Stress and Strain

1. STRESS AND STRAIN

2. 2

3. Topics:
• Introduction
• Main Principles of Statics
Stress
• Normal Stress
• Shear Stress
• Bearing Stress
• Thermal Stre

4. 41.1 Introduction
Mechanics : The study of how bodies react to forces acting on them
RIGID BODIES
(Things that do not change shape)
Statics : The study of bodies
in an equilibrium
DEFORMABLE BODIES
(Things that do change shape)
FLUIDS
Mechanics of Materials :
The study of the relationships
between the external loads
applied to a deformable body and
the intensity of internal forces
acting within the body.
Incompressible Compressible
Dynamics :
1. Kinematics – concerned
with the geometric aspects
of the motion
2. Kinetics – concerned
with the forces causing the
motion.

5. 51.2 Main Principles of Statics
External Loads
Surface Forces
- caused by direct contact of one body with
the surface of another.
Body Force
- developed when one body exerts a force on
another body without direct physical contact
between the bodies.
- e.g earth’s gravitation (weight)
concentrated force
linear distributed load, w(s)

6.  Axial Load
 Normal Stress
 Shear Stress
 Bearing Stress
 Allowable Stress
 Deformation of Structural under Axial Load
 Statically indeterminate problem
 Thermal Stress

7.  Mechanics of material is a study of the
relationship between the external loads applied
to a deformable body and the intensity of
internal forces acting within the body.
 Stress = the intensity of the internal force on a
specific plane (area) passing through a point.
 Strain = describe the deformation by changes in
length of line segments and the changes in the
angles between them

8. 81.1 Introduction
• Normal Stress : stress which acts perpendicular, or normal to, the
(σ) cross section of the load-carrying member.
: can be either compressive or tensile.
• Shear Stress : stress which acts tangent to the cross section of
(τ) the load-carrying member.
: refers to a cutting-like action.

9.  Normal Stress, 
the intensity of force, or force per unit area, acting
normal to A
A positive sign will be used to indicate a tensile stress
(member in tension)
A negative sign will be used to indicate a compressive
stress (member in compression)
 = P / A

10. (a)
(b)
Stress (  ) = Force (P)
Cross Section (A)
•Unit: Nm -²
•N/mm2 or MPa
N/m2 or Pa

11. 111.4 Axial Loading – Normal Stress
Assumptions :
1. Uniform deformation: Bar
remains straight before and
after load is applied, and
cross section remains flat or
plane during deformation
2. In order for uniform
deformation, force P be
applied along centroidal axis
of cross section C

12. A
P
AP
AFFF
A
zRz


  


 dd;
12
σ = average normal stress at any point
on cross sectional area
P = internal resultant normal force
A = cross-sectional area of the bar
1.4 Axial Loading – Normal Stress

13. 13
• Use equation of σ = P/A for cross-sectional area of a member when
section subjected to internal resultant force P
Internal Loading
• Section member perpendicular to its longitudinal axis at pt
where normal stress is to be determined
• Draw free-body diagram
• Use equation of force equilibrium to obtain internal axial
force P at the section
• Determine member’s x-sectional area at the section
• Compute average normal stress σ = P/A
Average Normal Stress
1.4 Axial Loading – Normal Stress

14. Example 1.1:
Two solid cylindrical rods AB and BC are welded
together at B and loaded as shown. Knowing
that d1=30mm and d2=20mm, find average
normal stress at the midsection of (a) rod AB,
(b) rod BC.

16. Example 1.2
Two solid cylindrical roads AB and BC are welded
together at B and loaded as shown. Knowing
that d1 = 30 mm and d2 = 50 mm, find the
average normal stress in the mid section of (a)
rod AB, (b) rod BC.

17.  Normal strain,  is the elongation or
contraction of a line segment per unit of
length
L = elongation
Lo = length
 = L / Lo
strainnormal
L



* L= 

18. Example 1.3:
Determine the corresponding strain for a bar of
length L=0.600m and uniform cross section
which undergoes a deformation =15010-6m.
6
6
6
150 10 m
250 10 m m
L 0 600m
250 10 250
/
.
@



 
    
  

19. 1.4 A cable and strut assembly ABC supports a vertical load
P=12kN. The cable has an effective cross sectional area of
160mm², and the strut has an area of 340mm².
(a) Calculate the normal stresses in the cable and strut.
(b) If the cable elongates 1.1mm, what is the strain?
(c) If the strut shortens 0.37mm, what is the strain?

