# 12.3 Stress, strain, and elastic modulus  (Page 3/26)

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In either of these situations, we define stress as the ratio of the deforming force ${F}_{\perp }$ to the cross-sectional area A of the object being deformed. The symbol ${F}_{\perp }$ that we reserve for the deforming force means that this force acts perpendicularly to the cross-section of the object. Forces that act parallel to the cross-section do not change the length of an object. The definition of the tensile stress is

$\text{tensile stress}=\frac{{F}_{\perp }}{A}.$

Tensile strain is the measure of the deformation of an object under tensile stress and is defined as the fractional change of the object’s length when the object experiences tensile stress

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

Compressive stress and strain are defined by the same formulas, [link] and [link] , respectively. The only difference from the tensile situation is that for compressive stress and strain, we take absolute values of the right-hand sides in [link] and [link] .

Young’s modulus Y is the elastic modulus when deformation is caused by either tensile or compressive stress, and is defined by [link] . Dividing this equation by tensile strain, we obtain the expression for Young’s modulus:

$Y=\frac{\text{tensile stress}}{\text{tensile strain}}=\frac{{F}_{\perp }\text{/}\phantom{\rule{0.1em}{0ex}}A}{\text{Δ}L\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}{L}_{0}}=\frac{{F}_{\perp }}{A}\phantom{\rule{0.2em}{0ex}}\frac{{L}_{0}}{\text{Δ}L}.$

## Compressive stress in a pillar

A sculpture weighing 10,000 N rests on a horizontal surface at the top of a 6.0-m-tall vertical pillar [link] . The pillar’s cross-sectional area is $0{\text{.20 m}}^{2}$ and it is made of granite with a mass density of ${2700\phantom{\rule{0.2em}{0ex}}\text{kg/m}}^{3}.$ Find the compressive stress at the cross-section located 3.0 m below the top of the pillar and the value of the compressive strain of the top 3.0-m segment of the pillar.

## Strategy

First we find the weight of the 3.0-m-long top section of the pillar. The normal force that acts on the cross-section located 3.0 m down from the top is the sum of the pillar’s weight and the sculpture’s weight. Once we have the normal force, we use [link] to find the stress. To find the compressive strain, we find the value of Young’s modulus for granite in [link] and invert [link] .

## Solution

The volume of the pillar segment with height $h=3.0\phantom{\rule{0.2em}{0ex}}\text{m}$ and cross-sectional area $A=0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ is

$V=Ah=\left(0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)\left(3.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)=0.60\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}.$

With the density of granite $\rho =2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3},$ the mass of the pillar segment is

$m=\rho V=\left(2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(0.60\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}\right)=1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{kg}.$

The weight of the pillar segment is

${w}_{p}=mg=\left(1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=1.568\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N.}$

The weight of the sculpture is ${w}_{s}=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N},$ so the normal force on the cross-sectional surface located 3.0 m below the sculpture is

${F}_{\perp }={w}_{p}+{w}_{s}=\left(1.568+1.0\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N}=2.568\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N.}$

Therefore, the stress is

$\text{stress}=\frac{{F}_{\perp }}{A}=\frac{2.568\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{N}}{0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}}=1.284\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\text{Pa}=\text{128.4 kPa.}$

Young’s modulus for granite is $Y=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\text{Pa}=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{kPa}.$ Therefore, the compressive strain at this position is

$\text{strain}=\frac{\text{stress}}{Y}=\frac{128.4\phantom{\rule{0.2em}{0ex}}\text{kPa}}{4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{kPa}}=2.85\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}.$

## Significance

Notice that the normal force acting on the cross-sectional area of the pillar is not constant along its length, but varies from its smallest value at the top to its largest value at the bottom of the pillar. Thus, if the pillar has a uniform cross-sectional area along its length, the stress is largest at its base.

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