Compressive Stress Strain

A test specimen under loading conditions located between the two flat, parallel faces of a testing machine is compressed at a specified rate (ASTM D 695). Stress and strain are computed from the measured compression test, and these are plotted as a compressive stress-strain curve for the material at the temperature and strain rate employed for the test.

Procedures in compression testing are similar to those in tensile testing. However in compression testing particular care must be taken to specify the specimen's dimensions and relate test results to these dimensions. If a sample is too long and narrow buckling may cause premature failure resulting in inaccurate compression test results. Buckling can be avoided by examining different size specimens. Consider a test specimen with a square cross-section and a longitudinal dimension twice as long as a side of the cross-section.

At high stress levels, compressive strain is usually less than tensile strain. Unlike tensile loading, which usually results in failure, stressing in compression produces a slow, indefinite yielding that seldom leads to failure. Where a compressive failure does occur, the designer should determine the material's strength by dividing the maximum load the sample supported by its initial cross-sectional area. When the material does not exhibit a distinct maximum load prior to failure, the designer should report the strength at a given level of strain that is usually at 10%.

The compression specimen's ends usually do not remain rigid. They tend to spread out or flower at its ends. Test results are usually very scattered requiring close examination as to what the results mean in reference to the behavior of the test specimens. Different clamping devices (support plates on the specimen sides, etc.) are used to

Figure 2 Stress-strain tensile and compressive response tends to be similar


Figure 2 Stress-strain tensile and compressive response tends to be similar


eliminate the flowering action that could provide inaccurate readings that in turn influence results by usually making them stronger.

The majority of tests to evaluate the characteristics of plastics are performed in tension and/or flexure because compression data could be misleading. The result is that compressive stress-strain behavior of many plastics is not well described. Generally, the behavior in compression is different from that in tension, but the S-S response in compression is usually close enough to that of tension so that possible differences can be neglected. Fig. 2.8 compares tensile and compression data of TPs.

The compressive strength of a non-reinforced plastic or a mat-based RP laminate is usually greater than its tensile strength (unidirectional fiber-reinforced plastic is usually slightly lower than its tensile strength). However, this is not generally true for reinforced TSs (RTSs). Different results are obtained with different plastics. As an example the compression testing of foamed plastics provides the designer with the useful recovery rate. A compression test result for rigid foamed insulating polyurethane (3.9 lb/ft3) resulted in almost one-half of its total strain recovered in one week.

Shear Stress-Strain

Shear deformation occurs in structural elements such as those subjected to torsional loads and in short beams subjected to transverse loads. Shear S-S data can be generated by twisting (applying torque) a specimen at a specified rate while measuring the angle of twist between the ends of the specimen and the torque load exerted by the specimen on the testing machine (ASTM D 732). Maximum shear stress at the surface of the specimen can be computed from the measured torque that is the maximum shear strain from the measured angle of twist.

The shear mode involves the application of a load to a material specimen in such a way that cubic volume elements of the material comprising the specimen become distorted, their volume remaining constant, but with opposite faces sliding sideways with respect to each other. Basically, shearing stresses are tangential stresses that act parallel to the planes they stress. The shearing force in a beam provides shearing stresses on both the vertical and horizontal planes within the beam. The two vertical stresses must be equal in magnitude and opposite in direction to ensure vertical equilibrium. However, under the action of those two stresses alone the element would rotate.

Another couple must counter these two stresses. If the small element is taken as a differential one, the magnitude of the horizontal stresses must have the value of the two vertical stresses. This principle is sometimes phrased as cross-shears are equal that refers to a shearing stress that cannot exist on an element without a like stress being located 90 degrees around the corner.

Fig. 2.9 schematic diagram is subjected to a set of equal and opposite shearing force/load (Q). The top view represents a material with equal and opposite shearing forces and the bottom view is a schematic of a theoretical infinitesimally thin layers subject to shear stress. As shown at the bottom with the infinitesimally thin layers there is a tendency for one layer of the material to slide over another to produce a shear form of deformation or failure if the force is great enough. The shear stress will always be tangential to the area upon which it acts. The shearing strain is the angle of deformation as measured in radians. For materials that behave according to Hooke's Law, shear strain is proportional to the shear stress. The constant G is called the shear modulus, modulus of rigidity, or torsion modulus.

