Bài giảng Strength of materials - Chapter 2: Axial Force, Shear Force and Bending Moment - Trần Minh Tú

STRENGTH OF MATERIALS
TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
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2
CHAPTER
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Axial Force, Shear Force and 
Bending Moment
2.1. Introduction
2.2. Internal Stress Resultants
2.3. Example
2.4. Relationships between loads, 
shear forces, and bending moments
2.5. Graphical Method for Constructing Shear
and Moment Diagrams
2.6. Normal, Shear force and 
bending moment diagram of frame
Contents
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2.1. Introduction 
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- Structural members are usually classified
according to the types of loads that they
support.
- Planar structures: if they lie in a single
plane and all loads act in that same plane.
2.1.1. Support connections.
- Structural members are joined together in various ways depending on
the intent of the designer. The three types of joint most often specified
are the pin connection, the roller support, and the fixed joint
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- Types of supports
2.1. Introduction 
Idealized
A
A
V
A
H
- Pin support: prevents the translation at the
end of a beam but does not prevent the
rotation.
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- Roller support: prevents the translation in
the vertical direction but not in the horizontal
direction, and does not prevent the rotation.
2.1. Introduction 
A
A
V
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2.1. Introduction 
- Fixed (clamped) support: the bar can neither translate nor rotate.
A
A
V
A
H
AM
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2.1. Introduction 
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2.1.2. Types of beams
2.1. Introduction 
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2.2. Internal Stress Resultants 
y
z
xMx
My
Mz Qx
NZ
Qy
In general, internal stress resultants
(internal forces) consist of 6 components
• Nz – Normal force
• Qx, Qy – Shear forces
• Mx, My – Bending moments
• Mz – Torsional moment
 Planar structures: if they lie in a
single plane and all loads act in that
same plane => Only 3 internal stress
resultants exert on this plane (zoy) .
y
z
xMx
NZ
Qy
• Nz – axial force (N);
• Qy – shear force (Q);
• Mx - bending moment (M)
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2.2. Internal force Resultants 
 To determine the internal force resultants => Using the method of
sections.
Q
N
Q
N NN N
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 Sign convention:
2.2. Internal force Resultants 
N N NN
• Axial force: positive when outward
of an element, negative when
inward of an element
• Shear force: positive when acts
clockwise against an element,
negative when acts counterclockwise
against an element
• Bending moment: positive when
compresses the upper part of the
beam and negative when
compresses the lower part of the
beam
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2.2. Axial, Shear and Moment diagram
• Because of the applied loadings, the beams develop an internal
shear force and bending moment that, in general, vary from point to
point along the axis of the beam. In order to properly design a beam it
therefore becomes necessary to determine the maximum shear and
moment in the beam.
• One way to do this is expressing N, Q and M as the functions of their
arbitrary position z along the beam’s axis. These axial, shear and
moment functions can then be plotted and represented by graph called
the axial, shear and moment diagram
2.1.3. Axial, Shear and Moment diagram
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2.2. Axial, Shear and Moment diagram
N, Q
z
M
z
Moment diagram to be plotted by side, which it is in tension
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Example 2.1: Draw the shear
and moment diagram for
the beam shown in the
figure.
Solution:
1. Support reactions VA VB
F
a b
C
 0A BM V a b Fa 
 0B AM V a b Fb 
 B
Fa
V
a b
 A
Fb
V
a b
Recheck: 0Y 
2.3. Example
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F
a b
VA VB
C
1
1
Section 1 – 1:
VA
z1
Q
M
N
0N 
10 z a 
0A A
Fb
Y Q V Q V
a b

