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Physics 6A
Stress, Strain and
Elastic Deformations
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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.
.)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
StressY
*Bonus Question – who is this formula named for? Click here for the answer
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
ForceStress
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
ForceStress
STRAIN is a measure of how much the object deforms. We divide the change in the length by the original length to get strain:
0L
LStrain
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
ForceStress
STRAIN is a measure of how much the object deforms. We divide the change in the length by the original length to get strain:
0L
LStrain
Now we can put these together to get our formula for the Young’s Modulus:
0LL
AF
Y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
ΔL=1.1m
L0=45m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
ΔL=1.1m
L0=45m
A couple of quick calculations and we can just plug in to our formula:
0LL
AF
Y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
ΔL=1.1m
L0=45m
7mm
25232 m1085.3)m105.3(rA
N6378.9kg65mgF2s
m
A couple of quick calculations and we can just plug in to our formula:
0LL
AF
Y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Don’t forget to cut the diameter in half.
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?
ΔL=1.1m
L0=45m
7mm
N6378.9kg65mgF2s
m
A couple of quick calculations and we can just plug in to our formula:
2
225
mN8m
N7
m45m1.1
m1085.3N637
1088.6024.0
1065.1Y
0LL
AF
Y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
25232 m1085.3)m105.3(rA
Don’t forget to cut the diameter in half.
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?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=0.25cm
400N
diam=?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=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:
2mN11
steel 102Y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=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:
2mN11
steel 102Y
0LL
AF
Y The only piece missing is the area – we can rearrange the formula
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=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:
2mN11
steel 102Y
0LL
AF
Y The only piece missing is the area – we can rearrange the formula
LY
LFA 0
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=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:
2mN11
steel 102Y
0LL
AF
Y The only piece missing is the area – we can rearrange the formula
26
mN11
0
m106.1m0025.0102
m2N400A
LY
LFA
2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=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:
2mN11
steel 102Y
0LL
AF
Y The only piece missing is the area – we can rearrange the formula
26
mN11
0
m106.1m0025.0102
m2N400A
LY
LFA
2
One last step – we need the diameter, and we have the area:
circle2 Ar
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=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:
2mN11
steel 102Y
0LL
AF
Y The only piece missing is the area – we can rearrange the formula
26
mN11
0
m106.1m0025.0102
m2N400A
LY
LFA
2
One last step – we need the diameter, and we have the area:
m1014.7m106.1
rAr 426
circle2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
L0=2m
ΔL=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:
2mN11
steel 102Y
0LL
AF
Y The only piece missing is the area – we can rearrange the formula
26
mN11
0
m106.1m0025.0102
m2N400A
LY
LFA
2
One last step – we need the diameter, and we have the area:
m1014.7m106.1
rAr 426
circle2
double the radius to get the diameter:
mm4.1m104.1d 3
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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?
x4)e
x2)d4
x)c
2
x)b
2
x)a
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
We can do this one just by staring at the formula for stress:
Area
ForceStress
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
x4)e
x2)d4
x)c
2
x)b
2
x)a
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
We can do this one just by staring at the formula for stress:
Area
ForceStress
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
x4)e
x2)d4
x)c
2
x)b
2
x)a
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
We can do this one just by staring at the formula for stress:
Area
ForceStress
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
x4)e
x2)d4
x)c
2
x)b
2
x)a
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
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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Bulk Modulus and Volume Changes
0VV
Bp
Bulk Modulus
Volume change
Pressure change
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.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
Bulk Modulus and Volume Changes
Answser: 6.6x108 N/m2
