1
Mechanics of Materials
2
3
§ 6–4 Determine deflections and angles of rotation of the beam by the
principle of superposition
§ 6–5 Ckeck the rigidity of the beam
CHAPTER 6 DEFORMATION IN
BENDING
§ 6–6 Strain energy of the beam in bending
§ 6–7 Method to solve simple statically indeterminate problems of
the beam
§ 6–8 How to increase the load-carrying capacity of the beam
§ 6–1 Summary
§ 6–2 Approximate differential equation of the deflection curve of
the beam and its integration
§ 6–3 Method of conjugate beam to determine the deflection and
the rotational angle of the beam
4
§ 6–1 概述
§ 6–2 梁的挠曲线近似微分方程及其积分
§ 6–3 求梁的挠度与转角的共轭梁法
§ 6–4 按叠加原理求梁的 挠度与转角
§ 6–5 梁的刚度校核
第六章 弯曲变形
§ 6–6 梁内的弯曲应变能
§ 6–7 简单超静定 梁的求解方法
§ 6–8 如何提高梁的承载能力
§ 6－１ SUMMARY
Study range,Calculation of the displacement of the straight beam with
equal sections in symmetric bending,
Study object：① checking rigidify of the beam；② Solving problems about
statically indeterminate beams（ to provide complementary equations for the
geometric-deformation conditions of the beam )
§ 6－１ 概 述
研究范围：等直梁在对称弯曲时位移的计算。
研究目的：①对梁作刚度校核；
②解超静定梁（为变形几何条件提供补充方
程）。
1).Deflection,The displacement of the centroid of a section in a direction perpendicular
to the axis of the beam,It is designated by v, It is positive if its direction is the same as
f,otherwise it is negative,
3,The relation between the angle of
rotation and the deflection curve,
1,Two basic displacement quantities of to measure deformation of the beam
( 1 )
d
dtg f
x
f ???? ??
Small deflection
P
x
v
C
?
C1 f
2),Angle of rotation,The angle by
which cross section turns with respect
to its original position about the neutral
axis,it is designated by ?,It is
positive if the angle of rotation rotates
in the clockwise direction,otherwise it
is negative,
2,deflection curve,The smooth curve that the axis of the beam is
transformed into after deformation is called the deflection curve,Its equation is
v =f (x)
1.挠度：横截面形心沿垂直于轴线方向的线位移。 用 v表示。
与 f 同向为正，反之为负。
2.转角：横截面绕其中性轴转
动的角度 。用 ? 表示，顺时
针转动为正,反之为负。
二、挠曲线：变形后，轴线变为光滑曲线，该曲线称为挠曲线。
其方程为,v =f (x)
三、转角与挠曲线的关系,
一、度量梁变形的两个基本位移量
( 1 )
d
dtg f
x
f ???? ??
小变形
P
x
v
C
?
C1 f
§ 6-2 APPROXIMATE DIFFERENTIAL EQUATION OF THE
DEFLECTION CURVE OF THE BEAM AND ITS INTEGTION
z
z
EI
xM )(1 ?
?
1,Approximate differential
equation of the deflection curve
z
z
EI
xMxf )()( ?????
Formula (2) is the approximate differential equation
of the deflection curve,
EI
xMxf )()( ???? ? …… （ 2 ）
)(
)1(
)(1
2
32 xf
f
xf ????
??
??
??
?
Small
deformation f
x M>0
0)( ??? xf
f
x
M<0
0)( ??? xf
(1)
§ 6-2 梁的挠曲线近似微分方程及其积分
z
z
EI
xM )(1 ?
?
一、挠曲线近似微分方程
z
z
EI
xMxf )()( ?????
式（ 2）就是挠曲线近似微分方程。
EI
xMxf )()( ???? ? …… （ 2 ）
)(
)1(
)(1
2
32 xf
f
xf ??
??
??
??
??
?
小变形
f
x M>0
0)( ??? xf
f
x
M<0
0)( ??? xf
(1)
)()( xMxfEI ????
For the straight beam with the same shape and equal section area,the approximate
differential equation of the deflection curve may be written as the following form,
2,Determine the equation of the deflection curve (elastic curve)
)()( xMxfEI ????
1d))(()( CxxMxfEI ???? ?
21d)d))((()( CxCxxxMxE I f ???? ? ?
1).Integration of the differential equation
2).Boundary conditions of the displacement
P
A B C P D
)()( xMxfEI ????
对于等截面直梁，挠曲线近似微分方程可写成如下形式,
二、求挠曲线方程（弹性曲线）
)()( xMxfEI ????
1d))(()( CxxMxfEI ???? ?
21d)d))((()( CxCxxxMxE I f ???? ? ?
1.微分方程的积分
2.位移边界条件
P
A B C P D
Discussion,① Applying to the planar bending of slender beams with elastic materials and
small deformations,
② May be applied to determine the displacements of beams with equal or variable sections
under all types of loads,
③ Integral constants may be determined by the geometric compatible conditions（ boundary
conditions,continuity conditions),
④ Advantages,Wide applications and accurate solutions by the direct-solving method
Disadvantages Complicated calculation,
?Displacement conditions at the supports,
?Continuity conditions,
?Smooth conditions,
0?Af 0?Bf 0?Df 0?D?
?? ? CC ff
?? ? CC ?? C ri g h tC l e f tor ?? ?
C ri g h tC l e f t ffor ?
讨论,
①适用于小变形情况下、线弹性材料、细长构件的平面弯曲。
②可应用于求解承受各种载荷的等截面或变截面梁的位移。
③积分常数由挠曲线变形的几何相容条件（边界条件、连续条
件）确定。
④优点：使用范围广，直接求出较精确； 缺点：计算较繁。
?支点位移条件,
?连续条件,
?光滑条件,
0?Af 0?Bf 0?Df 0?D?
?? ? CC ff
?? ? CC ?? 右左或写成 CC ?? ?
右左或写成 CC ff ?
Example 1 Determine the elastic curves, maximum deflections and maximum
angles of rotation of the following equal-section straight beams,
?Set up the coordinates and write out the
bending-moment equation
)()( LxPxM ??
?Write out the differential
equation and integrate it
?Determinate the integral constants by
the boundary conditions
)()( xLPxMfEI ??????
1
2)(
2
1 CxLPfEI ?????
21
3)(
6
1 CxCxLPE I f ????
0
6
1)0(
2
3 ??? CPLEI f
021)0()0( 12 ?????? CPLfEIEI ?
3
2
2
1 6
1 ;
2
1 PLCPLC ????
Solution,
P
L
x
f
x
[例 1] 求下列各等截面直梁的弹性曲线、最大挠度及最大转角。
?建立坐标系并写出弯矩方程
)()( LxPxM ??
?写出 微分方程并积分 ?应用位移边界条件 求积分常数
)()( xLPxMfEI ??????
1
2)(
2
1 CxLPfEI ?????
21
3)(
6
1 CxCxLPE I f ????
0
6
1)0(
2
3 ??? CPLEI f
021)0()0( 12 ?????? CPLfEIEI ?
3
2
2
1 6
1 ;
2
1 PLCPLC ????
解,P
L
x
f
x
?Write out the equation of the elastic curve and plot its curve
? ?323 3)(
6
)( LxLxL
EI
Pxf ????
EI
PLLff
3
)(
3
m a x ??EI
PLL
2
)(
2
m a x ?? ??
?The maximum deflection and the maximum angle of rotation
x
f
P
L
?写出弹性曲线方程并画出曲线
? ?323 3)(
6
)( LxLxL
EI
Pxf ????
EI
PLLff
3
)(
3
m a x ??EI
PLL
2
)(
2
m a x ?? ??
