PP. Strength of materials dr.Salem.pptx civil

torsion lecture

1. INTRODUCTION TO
STRENGTH OF MATERIALS
COURSE
INTRODUCTION

2. Details of Lecturer
Course Lecturer: Dr. Salem Alizi
RoomNumber:2Faculty of Engineering
Dep. Of Engineering Materials
Email: [email protected]
Tel. No. : 07722050281
Office Hours: 8.30 a.m. to 13 Noon.
(Sun, Mon, Tue, and Thu.)

3. COURSE GOALS
       
This course has two specific goals:
(i) 
 
To introduce students to concepts of
stresses and strain; shearing force and
bending; as well as torsion and deflection of
different structural elements.
(ii) To develop theoretical and analytical
skills relevant to the areas mentioned in (i)
above.

4. COURSE OUTLINE

5.         

Course Objectives
  Upon successful completion of this course,
students should be able to:
(i) 
 
Understand and solve simple problems
involving stresses and strain in two and three
dimensions.
(ii) Understand the difference between statically
determinate and indeterminate problems.
(iii) Understand and carry out simple experiments
illustrating properties of materials in tension,
compression as well as hardness and impact tests.

6. COURSE OBJECTIVES CONTD.
(iv) Analyze stresses in two dimensions and
understand the concepts of principal stresses
and the use of Mohr circles to solve two-
dimensional stress problems.
(v) Draw shear force and bending moment
diagrams of simple beams and understand the
relationships between loading intensity,
shearing force and bending moment.
 
(vi) Compute the bending stresses in beams
with one or two materials.

7. OBJECTIVES CONCLUDED
(vii) Calculate the deflection of beams
using the direct integration and moment-
area method.
 
(viii) 
 
Apply sound analytical techniques
and logical procedures in the solution of
engineering problems.
 

8. Teaching Strategies
The course will be taught via
Lectures. Lectures will also involve
the solution of tutorial questions.
Tutorial questions are designed to
complement and enhance both the
lectures and the students
appreciation of the subject.
 Course work assignments will be
reviewed with the students.

9. Lecture Times
Sunday: 8.30 to 10.30 a.m.
Thursday: 11.30 a.m. to 12.30 P.m.
Lab Sessions: Two Labs per student on
Mondays (Details to be Announced Later)
Attendance at the Lectures and Labs is
Compulsory.

10. Time-Table For Labs




MONDAY 1:00 - 4:00 P.M.

Week
Group

1,5,9

2,6,10

3,7,11,

4,8,12

K


-

ME13A

ME16A
(3,7)

ME13A

L

ME13A

-

ME13A

ME16A
(4,8)

M

ME16A
(5,9)

ME13A

-

ME13A

N


ME13A

ME16A
(6,10)

ME13A

-

11. More Course Details
BOOK – Strength of Materials 4th edition
(Solutions Manual) Singer Pytel (Legibly)
 COURSE WORK
1.Two Mid-Semester Test (40%);
2. Practical report (10%) and
3. End of Semester 1 Examination (50%).
 

12. MAE207: CHAPTER ONE
STRESS AND STRAIN
RELATIONS

13. 1.1 DIRECT OR NORMAL
STRESS
When a force is transmitted through a
body, the body tends to change its shape
or deform.The body is said to be strained.
Direct Stress = Applied Force (F)
Cross Sectional Area (A)
Units: Usually N/m
2
(Pa), N/mm
2
, MN/m
2
,
GN/m
2
or N/cm
2
Note: 1 N/mm
2
= 1 MN/m
2
= 1 MPa

14. Direct Stress Contd.
Direct stress may be tensile, t or
compressive,c and result from forces
acting perpendicular to the plane of the
cross-section


Tension
Compression

15. 1.2
Direct or Normal Strain
When loads are applied to a body,
some deformation will occur resulting to
a change in dimension.
Consider a bar, subjected to axial
tensile loading force, F. If the bar
extension is dl and its original length
(before loading) is L, then tensile strain
is:

16. Direct or Normal Strain Contd.
Direct Strain ( ) = Change in Length
Original Length
i.e. = dl/L
dl
F
F
L



17. Direct or Normal Strain Contd.
As strain is a ratio of lengths, it is
dimensionless.
Similarly, for compression by amount,
dl: Compressive strain = - dl/L
Note: Strain is positive for an increase
in dimension and negative for a
reduction in dimension.

18. 1.3 Shear Stress and Shear Strain
Shear stresses are produced by
equal and opposite parallel forces
not in line.
 The forces tend to make one part
of the material slide over the other
part.
Shear stress is tangential to the
area over which it acts.

