Objectives:
Engineering Stress and Strain; modulus of elasticity:
where Ao is the original cross-sectional
area of the test specimen before the stress is applied and 
is
the original distance between two gage marks on the specimen (see Fig.
6-1, page 136). Typical data from a tensile test are given in Table 6-1
(page 136) and the calculations from the table and subsequent plotting
of the stress-strain curve are shown on page 137. You should befamiliar
with the units listed at the bottom of page 137. A design problem based
only on stress-strain considerations is Example 6-2, page 138.
Yield strength is a stress, not a force! It is the stress at which plastic deformation becomes "noticeable." Since this is a relative term, engineers often arbitrarily choose a 0.2 % offset (which equals a strain of 0.002 in/in). This applies most often to ductile materials such as the nonferrous alloys of aluminum and copper, whose stress-strain curve looks like Figure 6-3a, page 139. Ferrous materials often have a stress-strain curve like Figure 6-3b, page 139.
Referring to Figure 6-3a, the material is stressed to approximately 40 ksi. (I often use kips per square inch to reduce the need to use commas, which can be confusing if you're talking with European engineers.) Then the stress is removed. The material returns elastically and thus ends up with a permanent deformation - a strain of 0.002 in/in. It's important to realize that this elastic return occurs. For example, suppose you need to draw down 2 inch (diameter) rod stock to 1.5 inches. Obviously, the rod has to enter the plastic deformation section of the stress-strain curve. If you use a die which has a diameter of 1.5 inches, the rod will have a diameter of 1.5 inches inside the die, where it's under stress. But as soon as it emerges from the die, it will expand elastically to some diameter a little larger. The die has to be smaller than 1.5 inches! One of you homework problems requires you to compute the diameter of a die.
The modulus of resilience is a measure of the elastic energy (energy per unit volume) that the material absorbs during loading (or releases during unloading). It is the area under the stress-strain curve up to the point of maximum stress applied. It is not difficult to see this from the units:
Poisson's ratio is the negative of the ratio of the lateral strain to the longitudinal strain:
As stress is applied to a test specimen,
necking occurs and the lateral dimension decreases whilethe longitudinal
dimension increases. The minus sign makes Poisson's ratio a positive number.
While your text does not mention it, the three moduli (elastic or Young's,
shear, and bulk) can be related through this ratio; they are not independent
of each other. You'll find values of 
in
Table 6-3 on page 141.
Brittle materials cannot be subjected to the same tensile tests that are used for ductile materials. Your text explains the bend test (page 143), and the flexural strength and modulus. Since these materials often have elastic moduli which are much smaller in tension than in compression, structures which incorporate brittle materials (like concrete) are usually designed so that the materials are in compression rather than tension. This is the reason for prestressing concrete.
True stress and strain are explained on page 146. These are often irrelevant terms, since most structural design does not exceed the yield point. But this is not true for the production of such materials; rolling, extruding, etc. require plastic deformation (i.e., exceeding the yield stress). And it may not be true for the production of some force-fitted components. For example, a very tight fit of a cylindrical sleeve onto a rod can be obtained by starting with a sleeve of the same diameter (or eve a little less) as the rod, heating it so that it expands, sliding it onto the rod, and then cooling it. Since it cannot contract back to its original size, there is residual stress in both components which may be large enough to cause one component to enter the plastic region.
The Brinell and Rockwell hardness tests are very commonly used (see page 351 for a TTT diagram used to determine the required quench rate to produce a certain hardness).
The Charpy and Izod impact tests are used to determine the response of materials to sudden impacts. Ductile materials often behave as if they were brittle under sudden impacts. And the reverse is sometimes true. Have you ever tried to squash a golf ball? Yet high speed photography of the moment of impact of a golf club with the ball shows that the ball changes shape considerably - momentarily. Figure 6-13 (page 150) plots the impact strength vs the temperature for a nylon polymer. Note the sharp drop in strength at the transition temperature? Compare this with Figure 6-14 (page 151). Stainless steel, being FCC, does not have a transition temperature, but an ordinary carbon steel does. This brings to mind the experience of the Navy with certain steel ships during World War II. Some were cracking in the cold waters of the North Atlantic! A good impact design problem is explained in Example 6-7, page 151.
Flaws in a material concentrate stress; i.e., the stress is larger there than in the bulk of the material. Most of us have noticed that a crack in glass quickly propagates. Sometimes people stop this by drilling a hole at the end of the crack. The hole has a much larger radius than the end of the crack did, and so the stress is reduced. Your text discusses the stress intensity factor; you should be able to do a problem like Example 6-8, page 155.
If you'll look at Equation 6-14, and solve it for the stress, you can see that the smaller the flaw size (a), the larger is the stress that can be applied.
Material fatigue, if it occurs, happens after repeated (cyclical) application of a stress. The stress may be applied by rotation, repeated bending, or vibration. For example, most of us haveremoved a tin can top by bending it back & forth until it broke. Figure 6-19 on page 157 shows that tool steel can withstand 100,000 cycles if the stress is 90,000 psi (90 ksi - kips per square inch). The endurance limit of 60 ksi means that there is a 50 % probability that fatigue failure will never occur for tool steel if the stress is less than 60 ksi; this is the preferred design criterion. Nonferrous alloys (like aluminum) do not have an endurance limit. Please review the design problem on page 158.