an incompressible elastomer is equal to 3. This modulus being known, the stress-strain behavior in uniaxial tension, biaxial tension, shear, or compression can be calculated in a simple manner. (If compressibility is taken into account, two moduli are required and the bulk modulus. ) The relation between elastic properties and molecular architecture becomes a simple relation between two numbers the shear modulus and the cross-link density (or the... 

The residual stresses in the CD are—equally biaxial compression near the surface and equally biaxial tensions in the core. According to the stress optic law (Eq. 8.9), the birefringence is proportional to the difference between the principal stresses. Therefore, the birefringence due to residual stresses should be nearly zero. 

Hivet G, Boisse P. Consistent mesoscopic mechanical behaviour model for woven composite reinforcements in biaxial tension. Composites Part B 2008 39(2) 345—61. 

Amijima S, Fujii T, Hamaguchi M. Static and fatigue tests of woven glass fabric composite under biaxial tension—torsion loading. Composites 1991 22(4) 281—9. 

Amijima S, Fujii T, Sagami T. Non-linear behavior of plain woven G.F.R.P. under repeated biaxial tension/torsion loading. J Energy Resour Technol 1991 113 ... 

Residual stresses can also play an important role in layered structures. For example, consider the sandwich structure shown in Fig. 4.19. If the outside faces (subscript F) have a lower thermal expansion coefficient than the core (subscript C), they will be placed in biaxial compression as the structure cools after fabrication. Conversely, a higher expansion coefficient for the faces will place them in biaxial tension. As with fiber composites, the forces between the layers must balance, so a- h+2(Tpt=0. Using Hooke s Law, the force balance can be written as... 

FLDs must cover as much as possible the strain domain which occurs in industrial sheet metalforming processes. The diagrams are established by experiments that provide pairs of values of the limit strains and obtained for various loading patterns (equi-biaxial, biaxial, uniaxial, etc.). In order to determine an FLD, one must generate load paths ranging from equi-biaxial tension (si = S2) to pure shear (si =—82). In practice the state of simple tension (Si =—82 for isotropic... 

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