12.3 Stress, strain, and elastic modulus  (Page 4/26)

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Check Your Understanding Find the compressive stress and strain at the base of Nelson’s column.

$206.8\phantom{\rule{0.2em}{0ex}}\text{kPa};\phantom{\rule{0.2em}{0ex}}4.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

Stretching a rod

A 2.0-m-long steel rod has a cross-sectional area of $0.30\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}.$ The rod is a part of a vertical support that holds a heavy 550-kg platform that hangs attached to the rod’s lower end. Ignoring the weight of the rod, what is the tensile stress in the rod and the elongation of the rod under the stress?

Strategy

First we compute the tensile stress in the rod under the weight of the platform in accordance with [link] . Then we invert [link] to find the rod’s elongation, using ${L}_{0}=2.0\phantom{\rule{0.2em}{0ex}}\text{m}.$ From [link] , Young’s modulus for steel is $Y=\phantom{\rule{0.2em}{0ex}}2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}\text{Pa}.$

Solution

Substituting numerical values into the equations gives us

$\begin{array}{ccc}\hfill \frac{{F}_{\perp }}{A}& =\hfill & \frac{\left(550\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.8\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)}{3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\text{Pa}\hfill \\ \hfill \text{Δ}L& =\hfill & \frac{{F}_{\perp }}{A}\phantom{\rule{0.2em}{0ex}}\frac{{L}_{0}}{Y}=\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\text{Pa}\right)\phantom{\rule{0.2em}{0ex}}\frac{2.0\phantom{\rule{0.2em}{0ex}}\text{m}}{2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}\text{Pa}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\text{m}=\phantom{\rule{0.2em}{0ex}}1.8\phantom{\rule{0.2em}{0ex}}\text{mm.}\hfill \end{array}$

Significance

Similarly as in the example with the column, the tensile stress in this example is not uniform along the length of the rod. Unlike in the previous example, however, if the weight of the rod is taken into consideration, the stress in the rod is largest at the top and smallest at the bottom of the rod where the equipment is attached.

Check Your Understanding A 2.0-m-long wire stretches 1.0 mm when subjected to a load. What is the tensile strain in the wire?

$5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}$

Objects can often experience both compressive stress and tensile stress simultaneously [link] . One example is a long shelf loaded with heavy books that sags between the end supports under the weight of the books. The top surface of the shelf is in compressive stress and the bottom surface of the shelf is in tensile stress. Similarly, long and heavy beams sag under their own weight. In modern building construction, such bending strains can be almost eliminated with the use of I-beams [link] .

A heavy box rests on a table supported by three columns. View this demonstration to move the box to see how the compression (or tension) in the columns is affected when the box changes its position.

Bulk stress, strain, and modulus

When you dive into water, you feel a force pressing on every part of your body from all directions. What you are experiencing then is bulk stress, or in other words, pressure    . Bulk stress always tends to decrease the volume enclosed by the surface of a submerged object. The forces of this “squeezing” are always perpendicular to the submerged surface [link] . The effect of these forces is to decrease the volume of the submerged object by an amount $\text{Δ}V$ compared with the volume ${V}_{0}$ of the object in the absence of bulk stress. This kind of deformation is called bulk strain and is described by a change in volume relative to the original volume:

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