Load-bearing structures in engineering applications need some attributes in response to some particular forms of mechanical loading conditions. There are mechanical properties like strength and ductility that can be evaluated from simple uniaxial ten­sile testing, whereas dynamic mechanical properties like fatigue could only be eval­uated from cyclic loading testing. Here, we will assume that the readers have been familiar with a basic engineering mechanics course. In this chapter, tensile, impact, fracture toughness, and fatigue and creep properties will be briefly discussed. The American Society for Testing of Materials (ASTM) has developed specific standards for each test procedure. These will be referred to here from time to time. It is impor­tant to understand them before we start deliberations on the effects of radiation on these properties in Chapter 6.

Before we go into the details of the various sections of the mechanical properties, we first describe the modes and types of deformation. There are, in general, four ways that a load may be applied: tension, compression, shear, and torsion, as depicted in Figure 5.1; we will not be concerned here of torsion. Force applied per unit area is known as stress (s or t) and the sample displacement per unit length is the strain (e or y); tension is regarded as positive while compression negative.

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Thus, from Figure 5.1, we note the following relations:

Note that the stress has units of N m~2 known as Pascal (Pa), and we generally talk about MPa (=106 Pa = Nmm~2), while the strain has no units and often referred to as %.

There are three types of deformation: elastic, anelastic, and plastic. Elastic defor­mation is instantaneous completely recoverable deformation where the body returns to the original shape once the external forces are removed. Anelastic defor­mation is time-dependent completely recoverable deformation, meaning that once the forces are removed, the body returns to the original shape in a time-dependent fashion. Plastic deformation is permanent deformation; i. e., it does not recover neither instantaneously nor as a function of time. During elastic deformation, force is directly proportional to the resulting strain as in Hooke’s law (force proportional

to displacement), and during uniaxial tension or compression, the proportionality constant is the Young’s modulus (E) also referred to as elastic modulus:

E = -. (5.2a)

e

Similarly, for shear loading,

—

G = -. (5.2b)

c

Here, for shear loading, G in Eq. (5.2b) is known as shear modulus or modulus of rigidity. In Section 5.1.1, we describe tensile properties, while shear character­istics will be mentioned as appropriate.

We note that when tensile stress is applied (Figure 5.1a), the diameter of the specimen decreases and the ratio of transverse contraction to longitudinal elonga­tion is given by the Poisson’s ratio (n), which is positive and has a value close to 0.3.

This implies that during elastic deformation, volume increases, which is known as elastic dilatation. For rectangular cross section in a tensile bar of an isotropic mate­rial with loading along the z-direction, the Poisson’s ratio is given by

Some materials such as Zr and Ti alloys in general exhibit anisotropy in which case the ratios of contractile width to axial elongation and contractile thickness to axial elongation may not be the same.

The volume change is given by the sum ofthe three orthogonal strains,

DV, r „ .

— = ex + ey + ez. (5.3a)

Under the application of hydrostatic pressure (P), the volume change is related to the bulk modulus (k), the inverse of which is known as compressibility (b):

The above relationships are usually derived in mechanical property courses and it is sufficient here to know that such interrelationships exist.

Example Problem

A cylindrical steel specimen (10 mm in diameter and 40 mm in length) is subjected to a stress of 100 MPa resulting in the length and diameter of the elastically deformed specimen of 40.019 and 9.9986 mm, respectively. Evaluate (i) elastic modulus, (ii) Poisson’s ratio, (iii) shear modulus, and (iv) change in volume.

184 I 5 Properties of Materials Solution

We first need to calculate the longitudinal (ej) and diametral (eD) strains:

Є = —- =————- 9 = 0.000475 = 0.0475% and eD =—-

‘ l 40 D D

= -0.00014 = -0.014%.

= 210 526 = 210.5 GPa.

2(1 + n) 2(1.295)

Change in volume can be found either by finding the final volume (pD2l/4) and subtracting from the initial volume or by adding the strains (el + 2eD) to be 1.953 x 10-4 or 0.0195%, an increase.

5.1.1