Deformation of Solids
• So far, we have assumed that solids always retain
their original shape and cannot be deformed (“rigid
bodies”)
• However, the stretching, squeezing, and twisting of
real objects when forces are applied are often too
important to ignore (causing deformations)
• Stress characterizes the strength of the forces
causing the deformation (force per unit area)
• Strain describes the resulting deformation
• When the stress and strain are small enough, the
two are directly proportional to each other:
Stress / Strain = Elastic modulus (Hooke’s Law)
• There are three general types of stresses and
strains: tensile, bulk, and shear
Tensile Stress and Strain
• A solid under tensile stress (or tension) produces a
distortion due to stretching (i.e. rope under tension):
A
l0
• Tensile stress = F / A
F
A
F
l
– Scalar quantity since F is the magnitude of the force
– SI unit of stress is the pascal: 1 pascal = 1 Pa = 1 N/m2
– In the British system, stress is measured by the pound per
square inch (psi): 1 psi = 6895 Pa
– Note that units of stress are same as pressure
• Tensile strain (stretch / unit length) = (l – l0) / l0
Dl / l0
– Pure (dimensionless) number with no units
=
Tensile/Compressive Stress and Strain
• For sufficiently small tensile stress, stress and strain
are proportional through Young’s
modulus Y:
Tensile stress F / A Fl
Y
Tensile strain

Dl / l0

0
ADl
• The stress can compress the object:
A
l0
F
A F
l
• Compressive strain defined same way as tensile
strain, but then Dl corresponds to decrease in length
• For many materials, Young’s modulus has same
value for both tensile and compressive stresses
– Materials with large Y are relatively unstretchable
Bulk Stress and Strain
• When an object experiences uniform
pressure on all sides, its shape
remains the same but its volume
changes
(The object is being squeezed!
Such is the case with an object
under water.)
• We characterize a volume compression with bulk
modulus B:
Volume stress
DF / A
DP
B


– P = pressure = F / A
Volume strain
DV / V
DV / V
– Minus sign included because increase in pressure always
causes decrease in volume (if DP is positive, DV is negative)
– Reciprocal of bulk modulus is compressibility (a material
having a large bulk modulus does not compress easily)
Shear Stress and Strain
• Another kind of deformation results when forces are
applied tangent to opposite surfaces of an object
(Examples of shear stress are a
ribbon being deformed by scissors
and a book being deformed while
pushing across the top of one cover)
• The ratio of shear stress to shear strain is called the
shear modulus S:
Sheer stress F / A
S
Sheer strain

Dx / h
– Material having large shear modulus is difficult to bend
– Shear stress, shear strain, and shear modulus apply to
solid materials only
– Liquids and gasses, not having definite shape or elastic
behavior, would not respond the same way as solids
CQ1: A single steel column is to support a mass
of 1.5 × 108 kg. If the yield strength for steel is
2.5 × 108 N/m2 and safety regulations require the
column to withstand five times the weight it
presently holds, what should be the approximate
cross-sectional area of the base of the column?
A)
B)
C)
D)
0.6 m2
3 m2
6 m2
30 m2
CQ2: The sole of a certain tennis shoe has a
shear modulus of 4 × 107. If the height of the
sole is doubled, the strain will:
A)
B)
C)
D)
decrease by a factor of two.
remain the same.
increase by a factor of two.
increase by a factor of four.
Example Problem #9.6
F
Wire:
F
F
A
F
A
l0
l
A stainless-steel orthodontic wire is applied to a tooth as shown above. The
wire has an unstretched length of 3.1 cm and a diameter of 0.22 mm. If the
wire is stretched 0.10 mm, find the magnitude and direction of the force on the
tooth. Disregard the width of the tooth, and assume that Young’s modulus for
stainless steel is 18  1010 Pa.
Solution (details given in class): 22 N (directed downward in the diagram)
Density and Pressure
m