20. 1.5 The bar shown has a square cross section
(20mm x 40mm) and length, L=2.8m. If an
axial force of 70kN is applied along the
centroidal axis of the bar cross sectional area,
determine the stress and strain if the bar end
up with 4m length.
70kN 70kN
2.8m

21.  Tensile test is an experiment to determine
the load-deformation behavior of the
material.
 Data from tensile test can be plot into stress
and strain diagram.
 Example of test specimen
- note the dog-bone geometry
28

22.  Universal Testing Machine - equipment used
to subject a specimen to tension,
compression, bending, etc. loads and
measure its response
29

23. Stress-Strain Diagrams
A number of important mechanical
properties of materials that can be deduced
from the stress-strain diagram are illustrated
in figure above.
30

24.  Point O-A = linear relationship between stress
and strain
 Point A = proportional limit (PL)
The ratio of stress to strain in this linear region
of stress-strain diagram is called Young Modulus
or the Modulus of Elasticity given
At point A-B, specimen begins yielding.
 Point B = yield point
 Point B-C = specimen continues to elongate without any increase in
stress. Its refer as perfectly plastic zone
 Point C = stress begins to increase
 Point C-D = refer as the zone of strain hardening
 Point D = ultimate stress/strength ; specimen
begins to neck-down
 Point E = fracture stress





 < PL
Unit: MPa
31

25. Point O to A
Point C to D
Point D to E
At point E
Normal or engineering stress can be determined
by dividing the applied load by the specimen
original cross sectional area.
True stress is calculated using the actual cross
sectional area at the instant the load is
measured.
31

26. Some of the materials like aluminum (ductile),
does not have clear yield point likes
structural steel. Therefore, stress value
called the offset yield stress, YL is used
in line of a yield point stress.
As illustrated, the offset yield stress is
determine by;
 Drawing a straight line that best fits the data in initial (linear)
portion of the stress-strain diagram
 Second line is then drawn parallel to the original line but offset
by specified amount of strain
 The intersection of this second line with
the stress-strain curve determine the
offset yield stress.
 Commonly used offset value is 0.002/0.2%
32

27. Brittle material such as ceramic and glass
have low tensile stress value but high in
compressive stress. Stress-strain diagram for
brittle material.
33

28.  Elasticity refers to the property of a material such that
it returns to its original dimensions after unloading .
 Any material which deforms when subjected to load
and returns to its original dimensions when unloaded
is said to be elastic.
 If the stress is proportional to the strain, the material
is said to be linear elastic, otherwise it is non-linear
elastic.
 Beyond the elastic limit, some residual strain or
permanent strains will remain in the material upon
unloading .
 The residual elongation corresponding to the
permanent strain is called the permanent set .
34

29. • The amount of strain which is recovered upon unloading is
called the elastic recovery.
35

30.  When an elastic, homogenous and isotropic material
is subjected to uniform tension, it stretches axially
but contracts laterally along its entire length.
 Similarly, if the material is subjected to axial
compression, it shortens axially but bulges out
laterally (sideways).
 The ratio of lateral strain to axial strain is a constant
known as the Poisson's ratio,
where the strains are caused by uniaxial stress only
 -ve sign is used since longitudinal elongation
(positive strain) causes lateral contraction (negative
strain) and vice versa.
axial
lateral
v



36
radial
longitudinal

31. Example 1.6
A 10 cm diameter steel rod is loaded with 862 kN by
tensile forces. Knowing that the E=207 GPa and =
0.29, determine the deformation of rod diameter
after being loaded.
Solution
 in rod,  =
Lateral strain,