G is directly comparable to the modulus of elasticity used in direct-stress applications. Only two material constants are required to characterize a material if one assumes the material to be linearly elastic, homogeneous, and isotropic. However, three material constants exist: the tensile modulus of elasticity (E), Poisson's ratio (v), and the shear modulus (G). An equation relating these three constants, based on engineering's elasticity principles is as follows:

This calculation that is true for most metals, is generally applicable to

Figure 2 9 Theoretical approaches to shear stress behavior






plastics. However, this calculation does not apply with the inherendy nonlinear, anisotropic nature of most plastics, particularly the fiber-reinforced and liquid crystal ones.

Torsion Stress-Strain

Shear modulus can be determined by a static torsion test or by a dynamic test using primarily a torsional pendulum (ASTM D 2236). Also used is an oscillatory rheometer test. The torsional pendulum is applicable to virtually all plastics and uses a simple specimen readily fabricated by all commercial fabricating processes or easily cut from fabricated part. The moduli of elasticity, G for shear and E for tension, are ratios of stress to strain as measured within the proportional limits of the material. Thus the modulus is really a measure of the rigidity for shear of a material or its stiffness in tension and compression. For shear or torsion, the modulus analogous to that for tension is called the shear modulus or the modulus of rigidity, or sometimes the transverse modulus.

Direct Load Shear Strength

Unlike the methods for tensile, flexural, or compressive testing, the typical procedure used for determining shear properties is intended only to determine the shear strength. It is not the shear modulus of a material that will be subjected to the usual type of direct loading (ASTM D 732). When analyzing plastics in a pure shear situation or when the maximum shear stress is being calculated in a complex stress environment, a shear strength equal to half the tensile strength or that from shear tests is generally used, whichever is less.

The shear strength values are obtained by such simple tests using single or double shear actions. In these tests the specimen to be tested is sheared between the hardened edges of the supporting block and the block to which the load is applied. The shearing strength is calculated as the load at separation divided by the total cross-sectional area being sheared. The maximum short-term shear stress (strength) of a material can also be determined from a punch shear test.

The test fixtures in these test devices indicates that bending stresses exist and the stress cannot be considered as being purely that of shear. Therefore, the shearing stress calculated must be regarded as an average stress. This type of calculation is justified in analyzing bolts, rivets, and any other mechanical member whose bending moments are considered negligible. Note that tests of bolts and rivets have shown that their strength in double shear can at times be as much as 20% below that for single shear. The values for the shear yield point (MPa or psi) are generally not available; however, the values that are listed are usually obtained by the torsional testing of round test specimens.

The data obtained using the test method reviewed should be reported as direct shear strength. These data can only be compared to data determined by the same direct-shear methods. This test cannot be used to develop shear S-S curves or determine a shear modulus, because bending or compression rather than pure shear transfer a considerable portion of the load. The test results depend on the susceptibility of the material to the sharpness of load faces.

It is important to note material such as certain plastics, concrete, or wood that are weak in either tension or compression will also be basically weak in shear. For example, concrete is weak in shear because of its lack of strength in tension. Reinforced bars in the concrete are incorporated to prevent diagonal tension cracking and strengthen concrete beams. Similar action occurs with RPs using fiber filament structures.

Although no one has ever been able to determine accurately the resistance of concrete to pure shearing stress, the matter is not very important, because pure shearing stress is probably never encountered in concrete structures. Furthermore, according to engineering mechanics, if pure shear is produced in one member, a principal tensile stress of an equal magnitude will be produced on another plane. Because the tensile strength of concrete is less than its shearing strength, the concrete will fail in tension before reaching its shearing strength. This action also occurs with plastics.

Residual Stress

It is the stress existing in a body at rest, in equilibrium, at uniform temperature, and not subjected to external forces. Often caused by the stresses remaining in a plastic part as a result of thermal and/or mechanical treatment in fabricating parts. Usually they are not a problem in the finished product. However, with excess stresses, the product could be damaged quickly or after in service from a short to long dme depending on amount of stress and the environmental conditions around the product.

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