Shear Force And Bending Moment In Beams22
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1
Mechanics of Materials
Chaper4
Shear Force and Bending Moment in Beams
HENAN UNIVERSITY OF SCIENCE & TECHNOLOGY
Edited by YANG MIN-xian
2
§4–1 Concepts of planar bending and calculation sketch of the beam
§4–2 The shearing force and bending moment of the beam
§4–3 The shearing-force and bending-moment equations · the shearing-
force and bending-moment diagrams
§4–4 Relations among the shearing force 、 the bending moment and the
density of the distributed load and their applications
§4–5 Plot the bending-moment diagram by the theorem of
superpositiom
§4–6 The internal-force diagrams of the planar rigid theorem frames
CHAPTER 4 Shear Force and Bending Moment in BeamsCHAPTER 4 Shear Force and Bending Moment in Beams
3
§4–1 CONCEPTS OF PLANAR BENDING AND
CALCULATION SKETCH OF THE BEAM
§4–1 CONCEPTS OF PLANAR BENDING AND
CALCULATION SKETCH OF THE BEAM
1 、 CONCEPTS OF BENDING
1). BENDING: The action of the external force or external the couple vector
perpendicular to the axis of the rod makes the axis of the rod change into
curve from original straight lines, this deformation is called bending.
2).BEAM ： The member of
which the deformation is
mainly bending is generally
called beam.
4
3).Practical examples in engineering about bending
5
6
4).Planar bending ： After deformation the curved axis of the beam is still in the same plane with the external forces.
Symmetric bending （ as shown in the following figure ）— a special example of the planar bending.
The plane of symmetry
M
P1 P2q
7
Unsymmetrical bending— if a beam does not possess any plane of
symmetry, or the external forces do not act in a plane of symmetry of the
beam with symmetric planes, this kind of bending is called unsymmetrical
bending. In later chapters we will mainly discuss the bending stresses and
deformations of the beam under symmetric bending.
8
2 、 Calculation sketch of the beam
In general supports and external forces of the beam are very complex.
We should do some necessary simplification for them for our convenient
calculation and obtain the calculation sketch.
1). Simplification of the beams
In general case we take the place of the beam by its axis.
2). Simplification of the loads
The loads (including the reaction) acting on the beam may be reduced into
three types ： concentrated force 、 concentrated force couple and distributed
force.3). Simplification of the supports
9
①Fixed hinged support
2 constraints ， 1 degree of
freedom. Such as the fixed
hinged support under bridges ，
thrust ball bearing etc.
②Movable hinged support
1 constraint ， 2 degree of
freedom. Such as the movable
hinged support under the
bridge ， ball bearing etc.
10
③Rigidly fixed end
3 constraints ， 0 degree of
freedom. Such as the support of
diving board at the swimming pool ，
support of the lower end of a
wooden pole.
XA
YA
MA
4) Three basis types of beams
①Simple beam(or simply
supported beam)
M —Concentrated force couple
q(x)—Distributed force
②Cantilever beam
A
11
③Overhanging beam—Concentrated
forcePq— Uniformly
distributed force
5). Statically determinate and statically indeterminate beams
Statically determinate beams ： Reactions of the beam can be determined only
by static equilibrium equations ， such as the above three kinds of basic beams.
Statically indeterminate beams ： Reactions of the beam cannot be
determined or only part of reactions can be determined by static equilibrium
equations.
12
Example 1 A stock tank is shown in the figure. Its length is L=5m ， its
inside diameter is D=1m ， thickness of its wall is t =10mm. Density of steel is
7.8g/cm³. Density of the liquid is 1g/cm³. Height of the liquid is 0.8m. Length of
overhanging end is 1m. Try to determine the calculation sketch of the stock tank.
Solution ：
q— UniformlyDistributed force
13
LgLAgLA
LgV
Lmgq 2211
rad85513106 0 ..
gRRgDt 222
1 )]sin(2
1[
gAgA 2211 (kN/m) 9
9.81000)]sin106.3(1.8550.521
0.5[3.148978000101143
2
2
...
q— UniformlyDistributed force
14
§4–2 THE SHEARING FORCE AND BENDING MOMENT OF THE BEAM
§4–2 THE SHEARING FORCE AND BENDING MOMENT OF THE BEAM
1 、 Internal force in bending ： Example Knowing conditions are P，
a ， l , as shown in the figure.