?最大挠度及最大转角
x
f
P
L
Solution,?Set up the coordinates and write out the bending-
moment equation
?
?
?
??
???
?
)( 0
)0( )(
)(
Lxa
axaxP
xM
?
?
?
?
?
???
??
1
1
2
)(
2
1
D
CxaP
fEI
?
?
?
?
?
?
???
?
21
21
3)(
6
1
DxD
CxCxaP
E I f
?
?
?
??
???
???
)( 0
)0( )(
Lxa
axxaP
fEI
x
f
P
L
a
? Write out the differential
equation and integrate it
解,?建立坐标系并写出弯矩方程
?
?
?
??
???
?
)( 0
)0( )(
)(
Lxa
axaxP
xM
?写出 微分方程并积分
?
?
?
?
?
???
??
1
1
2
)(
2
1
D
CxaP
fEI
?
?
?
?
?
?
???
?
21
21
3)(
6
1
DxD
CxCxaP
E I f
?
?
?
??
???
???
)( 0
)0( )(
Lxa
axxaP
fEI
x
f
P
L
a
?Determine the integral constants by boundary conditions
0
6
1)0(
2
3 ??? CPaEI f
0
2
1)0(
1
2 ???? CPaEI ?
3
22
2
11 6
1 ;
2
1 PaDCPaDC ??????
)()( ?? ? afaf
)()( ?? ? aa ?? 11 DC ??
2121 DaDCaC ????
P
L
a
x
f
?应用位移边界条件 求积分常数
0
6
1)0(
2
3 ??? CPaEI f
0
2
1)0(
1
2 ???? CPaEI ?
3
22
2
11 6
1 ;
2
1 PaDCPaDC ??????
)()( ?? ? afaf
)()( ?? ? aa ?? 11 DC ??
2121 DaDCaC ????
P
L
a
x
f
?Write out the equation of the elastic curve and plot its curve
3 2 3
23
( ) 3 ( 0 )
6
()
3 ( a )
6
P
a x a x a x a
EI
fx
P
a x a x L
EI
?
??? ? ? ? ?
????
? ?
? ?? ? ? ?
???
?
? ?aL
EI
PaLff ??? 3
6
)(
2
m a x
EI
Paa
2
)(
2
m a x ?? ??
?The maximum deflection and the maximum angle of rotation
P
L
a
x
f
?写出弹性曲线方程并画出曲线
? ?
? ?
?
?
?
??
?
?
???
?????
?
)(a 3
6
)0( 3)(
6
)(
32
323
Lx axa
EI
P
ax axaxa
EI
P
xf
? ?aL
EI
PaLff ??? 3
6
)(
2
m a x
EI
Paa
2
)(
2
m a x ?? ??
?最大挠度及最大转角
P
L
a
x
f
§ 6-3 METHOD OF CONJUGATE BEAM TO DETERMINE THE DEFLECTION
AND THE ROTATIONAL ANGLE OF THE BEAM
)()( xMxfEI ????
1,Usage of the method,Determine the deflection and the
rotational angle of the designated point in the beam
2,Theoretical basis of the method,Similar analogy,
)()( xqxM ???
Differential equation of the deflection
curve of the beam,
Relation between the external load the
internal force,
§ 6-3 求梁的挠度与转角的共轭梁法
)()(:梁的挠曲线微分方程 xMxfEI ????
一、方法的用途：求 梁上指定点的挠度与转角。
二、方法的理论基础：相似比拟。
)()(:为梁的外载与内力的关系 xqxM ???
上二式形式相同，用类比法，将微分方程从形式上转化为
外载与内力的关系方程。从而把求挠度与转角的问题转化为求
弯矩与剪力的问题。
3,Conjugate beam（ relations between the real beam and the
imaginary beam）,
① The direction of the axis x and the origin of coordinates
② Having same geometric shapes,
③ Corresponding equation of the real beam,)()( xMxfEI ????
)()( xqxM ???
)()( xMxfEI ??????⑤ Integral of the ―force‖ differential
equation of the imaginary beam,
00 d)()()( QxxqxMxQ
x ???? ?
000 0 d)d)(()( MxQxxxqxM
x x ??? ? ?
( ) ( ),
q x M x a u o r d i n g t o t h i s e q u a t i o n e s t a b l i s h
d i s t r i b u t e d l o a d s o n t h e i m a g i n a r y b e a m
??let,
。
④
)()( xqxM ???
Corresponding equation of the imaginary
beam,
三、共轭梁（实梁与虚梁的关系）,
① x轴指向及 坐标原点完全相同 。
② 几何形状完全相同。
③ 实梁对应方程,)()( xMxfEI ????
)()( xqxM ???
)()( xMxfEI ??????
⑤ 虚梁“力”微分方程的积分
00 d)()()( QxxqxMxQ
x ???? ?
000 0 d)d)(()( MxQxxxqxM
x x ??? ? ?
载荷。依此建立虚梁上的分布令,)()( xMxq ??
④
)()( xqxM ???
虚梁对应方程,
)()( xMxEI f ??
The quantities with subscripts ―0‖dentte they are in the coordinate
origin coordinate origin,
Integral of the―displacement‖ differential equation of the real beam
)()( xMxfEI ????
00 ))(()( ?? EIdxxMxfEIEI
x ????? ?
000 0 d)d))((()( E I fxEIxxxMxE I f
x x ???? ? ? ?
)()( xQxEI ??
⑥ Set up the ―force‖ boundary conditions of the imaginary beam according to
the ―displacement‖ boundary conditions of the real beam,
AAAA QEIMEI f ?? ? ;
)()( xMxEI f ??
下脚标带,0‖的量均为坐标原点的量。
实梁“位移”微分方程的积分
)()( xMxfEI ????
00 ))(()( ?? EIdxxMxfEIEI
x ????? ?
000 0 d)d))((()( E I fxEIxxxMxE I f
x x ???? ? ? ?
)()( xQxEI ??
⑥ 依实梁的“位移”边界条件建立虚梁的“力”边界条件。
AAAA QEIMEI f ?? ? ;
中间铰
支座 A
0?Af
0?A?
0?Af
0?A?
右左 AA ?? ?
右左 AA ff ? 右左 AA MM ?
右左 AA QQ ?
固定端 A
A
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0?Af
0?A?
0?? 右左 AA ??
0?Af
0?? 右左 AA QQ
0 ?AM
固定端 A
A
自由端 A A
自由端 A A
铰支端 A
A
铰支端 A
A
中间铰
支座 A
中间铰 A
中间铰 A
Imaginary beam Real beam Conjugated beam
Supports and ends Displacement boundary Supports and ends Force boundary
Fixed
end A
Free
end A
Hinged
support A
Middle
hinged
support A
Middle
hinge A
Free
end A
Fixed
end A
Hinged
support A
Middle
hinge A
Middle hinged
support A
中间铰
支座 A
虚虚 梁梁
实实 梁梁
共 轭 梁
支 承 和 端 部 情 况
位移边界 相应的支承和端部情况 力边界
0?Af
0?A?
0?Af
0?A?
右左 AA ?? ?
右左 AA ff ? 右左 AA MM ?
右左 AA QQ ?
固定端 A
A
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0?Af
0?A?
0?? 右左 AA ??
