ELASTISCITYName’s of

Groups:1. Arventa2. Atika3. Aulia4. Imam5. Sebma6. 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

ElasticityIn 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)

StrainWhen 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 :

e = ∆L L

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

L

F

L

Modulus of ElasticityModulus 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 :

Dengan :E : modulus of elasticity (N/m2 or Pa)

L

LE

A

F

LLAF

eE

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-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 N/m2, calculate the stress, strain, and length increment of the steel!

Sample problem :

1. A bar of steel 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 steel is 2 x 1011 N/m2, calculate the stress, strain, and length increment of the steel!

Solution :Because :A = 4 mm2 = 4 x 106 m2

L = 40 cm = 0,04 mF = 100 NE = 2 x 1011 N/m2

Then :Stress

2726

/105,2104

100mN

m

N

A

F

Strain

Length increment

4211

27

1025,1/102

/105,2

mN

mN

Ee

eE

cmmmLeLL

Le 000005,010504,01025,1 64

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)