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Lecturer; Dr. Dawood S. Atrushi
November 2014
Shearing Force and Bending Moment
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
10
for a < x < a + b
V = RA - q (x - a)
M = RA x - q (x - a)2 / 2
for a + b < x < L
V = - RB M = RB (L - x)
maximum moment occurs where V = 0
i.e. x1 = a + b (b + 2c) / 2L
Mmax = q b (b + 2c) (4 a L + 2 b c + b2) / 8L2
for a = c, x1 = L / 2
Mmax = q b (2L - b) / 8
for b = L, a = c = 0 (uniform loading over the entire span)
Mmax = q L2 / 8
Example 4-5
construct the V- and M-dia for the
cantilever beam supported to P1 and P2
RB = P1 + P2 MB = P1 L + P2 b
for 0 < x < a
V = - P1 M = - P1 x
for a < x < L
V = - P1 - P2 M
= - P1 x - P2 (x - a)
Content
¢ Beams ¢ Type of beams, loadings, and
reactions ¢ Shear Forces and Bending Moments ¢ Relationships Between Loads,
Shear Forces, and Bending Moments
¢ Shear-Force and Bending-Moment Diagrams
November, 2014 Shearing Force and Bending Moment - DAT 2
Beams
Members which are subjected to loads that are transverse to the longitudinal
axis are termed as the beam.
November, 2014
The members are either subjected to the forces or the moment having their
vectors perpendicular to the axis of the bar.
Shearing Force and Bending Moment - DAT 3
Types of Beams
① Statically Determinate beams
November, 2014
316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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Shearing Force and Bending Moment - DAT 4
Types of Beams
② Statically Indeterminate beams
November, 2014
316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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Shearing Force and Bending Moment - DAT 5
316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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Type of Loads
November, 2014
316
Chapter 5 Analysis and Design of Beams for Bending
5.1 Introduction 5.2 Shear and Bending-Moment
Diagrams 5.3 Relations Among Load, Shear,
and Bending Moment 5.4 Design of Prismatic Beams for
Bending *5.5 Using Singularity Functions to
Determine Shear and Bending Moment in a Beam
*5.6 Nonprismatic Beams
5.1 INTRODUCTIONThis chapter and most of the next one will be devoted to the analysis and the design of beams, i.e., structural members supporting loads applied at various points along the member. Beams are usually long, straight prismatic members, as shown in the photo on the previous page. Steel and aluminum beams play an important part in both struc-tural and mechanical engineering. Timber beams are widely used in home construction (Photo 5.1). In most cases, the loads are perpen-dicular to the axis of the beam. Such a transverse loading causes only bending and shear in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Photo 5.1 Timber beams used in residential dwelling.
CB
P1
(a) Concentrated loads
w
P2
A D
(b) Distributed load
AB
C
Fig. 5.1 Transversely loaded beams.
The transverse loading of a beam may consist of concentrated loads P1, P2, . . . , expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 5.1a), of a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 5.1b), or of a combination of both. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 5.1b), the load is said to be uniformly distributed over that part of the beam. Beams are classified according to the way in which they are supported. Several types of beams frequently used are shown in Fig. 5.2. The distance L shown in the various parts of the figure is
Fig. 5.2 Common beam support configurations.
L
(a) Simply supported beam
StaticallyDeterminateBeams
StaticallyIndeterminateBeams
L2L1
(d) Continuous beam
L
(b) Overhanging beam
L
Beam fixed at one endand simply supported
at the other end
(e)
L
(c) Cantilever beam
L
( f ) Fixed beam
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a) Concentrated load (single force)
b) Distributed load (measured by their intensity):
i. Uniformly distributed load (uniform load)
ii. Linearly varying load c) Couple
Shearing Force and Bending Moment - DAT 6
¢ Pinned support or hinged support.
November, 2014
Type of Reactions
Solid Mechanics
force and bending moment along the length of the beam.
These diagrams are extremely useful while designing the beams for various applications.
