Materi fisika bab elastisitas

2. Elasticity
• In physics, elasticity is
defined as the ability of an
object to return to its initial
form immediately after the
external force given to it is
removed (released)
• Stress ( Tegangan)
In this case, stress is defined
as the result of division
between the force acting
upon an object and its cross
section area.
Mathematically, stress can
be determined as follows:
Where:
• F = force (N)
• A = cross-section (m2)
•  = stress (N/m2)
A
F



3. • Strain (Regangan)
F
F
L L
Mathematically, strain can be determined as follows:
L
L
e


Where:
L = length increment (m)
L = initial length (m)
e = strain (has no unit)

4. • Modulus of Elasticity
A
B
c
D
E
0 X
stress
strain
Elastic
deformation
Plastic deformation
Hooke’s law limit
Elasticity limit
Bending point
Breaking point
Permanent deformation

5. • Mathematically, it can
be determined as
follows:
• Where:
• E = modulus of
elasticity or Young’s
modulus (Nm-2 or Pa)
e
E



6. • According to the stress and strain equation,
then the relation of modulus of elasticity ( E )
and force ( F ) can be determined as follows:
L
L
E
A
F
L
L
A
F
E
e
E







7. Sample Problem
1. A bar of stell which is 4 mm2 in
cross-sectional area and 4 cm in
length is pulled by a force of 100
N. If the modulus of elasticity of
stell is 2 x 1011 N/m2, calculate the
stress, strain, and length increment
of the stell.

8. 2. A cylinder made of stell has a length
of 10 m and diameter of 4 cm.
Calculate the length increment of the
cylinder if it is given a load of 105 N. (
E = 2 x 1011 N/m2)

9. Hooke’s Law
• Length increment in spring
L
L = x
F

10. • For that purpose, we
will prove the
relationship between
force and length
increment. For that
purpose, recall the
following equation:
•
L
A
E
k
where
x
k
F
x
L
EA
F
x
L
L
A
FL
E
.
:

















K = spring force
constanta
(N/m)

11. Sample Problem
• A spring will increase by 10 cm in length
if given a force of 10 N. What is the
length increment of the spring if gifen a
force of 7 N ?
• A metal has the Young’s modulus of 4 x
106 N/m2, its cross-sectional area is 20
cm2 and its length 5 m. What is the
force constant of the metal ?

12. Series Arrangement
Two or more spring that are arranged in
series comply with the following principles:
a. The pulling force on the series net
spring is equal to that experienced by
each spring.
F1 = F2 = F
b. The length increment of the series net
spring is equal to the sum of lengt
increment of ecah spring, then:
x = x1 + x2

13. • Then:
k1
k2
ks
F
2
2
2
2
2
2
1
1
1
1
1
1
k
F
x
x
k
F
k
F
x
x
k
F
k
F
x
x
k
F
s
s
















14. • Because x = x1 +
x2 , then :
• For n identical spring
of which the
constant of each is k
that are arranged in
series, the following
equation is valid:
2
2
1
1
1
k
k
ks


n
k
ks 

15. Parallel Arrangement
Two or more springs that are arranged in parrallel
comply with the following principles:
a. The pulling force on the parralel net spring
is equal to the total of the pulling force on
each spring:
F = F1 + F2
b. The length increment of the parallel net
spring is equal to lengt increment of ecah
spring, then:
x = x1 = x2

16. • • Then:
k1 k2
2
2
2
1
1
1
x
k
F
x
k
F
x
k
F p






F

17. • Because F = F1 + F2 , then :
kp = k1 + k2
For n indentical springs that are
arranged in parallel, of which each
constant is k, the following equation
applies:
kp = n k

18. Sample Problem
• Two springs having constants of 200 N/m and
300 N/m are arranged in series, then given
force of 30 N. What is the length increment
of that arrangement ?
• A load is hung on two springs that are
arranged in parallel where each springs
constant is 40 N/m and 60 N/m. What is the
length increment of that spring if the mass of
the load is 2 kg ? (g = 10 m/s2)

19. Simple Harmonic Motion (GHS)
• Pegas yang diberi beban digantung

20. • Pada GHS besar gaya pemulih pada pegas
sebanding dengan jarak benda dari titik
keseimbangan. Secara matematis dirumuskan:
• Tanda negatif pada persamaan di atas
menunjukkan bahwa arah F selalu berlawanan
arah dengan arah x
kx
F 


21. • Arah F selalu berlawanan dengan arah x
equilibrium
-x +F
+x -F

22. • Selain pada pegas, gaya pemulih juga bekerja
pada gerak harmonis pada bandul sederhana.
L
s


mg cos 
mg

23. • The restored force acting upon the harmonic
motion of a simple pendulum is component of
weight that is perpendicular to the rope. Thus,
the magnitude of the restored force on the
simple pendulum can be determined as
follows:
F = mg sin 

24. Period and Frequency on
spring
Period is the time
required to do one
complete vibration
or oscillation
(symbolized by T)
A – O – B – O – A
Frequency is the
number of
oscillations made
by the load in one
second
(symbolized by f)
•
B
O
A

25. • We have learnt known that there is a
relation between period (T) and frequency
(f) that is expressed by the equation as
follows:
f
T
1


26. • If the deviation of object that does the
harmonic motion is expressed by y, the
equation can be expressed as follows:
F = -ky
The restored force acted upon an object can
also be calculated from the acceleration of a
body based on Newton’s second law, that
is:
F =may

27. • Because ay = -2y, then :
F = -m 2 y
From the equations F = -ky and F = -
m 2 y, we can obtain an equation to
determine period and frequency of
object in harmonic motion on the
spring, is as follows:

28.  Because  = 2/T, then
the equation above
becomes:
m
k
m
k



2
k
m
T
m
k
T


2
2



29.  Where:
 T = period of the spring
vibration (s)
 m = the load mass (kg)
 k = spring force constant
(N/m)
 From the equation of
period above, we can
determine the equation of
frequency as follows:
m
k
f

2
1


30. Sample problem
 An object of 4 kg in mass is hung on spring
having a constant of 100 N/m. what is the period
and frequency of spring if the object is given a
small deviation (pulled then released) ?
 A spring is hung with a load of 1,8 kg, so that
the spring length increment is 2 cm. what is the
period and frequency of the spring oscillation ?
(g = 10 m/s2)

31. The Deviation of Simple Harmonic
Motion
 The deviation – time graph of the simple
harmonic motion.
y (m)
y = A
y = 0
amplitude
Equilibrium state

32.  Mathematically, the equation of deviation for y-t
sinusoidal graph as in the figure above can be
expressed by the following equation:
y= A sin t
Because  = 2/T = 2f, then:
ft
A
y
t
T
A
y


2
sin
2
sin

