1206 - Concepts in
Physics
Wednesday, November 04th
Request from Mark Brown
• Those of you who will miss the LAB
tomorrow, please let Mark B. know - either
stopping by or a quick email will be fine.
• State you name and the reason
Just wondering ...
• The solutions for assignment #3 and #4
have been taken off the wall across my
office during reading week
• This left some students without the
possibility to look at them
• This is not a good way to do things, please
think about others as well
Notes
• You will get the midterm exams back on
Friday (after class)
• Any other notes about that will be on
Friday as well
• After today’s class you have everything you
need to complete assignment #5
• Remember that only 1 point for #5 (8) is
part of the 25 points that count as 100%
We will talk about deformation today
deformations
magnified
100×
Elastic deformation and Young’s Modulus
A spring returns to its original shape when the force compressing or stretching it is
removed. In fact, all materials become distorted in some way when they are squeezed or
stretched and many of them return to their original shape (rubber). We call these material
“elastic”. To understand this a little deeper, we can make a picture (model) of how a certain
material looks like at the atomic level. For a solid, we can assume, that the forces between
atoms act like springs. Elastic behavior has its origin in the forces that atoms exert on each
other. These atomic-level “springs” help the material to return to its original shape, once the
forces that caused the deformation are removed. The force that holds atoms of a solid
together is particularly strong, therefore considerable force must be applied to stretch it.
Experimentally it has been shown, that the magnitude of the force can be expressed by the
following relation, provided that the amount of stretching is small compared to the original
length:
F = Y (ΔL/L0)A
where A is the cross-sectional area of (for example) a rod, ΔL the increase in length, L0
the original length and F the magnitude of the stretching force applied perpendicularly to
the surface at the end. The term Y is a proportionality constant called Young’s modulus.
Thomas Young lived (1773-1829).Young’s moduli are given in units of N/m2
Typical values for solids are 108 - 1011
Example
In a balancing act, a performer supports the combined weight
(1640 N) of a number of colleagues. Each thighbone of this
performer has a length of 0.55m and an effective crosssectional area of 7.7 x 10-4 m2. The Young’s Modulus for bone
compression is 9.4 x 109 N/m2. Determine the amount by
which each thighbone compresses under the extra weight.
Each thighbone supports half the weight, therefore F = 0.5 * 1640N = 820 N.
The Young’s modulus for bone compression is given as well as the length and the crosssectional area. So, we can find the amount by which the additional weight compresses
the thighbone ΔL.
F = Y (ΔL/L0)A ---> ΔL = FL0/YA
ΔL = (820N)(0.55m)/(9.4 x 109 N/m2 * 7.7 x 10-4 m2) = 6.2 x 10-5 m
This is a very small change.
Shear deformation and the shear modulus
It is possible to deform a solid object in a way other than stretching or compressing.
For example, imagine a thick book with a hardcover laying on a table. When you use
your hand to push it along the top cover will be shifted relative to the pages below and
the stationary bottom cover. The resulting deformation is called shear deformation and
occurs because of the combined effect of the force F applied (by the hand) to the top
of the book and the force -F applied (by the table) to the bottom of the book. The
direction of the forces are parallel to the covers of the book, each of which has an area
A. These two forces have equal magnitudes, but opposite direction, so the book
remains in equilibrium. The magnitude F of the force needed to produce an amount of
shear ΔX for an object with thickness L0
ΔX
L
F = S (ΔX/L0)A
where A is the cross-sectional area and S is the constant of
proportionality called the shear modulus with units N/m2
(like the Young’s modulus). Typical shear modulus are of the
order of 109 - 1011.
Example:
A block of Jell-O is resting on a plate. The block is 0.030 m
high, 0.070 m wide and 0.070 m long. A person is pushing
tangentially across the top surface with a force of F = 0.45 N
The top surface moves a distance ΔX = 6.0 x 10-3 m relative
to the bottom surface. What is the shear modulus of Jell-O?
The person applies a force that is parallel to the top surface of the Jell-O block. The
shape of the block changes. We can use the formula that give the magnitude of the
force to determine the shear modulus, since everything else is given.
F = S (ΔX/L0)A
--> S = FL0/ΔXA
S = [(0.45N)(0.030m)]/[(6.0 x 10-3m)(0.070m)(0.070m)] = 460 N/m2
This is a fairly small number, reflecting the fact that Jell-O can be deformed easily.