Supports and various types of beams
(a) Roller Support – resists vertical forces only
(b) Hinge support or pin connection – resists horizontal and vertical forces
Hinge and roller supports are called as simple supports
(c) Fixed support or built-in end
Solid Mechanics
force and bending moment along the length of the beam.
These diagrams are extremely useful while designing the beams for various applications.
Supports and various types of beams
(a) Roller Support – resists vertical forces only
(b) Hinge support or pin connection – resists horizontal and vertical forces
Hinge and roller supports are called as simple supports
(c) Fixed support or built-in end
Shearing Force and Bending Moment - DAT 7
¢ Roller support
November, 2014
Type of Reactions
Solid Mechanics
force and bending moment along the length of the beam.
These diagrams are extremely useful while designing the beams for various applications.
Supports and various types of beams
(a) Roller Support – resists vertical forces only
(b) Hinge support or pin connection – resists horizontal and vertical forces
Hinge and roller supports are called as simple supports
(c) Fixed support or built-in end
Shearing Force and Bending Moment - DAT 8
¢ Fixed support
November, 2014
Type of Reactions
Solid Mechanics
force and bending moment along the length of the beam.
These diagrams are extremely useful while designing the beams for various applications.
Supports and various types of beams
(a) Roller Support – resists vertical forces only
(b) Hinge support or pin connection – resists horizontal and vertical forces
Hinge and roller supports are called as simple supports
(c) Fixed support or built-in end
Solid Mechanics
force and bending moment along the length of the beam.
These diagrams are extremely useful while designing the beams for various applications.
Supports and various types of beams
(a) Roller Support – resists vertical forces only
(b) Hinge support or pin connection – resists horizontal and vertical forces
Hinge and roller supports are called as simple supports
(c) Fixed support or built-in end
Shearing Force and Bending Moment - DAT 9
1
Chapter 4 Shear Forces and Bending Moments
4.1 Introduction
Consider a beam subjected to
transverse loads as shown in figure, the
deflections occur in the plane same as
the loading plane, is called the plane of
bending. In this chapter we discuss shear
forces and bending moments in beams
related to the loads.
4.2 Types of Beams, Loads, and Reactions
Type of beams
a. simply supported beam (simple beam)
b. cantilever beam (fixed end beam)
c. beam with an overhang
Calculate the reaction forces of on of the below members
November, 2014
1
Chapter 4 Shear Forces and Bending Moments
4.1 Introduction
Consider a beam subjected to
transverse loads as shown in figure, the
deflections occur in the plane same as
the loading plane, is called the plane of
bending. In this chapter we discuss shear
forces and bending moments in beams
related to the loads.
4.2 Types of Beams, Loads, and Reactions
Type of beams
a. simply supported beam (simple beam)
b. cantilever beam (fixed end beam)
c. beam with an overhang
1
Chapter 4 Shear Forces and Bending Moments
4.1 Introduction
Consider a beam subjected to
transverse loads as shown in figure, the
deflections occur in the plane same as
the loading plane, is called the plane of
bending. In this chapter we discuss shear
forces and bending moments in beams
related to the loads.
