Physics 6B - CLAS

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Nov 29, 2013 (3 years and 6 months ago)

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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

Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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


Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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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For Campus Learning
Assistance Services at UCSB

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?

Prepared by Vince Zaccone

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Assistance Services at UCSB

Δ
L=1.1m

L
0
=45m

Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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


Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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


Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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


Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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.

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=?

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


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?

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




Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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

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









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


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









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










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









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




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?

x
4
)
e
x
2
)
d
4
x
)
c
2
x
)
b
2
x
)
a
Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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

Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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.

Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

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)

Prepared by Vince Zaccone

For Campus Learning
Assistance Services at UCSB

x
4
)
e
x
2
)
d
4
x
)
c
2
x
)
b
2
x
)
a