19. Shear Stress and Shear Strain
Contd.
P Q
S
R
F
D D’
A B
C
C’
L
x

Shear strain is the distortion produced by shear stress on
an element or rectangular block as above. The shear
strain, (gamma) is given as:
= x/L = tan  


20. Shear Stress and Shear Strain
Concluded
For small ,
 Shear strain then becomes the change
in the right angle.
It is dimensionless and is measured in
radians.


21. 1.3 Complementary Shear Stress
a

1

1

2

2
P
Q
S R
Consider a small element, PQRS of the material in the
last diagram. Let the shear stress created on faces PQ
and RS be 
1

22. Complimentary Shear Stress
Contd.
The element is therefore subjected to a
couple and for equilibrium, a balancing
couple must be brought into action.
 This will only arise from the shear stress on
faces QR and PS.
Let the shear stresses on these faces be


. 
2

23. Complimentary Shear Stress
Contd.
Let t be the thickness of the material at
right angles to the paper and lengths of
sides of element be a and b as shown.
For equilibrium,clockwise coupl =
anticlockwise couple
i.e. Force on PQ (or RS) x a = Force on
QR (or PS) x b
 
 
1 2
1 2
xbtxa xatxb
ie

..

24. Complimentary Shear Stress
Concluded
Thus: Whenever a shear stress occurs on a
plane within a material, it is automatically
accompanied by an equal shear stress on the
perpendicular plane.
 The direction of the complementary shear
stress is such that their couple opposes that of
the original shear stresses.

25. 1.4
Volumetric Strain

Hydrostatic stress refers to tensile or
compressive stress in all dimensions
within or external to a body.
Hydrostatic stress results in change in
volume of the material.
Consider a cube with sides x, y, z. Let
dx, dy, and dz represent increase in
length in all directions.
i.e. new volume = (x + dx) (y + dy) (z +
dz)

26. Volumetric
Strain Contd.
Neglecting products of small quantities:
New volume = x y z + z y dx + x z dy + x y dz
 Original volume = x y z
 = z y dx + x z dy + x y dz

Volumetric strain, =z y dx + x z dy + x y dz
x y z
 = dx/x + dy/y + dz/z
V
 
v x y z 

27. Strains Contd.
Note: By similar reasoning, on area x y
Also: (i) The strain on the diameter of a circle
is equal to the strain on the circumference.
 (ii) The strain on the area of a circle, is equal
to twice the strain on its diameter.
(iii) Strain on volume of a sphere, is equal to
three times the strain on its diameter.
  
a x y
 

28. Strains Contd.
()
,
ivGiven and asstrainsonthediameter
andlengthofacylinder
Strainonthevolumeis
D L
v D L
 
   2
These can be proved using the theorem
of small errors

29. Examples
(i) Diameter, D = 2 x radius, r i.e. D = 2 r
Taking logs: log D = log 2 + log r
Taking differentials: dD/D = dr/r
Also: Circumference, C = 2r
 i.e. log C = Log 2 + log r
 dC/C = dr/r = dD/D
i.e. the strain on the circumference,
= strain on the diameter,


c

D

30. Strains Contd.

Required: Prove the othertwo statements.
(iv) Volume of a cylinder, V =  r
2
L where L is the length
Taking logs: log V = log + 2 log r + log L
Taking differentials: dV/V = 2 dr/r + dL/L
i.e. 
v D L2

31. 1.5
Elasticity and Hooke’s Law
All solid materials deform when they are
stressed, and as stress is increased,
deformation also increases.
If a material returns to its original size and
shape on removal of load causing deformation,
it is said to be elastic.
 If the stress is steadily increased, a point is
reached when, after the removal of load, not all
the induced strain is removed.
This is called the elastic limit.

32. Hooke’s
Law
States that providing the limit of proportionality
of a material is not exceeded, the stress is
directly proportional to the strain produced.
If a graph of stress and strain is plotted as load
is gradually applied, the first portion of the
graph will be a straight line.
 The slope of this line is the constant of
proportionality called modulus of Elasticity, E or
Young’s Modulus.
It is a measure of the stiffness of a material.

33. Hooke’s Law
Modulus of Elasticity, E =
Directstress
Directstrain





Also: For Shear stress: Modulus of rigidity or shear modulus, G =
Shearstress
Shearstrain




Also: Volumetric strain,

is proportional to hydrostatic
stress, within the elastic range
i.e. : called bulk modulus.