0?Af
0?? 右左 AA QQ
0 ?AM
固定端 A
A
自由端 A A
自由端 A A
铰支端 A
A
铰支端 A
A
中间铰
支座 A
中间铰 A
中间铰 A
Summary,Relations between the equal-section real beam and the
imagine beam as following,
① Coordinate origin and direction
of axis x are all completely same,
② Geometric shapes are
completely same,
④ Set up the ―force‖ boundary conditions of the imaginary beam according to
the ―displacement‖ boundary conditions of the real beam,
AAAA QEIMEI f ?? ? ;
EI
Q
EI
M
f xxxx ?? ? ;
⑤ Determine the ―displacement‖ of the real beam according to the ―internal
force‖ of the imaginary beam,
a, Fixed end Free end
b,
c, middle hinged supports
middles hinge
L e t ( ) ( )
,
q x M x a c c o r d i n g t o t h i s e q u a t i o n e s t a b l i s h
u n i f o r m l o a d s o n t h e i m a g i n a r y b e a m
??：
③
Hinged
supports
Hinged
supports
总结：等截面实梁与虚梁的关系如下,
① x 轴指向及 坐标原点完全相同。 ② 几何形状完全相同。
④ 依实梁的“位移”边界条件，建立虚梁的“力”边界条件。
AAAA QEIMEI f ?? ? ;
EI
Q
EI
M
f xxxx ?? ? ;
⑤ 依虚梁的“内力”，求实梁的“位移”。
a,固定端 自由端
b,铰支座 铰支座
c,中间铰支座 中间铰链
载荷。依此建立虚梁上的分布令,)()( xMxq ??
③
Solution,? Set up coordinates and
the imaginary beam
Example 2 Determine the displacement at point B (deflection and angle of
rotation) of the following straight beams with equal sections in shape and area
2)(
2
q)( xLxM ???
2)(
2
q)( )( xLxMxq ????
q
L
A
B
f
x
2
2
0
qLq ?
)(xq
A B
L
Determine the bending moment of
the real beam in order to obtain loads
of the imaginary beam
? Determine the shearing force and
the bending moment at the point B of
the imaginary beam in order to obtain
the deflection and the rotational angle
at the same point of the real beam,
解,? 建立坐标和虚梁
[例 2] 求下列等截面直梁 B点的位移（挠度和转
角）。
?求虚梁 B点的剪力和弯矩，
以求实梁 B点的转角和挠度
求实梁的弯矩方程
以确定虚梁荷载
2)(
2
q)( xLxM ???
2)(
2
q)( )( xLxMxq ????
q
L
A
B
f
x
2
2
0
qLq ?
)(xq
A B
L
3
0 LqQEI
BB ???
84
3 4qLLAMEI f
qBB ????
EI
qL
B 6
3
?? ?
EI
qLf
B 8
4
??
)( xqQ B ?
2
2
0
qLq ?
)(xq
A B
L
( ) i n t
i n t
B B M o m e u t o f q x w i t h r e s p e c t t o P o B i n
t h e l e f t s i d e o f t h i s P o
M ?
Area of in the left side of the
point B
The moment of the area of
of left side of the point B about
point B
?求虚梁 B点的剪力和弯矩，以求实梁 B点的转角和挠度
3
0 LqQEI
BB ???
84
3 4qLLAMEI f
qBB ????
EI
qL
B 6
3
?? ?
EI
qLf
B 8
4
??
面积) 的(点左侧 xqBQ B ?
2
2
0
qLq ?
)(xq
A B
L
面积对) 的(点左侧 xqBBM ?
B点之矩
Solution,? Set up coordinates
and the imaginary beam
?Determine the shearing force and
bending moment of point B in
the imaginary beam,
Find out the bending-moment
equation to determine loads of the
imaginary beam,
) )( M ( xxq ??
???? 2 ; 2 qaRqaR DA
q qa2
qa
A B
C D
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
a a a
f
x
D
解,? 建立坐标和虚梁
?求虚梁 B点的剪力和弯矩
求实梁的弯矩方程以确定
虚梁荷载
) )( M ( xxq ??
???? 2 ; 2 qaRqaR DA
q qa2
qa
A B
C D
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
a a a
f
x
D
72
13 3 ?? qaR
A
3
23
72
5
22
1
72
13 qaaqaqa
BQ ????
?How about the displacement near
point C？
4
23
72
7
322
1
72
13 qaaaqaaqa
BM ?????
EI
qa
B 72
5 3
?? EI
qaf
B 72
7 4??
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
A B C
a a a
D
qa2/2
3qa2/8
?求虚梁 B点的剪力和弯矩
72
13 3 ?? qaR
A
3
23
72
5
22
1
72
13 qaaqaqa
BQ ????
?C点左右位移怎样？
4
23
72
7
322
1
72
13 qaaaqaaqa
BM ?????
EI
qa
B 72
5 3
?? EI
qaf
B 72
7 4??
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
A B C
a a a
D
qa2/2
3qa2/8
① Convert the change of section into the bending moment,
② Geometric shape,the length is not change,moment of inertia
changed into I0,
③ Corresponding equation of
the real beam,
Corresponding equation of
the imaginary beam,
)()(0 xMxfEI ?????
)()( xqxM ???
4,Method of conjugate beam for non-constant beam,
00
0 )(
)(
)()(
EI
xM
I
I
xEI
xMxf ????????
)()(0 xMxfEI ?????
)(
)()( 0
xI
IxMxM ??
L e t ( ) ( )q x M x???：④ According to this establish the uniform load on the imaginary
beam,They are all the same as the equal section straight beam
in other aspects,
① 将截面的变化折算到弯矩之中去。
② 几何形状：长度不变，惯性矩变为 I0 。
③ 实梁对应方程,
虚梁对应方程,
)()(0 xMxfEI ?????
)()( xqxM ???
四、变截面直梁的共轭梁法,
00
0 )(
)(
)()(
EI
xM
I
I
xEI
xMxf ????????
)()(0 xMxfEI ?????
)(
)()( 0
xI
IxMxM ??
其它与等截面直梁完全相同。
载荷。依此建立虚梁上的分布令,)()( xMxq ???
④
Example 3 Determine the displacement
of point C on the following beam with
non-constant section,Knowing：
IDE=2IEB =2IAD,
Solution,? Set up coordinates and
the imaginary beam,
) )( (xMxq ???
ADADAD xMxI
IxMxM )(
)(
)()( 0 ???
2
)()()( DE
DE
AD
DEDE
xM
I
IxMxM ???
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
[例 3] 求下列变截面直梁 C点的
位移，已知,IDE =2IEB =2IAD 。
解,? 建立坐标和虚梁
) )( (xMxq ???
ADADAD xMxI
IxMxM )(
)(
)()( 0 ???
2
)()()( DE
DE
AD
DEDE
xM
I
IxMxM ???
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
?Determine the shearing force and
the bending moment at point C
of the imaginary beam,
32
5 2 ?? PaR
A
0?CQ
3
32
3
6282
1
428 Pa
aaPaaaPa ??
0?C?
AD
C EI
Pa
f
32
3 3
?
???
3
2
242
1
32
5 2 aaPaaPa
CM
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
?求虚梁 C点的剪力和弯矩
32
5 2 ?? PaR
A
0?CQ
3
32
3
6282
1
428 Pa
aaPaaaPa ??
0?C?
AD
C EI
Pa
f
32
3 3
?
???
3
2
242
1
32
5 2 aaPaaPa
CM
§ 6-4 DETERMINE DEFLECTIONS AND ROTATIONAL ANGLES
OF THE BEAM BY THE PRINCIPLE OF SUPERPOSITION
1,Superposition of loads,The deformation of a structure due
to the action of multi-loads is equal to the algebraic sum of
deformation resulting from the separate action of each load,
1 2 1 1 2 2( ) ( ) ( ) ( )n n nP P P P P P? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
1 2 1 1 2 2( ) ( ) ( ) ( )n n nf P P P f P f P f P? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
2,Superposition of structural forms（ rigidization method of
segment by segment）,
§ 6-4 按叠加原理求梁的 挠度与转角
一、载荷叠加,多个载荷同时作用于结构而引起的变形
等于每个载荷单独作用于结构而引起的变形的代数和。
1 2 1 1 2 2( ) ( ) ( ) ( )n n nP P P P P P? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
1 2 1 1 2 2( ) ( ) ( ) ( )n n nf P P P f P f P f P? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
二、结构形式叠加（逐段刚化法）,
Solution:① Decompose the loads as
shown in the figure
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
q L af
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
② Determine the deformations of the
beam under the action of simple loads
by looking up the table,
Example 4 Determine the angle
of rotation of point A and the
deflection of point C by the principle of
superposition,
[例 4] 按叠加原理求 A点转角和 C
点
挠度。
解,① 载荷分解如图
② 由梁的简单载荷变形表,
查简单载荷引起的变形。
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
qaf
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
qaf
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
③ Sum up
qAPAA ??? ??
