File

Dr. Ahmed M. Maghraby
General Physics
Chapter (2):
Mechanical Properties of Materials
- When applying an external force on a material, some materials response is the
deformation.
- Deformation of a body involves the change in its size, or shape, or both of them.
- Deformation is of two types:
1- Elastic deformation: A temporary change in the size or shape of
a body and disappears after stopping the external force.
2- Plastic deformation: A permanent change in the size or shape of
a body even after stopping the external force.
- Materials can be classified into two categories:
•
Elastic material: Is the material which is able to restore its original size and
shape after removing the external force. Examples: Rubber, Steel
•
Plastic material: Is the material which fails to restore its original size and
shape after removing the external force. Examples: Clay, Mud, Dough
-
Elasticity of materials can be compared in terms of “elasticity moduli”.

Elasticity modulus is a property for each material.
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Dr. Ahmed M. Maghraby
General Physics

Elasticity modulus is the ratio of stress to strain.

Stress (σ): It is the instantaneous perpendicular force (F) per unit cross sectional area (A).

Strain (ε): is a measure of deformation as a result of stress.
- Types of Elasticity Moduli:
1 – Young’s modulus (Y): for the change in length of a linear body.
2- Shear modulus (S): for the shear effect in a body.
3- Bulk modulus (B): for the change in the volume of a bulk body.
===========================
1- Young’s modulus (Y):
-
It is the ratio of tensile stress to tensile strain.
-
Tensile stress (σ): it is the ratio of perpendicular applied force (F) to the
area (A).
(σ) =
F
A
-
σ has the unit of N/m2.
-
Tensile strain (ε): is the ratio of the change in length (ΔL) to the original
length (Lo).
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Dr. Ahmed M. Maghraby
General Physics
ε=
-
L
Lo
Tensile strain is a ratio of similar quantities and has no unit.
N/m2 is called “Pascal” or “Pa”
-
Strain can express elongation (increase
in length) or compression (decrease in
length).
-
Hooke’s law:
-
Hooke’s law represents the relation between the stress and strain.
Hooke’s law states that “The Stress is directly proportional to the Strain”
σ=Yε
-
If a rod is stretched by a force F1 , change in length is ∆L, then
Since Y, A and L is constant for each material,
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Dr. Ahmed M. Maghraby
-
So we can write:
So..
-
General Physics
F1 = K ΔL
According to Hooke’s law, “the change in length (ΔL) is directly
proportional to the applied force (F)”.
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Dr. Ahmed M. Maghraby

General Physics
The figure shows the Hooks law where the relation between the applied
force and the extension is linear with slope of force constant, k.
-
when an elastic body is stretched by external agent, the body exerts force
proportional to the distance and in opposite direction to external force:
F2 = - F1
-
Stress – Strain Relationship:
-
Hook’s law does not describe the whole relationship between the stress
and strain; it only describes the first linear part.
The above figure shows the stress-strain relationship.
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Dr. Ahmed M. Maghraby
General Physics
1. Point A is called “Proportional Limit”: it represents the limit of linear range
between stress and strain.
2. Point B is called “elasticity limit”: represents the limit of elasticity region,
beyond which the material deformation is plastic.
3. Point C is called “Yield point”.
4. Point D is called “Maximum stress point” represents the maximum value of
applied stress.
5. Point E is called “Fracture point”: is the point at which the body is fractured
(broken).
In region (OA)
- There is a linear relationship between stress and strain and this region called
Hookean behavior because the material obeys Hook’s law.
- The slope of straight line give modulus of elasticity define by
Modulus of elasticity=
Stress
Strain
(Pascal or N/m2).
In region (AB)
-
The stress increases in proportional to strain but not linearly.
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Dr. Ahmed M. Maghraby
-
General Physics
The point b is the elastic limit ( it is maximum stress, which a material can
withstand without undergoing some permanent deformation.
In region (BDE)
-
This region, Indicates the degree of permanent deformation in which a
material up to the point of fracture (E).
-
Example (2-1):
The bar shown has a square cross section for which the depth and thickness are
40 mm. If an axial force of 800 N is applied along the centroidal axis of the
bar’s cross sectional area, determine the average normal stress acting on the bar?
Solution:
Stress is  
-
F
800

 500 * 10 3 Pa
3 2
A
(40 * 10 )
Example (2-2):
A 80 Kg mass is hung on a steel wire having 18m long and 3mm diameter. What
is the elongation of the wire, knowing Young's modulus for steel is 21 x 1010
N/m2?
Solution:
m=80 kg
2r=3 mm= 0.003m
L 
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Lo= 18 m
r=0.0015m
Y= 21 x 1010 N/m2
80x9.8
18
x
 0.0095m  9.5mm
2
 (0.0015) 21x1010
Dr. Ahmed M. Maghraby
-
General Physics
Example (2-3):
A piece of copper originally 305 mm long is pulled in tension with a stress of
276 MPa. If the deformation is entirely elastic, what will be the resultant
elongation?
Solution
Young's modulus is given by:
Then, L 
-
(F A )L  L (276x 106 )(305x 103 )


 0.76x 103 m
11
Y
Y
(11x 10 )
Example (2-4):
A person carries a 21 kg suitcase in one hand. Assuming the humorous bone supports
the entire weight, determine how much it stretches, assume the humorous is 33 cm in
length and has an effective cross-sectional area of 5.2 x 10-4 m2. Where Y= 1.6x 1010
N/m2
Solution
Young,s modulus is
Y 
F /A
FL mgL
(21)(9.8)(0.33)
 L 