Section 2 – 2:
1
0 1 10A A
Fbz
M M V z M V z
a b

0N 
20 z b 
0B B
Fa
Y Q V Q V
a b

2
0 2 20B B
Faz
M M V z M V z
a b

2
2
VB
z2
Q
M
N
Segment AC
Segment BC
A B
2.3. Example
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2.3. Example
Comment 1:
F
a b
VA VB
Fb
a+b
a+b
Fa
+
N
M
Q
Fab
a+b
F
C
:
Fb
AC Q
a b
:
Fa
BC Q
a b
1:
Fbz
AC M
a b
2:
Faz
BC M
a b
The section on which the
concentrated force acts, the
shear force diagram has
“jump”
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2.3. Example 
L
q
VA VB
Example 2.2: Draw the shear and
moment diagram for the beam
shown in the figure.
Solution:
1. Support reactions
2
. 0
2
A B
ql
M V l 
.
2
A
q l
V 
2
. 0
2
B A
ql
M V l 
.
2
B
q l
V 
Symmetrically:
.
2
A B
q l
V V 
Or:
2. Internal force resultant’s functions:
Segment 1-1 
(0 ≤ z L)
.
2
ql
Q q z 
2. .
2 2
ql q
M z z 
1
1
Q
zVA
M
N
q
0AY Q qz V 
2
0 0
2
A
qz
M M V z 
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2.3. Example 
 Comment 2
L
q
VA VBqL/2
qL/2
+
Q
L/2
qL2/8
M
.
2
ql
Q q z 
2. .
2 2
ql q
M z z 
0
2
A
qL
z Q 
2
B
qL
z L Q 
0 0Az M 
0Bz L M 
2
qL
M ' qz 0
2
L
M ' z 
0M'' q 
2
2
8
max z L/
qL
M M
The section on which the shear
force is equal to zero then the
bending moment is maximum.
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2.3. Example – example 2.3
1. Support reactions:
.( ) 0A BM V a b M 
.( ) 0B AM V a b M 
B
M
V
a b
A
M
V
a b
2. Stress resultants:
AC: Section 1-1 ( 0 ≤ z1 a)
y A
M
Q V
a b
VA VB
a b
C
M
.x AM V z 
Q
VA
M
z1 VB
M
Q z2
1
1
2
2
Section 2-2 ( 0 ≤ z2 b)
y A
M
Q V
a b
2.x BM V z 
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2.3. Example – example 2.3
a b
VA VB
M
(a+b) M
(a+b)Ma
(a+b)
Mb
(a+b)
M
Q
M
C
M
y A
M
Q V
a b
1.x AM V z 
AC: ( 0 ≤ z1 a)
y A
M
Q V
a b
BC: ( 0 ≤ z2 b)
2.x BM V z 
Comment 3
The section on which the
concentrated moment acts,
the bending moment diagram
has “jump”
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2.4. Relationships between transverse loads, shear forces,
and bending moments
- Consider the beam shown in the figure,
which is subjected to an arbitrary loading.
The free-body diagram for a small
segment dz of the beam:
( ) 0Y Q dQ Q q z dz 
( )
dQ
q z
dz
( ) 0
2 2
dz dz
M M dM M Q dQ Q 
dM
Q
dz
2
2
( )
d M dQ
q z
dz dz
- Positive distributed load q(z): acts
upward on the beam
q(z) > 0
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2.4. Relationships between loads, shear forces,
and bending moments
Application: 
- Recognizing the type of Q and M diagrams when the distributed
load’s function is known, i. e if the distributed load’s function is n-
degree, then the shear force’s function will be (n+1)-degree and the
bending moment function will be (n+2)-degree.
- The section, on which the shear force is equal to zero then the
bending moment is maximum.
- Determining Q, M on the arbitrary section, when knowing the value of
Q and M on the specific section.
•Qright = Qleft + Sq ( Sq – Area of distributed load diagram q)
•Mright = Mleft + SQ ( SQ – Area of shear force diagram Q)
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2.4. Relationships between loads, shear forces,
and bending moments
q
z
q(z)
A B
( )
B B
A A
dQ q z dz 
B A qQ Q S 
Sq
Q
z
Q(z)
A B
SQ
( )
B B
A A
dM Q z dz 
B A QM M S 
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2.4. Relationships between loads, shear forces,
and bending moments
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2.5. Graphical Method for Constructing Shear 
and Moment Diagrams
- Base on the relationships between loads, shear force, and bending
moments
- Knows the distributed load q(z) => Predict the types of shear force
and bending moment diagram => Indentify the necessary number of
points to construct the diagram.
• q=0 => Q=const => QA=? (or QB)
M linear => MA=? and MB=?
• q=const => Q linear => QA=? QB=?
M quadratic => MA=?; MB=?; maximum?
convex, concave,..?
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• Support reactions
