### Document

```Physics 7C lecture 16
Stress, Strain, and elastic
moduli
Tuesday November 26, 8:00 AM – 9:20 AM
Engineering Hall 1200
Strain, stress, and elastic moduli
• Stretching, squeezing, and twisting a real body causes it to deform,
as shown in figure below. We shall study the relationship between
forces and the deformations they cause.
• Stress is the force per unit area and strain is the fractional
deformation due to the stress. Elastic modulus is stress divided by
strain.
• The proportionality of stress and strain is called Hooke’s law.
Strain, stress, and elastic moduli
Elastic modulus
=
molecular basis for Hook’s law:
Stress
force per unit dimension
Strain
fractional deformation
Tensile and compressive stress and strain
• Tensile stress = F /A and tensile strain = l/l0. Compressive stress
and compressive strain are defined in a similar way.
• Young’s modulus is tensile stress divided by tensile strain, and is
given by Y = (F/A)(l0/l).
Tensile and compressive stress and strain
Young’s modulus = tensile stress / tensile strain,
Y = (F/A)(l0/l).
setup for measuring Young’s modulus
This is pressure!
tiny!
Some values of elastic moduli
Tensile and compressive stress and strain
setup for measuring Young’s modulus
1 meter long steel string
with 1 mm diameter,
pulled by 10 N weight!
Ysteel = 20 X 1010 P
the elongation is
dL = L * F / (A Y)
= only 0.5 mm!
Tensile stress and strain
• In many cases, a body can experience both tensile and
compressive stress at the same time.
Q11.5
Two rods are made of
the same kind of steel
and have the same
diameter.
F
F
length L
length 2L
A force of magnitude F is applied to the end of each rod.
Compared to the rod of length L, the rod of length 2L has
A. more stress and more strain.
B. the same stress and more strain.
C. the same stress and less strain.
D. less stress and less strain.
E. the same stress and the same strain.
F
F
A11.5
Two rods are made of
the same kind of steel
and have the same
diameter.
F
F
length L
length 2L
A force of magnitude F is applied to the end of each rod.
Compared to the rod of length L, the rod of length 2L has
A. more stress and more strain.
B. the same stress and more strain.
C. the same stress and less strain.
D. less stress and less strain.
E. the same stress and the same strain.
F
F
Q11.6
Two rods are made of
the same kind of steel.
The longer rod has a
greater diameter.
F
F
length L
length 2L
A force of magnitude F is applied to the end of each rod.
Compared to the rod of length L, the rod of length 2L has
A. more stress and more strain.
B. the same stress and more strain.
C. the same stress and less strain.
D. less stress and less strain.
E. the same stress and the same strain.
F
F
A11.6
Two rods are made of
the same kind of steel.
The longer rod has a
greater diameter.
F
F
length L
length 2L
A force of magnitude F is applied to the end of each rod.
Compared to the rod of length L, the rod of length 2L has
A. more stress and more strain.
B. the same stress and more strain.
C. the same stress and less strain.
D. less stress and less strain.
E. the same stress and the same strain.
F
F
Bulk stress and strain
Bulk stress and strain
• Pressure in a fluid is force
per unit area: p = F/A.
• Bulk stress is pressure
change p and bulk strain is
fractional volume change
V/V0.
• Bulk modulus is bulk stress
divided by bulk strain and is
given by B = –p/(V/V0).
Anglerfish
1000 m deep sea
Table 11.2
1/ bulk modulus = compressibility
k = 1/B
In 1000 meter deep sea, how much
is water compressed?
dV/V = k * p
= 46.4E-6 atm-1* 100 atm
= 0.464 % !
water is very hard to compress!
© 2012 Pearson Education, Inc.
Sheer stress and strain
Sheer stress and strain
• Sheer stress is F||/A and
sheer strain is x/h, as
shown in figure.
• Sheer modulus is sheer
stress divided by sheer
strain, and is given by
S = (F||/A)(h/x).
Sheer stress and strain of brass
The left piece experience shear force in an earthquake. It is 0.8
m square and 0.5 cm thick. What is the force exerted on each of
its edges if the displacement is x = 0.16 mm?
sol:
shear strain= x/h = 0.16 mm / 0.8 m = 0.0002
shear stress = shear strain * S = 0.0002 * 3.5E10 Pa = 7E6 Pa
force F = shear stress * A = 7E6 Pa * 0.8 m* 0.5 cm = 2.8E4 N
This is a huge force, even the displacement (0.16) is tiny!
Elasticity and plasticity
• Hooke’s law applies up to point a in Figure 11.18 below.
• Table 11.3 shows some approximate breaking stresses.
Midterm problem
A solid, uniform cylinder (radius r; mass m; moment of inertia around central axis I = ½ m r2) rolls
without slipping up a hill, as shown in the figure. The initial speed is v0 and the height of the hill is h.
At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff.
(a) How far from the foot of the cliff does the cylinder land? (20 points)
(b) Calculate the linear speed v2 and angular speed ω2 when the cylinder lands. Is the total kinetic
energy at that moment equal or smaller than the initial kinetic energy when the cylinder was at the
initial speed v0? (20 points)