38
MPa
m
Nx
A
p
7.109
)1.0(
4
1
10862
22
3


00053.0
10207
7.109
3

MPax
MPa
E
a


)00053.0(29.)( oal  
000154.0
)1.0)(000154.0()(  Dd l
cm00154.0

32. Exercises 1
1. A steel pipe of length L=1.2 m, outside diameter d2=150mm and
inside diameter d1=110mm is compressed by an axial force P=
620kN.The material has modulus of elasticity E= 200GPa and
Poisson’s Ratio v = 0.30.Determine :
a) the shortening, δ ( ans :-0.455 mm)
b) the lateral strain,ε lateral (ans: 113.9x10-6)
39

33. 2. A hollow circular post ABC as shown in Figure 2 supports a load
P1=7.5 kN acting at the top. A second load P2 is uniformly
distributed around the cap plate at B. The diameters and
thicknesses of the upper and lower parts of the post are dAB=32
mm, tAB= 12mm, dBC 57 mm and tBC=9mm, respectively.
a) Calculate the normal stress, σAB in the upper part
of the post. (ans: 9.95 MPa)
b) If it is desired that the lower part of the post
have the same compressive stress as the upper
part, what should be the magnitude of the load P2?
(ans : P2=6kN)
40

34. 3. A standard tension test is used to determine the
properties of an experimental plastic. The test
specimen is a 15 mm diameter rod and it is
subjected to a 3.5 kN tensile force. Knowing that an
elongation of 11 mm and a decrease in diameter of
0.62 mm are observed in a 120 mm gage length.
Determine the modulus of elasticy, the modulus of
rigidity, and Poisson’s ratio of the material.
49

35.  A force acting parallel or tangential to a section taken
through a material (i.e. in the plane of the material) is called a
shear force
 The shear force intensity, i.e. shear force divided by the area
over which it acts, is called the average shear stress, 
 = shear stress
V = shear force
A = cross-sectional area
 Shear stress arises as a result of the direct action of forces
trying to cut through a material, it is known as direct shear
force
Shear stresses can also arise indirectly as a result of tension,
torsion or bending of a member.
A
V

41

36.  Depending on the type of connection, a connecting
element (bolt, rivet, pin) may be subjected to single
shear or double shear as shown.
Rivet in Single Shear
4
2
d
P
A
V

 
42

37. Rivet in Double Shear
Example 1.9
For the 12 mm diameter bolt shown in the bolted joint below,
determine the average shearing stress in the bolt.
22
2
)
4
(2
d
P
d
P
A
V


 
43

38. A
F
A
P
ave
Single Shear
A
F
A
P
2
ave 
Double Shear

39.  The effect of shear stress is to distort the shape of a
body by inducing shear strains
 The shear strain,  is a measure of the angular
distortion of the body.
(units: degrees, radians)
L
Vx

L
x

44

40.  Bearing stress is also known as a contact stress
Bearing stress in shaft key;
Bearing stress in rivet and plat;
rhL
M
Lh
rM
A
P
b
b
2
)2(

td
P
b 
45

41. Example 2.0
A punch for making holes in steel plates is shown in
the figure. Assume that a punch having diameter
d=20 mm is used to punch a hole in an 8 mm
plates, what is the average shear stress in the plate
and the average compressive stress in the punch if
the required force to create the hole is P = 110kN.
. P
20 mm
8 mm
46

42.  It also known as Shear Modulus of Elasticity or the
Modulus of Rigidity.
 Value of shear modulus can be obtained from the linear
region of shear stress-strain diagram.
 The modulus young (E), poisson’s ratio() and the
modulus of rigidity (G) can be related as
 G Unit : Pa
)1(2 

E
G
48

43. Because of the change in the dimensions of a body as
a result of tension or compression, the volume of the
body also changes within the elastic limit.
Consider a rectangular parallel piped having sides a,
b and c in the x, y and z directions, respectively.
58

44.  The tensile force P causes an axial elongation of a
and lateral contractions of b and c in the x, y, and
z directions respectively. Hence,
Initial volume of body, Vo = abc
Final volume, Vf = (a + a)(b - b)(c - c)
= abc(1 + )(1 - )2
Initial
body
59