Determine the internal forces on the
section at the distance x to the end A.
Pa
P
l
YA
XA
RB
A
A B
B
Solution ：① Determine external forces
l
alPYY
l
PaRm
XX
ABA
A
)( , 0 , 0
0 , 0
15
A BP
YA
XA
RBm
m
x
②Determine internal forces— method of section
xYMm
l
alPYQY
AC
A
, 0
)( , 0
A
YA
Q
M
RB
P
M
Q
Internal forces of the beam in bending
Shearing force
Bending moment
1). Bending moment ： M
Moment of the internal force couple with
the acting plane in the cross-section
perpendicular to the section when the beam is
bending.
C
C
16
2). Shearing force ： Q
Internal force which the acting line in the cross-section parallel to the section, when the beam is
bending.
3).Sign conventions for the internal forces:
① Shearing force Q: It is positive when it results in a clockwise rotation with respect to the object under consideration, otherwise it is negative.
②Bending moment M ： It is positive when it tends to bend the portion concave upwards, otherwise it is negative.
Q(+) Q(–)
Q(–)Q(+)
M(+) M(+)
M(–) M(–)
17
Example 2 ： Determine the internal forces acting on sections 1—1 and 2—2
section as shown in fig.(a).
qLQ
QqLY
1
1
0
Solution ： Determine internal
forces by the method of section.
Free body diagram of the left portion of
section 1—1 is shown in fig. （ b ） .
Fig. （ a ）
11
11
0)(
qLxM
MqLxFm iA
2 、 Examples
qqL
a b1
12
2
qL
Q1
A
M1
Fig. （ b ）x1
18
L)axq Q 22 (
axqMqLx
Fm iB
0)(21
, 0)(
2222
Free body diagram of the left portion of section 2—2 is shown in fig. （ b ） .
)ax(qQqLY 022
22
22 )(21 qLxaxqM
xy
图（ a ）
qqL
a b1
12
2
qL
Q2
B
M2
x2
图（ c ）
19
1. Internal-force equations: Expressions that show the internal forces as functions of the position x of the section..
2. The shearing-force and bending-moment diagrams:
)(xQQ Shearing force equation
)(xMM Bending moment equation
)(xQQShearing-force diagram sketch of the shearing-force equation
)(xMM
Bending Moment diagram sketch of the bending-moment equation
§4–3 THE SHEARING-FORCE AND BENDING-MOMENT EQUATIONS THE SHEARING-FORCE AND BENDING-MOMENT DIAGRAMS§4–3 THE SHEARING-FORCE AND BENDING-MOMENT EQUATIONS THE SHEARING-FORCE AND BENDING-MOMENT DIAGRAMS
20
Example 3 Determine the internal-force equations and plot the diagrams of the
beam shown in the following figure.
PY)x(Q O
Solution ：① Determine the reactions of the supports
)Lx(P
MxY)x(M OO
②Write out the internal- force equations
PL MPY OO ;
P
③Plot the internal- force diagrams
Q(x)
M(x)
x
x
P
–PL
YO
L
M(x)
xQ(x)
MO
⊕
○
21
Solution ：① Write out the internal-force equations
②Plot the internal-force diagram
qx)x(Q
2
21 qx)x(M
L
q
M(x)
xQ(x)
Q(x)
x
– qL
2
2qL
M(x) x
⊕
○
22
)3(6
220 xLL
q)x(Q
Solution ：① Determine the reactions of the supports
② Write out the internal- force equations
3 ;
600 Lq
RLq
R BA
q0
RA
③ Plot the internal- force diagrams
RB
L
)xL(Lxq
xM 220
6)(
x
L33
Q(x)
x6
20Lq
30Lq
273 2
0Lq
M(x)
⊕
⊕
○
23
1 、 Relations among the shearing force 、the bending moment and the the distributed load
By analysis of the equilibrium of the
infinitesimal length dx ， we can get
0dd
0
)x(Q)x(Qx)x(q)x(Q
Y
)x(Qx)x(q dd
§4–4 RELATIONS AMANG THE SHEARING FORCE, THE BENDING MOMENT
AND THE INDENSITY OF THE DISTRIBUTED LOAD AND THEIR APPLICATIONS §4–4 RELATIONS AMANG THE SHEARING FORCE, THE BENDING MOMENT
AND THE INDENSITY OF THE DISTRIBUTED LOAD AND THEIR APPLICATIONS
dxx
q(x)
q(x)
M(x)+d M(x)
Q(x)+d Q(x)
Q(x)
M(x)
dx
A
y xqx
xQ
d
d
Slope of the tangential line at a point in the
shearing-force diagram is equal to the intensity
of the distributed load at the same point.
