Fiddle Faddle« & Screaming Yellow Zonkers«, przepisy, Przepisy Pizza Hut KFC MC DONALDS, Przepisy po angielsku
Fatburger«, przepisy, Przepisy Pizza Hut KFC MC DONALDS, Przepisy po angielsku
Food 20 adjectives, ANGIELSKI, !!!!!!!!!!pomoce, słownictwo, food
Frommer's Naples and the Amalfi Coast day BY Day, Travel Guides- Przewodniki (thanx angielski i stuff)
Frommer s Sicily Day By Day, Travel Guides- Przewodniki (thanx angielski i stuff)
Fuyumi Ono - Twelve Kingdoms 01 - Shadow of the Moon a Sea of Shadows, Angielskie [EN](4)(2)
Farago&Zwijnenberg (eds) - Compelling Visuality ~ The work of art in and out of history, sztuka i nie tylko po angielsku
Forex Study Book For Successful Foreign Exchange Dealing, gielda walutowa, Angielskie
Fabulous Creatures Mythical Monsters and Animal Power Symbols-A Handbook, Angielski
Furocious Lust Shorts 4 - Stripe Tease - Milly Taiden,
Fundamentals of College Physics - Chapter 10, Angielskie [EN](4)(2)
[ Pobierz całość w formacie PDF ]Chapter 10 Elasticity
“
If I have seen further than other men, it is because I stood on the shoulders of giants.”
Isaac Newton
10.1 The Atomic Nature of Elasticity
Elasticity
is that property of a body by which it experiences a change in size or shape whenever a deforming force
acts on the body.
When the force is removed the body returns to its original size and shape. Most people are
familiar with the stretching of a rubber band. All materials, however, have this same elastic property, but in most
materials it is not so pronounced.
The explanation of the elastic property of solids is found in an atomic description of a solid. Most solids are
composed of a very large number of atoms or molecules arranged in a fixed pattern called the
lattice structure of
a solid
and shown schematically in figure 10.1(a). These atoms or molecules are held in their positions by
electrical forces. The electrical force between the molecules is attractive and tends to pull the molecules together.
Thus, the solid resists being pulled apart. Any one molecule in figure 10.1(a) has an attractive force pulling it to
the right and an equal attractive force pulling it to the left. There are also equal attractive forces pulling the
molecule up and down, and in and out. A repulsive force between the molecules also tends to repel the molecules if
they get too close together. This is why solids are difficult to compress. To explain this repulsive force we would
need to invoke the Pauli exclusion principle of quantum mechanics (which we discuss in section 32.8). Here we
simply refer to all these forces as molecular forces.
Figure 10.1
(a) Lattice structure of a solid. (b) Actual pictures of atoms in a solar cell.
The net result of all these molecular forces is that each molecule is in a position of equilibrium. If we try to
pull one side of a solid material to the right, let us say, then we are in effect pulling all these molecules slightly
away from their equilibrium position. The displacement of any one molecule from its equilibrium position is quite
small, but since there are billions of molecules, the total molecular displacements are directly measurable as a
change in length of the material. When the applied force is removed, the attractive molecular forces pull all the
molecules back to their original positions, and the material returns to its original length.
If we now exert a force on the material in order to compress it, we cause the molecules to be again
displaced from their equilibrium position, but this time they are pushed closer together. The repulsive molecular
force prevents them from getting too close together, but the total molecular displacement is directly measurable as
a reduction in size of the original material. When the compressive force is removed, the repulsive molecular force
causes the atoms to return to their equilibrium position and the solid returns to its original size.
Hence, the elastic
properties of matter are a manifestation of the molecular forces that hold solids together.
Figure 10.1(b) shows a
typical lattice structure of atoms in a solar cell analyzed with a scanning tunneling microscope.
10.2 Hooke’s Law--Stress and Strain
If we apply a force to a rubber band, we find that the rubber band stretches. Similarly, if we attach a wire to a
support, as shown in figure 10.2, and sequentially apply forces of magnitude
F,
2
F,
and 3
F
to the wire, we find
that the wire stretches by an amount ∆
L,
2∆
L,
and 3∆
L,
respectively. (Note that the amount of stretching is
greatly exaggerated in the diagram for illustrative purposes.) The
deformation, ∆
L,
is directly proportional to the magnitude of the
applied force
F
and is written mathematically as
∆
L
∝
F
(10.1)
This aspect of elasticity is true for all solids. It would be
tempting to use equation 10.1 as it stands to formulate a theory of
elasticity, but with a little thought it becomes obvious that although it
is correct in its description, it is incomplete.