45. Expanding and neglecting higher orders of  (since  is
very small),
Final volume, Vf = abc(1 +  - 2)
Change in volume,
V = Final Volume - Initial Volume
= abc(1 +  - 2 ) - abc
= abc(1 +  - 2  - 1)
= abc( - 2 )
= Vo (1 - 2 )
Hence,
60
)21(
)21(





E
V
V
o

46.  Isotropic material is subjected to general triaxial
stress x, y and z.
 Since all strain satisfy  << 1, so v = x + y + z
x =
y =
z =
 )(
1
zyx
E
 
 )(
1
zxy
E
 
 )(
1
yxz
E
 
)(
21
zyxv
E
 


61

47. Example 2.1
A titanium alloy bar has the following original dimensions: x =
10cm; y = 4cm; and z = 2cm. The bar is subjected to stresses x
= 14 N and y = - 6 N, as indicated in figure below. The
remaining stresses (z, xy, xz and yz) are all zero. Let E = 16
kN and  = 0.33 for the titanium alloy.
(a)Determine the changes in the length for
x, y and z.
(b) Determine the dilatation, v.
z
x
y
14 N14 N
6 N
6 N
62

48.  Applied load that is less than the load the member can fully support.
(maximum load)
 One method of specifying the allowable load for the design or
analysis of a member is use a number called the Factor of Safety (FS).
Allowable-Stress Design
allow
fail
F
F
FS 
FS > 1
FS
or
FS
yield
allow
yield
allow



 
63

49.  If a bar is fixed at both ends, as shown in
fig. (a), two unknown axial reactions
occurs, and the force equilibrium equation
becomes;
0PFF
;0F
AB
y


0B/A 
• In this case, the bar is called statically
indeterminate, since the equilibrium
equation are not sufficient to determine
the reactions.
• the relationship between the forces acting on
the bar and its changes in length are known as
force-displacement relations
• the relative displacement of one end of the bar
with respect to the other end is equal to zero
since the ends supports are fixed. Hence;

50. AE
PL
,0B/A  0BA 
AC
CBB
A
AC
CBB
A
CBBACA
CBBACA
L
LF
F
L
AE
AE
LF
F
AE
LF
AE
LF
0
AE
LF
AE
LF















1
L
L
FP
F
L
LF
P
L
LF
FP
AC
CB
B
B
AC
CBB
AC
CBB
B
• Realizing that the internal force in segment AC is +FA, and in segment CB,
the internal force is –FB. Therefore, the equation can be written as;























 










L
L
PF
L
L
FP
L
LL
FP
L
L
L
L
FP
AC
B
AC
B
AC
ACCB
B
AC
AC
AC
CB
B
BAAB FPF,0PFF 

51. Example 2.2:

52. 3
X A B
3
B A
F 0 F F 20 10 N 0 1
F 20 10 F
, ( ) ................( )
( )
       
 
   
B A
A B
A AC B CB
A
2 29 2 9 2
A B
0 001m
0 001m
F L F L
0 001m
AE AE
F 0 4m FB 0 8m
0 001m
0 0025m 200 10 Nm 0 0025m 200 10 Nm
or
F 0 4m F 0 8m 3927 0N 2
Substitute eq 1 oeq 2
F
/ .
.
.
( . ) ( . )
.
. .
( . ) ( . ) . ................( )
( )int ( )
 
 
   
 
 
                  
 
A A
A
0 4m 20 000N F 0 8m 3927 0N
F 16 6kN
FB 3 39kN
( . ) ( , )( . ) .
.
.
  



53. Example 2.3:

54. Solution: 3
y A C E
C
3
A E
F 0 F F F 15 10 N 0 1
CCW M 0
F 0 4 15 10 0 2 F 0 4 0 2
, ( ) ................( )
( . ) ( )( . ) ( . ) ...........( )
       
 
   
The applied load will cause the horizontal line
ACE move to inclined line A’C’E’
C EA E
C E A E
A E
C E
A E
C E
C A E
C CD A AB E EF
5 5 5
st st st
C A
5 5
st st
0 8 0 4
0 4 0 8
0 4
0 8
0 4 0 4
0 8
0 5 0 5
F L F L F L
0 5 0 5
1 5 10 E 2 5 10 E 2 5 10 E
F 0 5 F 0 5
0 5
1 5 10 E 2 5 10 E
. .
. .
.
.
. .
.
. .
. .
. . .
( . ) ( . )
.
. .
  