Volume deformation and the bulk modulus
So far we have been talking about compressive forces along one dimension. It is also
possible to apply compressive forces so that the size of every dimension (length,
width, and depth) decreases, leading to a decrease in volume.
This over compression occurs, for example, when an object is
submerged in a liquid, and the liquid presses inward everywhere
on the object. The forces acting in such situation are applied
perpendicular to every surface, and we speak of the
perpendicular force per unit area. The magnitude of the
perpendicular force per unit area is called the pressure P.
This should be familiar, since we have discussed this for fluids before.
P = F/A
(unit for pressure is N/m2 = Pa)
Suppose we change the pressure on an object by an amount ΔP (final - initial). This
change in pressure causes a change in volume of the object by an amount ΔV.
Experimentally one finds that the change in pressure is directly proportional to the
fractional change in volume ΔV/V0:
ΔP = -B (ΔV/V0)
The proportionality constant B is known as the bulk modulus. The minus sign is
necessary because and increase in pressure always causes a decrease in volume.
Stress, Strain and Hooke’s law
Over the last couple of slides we have defined amounts of forces that are needed for a
given amount of deformation. They have common features, even so they describe
different types of deformation. We will re-write them here:
F/A
=Y
ΔL/L0
F/A
=S
ΔX/L0
ΔP
= -B
ΔV/V0
Stress
is
proportional
to
The left side of each equation is the
magnitude of the force per unit area
required to cause an elastic
deformation and is in general called
stress. The right side of each equation
involves the change in a quantity divided
by a quantity (unitless ratio) and is
referred to as the strain that results
from the applied stress.
Strain
Stress is directly proportional to strain, this
relationship was first discovered by Robert Hooke
(1635-1703) and is knows as Hooke’s law.
There are limits to the proportionality
We had this before for the specific
case of springs
Stress
Strain
Springs are also a good example for the
proportionality limit. Once you have stressed (applied
force) to a spring beyond its elastic limit, it will not
return to its original shape and can’t be used as a
spring anymore.
Thermal Stress
Do you remember the example of the buckling sidewalk?
If the concrete slabs had not buckled upward, they would have been
subjected to immense forces from the buildings. The forces needed
to prevent a solid object from expanding must be strong enough to
counteract any change in length that would occur due to a change in
temperature. Although the change in temperature maybe small, the
forces - and hence the stresses - can be enormous.
Example
A steel beam is used in the roadbed of a bridge. The beam is mounted between two
concrete supports when the temperature is 23 °C, with no room provided for thermal
expansion. What compressional stress must the concrete supports apply to each end of the
beam, if they are to keep the beam from expanding when the temperature rises to 42 °C.
Beam
Concrete
support
Stress = F/A = Y ΔL/L0
where Y is the Young’s modulus. If the steel beam were
free to expand because of the change in temperature,
the length would change by ΔL = αL0ΔT. Because the
concrete supports do not permit any expansion, they
Concrete
support
must supply a stress to compress the beam by an
amount ΔL.
Stress = Y ΔL/L0 = Y (αL0ΔT)/L0 = YαΔT
For steel, the values of Young’s modulus and the coefficient of linear expansion are
Y = 2.0 x 1011 N/m2 and α = 12 x 10-6(°C)-1 respectively. The change in temperature is
ΔT = 42C - 23C = 19C. The thermal stress is
Stress = YαΔT = (2.0 x 1011 N/m2)(12 x 10-6(°C)-1)(19 C) = 4.6 x 107 N/m2
Bimetallic strip
How does your electric water boiler “know” when to turn itself off when the water
has reached a certain temperature?
Bimetallic strips are used for that purpose (and many others). A bimetallic strip is
made from two think strips of metal that have different coefficients of linear
expansion - therefore they behave differently when heat (or cooling) is applied.
Often brass (α = 19 x 10-6(C°)-1) and steel
(α = 12 x 10-6(C°)-1) are selected. The two
pieces are welded together. When the
bimetallic strip is heated, the brass
(green), having the larger value of α,
expands more than the steel (red). Since
the two metals are bonded together, the
bimetallic strip bends into an arc. When
the strip is cooled, the bimetallic strip
bends into the opposite direction
Expansion of holes - Example
When we have a hole in a piece of solid that is heated, does is expand as well?