Let us consider two wires, one of cross-sectional area
A,
and
another with twice that area, namely 2
A,
as shown in figure 10.3.
When we apply a force
F
to the first wire, that force is distributed over
all the atoms in that cross-sectional area
A.
If we subject the second
wire to the same applied force
F,
then this same force is
Figure 10.2
Stretching an object.
distributed over twice as many atoms in the area 2
A
as it was in the area
A.
Equivalently we can say that each atom receives only half the force in the area
2
A
that it received in the area
A.
Hence, the total stretching of the 2
A
wire is
only 1/2 of what it was in wire
A.
Thus, the elongation of the wire ∆
L
is
inversely proportional to the cross-sectional area
A
of the wire, and this is
∆
L
∝
1
(10.2)
A
Note also that the original length of the wire must have something to
do with the amount of stretch of the wire. For if a force of magnitude
F
is
applied to two wires of the same cross-sectional area, but one has length
L
0
Figure 10.3
The deformation is inversely
proportional to the cross-sectional
area of the wire.
and the other has length 2
L
0
, the same force is transmitted to every
molecule in the length of the wire. But because there are twice as
many molecules to stretch apart in the wire having length 2
L
0
, there is
twice the deformation, or 2∆
L,
as shown in figure 10.4. We write this
as the proportion
∆
L
∝
L
0
(10.3)
The results of equations 10.1, 10.2 and 10.3 are, of course, also
deduced experimentally.
The deformation
∆
L of the wire is thus
directly proportional to the magnitude of the applied force F (equation
10.1), inversely proportional to the cross-sectional area A (equation
10.2), and directly proportional to the original length of the wire L
0
(equation 10.3).
These results can be incorporated into the one
proportionality
∆
L
∝
FL
0
A
which we rewrite in the form
F
∝
∆
L
(10.4)
A
L
0
Figure 10.4
The deformation is directly
proportional to the original length of the wire.
The ratio of the magnitude of the applied force to the cross-sectional area of the wire is called the
stress
acting on the wire, while the ratio of the change in length to the original length of the wire is called the
strain
of the
wire.
Equation 10.4 is a statement of
Hooke’s law of elasticity,
which says that in an elastic body the stress is
directly proportional to the strain,
that is,
stress ∝ strain (10.5)
The stress is what is applied to the body, while the resulting effect is called the strain.
To make an equality out of this proportion, we must introduce a constant of proportionality (see appendix
C on proportionalities). This constant depends on the type of material used, since the molecules, and hence the
molecular forces of each material, are different. This constant, called
Young’s modulus of elasticity
is denoted
by the letter
Y.
Equation 10.4 thus becomes
F
=
Y
∆
L
(10.6)
A
L
0
The value of
Y
for various materials is given in table 10.1.
Table 10.1
Some Elastic Constants
Substance
Young’s
Modulus
Shear
Modulus
Bulk Modulus
Elastic Limit
Ultimate
Tensile
Stress
N/m
2
× 10
10
N/m
2
× 10
10
N/m
2
× 10
10
N/m
2
× 10
8
N/m
2
× 10
8
Aluminum
Bone
Brass
Copper
Iron
Lead
Steel
7.0
1.5
9.1
11.0
9.1
1.6
21
3
8.0
3.6
4.2
7.0
0.56
8.4
7
1.4
1.4
1.30
4.5
4.1
3.2
0.2
4.8
6
14
10
0.77
16
3.5
1.6
1.7
2.4
Example 10.1
Stretching a wire.
A steel wire 1.00 m long with a diameter
d
= 1.00 mm has a 10.0-kg mass hung from it. (a) How
much will the wire stretch? (b) What is the stress on the wire? (c) What is the strain?
Solution
a.
The cross-sectional area of the wire is given by
A
= π
d
2
= π(1.00 × 10
−
3
m)
2
= 7.85 × 10
−
7
m
2
4 4
We assume that the cross-sectional area of the wire does not change during the stretching process. The force
stretching the wire is the weight of the 10.0-kg mass, that is,
F
=
mg
= (10.0 kg)(9.80 m/s
2
) = 98.0 N
Young’s modulus for steel is found in table 10.1 as
Y
= 21 × 10
10
N/m
2
. The elongation of the wire, found from
modifying equation 10.6, is
∆
L
=
FL
0
AY
= (98.0 N)(1.00 m)
(7.85 × 10
−
7
m
2
)(21.0 × 10
10
N/m
2
)
= 0.594 × 10
−
3
m = 0.594 mm
b.