4.2 Types of Beams, Loads, and Reactions
Type of beams
a. simply supported beam (simple beam)
b. cantilever beam (fixed end beam)
c. beam with an overhang
Shearing Force and Bending Moment - DAT 10
a)
November, 2014
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
2
Type of loads
a. concentrated load (single force)
b. distributed load (measured by their intensity) :
uniformly distributed load (uniform load)
linearly varying load
c. couple
Reactions
consider the loaded beam in figure
equation of equilibrium in horizontal direction
� Fx = 0 HA - P1 cos � = 0
HA = P1 cos �
� MB = 0 - RA L + (P1 sin �) (L - a) + P2 (L - b) + q c2 / 2 = 0
(P1 sin �) (L - a) P2 (L - b) q c2 RA = CCCCCCC + CCCC + CC L L 2 L
(P1 sin �) a P2 b q c2 RB = CCCCC + CC + CC L L 2 L
for the cantilever beam
� Fx = 0 HA = 5 P3 / 13
12 P3 (q1 + q2) b � Fy = 0 RA = CC + CCCCC 13 2
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
2
Type of loads
a. concentrated load (single force)
b. distributed load (measured by their intensity) :
uniformly distributed load (uniform load)
linearly varying load
c. couple
Reactions
consider the loaded beam in figure
equation of equilibrium in horizontal direction
� Fx = 0 HA - P1 cos � = 0
HA = P1 cos �
� MB = 0 - RA L + (P1 sin �) (L - a) + P2 (L - b) + q c2 / 2 = 0
(P1 sin �) (L - a) P2 (L - b) q c2 RA = CCCCCCC + CCCC + CC L L 2 L
(P1 sin �) a P2 b q c2 RB = CCCCC + CC + CC L L 2 L
for the cantilever beam
� Fx = 0 HA = 5 P3 / 13
12 P3 (q1 + q2) b � Fy = 0 RA = CC + CCCCC 13 2
b)
c)
Shearing Force and Bending Moment - DAT 11
Shear Forces and Bending Moments
At any point (x) along its length, a beam can transmit a bending moment M(x) and a shear force V(x).
November, 2014
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
Shearing Force and Bending Moment - DAT 12
¢ Consider a cantilever beam with a concentrated load P applied at the end A, at the cross section mn, the shear force and bending moment are found
November, 2014
Lecture Notes of Mechanics of Solids, Chapter 5 2
Look at the FBD of an elemental length dx of the above loaded beam (Fig. 5.2). As it has an
infinitesimal length, the distributed load can be considered as a Uniformly Distributed Load
(UDL) with constant magnitude w(x) over the differential length dx.
It is now necessary to equate the equilibrium of the element. Starting with vertical equilibrium
� � � � � � � �00 ¸
¹·
¨©§ ��� n� ¦ dx
dx
xdVxVdxxwxVF
y (5.1)
Dividing by dx in the limit as dxĺ0,
� � � �xwdx
xdV� (5.2)
Taking moments about the right hand edge of the element:
� � � � � � � � � �0
20 ¸
¹·
¨©§ ����� ¦ dx
dx
xdMxM
dxdxxwdxxVxMM
Edge.H.R (5.3)
Dividing by dx in the limit as dxĺ0,
� � � �xVdx
xdM (5.4)
Eqs. (5.2) & (5.3) are important when we have found one and want to determine the others.
5.3 BENDING MOMENT AND SHEAR FORCE EQUATIONS Introductionary Example - Simply Supported Beam By using the free body diagram technique, the bending moment and shear force distributions
can be calculated along the length of the beam. Let’s take a simply supported beam, Fig. (5.3),
as an example to shown the solutions:
P
L
a
RAY
=(1-a/L)P RBY
=Pa/L
I
I
RAY
=(1-a/L)P
x
F.B.D. (method of section I-I)
o
F.B.D. (global equilibrium)
II
II M(x)
V(x)
A B A
Fig. 5.3 FBD of beam cut before force P
Step A: Cut beam just before the force P (i.e. Section I-I), and draw a free body diagram
including the unknown shear force and bending moment as in Fig. 5.3.
Take moments about the right hand end (O):
� � 010 �¸¹·
¨©§ �� ¦ xMx
L
aPM o ĺ � � x
L
aPxM ¸
¹·
¨©§ � 1 (5.5)
To determine the shear force, use Eq. (5.4), giving that:
� � � �P
L
a
dx
xdMxV ¸
¹·
¨©§ � 1 (5.6)
To verify Eq. (5.6), equate vertical equilibrium:
++
++
Lecture Notes of Mechanics of Solids, Chapter 5 2
Look at the FBD of an elemental length dx of the above loaded beam (Fig. 5.2). As it has an
infinitesimal length, the distributed load can be considered as a Uniformly Distributed Load
(UDL) with constant magnitude w(x) over the differential length dx.
