STRENGTH OF MATERIALS
TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
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3
CHAPTER
Axially loaded
members
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Contents
3.1. Introduction
3.2.Normal Stress and Normal Strain
3.3. Tension and Compression Test
3.4. Poisson’s ratio
3.5. Shearing Strain
3.6. Allowable Stress – Factor of Safety
3.7. Statically Indeterminate Problem
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3.1. Introduction
• Suitability of a structure or machine may depend

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on the
deformations in the structure as well as the stresses induced
under loading. Statics analyses alone are not sufficient.
• Considering structures as deformable ones allows us to
determinate the member forces and reactions which are statically
indeterminate.
• Determination of the stress distribution within a member also
requires the consideration of deformations in the member.
• Chapter 3 is concerned with the stress and deformation of a
structural member under axial loading. Later chapters will deal
with torsional and pure bending loads.
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3.1. Introduction
Axially loaded members are structural components subjected only to
axial force (tension or compression)
• Prismatic bar: Straight structural member with the same cross-
section throughout its length
• Axial force: Load directed along the axis of the member
• Axial force can be tensile or compressive
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3.1. Introduction
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3.1. Introduction
Axial force diagram
0 ...zZ N
Using the method of section , the internal axial force is obtained from the
equilibrium as a function of coordinate z
Kinematic assumptions Before deformation
After deformation
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Kinematic assumptions
3.2. Normal stress and normal strain
Normal stress
z
z
N
A
z const
– normal stress at any point on
the cross-sectional area
Nz – internal resultant normal force
A – cross-sectional area
1. The axis of the member remains straight
2. Cross sections which are plane and are perpendicular to the axis before
deformation, remain plane and remain perpendicular to the axis after
deformation. And the cross sections do not rotate about the axis
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3.2. Normal stress and normal strain
Elongation of the bar:
Consider the bar, which has a cross-sectional area that gradually varies
along its length L. The bar is subjected to concentrated loads at its ends
and variable external load distributed along its length.
0
( )
( )
L
N z dz
L
EA z
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3.2. Normal stress and normal strain
Elongation of the bar – constant load
and cross-sectional area:
zN LL
EA
zN
L EA
Normal Strain – elongation per unit
length
EA – stiffness of axially loaded bar
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zi
i
N
EA
const
n
zi i
i 1 i
N L
L
EA
3.2. Normal stress and normal strain
3.4. Poisson’s Ratio
• For a slender bar subjected to axial
loading:
0zz x y
E
• The elongation in the x-direction is
accompanied by a contraction in the other
directions. Assuming that the material is
isotropic (no directional dependence),
0x y
• Poisson’s ratio is defined as
lateral strain
axial strain
yx
z z
Poisson’s Ratio
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• Strength of a material can only be determined by experiment
• One test used by engineers is the tension or compression test
• This test is used primarily to determine the relationship between the
average normal stress and the average normal strain in common
engineering materials, such as metals, ceramics, polymers and
composites
3.3. Tension and Compression Test
Performing the tension or compression test
• Specimen of material is made into “standard” shape and size
• Before testing, 2 small punch marks are identified along the specimen’s
length
• Measurements are taken for both the specimen’s initial x-sectional area
A0 and the gauge-length distance L0; between the two marks
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1Performing the tension or compression test
• Seat the specimen into a testing machine as shown below
- The machine will stretch the specimen
at a slow constant rate until the breaking
point
- At frequent intervals during test, the
data is recorded of the applied load P.
3.3. Tension and Compresion Test
• The Elongation δ = L − L0 is measured
by using either a caliper or an
extensometer
• δ is used to calculate the normal strain
in the specimen
• Sometimes, the strain can also be read
directly by using an electrical-resistance
strain gauge
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3.3. Tension and Compresion Test
• A stress-strain diagram is obtained by plotting the various values of
the stress and corresponding strain in the specimen
Conventional stress-strain diagram
• Using recorded data, we can determine the nominal or engineering
stress by
P
A0
σ =
• Likewise, the nominal or engineering strain is found directly from strain
gauge reading, or by
δ
L0
=
By plotting σ (ordinate) against (abscissa), we get a conventional
stress-strain diagram
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3.3. Tension and Compression Test
Conventional stress-strain diagram
• This Figure shows the characteristic stress-strain diagram for steel, a
commonly used material for structural members and mechanical
elements
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3.3. Tension and Compresion Test
Conventional stress-strain diagram
Elastic behavior.
• the straight line
• The stress is proportional to the
strain, i.e., linearly elastic
• Upper stress limit, or proportional
limit; σpl
• If the load is removed upon
reaching the elastic limit , the
specimen will return to its original
shape
Yielding.