 
    

     

  
    
  
   
    
   
    
        


 
E
5
st
3 3 3
C A E
3 3
A E
C 3
C A E
F 0 5
0 5
2 5 10 E
33 33 10 F 10 10 F 10 10 F
10 10 F 10 10 F
F
33 33 10
F 0 3F 0 3F eq 3
( . )
.
.
.
.
. . ................. ( )

  
   
     
    
  


 

55. 3
y A C E
C
3
A E
C A E
3
A C E
A
F 0 F F F 15 10 N 0 1
CCW M 0
F 0 4 15 10 0 2 F 0 4 0 2
F 0 3F 0 3F eq 3
Substituteeq 3 oeq 1
F F F 15 10 N 0 1
F
, ( ) ................( )
( . ) ( )( . ) ( . ) ...........( )
. . ................. ( )
( ) int ( )
( ) ................( )
       
 
   
 
   
3
A E E
3
A E
3
A
E
3
E A
0 3F 0 3F F 15 10 0
1 3F 1 3F 15 10
15 10 1 3F
F
1 3
F 11 538 10 F eq 4
( . . ) ( )
. . ( )
( ) .
.
. ( ) ....................... ( )
   
 


 
3
A E
3 3
A A
3 3
A A
3
A
3
Substituteeq 4 oeq 2
F 0 4 15 10 0 2 F 0 4 0
F 0 4 3 10 0 4 11 538 10 F 0
F 0 4 3 10 4 615 10 0 4F 0
7 61510
F
0 8
9 519 10
9 52kN
( ) int ( )
( . ) ( )( . ) ( . )
( . ) ( ) ( . ) . ( )
( . ) ( ) . ( ) .
.
.
. ( )
.
   
     
 
    





A
3
E A
3 3
E
C A E
3 3
place F 9 52kN o eq 4
9 52kN
F 11 538 10 F
11 538 10 9 52 10
2 02kN
place F 2 02kN o eq 3
F 0 3F 0 3F
0 3 9 519 10 0 3 2 02 10
3 462 kN
Re . int ( )
.
. ( )
. ( ) . ( )
.
Re . int ( )
. .
. ( . ( ) . ( . )
.


 
 


 
  


56.  A change in temperature can cause material to change its
dimensions.
 If the temperature increases, generally a material expands,
whereas if the temperature decreases, the material will
contract.
 If this is the case, and the material is homogenous and
isotropic, it has been found from experiment that the
deformation of a member having a length L can be calculated
using the formula;
T=TL
Where
=linear coefficient of thermal expansion (unit:
1/C)
T=change in temperature
L=original length of the member
T=change in length of the member

57. Given: =12x10-6/C
Example 2.4:

58. FFF
0F
BA
Y


Solution:
AB 0 The change in length of the bar
is zero (because the supports do
not move)
AB T F( )     
To determine the change in
length, remove the upper support
of the bar and obtain a bar is
fixed at the base and free to
displace at the upper end.
So the bar will elongate by an
amount δT when only
temperature change is acting
And the bar shortens by an
amount δF when only the reaction
is acting

59. AB T F
T F
6
2 9
4
2 9
4 2 9
0
FL
TL 0
AE
F 1
12 10 60 30 1 0
0 01 200 10
F 1
3 6 10
0 01 200 10
F 3 6 10 0 01 200 10
7 2kN
( )
( )
( )( )
. ( )
( )
.
. ( )
. . ( )
.



     
  
  
     

 

   

2
F 7 2kN
72MPa
A 0 01
.
;
.
 
Average normal thermal stress:

60. Example 2.5
Given:
6
st
6
al
9
st
9
al
12 10 C
23 10 C
E 200 10 Pa
E 73 1 10 Pa
/
/
.



   
  
 
 

61. 3
y st alF 0 2F F 90 10 N 0 eq 1, ( ) ......... ( )      
st al
st st T st F
al al T al F
st T st F al T al F
st al
st al
6 st
2 9
6
eq 2
F L F L
TL TL
A E A E
F 0 25
12 10 80 20 0 25
0 02 200 10
23 10 80 20
............................... ( )
( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( . )
( )( . )
( . ) ( )
(


  
      
    
      
   
     
 
    al
2 9
4 4st al
6 6
4 10 4 9
st al
10 4 9 4
st al
st
F 0 25
0 25
0 03 73 1 10
F 0 25 F 0 25
1 8 10 3 45 10
251 327 10 206 685 10
1 8 10 9 947 10 F 3 45 10 1 21 10 F
9 947 10 F 3 45 10 1 21 10 F 1 8 10
1 65
F
( . )
)( . )
( . ) ( . )
( . ) ( . )
. .
. ) ( . )
. . . .
. . . .
.
 
   
   

 
    
 
      
       


4 9
al
10
3
al
10 1 21 10 F
9 947 10
165 88 10 1 216F eq 3
.
.
. . ............... ( )
 

 
 
   

62. 3
st al
3 3
al al
3 3
al al
3
al
al
al
3
st
Substituteeq 3 oeq 1
2F F 90 10 N 0
2 165 88 10 1 216F F 90 10 N 0
331 76 10 2 432F F 90 10 N 0
3 432F 421 76 10
F 122 89kN
Substitute F 122 89kN o eq 3
F 165 88 10 1 216F
( )int ( )
( )
( . . ) ( )
. . ( )
. .
.
. int ( )
. .
  
     
     
 


    al
3 3
165 88 10 1 216 122 89 10
16 445kN
. . ( . )
.
    
 
The negative value for F steel indicates that the
force acts opposite to arrow shown.
THE STEEL POSTS ARE IN TENSION and
ALUMINIUM POSTS IS IN COMPRESSION

63. TUTORIAL 1
Determine the reactions at A and B for the steel bar
and loading shown, assuming a close fit at both
supports before the loads are applied.
66
Answer, RA= 323 kN, Rb= 577kN

64. Determine the reactions at A and B for
the steel bar and loading shown, assuming
a close fit at both supports before the
loads are applied.
• Solve for the reaction at A due to
applied loads and the reaction found at B.
• Require that the displacements due to
the loads and due to the redundant
reaction be compatible, i.e., require that
their sum be zero.
• Solve for the displacement at B due to
the redundant reaction at B.
SOLUTION:
• Consider the reaction at B as redundant,
release the bar from that support, and
solve for the displacement at B due to
the applied loads.

65. SOLUTION:
• Solve for the displacement at B due to the applied
loads with the redundant constraint released,
EEA
LP
LLLL
AAAA
PPPP
i ii
ii
9
L
4321
26
43
26
21
3
4
3
321
10125.1
m150.0
m10250m10400
N10900N106000







• Solve for the displacement at B due to the
redundant constraint,
 






i
B
ii
ii
R
B
E
R
EA
LP
δ
LL
AA
RPP
3
21
26
2
26
1
21
1095.1
m300.0
m10250m10400

66. • Require that the displacements due to the loads and due to
the redundant reaction be compatible,
 
kN577N10577
0
1095.110125.1
0
3
39







B
B
RL
R
E
R
E


• Find the reaction at A due to the loads and the reaction at
B
kN323
kN577kN600kN3000

 
A
Ay
R
RF
kN577
kN323


B
A
R
R

67. TUTORIAL 2
Two cylindrical rods, CD made of steel (E=200 GPa) and
AC made of aluminum (E=72 GPa), are joined at C and
restrained by rigid supports at A and D. Determine
(a) the reactions at A and D (RA=52.9kN, RD= 87.1 kN)
(b) The deflection of point C (0.086 mm)
67

70. TUTORIAL 3
At room temperature (21oC) a 0.5 mm gap exists
between the ends of the rods shown. At a later time when
the temperature has reached 1600C, determine
(a)The normal stress in the aluminum rod (σa =-150.6
MPa)
(b)The change in length of the aluminum rod (δa= 0.369
mm)
69