)43(
12
2
qaP
EI
a ??
EI
Pa
EI
qaf
C 624
5 34 ??
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
qaf
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
③ 叠加
qAPAA ??? ??
)43(
12
2
qaP
EI
a ??
EI
Pa
EI
qaf
C 624
5 34 ??
Example 5 Determine the deflection of point C by the principle of
superposition,
Solution,① Divide the load infinitely
as shown in the figure,
② Determine the deformations of the
beam under the action of simple loads by looking up the table,
③ Sum up
22( d ) ( 3 4 )
48d P C
P b L bf
EI
??
bLbqxxqP d2d)(d 0??
2 2 2
0 ( 3 4 ) d
24
q b L b b
E I L
??
?? ? d P CqC ff
2 2 2 40, 5
00
0
( 3 4 ) d
2 4 2 4 0
L q b L b q L
b
E I L E I
? ??
q0
0.5L 0.5L
x dx x
f
C
b
[例 5] 按叠加原理求 C点挠度。 解,?载荷无限分解如图
?由梁的简单载荷变形表,
查简单载荷引起的变形。
?叠加
22( d ) ( 3 4 )
48d P C
P b L bf
EI
??
bLbqxxqP d2d)(d 0??
2 2 2
0 ( 3 4 ) d
24
q b L b b
E I L
??
?? ? d P CqC ff
2 2 2 40, 5
00
0
( 3 4 ) d
2 4 2 4 0
L q b L b q L
b
E I L E I
? ??
q0
0.5L 0.5L
x dx
b
x
f
C
Example 6 Explanation of the superposition of structural forms (digitization
method of segment by segment ),
=
+
P L1 L2
A B C
B C
P L2
f1
f2
Equivalence
x
f
x
f
21 fff ??
f
P L1 L2
A B C Rigidize the
segment AC
P L1 L2
A B C
Rigidize the
segment BC
P L1 L2
A B C M x
f
Equivalence
[例 6] 结构形式叠加（逐段刚化法 ) 原理说明。
=
+
P L1 L2
A B C
B C
P L2
f1
f2
等价
等价
x
f
x
f
21 fff ??f
P L1 L2
A B C 刚化 AC段
P L1 L2
A B C
刚化 BC段
P L1 L2
A B C M x
f
§ 6-5 RIGIDITY CHECK OF THE BEAM
))1000 1~250 1( ( m a x ??
?
?
??
?
??
?
??
??
L
f
L
f
L
f
? ??? ?m a x
1,Rigidity conditions of the beam
Where [?] is called the permissible angle of rotation； [f/L] is called the
permissible ratio of the deflection and the span,In general we can do three
kinds of calculations about the rigidity by theses conditions,
?,Check the rigidity,
??
?
??
??
L
f
L
f m a x
? ??? ?m a x
(In civil engineering,the strength often-plays an
important role and the rigidity plays a secondary role
except special cases.）
For the civil engineering,
?,Design the dimension
of the section；
?,Determine the
permissible load。
§ 6-5 梁的刚度校核
))
1000
1~
250
1(,对土建工程( m a x ?
??
?
??
?
??
?
??
??
L
f
L
f
L
f
一、梁的刚度条件
其中 [?]称为许用转角； [f/L]称为许用挠跨比。通常 依此条件
进行如下三种刚度计算,
?,校核刚度,
?,设计截面尺寸；
?,设计载荷。
(但：对于土建工程，强度常处于主要地位，
刚度常处于从属地位。特殊构件例外）
? ??? ?m a x
??
?
??
??
L
f
L
f m a x
? ??? ?m a x
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
Example 7 A hollow circular-section beam is shown in the following figure,Its
inside and outside diameter is respectively,d=40mm,D=80mm,Knowing
E=210GPa,[f/L]=0.00001 at point C,[?]=0.001rad at point B,Try to check the
rigidity of the beam,
=
+ +
=
P1=1kN
A B D C
P2
B C D A
P2=2kN
B C D A
P2
B C
a
P2
B C D A
M
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
[例 7] 下图为一空心圆截面梁，内外径分别为,d=40mm、
D=80mm，梁的 E=210GPa，工程规定 C点的 [f/L]=0.00001,B点
的 ??]=0.001弧度,试校 核此梁的刚度。
=
+ +
=
P1=1kN
A B D C
P2
B C D A
P2=2kN
B C D A
P2
B C
a
P2
B C D A
M
P2
B C
a
=
+
+
图 1
图 2
图 3
EI
aLPaf
BC 16
2
1
11 ?? ?
EI
LP
B 16
2
1
1 ??
EI
L a P
EI
ML
B 33
2
3 ?????
EI
LaPaf
BC 3
2
2
33 ??? ?
Solution,?Transform the structure
and determine the deformations under
the actions of each simple load by
looking up the table,
02 ?B?
EI
aPf
C 3
3
2
2 ??
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
P2
B C
a
=
+
+
图 1
图 2
图 3
EI
aLPaf
BC 16
2
1
11 ?? ?EI
LP
B 16
2
1
1 ??
EI
L a P
EI
ML
B 33
2
3 ?????
EI
LaPaf
BC 3
2
2
33 ??? ?
解,?结构变换，查表求简单
载荷变形。
02 ?B? EI
aPf
C 3
3
2
2 ??
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
P2
B C
a
=
+
+
图 1
图 2
图 3
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
EI
LaP
EI
aP
EI
aLPf
C 3316
2
2
3
2
2
1 ???
EI
LaP
EI
LP
B 316
2
2
1 ???
?Determine the deformation under the action of complex loads by superposition
48
1244
44
m10188
10)4080(
64
14.3
)(
64
?
?
??
???
?? dDI
?
P2
B C
a
=
+
+
图 1
图 2
图 3
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
EI
LaP
EI
aP
EI
aLPf
C 3316
2
2
3
2
2
1 ???
EI
LaP
EI
LP
B 316
2
2
1 ???
?叠加求复杂载荷下的变形
48
1244
44
m10188
10)4080(
64
14.3
)(
64
?
?
??
???
?? dDI
?
m1019.5
3316
6
2
2
3
2
2
1 ???????
EI
LaP
EI
aP
EI
aLPf
C
42
2
1 10423.0)
3
200
16
400(
1880210
4.0
316
?????
???? EI
LaP
EI
LP
B?
? ? 001.010423.0 4m a x ???? ? ??
??
?
??
??
L
f
L
f m a x
? ? m101m1019.5 56m a x ?? ????? ff
?Check the rigidity
(rad)
m1019.5
3316
6
2
2
3
2
2
1 ???????
EI
LaP
EI
aP
EI
aLPf
C
)(10423.0)
3
200
16
400(
1880210
4.0
316
42
2
1 弧度?????
?
???
EI
LaP
EI
LP
B?
? ? 001.010423.0 4m a x ???? ? ??
??
?
??
??