• Density of uniform substance is defined as:
V
– Under normal conditions, densities of solids and liquids are
about 1000 times greater than the densities of gases
– Some densities vary within the material  above formula
then gives average density
– In general, density depends on environmental factors such
as temperature and pressure
– Specific gravity is ratio of material’s density to density of
water at 4°C (1000 kg/m3)
• Pressure = the amount of force per unit area
F
P
exerted on an object
A
– Force exerted by fluid (at rest) on
object is always perpendicular to the
surfaces of the object
Fluids at Rest
• The force associated with fluid pressure is due to
molecules of fluid colliding with their surroundings
• SI units of pressure are the pascal (1 Pa = 1 N/m2)
• Although pressure and force are used
interchangeably in everyday life, they are quite
different in physics
– Pressure is a scalar quantity
– Fluid pressure acts perpendicular to any surface in the
fluid, no matter how the surface is oriented
– Pressure depends on force and area over which force is
applied
• In order for a fluid to be in equilibrium, all points at
the same depth must be at the same pressure
(otherwise a given portion of fluid would accelerate
left or right) F
F
1
2
Fluids at Rest
• The pressure P at a depth h below the surface of a
fluid is greater than the external pressure P0 at the
surface (independent of the shape of the container):
P  P0  gh (assumes uniform density)
– Normal atmospheric pressure at sea level is
P0 = 1.013  105 Pa
– Liquids have nearly uniform densities, since
their density is nearly independent of pressure
– Gases only have uniform densities over short vertical
distances
• Gauge pressure measures P – P0
– Tire pressure gauges typically measure gauge pressure
– Negative for partial vacuums
• Total pressure is called absolute pressure
Pressure Measurements
• Simplest pressure gauge: open-tube manometer
– Unknown pressure given by P = P0 + gh
• Mercury barometer is another common
pressure gauge
– Measures atmospheric pressure P0 = gh,
where P ≈ 0 (top of tube filled only with
some mercury vapor),  = density of mercury,
h = height of mercury column
• Sphygmomanometer used as bloodpressure gauge
– Specialized mercury-filled manometer
– Measurements taken just below maximum
(systolic pressure) and minimum (diastolic
pressure) value produced by heart
– Depends on blood flow through brachial artery
CQ3: Mercury has a specific gravity of 13.6. The
column of mercury in the barometer below has a
height h = 76 cm. If a similar barometer were
made with water, what would be the approximate
height h of the column of water?
A)
B)
C)
D)
5.6 cm
76 cm
154 cm
1034 cm
CQ4: Two identical discs sit at the bottom of a 3 m pool
of water whose surface is exposed to atmospheric
pressure. The first disc acts as a plug to seal the drain
as shown. The second disc covers a container
containing nearly a perfect vacuum. If each disc has an
area of 1 m2, what is the approximate difference in force
necessary to open the containers? (Note: 1 atm =
101,300 Pa)
A)
B)
C)
D)
There is no difference.
3000 N
101,300 N
104,300 N
Pascal’s Law
• If we increase P0 at top surface (say through use of
piston pushing down on fluid), P at any depth
increases by exactly the same amount
• Pascal’s Law: Pressure applied to an enclosed fluid
is transmitted undiminished to every portion of the
fluid and the walls of the containing vessel.
F1 F2
DP 

A1 A2
(DP is applied pressure increase)
• This concept is used in hydraulic
lifts to multiply force (dentist’s
chairs, car lifts and jacks,
elevators, hydraulic brakes)
Example Problem #9.28
Piston 1 in the figure at right has a
diameter of 0.25 in; piston 2 has a
diameter of 1.5 in. In the absence
of friction, determine the force F
necessary to support the 500-lb
weight.
Solution (details given in class):
2.3 lb
Buoyancy
• Archimedes’s principle: When a body is completely
or partially immersed in a fluid, the fluid exerts an
upward force (“buoyant force”) on the body equal to
the weight of the fluid displaced by the body (the
fluid that makes way for the body)
• Buoyant force arises because fluid pressure below
an object is larger than pressure above it
• Buoyant force is the same for
objects with different densities
but same volume
(Force B is the same whether cube is
steel, aluminum, lead, or water!)
Buoyancy
• Case I: Totally submerged object
B  mdispfluid g  fluidVdispfluid g  fluidVobjg
Wobj  mobjg   objVobjg


