14.7.1 Tensile Stress and Young’s Modulus

Suppose you clamp one end of a solid rod while using a strong machine to pull on the other with force F⃗. FIGURE 14.37a shows the experimental arrangement. We usually think of solids as being, well, solid. But any material, be it plastic, concrete, or steel, will stretch as the spring-like molecular bonds expand.

FIGURE 14.37 Stretching a solid rod.

The figure labeled ("a") illustrates molecular bonds in a stretched solid rod.
The figure labeled (b) illustrates a force versus change in length graph for the stretching of a solid rod.
Full Alternative Text

FIGURE 14.37b shows graphically the amount of force needed to stretch the rod by the amount ΔL. This graph contains several regions of interest. First is the elastic region, ending at the elastic limit. As long as ΔL is less than the elastic limit, the rod will return to its initial length L when the force is removed. Just such a reversible stretch is what we mean when we say a material is elastic. A stretch beyond the elastic limit will permanently deform the object; it will not return to its initial length when the force is removed. And, not surprisingly, there comes a point when the rod breaks.

For most materials, the graph begins with a linear region, which is where we will focus our attention. If ΔL is within the linear region, the force needed to stretch the rod is

F=k ΔL (14.39)

where k is the slope of the graph. You’ll recognize Equation 14.39 as none other than Hooke’s law.

The difficulty with Equation 14.39 is that the proportionality constant k depends both on the composition of the rod—whether it is, say, steel or aluminum—and on the rod’s length and cross-section area. It would be useful to characterize the elastic properties of steel in general, or aluminum in general, without needing to know the dimensions of a specific rod.

We can meet this goal by thinking about Hooke’s law at the atomic scale. The elasticity of a material is directly related to the spring constant of the molecular bonds between neighboring atoms. As FIGURE 14.38 shows, the force pulling each bond is proportional to the quantity F/A. This force causes each bond to stretch by an amount proportional to ΔL/L. We don’t know what the proportionality constants are, but we don’t need to. Hooke’s law applied to a molecular bond tells us that the force pulling on a bond is proportional to the amount that the bond stretches. Thus F/A must be proportional to ΔL/L. We can write their proportionality as

FA=YΔLL (14.40)

FIGURE 14.38 A material’s elasticity is directly related to the spring constant of the molecular bonds.

The figure illustrates a horizontal cylindrical object of length L.
FIGURE 14.38 Full Alternative Text

The proportionality constant Y is called Young’s modulus. It is directly related to the spring constant of the molecular bonds, so it depends on the material from which the object is made but not on the object’s geometry.

A comparison of Equations 14.39 and 14.40 shows that Young’s modulus can be written as Y=k L/A. This is not a definition of Young’s modulus but simply an expression for making an experimental determination of the value of Young’s modulus. This k is the spring constant of the rod seen in Figure 14.37. It is a quantity easily measured in the laboratory.

The quantity F/A, where A is the cross-section area, is called tensile stress. Notice that it is essentially the same definition as pressure. Even so, tensile stress differs in that the stress is applied in a particular direction whereas pressure forces are exerted in all directions. Another difference is that stress is measured in N/m2 rather than pascals. The quantity ΔL/L, the fractional increase in the length, is called strain. Strain is dimensionless. The numerical values of strain are always very small because solids cannot be stretched very much before reaching the breaking point.

With these definitions, Equation 14.40 can be written

stress=Y×strain (14.41)

Because strain is dimensionless, Young’s modulus Y has the same dimensions as stress, namely N/m2. TABLE 14.4 gives values of Young’s modulus for several common materials. Large values of Y characterize materials that are stiff and rigid. “Softer” materials, at least relatively speaking, have smaller values of Y. You can see that steel has a larger Young’s modulus than aluminum.

TABLE 14.4 Elastic properties of various materials

Substance Young’s modulus (N/m2) Bulk modulus (Pa)
Steel 2.0×1011 1.6×1011
Copper 1.1×1011 1.4×1011
Aluminum 7.0×1010 7.6×1010
Concrete (typical) 3.0×1010 –
Wood (Douglas fir) 1.0×1010 –
Plastic (polystyrene) 3.5×109 –
Mercury – 2.9×1010
Water – 2.2×109
A photograph of a concrete flyover.

Concrete is a widely used building material because it is relatively inexpensive and, with its large Young’s modulus, it has tremendous compressional strength.

We introduced Young’s modulus by considering how materials stretch. But Equation 14.41 and Young’s modulus also apply to the compression of materials. Compression is particularly important in engineering applications, where beams, columns, and support foundations are compressed by the load they bear. Concrete is often compressed, as in columns that support highway overpasses, but rarely stretched.

NOTE   Whether the rod is stretched or compressed, Equation 14.41 is valid only in the linear region of the graph in Figure 14.37b. The breaking point is usually well outside the linear region, so you can’t use Young’s modulus to compute the maximum possible stretch or compression.

EXAMPLE 14.15 Stretching a wire

A 2.0-m-long, 1.0-mm-diameter wire is suspended from the ceiling. Hanging a 4.5 kg mass from the wire stretches the wire’s length by 1.0 mm. What is Young’s modulus for this wire? Can you identify the material?

MODEL The hanging mass creates tensile stress in the wire.

SOLVE The force pulling on the wire, which is simply the weight of the hanging mass, produces tensile stress

FA=mgπr2=(4.5 kg)(9.80 m/s2)π(5.0×10−4 m)2=5.6×107 N/m2

The resulting stretch of 1.0 mm is a strain of ΔL/L =(1.0 mm)/(2000 mm)=5.0×10−4. Thus Young’s modulus for the wire is

Y=F/AΔL/L=11×1010 N/m2

Referring to Table 14.4, we see that the wire is made of copper.