Definition of Strain
Define variables the cause deformation of a material and the response of
the material.
• System Response – Linear Deformation:
L
∆L
F
Suppose that a wire of length L is stretched
by a force F then:
Linear Strain =
∆L
L
• System Response – Volume Deformation:
Consider an object immersed in a fluid. The fluid
will exert a pressure P from all sides causing a
volume deformation.
Volume Strain =
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University of Florida
∆V
V
PHY 2053
Page 1
Definition of Stress
Define variables the cause deformation of a material and the response of
the material.
L
• Linear Deformation:
∆L
F
Suppose that a wire of length L is stretched
by a force F then:
Linear Stress =
Area A
F
A
• Volume Deformation:
Consider an object immersed in a fluid. The fluid
will exert a pressure P from all sides causing a
volume deformation.
Volume Stress =
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University of Florida
F
=P
A
PHY 2053
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Hooke’s Law Spring
• Linear Restoring Force: Ideal Spring
Fext
Fspring = − k∆x
Spring Constant k
Spring Force
Fext = k∆L
Fext  kL  ∆L
∆L
= 
=C
A  A L
L
Stress = C × Strain
Generalized Hooke’s Law!
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University of Florida
Constant
PHY 2053
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Stress Proportional to Strain
• Linear Deformation:
L
F
Suppose that a wire of length L is stretched
by a force F then:
Stress = Y × Strain
Young’s Modulus
Units = Pa
F
∆L
=Y
A
L
∆L
Area A
• Volume Deformation:
Consider an object immersed in a fluid. The fluid
will exert a pressure P from all sides causing a
volume deformation.
Stress = -B × Strain
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University of Florida
F
∆V
= P = −B
A
V
Bulk Modulus
Units = Pa
PHY 2053
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Example Problem: Young’s Modulus
• Linear Deformation:
A wrecking ball with mass M is to be lifted by a crane with a
steel cable that has a diameter of 1.5 cm and an unstretched
length of 30 m. The Young’s modulus of steel is 2.0 × 1011
Pa. Ignoring the weight of the cable itself, when the ball is
lifted and held at rest the cable stretches by 1.66 cm, what is
the mass M of the wrecking ball?
F = FT = Mg
A = πr 2 = π (D / 2) 2
∆L
F = Mg = AY
L
F
∆L
=Y
A
L
L
FT
∆L
Mg
∆L
∆L
2
M = AY
= π ( D / 2) Y
gL
gL
0.0166m
= π (0.0075m) (2 × 10 N / m )
≈ 2,000kg
2
(9.8m / s )(30m)
2
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University of Florida
11
2
PHY 2053
Page 5
Exam 2 Fall 2013: Problem 59
L
• Near the surface of the Earth, a 2,000 kg wrecking ball is
×
connected to a steel cable that has a diameter of 2.5 cm and an
unstretched length of L = 40 m. The other end of the cable is fixed
in position and the ball is initially held at rest horizontally, as
shown in the figure. When the ball is released from rest it swings
down. The Young’s modulus of steel is 2.0 × 1011 Pa. Ignoring the
weight of the cable itself, when the ball-cable system swing
through vertical (i.e. ball at its lowest point) how much does the
cable stretch (in cm)?
v2
Mv 2
FT − Mg = Maradial = M
=
Answer: 2.40
R L + ∆L
2
% Right: 6%
Mv
1
2
Mv = Mg ( L + ∆L)
2
Mv 2 = 2 Mg ( L + ∆L)
FT = Mg +
L + ∆L
= 3Mg
A = πr = π (D / 2)
2
∆L FT
3Mg
=
=
L
AY π ( D / 2) 2 Y
PHY 2053
L
FT
2
3MgL
3(2000kg )(9.8m / s 2 )(40m)
∆L =
=
≈ 2.4cm
2
2
11
2
π ( D / 2) Y π (0.0125m) (2 ×10 N / m )
R. Field 3/25/2014
University of Florida
M
∆L
v
Mg
Page 6
Example Problem: Bulk Modulus
• Volume Deformation:
The atmospheric pressure at the surface of a clear lake
is 101 kPA. By what percentage does the density of lake
water increase at a depth of 1.0 km below the surface of
the lake? The bulk modulus for water is 2.2×109 Pa?
F
∆V
= P = −B
A
V
Pd = P0 + ρgd
− VPd
∆Vd =
B
ρgd = (1,000kg / m 3 )(9.8m / s 2 )(1,000m) = 9.8 × 106 Pa
M
M
M
M /V
ρ0 =
=
=
=
V0 V + ∆V0 V − (VP0 / B) 1 − P0 / B
ρd =
M
M
M
M /V
=
=
=
Vd V + ∆Vd V − (VPd / B ) 1 − Pd / B
0.447%
ρ d − ρ 0 1 − P0 / B
ρgd
Pd − P0
9.8 × 106
=
−1 =
=
=
≈ 0.00447
9
6
ρ0
1 − Pd / B
B − Pd B − ( P0 + ρgd ) 2.2 × 10 − 9.901× 10
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University of Florida
PHY 2053
Page 7