1
Mechanics of Materials
2
3
§ 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
and curved rods
Exercise lessons about the internal force of bending
CHAPTER 4 INTERNAL FORCES IN BENDING
4
§ 4–1 平面弯曲的概念及梁的计算简图
§ 4–2 梁的剪力和弯矩
§ 4–3 剪力方程和弯矩方程 ·剪力图和弯矩图
§ 4–4 剪力、弯矩与分布荷载集度间的关系及应用
§ 4–5 按叠加原理作弯矩图
§ 4–6 平面刚架和曲杆的内力图

5
§ 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,
6
§ 4–1 平面弯曲的概念及梁的计算简图

1,弯曲, 杆受垂直于轴线的外力或外力偶矩矢的作用时,轴

2,梁,以 弯曲变形为主的

7
3).Practical examples in engineering about bending
8
3,工程实例
9
10
11
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 P2 q
12
4,平面弯曲,杆发生弯曲变形后，轴线仍然和外力在同一

P1 P2 q
13
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,
14

15
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,
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
16

1,构件本身的简化

2,载荷简化

3,支座简化
17
① Fixed hinged support
2 constraints,1 degree of
freedom,Such as the fixed
hinged support under bridges，
thrust ball bearing etc,
② Movable hinged support
1 constraint,2 degree of
freedom,Such as the movable
hinged support under the
bridge,ball bearing etc,
18
① 固定铰支座
2个约束,1个自由度。

② 可动铰支座
1个约束,2个自由度。

19
③ Rigidly 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
① Simple beam(or simply
supported beam)
M — Concentrated
force couple
q(x) — Distributed force
② Cantilever beam
A
20
③ 固定端
3个约束,0个自由度。

XA
YA
MA
4,梁的三种基本形式
① 简支梁
M — 集中力偶
q(x) — 分布力
② 悬臂梁
21
③ Overhanging beam
— Concentrated force P q — 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,
22
③ 外伸梁
— 集中力 P q — 均布力
5,静定梁与超静定梁

23
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 — Uniformly Distributed
force
24
[例 1]贮液罐如图示，罐长 L=5m，内径 D=1m，壁厚 t =10mm,

0.8m，外伸端长 1m，试求贮液罐 的计算简图。

25
L
gLAgLA
L
gV
L
mgq 2211 ??? ??????
r a d8 5 5131 0 6 0,,???
gRRgDt 2221 )]s i n(
2
1[ ?????? ????
gAgA 2211 ?? ??
(k N / m ) 9?
9, 81 0 0 0)]s i n 1 0 6, 3( 1, 8 5 50, 5
2
1
0, 5[ 3, 1 4897 8 0 00101143
2
2
?????
????????,..
q — Uniformly Distributed
force
26
L
gLAgLA
L
gV
L
mgq 2211 ??? ??????
r a d8 5 5131 0 6 0,,???
gRRgDt 2221 )]s i n(
2
1[ ?????? ????
gAgA 2211 ?? ??
(k N / m ) 9?
9, 81 0 0 0)]s i n 1 0 6, 3( 1, 8 5 50, 5
2
1
0, 5[ 3, 1 4897 8 0 00101143
2
2
?????
????????,..
q — 均布力
27
§ 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,
P a
P
l
YA
XA
RB
A
A B
B
Solution：① Determine
external forces
l
alP
YY
l
Pa
Rm
XX
ABA
A
)(
,0,0
0,0
?
????
??
??
?
28
§ 4–2 梁的剪力和弯矩

[举例 ]已知：如图,P,a,l。

P a
P
l
YA
XA
RB
A
A B
B

l
alP
YY
l
Pa
Rm
XX
A
BA
A
)(
,0
,0
0,0
?
?? ??
?? ??
?? ??
29
A B
P
YA
XA
RB m
m
x
② Determine internal forces—
method of section
xYMm
l
alP
YQY
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
30
A B
P
YA
XA
RB m
m
x
② 求内力 ——截面法
xYMm
l
alP
YQY
AC
A
??? ??
?
??? ??
,0
)(
,0
A
YA
Q
M
RB
P
M
Q
∴ 弯曲构件内力