24
q(x)
M(x)+d M(x)
Q(x)+d Q(x)
Q(x)
M(x)
dx
A
y
0)](d)([)())(d(2
1)d( 0,)( 2 xMxMxMxxqxxQFm iA
)(d
)(dxQ
x
xM
Slope of the tangential line at a point in the bending-moment diagram is
equal to the magnitude of the shearing force at the same point.
)(d
)(d2
2
xqx
xM
Relation between the bending moment and the indensity of the distributed load ：
25
2 、 Relations between the shearing force 、 the bending moment and the external load
Extern
al force
No external-force segment
Uniform-load segment
Concentrated force Concentrated couple
q=0 q>0 q<0
Ch
aracteristics of Q-
diagram
Ch
aracteristics of M-
diagram
C
P
C
m
Horizontal straight line
x
Q
Q>0
Q
Q<0
x
Inclined straight line
Increasing function
x
Q
x
Q
Decreasing function
x
Q
CQ1
Q2
Q1–Q2=P
Sudden change from the left to right
x
Q
C
No change
Inclined straight line
x
MIncreasing function
x
MDecreasing function
curves
x
MTomb-like
x
MBasin-like
Flex from the left to the right
Sudden change from the left to the right
Op
posite tom
x
MFlex opposite to P
M
xM1M
2
mMM 21
26
Simple method to plot the diagram: The method to plot the diagrams by
using the relation between the internal forces and the external forces and values
of the internal forces at some special points.
Example 4 Plot the internal force diagrams of the beams shown in the
following figures by the simple method to plot the diagram.
Solution:
Special points:a a
qa q
A
Plot the diagram by using the relation between the internal forces and the external forces and the internal force values at
some special points of the beam.
End point 、 partition point （ the point at which external forces changed ） and
stationary point etc.
27
2
2
30 qaM;Q
0 ; MqaQ
2 ; qaMqaQ
2
2
3; 0 qaMQ
a a
qa q
ALeft end ：Shape of the curve is determined according to
)(d
)(d xQx
xM )(d
)(d2
2
xqx
xM
； xqxxQ
dd
；
And the law of the point acted by concentrated force.
Partition
point A ：Stationary
point of M ：
Right end ：
Q x
2
23 qa
qa2
–
qa
–
xM
28
Example 5 Plot the internal-force diagrams of the beams shown in the
following figures by the simple method to plot the diagram.
Solution ：Determine reactions
2
; 2
qaR
qaR DA
0;2
Mqa
QLeft end A ：
2
2
1;
2qaM
qaQ
2
2
1;
2qaM
qaQ Right of
point B ：2
2
1;
2qaM
qaQ Left of
point C ：Stationary
point of M ：2
8
3; 0 qaMQ
2
2
1;
2qaM
qaQ Right of
point C ：0 ; 2
1 MqaQRight end D ：
q qa2
qaRA RD
Q x
qa/2 qa/2
qa/2
– –
+
A B
C D
qa2/2
xM qa2/2
qa2/23qa2/8–
+
Left of point B ：
29
§4–5 PLOT THE DIAGRAM OF BENDING MOMENT BY THE THEOREM OF SUPERPOSITIOM
§4–5 PLOT THE DIAGRAM OF BENDING MOMENT BY THE THEOREM OF SUPERPOSITIOM
1 、 Theorem of superposition ： Internal forces in the structure due to simultaneous action of many forces are
equal to algebraic sum of the internal forces due to separate action of each force.
