Changes in Length—Tension and Compression: Elastic Modulus

A change in length [latex]\Delta L[/latex] is produced when a force is applied to a wire or rod parallel to its length [latex]{L}_{0}[/latex], either stretching it (a tension) or compressing it. (See Figure 32.3.)

Experiments have shown that the change in length ([latex]\Delta L[/latex]) depends on only a few variables. As already noted, [latex]\Delta L[/latex] is proportional to the force [latex]F[/latex] and depends on the substance from which the object is made. Additionally, the change in length is proportional to the original length [latex]{L}_{0}[/latex] and inversely proportional to the cross-sectional area of the wire or rod. For example, a long guitar string will stretch more than a short one, and a thick string will stretch less than a thin one. We can combine all these factors into one equation for [latex]\Delta L[/latex]:

where [latex]\Delta L[/latex] is the change in length, [latex]F[/latex] the applied force, [latex]Y[/latex] is a factor, called the elastic modulus or Young’s modulus, that depends on the substance, [latex]A[/latex] is the cross-sectional area, and [latex]{L}_{0}[/latex] is the original length. Table 32.1 lists values of [latex]Y[/latex] for several materials—those with a large [latex]Y[/latex] are said to have a large tensile stiffness because they deform less for a given tension or compression.

Young’s moduli are not listed for liquids and gases in Table 32.1 because they cannot be stretched or compressed in only one direction. Note that there is an assumption that the object does not accelerate, so that there are actually two applied forces of magnitude [latex]F[/latex] acting in opposite directions. For example, the strings in Figure 32.3 are being pulled down by a force of magnitude [latex]w[/latex] and held up by the ceiling, which also exerts a force of magnitude [latex]w[/latex].

Example 32.1: The Stretch of a Long Cable

Suspension cables are used to carry gondolas at ski resorts. (See Figure 32.4) Consider a suspension cable that includes an unsupported span of 3 km. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is [latex]3\text{.}0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex].

Strategy

The force is equal to the maximum tension, or [latex]F=3\text{.}0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex]. The cross-sectional area is [latex]{\mathrm{\pi r}}^{2}=2\text{.}\text{46}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}[/latex]. The equation [latex]\Delta L=\frac{1}{Y}\frac{F}{A}{L}_{0}[/latex] can be used to find the change in length.

Solution

All quantities are known. Thus,

[latex]\begin{array}{lll}\Delta L& =& \left(\frac{1}{\text{210}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}}\right)\left(\frac{3\text{.}0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N}}{2.46×{10}^{–3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}}\right)\left(\text{3020 m}\right)\\ & =& \text{18 m}.\end{array}[/latex]

Discussion

This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important in these environments.

Biological Applications of Elasticity

Bones Under Load

Bones, while strong under tension and compression, are more likely to fracture under shear or bending forces. When a sideways impact or torsional stress occurs—like during a fall—the bone can snap or shear, rather than compress. Bones serve as weight-bearing columns, like those found in architecture or tree trunks. They are reinforced naturally with fibrous structures, just as buildings use steel rods.

The internal architecture of bones varies depending on their location and function. For instance:

• Femur head: Made of thin, sheet-like bone separated by marrow to manage compressive stress.

• Long bones (e.g., tibia, humerus): Cylindrical, solid or marrow-filled, optimized for longitudinal stress.

• Joint bones: Especially sensitive to sustained compression, which is why overweight individuals are prone to joint and tendon damage.

Tendons: Nonlinear Elasticity

Tendons connect muscle to bone and exhibit nonlinear elasticity—a hallmark of complex biological materials. This means that tendons:

• Stretch easily at first (low stiffness in the toe region).

• Become significantly stiffer under larger forces (in the linear region).

• Eventually fail if the force continues (in the failure region).

Figure 32.5 shows the stress-strain curve for mammalian tendon:

This behavior is often modeled by parallel springs, where each spring activates at different lengths of stretch—mimicking how different collagen fibers respond.

Ligaments, Arteries, Lungs, and Skin

• Ligaments (connecting bone to bone) behave similarly to tendons, though often with lower strain tolerance.

• Arteries must be highly elastic to accommodate blood pressure pulses. When blood is pumped out of the heart, the arterial walls expand. When the aortic valve closes, elastic recoil helps maintain blood flow.

• Lungs expand with muscular effort and recoil passively—another beautiful example of elasticity in function.

• Skin, especially in young individuals, exhibits remarkable elasticity. Youthful skin can accommodate large body size changes (e.g., pregnancy, weight loss) with little sag.

• Unfortunately, elasticity diminishes with age, beginning subtly in our 20s. This affects not just cosmetic properties, but also organ function.

Stress, Strain, and Shear: Extensions of Hooke’s Law

The equation for change in length due to an applied force can be rearranged and written as:

[latex]\frac{F}{A} = Y \frac{\Delta L}{L_0},[/latex]

where:

• [latex]\frac{F}{A}[/latex] is the stress (in [latex]\text{N/m}^2[/latex]),

• [latex]\frac{\Delta L}{L_0}[/latex] is the strain (dimensionless), and

• [latex]Y[/latex] is Young’s modulus (a material-specific constant).