1,弯矩,M

C
C
31
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,
② Bending 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(–)
32
2,剪力,Q

3.内力的正负规定,
① 剪力 Q,绕研究对象顺时针转为正剪力；反之为负。
② 弯矩 M：使梁变成凹形的为正弯矩；使梁变成凸形的为负弯矩。
Q(+) Q(–)
Q(–) Q(+)
M(+) M(+)
M(–) M(–)
33
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)(
q L xM
Mq L xFm iA
???
????
2,Examples
q qL
a b 1
1
2
2
qL
Q1
A
M1
Fig,（ b）
x1
34
[例 2],求图 （ a）所 示梁 1--1,2--2截面处的内力。
x y
qLQ
QqLY
???
????
1
1
0

1--1截面处截取的分离体

11
11
0)(
q L xM
Mq L xFm iA
???
????

q qL
a b 1
1
2
2
qL
Q1
A
M1

35
L)axq Q ???? 22 (
axqMq L x
Fm
iB
0)(
2
1
,0)(
2
222
????
??
Free body diagram of the left portion of
section 2—2 is shown in fig.（ b）,
)ax(qQqLY 022 ??????
2
2
22 )(2
1 q L xaxqM ???
x y 图（ a）
q qL
a b 1
1
2
2
qL
Q2
B
M2
x2

36
L)axq Q ???? 22 (
axqMq L x
Fm
iB
0)(
2
1
,0)(
2
222
????
??
2--2截面处截取的分离体如图（ c）
)ax(qQqLY 022 ??????
2
2
22 )(2
1 q L xaxqM ???
x y 图（ a）
q qL
a b 1
1
2
2
qL
Q2
B
M2
x2

37
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,
) ( x Q Q ? Shearing force equation
) ( x M M ? Bending moment equation
) ( x Q Q ?
Shearing-force diagram sketch of the shearing-force equation
) ( x M M ?
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
38
1,内力方程：内力与截面位置坐标（ x）间的函数关系式。
2,剪力图和弯矩图,
) ( x Q Q ? 剪力方程
) ( x M M ? 弯矩方程
) ( x Q Q ? 剪力图 的图线表示
) ( x M M ? 弯矩图 的图线表示
§ 4–3 剪力方程和弯矩方程 · 剪力图和弯矩图
39
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
??
??
② Write out the internal-
force equations
PL MPY OO ?? ;
P
③ Plot the internal-
force diagrams
Q(x)
M(x)
x
x
P
–PL
YO
L
M(x)
x Q(x)
MO

40
[例 3] 求下列各图示梁的内力方程并画出内力图。
PY)x(Q O ??

)Lx(P
MxY)x(M OO
??
??
② 写出内力方程
PL MPY OO ?? ;
P
YO
L
③ 根据方程画内力图
M(x)
x Q(x)
Q(x)
M(x)
x
x
P
–PL
MO

41
Solution：① Write out the
internal-force equations
② Plot the internal-
force diagram
qx)x(Q ??
2
2
1 qx)x(M ??
L
q
M(x)
x Q(x)
Q(x)
x
– qL
2
2qL
?
M(x) x

42

② 根据方程画内力图
qx)x(Q ??
2
2
1 qx)x(M ??
L
q
M(x)
x Q(x)
Q(x)
x
M(x) x
– qL
2
2qL
?

43
)3(6 220 xLLq)x(Q ??
Solution：① Determine the
reactions of the supports
② Write out the internal-
force equations
3 ; 6
00 Lq RLqR
BA ??
q0
RA
③ Plot the internal-
force diagrams
RB
L
)xL(LxqxM 2206)( ??x
L33
Q(x)
x
6
2
0Lq
3
0Lq
27
3 20 Lq
M(x)

44
)3(6 220 xLLq)x(Q ??