)()()()( 221121 nnn PQPQPQPPPQ
)()()()( 221121 nnn PMPMPMPPPM
Applying condition ： Relation between the parameters
(internal forces 、 stresses 、 displacements ） and the
external forces must be linear, that is they satisfy Hooke’s
law.
30
2 、 Structural members in mechanics of material is of small
deformation and linear elasticity, and must obey this principle
—— method of superposition
Steps ： ①Plot respectively the diagram of the bending moment of
the beam under the separate action of each external load ； ②Sum up the corresponding longitudinal coordinates
(Attention: do not simply piece together figures. ）
31
Example 6 Plot the diagram of bending moment by the principle of superposition.
(AB=2a ， force P is acting at the middle point of the beam AB. ） P
q
q
P =+
A
A
A
B
B
B x
M2
x
M1
x
M
2Pa
+
+
+
2
2qa
22
2qaPa =+
32
3 、 Applications of symmetry and antisymmetry ： For the symmetric structure under the action of symmetric loads
the diagram of its shearing stress Q is antisymmetric and the
diagram of the bending moment M is symmetric. For the symmetric
structure under the action of antisymmetric loads the diagram of its
shearing stress Q is symmetric and the diagram of the bending
moment M is antisymmetric.
33
Example 7 Plot internal-force diagrams of the beams shown in the
following figure.PPL
P
PL
L L
L L
L L
0.5P
0.5P
0.5P
0.5P
P0
Qx
Q1 x
Q2 x
–
0.5P
0.5P
0.5P
–+
–P
34
PPL
P
PL
L L
L L
L L
0.5P
0.5P
0.5P
0.5P
P0 M
x
M1
x
M2
x
0.5PL
PL
0.5PL–
+
+
0.5PL
+
35
Example 8 Correct the mistakes in the following internal-force diagrams.
a 2aa
qqa2
AB
Qx
x
M
– –
+
+qa/4 qa/4
3qa/4
7qa/4
qa2/4
49qa2/32
3qa2/2
5qa2/4
4
7
;4qa
R
qaR
B
A
RA RB
36
Example 9 Knowing Q-diagram, determine external loads and M-diagram (Therefore no concentrated force couples acted on the beam).
M(kN·m)
Q(kN)x
1m 1m2m
2
3
1
5kN 1kN
q=2kN/m
+
–
+
x+
1
1
1.25
–
37
§4–6 THE INTERNAL-FORCE DIAGRAMS OF THE PLANAR RIGID FRAMES§4–6 THE INTERNAL-FORCE DIAGRAMS OF THE PLANAR RIGID FRAMES
1 、 Planar rigid frame
1). Planar rigid frame ： Structure made from rods of different direction
that are mutually connected in rigidity at their ends in the same plane.
Characteristics ： There are internal forces Q, M and N in each rod.2). Conventions to plot diagram of internal forces ： Bending-moment diagram ： Plot it at the side where fibers are
elongated and not mark the sign of positive or negative.
Shearing-force and axial-force diagrams ： May be plotted
at any side of the frame （ In common the diagram with positive value is
plotted outside the frame ）， but must mark the signs of positive and
negative.
38
Example 10 Try to plot the internal-force diagrams of the rigid frame
shown in the figure.P1P2
a
l
A
B C
– N-diagram
Q -diagram
P2 +
P1
+
P1
M -diagram
P1a
P1a
P1a
+ P
2 l
39
1 、 Method to determine directly the internal forces ： When we determine the internal forces in an arbitrary section A, we can take
the left part of section A as our study object and use the following formulas to
calculate internal forces. where Pi and Pj are respectively upward and
downward external forces acted on the left part.