Net force  B  Wobj  fluid  obj Vobjg
– Unsupported object will sink or float depending on
whether obj is greater than or less than fluid
– Explains how hot air balloons work
Buoyant Force Demo
• Case II: Floating Object
– Object is partially submerged and in equilibrium (B = Wobj)
B  fluidVdispfluid g

  obj Vdispfluid

  fluidVdispfluid g  objVobj g 
Vobj
Wobj  mobjg   objVobjg 
  fluid
– Explains why icebergs are mostly (≈ 90%) submerged,
and why brain is mostly supported by cerebrospinal fluid
CQ5: A helium balloon will rise into the
atmosphere until:
A) The temperature of the helium inside the
balloon is equal to the temperature of the air
outside the balloon.
B) The mass of the helium inside the balloon is
equal to the mass of the air outside the
balloon.
C) The weight of the balloon is equal to the
force of the upward air current.
D) The density of the helium in the balloon is
equal to the density of the air surrounding
the balloon.
Example Problem #9.41
A sample of an unknown material appears to
weigh 300 N in air and 200 N when immersed
in alcohol of specific gravity 0.700. What are
(a) the volume and
(b) the density of the material?
Solution (details given in class):
(a) 1.46  10–2 m3
(b) 2.10  103 kg/m3
CQ6: A brick with a density of 1.4 × 103 kg/m3 is
placed on top of a piece of Styrofoam floating on
water. If one half the volume of the Styrofoam
sinks below the water, what is the ratio of the
volume of the Styrofoam compared to the
volume of the brick? (Assume the Styrofoam is
massless.)
A)
B)
C)
D)
0.7
1.4
2.8
5.6
Fluid Flow
• We will use the simple idealized model of an ideal
fluid to describe fluid motion
– Fluid is incompressible (its density doesn’t change)
– Fluid has no internal friction (viscosity)
– Fluid motion is steady (no change in velocity, density, and
pressure at each point in fluid with time)
– Fluid flow is streamline (or laminar), i.e. not turbulent (a
small wheel placed in fluid would translate but not rotate)
• Since the mass of a moving fluid doesn’t change as
it flows, and since the flow is steady:
A1v1  A2v2  constant (equation of continuity)
– Amount of fluid entering pipe in given time interval =
amount of fluid leaving pipe in same interval
– Explains flow from garden hoses, faucets, even winds
downtown with tall buildings
Fluid Flow
• Energy conservation applied to an ideal fluid gives
Bernoulli’s equation:
1 2
1 2
P1  v1  gy1  P2  v2  gy2  constant
2
2
– P1 , P2 : Pressure at points 1 and 2
– ½v2 : Kinetic energy per unit volume
– gy : Potential energy per unit volume
• Bernoulli’s equation can be used to
understand:
–
–
–
–
–
Lift on airplane wings
Lift of golf balls/drop of curveballs
Household plumbing
Blood flow in arteries
Consistent with Newton’s 3rd law/cons. of momentum
CQ7: If the container pictured below is filled with
an ideal fluid, which point in the fluid most likely
has the greatest pressure?
A)
B)
C)
D)
Point A
Point B
Point C
Point D
Example Problem #9.54
A large storage tank, open to the atmosphere at the top
and filled with water, develops a small hole in its side at
a point 16.0 m below the water level. If the rate of flow
from the leak is 2.50  10–3 m3/min, determine
(a) the speed at which the water leaves the hole and
(b) the diameter of the hole.
Solution (details given in class):
(a) 17.7 m/s
(b) 1.73 mm
CQ8: A spigot is to be placed on a water tank
below the surface of the water. Which of the
following gives the distance of the spigot below
the surface h compared to the velocity with which
the water will run through the spigot?
A)
B)
C)
D)
Figure A
Figure B
Figure C
Figure D