Sideways Stress: Shear Modulus

Sideways Stress: Shear Modulus

The figure below illustrates what is meant by a sideways stress or a shearing force. Here the deformation is called \(\Delta x\) and it is perpendicular to \({L}_{0}\), rather than parallel as with tension and compression. Shear deformation behaves similarly to tension and compression and can be described with similar equations. The expression for shear deformation is

\(\Delta x=\frac{1}{S}\frac{F}{A}{L}_{0},\)

where \(S\) is the shear modulus (see this table from the previous lesson) and \(F\) is the force applied perpendicular to \({L}_{0}\) and parallel to the cross-sectional area \(A\). Again, to keep the object from accelerating, there are actually two equal and opposite forces \(F\) applied across opposite faces, as illustrated in the figure below. The equation is logical—for example, it is easier to bend a long thin pencil (small \(A\)) than a short thick one, and both are more easily bent than similar steel rods (large \(S\)).

Shearing forces are applied perpendicular to the length \({L}_{\text{0}}\) and parallel to the area \(A\), producing a deformation \(\text{Δx}\). Vertical forces are not shown, but it should be kept in mind that in addition to the two shearing forces, \(\mathbf{\text{F}}\), there must be supporting forces to keep the object from rotating. The distorting effects of these supporting forces are ignored in this treatment. The weight of the object also is not shown, since it is usually negligible compared with forces large enough to cause significant deformations.

Examination of the shear moduli in this table from the previous lesson reveals some telling patterns. For example, shear moduli are less than Young’s moduli for most materials. Bone is a remarkable exception. Its shear modulus is not only greater than its Young’s modulus, but it is as large as that of steel. This is why bones are so rigid.

The spinal column (consisting of 26 vertebral segments separated by discs) provides the main support for the head and upper part of the body. The spinal column has normal curvature for stability, but this curvature can be increased, leading to increased shearing forces on the lower vertebrae. Discs are better at withstanding compressional forces than shear forces. Because the spine is not vertical, the weight of the upper body exerts some of both.

Pregnant women and people that are overweight (with large abdomens) need to move their shoulders back to maintain balance, thereby increasing the curvature in their spine and so increasing the shear component of the stress. An increased angle due to more curvature increases the shear forces along the plane. These higher shear forces increase the risk of back injury through ruptured discs. The lumbosacral disc (the wedge shaped disc below the last vertebrae) is particularly at risk because of its location.

The shear moduli for concrete and brick are very small; they are too highly variable to be listed. Concrete used in buildings can withstand compression, as in pillars and arches, but is very poor against shear, as might be encountered in heavily loaded floors or during earthquakes. Modern structures were made possible by the use of steel and steel-reinforced concrete. Almost by definition, liquids and gases have shear moduli near zero, because they flow in response to shearing forces.

Example: Calculating Force Required to Deform: That Nail Does Not Bend Much Under a Load

Find the mass of the picture hanging from a steel nail as shown in the figure below, given that the nail bends only \(\text{1.80 µm}\). (Assume the shear modulus is known to two significant figures.)

Side view of a nail with a picture hung from it. The nail flexes very slightly (shown much larger than actual) because of the shearing effect of the supported weight. Also shown is the upward force of the wall on the nail, illustrating that there are equal and opposite forces applied across opposite cross sections of the nail. This example shows a calculation of the mass of the picture.

Strategy

The force \(F\) on the nail (neglecting the nail’s own weight) is the weight of the picture \(w\). If we can find \(w\), then the mass of the picture is just \(\frac{w}{g}\) . The equation \(\Delta x=\frac{1}{S}\frac{F}{A}{L}_{0}\) can be solved for \(F\).

Solution

Solving the equation \(\Delta x=\frac{1}{S}\frac{F}{A}{L}_{0}\) for \(F\), we see that all other quantities can be found:

\(F=\frac{\mathrm{SA}}{{L}_{0}}\Delta x.\)

S is found in this table from the previous lesson and is \(S=\text{80}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\). The radius \(r\) is 0.750 mm (as seen in the figure), so the cross-sectional area is

\(A={\mathrm{\pi r}}^{2}=1\text{.}\text{77}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}.\)

The value for \({L}_{0}\) is also shown in the figure. Thus,

\(F=\frac{\left(\text{80}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}{\text{N/m}}^{2}\right)\left(1\text{.}\text{77}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)}{\left(5\text{.}\text{00}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}\left(1\text{.}\text{80}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{m}\right)=\text{51 N.}\)

This 51 N force is the weight \(w\) of the picture, so the picture’s mass is

\(m=\frac{w}{g}=\frac{F}{g}=5\text{.2 kg}.\)

Discussion

This is a fairly massive picture, and it is impressive that the nail flexes only \(\text{1.80 µm}\)—an amount undetectable to the unaided eye.