② 内力方程
3 ; 6
00 Lq RLqR
BA ??
q0
RA
③ 根据方程画内力图
RB
L
)xL(LxqxM 2206)( ??x
L33
Q(x)
x
6
2
0Lq
3
2
0Lq
27
3 20 Lq
M(x)

45
1,Relations among the shearing force、
the bending moment and the the
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
dx x
q(x)
q(x)
M(x)+d M(x)
Q(x)+d Q(x)
Q(x)
M(x)
dx
A
y
? ? ? ?xq
x
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,
46

? ? 0dd
0
????
??
)x(Q)x(Qx)x(q)x(Q
Y
)x(Qx)x(q dd ?
§ 4–4 剪力、弯矩与分布荷载集度间的关系及应用
dx x
q(x)
q(x)
M(x)+d M(x)
Q(x)+d Q(x)
Q(x)
M(x)
dx
A
y ? ?
? ?xqx xQ ?dd

47
q(x)
M(x)+d M(x)
Q(x)+d Q(x)
Q(x)
M(x)
dx
A
y
0)](d)([)()) ( d(21)d( 0,)( 2 ??????? xMxMxMxxqxxQFm iA
)(d )(d xQx 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 )(d 2
2
xqx xM ?
Relation between the bending
moment and the indensity of the
48
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 ?????
??
xMxMxMxxqxxQ
Fm iA
)(d )(d xQx xM ?

)(d )(d 2
2
xqx xM ?

49
2,Relations between the shearing force,the bending moment and the external load
Ex
ter
na
l fo
rce
No external-force
segment
segment Concentrated force Concentrated couple
q=0 q>0 q<0
Ch
ar
ac
ter
ist
ics
of
Q
-
dia
gr
am
Ch
ar
ac
ter
ist
ics
of
M
-
dia
gr
am
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
C
Q1
Q2
Q1–
Q2=P
Sudden change from
the left to right
x
Q
C
No change
Inclined straight line
x
M
Increasing function
x
M
Decreasing function
curves
x
M
Tomb-like
x
M
Basin-like
Flex from the left to
the right
Sudden change from the
left to the right O
pp
os
ite
to
m
x
M
Flex opposite to P
M
x
M1
M2
mMM ?? 21
50

q=0 q>0 q<0
Q

M

C
P
C
m

x
Q
Q>0
Q
Q<0
x

x
Q
x
Q

x
Q
C
Q1
Q2
Q1–
Q2=P

x
Q
C

x
M

x
M

x
M

x
M

m

x
M

M
x
M1
M2
mMM ?? 21
51
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,
52

[例 4] 用简易作图法画图示梁的内力图。

a a
qa q
A
53 2
2
3
0 qaM;Q ???
0 ; ??? MqaQ
2 ; qaMqaQ ????
2
2
3; 0 qaMQ ???
a a
qa q
A
Left end,
Shape of the curve is determined
according to
)(d )(d xQx xM ? )(d )(d 2
2
xqx xM ?；
? ? ? ?xq
x
xQ ?
d
d ；
And the law of the point acted
by concentrated force,
Partition
point A,
Stationary
point of M,
Right end,
Q x
2
2
3 qa
qa2

qa

x
M
54
2
2
3
0 qaM;Q ???
0 ; ??? MqaQ
2 ; qaMqaQ ????
2
2
3; 0 qaMQ ???
a a
qa q
A

)(d )(d xQx xM ? )(d )(d 2
2
xqx xM ?；
? ? ? ?xq
x
xQ ?
d
d ；

M 的驻点,

Q x
2
2
3 qa
qa2

qa

x
M
55
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
qaRqaR
DA
0;
2
??? MqaQ
Left end A,
2
2
1;
2
qaMqaQ ????
2
2
1;
2
qaMqaQ ???
Right of
point B,
2
2
1;
2
qaMqaQ ????
Left of
point C,
Stationary
point of M,283; 0 qaMQ ???
2
2
1;
2
qaMqaQ ???
Right of
point C,
0 ;
2
1 ??? MqaQRight end D,
q qa2
qa RA RD
Q x
qa/2 qa/2
qa/2
– –
+
A B
C D
qa2/2
x M qa2/2
qa2/2 3qa2/8