The stress acting on the wire is
F
= 98.0 N = 1.25 × 10
8
N/m
2
A
7.85 × 10
−
7
m
2
c.
The strain of the wire is
∆
L
=
0.594 × 10
−
3
m
= 0.594 × 10
−
3
L
0
1.00 m
To go to this Interactive Example click on this sentence.
The applied stress on the wire cannot be increased
indefinitely if the wire is to remain elastic. Eventually a point is
reached where the stress becomes so great that the atoms are pulled
permanently away from their equilibrium position in the lattice
structure. This point is called the
elastic limit
of the material and is
shown in figure 10.5. When the stress exceeds the elastic limit the
material does not return to its original size or shape when the stress
is removed. The entire lattice structure of the material has been
altered. If the stress is increased beyond the elastic limit, eventually
the ultimate stress point is reached. This is the highest point on the
stress-strain curve and represents the greatest stress that the
material can bear. Brittle materials break suddenly at this point,
while some ductile materials can be stretched a little more due to a
decrease in the cross-sectional area of the material. But they too
break shortly thereafter at the breaking point.
Hooke’s law is only
Figure 10.5
Stress-strain relationship.
valid below the elastic limit,
and it is only that region that will concern us.
Although we have been discussing the stretching of an elastic body, a body is also elastic under
compression. If a large load is placed on a column, then the column is compressed, that is, it shrinks by an amount
∆
L.
When the load is removed the column returns to its original length.
Example 10.2
Compressing a steel column.
A 445,000-N load is placed on top of a steel column 3.05 m long and 10.2 cm in
diameter. By how much is the column compressed?
Solution
The cross-sectional area of the column is
A
= π
d
2
= π(10.2 cm)
2
= 81.7 cm
2
4 4
The change in length of the column, found from equation 10.6, is
∆
L
=
FL
0
AY
(
)(
)
2
445,000 N
3.05 m
100 cm
1 m
=
(
)(
)
2
10
2
81.7 cm
21
×
10
N/m
= 7.91 × 10
−
4
m = 0.0791 cm = 0.791 mm
Note that the compression is quite small (0.791 mm) considering the very large load (445,000 N). This is indicative
of the very strong molecular forces in the lattice structure of the solid.
To go to this Interactive Example click on this sentence.
Example 10.3
Exceeding the ultimate compressive strength.
A human bone is subjected to a compressive force of 5.00 × 10
5
N/m
2
.
The bone is 25.0 cm long and has an approximate area of 4.00 cm
2
. If the ultimate compressive strength for a bone
is 1.70 × 10
8
N/m
2
, will the bone be compressed or will it break under this force?
Solution
The stress acting on the bone is found from
F
= 5.00 × 10
5
N = 12.5 × 10
8
N/m
2
A
4.00 ×10
−
4
m
2
Since this stress exceeds the ultimate compressive stress of a bone, 1.70 × 10
8
N/m
2
, the bone will break.
To go to this Interactive Example click on this sentence.
10.3 Hooke’s Law for a Spring
A simpler formulation of Hooke’s law is sometimes useful and can be found from equation 10.6 by a slight
rearrangement of terms. That is, solving equation 10.6 for
F
gives
F
=
AY
∆
L
L
0
Because
A,
Y,
and
L
0
are all constants, the term
AY
/
L
0
can be set equal to a new constant
k,
namely
k
=
AY
(10.7)
L
0
We call
k
a force constant or a spring constant. Then,
F
=
k
∆
L
(10.8)
The change in length ∆
L
of the material is simply the final
length
L
minus the original length
L
0
. We can introduce a new
reference system to measure the elongation, by calling the location of
the end of the material in its unstretched position,
x
= 0. Then we
measure the stretch by the value of the displacement
x
from the
unstretched position, as seen in figure 10.6. Thus, ∆
L
=
x,
in the new
reference system, and we can write equation 10.8 as
F
=
kx
(10.9)
Figure 10.6
Changing the reference system.
Equation 10.9 is a simplified form of Hooke’s law that we use in vibratory motion containing springs. For a helical
spring, we can not obtain the spring constant from equation 10.7 because the geometry of a spring is not the same
as a simple straight wire. However, we can find
k
experimentally by adding various weights to a spring and
measuring the associated elongation
x,
as seen in figure 10.7(a). A plot of the magnitude of the applied force
F
versus the elongation
x
gives a straight line that goes through the origin, as in figure 10.7(b). Because Hooke’s law
for the spring, equation 10.9, is an equation of the form of a straight line passing through the origin, that is,
y
=
mx
[ Pobierz całość w formacie PDF ]