q
F=qa
VA VB
.3 2 .2 . 0B AM V a qa a F a 
5
3
AV qa 
.3 2 . .2 0A BM V a qa a F a 
4
3
BV qa 
Segment AC:
2a a
C
q=const Q linear
QA=VA
QC=VA+Sq=5qa/3-2qa=-qa/3
M quadratic: MA=0
MC=MA+SQ=4qa
2/3; 
Mmax=25qa2/18 
5
3
qa
1
3
qa
+
5a/3
Mmax=25qa
2/18 
4qa2/3 
2.5. Graphical Method for Constructing Shear 
and Moment Diagrams -Example
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2.5. Graphical Method for Constructing Shear 
and Moment Diagrams -Example
2a a
VA VB
C
5
3
qa
4
3
qa
1
3
qa
+
5a/3
Mmax=25qa
2/18 
4qa2/3 
Segment BC: q
F=qa
q= 0
Q = const
QB= - VB
M linear:
MB=0
MC = MB - SQ=4qa
2/3
Q
M
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F
a aa
A B
C
D
2.5. Graphical Method for Constructing Shear 
and Moment Diagrams - Example
a aa
A R
B
B
C
D
VD
F
R
Example 2.5: Draw the shear force
and bending moment diagram for the
compound beam shown in the figure:
Solution:
System of beams ABCD consists of:
+ Secondary beam BCD
+ Main beam AB
1) Secondary beam BCD:
- Support reactions:
2
D
F
V R 
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2.5. Graphical Method for Constructing Shear 
and Moment Diagrams -Example
0 ( )
2 2
C B Q
F Fa
M M S a 
(Q)2
F
2
F
(M)
2
Fa
0 ( )
2 2
C D Q
Fa Fa
M M S 
F
a aa
A B
C
D
B
C
D
VD
F
R
a. Segment BC: q(z)=0
=> Q=const => QB= R = F/2
=> M linear 0BM 
b. Segment CD: q(z)=0
=> Q=const => QD= -VD = - F/2
=> M linear 0DM 
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2.) Mean beam AB:
2
F
(Q)
2
Fa
(M)
F
a aa
A B
C
D
A R
B
2.5. Graphical Method for Constructing Shear 
and Moment Diagrams -Example
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2.5. Graphical Method for Constructing Shear 
and Moment Diagrams -Example
a aa
A B
C
D
2
Fa
2
Fa
2
F
2
F
2
F
(Q)
(M)
3.) The Shear force and bending
moment diagram of a system of
the beams
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2.6. Normal, Shear force and 
bending moment diagram of frame
- The frame is composed of several connected members that are fixed
connection. The design of these structures often requires drawing the
shear and moment diagram for each of the members
- Using a method of section, we determine the axial force, the shear
force, and the bending moment acting on each members.
- Always draw the moment diagram on the tensile side of the member.
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2.6. Normal, Shear force and 
bending moment diagram of frame
VD
VA
HA
a
a
a
F
q
A
D
B
C
Draw the axial force, shear force and bending moment diagram of the frame:
with q=8kN/m, F=5kN, a=1m
Solution:
1. Support reactions:
x
y
0 5( )AX H F kN 
1
.1 .1. .1 0
2
A DM V q F  9( )DV kN 
1
.1 .1. .1 .2 0
2
D A AM V q F H  1( )AV kN 
2. Axial force diagram N
AB: 1AB AN V kN 
BC: 1BC AN V kN 
CD: 0CDN 
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2.6. Normal, Shear force and 
bending moment diagram of frame
1
1
1
+
+
N
kN
3. Shear force and bending moment diagram
AB: q=0 Q const 5A AQ H kN 
5
5
+
 M linear: 0AM 
0 5.1 5B A QM M S kNm 
5
Q
kNM
kNm
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2.6. Normal, Shear force and 
bending moment diagram of frame
BC: q=0 Q const 0BQ 
 M linear: 5BM kNm 5 0 5C B QM M S kNm 
1
1
1
+
+
N
kN
5
5
+
5
Q
kNM
kNm
5
CD: q=const 9D DQ V kN 
 M quadratic: 0DM 
0 ( 1 9).1/ 2 5C D QM M S kNm 
 Q linear: 9 ( 8.1) 1C D qQ Q S kN 
9
1 -
5
1kN
Equilibrium of joint
5kNm
1kN
5kNm
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Draw the shear and moment diagram for the beam shown in the
figure.
2.7. Home works
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2.7. Homework
Draw the shear and moment diagram for the beam shown in the
figures.
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Draw the shear and moment diagram for the compound beam shown
in the figures.
2.7. Homework
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THANK YOU FOR 
YOUR ATTENTION !
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