Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
Chapter (2)
Elasticity
Mechanical Properties of Metals
Many materials may be deformed when external forces exert on them.
‫او‬
‫تا قر‬
‫ندادم‬
‫معظم المواد يمكن إحداث تشويه (تغيير) بها‬
. ‫خ رجية نليه‬
Elasticity materials:‫ المواد المرنة‬: If the material restore to its original
shape and size after removing the load from it, it's said to be elastic.
‫الجسم المرن يستعيد شكله األصلى بزوال اإلجه د الم قر نليه م لم‬
‫يتجوز حد المرونة‬
Plastic materials:‫المواد الغير مرنة‬
If the material fails to restore its original dimensions after removing the applied
stress, it's said to plastic. ‫الجسم الغيار مارن سيساتعيد شاكله األصالى‬
‫(يتشوه) بزوال اإلجه د الم قر نليه‬
Elastic Modulus: Is the constant of each matter and equal ratio
between stress and strain ‫و‬
‫مع مل المرونة هو ق بتة لكل م د مدفرد‬
‫الدسبة بين اسجه د واسنفع ل‬
Concepts of Stress and Strain
(1) Stress () ( ‫)اإلجهاد‬
It is the instantaneous perpendicular force (F) per unit cross - sectional area (Ao).
.i.e. is related to the force causing the deformation
‫نلاى‬
‫(ى‬
‫المسا ح‬
‫نمودي ً نلى وحد‬
))‫المسببة لعملية تشويه (تغيير‬
σ
F
A
‫الم قر‬
‫ هو مقدار القو‬-
‫مس حة المقطع ( هو متعلق بقو‬
N/m 2
or
Ib/in 2
(2) Strain () ( ‫)اإلنفعال‬
Is a measure of the degree of deformation?
)‫هو مقي س لعملية تشويه (تغيير‬
-Chapter(2- 1 (-
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
Elastic Modulus
Young's
(tensile )
modulus
Shear
( rigidity)
modulus
Bulk (volume)
modulus
Y=stress/strain
S=stress/strain
B=stress/strain
Stress Different Types
Strain
Elastic Deformation
Plastic Deformation
‫تغير دائم س يزول بزوال القو الم قر‬
‫تغير م ت يزول بزوال القو الم قر‬
a)Young,s modulus: (Y) It is the ratio between the stress and the strain.
‫ الدسبة بين اإلجه د واإلنفع ل (الح دقين فى نفس إتجا ه القاو‬.)‫ ويكون ذلك فى مدطقة المرونة فقط‬- ‫الم قر‬
Y 
F /A
σ

L / L
ε
N/m 2
or
-Chapter(2- 2 (-
Ib/in 2
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
where