/
v
K

34. Stress-Strain Relations of Mild
Steel

35. Equation For Extension
From the above equations:

E
FA
dlL
FL
Adl
dl
FL
AE




/
/

This equation for extension is
very important

36. Extension For Bar of Varying Cross
Section

F o r a b a r o f v a r y in g c r o s s s e c t i o n :

P
A 1 A 2 A 3 P

L 1 L 2 L 3


d l
F
E
L
A
L
A
L
A
  
L
N
M
O
Q
P
1
1
2
2
3
3

37. Factor of Safety
The load which any member of a machine
carries is called working load, and stress
produced by this load is the working stress.
 Obviously, the working stress must be less
than the yield stress, tensile strength or the
ultimate stress.
This working stress is also called the
permissible stress or the allowable stress or the
design stress.

38. Factor of Safety Contd.
Some reasons for factor of safety
include the inexactness or inaccuracies
in the estimation of stresses and the
non-uniformity of some materials.
Factor of safety =
Ultimateoryieldstress
Designorworkingstress

Note: Ultimate stress is used for materials e.g.
concrete which do not have a well-defined yield point,
or brittle materials which behave in a linear manner up
to failure. Yield stress is used for other materials e.g.
steel with well defined yield stress.

39. 1.7 Practical Class Details
Each Student will have two practical
classes: one on :
Stress/strain characteristics and
 Hardness and impact tests.
(i)The stress/strain characteristics
practical will involve the measurement
of the characteristics for four metals,
copper, aluminium, steel and brass
using a tensometer.

40. Practical Class Details Contd.
The test will be done up to fracture of the
metals.
This test will also involve the accurate
measurement of the modulus of elasticity for
one metal.
There is the incorporation of an
extensometer for accurate measurement of
very small extensions to produce an
accurate stress-strain graphs.
 The test will be done up to elastic limit.

41. Practical Class Details Contd.
(ii)The hardness test will be done
using the same four metals and
the Rockwell Hardness test.
The impact test with the four
metals will be carried out using the
Izod test.

42. 1.8MATERIALS
TESTING
1.8.1.Tensile Test: This is the most common
test carried out on a material.
It is performed on a machine capable of
applying a true axial load to the test
specimen. The machine must have:
(i) A means of measuring the applied load
and
(ii)An extensometer is attached to the test
specimen to determine its extension.

43. Tensile Test Contd.
Notes:1.For iron or steel, the limit of
proportionality and the elastic limit are
virtually same but for other materials like
non-ferrous materials, they are different.
2. Up to maximum or ultimate stress, there is
no visible reduction in diameter of specimen
but after this stress, a local reduction in
diameter called necking occurs and this is
more well defined as the load falls off up to
fracture point.
Original area of specimen is used for
analysis.

44. Results From a Tensile Test
(a) Modulus of Elasticity, E
Stressuptoitofproportionality
Strain

lim

(b) Yield Stress or Proof Stress (See below)
(c) Percentage elongation =
Increaseingaugelength
Originalgaugelength
x100
(d) Percentage reduction in area =
Originalareaareaatfracture
Originalarea
x

100
(e) Te nsile Strength =
Maximumload
Originalcrosstionalareasec

The percentage of elongation and percentage reduction in area give an indication of th e
ductility of the material i.e. its ability to withstand strain without fracture occurring.

45. Proof Stress
High carbon steels, cast iron and most of the
non-ferrous alloys do not exhibit a well defined
yield as is the case with mild steel.
 For these materials, a limiting stress called
proof stress is specified, corresponding to a
non-proportional extension.
The non-proportional extension is a specified
percentage of the original length e.g. 0.05,
0.10, 0.20 or 0.50%.

46. Determination of Proof Stress
PProof Stress
Stress
The proof stress is obtained by drawing AP parallel to the initial
slope of the stress/strain graph, the distance, OA being the strain
corresponding to the required non-proportional extension e.g. for
0.05% proof stress, the strain is 0.0005.
A
Strain

47. 1.8.2Hardness
Test
The hardness of a material is determined by
its ability to withstand indentation. There are
four major hardness tests.
(i)Rockwell Hardness Test: This uses
an indentor with a 120
o
conical diamond with a
rounded apex for hard materials, or steel ball
for softer materials.
A minor load, F is applied to cause a small
indentation as indicated in Fig. (a) below.

The major load, F
m
is then applied and
removed after a specified time to leave load F
still acting. The two stages are shown as (b)
and (c).

48. Rockwell Hardness Test

49. Hardness Test Contd.
Thus the permanent increase in the
depth of penetration caused by the
major loadis d mmThe Rockwell
hardness number, H
R
is:

H
R
= K - 500 d
Where: K is a constant with value of 100
for the diamond indentor and 130 for the
steel indentor.

