Physics 6B
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 : F
spring
=

k
•
Δ
x
Hooke’s Law is a special case of a more general rule involving stress and strain.
.)
const
(
Strain
Stress
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:
Strain
Stress
Y
*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:
Area
Force
Stress
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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:
Area
Force
Stress
STRAIN is a measure of how much the object deforms. We divide the change in the
length by the original length to get strain:
0
L
L
Strain
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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:
Area
Force
Stress
STRAIN is a measure of how much the object deforms. We divide the change in the
length by the original length to get strain:
0
L
L
Strain
Now we can put these together to get our formula for the Young’s Modulus:
0
L
L
A
F
Y
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Problem 11.6 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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Δ
L=1.1m
L
0
=45m
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Problem 11.6 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?
Δ
L=1.1m
L
0
=45m
A couple of quick calculations and we can just plug in to our formula:
0
L
L
A
F
Y
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Problem 11.6 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?
Δ
L=1.1m
L
0
=45m
7mm
2
5
2
3
2
m
10
85
.
3
)
m
10
5
.
3
(
r
A
N
637
8
.
9
kg
65
mg
F
2
s
m
A couple of quick calculations and we can just plug in to our formula:
0
L
L
A
F
Y
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Problem 11.6 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?
Don’t forget to cut
the diameter in half.
Δ
L=1.1m
L
0
=45m
7mm
N
637
8
.
9
kg
65
mg
F
2
s
m
A couple of quick calculations and we can just plug in to our formula:
2
2
2
5
m
N
8
m
N
7
m
45
m
1
.
1
m
10
85
.
3
N
637
10
88
.
6
024
.
0
10
65
.
1
Y
0
L
L
A
F
Y
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Problem 11.6 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?
2
5
2
3
2
m
10
85
.
3
)
m
10
5
.
3
(
r
A
Don’t forget to cut
the diameter in half.
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Problem 11.7 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?
L
0
=2m
Δ
L=0.25cm
400N
diam=?
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Problem 11.7 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?
L
0
=2m
Δ
L=0.25cm
400N
diam=?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table:
2
m
N
11
steel
10
2
Y
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Problem 11.7 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?
Problem 11.7 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?
L
0
=2m
Δ
L=0.25cm
400N
diam=?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table:
2
m
N
11
steel
10
2
Y
0
L
L
A
F
Y
The only piece missing
is the area
–
we can
rearrange the formula
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 11.7 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?
L
0
=2m
Δ
L=0.25cm
400N
diam=?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table:
2
m
N
11
steel
10
2
Y
0
L
L
A
F
Y
The only piece missing
is the area
–
we can
rearrange the formula
L
Y
L
F
A
0
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L
0
=2m
Δ
L=0.25cm
400N
diam=?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table:
2
m
N
11
steel
10
2
Y
0
L
L
A
F
Y
The only piece missing
is the area
–
we can
rearrange the formula
2
6
m
N
11
0
m
10
6
.
1
m
0025
.
0
10
2
m
2
N
400
A
L
Y
L
F
A
2
Prepared by Vince Zaccone
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Assistance Services at UCSB
Problem 11.7 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?
L
0
=2m
Δ
L=0.25cm
400N
diam=?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table:
2
m
N
11
steel
10
2
Y
0
L
L
A
F
Y
The only piece missing
is the area
–
we can
rearrange the formula
2
6
m
N
11
0
m
10
6
.
1
m
0025
.
0
10
2
m
2
N
400
A
L
Y
L
F
A
2
One last step
–
we need the diameter, and we have the area:
circle
2
A
r
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Problem 11.7 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?
L
0
=2m
Δ
L=0.25cm
400N
diam=?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table:
2
m
N
11
steel
10
2
Y
0
L
L
A
F
Y
The only piece missing
is the area
–
we can
rearrange the formula
2
6
m
N
11
0
m
10
6
.
1
m
0025
.
0
10
2
m
2
N
400
A
L
Y
L
F
A
2
One last step
–
we need the diameter, and we have the area:
m
10
14
.
7
m
10
6
.
1
r
A
r
4
2
6
circle
2
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Assistance Services at UCSB
Problem 11.7 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?
L
0
=2m
Δ
L=0.25cm
400N
diam=?
We have most of the information for our formula. We can look up Young’s
modulus for steel in a table:
2
m
N
11
steel
10
2
Y
0
L
L
A
F
Y
The only piece missing
is the area
–
we can
rearrange the formula
2
6
m
N
11
0
m
10
6
.
1
m
0025
.
0
10
2
m
2
N
400
A
L
Y
L
F
A
2
One last step
–
we need the diameter, and we have the area:
m
10
14
.
7
m
10
6
.
1
r
A
r
4
2
6
circle
2
double the radius to get the diameter:
mm
4
.
1
m
10
4
.
1
d
3
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Assistance Services at UCSB
Problem 11.7 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?
x
4
)
e
x
2
)
d
4
x
)
c
2
x
)
b
2
x
)
a
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11.12 (MC) 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
We can do this one just by staring at the formula for stress:
Area
Force
Stress
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x
4
)
e
x
2
)
d
4
x
)
c
2
x
)
b
2
x
)
a
11.12 (MC) 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
We can do this one just by staring at the formula for stress:
Area
Force
Stress
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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x
4
)
e
x
2
)
d
4
x
)
c
2
x
)
b
2
x
)
a
11.12 (MC) 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
11.12 (MC) 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
We can do this one just by staring at the formula for stress:
Area
Force
Stress
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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x
4
)
e
x
2
)
d
4
x
)
c
2
x
)
b
2
x
)
a
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