L
f
L
f m a x
?校核刚度
? ? m101m1019.5 56m a x ?? ????? ff
dx
x
Q Q+dQ
M M+dM
1,Calculation of strain energy in bending,
§ 6–6 STRAIN ENERGY OF THE BEAM IN BENDING
EI
xM )(1 ?
?
?d)(
2
1dd ??? xMWU
x
EI
xMU d
2
)(d 2?
?? L xEI xMU d2 )(
2
?
?
xd
d ?
Strain energy is equal to the work of external forces.Neglect the shearing
strain energy an omit
?dd ?M
d?
M(x)
P1
M x
f
P2 dx
d? ?
dx
x
Q Q+dQ
M M+dM
一、弯曲应变能的计算,
§ 6–6 梁内的弯曲应变能
EI
xM )(1 ?
?
?d)(
2
1dd ??? xMWU
x
EI
xMU d
2
)(d 2?
?? L xEI xMU d2 )(
2
?
?
xd
d ?
应变能等于外力功。不计剪切应变能并略去 ?dd ?M
d?
M(x)
P1
M x
f
P2 dx
d? ?
Example 8 Determine the deflection of point C on the beam by the energy
method,The beam is of equal section and straight,
CPfW 2
1?
Solution,Strain energy is equal to
the work of external forces,
?? L xEI xMU d2 )(
2
)0(;
2
)( axxPxM ???
Using the symmetry we get,
EI
aPxxP
EI
U a
12
d)
2
(
2
12 32
0
2 ?? ?
EI
PafUW
C 6 h a v e W eL o t
3
??Thinking,Can we use this method
to determine the displacement of point
C when distributed loads are applied?
P
a a
q
x
f
A
C
B
[例 8] 用能量法求 C点的挠度。梁为等截面直梁。
CPfW 2
1?
解：外力功等于应变能
?? L xEI xMU d2 )(
2
)0(;
2
)( axxPxM ???
在应用对称性，得,EIaPxxPEIU a 12d)2(2 12 32
0
2 ?? ?
EI
PafUW
C 6
3
????
思考：分布荷载时，可否用此法求 C点位移？
q
x
f
P
a a
A
C
B
2,Impact problems of the beam
1).Assumptions,
?Impact bodies are rigid bodies；
?Neglecting the potential energy of weight
and the dynamic energy of the impacted
bodies；
?Impact bodies do not rebound；
?Neglecting the loss of energy of sound、
light,heat and so on(energy is conservative),
0)(21 2111 ?????? dfhmgmvUVT
mg
L
h A B C
A B C x
f
fd Before impact,
二,梁的冲击问题 1.假设,
?冲击物为刚体；
?不计被冲击物的重力势能和动能；
?冲击物不反弹；
?不计声、光、热等能量损耗（能
量守恒）。
0)(
2
1
冲击前
2
111
???
???
dfhmgmv
UVT
mg
L
h A B C
A B C x
f
fd
A B C x
f
fd
22
2
)(
2
1
)(
2
1
)(
2
1
2
1
00
d
j
d
j
j
ddd
f
f
mg
f
f
P
fkfP
??
????
After impact
222 UVT ??
The energy before and after impact
is conservative,therefore,
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgvf ????? )2)(11( 2
d
jj
d
d f
hgv
f
fK 2)2(11 ?????
jf
h
dK
211 ??? 2?dK
Coefficient of
dynamic load
(2)Sudden load,(1)Freely falling body
22
2
222
)(
2
1
)(
2
1
)(
2
1
2
1
00
冲击后
d
j
d
j
j
ddd
f
f
mg
f
f
P
fkfP
UVT
??
????
??
冲击前、后，能量守恒，所以,
A B C x
f
fd
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgvf ????? )2)(11( 2
d
jj
d
d f
hgv
f
f
K
2)2(
11:
?
????动荷系数
jf
h
dK
211:)1( ???自由落体 2:)2( ?
dK突然荷载
h
B A C
mg E =P
3,Calculation of dynamic response,
Solution,?Determine the
static deflection of point C,
2;
2 21
1 PRCCAAf
ACj ???
The dynamic response is equal to the product of the static response and the
coefficient of dynamic load,
ABDE
A
EI
PL
EI
LR
4848
33
?? EI
PL
32
3
?
C1
A1
D
EIEIEI DEAB ??
L
C2
Example 9 A structure is shown
in the figure,AB=DE=L,A,C are
respectively the middle points of AB
and DE, Determine the dynamic
`stress of section C under
the impact of weight mg,
h
B A C
mg E =P
三、动响应计算,
解,?求 C点静挠度
2;2 21
1 PRCCAAf
ACj ???
动响应计算等于静响应计算与动荷系数之积,
[例 9] 结构如图,AB=DE=L,A,C 分别为 AB 和 DE 的中点，
求梁在重物 mg 的冲击下,C 面的动应力。
ABDE
A
EI
PL
EI
LR
4848
33
??
EI
PL
32
3
?
C1
A1
D
EIEIEI DEAB ??
L
C2
?Coefficient of dynamic load
3
64
11
2
11
PL
E I h
f
h
d
K
Cj
???
???
?Determine the dynamic stress in the section C
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
64
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
?动荷系数
3
64
11
2
11
PL
E I h
f
h
d
K
Cj
???
???
?求 C面的动应力
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
64
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
§ 6-7 METHOD TO SOLVE SIMPLE
STATICALLY INDETERMINATE
PROBLEMS OF THE BEAM
1,Treatment method,
Combining the compatibility equation of
deformation,physical equation with
equilibrium equations to determine the
whole unknown forces,
Solution,? Set up the primary beam
Determine the degree of statically
indeterminacy,The structure in which
redundant constraints are substituted by
reactions—primary structure。
=
EI
q0
L A
B
L
q0 MA
B A
q0
L RB A
B
x
f
§ 6-7 简单超静定 梁的求解方法
1、处理方法：变形协调方程、物理
方程与平衡方程相结合，求全部未
知力。
解,?建立静定基
确定超静定次数，用反力
代替多余约束所得到的结构 —
—静定基。
=
EI
q0
L A
B
L
q0 MA
B A
q0
L RB A
B
x
f
?Geometric equation—compatibility
equation of deformation
0??? BBRBqB fff
+
q0
L RB A
B
=
RB
A B
q0
A B
?Physical equation—relation between
the deformation and forces
?Complementary equation
EI
LRf
EI
qLf B
BRBq B 3;8
34
???
0
38
34
??
EI
LR
EI
qL B
8
3 qLR
B ??
?Solve other problems（ reaction、
stress,deformation etc.）
?几何方程 ——变形协调方程
0??? BBRBqB fff
+
q0
L RB A
B
=
RB
A B
q0
A B
?物理方程 ——变形与力的关系
?补充方程
EI
LRf
EI
qLf B
BRBq B 3;8
34
???
0
38
34
??
EI
LR
EI
qL B
8
3 qLR
B ??
?求解其它问题（反力、应力,
变形等）
?Geometric equation—compatibility
equation of deformation
Solution,? Set up the primary beam
BCBRBqB Lfff B ????
=
Example 10 Determine the
reaction at end B in the structure as
shown in the figure,
LBC EA
x
f
q0
L RB A
B
C
q0
L RB A
B
EI =
RB
A B
+ q0
A B
EI
?几何方程
——变形协调方程,
解,?建立静定基
BCBRBqB Lfff B ????
=
[例 10] 结构如图，求 B点反
力。 LBC EA
x
f
q0
L RB A
B
C
q0
L RB A
B
EI =
RB
A B
+ q0
A B
EI
=
LBC EA
x
f
q0
L RB A
B
C
RB
A B +
q0
A B
?Physical equation—relation between the
deformation and forces
?Complementary equation
? Solve other problems（ reaction、
stress,deformation etc.）
EI
LRf
EI
qLf B
BRBq B 3; 8
34
???