50. 1.8.3Impact
Testing
The toughness of a material is defined
as its ability to withstand a shock
loading without fracture. Two principal
impact tests are the:
 Izod and the
Charpy tests.
A test specimen is rigidly supported
and is impacted by a striker attached to
a pendulum.

52. Impact Test Concluded
The difference in height from which a
pendulum is released and the height to
which it rises after impact gives a
measure of the energy absorbed by the
specimen and this is recorded on a dial
mounted on a tester.

54. Steel Diagram of a Trapezoidal l Plate
dx
P
P
x
L
B1
B2
t
A flat plate of steel, 1 cm thick, and of trapezoidal form tapers
from 5 cm width to 10 cm width in a length of 40 cm. Determine
the elongation under an axial force of 50 kN. E = 2 x 10
7
N/cm
2
.

55. Solution
Consider a length, dx at a distance, x from width, B1,
Width at that section






B
BB
L
xBKx
whereK
BB
L
1
2 1
1
2 1

Area (Ax) of chosen c/section = ( B1 + K x ) t. If the length ‘dx’
elongates an amount du under load, its strain is:

du
dx
P
AE
.
1

56. Solution Contd.
Total extension of bar, u
u
P
AE
dx
P
BKxtE
dx
u
P
tE
dx
Bkx
P
KtE
BKx
L
u
P
KtE
BKL
B
x
L L
L
 



 


zz
z
0
1
0
1
0
1
1
1
0
()
ln
ln

57. Solution Contd.
S u b s t i t u t i n g b a c k f o r K ,
u
P
B B
L
tE
B B B
B
u
P
B B
L
tE
B
B


 


( )
l n
( )
l n
2 1
1 2 1
1
2 1
2
1

I n p r o b l e m , t = 1 c m , B 1 = 5 c m , B 2 = 1 0 c m , L = 4 0 c m , P = 5 0 , 0 0 0 N , E = 2 x 1 0 7 N / c m
2

u
N
x c m x x
c m


5 00 0 0
1 0 5
4 0
1 2 1 0
1 0
5
00 1 3 8 6
7
,
( )
l n .

58. Solution Concluded
S u b s t i t u t i n g b a c k f o r K ,
u
P
B B
L
tE
B B B
B
u
P
B B
L
tE
B
B


 


( )
l n
( )
l n
2 1
1 2 1
1
2 1
2
1

I n p r o b l e m , t = 1 c m , B 1 = 5 c m , B 2 = 1 0 c m , L = 4 0 c m ,
P = 5 0 , 0 0 0 N , E = 2 x 1 0 7 N / c m
2

u
N
x c m x x
c m


5 0 0 0 0
1 0 5
4 0
1 2 1 0
1 0
5
00 1 3 8 6
7
,
( )
l n .

59. 1.9
Lateral Strain and Poisson’s Ratio
 
Under the action of a longitudinal stress,
a body will extend in the direction of the
stress and contract in the transverse or
lateral direction
(see Fig. below).
The reverse occurs under a compressive
load.

60. Stress Effects
P
P
Longitudinal Tensile Stress Effect
Longitudinal Compressive Stress Effect
PP

61. Poisson’s Ratio
Lateral strain is proportional to the longitudinal strain,
with the constant of proportionality called ‘Poisson’s ratio’ with symbol, .

Mathematically, 
Lateralstrain
Directorlongitudinalstrain



For most metals, the range of  is 0.28 to 0.33.

62. 1.10
Thermal Strain
Most structural materials expand when heated,
in accordance to the law: T
where  is linear strain and
 is the coefficient of linear expansion;
T is the rise in temperature.
That is for a rod of Length, L;
if its temperature increased by t, the extension,
dl =  L T.

63. Thermal Strain Contd.
As in the case of lateral strains, thermal strains
do not induce stresses unless they are constrained.
The total strain in a body experiencing thermal stress
may be divided into two components:
Strain due to stress, 

and
That due to temperature, 
T
.
Thus:  = 

+ 
T

 =


E
T

64. 1.11.

Principle
of Superposition
It states that the effects of several actions
taking place simultaneously can be
reproduced exactly by adding the effect of
each action separately.
The principle is general and has wide
applications and holds true if:
(i) The structure is elastic
(ii) The stress-strain relationship is linear
(iii) The deformations are small.

65. 1.12 General Stress-Strain
Relationships

66. 1.12 General Stress-Strain
Relationships
F o r t h e e l e m e n t o f m a t e r i a l a s i n F i g u r e a b o v e
s u b j e c t e d t o u n i a x i a l s t r e s s , 
x
, t h e e n s u i n g s t r a i n
i s a s s h o w n i n ( b ) .
S t r a i n i n x d i r e c t i o n , 

x
x
E

S t r a i n s i n y a n d z d i r e c t i o n s a s a
r e s u l t o f s t r a i n i n x –d i r e c t i o n
=     

x x
x
a n d
E
e a c h
N o t e : T h e n e g a t i v e s i g n i n d i c a t e s c o n t r a c t i o n .