+
Left of
point B,
56
[例 5] 用简易作图法画下列各图示梁的内力图。

2 ; 2
qaRqaR
DA
0;
2
??? MqaQ

2
2
1;
2
qaMqaQ ????
B点左,
2
2
1;
2
qaMqaQ ???
B点右,
2
2
1;
2
qaMqaQ ????
C点左,
M 的驻点,
2
8
3; 0 qaMQ ???
2
2
1;
2
qaMqaQ ???
C点右,
0 ;
2
1 ??? MqaQ右端点 D,
q qa2
qa RA RD
Q x
qa/2 qa/2
qa/2
– –
+
A B
C D
qa2/2
x M qa2/2
qa2/2 3qa2/8

+
57
§ 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,
58
§ 4–5 按叠加原理作弯矩图

)()()()( 221121 nnn PQPQPQPPPQ ????????????????
)()()()( 221121 nnn PMPMPMPPPM ????????????????

59
2,Structural members in mechanics of material is of small
deformation and linear elasticity,and must obey this principle
—— method of superposition
Steps,
① Plot respectively the diagram of the bending moment of
the beam under the separate action of each external load；
② Sum up the corresponding longitudinal coordinates
(Attention,do not simply piece together figures.）
60

——叠加方法

①分别作出各项荷载单独作用下梁的弯矩图；
②将其相应的纵坐标叠加即可（注意：不是图形的简单

61
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
2
Pa
+
+
+
2
2qa
22
2qaPa
?
=
+
62
[例 6]按叠加原理作弯矩图 (AB=2a，力 P作用在梁 AB的中点处）。
q
q
P
P =
+
A
A
A
B
B
B x
M2
x
M1
x
M
2
Pa
+
+
+
2
2qa
22
2qaPa
?
=
+
63
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,
64

65
Example 7 Plot internal-force diagrams of the beams shown in the
following figure,P PL
P
PL
L L
L L
L L
0.5P
0.5P
0.5P
0.5P
P 0
Q x
Q1 x
Q2 x

0.5P
0.5P
0.5P

+

P
66
[例 7] 作下列图示梁的内力图。
P PL
P
PL
L L
L L
L L
0.5P
0.5P
0.5P
0.5P
P 0
Q x
Q1 x
Q2 x

0.5P
0.5P
0.5P

+

P
67
P PL
P
PL
L L
L L
L L
0.5P
0.5P
0.5P
0.5P
P 0 M
x
M1
x
M2
x
0.5PL
PL
0.5PL

+
+
0.5PL
+
68
P PL
P
PL
L L
L L
L L
0.5P
0.5P
0.5P
0.5P
P 0 M
x
M1
x
M2
x
0.5PL
PL
0.5PL

+
+
0.5PL
+
69
Example 8 Correct the mistakes in the following internal-force diagrams,
a 2a a
q qa2 A
B
Q x
x
M
– –
+
+
qa/4 qa/4
3qa/4
7qa/4
qa2/4
49qa2/32
3qa2/2
5qa2/4
4
7;
4
qa
R
qa
R
B
A
?
?
RA RB
70
[例 8] 改内力图之错。
a 2a a
q qa2 A
B
Q x
x
M
– –
+
+
qa/4 qa/4
3qa/4
7qa/4
qa2/4
49qa2/32
3qa2/2
5qa2/4
4
7;
4
qaRqaR
BA ??
71
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 1m 2m
2
3
1
5kN 1kN
q=2kN/m
+

+
x
+
1
1
1.25

72
[例 9] 已知 Q图，求外载及 M图（梁上无集中力偶）。
Q(kN)
x
1m 1m 2m
2
3
1
5kN 1kN
q=2kN/m
+

+
M(kN·m)
x
+
1
1
1.25

73
§ 4–6 THE INTERNAL-FORCE DIAGRAMS OF
THE PLANAR RIGID FRAMES AND CURVED RODS
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,
74
§ 4–6 平面刚架和曲杆的内力图