EA
LR
EI
LR
EI
qL BCBB ??
38
34
)
3
(8
3
4
I
L
A
L
I
qL
R
BC
B
?
??
EA
LRL BCB
BC ??
=
LBC EA
x
f
q0
L RB A
B
C
RB
A B +
q0
A B
?物理方程 ——变形与力的关系
?补充方程
?求解其它问题（反力、应力,
变形等）
EI
LRf
EI
qLf B
BRBq B 3; 8
34
???
EA
LR
EI
LR
EI
qL BCBB ??
38
34
)
3
(8
3
4
I
L
A
L
I
qL
R
BC
B
?
??
EA
LRL BCB
BC ??
§ 6-8 HOW TO INCREASE THE LOAD-CARRYING
CAPACITY OF THE BEAM
Strength,
Normal stress,
Shearing stress,
? ? m a x ?? ??
zW
M
? ??? ??
z
z
bI
QS *
zEI
xMf )(????Rigidity,
Stability,
Those parameters are all relative to the internal forces and the
properties of the section,
§ 6-8 如何提高梁的承载能力
强度：正应力,
剪应力,
? ? m a x ?? ??
zW
M
? ??? ??
z
z
bI
QS *
zEI
xMf )(????
刚度,
稳定性,
都与内力和截面性质有关 。
R
b
h
（ 1） Reasonable ration between the height and the width of the section of
the wooden beam with rectangular section
Li Jie of north-Song dynasty pointed out in the book
of written in the year 1100 that the reasonable ratio
of height and width (h/b) of a rectangular wooden
beam is 1.5,
T·Young pointed out in the book of written in 1807 that the strength
reaches the maximum when the reasonable ratio of height to width
of a rectangular woden beam is h/b= and the rigidity reaches
the maximum when h/b=,
2
3
一、选择梁的合理截面
矩形木梁的合理高宽比
北宋李诫于 1100年著 ?营造法式 ?一书中指出,
矩形木梁的合理高宽比 ( h/b)为 1.5
英 (T.Young)于 1807年著 ?自然哲学与机械技术讲义 ?一书中指出,
矩形木梁的合理高宽比 为
刚度最大。时强度最大时,3 ;,2 ?? bhbh
R
b
h
Reasonable section in general case
A
Q
3
433.1
mm a x ?? ?? 32
3
1
1
DW
z
??
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 ?? ?
)2/( ;,
4
w h e n 12
2
1 DRRaaD ??? ??
1,If areas of sections are the same,select the section with a larger
modulus in bending,
z
D1
z a
a
1, 0 5
12
1
3
2 zz I
bhI ??
一般的合理截面
A
Q
3
433.1
mm a x ?? ??
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 ?? ?
)2/( ;,
4 1
2
2
1 DRRaaD ??? ?? 时当
1、在面积相等的情况下，选择抗弯模量大的截面
z
D1
z a
a
1, 0 5
12
1
3
2 zz I
bhI ??
32
3
1
1
DW
z
??
mm ax 2 ?? ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ? 1
222
1 67.1,
4
])8.0([
4w h e n DD
DDD ??? ??
4/2,2
4 11
2
1
2
1w h e n DaaD ?? ??
1
3
1
2
4 67.1 6
4
6 zz
WabhW ???
mm ax 5.1 ?? ?
z D
0.8
D
a1
2a
1 z
59.4)8.01(
64
14
4
3 zz I
DI ??? ?
2, 0 912812 z1
4
1
3
4 I
abh I
z ???
mm ax 2 ?? ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ? 1
222
1 67.1,
4
])8.0([
4 DD
DDD ??? 时当 ??
4/2,2
4 11
2
1
2
1 DaaD ?? ?? 时当
1
3
1
2
4 67.1 6
4
6 zz
WabhW ???
mm ax 5.1 ?? ?
z D
0.8
D
a1
2a
1 z
59.4)8.01(
64
14
4
3 zz I
DI ??? ?
2, 0 912812 z1
4
1
3
4 I
abh I
z ???
55.9 15 zz II ?
)(= 3.2 mm a x
fA
Q?? ?
For the I-shape section the method
to determine the shearing stress is
similar to that for the –shape
section,
15 57.4 zz WW ?
,6.18.024 2222
2
1 aaD ????
0.8a2
a2
1.6
a 2 2a 2 z
when 12 05.1 Da ?
55.9 15 zz II ?
)(= 3.2 mm a x
fA
Q?? ?
工字形截面与框形截面类似。
15 57.4 zz WW ?
12
2
2
2
2
2
1 05.1,6.18.02
4
DaaaD ???? 时当
?
0.8a2
a2
1.6
a 2 2a 2 z
2,Select shapes of the section according to properties material
?
G
z
For example,for the material of cast iron,the T-shape section is often
used as shown in the following figure,
2,Select the beam with non-constant sections,
The best is to select equal strength beam,that is
][
)(
)()(
m a x ?? ?? xW
xMx
If it is an equal-strength rectangular section,its height is
][
)(6)(
?b
xMxh ?
At the same time
][
)(
5.1m a x ?? ??
xbh
Q
][
5.1)(
?b
Qxh ?
P
x
)(xh
2、根据材料特性选择截面形状
?
G
z
如铸铁类材料，常用 T字形类的截面，如下图,
二、采用变截面梁
最好是等强度梁，即
][
)(
)()(
m a x ?? ?? xW
xMx
若为等强度矩形截面，则高为
][
)(6)(
?b
xMxh ?
同时
][
)(
5.1m a x ?? ??
xbh
Q
][
5.1)(
?b
Qxh ??
P
x
)(xh
EI
PLy 3
m a x 0 2 1.0?
EI
PLy 3
m a x 0 1 4.0?
EI
PLy 3
m a x 0 0 7 3.0?
3,Arrange reasonably external forces（ include reactions of supports）
and make the maximum bending moment M max as small as possible,P
L/2 L/2
x
M
+
PL/4
P
L/4 3L/4
P=qL
L/5 4L/5
Symmetry
M
x
3PL/16
+
M
x
qL2/10
+
EI
PLy 3
m a x 0 2 1.0?
EI
PLy 3
m a x 0 1 4.0?
EI
PLy 3
m a x 0 0 7 3.0?
三、合理布置外力（包括支座），使 M max 尽可能小。
P
L/2 L/2 M
x
+
PL/4
P=qL
L/5 4L/5
对称
P
L/4 3L/4 M
x
3PL/16
+
M
x
qL2/10
+
EI
qLy 4
m a x 013.0?
EI
qLy 43
m a x 107 8 7 5.0
???
EI
qLy 43
m a x 10326.0
???
q
L
L/5
q
L/5
q
L/2 L/2
M
x
8
2qL
+
40 2 qL
50 2 qL ?
M
x -
+ -
32 2 qL ?
M
x
32
9
16
9 2qL?
32
9
16
9 2qL?
+ +
-
EI
qLy 4
m a x 013.0?
EI
qLy 43
m a x 107 8 7 5.0
???
EI
qLy 43
m a x 10326.0
???
q
L
L/5
q
L/5
q
L/2 L/2
M
x
8
2qL
+
40 2 qL
50 2 qL ?
M
x -
+ -
32 2 qL ?
M
x
32
9
16
9 2qL?
32
9
16
9 2qL?
+ +
-
Z
Y
cr
I
I
L
GEb ?
? ?
4,Lateral buckling of the beam
1),Critical load of the beam with rectangular section in pure bending
M
M
x
y
z
Z
Y
cr
I
I
L
GEb ?
? ?