Lets assume a pattern made out of tiles, 3 up and 3 to the side, where the central
one has been left out. We can then think about this question on a single tile basis
first and then put it back together after the heat has been applied. If every single
tile expands when heated, which we know to be true, then the hole must have
expanded as well. It will still have the size of one (now expanded tile) after
heating. So holes expand in exactly the same way as the surrounding material.
This conclusion is true for any solid and for any shape of hole. Thus, if follows that
a hole in a piece of solid material expands when heated and contract when
cooled, just as if it were filled with the material that surrounds it. If the hole is
circular, we can use equation ΔL = αL0ΔT to find the change in any linear
dimension.
A gold ring (α=14 x 10-6(°C)-1) has an inner diameter of 1.5 x 10-2 m and a
temperature of 27 °C. The ring falls into a sink of hot water with a temperature
of 49 °C. What is the change in the diameter of the hole in the ring?
Change in temperature ΔT = 49 °C - 27 °C = 22°C
α and L0 are given, therefore
ΔL = αL0ΔT = (1.5 x 10-2 m)(14 x 10-6(°C)-1)(22°C) = 4.6 x 10-6 m
Volume thermal expansion
So far we have been looking at linear expansion for one- or two-dimensional changes.
The volume of a normal material increases as the temperature increases. Analog to linear
thermal expansion, the change in volume ΔV is proportional to the change in
temperature ΔT and to the initial volume V0, provided the change in temperature is not
too large. The proportionality constant β, known as the coefficient of volume
expansion can be used to write:
ΔV = βV0ΔT
Common unit for the coefficient of the volume expansion is (C°)-1
Thermal expansion usually will be measured at room temperature (20 °C). The
values for β for liquids are substantially larger than those for solids, because liquids
typically expand more than solids for the same initial volumes and temperature
changes. For most solids the coefficient of volume expansion is three times as much
as the coefficient for linear expansion. If a cavity exists within a solid object, the
volume of the cavity increases when the object expands, just as if the cavity were
filled with the surrounding material. This is due to the same argument we have used
to derive at the conclusion that a hole expands the same as the surrounding
material.
Example - an automobile radiator
A small plastic container, called the coolant reservoir, catches the radiator fluid that
overflows when an automobile engine becomes hot. The radiator is made of copper
(β = 51 x 10-6(C°)-1), and the coolant has a coefficient of volume expansion of β =
4.10 x 10-4(C°)-1. If the radiator is filled to its 15-quart capacity when the engine is
cold (6.0 °C), how much overflow from the radiator will spill into the reservoir when
the coolant reaches its operating temperature of 92 °C?
When the temperature increases, both the coolant and the radiator expand. If they
were to expand by the same amount, there would be no overflow. However, the
liquid coolant expands more than the radiator, and the overflow volume is the
amount of coolant expansion minus the amount of the radiator cavity expansion.
When the temperature increases by 86 C°, the coolant expands by:
ΔV = βV0ΔT = (4.10 x 10-4(C°)-1)(15 quarts)(86 C°) = 0.53 quart
When the temperature increases by 86 C°, the radiator expands by:
ΔV = βV0ΔT = (51 x 10-6(C°)-1)(15 quarts)(86 C°) = 0.066 quart
Therefore the overflow volume is 0.53 quart - 0.07 quart = 0.45 quart
Special case - water
Although most substances expand when heated, a few do not. For instance, if water is
at its freezing point of 0 °C, its volume decreases until the temperature reaches 4 °C.
Above 4 °C water behaves normally (meaning as other solids and fluids) and its volume
increases as the temperature increases. Because a given mass of water has a minimum
volume at 4°C, see figure.
The fact that water has its greatest density at
4 °C, rather than at 0 °C, has important
consequences for the way in which a lake
freezes. When the air temperature drops, the
surface layer of water is chilled. As the
temperature of the surface layer drops toward
4 °C, this layer becomes more dense than the
warmer water below. The denser water sinks
and pushes up the deeper and warmer water,
which in turn is chilled at the surface.
This process continues until the temperature of the entire lake reaches 4 °C. Further
cooling of the surface water below 4 °C makes it less dense than the deeper layers;
consequently, the surface does not sink but stays on top. Once the cooling reaches
0 °C, the formation of ice that floats on the water starts. Ice has a smaller density
than water. Below the ice the temperature of the water however remains above
0 °C. The sheet of ice acts as an insulator that reduces the loss of heat from the
lake, especially if the ice is covered with a blanket of snow. As a result, lakes usually
do not freeze solid and therefore fish and other aquatic life can survive.