Compression. Our material properties considerations have, until now, been concer

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Compression. Our material properties considerations have, until now, been concerned with tensile stress and strain, where the applied force increases the material's dimension along the axis of application. If, however, a force is applied to a material such that it reduces the material's dimension along the axis of application, then it is a compressive force and the material is said to be in compression. As with tensile testing, a stress-strain curve can be constructed for a material in compression, and properties such as modulus, elastic and plastic deformation, ultimate strength, malleability (the compression counterpart of ductility), failure strength, etc. can be similarly determined (by convention, negative values are used for compressive force and deformation). For many materials, the stress-strain curve transitions smoothly from tension in quadrant 1 to compression in quadrant 3 of the stress-strain relationship (see Fig. 1). Said another way, the modulus of many materials is the same under compression as it is under tension for suitably small deformations about the strain-0 axis tension compression Figure 1 Shear. Another type of stress that a material can be subjected to is shear stress. Shear stress occurs when opposing parallel forces are applied to a material and the forces do not share the same line of action (for example, when compressive forces are distributed unevenly on opposing surfaces of a material). Shear stresses result in a tendency for one part of the material to slide past (or "shear" from) another part as illustrated in Fig. 2 You will be compression-testing sections of mezzi rigatoni fresh chicken tibia, and dried chicken tibia. Bone specimens will be tested in the "longitudinal" direction (i.e. force will be applied in the direction of the long axis of the bone from which the section came). Testing is performed by supporting the material from below with a fixed, flat surface while controlling a second flat surface, above and parallel to the supporting surface, to move in the direction of the lower surface, thereby compressing the material 1. For constructing stress-strain curves from your raw data, describe how you will determine compressional stress tensional stress shear stress Figure 2 the relevant cross-sectional areas for the bone samples. Perform an example cross-sectional area calculation for a typical chicken tibia using bone dimensions of 8 mm for the outer diameter and 5 mm for the inner diameter. Do you expect the material properties of the bones that you isolated and stored for a week in the fume hood to differ from the material properties of freshly-isolated bone? Why? Why is it important that the test specimens have completely flat ends (i.e. the ends that will be in contact with the compression plates) and that the axis of loading be perpendicular to the flat ends? Describe what you would do to intentionally create shear stresses within the test specimens 2. 3.

Explanation / Answer

1) To determine the relevant cross sectional area of the bone samples, considering an example of a bone dimensions,

Di = 5mm, Do = 8mm

Thus the cross-sectional area can be calculated as,

A = /4 × ( Do2 - Di2 )

For the above example, A = /4 × (82 - 52) = 30.6305 mm2

2) Considering two samples of the bone:

Sample1: Freshly isolated bone

In this case the body fluids are still present in the bone. Thus the bone fibers of the bone are somewhat flexible. This causes the bone to exhibit more elastic properties , i.e. the tensile strength of the bone is higher and it can absorb more energy before failure.

Sample2 : Bone sample stored for a week in a fume hood

In this case the body fluids in the bone fibers dry off. The resulting sample is a hard solid with low flexibility. Thus the hardness of the bone increases but it becomes brittle.

3) For undergoing a compression test the specimen should face the compressive load uniformly distributed throughout its cross sectional area and the compressive load should be perpendicular to the area of cross-section.

For these reasons the specimen should have completely flat ends ( for uniform load distribution) and the axis of loading should be perpendicular to flat ends ( i.e. perpendicular to area of cross-section).

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