• The material deforms permanently;
yielding; plastic deformation
• Yield stress, σY
• Once the yield point is reached, the specimen continues to elongate
(strain) without any increase in load 17
3.3. Tension and Compresion Test
Conventional stress-strain diagram
Figure 3-4
Strain hardening.
• Ultimate stress, σu
• While the specimen is elongating,
its z-sectional area will decrease
• Decrease in area is fairly uniform
over entire gauge length
Necking.
• At ultimate stress, x-sectional
area begins to decrease in a
localized region
• As a result, a constriction or
“neck” tends to form in this region
as the specimen elongates further
• The Specimen finally breaks at fracture stress, σf
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3.3. Tension and Compresion Test
Stress – Strain Diagram: Ductile materials
• Defined as any material that can be subjected to large strains before it
rupture, e.g., mild steel
• Such materials are used because of its capacity of absorbing shock or
energy so that it, will exhibit a large deformation before failing
• Ductility of material is to report its percent elongation or percent
reduction in area at time of fracture
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1Ductile materials
• Percent elongation is the specimen’s fracture strain expressed as a
percent
• Percent reduction in area is defined within necking region as
• Most metals do not exhibit constant yielding behavior beyond the
elastic range, e.g. aluminum
• It does not have a well-defined yield point, thus it is standard practice
to define its yield strength using a graphical procedure called the offset
method
Percent elongation =
Lf − L0
L0
(100%)
Percent reduction in area =
A0 − Af
A0
(100%)
3.3. Tension and Compression Test
1Stress – Strain Diagram: Brittle Materials
• Materials that exhibit little or no yielding before failure are referred to as
brittle materials, e.g., gray cast iron
• Brittle materials do not have a well-defined tensile fracture stress, since
the appearance of an initial cracks in a specimen is quite random
3.3. Tension and Compression Test
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3.3. Tension and Compresion Test
• Most engineering materials
exhibit a linear relationship
between the stress and strain
with the elastic region
• Discovered by Robert Hooke in
1676 using springs, known as
Hooke’s law
σ = E
• E represents the constant of
proportionality, also called the
modulus of elasticity or Young’s
modulus
• E has units of stress, i.e., pascals,
MPa or GPa.
Hooke’s Law: Modulus of Elasticity
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3.3. Tension and Compresion Test
• If the strain disappears when
the stress is removed, the
material is said to behave
elastically.
• 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.
Elastic vs. Plastic Behavior
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3.3. Tension and Compresion Test
3.5. Allowable Stress – Factor of Safety
• When designing a structural member or mechanical element, the
stress in it must be restricted to safe level
• Choose an allowable load that is less than the load the member can
fully support
• One method used is the factor of safety (F.S.)
F.S. =
Ffail
Fallow
• If the load applied is linearly related to the stress developed within
the member, then F.S. can also be expressed as:
F.S. =
σfail
σallow F.S. =
τfail
τallow
• In all the equations, F.S. is chosen to be greater than 1, to avoid
potential for failure
• Specific values will depend on the type of material used and its
intended purpose
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3.5. Allowable Stress – Factor of Safety
• Fail Stress: 0 – stress coresponding lossing ability to sustain a load
0
.
fail
allow
n F S
• Allowable Stress:
• Axially Loaded Members –True stress must satisfy this condition:
allow
P
A
• Three main problems:
1. Determine if the condition of strength is satisfied or not.
2. Determine the required cross-sectional area.
3. Determine the maximum applied load.
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3.6. Statically Indeterminate Problem
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Sample Problem 3.1
The A-36 steel bar shown in the Figure is made from two segments having the
cross-sectional areas of AAB=1cm
2; ABD=2cm
2. Draw the axial force diagram and
determine the maximum normal stress and the vertical displacement of end A
1. Use the method of section
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Sample Problem 3.1
2. Normal stress
215 15( / )
1
AB
AB
AB
N
kN cm
A
27 3.5( / )
2
BC
BC
BC
N
kN cm
A
29 4.5( / )
2
CD
CD
CD
N
kN cm
A
2
max
15( / )kN cm
3. Displacement
0.0127( )BC BC CD CDAB ABA AD AB BC CD
AB BC CD
N l N lN l
L cm
EA EA EA
Since the result is positive, the bar elongates and so the displacement at A is upward
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The rigid bar BDE is supported by two
links AB and CD.
Link AB is made of aluminum (E = 70
GPa) and has a cross-sectional area
of 500 mm2. Link CD is made of steel
(E = 200 GPa) and has a cross-
sectional area of (600 mm2).
For the 30-kN force shown, determine
the deflection a) of B, b) of D, and c) of
E.
