Elasticity

E AST
L ISCITY
Name’s of
Groups:
1. Arventa
2. Atika
3. Aulia
4. Imam
5. Sebma
6. M. Ababil

Basic Competition :
• Analyzing the effect of forces on the
elastic properties of materials.

Indicator :
 Determining connection with the nature
of the concept of elasticity of the
spring force material
 Analyze the motion under the influence
of spring force

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
If a wire that has a cross sectional area A
experiences pulling force on both ends, then the
wire will experiences a stress. Stress is defined
as the result of devinision between the force
acting upon on object and its cross sectional area.
Mathematically, stress can be determined as
follows:
σ=F
A
Where :
F : force (N)
A : cross-sectional area (m2)
σ : stress (N/m2)

Strain
When a wire is pulled at both ends, then
besides experiencing stress, the wire increase in
length. In this case, the ratio between length
increment of the wire and the initial length is
called strain. Mathematically, strain can be
determined as follows : L
e = ∆L
L
Where : ∆L

L : initial length (m)
∆L : length increment (m)
F
e : strain (hasn’t unit)

Modulus of Elasticity
Modulus of Elasticity can be defined as the
ratio between stress and strain experienced by an
object. Modulus of elasticity is often called as
Young’s modulus. Mathematically, it can be
determined as follows :
F
E =
σ= A
e ∆L
L
F ∆L
=E
A L
Dengan :
E : modulus of elasticity (N/m2 or Pa)

Modulus of elasticity of several objects :
Object Modulus of elasticity
Aluminium 70 x 109
Steel 200 x 109
Iron 100 x 109
Concrete 20 x 109
Coal 14 x 109
Granite 45 x 109
Wood 10 x 109
Marble 50 x 109
Nylon 5 x 109
Bronze 100 x 109

Sample problem :
1. A bar of steel which is 4 mm2 in cross-
2

sectional area and 4 cm in length is
pulled by a force of 100 N. If the
modulus of elasticity of steel is 2 x 1011
1011 N/m2, calculate stress, strain, and
N/m2, calculate the the stress, strain,
and length increment of the steel!
length increment of the steel!

Solution :
Because :
A = 4 mm2 = 4 x 106 m2
L = 40 cm = 0,04 m
F = 100 N
E = 2 x 1011 N/m2
Then :
Stress
F 100 N
σ= = = 2,5 × 107 N / m 2
A 4 × 10 − 6 m 2

Strain

σ σ 2,5 ×10 7 N / m 2
E= ⇒e= = = 1,25 ×10 − 4
e E 2 ×1011 N / m 2

Length increment

∆L
e= ⇒ ∆L = e × L = 1,25 ⋅10 − 4 × 0,04m = 5 ⋅10 −6 m = 0,000005cm
L

Exercise :
1. A cylinder made of steel 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)
2. A metal wire having a diameter of 0,125 cm and
length of 80 cm is given a load of 100 N, and
the wire increases 0,51 mm in length. Calculate
the stress, strain, and Young’s modulus of
substance forming the wire!
3. A small block of aluminium of 2,5 m in length, 1
cm in width and 1,5 mm in thickness is hung and
given load of 50 kg, and the block increases 1,2
mm in length. Calculate the Young’s modulus of
that aluminium!

4. A metal wire having a length of 4 m and 2 x
10-6 N/m2 in cross-sectional area. A force
exerted to pulled this metal wire so that the
wire increases 0,3 m. Calculate the force
acting upon a metal wire!
5. For the safety, a climber used a nylon rope
which 50 m in length and 1 cm in diameter.
When it shore up a climber whose mass is 80
kg, the rope increases 1,6 m i length.
Determine the modulus of elasticity of nylon!
(π = 3,15 and g = 9,86 m/s2)

Elasticity

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