Physics 6A
Stress, Strain and
Elastic Deformations
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When a force is applied to an object, it will deform. If it snaps back to its original shape
when the force is removed, then the deformation was ELASTIC.
We already know about springs - remember Hooke’s Law : Fspring = -k•Δx
Hooke’s Law is a special case of a more general rule involving stress and strain.
Stress
 (const.)
Strain
The constant will depend on the material that the object is made from, and it is called an
ELASTIC MODULUS. In the case of tension (stretching) or compression we will call it
Young’s Modulus*. So our basic formula will be:
Stress
Y
Strain
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*Bonus Question – who is this formula named for?
Click here for the answer
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To use our formula we need to define what we mean by Stress and Strain.
STRESS is the same idea as PRESSURE. In fact it is the same formula:
Stress 
Force
Area
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To use our formula we need to define what we mean by Stress and Strain.
STRESS is the same idea as PRESSURE. In fact it is the same formula:
Stress 
Force
Area
STRAIN is a measure of how much the object deforms. We divide the change in the
length by the original length to get strain:
L
Strain 
L0
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To use our formula we need to define what we mean by Stress and Strain.
STRESS is the same idea as PRESSURE. In fact it is the same formula:
Stress 
Force
Area
STRAIN is a measure of how much the object deforms. We divide the change in the
length by the original length to get strain:
L
Strain 
L0
Now we can put these together to get our formula for the Young’s Modulus:
Y
F
L
A
L0
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EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
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EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
L0=45m
ΔL=1.1m
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EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
A couple of quick calculations and we can just plug in to our formula:
Y
F
L
A
L0=45m
L0
ΔL=1.1m
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EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
A couple of quick calculations and we can just plug in to our formula:
Y
F
L
A
L0=45m
L0
ΔL=1.1m
F  mg  65kg 9.8 m2   637N
s 

7mm
A  r 2  (3.5  103 m)2  3.85  105 m2
Don’t forget to cut
the diameter in half.
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EXAMPLE: A nylon rope used by mountaineers elongates 1.10m under the weight of a
65.0kg climber. If the rope is initially 45.0m in length and 7.0mm in diameter, what is
Young’s modulus for this nylon?
A couple of quick calculations and we can just plug in to our formula:
Y
637N
Y
5
2
F
L
3.8510 m 
1.1m
45m
L0=45m
A
L0
1.65  107 N2
m
0.024
ΔL=1.1m
 6.88  108 N2
m
F  mg  65kg 9.8 m2   637N
s 

7mm
A  r 2  (3.5  103 m)2  3.85  105 m2
Don’t forget to cut
the diameter in half.
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
diam=?
L0=2m
ΔL=0.25cm
400N
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
diam=?
L0=2m
ΔL=0.25cm
400N
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
A
L0
The only piece missing
is the area – we can
rearrange the formula
diam=?
L0=2m
ΔL=0.25cm
400N
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
A
F
L
A
L0
The only piece missing
is the area – we can
rearrange the formula
diam=?
L0=2m
F  L0
Y  L
ΔL=0.25cm
400N
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

ΔL=0.25cm
400N
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

One last step – we need the diameter, and we have the area:
ΔL=0.25cm
400N
r 2  Acircle
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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

One last step – we need the diameter, and we have the area:
r 2  A circle  r 
ΔL=0.25cm
400N
1.6  106 m2
 7.14  10 4 m

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EXAMPLE: A steel wire 2.00 m long with circular cross-section must stretch no more
than 0.25cm when a 400.0N weight is hung from one of its ends. What minimum
diameter must this wire have?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table: Ysteel  2  1011 N2
m
Y
F
L
The only piece missing
is the area – we can
rearrange the formula
A
L0
F  L0
Y  L
400N2m
A
 1.6  10 6 m2
11 N
0.0025m
2  10
2
diam=?
L0=2m
A

m

One last step – we need the diameter, and we have the area:
r 2  A circle  r 
ΔL=0.25cm
400N
1.6  106 m2
 7.14  10 4 m

double the radius to get the diameter:
d  1.4  103 m  1.4mm
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EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2 x
e )4 x
a)
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EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2 x
e )4 x
a)
We can do this one just by staring at the formula for stress:
Force
Stress 
Area
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EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2 x
e )4 x
a)
We can do this one just by staring at the formula for stress:
Force
Stress 
Area
The force is the same in both cases because it says they use the same weight.
The area is related to the square of the radius (or diameter), so when the
diameter doubles the area goes up by a factor of 4.
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EXAMPLE: When a weight is hung from a cylindrical wire of diameter D, it produces a
tensile stress X in the wire. If the same weight is hung from a wire having twice the
diameter as the first one, the tensile stress in this wire will be
x
2
x
b)
2
x
c)
4
d)2 x
e )4 x
a)
We can do this one just by staring at the formula for stress:
Force
Stress 
Area
The force is the same in both cases because it says they use the same weight.
The area is related to the square of the radius (or diameter), so when the
diameter doubles the area goes up by a factor of 4.
Thus the stress should go down by a factor of 4 (area is in the denominator)
Answer c)
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Bulk Modulus and Volume Changes
When pressure is applied to an object from all directions, its
volume will change accordingly. Think of squishing a foam ball, or
inflating a balloon.
In this case we use a 3-dimensional version of Young’s modulus.
We call it BULK MODULUS, and it is defined in a similar way:
Bulk Modulus
Pressure change
V
p  B
V0
Volume change
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Bulk Modulus and Volume Changes
Example: When water freezes into ice it expands in volume by 9.05 percent. Suppose a
volume of water is in a household water pipe or a cavity in a rock. If the water freezes, what
pressure must be exerted on it to keep its volume from expanding? (If the pipe or rock
cannot supply this pressure, the pipe will burst or the rock will split.)
The bulk modulus for ice is 8x109 N/m2.
Answser: 6.6x108 N/m2
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