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 : 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

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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

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

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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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For Campus Learning Assistance Services at UCSB

ΔL=1.1m

L0=45m

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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?

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

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?

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

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.

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

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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?

25232 m1085.3)m105.3(rA

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?

L0=2m

ΔL=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?

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

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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?

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

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?

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

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For Campus Learning Assistance Services at UCSB

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

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?

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

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?

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

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?

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

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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?

x4)e

x2)d4

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

ForceStress

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For Campus Learning Assistance Services at UCSB

x4)e

x2)d4

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

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

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

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)

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x4)e

x2)d4

x)c

2

x)b

2

x)a