四、梁的侧向屈曲
1.矩形纯弯梁的临界载荷
M
M
x
y
z
2,Critical load of the beam with the I-shape section in pure bending
M
M
x
y
z
h
??
?
?
??
?
?
???
?
?
?
?
??
Z
Y
Z
Y
Z
Y
cr I
I
I
I
EG
I
I
L
E
L 2
2
2
2
)(
2
h ??
?
From the above analysis we know if I y is too small,the strength and rigidity
of the structure are higher,but the possibility of lateral loss of stability is also
increasing,About this point we must pay enough attention,
2.工字钢形截面纯弯梁的临界载荷
M
M
x
y
z
h
??
?
?
??
?
?
???
?
?
?
?
??
Z
Y
Z
Y
Z
Y
cr I
I
I
I
EG
I
I
L
E
L 2
2
2
2
)(
2
h ??
?
由上可见,I y过小时，虽然强度和刚度较高，但侧向失稳
的可能性却增大了，这点应引起注意。
5,Select the material of high strength to increase the
permissible stress
For the same kind of materials,differences of their“E” are small
,but those of their,?jx” are larger,So the strength can be improved by
changing another same kind of material,but the rigidity and stability
can not be improved,
For the different kind of materials,differences of their E and G
are very large（ steel, E=200GPa,copper, E=100GPa）,So the
rigidity and stability can be improved by selecting different Rind of
materials, But This will cause an obvious change in the cost of
materials !
五、选用高强度材料，提高许用应力值
同类 材料,, E”值相差不多,, ?jx”相差较大, 故换
用同类材料只能提高强度，不能提高刚度和稳定性 。
不同类材料,E和 G都相差很多（钢 E=200GPa,铜
E=100GPa），故可选用不同的材料以达到提高刚度和稳
定性的目的。但是，改换材料，其 原料费用 也会随之发生
很大的改变！
111
Chapter 6 Exercises
1,Which aspects can reflect the approximation of the
approximate differential equation of the deflective
curve?
2,What are the support conditions and the continuity
conditions when the integration method is used to determine
the equation of the deflective curve of the composite beam
shown in the figure?
3,An equal-section beam with the length L and the
weight P is put on the horizontal rigid plane,If one end of
the beam is lifted by the force P/3,but the other part still
stick on the plane,Try to determine the length of the lifted
part,
EI
xM
dx
vd )(
2
2
?
112
第六章 练习题
一、挠曲线近似微分方程 的近似性
反映在哪几方面?
二、用积分法求图示组合梁的挠曲线方程时,
需应用的支承条件和连续条件是什么?
三、长度为 L,重量为 P的等截面直梁,放置在水平
刚性平面上。若在端点施力 P/3上提,未提起部分仍
保持与平面密合,试求提起部分的长度。
EI
xM
dx
vd )(
2
2
?
113
Solution,The radius of curvature at point A of the
beam is,that is
?
01 ?? EIM A
A?
? 0?AM
0213 2 ?????? aLPaPM A
La 32?
114
解,A点处梁的曲率半径为,即
?
01 ?? EIM A
A?
? 0?AM
0213 2 ?????? aLPaPM A
La 32?
115
116
Mechanics of Materials
2
3
§ 6–4 Determine deflections and angles of rotation of the beam by the
principle of superposition
§ 6–5 Ckeck the rigidity of the beam
CHAPTER 6 DEFORMATION IN
BENDING
§ 6–6 Strain energy of the beam in bending
§ 6–7 Method to solve simple statically indeterminate problems of
the beam
§ 6–8 How to increase the load-carrying capacity of the beam
§ 6–1 Summary
§ 6–2 Approximate differential equation of the deflection curve of
the beam and its integration
§ 6–3 Method of conjugate beam to determine the deflection and
the rotational angle of the beam
4
§ 6–1 概述
§ 6–2 梁的挠曲线近似微分方程及其积分
§ 6–3 求梁的挠度与转角的共轭梁法
§ 6–4 按叠加原理求梁的 挠度与转角
§ 6–5 梁的刚度校核
第六章 弯曲变形
§ 6–6 梁内的弯曲应变能
§ 6–7 简单超静定 梁的求解方法
§ 6–8 如何提高梁的承载能力
§ 6－１ SUMMARY
Study range,Calculation of the displacement of the straight beam with
equal sections in symmetric bending,
Study object：① checking rigidify of the beam；② Solving problems about
statically indeterminate beams（ to provide complementary equations for the
geometric-deformation conditions of the beam )
§ 6－１ 概 述
研究范围：等直梁在对称弯曲时位移的计算。
研究目的：①对梁作刚度校核；
②解超静定梁（为变形几何条件提供补充方
程）。
1).Deflection,The displacement of the centroid of a section in a direction perpendicular
to the axis of the beam,It is designated by v, It is positive if its direction is the same as
f,otherwise it is negative,
3,The relation between the angle of
rotation and the deflection curve,
1,Two basic displacement quantities of to measure deformation of the beam
( 1 )
d
dtg f
x
f ???? ??
Small deflection
P
x
v
C
?
C1 f
2),Angle of rotation,The angle by
which cross section turns with respect
to its original position about the neutral
axis,it is designated by ?,It is
positive if the angle of rotation rotates
in the clockwise direction,otherwise it
is negative,
2,deflection curve,The smooth curve that the axis of the beam is
transformed into after deformation is called the deflection curve,Its equation is
v =f (x)
1.挠度：横截面形心沿垂直于轴线方向的线位移。 用 v表示。
与 f 同向为正，反之为负。
2.转角：横截面绕其中性轴转
动的角度 。用 ? 表示，顺时
针转动为正,反之为负。
二、挠曲线：变形后，轴线变为光滑曲线，该曲线称为挠曲线。
其方程为,v =f (x)
三、转角与挠曲线的关系,
一、度量梁变形的两个基本位移量
( 1 )
d
dtg f
x
f ???? ??
小变形
P
x
v
C
?
C1 f
§ 6-2 APPROXIMATE DIFFERENTIAL EQUATION OF THE
DEFLECTION CURVE OF THE BEAM AND ITS INTEGTION
z
z
EI
xM )(1 ?
?
1,Approximate differential
equation of the deflection curve
z
z
EI
xMxf )()( ?????
Formula (2) is the approximate differential equation
of the deflection curve,
EI
xMxf )()( ???? ? …… （ 2 ）
)(
)1(
)(1
2
32 xf
f
xf ????
??
??
??
?
Small
deformation f
x M>0
0)( ??? xf
f
x
M<0
0)( ??? xf
(1)
§ 6-2 梁的挠曲线近似微分方程及其积分
z
z
EI
xM )(1 ?
?
一、挠曲线近似微分方程
z
z
EI
xMxf )()( ?????
式（ 2）就是挠曲线近似微分方程。
EI
xMxf )()( ???? ? …… （ 2 ）
)(
)1(
)(1
2
32 xf
f
xf ??
??
??
??
??
?
小变形
f
x M>0
0)( ??? xf
f
x
M<0
0)( ??? xf
(1)
)()( xMxfEI ????
For the straight beam with the same shape and equal section area,the approximate
differential equation of the deflection curve may be written as the following form,
2,Determine the equation of the deflection curve (elastic curve)
)()( xMxfEI ????
1d))(()( CxxMxfEI ???? ?
21d)d))((()( CxCxxxMxE I f ???? ? ?
1).Integration of the differential equation
2).Boundary conditions of the displacement
P
A B C P D
)()( xMxfEI ????
对于等截面直梁，挠曲线近似微分方程可写成如下形式,
二、求挠曲线方程（弹性曲线）
)()( xMxfEI ????
1d))(()( CxxMxfEI ???? ?
21d)d))((()( CxCxxxMxE I f ???? ? ?