It is now necessary to equate the equilibrium of the element. Starting with vertical equilibrium
� � � � � � � �00 ¸
¹·
¨©§ ��� n� ¦ dx
dx
xdVxVdxxwxVF
y (5.1)
Dividing by dx in the limit as dxĺ0,
� � � �xwdx
xdV� (5.2)
Taking moments about the right hand edge of the element:
� � � � � � � � � �0
20 ¸
¹·
¨©§ ����� ¦ dx
dx
xdMxM
dxdxxwdxxVxMM
Edge.H.R (5.3)
Dividing by dx in the limit as dxĺ0,
� � � �xVdx
xdM (5.4)
Eqs. (5.2) & (5.3) are important when we have found one and want to determine the others.
5.3 BENDING MOMENT AND SHEAR FORCE EQUATIONS Introductionary Example - Simply Supported Beam By using the free body diagram technique, the bending moment and shear force distributions
can be calculated along the length of the beam. Let’s take a simply supported beam, Fig. (5.3),
as an example to shown the solutions:
P
L
a
RAY
=(1-a/L)P RBY
=Pa/L
I
I
RAY
=(1-a/L)P
x
F.B.D. (method of section I-I)
o
F.B.D. (global equilibrium)
II
II M(x)
V(x)
A B A
Fig. 5.3 FBD of beam cut before force P
Step A: Cut beam just before the force P (i.e. Section I-I), and draw a free body diagram
including the unknown shear force and bending moment as in Fig. 5.3.
Take moments about the right hand end (O):
� � 010 �¸¹·
¨©§ �� ¦ xMx
L
aPM o ĺ � � x
L
aPxM ¸
¹·
¨©§ � 1 (5.5)
To determine the shear force, use Eq. (5.4), giving that:
� � � �P
L
a
dx
xdMxV ¸
¹·
¨©§ � 1 (5.6)
To verify Eq. (5.6), equate vertical equilibrium:
++
++
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
Shearing Force and Bending Moment - DAT 13
Sign Convention
¢ The shear force tends to rotate the material clockwise is defined as positive
¢ The bending moment tends to compress the upper part of the beam and elongate the lower part is defined as positive
November, 2014
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
3
12 P3 q1 b q1 b � MA = 0 MA = CC + CC (L – 2b/3) + CC (L – b/3) 13 2 2 for the overhanging beam � MB = 0 - RA L + P4 (L– a) + M1 = 0
� MA = 0 - P4 a + RB L + M1 = 0
P4 (L– a) + M1 P4 a - M1 RA = CCCCCC RB = CCCC
L L 4.3 Shear Forces and Bending Moments
Consider a cantilever beam with a
concentrated load P applied at the end
A, at the cross section mn, the shear
force and bending moment are found
� Fy = 0 V = P
� M = 0 M = P x
sign conventions (deformation sign conventions)
the shear force tends to rotate the
material clockwise is defined as positive
the bending moment tends to compress
the upper part of the beam and elongate the
lower part is defined as positive
Shearing Force and Bending Moment - DAT 14
Example 1
A simple beam AB supports a force P and a couple M0, find the shear V and bending moment M at
a) x = L/2 b) x = (L/2)+
November, 2014
4
Example 4-1
a simple beam AB supports a force P
and a couple M0, find the shear V and
bending moment M at
(a) at x = (L/2)_ (b) at x = (L/2)+
3P M0 P M0 RA = C - C RB = C + C 4 L 4 L
(a) at x = (L/2)_
� Fy = 0 RA - P - V = 0
V = RA - P = - P / 4 - M0 / L
� M = 0 - RA (L/2) + P (L/4) + M = 0
M = RA (L/2) + P (L/4) = P L / 8 - M0 / 2
(b) at x = (L/2)+ [similarly as (a)]
V = - P / 4 - M0 / L
M = P L / 8 + M0 / 2
Example 4-2
a cantilever beam AB subjected to a
linearly varying distributed load as shown, find
the shear force V and the bending moment M
q = q0 x / L
� Fy = 0 - V - 2 (q0 x / L) (x) = 0
V = - q0 x2 / (2 L)
Vmax = - q0 L / 2
Shearing Force and Bending Moment - DAT 15
Example 2
A cantilever beam AB subjected to a linearly varying distributed load as shown, find the shear force V and the bending moment M
November, 2014
4
Example 4-1
a simple beam AB supports a force P
and a couple M0, find the shear V and
bending moment M at
(a) at x = (L/2)_ (b) at x = (L/2)+
3P M0 P M0 RA = C - C RB = C + C 4 L 4 L
(a) at x = (L/2)_
� Fy = 0 RA - P - V = 0
V = RA - P = - P / 4 - M0 / L
