# 12.3 Stress, strain, and elastic modulus  (Page 5/26)

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$\text{bulk strain}=\frac{\text{Δ}V}{{V}_{0}}.$ An object under increasing bulk stress always undergoes a decrease in its volume. Equal forces perpendicular to the surface act from all directions. The effect of these forces is to decrease the volume by the amount Δ V compared to the original volume, V 0 .

The bulk strain results from the bulk stress, which is a force ${F}_{\perp }$ normal to a surface that presses on the unit surface area A of a submerged object. This kind of physical quantity, or pressure p , is defined as

$\text{pressure}=p\equiv \frac{{F}_{\perp }}{A}.$

We will study pressure in fluids in greater detail in Fluid Mechanics . An important characteristic of pressure is that it is a scalar quantity and does not have any particular direction; that is, pressure acts equally in all possible directions. When you submerge your hand in water, you sense the same amount of pressure acting on the top surface of your hand as on the bottom surface, or on the side surface, or on the surface of the skin between your fingers. What you are perceiving in this case is an increase in pressure $\text{Δ}p$ over what you are used to feeling when your hand is not submerged in water. What you feel when your hand is not submerged in the water is the normal pressure     ${p}_{0}$ of one atmosphere, which serves as a reference point. The bulk stress is this increase in pressure, or $\text{Δ}p,$ over the normal level, ${p}_{0}.$

When the bulk stress increases, the bulk strain increases in response, in accordance with [link] . The proportionality constant in this relation is called the bulk modulus, B , or

$B=\frac{\text{bulk stress}}{\text{bulk strain}}=-\frac{\text{Δ}p}{\text{Δ}V\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}{V}_{0}}=\text{−}\text{Δ}p\phantom{\rule{0.2em}{0ex}}\frac{{V}_{0}}{\text{Δ}V}.$

The minus sign that appears in [link] is for consistency, to ensure that B is a positive quantity. Note that the minus sign $\left(–\right)$ is necessary because an increase $\text{Δ}p$ in pressure (a positive quantity) always causes a decrease $\text{Δ}V$ in volume, and decrease in volume is a negative quantity. The reciprocal of the bulk modulus is called compressibility     $k,$ or

$k=\frac{1}{B}=-\frac{\text{Δ}V\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}{V}_{0}}{\text{Δ}p}.$

The term ‘compressibility’ is used in relation to fluids (gases and liquids). Compressibility describes the change in the volume of a fluid per unit increase in pressure. Fluids characterized by a large compressibility are relatively easy to compress. For example, the compressibility of water is $4.64\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{/atm}$ and the compressibility of acetone is $1.45\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{/atm}.$ This means that under a 1.0-atm increase in pressure, the relative decrease in volume is approximately three times as large for acetone as it is for water.

## Hydraulic press

In a hydraulic press [link] , a 250-liter volume of oil is subjected to a 2300-psi pressure increase. If the compressibility of oil is $2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\text{atm},$ find the bulk strain and the absolute decrease in the volume of oil when the press is operating. In a hydraulic press, when a small piston is displaced downward, the pressure in the oil is transmitted throughout the oil to the large piston, causing the large piston to move upward. A small force applied to a small piston causes a large pressing force, which the large piston exerts on an object that is either lifted or squeezed. The device acts as a mechanical lever.

## Strategy

We must invert [link] to find the bulk strain. First, we convert the pressure increase from psi to atm, $\text{Δ}p=2300\phantom{\rule{0.2em}{0ex}}\text{psi}=2300\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}14.7\phantom{\rule{0.2em}{0ex}}\text{atm}\approx \phantom{\rule{0.2em}{0ex}}160\phantom{\rule{0.2em}{0ex}}\text{atm},$ and identify ${V}_{0}=\phantom{\rule{0.2em}{0ex}}250\phantom{\rule{0.2em}{0ex}}\text{L}.$

## Solution

Substituting values into the equation, we have

$\begin{array}{}\\ \\ \text{bulk strain}=\frac{\text{Δ}V}{{V}_{0}}=\frac{\text{Δ}p}{B}=k\text{Δ}p=\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{/atm}\right)\left(160\phantom{\rule{0.2em}{0ex}}\text{atm}\right)=0.0032\hfill \\ \text{answer:}\phantom{\rule{0.5em}{0ex}}\text{Δ}V=0.0032\phantom{\rule{0.2em}{0ex}}{V}_{0}=0.0032\left(250\phantom{\rule{0.2em}{0ex}}\text{L}\right)=0.78\phantom{\rule{0.2em}{0ex}}\text{L.}\hfill \end{array}$

## Significance

Notice that since the compressibility of water is 2.32 times larger than that of oil, if the working substance in the hydraulic press of this problem were changed to water, the bulk strain as well as the volume change would be 2.32 times larger.

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