1,平面刚架,同一平面内，不同取向的杆件，通过杆端相

2,内力图规定,

75
Example 10 Try to plot the internal-force diagrams of the rigid frame
shown in the figure,
P1 P
2
a
l
A
B C
– N-diagram
Q -diagram
+
+
P 1
M -diagram
76
[例 10] 试作图示刚架的内力图。
P1 P
2
a
l
A
B C –
N 图
P2
+
Q 图
P 1
+
P1
P1a
M 图
P 1
a
P1a+ P2 l
77
Example 11 As shown in the figure,P and R are known,try to plot internal
force diagrams of Q,M and N,
O
P R
?
x
Solution,set up polar coordinates,O is
the pole and OB is polar axis,? denotes the
position of the section m-m,
)(0 )co s1()co s()( ????? ??????? PRRRPPxM
)(0 c o s)( 2 ???? ???? PPN
)(0 s i n)( 1 ???? ???? PPQ
A B
2,Planar rod,Rod that the axis is of a planar curve,
Method to plot internal-force diagram of a curved rod is the same as that of
the planar rigid frame,
78

[例 11] 已知：如图所示,P及 R 。试绘制 Q,M,N 图。
O
P R
?
x

)(0 )co s1()co s()( ????? ??????? PRRRPPxM
)(0 c o s)( 2 ???? ???? PPN
)(0 s i n)( 1 ???? ???? PPQ
A B
79
O
P R
?
x
)(0 )co s1()co s()( ????? ??????? PRRRPPxM
)(0 c o s)( 2 ???? ???? PPN
)(0 s i n)( 1 ???? ???? PPQ
A B
A B
O
M图
O O
+ Q图 N图
2PR
P P
– +
80
O
P R
?
x
)(0 )co s1()co s()( ????? ??????? PRRRPPxM
)(0 c o s)( 2 ???? ???? PPN
)(0 s i n)( 1 ???? ???? PPQ
A B
A B
O
M-diagram
O O
+ Q -diagram N-diagram
2PR
P P
– +
81
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
? ? ? ??? ???? jiA PPQ
? ? ? ??? ?? )( )( jAiAA PmPmM
82

? ? ? ?? ??? ?? iiA PPQ
? ? ? ????? )( )( iAiAA PmPmM
83
)(d )(d 2
2
xqx xM ?
Relations among the shearing force,the bending moment and
q(x) ? ? ? ?xqx xQ ?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,
84
)(d )(d 2
2
xqx xM ?

q(x) ? ? ? ?
xqx xQ ?dd
)(d )(d xQx xM ?

85
3,Principle 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
86

)()()()( 221121 nnn PQPQPQPPPQ ????????????????
)()()()( 221121 nnn PMPMPMPPPM ????????????????

87
5,Relations between the shearing force,the bending moment and the external load
Ex
ter
na
l fo
rce
No external-force
segment
segment Concentrated force Concentrated couple
q=0 q>0 q<0
Ch
ar
ac
ter
ist
ics
of
Q
-
dia
gr
am
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
C
Q1
Q2
Q1–
Q2=P
Sudden change from
the left to right
x
Q
C
No change
Inclined straight line
x
M
Increasing function
x
M
Decreasing function
curves
x
M
Tomb-like
x
M
Basin-like
Flex from the left to
the right
Sudden change from the
left to the right O
pp
os
ite
to
m
x
M
Flex opposite to P
M
x
M1
M2
mMM ?? 21
88

q=0 q>0 q<0
Q

M

C
P
C
m

x
Q
Q>0
Q
Q<0
x

x
Q
x
Q

x
Q
C
Q1
Q2
Q1–
Q2=P
x
Q
C

x
M

x
M

x
M
x
M
x
M
x
M

M1
M2

m

mMM ?? 21
89
Example 1 Plot the bending-moment diagrams of the beam shown in the
following figure,
2P
a a P
=
2P
P
+
x
M
x M
1
x
M2
=
+

+
+
2Pa
2Pa
Pa
(1)
90
[例 1] 绘制下列图示梁的弯矩图。
2P
a a P
=
2P
P
+
x
M
x M
1
x
M2
=
+

+
+
2Pa
2Pa
Pa
(1)
91
(2)
a
a
q
q
q
q
=
+ x M1
=
x M
+

+

x M
2
3qa2/2
qa2/2
qa2
92
(2)
a
a
q
q
q
q
=
+ x M1
=
x M
+

+

x M
2
3qa2/2
qa2/2
qa2
93
(3) P
L/2 L/2
PL/2
=
+
P
x M
2
x M
=
+
PL/2
PL/4
PL/2
x M
1

