Definition of Strain
Define variables the cause deformation of a material and the response of
the material.
• System Response – Linear Deformation:
L
ΔL
F
Suppose that a wire of length L is stretched
by a force F then:
Linear Strain =
ΔL
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 =
R. Field 10/31/2013
University of Florida
ΔV
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:
ΔL
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 =
R. Field 10/31/2013
University of Florida
F
=P
A
PHY 2053
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Hooke’s Law Spring
• Linear Restoring Force: Ideal Spring
Fext
Fspring = − kΔx
Spring Constant k
Spring Force
Fext = kΔL
Fext ⎛ kL ⎞ ΔL
ΔL
=⎜ ⎟
=C
A ⎝ A⎠ L
L
Stress = C × Strain
Generalized Hooke’s Law!
R. Field 10/31/2013
University of Florida
Constant
PHY 2053
Page 3
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
ΔL
=Y
A
L
ΔL
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
R. Field 10/31/2013
University of Florida
F
ΔV
= 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 = Mg
A = πr 2 = π (D / 2) 2
ΔL
F = Mg = AY
L
F
ΔL
=Y
A
L
M = AY
L
ΔL
Mg
ΔL
ΔL
= π ( D / 2) 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
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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
ΔV
= P = −B
A
V
Pd = P0 + ρgd
− VPd
ΔVd =
B
ρgd = (1,000kg / m 3 )(9.8m / s 2 )(1,000m) = 9.8 ×106 Pa
M
M
M
M /V
ρ0 =
=
=
=
V0 V + ΔV0 V − (VP0 / B ) 1 − P0 / B
ρd =
M
M
M
M /V
=
=
=
Vd V + ΔVd V − (VPd / B ) 1 − Pd / B
0.445%
ρ d − ρ 0 1 − P0 / B
Pd − P0
ρgd
9.8 × 10 6
≈ 0.00445
=
=
=
−1 =
9
6
ρ0
1 − Pd / B
B − Pd B − ( P0 + ρgd ) 2.2 × 10 − 9.91× 10
R. Field 10/31/2013
University of Florida
PHY 2053
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