� M = 0 - RA (L/2) + P (L/4) + M = 0
M = RA (L/2) + P (L/4) = P L / 8 - M0 / 2
(b) at x = (L/2)+ [similarly as (a)]
V = - P / 4 - M0 / L
M = P L / 8 + M0 / 2
Example 4-2
a cantilever beam AB subjected to a
linearly varying distributed load as shown, find
the shear force V and the bending moment M
q = q0 x / L
� Fy = 0 - V - 2 (q0 x / L) (x) = 0
V = - q0 x2 / (2 L)
Vmax = - q0 L / 2
Shearing Force and Bending Moment - DAT 16
5
� M = 0 M + 2 (q0 x / L) (x) (x / 3) = 0
M = - q0 x3 / (6 L)
Mmax = - q0 L2 / 6
Example 4-3
an overhanging beam ABC is
supported to an uniform load of intensity
q and a concentrated load P, calculate
the shear force V and the bending
moment M at D
from equations of equilibrium, it is found
RA = 40 kN RB = 48 kN
at section D
� Fy = 0 40 - 28 - 6 x 5 - V = 0
V = - 18 kN
� M = 0
- 40 x 5 + 28 x 2 + 6 x 5 x 2.5 + M = 0
M = 69 kN-m
from the free body diagram of the right-hand part, same results can be
obtained
4.4 Relationships Between Loads, Shear Forces, and Bending Moments
consider an element of a beam of length
dx subjected to distributed loads q
equilibrium of forces in vertical direction
Example 3
An overhanging beam ABC is supported to an uniform load of intensity q and a concentrated load P, calculate the shear force V and the bending moment M at D
November, 2014 Shearing Force and Bending Moment - DAT 17
Relationships Between Loads, Shear Forces, and Bending Moments
a) Consider an element of a beam of length dx subjected to distributed loads q.
November, 2014
5
� M = 0 M + 2 (q0 x / L) (x) (x / 3) = 0
M = - q0 x3 / (6 L)
Mmax = - q0 L2 / 6
Example 4-3
an overhanging beam ABC is
supported to an uniform load of intensity
q and a concentrated load P, calculate
the shear force V and the bending
moment M at D
from equations of equilibrium, it is found
RA = 40 kN RB = 48 kN
at section D
� Fy = 0 40 - 28 - 6 x 5 - V = 0
V = - 18 kN
� M = 0
- 40 x 5 + 28 x 2 + 6 x 5 x 2.5 + M = 0
M = 69 kN-m
from the free body diagram of the right-hand part, same results can be
obtained
4.4 Relationships Between Loads, Shear Forces, and Bending Moments
consider an element of a beam of length
dx subjected to distributed loads q
equilibrium of forces in vertical direction
Shearing Force and Bending Moment - DAT 18
Equilibrium of forces in vertical direction
November, 2014
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
Integrate between two points A and B;
= - (area of the loading diagram between A and B) Shearing Force and Bending Moment - DAT 19
Moment equilibrium of the element
November, 2014
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
Maximum (or minimum) bending-moment occurs at dM/dx = 0, i.e. at the point of shear force V = 0. Shearing Force and Bending Moment - DAT 20
Integrate between two points A and B
November, 2014
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
6
� Fy = 0 V - q dx - (V + dV) = 0
or dV / dx = - q
integrate between two points A and B
B B Н dV = -Н q dx A A i.e. B VB - VA = -Н q dx A = - (area of the loading diagram between A and B)
the area of loading diagram may be positive or negative
moment equilibrium of the element
� M = 0 - M - q dx (dx/2) - (V + dV) dx + M + dM = 0
or dM / dx = V
maximum (or minimum) bending-moment occurs at dM / dx = 0,
i.e. at the point of shear force V = 0
integrate between two points A and B
B B Н dM = Н V dx
A A
i.e. B MB - MA = Н V dx A
= (area of the shear-force diagram between A and B)
this equation is valid even when concentrated loads act on the beam
between A and B, but it is not valid if a couple acts between A
and B
= (area of the loading diagram between A and B)
Shearing Force and Bending Moment - DAT 21
b) Consider an element of a beam of length dx subjected to concentrated load P.