67. General Stress-Strain Relationships
Contd.
F o r a n e l e m e n t s u b j e c t e d t o t r i a x i a l s t r e s s e s ,
  
x y z
a n d, , t h e t o t a l s t r a i n i n x d i r e c t i o n w i l l b e
d u e t o 
x
a n d l a t e r a l s t r a i n s d u e to  
y za n d .
U s i n g t h e p r i n c i p l e o f s u p e r p o s i t i o n , t h e r e s u l t a n t s t r a i n i n x -d i r e c t i o n i s :


    
   
x
x y z
x x y z
E E E
ie
E
  
  .. { ( ) }
1

   
y y x z
E
  
1
{ ( ) } G e n e r a l i s e d H o o k e ’ s L a w i n t h r e e d i m e n s i o n s
   
z z x y
E
  
1
{ ( ) }

68. General Stress-Strain Relationships
Contd.

Note: In the case of shear strain,
there is no lateral strain, hence
the shear stress/shear strain
relationship is the same for both
unaxial and complex strain
systems.

69. Plain Stress and Plain Strain
A plain stress condition is said to exist when
stress in the z direction is zero.
The above equations may be applied for
but strain in the z direction is not zero.
Also plain strain condition exists when the
strain in z direction is zero.
Using strain in Z direction as zero in this
case does not mean that stress in the z
direction is zero.

70. Strain Caused by Stress and
Temperature
I n a d d i t i o n t o s t r a i n c a u s e d b y s t r e s s , t h e r e m a y a l s o b e t h e r m a l s t r a i n
d u e t o c h a n g e i n t e m p e r a t u r e . T h e g e n e r a l f o r m o f t h e s t r e s s / s t r a i n
r e l a t i o n s i s :

     
x x y z
E
t   
1
{ ( ) }
     
y y x z
E
t   
1
{ ( ) }
     
z z x y
E
t   
1
{ ( ) }






x y
x y
y z
y z
z x
z x
G G G
  ; ;

71. Try On Your Own

Show that : 


v x y z
E



12
( )

72. Example;
E x a m p l e : A p l a t e o f u n if o r m t h i c k n e s s 1 c m a n d d i m e n s i o n 3 x 2 c m is a c t e d u p o n b y
t h e lo a d s s h o w n . T a k in g E = 2 x 1 0 7 N / c m 2 , d e t e r min e  
x y
a n d. P o i s s o n ’s r a t io i s
0 . 3 . 4 2 k N

y
1 8 k N 2 c m 1 8 k N
x

4 2 k N
3 c m

73. Solution:

x
N
c m x c m
N c m 
1 8 0 0 0
2 1
9 0 0 0
2
/


y
N
c m x c m
N c m 
4 2 0 0 0
3 1
1 4 0 0 0
2
/

H o o k e ’ s l a w i n t w o d i m e n s i o n s s t a t e s t h a t :
   
x x y
E x
x    
1 1
2 1 0
9 0 0 0 0 3 1 4 0 0 0 2 4 0 1 0
7
6
[ ] [ . ( ]
a n d    
y y x
E x
x    
1 1
2 1 0
1 4 0 0 0 0 3 9 0 0 0 5 6 5 1 0
7
6
[ ] [ . ( ]

74. 1.13
Relationship between Elastic
Modulus
(E) and Bulk Modulus, K
It has been shown that : 
vxyz

 

 
















x x y z
x y z
x
y z
v x y z
v
v
v
E
Forhydrostaticstress
ie
E E
Similarlyandareeach
E
Volumetricstrain
E
E
BulkModulusK
Volumetricorhydrostaticstress
Volumetricstrain
ieEK andK
E
  

   

 
 
 
 
  

1
1
2 12
12
3
12
3
12
312
312
( )
,
..
,
,
..

