# 12.3 Stress, strain, and elastic modulus  (Page 6/26)

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Check Your Understanding If the normal force acting on each face of a cubical $1{\text{.0-m}}^{3}$ piece of steel is changed by $1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{N},$ find the resulting change in the volume of the piece of steel.

63 mL

## Shear stress, strain, and modulus

The concepts of shear stress and strain concern only solid objects or materials. Buildings and tectonic plates are examples of objects that may be subjected to shear stresses. In general, these concepts do not apply to fluids.

Shear deformation occurs when two antiparallel forces of equal magnitude are applied tangentially to opposite surfaces of a solid object, causing no deformation in the transverse direction to the line of force, as in the typical example of shear stress illustrated in [link] . Shear deformation is characterized by a gradual shift $\text{Δ}x$ of layers in the direction tangent to the acting forces. This gradation in $\text{Δ}x$ occurs in the transverse direction along some distance ${L}_{0}.$ Shear strain is defined by the ratio of the largest displacement $\text{Δ}x$ to the transverse distance ${L}_{0}$

$\text{shear strain}=\frac{\text{Δ}x}{{L}_{0}}.$

Shear strain is caused by shear stress. Shear stress is due to forces that act parallel to the surface. We use the symbol ${F}_{\parallel }$ for such forces. The magnitude ${F}_{\parallel }$ per surface area A where shearing force is applied is the measure of shear stress

$\text{shear stress}=\frac{{F}_{\parallel }}{A}.$

The shear modulus is the proportionality constant in [link] and is defined by the ratio of stress to strain. Shear modulus is commonly denoted by S :

$S=\frac{\text{shear stress}}{\text{shear strain}}=\frac{{F}_{\parallel }\text{/}\phantom{\rule{0.1em}{0ex}}A}{\text{Δ}x\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}{L}_{0}}=\frac{{F}_{\parallel }}{A}\phantom{\rule{0.2em}{0ex}}\frac{{L}_{0}}{\text{Δ}x}.$

## An old bookshelf

A cleaning person tries to move a heavy, old bookcase on a carpeted floor by pushing tangentially on the surface of the very top shelf. However, the only noticeable effect of this effort is similar to that seen in [link] , and it disappears when the person stops pushing. The bookcase is 180.0 cm tall and 90.0 cm wide with four 30.0-cm-deep shelves, all partially loaded with books. The total weight of the bookcase and books is 600.0 N. If the person gives the top shelf a 50.0-N push that displaces the top shelf horizontally by 15.0 cm relative to the motionless bottom shelf, find the shear modulus of the bookcase.

## Strategy

The only pieces of relevant information are the physical dimensions of the bookcase, the value of the tangential force, and the displacement this force causes. We identify ${F}_{\parallel }=50.0\phantom{\rule{0.2em}{0ex}}\text{N},\phantom{\rule{0.2em}{0ex}}\text{Δ}x=15.0\phantom{\rule{0.2em}{0ex}}\text{cm},$ ${L}_{0}=180.0\phantom{\rule{0.2em}{0ex}}\text{cm},$ and $A=\text{(30.0 cm)}\phantom{\rule{0.1em}{0ex}}\text{(90.0 cm)}=2700.0\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2},$ and we use [link] to compute the shear modulus.

## Solution

Substituting numbers into the equations, we obtain for the shear modulus

$S=\frac{{F}_{\parallel }}{A}\phantom{\rule{0.2em}{0ex}}\frac{{L}_{0}}{\text{Δ}x}=\frac{50.0\phantom{\rule{0.2em}{0ex}}\text{N}}{2700.0\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}}\phantom{\rule{0.2em}{0ex}}\frac{180.0\phantom{\rule{0.2em}{0ex}}\text{cm}\text{.}}{15.0\phantom{\rule{0.2em}{0ex}}\text{cm}\text{.}}=\frac{2}{9}\phantom{\rule{0.2em}{0ex}}\frac{\text{N}}{{\text{cm}}^{2}}=\frac{2}{9}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\frac{\text{N}}{{\text{m}}^{2}}=\frac{20}{9}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{Pa}=\text{2.222 kPa.}$

We can also find shear stress and strain, respectively:

$\begin{array}{c}\frac{{F}_{\parallel }}{A}=\frac{50.0\phantom{\rule{0.2em}{0ex}}\text{N}}{2700.0\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}}=\frac{5}{27}\phantom{\rule{0.2em}{0ex}}\text{kPa}=\text{185.2 Pa}\hfill \\ \frac{\text{Δ}x}{{L}_{0}}=\frac{15.0\phantom{\rule{0.2em}{0ex}}\text{cm}}{180.0\phantom{\rule{0.2em}{0ex}}\text{cm}}=\frac{1}{12}=0.083.\hfill \end{array}$

## Significance

If the person in this example gave the shelf a healthy push, it might happen that the induced shear would collapse it to a pile of rubbish. Much the same shear mechanism is responsible for failures of earth-filled dams and levees; and, in general, for landslides.

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