November, 2014
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
Shearing Force and Bending Moment - DAT 22
Equilibrium of force;
November, 2014
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
i.e. an abrupt change in the
shear force occurs at any point where a concentrated
load acts.
Since the length dx of the element is infinitesimally small, i.e. M1 is also infinitesimally small, thus, the bending moment does not change as we pass through the point of application of a concentrated load.
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
Equilibrium of moment;
Shearing Force and Bending Moment - DAT 23
c) Consider an element of a beam of length dx subjected to concentrated loads in form of couples, M0.
November, 2014
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
Equilibrium of force; V1 = 0
Shearing Force and Bending Moment - DAT 24
¢ i.e. no change in shear force at the point of application of a couple.
November, 2014
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
7
concentrated loads
equilibrium of force
V - P - (V + V1) = 0
or V1 = - P
i.e. an abrupt change in the shear force occurs
at any point where a concentrated load acts
equilibrium of moment
- M - P (dx/2) - (V + V1) dx + M + M1 = 0
or M1 = P (dx/2) + V dx + V1 dx M 0
since the length dx of the element is infinitesimally small, i.e. M1
is also infinitesimally small, thus, the bending moment does not change as
we pass through the point of application of a concentrated load
loads in the form of couples
equilibrium of force V1 = 0
i.e. no change in shear force at the point of
application of a couple
equilibrium of moment
- M + M0 - (V + V1) dx + M + M1 = 0
or M1 = - M0
the bending moment changes abruptly at a point of application of a
couple
Equilibrium of moment;
¢ The bending moment changes abruptly at a point of application of a couple.
Shearing Force and Bending Moment - DAT 25
Shear-Force and Bending-Moment Diagrams
Concentrated loads Consider a simply support beam AB with a concentrated load P
November, 2014
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
Shearing Force and Bending Moment - DAT 26
November, 2014
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
Shearing Force and Bending Moment - DAT 27
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
With Mmax = Pab/L
November, 2014
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
V - Diagram
M - Diagram
Shearing Force and Bending Moment - DAT 28
Uniform loads Consider a simply support beam AB with a uniformly distributed load of constant intensity q;
November, 2014
8
4.5 Shear-Force and Bending-Moment Diagrams
concentrated loads
consider a simply support beam AB
with a concentrated load P
RA = P b / L RB = P a / L
for 0 < x < a
V = RA = P b / L
M = RA x = P b x / L
note that dM / dx = P b / L = V
for a < x < L
V = RA - P = - P a / L
M = RA x - P (x - a) = P a (L - x) / L
note that dM / dx = - P a / L = V
with Mmax = P a b / L
uniform load
consider a simple beam AB with a
uniformly distributed load of constant
intensity q
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
Shearing Force and Bending Moment - DAT 29
November, 2014
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
V - Diagram M - Diagram
Shearing Force and Bending Moment - DAT 30
Several concentrated loads Consider a simply support beam AB with a several concentrated pi;
November, 2014
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x 9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
Shearing Force and Bending Moment - DAT 31
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
November, 2014
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x 9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
V - Diagram
M - Diagram
Shearing Force and Bending Moment - DAT 32
Example 4
Construct the shear force and bending moment diagrams for the simple beam AB.