SOLUTION:
• Apply a free-body analysis to the
bar BDE to find the forces
exerted by links AB and DC.
• Evaluate the deformation of links
AB and DC or the displacements
of B and D.
• Work out the geometry to find
the deflection at E given the
deflections at B and D.
Sample Problem 3.2
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Displacement of B:
m10514
Pa1070m10500
m3.0N1060
6
926-
3
AE
PL
B
mm 514.0B
Displacement of D:
m10300
Pa10200m10600
m4.0N1090
6
926-
3
AE
PL
D
mm 300.0D
Free body: Bar BDE
ncompressioF
F
tensionF
F
M
AB
AB
CD
CD
B
kN60
m2.0m4.0kN300
0M
kN90
m2.0m6.0kN300
0
D
SOLUTION:
Sample Problem 3.2
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Displacement of D:
mm 7.73
mm 200
mm 0.300
mm 514.0
x
x
x
HD
BH
DD
BB
mm 928.1E
mm 928.1
mm 7.73
mm7.73400
mm 300.0
E
E
HD
HE
DD
EE
Sample Problem 3.2
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b a
B
A2
F2
F1
A1
C D
Consider the bar made from two segments
having the cross-sectional areas of A1 and A2 .
1. Draw the axial force diagram.
2. Determine the max. normal stress.
3. Determine the displacement of end D.
with F1=10kN; F2=25kN; A1=5cm
2; A2=8cm
2
a=b=1m; E=2.104kN/cm2
SOLUTION
1. Using method of section, the internal axial
force in each segments are:
z1
F1
DNCD
1 10CDN F kN a
F2
F1
C D
z2
NBC
1 2 15BCN F F kN
Sample Problem 3.3
34
b a
B
A2
F2
F1
A1
C D
10
N
kN
1
5
Axial force diagram:
2. Determine the maximum normal stress
2
1
10
2( / )
5
CD
CD
N
kN cm
A
2
2
15
1,875( / )
8
BC
BC
N
kN cm
A
22( / )
max
kN cm
3. The displacement of point D
2 1
. .BC CD
D BD BC CD
N b N a
w L l l
EA EA
2 2
2
4
1 15.10 10.10
0,0625.10 ( )
2.10 8 5
Dw cm
=>displacement toward right
Sample Problem 3.3
35
EA EA
P
D
C
E
h
The steel bars CD an CE with Young’s modulus E,
each have a cross-sectional area of A, are joined
at C with a pin. Determine the axial forces in each
bars and the displacement of point C cause by
load P.
SOLUTION:
1. Determine axial forces:
Using method of join. FBD of join C.
1 2
1 2
0 sin sin 0X N N
N N
1 2
1
0 os o 0
2 os
Y N c N c P
N c P
(1)
(2)
1 2(1) (2)
2cos
P
N N
P
C
N1 N2
X
Y
Sample Problem 3.4
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2. Displacement of join C:
EA EA
D
C
E
h
C’
From geometry:
1L
yC
1
C
L
y
cos
1 1
1
N L
L
EA
1
2cos
P
N
And:
1
cos
h
L
1 22
Ph
L
EAcos
32
C
Ph
y
EAcos
Sample Problem 3.4
with:
37
Bài 2.4: Three steel rods, each have a cross-
sectional area of A=5cm2, jointly support the
load P= 50kN . Determine the axial forces in
these rods and the displacement of join C.
Using
E = 2.104kN/cm2, H=4m
SOLUTION:
1. Axial forces: FBD of join C
1 3
1 3
0 sin30 sin30 0o oX N N
N N
(1)
(2)
1 3 2
1 2
0 ( ). os30 0
3
oY N N c N P
N N P
N3
30o 30o
C
N1
N2
P
Compatibility
o
1 3 2 2
3
os30
2
L L L c L
1 2
1 2
2 .3 3
2 43
N H N H
N N
EAEA
A A
30o 30
o
C
A
P
H
(3)
A A
30o 30
o
C
A
P
H
2L
1L
Sample Problem 3.5
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Sample Problem 3.6
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Sample Problem 3.6
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Sample Problem 3.7
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Sample Problem 3.7
3
2
9
. 20.10 .0,04
0,0002
0,01
200.10 . .
4
AC
B AC
P l
l
EA
=> Appear reaction FB at fixed end B
• Compatibility condition:
0ABl
0.40.8
0.0002BB
F PF
EA EA
2
9 0.010.12 0.04 0,0002 200 10
4
BF P
4.05( )BF kN
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Sample Problem 3.7
THANK YOU FOR
YOUR ATTENTION !
1/10/2013 43

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