75. Maximum Value For Poisson’s
Ratio
F r o m t h e e q u a t i o n , i f v = 0 . 5 , t h e v a l u e o f K b e c o m e s i n f i n i t e l y l a r g e .
H e n c e t h e b o d y i s i n c o m p r e s s i b l e . I f v > 0 . 5 , K b e c o m e s n e g a t i v e
i . e . t h e b o d y w i l l e x p a n d u n d e r h y d r o s t a t i c p r e s s u r e w h i c h i s
i n c o n c e i v a b l e . I t m a y b e c o n c l u d e d t h a t t h e u p p e r l i m i t o f P o i s s o n ’ s r a t i o
i s 0 . 5 .
N o t e : K
G
a n d E G


 
2 1
31 2
2 1



W h e r e : G i s S h e a r M o d u l u s

76. 1.14 Compound Bars
A compound bar is one comprising two or more parallel elements, of different materials,
which are fixed together at their end. The compound bar may be loaded in tension or
compression.
1 2

F F

2

Section through a typical compound bar consisting of a circular bar (1) surrounded by a
tube (2)

77. 1.14.1
Stresses Due to Applied Loads
in
Compound Bars
I f a c o m p o u n d b a r i s l o a d e d i n c o m p r e s s i o n b y a f o r c e , F ,
Si n c e t h e r o d a n d t u b e a r e o f t h e s a m e l e n g t h a n d m u s t r e m a i n
t o g e t h e r , t h e t wo m a t e r i a l s m u s t h a v e t h e s a m e s t r a i n i . e .
 
 


1 2
1
1
2
2
2
1 2
1
1

  S t r a i n
S t r e s s
E
ie
E E
E
E
. , .....()

W h e r e E 1 a n d E 2 a r e t h e e l a s t i c m o d u l i o f m a t e r i a l s 1 a n d 2 r e s p e c t i v e l y .
A l s o : T h e t o t a l l o a d , F m u s t b e s h a r e d b y t h e t w o m a t e r i a l s , i . e . F = F 1 + F 2
W h e r e: F 1 a n d F 2 a r e t h e l o a d s i n t h e i n d i v i d u a l e l e m e n t s .

78. Compound Bars Contd.
N o w: a s fo rc e = s tre s s x a re a : T h e n : F =  
11 22
A A ...............(2 )
W h e re A1 a n d A2 a re th e a re a s o f m a te ria ls 1 a n d 2 re s p e c tiv e ly .

S u b s titu tin g fo r 
2
fro m E q n . 1 in to E q n 2 :


F A
EA
E
A
EA
E
FE
EA EA
a n d
FE
EA EA
   
L
N
M
O
Q
P







 
1 1
1 2 2
1
1 1
2 2
1
1
1
1 1 2 2
2
2
1 1 2 2

79. 1.14.2
Temperature stresses in
compound
bars

1 
1

2 
2

L
(a) L
1
T
1
L
2
T
2 {b}
FL
AE
11

F 1 F

F 2 F
(c)
FL
AE
22

80. Temperature
stresses in compound
bar
Contd.

Consider a compound bar, see (a) above of length, L consisting of 2
different materials (1) and (2) having coefficients of expansion

1
and 
2
respectively with 
1
>
2
. If the bar is subjected to a

1
uniform temperature rise, T and the right hand fixing released,

1
the bar (1) will expand more than (2) as shown in diagram (b).
However, because of the end fixing, free expansion cannot occur.
Diagram (c) shows that the end fixing must supply a force which
decreases the length of bar (1) and increases the length of bar (2)
until equilibrium is achieved at a common length.
As no external forces are involved, a self equilibrating
(balancing force system is created).

81. Temperature Stresses Contd.
Free expansions in bars (1) and (2) are LTandLT 
1 2
respectively.
Due to end fixing force, F: the decrease in length of bar (1) is
FL
AE
11
and the increase in length of (2) is
FL
AE
22
.
At Equilibrium:

LT
FL
AE
LT
FL
AE
ieF
AEAE
T
ieA
AEAE
EEAA
T
T AEE
AEAE
T AEE
AEAE
 

 




1
11
2
22
11 22
1 2
11
22 11
1212
1 2
1
1 2212
11 22
2
1 2112
11 22
11
 
 
L
N
M
O
Q
P
 






..[ ]( )
.. ( )
( )
( )

Note: As a result of Force, F, bar (1) will be in compression while (2) will be in tension.


1 
1

2 
2

L
(a) L
1
T
1
L
2
T
2 {b}
FL
AE
11

F 1 F

F 2 F
(c)
FL
AE
22

82. Example
A steel tube having an external diameter of 36
mm and an internal diameter of 30 mm has a
brass rod of 20 mm diameter inside it, the two
materials being joined rigidly at their ends when
the ambient temperature is 18
0
C. Determine
the stresses in the two materials: (a) when the
temperature is raised to 68
0
C (b) when a
compressive load of 20 kN is applied at the
increased temperature.