tensile Stress () ( ‫)اإلجهاد‬
It is the instantaneous perpendicular force (F) per unit cross - sectional area (Ao).
‫(ى نلاى‬
‫ هو مقدار القو الم قر نمودي ً نلى وحد المسا ح‬.)‫مس حة المقطع‬
F
A
σ
N/m 2
or
Ib/in 2
tensile Strain () ( ‫)اإلنفعال‬
It is the ratio between the change in length (L) and the original length (Lo=L).
‫ الدسبة بين التغير الح دث فى الطول (نتيجة تأقير القو ) إلى‬.‫الطول األصلى‬
ε
ΔL
Lo
,
L : is the elongation or stretch
Hooke’s Law ( ‫)قانون هوك‬
“The Stress is directly proportional to the Strain”
=Y
‫يتد سب اإلجه د تد سب ً طردي ً مع اإلنفعا ل (فاى مدطقاة المروناة‬
.)‫فقط‬
Y: is the Elastic modulus or Young’s modulus
 The greater the modulus, the stiffer the material, or, the smaller the strain
results from the application of the stress.
‫او‬
‫يمة مع مل المرونة تزداد‬
‫يمة اإلنفع ل الد تج نن‬
‫تقل‬
‫زاد‬
‫كلم‬
‫تحمل المعدن – ىو‬
.‫تأقير اإلجه د‬
 If a rod is stretched by a force F1 distance ∆L, then
Y 
F1L
YA
F 
L
A L
L
(1)
-Chapter(2- 3 (-
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
Since Y, A and L is constant for each material,
then we can write K 
YA
(Constant Force)
L
So, we can write Eq.(1) as F1  K L (2)
 As the rod stop stretching this means that there another internal force, F2
counterpart this force and we have F2  F1
 This is the Hooke's law which states that when an elastic body is stretched by
external agent, the body exerts force proportional to the distance and in
opposite direction to external force. The figure shows the Hooks law where the
relation between the applied force and the extension is linear with slope of
force constant, k.
‫يتم تمدد جسم مرون من‬
‫نون هوك التي تدص نلى ىنه نددم‬
‫الجسم متد سبة مع المس فة وفي استج ه‬
‫و‬
‫هو‬
‫ وتم رس‬،‫خ رجية‬
‫ ويوضح الشكل الق نون خط مستقيم حيث العال ة بين‬.‫خ رجية‬
.K ،‫ق بتة‬
‫و‬
‫هذا‬
‫و‬
‫بل ا‬
‫المع كس لقو‬
‫المستخدمة وتمديد خطي مع انحدار‬
‫القو‬
Note That;
1 Ib/in2 (psi)= 6891 N/m2 (Pa) 1 N/m2 (Pa)= 1.451*10-4 Ib/in2 (psi)
Stress – Strain Behavior
Elasticity and Plasticity
Stress
Elastic limit OR,
Yield point
b
Elastic behavior
Proportional
limit
‫حد التناسب أى‬
‫فناسب اإلجهاد‬
‫مع اإلنفعال‬
‫نقط أقصح إجهاد يتظيل الجسم‬
Necking point or Tensile Stress
‫نقط حد اليرون أو‬
‫نقط إذعان الجسم‬
‫للقوك اليؤثرك‬
d
c
Fracture
point
a
Plastic deformation
‫ةةة لها‬
‫منطقةةة اللدانةةة التةةةح فظةةةتفد اليةةةادك هاهةةةا‬
. ‫والفعود لألصل حتح عد زوال القوك اليؤثرك‬
o
-Chapter(2- 4 (-
Strain, OR,
Percent elongation
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
‫إلاى خا رد حادود‬
‫نلى الما د‬
. ‫تدخل فى مدطقة اللدونة‬
‫الم قر‬
‫تزداد القو‬
‫ نددم‬-
‫مدطقة المرونة فإن الم د‬
‫الزي د الطولية الح دقة فى الم د ستعتمد فقاط نلاى‬
‫القو الم قر ولكن ىيضا ً نلاى ناول الما د وطولها‬
‫األصلى ومس حة مقطعه‬
-
 In region (oa)
 There is a linear relationship between stress and strain and this region
called Hookean behavior because the material obey Hook’s law.
 The slope of straight line give modulus of elasticity (‫)معامل المرونة‬define by
Stress
Modulus of elasticity=
(Pascal or N/m2).
Strain
 In region (ab)
The stress increases in proportional to strain but not
linearly.
 The point b is the elastic limit ( it is maximum stress,
which a material can withstand (‫)تقاوم‬without
undergoing some permanent deformation(‫(تشوة دائم‬
 In region (bd)
 This region, Indicates the degree of permanent
deformation in which a material up to the point of
fracture (d).

Example (2.1)
The bar shown has a square cross section(‫ )مقطةع مر ةع‬for 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?(‫)عان متوسط االجهاد الواقع ال ريط‬
Answer
Stress is  
800 N
F
800

 500 * 10 3 Pa
3 2
A
(40 * 10 )
Example (2.2)
-Chapter(2- 5 (-
800 N
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
A 80 Kg mass is hung ( ‫ )علقت‬on 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? ‫مم مامقدار التمدد فى طوله‬3 ‫ متر و قطره‬81 ‫ كج علقت فى سلك طوله‬18 ‫كتلة‬
Answer
m=80 kg
2r=3 mm= 0.003m
Lo= 18 m
r=0.0015m
Young's modulus is given by Y
L 

F A
 L Lo
Y= 21 x 1010 N/m2
so the elongation is
L 
F Lo
AY
80 x9.8
18
x
 0.0095m  9.5mm
 (0.0015) 2 21x1010
Exemple (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?
Answer
Young's
modulus
is
given
by
Y 
F A
 L Lo
then
(F A )L  L (276x 106 )(305x 103 )
L 


 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 humerus bone ( ‫عضةد‬
‫)العظةم‬supports the entire weight(‫)وزن‬, determine how much it stretches( ‫)االستطالة‬. Assume
the humerus 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
Answer
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 )
-Chapter(2- 6 (-
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
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?
Answer
A =  r2 = 3.14 * (1.1*10-3)2 = 3.8 * 10-6 m2
F
380