 8.17x 106 m
10
4
L / L
YA YA
(1.6x 10 )(5.2x 10 )
Exemple (2.5)
A telephone wire 120 m long and 2.2 mm in diameter is stretched by a force of 380 N.
What is the longitudinal stress? If the length after stretching is 120.10 m, what is the
longitudinal strain? Determine Young’s modulus for the wire?
Solution
A = 𝜋 r2 = 3.14 * (1.1*10-3)2 = 3.8 * 10-6 m2
F
380

 100*106
-6
A
3.8*10
ΔL  120.1 - 120  0.1 m
σ 
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N / m 2  100 MPa
Dr. Ahmed M. Maghraby
ε 
General Physics
ΔL
0.1

 8.33*10-4
Lo
120
σ
100*106
Y 

 12*1010 N / m 2  12*10 4 MPa
-4
ε
8.33*10
Exemple (2.7)
A load of 102 kg is supported by a wire of length 2 m and cross sectional area 0.1 cm2.
The wire is stretched by 0.22 cm . Find the tensile stress, tensile strain, and Young’s
modulus of the wire ?
m = 102 kg
L=2m
A = 0.1 cm2
∆L = 0.22 cm
Tensile Stress () 
F m g 102 * 9.8


 999.6 *10 5 N/m 2
4
A A 0.1 *10
Tensile Strain () 
L
0.22

 11 * 10  4
L o 2 * 100
Young's modulus (Y) 

999.6 * 105

 90.87 * 109 N/m2
4

11 * 10
Exemple (2.8)
A structure steel rod has a radius R of 9.5 mm and a length L of 81 cm. A force F of 6.2
* 104 N stretches it axially. (Esteel = 2 * 1011 N/m2)
(a) What is the stress in the rod ?
(b) What is the elongation of the rod under this load ?
(c) What is the strain?
Radius = 9.5 mm
L = 81 cm
F = 6.2 * 104 N
F
F
6.2 * 10 4
Tensile Stress ()   2 
 2.19 * 10 8 N/m 2
3 2
A r
 (9.5 * 10 )
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Dr. Ahmed M. Maghraby
Tensile Strain ( ) 
General Physics
L
Lo
;


E 

2.19 * 108
L   * Lo 
* Lo 
* 81 * 102  8.86 * 104 m
11
Y
2 * 10
Tensile Strain ( ) 

Y

2.19 * 108
 1.1 * 103
11
2 * 10
2- Shear modulus (S):
Shear stress:
Shear strain:
Shear Strain 
Tan =∆x/h
but  is small so
Shear strain =  = ∆x/h
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x
h
tan  so
Dr. Ahmed M. Maghraby
General Physics
Exemple (2.9)
A horizontal force of 1.2 N is applied to the top of a pile of compact discs 13 cm in
diameter and 9 cm high. The result is a 2.5 cm shear. Find the shear modulus?
Solution:
Shear Modulus (S)   F / A  /  
S
F h
F h
(1.2)(0.09)
= 2
=
A x  r x  (0.13 / 2) 2 (0.025)
3- Bulk modulus (B):
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 F / A  / (x / h )
N/m 2
Dr. Ahmed M. Maghraby
General Physics
Note: Bulk modulus is negative if the stress results in the decrease in volume and
positive if the stress results in the increase in volume.
Bulk stress (Volume stress):
Bulk strain (Volume strain): Bulk strain is A measure for the change in Volume
Volume Strain 
The compressibility factor, K 
V
V
1
B
Note: Bulk modulus applies for both solids and liquids.
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Dr. Ahmed M. Maghraby
General Physics
Energy Stored in a stretched wire (Elastic potential energy)
1-The stored potential energy (U):
1
U= 𝜎𝑥𝜀
2
2-The work done = 𝜎 𝑥 𝜀
Exemple (2.10)
A wire of length 120cm and diameter 0.82mm, supported from one end, A 5.3kg in the
other end. Find:
a) The stress in the wire
b) The strain in the wire
c) The strain energy where Y = 1.2x 1012 N/m2
Solution
r
0.82
 0.041 cm and m = 5.3kg
20
𝐹
𝑚𝑔
𝐴
𝐴
The stress = =
=
(5.3 𝑥 9.8)
𝜋(0.41 𝑥 10−3 )
= 40344 𝑁/𝑚2
40344
The strain  Stress 
=3.36 x 10-8
1.2 𝑥 1012
Y
1
Strain Energy = (Stress)(Strain) = 0.5 x 40344 x 3.36 x 10-8 = 6.7 x 10-4 J
2
Exemple (2.11)
A uniform wire of length 20cm density 0.78g/cm3 and mass 16gm stretched by a
distance 1.2mm when 8kg is supported on it, Find:
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Dr. Ahmed M. Maghraby
General Physics
a) The stress in the wire
b) Young's modulus
c) The strain energy
Solution
Volume (V)
V=
𝑚
𝜌
=
16 𝑥 10−3
0.78 𝑥 103
V = 20.5 x 10-6 m3
But V=A . L
A = V/L = 20.5 x 10-6 / 0.16 =128 x 10-6 m2
Stress=
𝐹
𝐴
=
𝑚𝑔
𝐴
=
8 𝑥 9.8
128 𝑥 10−6
= 0.612 x 106 N/m2
Y = Stress/Strain = 0.612 x 106 / (1.2 x 10-3/20 x 10-2)
Y = 1.02 x 1012 N/m2
Strain energy = ½ (Stress) x (Strain)
Strain energy = ½ x 0.612 x 106 x (1.2 x 10-3/20 x 10-2)
Strain energy = 12.75 x 108 J
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