83. Example Contd.
For brass: Modulus of elasticity = 80 GN/m
2
;
Coefficient of expansion = 17 x 10 -6 /
0
C
For steel: Modulus of elasticity = 210 GN/m
2
;
Coefficient of expansion = 11 x 10 -6 /
0
C



3 0 B r a s s r o d 2 0 3 6




S t e e l t u b e

A r e a o f b r a s s r o d ( A b) =
x
m m
2 0
4
3 1 41 6
2
2
 .
A r e a o f s t e e l t u b e ( A s) =
x
m m
( )
.
3 6 3 0
4
3 1 10 2
2 2
2

AE x m x x N m x N
s s
 

3 1 10 2 1 0 2 1 0 1 0 06 5 3 1 4 2 1 0
6 2 9 2 8
. / .
1
15 3 1 0 6 1 0
8
AE
x
s s


.
Solution:

84. AE x mx x Nm x N
b b 

3 1 41 61 0 8 01 0 02 5 1 3 2 71 0
6 2 9 2 8
. / .
1
39 7 8 8 7 3 61 0
8
AE
x
b b


.
T x x
b s
( ) ( )    
 
5 01 71 1 1 0 31 0
6 4

W i t h i n c r e a s e i n t e m p e r a t u r e , b r a s s w i l l b e i n c o m p r e s s i o n w h i l e
s t e e l w i l l b e i n t e n s i o n . T h i s i s b e c a u s e e x p a n d s m or e t h a n s t e e l .
ie F
AE AE
T
s s b b
b s
.. [ ] ( )
1 1
   

i . e . F [ 1 . 5 3 1 0 6 + 3 . 9 7 8 8 7 3 6 ] x 1 0
-8
= 3 x 1 0
-4
F = 5 4 4 4 . 7 1 N
Solution Contd

86. Solution Concluded
S t r e s s i n s t e e l t u b e =
5 4 4 4 7 1
3 1 1 0 2
1 7 5 1 1 7 5 1
2
2 2.
.
. / . / ( )
N
m m
N m m M N m T e n s i o n 
S t r e s s i n b r a s s r o d =
5 4 4 4 7 1
3 1 4 1 6
1 7 3 3 1 7 3 3
2
2 2.
.
. / . / ( )
N
m m
N m m M N m C o m p r e s s i o n 
( b ) S t r e s s e s d u e t o c o m p r e s s i o n f o r c e , F ’ o f 2 0 k N

s
s
s s b b
F E
E A E A
x N x x N m
x
M N m C o m p r e s s i o n




' /
. .
. / ( )
2 0 1 0 2 1 0 1 0
0 6 5 3 1 4 2 0 2 5 1 3 2 7 1 0
4 6 4 4
3 9 2
8
2


b
b
s s b b
F E
E A E A
x N x x N m
x
M N m C o m p r e s s i o n




' /
. .
. / ( )
2 0 1 0 8 0 1 0
0 6 5 3 1 4 2 0 2 5 1 3 2 7 1 0
1 7 6 9
3 9 2
8
2

R e s u l t a n t s t r e s s i n s t e e l t u b e = - 4 6 . 4 4 + 1 7 . 5 1 = 2 8 . 9 3 M N / m
2
( C o m p r e s s i o n )
R e s u l t a n t s t r e s s i n b r a s s r o d = -1 7 . 6 9 - 1 7 . 3 3 = 3 5 . 0 2 M N / m
2
( C o m p r e s s i o n )

87. Example:
A composite bar, 0.6 m long comprises a steel
bar 0.2 m long and 40 mm diameter which is
fixed at one end to a copper bar having a
length of 0.4 m.
Determine the necessary diameter of the
copper bar in order that the extension of each
material shall be the same when the composite
bar is subjected to an axial load.
 What will be the stresses in the steel and
copper when the bar is subjected to an axial
tensile loading of 30 kN? (For steel, E = 210
GN/m
2
; for copper, E = 110 GN/m
2
)

88. Solution:
0.2 mm
0.4 mm
F 40 mm dia d F

Let the diameter of the copper bar be d mm
Specified condition: Extensions in the two bars are equal

dldl
dlL
E
L
FL
AE
c s
 

Thus:
FL
AE
FL
AE
cc
cc
ss
ss


89. Solution Concluded
A l s o : T o t a l f o r c e , F i s t r a n s m i t t e d b y b o t h c o p p e r a n d s t e e l
i . e . F c = F s = F
ie
L
A E
L
A E
c
c c
s
s s
. . 
S u b s t i t u t e v a l u e s g i v e n i n p r o b l e m :
0 4
4 1 1 0 1 0
0 2
4 0 0 4 0 2 1 0 1 0
2 2 9 2 2 9 2
.
/ /
.
/ . /
m
d m x N m
m
x x x N m 

