<< Chapter < Page Chapter >> Page >

where B is the bulk modulus (see [link] ), V 0 size 12{V rSub { size 8{0} } } {} is the original volume, and F A size 12{ { {F} over {A} } } {} is the force per unit area applied uniformly inward on all surfaces. Note that no bulk moduli are given for gases.

What are some examples of bulk compression of solids and liquids? One practical example is the manufacture of industrial-grade diamonds by compressing carbon with an extremely large force per unit area. The carbon atoms rearrange their crystalline structure into the more tightly packed pattern of diamonds. In nature, a similar process occurs deep underground, where extremely large forces result from the weight of overlying material. Another natural source of large compressive forces is the pressure created by the weight of water, especially in deep parts of the oceans. Water exerts an inward force on all surfaces of a submerged object, and even on the water itself. At great depths, water is measurably compressed, as the following example illustrates.

Calculating change in volume with deformation: how much is water compressed at great ocean depths?

Calculate the fractional decrease in volume ( Δ V V 0 size 12{ { {ΔV} over {V rSub { size 8{0} } } } } {} ) for seawater at 5.00 km depth, where the force per unit area is 5 . 00 × 10 7 N / m 2 size 12{5 "." "00" times "10" rSup { size 8{7} } N/m rSup { size 8{2} } } {} .

Strategy

Equation Δ V = 1 B F A V 0 is the correct physical relationship. All quantities in the equation except Δ V V 0 are known.

Solution

Solving for the unknown Δ V V 0 gives

Δ V V 0 = 1 B F A . size 12{ { {ΔV} over {V rSub { size 8{0} } } } = { {1} over {B} } { {F} over {A} } } {}

Substituting known values with the value for the bulk modulus B from [link] ,

Δ V V 0 = 5.00 × 10 7 N/m 2 2 . 2 × 10 9 N/m 2 = 0.023 = 2.3%.

Discussion

Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500 atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress.

Got questions? Get instant answers now!

Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so—which is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way.

Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here.

Section summary

  • Hooke’s law is given by
    F = k Δ L , size 12{F=kΔL} {}

    where Δ L size 12{ΔL} {} is the amount of deformation (the change in length), F size 12{F} {} is the applied force, and k size 12{k} {} is a proportionality constant that depends on the shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be written as

    Δ L = 1 Y F A L 0 , size 12{ΔL= { {1} over {Y} } { {F} over {A} } L rSub { size 8{0} } } {}

    where Y size 12{Y} {} is Young’s modulus , which depends on the substance, A size 12{A} {} is the cross-sectional area, and L 0 size 12{L rSub { size 8{0} } } {} is the original length.

  • The ratio of force to area, F A size 12{ { {F} over {A} } } {} , is defined as stress , measured in N/m 2 .
  • The ratio of the change in length to length, Δ L L 0 size 12{ { {ΔL} over {L rSub { size 8{0} } } } } {} , is defined as strain (a unitless quantity). In other words,
    stress = Y × strain . size 12{"stress"=Y times "strain"} {}
  • The expression for shear deformation is
    Δ x = 1 S F A L 0 , size 12{Δx= { {1} over {S} } { {F} over {A} } L rSub { size 8{0} } } {}

    where S is the shear modulus and F is the force applied perpendicular to L 0 and parallel to the cross-sectional area A .

  • The relationship of the change in volume to other physical quantities is given by
    Δ V = 1 B F A V 0 , size 12{ΔV= { {1} over {B} } { {F} over {A} } V rSub { size 8{0} } } {}

    where B is the bulk modulus, V 0 is the original volume, and F A size 12{ { {F} over {A} } } {} is the force per unit area applied uniformly inward on all surfaces.

Practice Key Terms 6

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask