The document focuses on stress and strain in materials under axial loading, detailing concepts such as normal strain, elastic vs. plastic behavior, and fatigue failure. It introduces Hooke's law and discusses the effects of alloying and temperature on material properties, along with methods for resolving statically indeterminate problems. The document also addresses generalized Hooke's law for multi-axial loading and relationships among normal and shear stresses and strains.

1. MECHANICS OF MATERIALS (I)
STRESS AND STRAIN – AXIAL LOADING
Dr. Jung-San Chen
Department of Engineering Science
National Cheng Kung University

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NORMAL STRAIN
(elongation per unit length)
https://www.youtube.com/watch?v=WJI3Lxk-z2c

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NORMAL STRAIN

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strain
normal
stress
=
=
=
=
L
A
P



L
A
P
A
P



=
=
=
2
2
L
L
A
P




=
=
=
2
2

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STRESS-STRAIN TEST
http://www.youtube.com/watch?v=67fSwIjYJ-E (Steel)
http://www.youtube.com/watch?v=hD_NJaZIpT0 (Al)

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STRESS-STRAIN DIAGRAM: DUCTILE MATERIALS

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STRESS-STRAIN DIAGRAM: BRITTLE MATERIALS

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STRESS-STRAIN DIAGRAM: DUCTILE MATERIALS

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STRESS-STRAIN DIAGRAM: DUCTILE MATERIALS
Yield Strength (Stress) = 0.2% Offset Yield Strength

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ELASTIC VS. PLASTIC BEHAVIOR
Residual strain or
permanent strain

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ELASTIC VS. PLASTIC BEHAVIOR
• If the strain disappears when the stress is removed,
the material is said to behave elastically. (before
point B)
• When the strain does not return to zero after the stress
is removed, the material is said to behave plastically.
• The largest stress for which this occurs is called the
elastic limit. (=proportional limit)

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FATIGUE
• When the stress is reduced below the endurance limit,
fatigue failures do not occur for any number of cycles.
• A member may fail due to fatigue at stress levels
significantly below the ultimate strength if subjected to
many loading cycles.
• Fatigue failure is of a brittle nature, even for materials
that are normally ductile.
[Note] : endurance limit is the stress for which failure
does not occur, even for an indefinitely large number
of loading cycles.

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FATIGUE
• Fatigue properties are shown on S-N diagrams.
Stress at failure

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HOOKE’S LAW: MODULUS OF ELASTICITY
• Strength is affected by alloying,
heat treating, and manufacturing
process but stiffness (Modulus
of Elasticity) is not.
• Below the yield stress
( ' )
Young's Modulus or
Modulus of Elasticity
E Hooke s Law
E
 
=
=
Note: Strength refers to the
capacity of a structure to resist loads;
stiffness is ability to resist a
deformation

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AXIALLY LOADED MEMBERS
• Recall: Spring L: natural length (unstressed length,
relaxed length, or free length)
In linearly region:
k: stiffness f: flexibility
P k or fP
 
= =
=
1
f k
P
k
P f


 =
=
L
P
L 
+
k
k

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DEFORMATIONS UNDER AXIAL LOADING
E
 
=
• From Hooke’s Law:
• From the definition of strain:
L

 =
• Equating and solving for the
deformation,
PL
k P EA L
AE
f P L EA
 

=  = =
= =
• From the definition of stress:
P
A
 =
Prismatic Bar
EA: axial rigidity of the bar

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DEFORMATIONS UNDER AXIAL LOADING
• With variations in loading, cross-section or material
properties, , :internal force
i i
i
i i i
PL
P
AE
 = 
1
2
3
A D C
D C
C
P P P P
P P P
P P
= − + +
= +
=
A
D
1
P
2
P
3
P
A
P
D
P
A
P
D
P D
P 3 3
1 1 2 2
1 2 3
1 2 3
, ,
P L
PL P L
EA EA EA
  
   
= = =
= + +
1
L
2
L
3
L
This equation is only valid for prismatic bar or bar consisting of
prismatic segments ( is the internal force)
i
P
(1)
(2)
(3)

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DEFORMATIONS UNDER AXIAL LOADING
Strains are not Uniformly Distributed:
PL
AE
 =
/
P A
E E

 = =

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BAR WITH VARYING CROSS-SECTIONAL AREA
P(x)
dx 0
( )
( )
( )
( )
L
N x dx
d
EA x
N x dx
EA x


=
= 

0
If is small error
  
It has been assumed that the stress distribution is uniform over cross
section

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https://www.youtube.com/watch?v=Caon9BRk_gQ

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AD: AB AE: AC DE:BC
= =
A
D E
B C
(obtained)
(obtained)
= ?
Similar figures:

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STATIC INDETERMINATE
Statically Determinate Problems:
The reaction and internal forces can be obtained
from the equilibrium equations.
Statically Indeterminate Problems:
The reaction and internal forces can not be
obtained from the equilibrium equations. It has to
consider the deformation geometry (boundary
condition).

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Statically Indeterminate Problem

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EXAMPLE 2.03 Statically Indeterminate Problem

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Statically Indeterminate Problem

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SOLUTION:
• Solve for the displacement at B due to
the applied loads with the redundant
constraint released,
E
E
A
L
P
L
L
L
L
A
A
A
A
P
P
P
P
i i
i
i
i
9
L
4
3
2
1
2
6
4
3
2
6
2
1
3
4
3
3
2
1
10
125
.
1
m
150
.
0
m
10
250
m
10
400
N
10
900
N
10
600
0

=

=
=
=
=
=

=
=

=
=

=

=
=
=
−
−


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• Solve for the displacement at B due to the
redundant constraint,
( )


−
=
=
=
=

=

=
−
=
=
−
−
i
B
i
i
i
i
R
B
E
R
E
A
L
P
δ
L
L
A
A
R
P
P
3
2
1
2
6
2
2
6
1
2
1
10
95
.
1
m
300
.
0
m
10
250
m
10
400

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• Require that the displacements due to the loads and due
to the redundant reaction be compatible,
( )
kN
577
N
10
577
0
10
95
.
1
10
125
.
1
0
3
3
9
=

=
=

−

=
=
+
=
B
B
R
L
R
E
R
E




• Find the reaction at A
kN
323
kN
577
kN
600
kN
300
0
=
 +
−
−
=
=
A
A
y
R
R
F
kN
577
kN
323
=
=
B
A
R
R

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THERMAL STRAIN
T
T
L

 =

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THE STRESS CHANGE DUE TO THE TEMPERATURE
RAISES

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POISSON’S RATIO
• For a slender bar subjected to axial loading:
0, 0 x
x y z x
E

   
 = =  =
• Assuming that the material is isotropic
(no directional dependence),
0

= z
y 

• Poisson’s ratio is defined as
if the material is linear elastic
lateral strain
axial strain
y z
x x
 

 
= − = − = −

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VOLUME CHANGE
Strain :
x direction:
y, z direction:


−
y


x
z
a
b
c
Length change :
x direction: a
y direction: b
−
z direction: c
−
Final length:
x direction:
y direction:
z direction:
(1 )
a 
+
(1 )
b 
−
(1 )
c 
−
length change
strain=
original length
Final length=original length +length change

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VOLUME CHANGE
1
2 2 2 2 3
(1 ) (1 ) (1 )
(1 2 2 )
(1 2 ) 1
V a b c
abc
abc if
  
      
  
= + − −
= + − − + +
 + − 
Original volume: 0
V abc
=
Final volume:
Unit volume change : (dilatation)
1 0
0 0
(1 2 )
(1 2 ) (1 2 ) 0
1 2 0 1/ 2
V V
V abc abc
e
V V abc
E
 

  
 
−
 + − −
= = =
= − = − 
 −   
Volume must increase in tension
condition

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GENERALIZED HOOKE’S LAW FOR MULTI-AXIAL
LOADING
• For an element subjected to multi-axial loading, the normal
strain components resulting from the stress components may be
determined from the principle of superposition.
E
E
E
E
E
E
E
E
E
z
y
x
z
z
y
x
y
z
y
x
x












+
−
−
=
−
+
−
=
−
−
+
=
0, 0, 0
x y z
  
  

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SHEARING STRAIN (3D)
normal stresses
shear stresses
length change
change in angle
• Sign conventions for shear stresses and strains
shear stress plane direction force direction


+ + +
− + −
+
−
−
+ − −
Shear strain The positive shear stresses are accompanied by
positive shear strains


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SHEARING STRAIN (2D)

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SHEARING STRAIN VS. SHEARING STRESS
• A plot of shear stress vs. shear strain is similar the
previous plots of normal stress vs. normal strain
except that the strength values are approximately half.
zx
zx
yz
yz
xy
xy G
G
G 




 =
=
=
where G is the modulus of rigidity or shear modulus.
Hooke’s law in shear

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SHEARING STRAIN VS. SHEARING STRESS
Generalized Hook’s Law for a Homogeneous and Isotropic
Material under Most General Stress Condition:

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RELATION AMONG E, , AND G
• An axially loaded slender bar will elongate in the axial
direction and contract in the transverse directions.
• An initially cubic element oriented as in figure will
deform into a rectangular parallelepiped. The axial
load produces a normal strain.


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• If the cubic element is oriented as in the figure, it will
deform into a rhombus. Axial load also results in a shear
strain.
RELATION AMONG E, , AND G


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Before After
tan tan
tan( )
1 tan tan
a b
a b
a b
−
− =
+
RELATION AMONG E, , AND G


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https://www.youtube.com/watch?v=9igkb
rYNN6I