d
x x
m d m m m
2
2
22 2 1 0 0 0 4 0
1 1 0
0 0 7 8 1 6 7 8 1 6  
.
; . . .
T h u s f o r a l o a d i n g o f 3 0 k N
S t r e s s i n s t e e l , 

s
x N
x x
M N m 

3 0 1 0
4 0 0 4 0 1 0
2 3 8 7
3
2 6
2
/ .
. /
S t r e s s i n c o p p e r , 

c
x N
x x
M N m 

3 0 1 0
4 0 0 7 8 1 6 1 0
9
3
2 6
2
/ .
/

90. 1.15
Elastic Strain Energy
If a material is strained by a gradually
applied load, then work is done on the
material by the applied load.
The work is stored in the material in the form
of strain energy.
 If the strain is within the elastic range of the
material, this energy is not retained by the
material upon the removal of load.

91. Elastic Strain Energy Contd.
Figure below shows the load-extension graph of a uniform bar.
The extension dl is associated with a gradually applied load, P
which is within the elastic range. The shaded area represents
the work done in increasing the load from zero to its value
Load
P


Extension
dl
Work done = strain energy of bar = shaded area

92. Elastic Strain Energy Concluded
W = U = 1/2 P dl (1)
Stress,  = P/A i.e P =  A
Strain = Stress/E
i.e dl/L = /E , dl = (L)/E L= original length
Substituting for P and dl in Eqn (1) gives:
W = U = 1/2  A . ( L)/E = 
2
/2E x A L
A L is the volume of the bar.

i.e U = 
2
/2E x Volume

The units of strain energy are same as those of work i.e. Joules. Strain energy
per unit volume, 
2
/2E is known as resilience. The greatest amount of energy that can
stored in a material without permanent set occurring will be when  is equal to the
elastic limit stress.

93. Torsion
Consider a bar to be rigidly
attached at one end and
twisted at the other end by a
torque or twisting moment
T equivalent to F × d, which
is applied perpendicular to
the axis of the bar, as shown
in the figure. Such a bar is
said to be in torsion.
TORSIONAL
SHEARING
STRESS,

τ

For a solid or hollow circular shaft
subject to a twisting moment T, the
torsional shearing stress τ at a distance
ρ from the center of the shaft is :

94. where J is the polar moment of inertia of the
section and r is the outer radius.
For solid cylindrical shaft:
For hollow cylindrical shaft:
ANGLE
OF TWIST
The angle θ through which the bar
length L will twist is
where

95. Solved Problems in Torsion
1-A steel shaft 3 ft long that has a diameter of 4 in. is subjected
to a torque of 15 kip·ft. Determine the maximum shearing stress
and the angle of twist. Use G = 12 × 106 psi.
Solution:

96. 2-What is the minimum diameter of a solid steel shaft that will not
twist through more than 3° in a 6-m length when subjected to a
torque of 12 kN·m? What maximum shearing stress is developed?
Use G = 83 GPa.
Solution:

97. Torsion of Thin-Walled Tubes
The torque applied to thin-walled tubes is expressed as :
where T is the torque in N·mm, A is the area enclosed by the
centerline of the tube (as shown in the stripefilled portion) in mm2,
and q is the shear flow in N/mm.
The average shearing stress across any
thickness t is :

98. Shear & Moment in Beams
DEFINITION OF A BEAM
A beam is a bar subject to forces or couples that lie in a plane containing
the longitudinal of the bar. According to determinacy, a beam may be
determinate or indeterminate.
STATICALLY
DETERMINATE BEAMS
Statically determinate beams are those beams in which the reactions of the
supports may be determined by the use of the equations of static equilibrium.
The beams shown below are examples of statically determinate beams.

99. STATICALLY
INDETERMINATE BEAMS
If the number of reactions exerted upon a beam exceeds
the number of equations in static equilibrium, the beam is
said to be statically indeterminate
The degree of indeterminacy is taken as the difference between the
umber of reactions to the number of equations in static equilibrium
that can be applied. In the case of the propped beam shown, there are
three reactions R1, R2, and M and only two equations (ΣM = 0 and
sum;Fv = 0) can be applied, thus the beam is indeterminate to the first
degree (3 – 2 = 1).

100. LOADS
APPLIED TO THE BEAM MAY CONSIST OF A
CONCENTRATED
LOAD (LOAD APPLIED AT A POINT), UNIFORM
LOAD,
UNIFORMLY VARYING LOAD, OR AN APPLIED COUPLE
OR
MOMENT. THESE LOADS ARE SHOWN IN THE FOLLOWING
FIGURES
TYPES
OF LOADING