 100*106
A
3.8*10-6
ΔL  120.1 - 120  0.1 m
ΔL
0.1
ε 

 8.33*10-4
Lo
120
σ 
Y 
N / m 2  100 MPa
σ
100*106

 12*1010 N / m 2  12*10 4 MPa
ε
8.33*10-4
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

11 * 104
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
-Chapter(2- 7 (-
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
F
F
6.2 *10 4
Tensile Stress ()   2 
 2.19 *10 8 N/m 2
3 2
A r
 (9.5 *10 )
Tensile Strain ( ) 
L   * Lo 

Y
L
Lo
E 
;
* Lo 


2.19 * 108
* 81 * 10 2  8.86 * 10 4 m
11
2 * 10

2.19 * 108
Tensile Strain ( ) 

 1.1 * 103
11
Y
2 * 10
Shear Modulus (Elasticity in Shape) (S) ( ‫)مع مل القص‬
Shear Modulus (S) 
Shear Stress
= F / A /  
Shear Strain
 F / A  / (x / h )
N/m 2
Ib/in 2
or
Where
Shear Stress ( ‫)إجه د القص‬
Shear Stress 
Ft
A
N/m 2
or
Ib/in 2
Shear Strain ( ‫)إنفع ل القص‬
The shear strain = tan=∆x/h but  is small so tan 
so
shear strain= =∆x/h.
Shear Strain 
‫در‬
‫يقيس ن‬
x
h
‫القص‬
x: is the deflection (‫)اإلنحراف‬
‫ومع مل‬
‫كالً من مع مل ي نج‬
‫نلى مق ومة التغير المرن‬
-Chapter(2- 8 (-
‫الحظ أن‬
‫الم د‬
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa

Example (2.9)
A horizontal force( ‫ )قتة اقق ت‬of 1.2 N is applied to the top of a stack of pancakes ( ‫كومة مة‬
‫)الفطةئر‬13 cm in diameter( ‫ )قات‬and 9 cm high( ‫)ارتفتل‬. The result is a 2.5 cm shear. Find the
shear modulus.
Answer
Shear Modulus (S)   F / A  /  
S
 F / A  / (x / h )
F h
F h
(1.2)(0.09)
= 2
=
A x  r x  (0.13 / 2) 2 (0.025)
N/m 2
Bulk Modulus (Elasticity in Volume) (B) ( ‫)مع مل حجمى‬
Where
Change in Volume
-Chapter(2- 9 (-
Phys 1010
Chapter (2) : Elasticity
Dr H. Gomaa
Volume Stress (P) ( ‫)اإلجه د‬
P 
Fn
A
N/m 2
or
Ib/in 2
Volume Strain ( ‫)اإلنفع ل‬
Volume Strain 
F
Volume Stress
Ao
B  
 
V
VolumeStra in
V
*** The compressibility factor, K 
V
V
N/m 2
1
B
-Chapter(2- 10 (-
or
Ib/in 2
Phys 1010
Dr H. Gomaa
Chapter (2) : Elasticity
Energy Stored in a stretched wire (Elastic potential energy)
1
1
The strord potentail energy(U) =thermal energy= kx 2  strain  stress
2
2
 The work done  kx 2
Exemple (2.12)
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 dyne/cm2 and g = 980cm/sec2
Solution
r
0.82
 0.041 cm
20
The stress 
and m = 5.3x103gm
F m g 5.3 x 10 3  980



2
A
A
 0.041
The strain 
Stress

Y
Strain Energy 
1
Stress Strain 
2
Exemple (2.13)
-Chapter(2- 11 (-
dyne /cm2
Phys 1010
Dr H. Gomaa
Chapter (2) : Elasticity
A uniform wire of length 20cm density 0.78gm / cm3 and mass 16gm
stretched by a distance 1.2mm when 8kg is supported on it , Find :
a) The stress in the wire
b) Young's modulus
c) The strain energy
Solution
Volume V 
m


16
 2.05 cm3
7.8
But V=A . L  A 
V 2.05

 0.1 cm 2
L
20
F m g 8 x 10 3  980


Stress  
A
A
0.1
But Stress = Y
L
Strain Energy 
L
 Y
Stress . 20

1.2 x 10  1
1
Stress Strain 
2
-Chapter(2- 12 (-
dyne / cm2
dyne / cm 2