In other words,

[latex]\text{stress} = Y \times \text{strain}.[/latex]

This form mirrors Hooke’s law, treating stress like force and strain like deformation. If we solve for force, we get:

[latex]F = Y A \frac{\Delta L}{L_0},[/latex]

and the proportionality constant resembles a spring constant:

[latex]k = \frac{YA}{L_0}.[/latex]

This general idea—that force and the deformation it causes are proportional for small deformations—applies to changes in length, sideways bending, and changes in volume.

Definitions

Stress
The ratio of force to area:

[latex]\text{stress} = \frac{F}{A}[/latex]
Units: [latex]\text{N/m}^2[/latex]

Strain
The ratio of change in length to the original length:

[latex]\text{strain} = \frac{\Delta L}{L_0}[/latex]
(Strain has no units.)

Sideways Stress: Shear Modulus

Sideways stress, or shear, is another important deformation mode. In this case, the deformation [latex]\Delta x[/latex] is perpendicular to the original length [latex]L_0[/latex]. This behavior follows a similar relation:

[latex]\Delta x = \frac{1}{S} \frac{F}{A} L_0,[/latex]

where:

• [latex]S[/latex] is the shear modulus,

• [latex]F[/latex] is the force applied parallel to the surface, and

• [latex]A[/latex] is the cross-sectional area across which the force is applied.

Shear Deformation

[latex]\Delta x = \frac{1}{S} \frac{F}{A} L_0[/latex]

As illustrated in Figure 32.6, two equal and opposite shearing forces [latex]F[/latex] are applied, and the object deforms by [latex]\Delta x[/latex] without rotating.

Shearing is easier in objects with longer lengths, smaller cross-sectional areas, or lower shear modulus—like rubber or soft tissues.

Biological Implications of Shear Stress

• Bone is a surprising case: its shear modulus exceeds its Young’s modulus and approaches that of steel, which explains its rigidity.

• The spinal column, comprised of 26 vertebrae separated by discs, is highly affected by shear forces—especially at its curved lower end.

  • Pregnancy or obesity increases curvature, hence increasing shear on the lumbosacral disc, which raises risk for disc rupture.

• Concrete and brick have very low shear moduli and resist compression far better than shear. This is why steel reinforcement is critical in modern buildings.

• Liquids and gases flow under shear and are considered to have zero shear modulus.

Changes in Volume: Bulk Modulus

An object will be compressed in all directions if inward forces are applied evenly across all its surfaces, as shown in Figure 32.8. Gases are relatively easy to compress, whereas liquids and solids are extremely difficult to compress. For example, the air in a wine bottle is compressed when corked, but a brim-full bottle of wine cannot be corked unless some liquid is removed—liquids are nearly incompressible.

The reason lies in the structure of matter: gases have atoms and molecules widely spaced, allowing them to be pushed closer together. In contrast, atoms and molecules in liquids and solids are already tightly packed, and further compression means pushing against strong electromagnetic forces between them.

Bulk Modulus Equation

A uniformly applied inward force generates the same stress, defined as force per unit area [latex]\frac{F}{A}[/latex], on all surfaces of an object. This results in a change in volume, [latex]\Delta V[/latex], given by:

[latex]\Delta V = \frac{1}{B} \frac{F}{A} V_0,[/latex]

where:

• [latex]B[/latex] is the bulk modulus (see Table 32.1),

• [latex]V_0[/latex] is the original volume,

• [latex]\frac{F}{A}[/latex] is the inward pressure applied uniformly on all sides.

This expression, like those for shear and tension, shows how deformation (in this case, volume change) depends on both the applied stress and the material’s resistance.

Note: Bulk modulus is typically not defined for gases, since they compress too easily to follow the same relationship.

Real-World Applications of Bulk Modulus

• Industrial diamonds are created by applying massive pressure to carbon. The atoms are forced into a tightly packed crystal structure, forming diamond.

• Natural diamond formation occurs under similar conditions, deep beneath Earth’s surface where overlying rock exerts extreme compressive forces.

• Deep-sea pressure is another example. Water pressure increases with depth and exerts inward forces on submerged objects—and even on the water itself. At sufficient depths, water itself becomes measurably compressed.

Conversely, very large internal forces are generated when liquids or solids attempt to expand but are constrained from doing so. This situation is mechanically equivalent to compressing them below their equilibrium volume. Such conditions frequently arise due to thermal expansion: most materials increase in volume as their temperature rises. When this expansion is restricted, the resulting internal stresses can cause significant deformation or even failure of the surrounding container.

A dramatic and common illustration of this occurs when water freezes. Unlike most substances, water expands upon freezing. This expansion can exert enough force to:

• Fracture rocks and boulders through freeze-thaw weathering,

• Rupture biological cells, causing frost damage in plants and tissues,

• Crack engine blocks or pipes if water inside them freezes without space to expand.

These phenomena emphasize the importance of understanding elastic and plastic deformation across materials under varying physical conditions.

Finally, it is worth noting that other types of deformation—such as torsion (twisting)—follow similar mechanical principles as those described for tension, shear, and bulk compression. These torsional deformations are characterized by their own modulus (the torsion modulus) and will be encountered in further sections or advanced studies.