+

PL/2
94
(3) P
L/2 L/2
PL/2
=
+
P
x M
2
x M
=
+
PL/2
PL/4
PL/2
x M
1

+

PL/2
95
(4) 50kN
2m 2m
20kNm
=
+
x
M2
x M
=
+
20kNm
50kNm
x M1
20kNm
50kN
20kNm 20kNm
+
+

20kNm
30kNm
20kNm
96
(4) 50kN
a a
20kNm
=
+
x
M2
x M
=
+
20kNm
50kNm
x M1
20kNm
50kN
20kNm 20kNm
+
+

20kNm
30kNm
20kNm
97 y
z h
b
)4(2 2
2
yhIQbIQS
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
??
98 y
z h
b )
4(2
2
2
yhIQbIQS
zz
z ???
?
?

[例 2]结构如图，试证明,
（ 1）任意横截面上的剪应力的合力等于该面的剪力；
（ 2）任意横截面上的正应力的合力矩等于该面的弯矩；
（ 3）过高度中点做纵截面，那么，此纵截面上的剪应力的

q

zI
My
??
99
??
?
??
h
h zA
yby
h
I
Q
A
5.0
5.0
2
2
d )
4
(
2
d?
MIIMAIMyM z
z
h.
h,z
z ????
?
50
50
2
d
(2) Resultant shearing force in the cross section is,
Q
hh
I
Qb
z
???? ])
2
(
3
2
4
[
2
3
3
(3) Resultant force couple
100
??
?
??
h
h zA
yby
h
I
Q
A
5.0
5.0
2
2
d )
4
(
2
d?
MIIMAIMyM z
z
h.
h,z
z ????
?
50
50
2
d
(2) 横截面上的合剪力为,
Q
hh
I
Qb
z
???? ])
2
(
3
2
4
[
2
3
3
(3) 合力偶
101
)(bhqx.A xQ,?????? 51)(51m a x??
h
qL
xqx
h
AQ
LL
AB 4
3
d)(
2
3
d
2
00
???? ?? ?
z
A W
AMAN
22
1 1m a x
1m a x 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
102
)(bhqx.A xQ,?????? 51)(51m a x??
h
qLx)qx(
hAQ
LL
AB 4
3d
2
3d 2
00
???? ?? ?
z
A W
AMAN
22
1 1m a x
1m a x 1 ?? ?
1AAB NQ ?
(4)中面上的剪应力为,

(5) 纵 截面上的合剪力 大小 为,
? max ??
h
qLbh
bh
qL
4
3
2
6
22
1 2
2
2
??
x L
103
Chapter 4 Exercises
1.Try to list some members in bending
according to your experiences and simplify them
into various kinds of beams,
2.Try to plot the bending-moment diagram of
the beam by the superposition method,
2
2
2
2
2
2
104

2
2
2
2
2
2
2
105
3.Plot the Q-diagram and the M-diagram of the
composite beam with a middle hinge shown in the
figure,
qa2
qa2 /2
106

qa2
qa2 /2
107
4,Lift an equal-section beam with the weight q(N/m),Ask,what is
the reasonable location x of the lift point?
Note,make the maximum positive bending moment equal to the absolute
value of the maximum negative bending moment,
Solution,
As
We have
Take
(It is meaningless for x to take the negative value.)
2
2qx
M A ??
)2(242 xLqLLLqM C ????
CA MM ?
044 22 ??? LLxx
Lx 2 21 ???
Lx 20 7.0?
ql/2 ql/2
108

（ x为负值无意义 ）
2
2qx
M A ??
)2(242 xLqLLLqM C ????
CA MM ??
044 22 ???? LLxx
Lx 2 21 ???
Lx 20 7.0?
ql/2 ql/2
109
110