DIAGRAMS OF SHEARING STRESSES AND BENDING MOMENTS
EXERCISE LESSONS ABOUT INTERNAL FORCES OF BENDING EXERCISE LESSONS ABOUT INTERNAL FORCES OF BENDING
jiA PPQ
)( )( jAiAA PmPmM
40
)(d
)(d2
2
xqx
xM
Relations among the shearing force 、 the bending moment and the external load ：
q(x)
xqxxQ
dd
)(d
)(d xQx
xM
2 、 Simple method to plot the diagram:
The method to plot the diagrams by using the relation between the internal forces
and the external forces and using values of the internal forces at some special
points.
41
33 、、 Principle of superpositionPrinciple of superposition ： Internal forces in the structure due to simultaneous action of many forces are
equal to the algebra sum of the internal forces due to separate action of each
force.
)()()()( 221121 nnn PQPQPQPPPQ
)()()()( 221121 nnn PMPMPMPPPM 4 、 Applications of symmetry and antisymmetry ： For the symmetric structure under the action of symmetric loads the diagram of its
shearing stress is antisymmetric and the diagram of bending moment is symmetric.
For the symmetric structure under the action of antisymmetric loads the diagram of
its shearing stress is symmetric and the diagram of bending moment is
antisymmetric
42
5 、 Relations between the shearing force, the bending moment and the external load
Extern
al force
No external-force segment
Uniform-load segment
Concentrated force Concentrated couple
q=0 q>0 q<0
Ch
aracteristics of Q-
diagram
C
P
C
m
Horizontal straight line
x
Q
Q>0
Q
Q<0
x
Inclined straight line
Increasing function
x
Q
x
Q
Decreasing function
x
Q
CQ1
Q2
Q1–Q2=P
Sudden change from the left to right
x
Q
C
No change
Inclined straight line
x
MIncreasing function
x
MDecreasing function
curves
x
MTomb-like
x
MBasin-like
Flex from the left to the right
Sudden change from the left to the right
Op
posite tom
x
MFlex opposite to P
M
xM1M
2
mMM 21
43
Example 1 Plot the bending-moment diagrams of the beam shown in the
following figure.2P
a aP
=
2P
P
+
xM
xM1
x
M2
=+
–
+
+
2Pa
2Pa
Pa
(1)
44
(2)a
a
q
q
q
q
=+
xM1
=
xM
+
–
+
–
xM2
3qa2/2
qa2/2
qa2
45
(3) P
L/2 L/2
PL/2
=+
P
xM2
xM
=+
PL/2
PL/4
PL/2
xM1
–
+
–PL/2
46
(4) 50kN
2m 2m
20kNm
=+
xM2
xM
=+
20kNm
50kNm
xM1
20kNm
50kN
20kNm20kNm
+
+
–
20kNm
30kNm
20kNm
47y
z
h
b )4
(2
22
yhI
QbIQS
zz
z
Solution: （ 1 ） Shearing stress on the cross section is
Example 2 The structure is shown in the figure. Try to prove ： (1 ） resultant of
the shearing stresses in an arbitrary cross section is equal to the shearing force in
the same section ；（ 2 ） Resultant moment of the normal stresses in an arbitrary
cross section is equal to the bending moment in the same section ；（ 3 ） which
force can balance the resultant of the shearing stress in the longitudinal section at
middle height balanced ?.
q
Normal stress on the cross section is
zI
My
48
h
h zA
ybyh
I
QA
5.0
5.0
22
d )4
(2
d
MIIMA
IMyM z
z
h.
h. zz
50
50
2
d
(2) Resultant shearing force in the cross section is ：
Qhh
I
Qb
z
])2
(3
2
4[
23
3
(3) Resultant force couple
49
)(bhqx.
AxQ. 51)(51max
h
qLxqx
hAQ
LL
AB 4
3d)(
2
3d
2
00
zA W
AMAN
22
1 1max1max 1
1AAB NQ
(4)Shearing stress in the
middle longitudinal section is ：
Resultant of the shearing stress in the longitudinal section is balanced by resultant of the normal stress in the right-side section.
(5)Resultant of the shearing stress in the longitudinal section is ：
max
h
qLbh
bh
qL
4
3
2
6
22
1 2
2
2
x L
50
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