Coupon Accepted Successfully!


Solid Properties

We have also been pretending that sticks and strings and such things are absolutely rigid, but we know that solid objects do stretch and bend and sometimes break. Solids bend when you exert different forces on two sides of them. To keep things simple, we will look at how a solid cylinder reacts to forces placed on it.

Figure 7-14 shows two rods, made of the same material and having the same cross-sectional area, but the first rod L1 is longer than the second L2. If we apply the same magnitude tensile force (that is, pulling) to the four ends, then we expect the rods to stretch. But the longer rod has more material to stretch than the shorter one, so the change in length ΔL will be greater. On the other hand, we can even things out by taking the ratio (ΔL)/L, which is called the strain.

..\art 7 jpg\figure 7-tn.jpg

Figure 7-14


Now let's consider two rods of the same length with different cross-sectional areas (Figure 7-15). Again we apply the same magnitude tensile force to the ends. The thicker rod stretches less than the thinner one. This time, to even things out, we introduce the quantity stress, which is F/A. This should remind you of pressure, since pressure is a kind of stress.

..\art 7 jpg\figure 7-to.jpg

Figure 7-15


As long as the forces involved are not too large, the resulting strain is proportional to the stress placed on the rod, so that we can write



where Y is the Young's modulus, having units of [N/m2]. You might think that it is easier to stretch a rod than to compress it, but it turns out not to be so (as long as you do not go too far). The shortening ΔL due to a compressive force (pushing) is the same as the lengthening ΔL due to an equal-sized tensile force. The Young's modulus depends only on the material.


A shear force is a force applied perpendicular to the surface. Figure 7-16 shows an example of four shear forces applied to a block. Note that the net force is zero and the net torque is zero. The block bends, a distance Δx, and the relationship between stress (F/A) and strain (Δx/L) is



where S is the shear modulus with units [N/m2].

..\art 7 jpg\figure 7-tp.jpg

Figure 7-16

Shear is a bit more complicated than compression and tension. In Figure 7-17, the compression force in the trunk is nearly uniform. The shear force in the large branch to the right is composed of tension at the top of the branch and compression at the bottom of the branch, as well as pure shear. If you want to weaken the branch, the most effective place to cut is on top and the least effective place is in the middle, which is called the neutral layer.

..\art 7 jpg\figure 7-tq.jpg

Figure 7-17


The proportionality in equations (6) and (7) holds for a large range of forces, but things will break if you pull them too hard. Before they break, they may go soft. The regime in which equations (6) and (7) holds is called elastic. The point at which the constituent particles of the material begin to flow and cause the material to go soft is the elastic limit. (See Figure 7-18.)

..\art 7 jpg\figure 7-tr.jpg

Figure 7-18


In this chapter we looked at using torques in order to solve for forces in certain static structures. In problems of this type we begin by drawing a force diagram, as we have always done, but now we need to be careful to locate the force at the right place. Generally, in a given problem, there will be forces we do not know and do not need, and it will be possible to choose a pivot so that the torques of all such forces are zero. Then if we write down the torque balance equation τnet = 0, we will be able to obtain the magnitude of the desired forces.

We also looked at the static properties of solids. It is helpful to think in terms of a stress (force per area) being applied to a solid, and this stress causes a strain (displacement per length). For many materials stress and strain are proportional. Just realizing this proportionality is the key to solving some problems.

Test Your Skills Now!
Take a Quiz now
Reviewer Name