November, 2014
9
RA = RB = q L / 2
V = RA - q x = q L / 2 - q x
M = RA x - q x (x/2) = q L x / 2 - q x2 / 2
note that dM / dx = q L / 2 - q x / 2 = V
Mmax = q L2 / 8 at x = L / 2
several concentrated loads
for 0 < x < a1 V = RA M = RA x
M1 = RA a1
for a1 < x < a2 V = RA - P1
M = RA x - P1 (x - a1)
M2 - M1 = (RA - P1 )(a2 - a1)
similarly for others
M2 = Mmax because V = 0 at that point
Example 4-4
construct the shear-force and bending
-moment diagrams for the simple beam
AB
RA = q b (b + 2c) / 2L
RB = q b (b + 2a) / 2L
for 0 < x < a
V = RA M = RA x
Shearing Force and Bending Moment - DAT 33
Example 5
Construct the V- and M-diagram for the cantilever beam supported to P1 and P2.
November, 2014
10
for a < x < a + b
V = RA - q (x - a)
M = RA x - q (x - a)2 / 2
for a + b < x < L
V = - RB M = RB (L - x)
maximum moment occurs where V = 0
i.e. x1 = a + b (b + 2c) / 2L
Mmax = q b (b + 2c) (4 a L + 2 b c + b2) / 8L2
for a = c, x1 = L / 2
Mmax = q b (2L - b) / 8
for b = L, a = c = 0 (uniform loading over the entire span)
Mmax = q L2 / 8
Example 4-5
construct the V- and M-dia for the
cantilever beam supported to P1 and P2
RB = P1 + P2 MB = P1 L + P2 b
for 0 < x < a
V = - P1 M = - P1 x
for a < x < L
V = - P1 - P2 M
= - P1 x - P2 (x - a)
Shearing Force and Bending Moment - DAT 34
Example 6
Construct the V- and M-diagram for the cantilever beam supporting to a constant uniform load of intensity q.
November, 2014
11
Example 4-6
construct the V- and M-dia for the
cantilever beam supporting to a constant uniform
load of intensity q
RB = q L MB = q L2 / 2
then V = - q x M = - q x2 / 2
Vmax = - q L Mmax = - q L2 / 2
alternative method
x V - VA = V - 0 = V = -Н q dx = - q x 0 x M - MA = M - 0 = M = -Н V dx = - q x2 / 2 0
Example 4-7
an overhanging beam is subjected to a
uniform load of q = 1 kN/m on AB
and a couple M0 = 12 kN-m on midpoint
of BC, construct the V- and M-dia for
the beam
RB = 5.25 kN RC = 1.25 kN
shear force diagram
V = - q x on AB
V = constant on BC
Shearing Force and Bending Moment - DAT 35
Example 7
An overhanging beam is subjected to a uniform load of q = 1 kN/m on AB and a couple M0 = 12 kN-m on midpoint of BC. Construct the V- and M-diagram for the beam.
November, 2014
11
Example 4-6
construct the V- and M-dia for the
cantilever beam supporting to a constant uniform
load of intensity q
RB = q L MB = q L2 / 2
then V = - q x M = - q x2 / 2
Vmax = - q L Mmax = - q L2 / 2
alternative method
x V - VA = V - 0 = V = -Н q dx = - q x 0 x M - MA = M - 0 = M = -Н V dx = - q x2 / 2 0
Example 4-7
an overhanging beam is subjected to a
uniform load of q = 1 kN/m on AB
and a couple M0 = 12 kN-m on midpoint
of BC, construct the V- and M-dia for
the beam
RB = 5.25 kN RC = 1.25 kN
shear force diagram
V = - q x on AB
V = constant on BC
Shearing Force and Bending Moment - DAT 36
Thank You
November, 2014 Shearing Force and Bending Moment - DAT 37