1.微分方程的积分
2.位移边界条件
P
A B C P D
Discussion,① Applying to the planar bending of slender beams with elastic materials and
small deformations,
② May be applied to determine the displacements of beams with equal or variable sections
under all types of loads,
③ Integral constants may be determined by the geometric compatible conditions（ boundary
conditions,continuity conditions),
④ Advantages,Wide applications and accurate solutions by the direct-solving method
Disadvantages Complicated calculation,
?Displacement conditions at the supports,
?Continuity conditions,
?Smooth conditions,
0?Af 0?Bf 0?Df 0?D?
?? ? CC ff
?? ? CC ?? C ri g h tC l e f tor ?? ?
C ri g h tC l e f t ffor ?
讨论,
①适用于小变形情况下、线弹性材料、细长构件的平面弯曲。
②可应用于求解承受各种载荷的等截面或变截面梁的位移。
③积分常数由挠曲线变形的几何相容条件（边界条件、连续条
件）确定。
④优点：使用范围广，直接求出较精确； 缺点：计算较繁。
?支点位移条件,
?连续条件,
?光滑条件,
0?Af 0?Bf 0?Df 0?D?
?? ? CC ff
?? ? CC ?? 右左或写成 CC ?? ?
右左或写成 CC ff ?
Example 1 Determine the elastic curves, maximum deflections and maximum
angles of rotation of the following equal-section straight beams,
?Set up the coordinates and write out the
bending-moment equation
)()( LxPxM ??
?Write out the differential
equation and integrate it
?Determinate the integral constants by
the boundary conditions
)()( xLPxMfEI ??????
1
2)(
2
1 CxLPfEI ?????
21
3)(
6
1 CxCxLPE I f ????
0
6
1)0(
2
3 ??? CPLEI f
021)0()0( 12 ?????? CPLfEIEI ?
3
2
2
1 6
1 ;
2
1 PLCPLC ????
Solution,
P
L
x
f
x
[例 1] 求下列各等截面直梁的弹性曲线、最大挠度及最大转角。
?建立坐标系并写出弯矩方程
)()( LxPxM ??
?写出 微分方程并积分 ?应用位移边界条件 求积分常数
)()( xLPxMfEI ??????
1
2)(
2
1 CxLPfEI ?????
21
3)(
6
1 CxCxLPE I f ????
0
6
1)0(
2
3 ??? CPLEI f
021)0()0( 12 ?????? CPLfEIEI ?
3
2
2
1 6
1 ;
2
1 PLCPLC ????
解,P
L
x
f
x
?Write out the equation of the elastic curve and plot its curve
? ?323 3)(
6
)( LxLxL
EI
Pxf ????
EI
PLLff
3
)(
3
m a x ??EI
PLL
2
)(
2
m a x ?? ??
?The maximum deflection and the maximum angle of rotation
x
f
P
L
?写出弹性曲线方程并画出曲线
? ?323 3)(
6
)( LxLxL
EI
Pxf ????
EI
PLLff
3
)(
3
m a x ??EI
PLL
2
)(
2
m a x ?? ??
?最大挠度及最大转角
x
f
P
L
Solution,?Set up the coordinates and write out the bending-
moment equation
?
?
?
??
???
?
)( 0
)0( )(
)(
Lxa
axaxP
xM
?
?
?
?
?
???
??
1
1
2
)(
2
1
D
CxaP
fEI
?
?
?
?
?
?
???
?
21
21
3)(
6
1
DxD
CxCxaP
E I f
?
?
?
??
???
???
)( 0
)0( )(
Lxa
axxaP
fEI
x
f
P
L
a
? Write out the differential
equation and integrate it
解,?建立坐标系并写出弯矩方程
?
?
?
??
???
?
)( 0
)0( )(
)(
Lxa
axaxP
xM
?写出 微分方程并积分
?
?
?
?
?
???
??
1
1
2
)(
2
1
D
CxaP
fEI
?
?
?
?
?
?
???
?
21
21
3)(
6
1
DxD
CxCxaP
E I f
?
?
?
??
???
???
)( 0
)0( )(
Lxa
axxaP
fEI
x
f
P
L
a
?Determine the integral constants by boundary conditions
0
6
1)0(
2
3 ??? CPaEI f
0
2
1)0(
1
2 ???? CPaEI ?
3
22
2
11 6
1 ;
2
1 PaDCPaDC ??????
)()( ?? ? afaf
)()( ?? ? aa ?? 11 DC ??
2121 DaDCaC ????
P
L
a
x
f
?应用位移边界条件 求积分常数
0
6
1)0(
2
3 ??? CPaEI f
0
2
1)0(
1
2 ???? CPaEI ?
3
22
2
11 6
1 ;
2
1 PaDCPaDC ??????
)()( ?? ? afaf
)()( ?? ? aa ?? 11 DC ??
2121 DaDCaC ????
P
L
a
x
f
?Write out the equation of the elastic curve and plot its curve
3 2 3
23
( ) 3 ( 0 )
6
()
3 ( a )
6
P
a x a x a x a
EI
fx
P
a x a x L
EI
?
??? ? ? ? ?
????
? ?
? ?? ? ? ?
???
?
? ?aL
EI
PaLff ??? 3
6
)(
2
m a x
EI
Paa
2
)(
2
m a x ?? ??
?The maximum deflection and the maximum angle of rotation
P
L
a
x
f
?写出弹性曲线方程并画出曲线
? ?
? ?
?
?
?
??
?
?
???
?????
?
)(a 3
6
)0( 3)(
6
)(
32
323
Lx axa
EI
P
ax axaxa
EI
P
xf
? ?aL
EI
PaLff ??? 3
6
)(
2
m a x
EI
Paa
2
)(
2
m a x ?? ??
?最大挠度及最大转角
P
L
a
x
f
§ 6-3 METHOD OF CONJUGATE BEAM TO DETERMINE THE DEFLECTION
AND THE ROTATIONAL ANGLE OF THE BEAM
)()( xMxfEI ????
1,Usage of the method,Determine the deflection and the
rotational angle of the designated point in the beam
2,Theoretical basis of the method,Similar analogy,
)()( xqxM ???
Differential equation of the deflection
curve of the beam,
Relation between the external load the
internal force,
§ 6-3 求梁的挠度与转角的共轭梁法
)()(:梁的挠曲线微分方程 xMxfEI ????
一、方法的用途：求 梁上指定点的挠度与转角。
二、方法的理论基础：相似比拟。
)()(:为梁的外载与内力的关系 xqxM ???
上二式形式相同，用类比法，将微分方程从形式上转化为
外载与内力的关系方程。从而把求挠度与转角的问题转化为求
弯矩与剪力的问题。
3,Conjugate beam（ relations between the real beam and the
imaginary beam）,
① The direction of the axis x and the origin of coordinates
② Having same geometric shapes,
③ Corresponding equation of the real beam,)()( xMxfEI ????
)()( xqxM ???
)()( xMxfEI ??????⑤ Integral of the ―force‖ differential
equation of the imaginary beam,
00 d)()()( QxxqxMxQ
x ???? ?
000 0 d)d)(()( MxQxxxqxM
x x ??? ? ?
( ) ( ),
q x M x a u o r d i n g t o t h i s e q u a t i o n e s t a b l i s h
d i s t r i b u t e d l o a d s o n t h e i m a g i n a r y b e a m
??let,
。
④
)()( xqxM ???
Corresponding equation of the imaginary
beam,
三、共轭梁（实梁与虚梁的关系）,
① x轴指向及 坐标原点完全相同 。
② 几何形状完全相同。
③ 实梁对应方程,)()( xMxfEI ????
)()( xqxM ???
)()( xMxfEI ??????
⑤ 虚梁“力”微分方程的积分
00 d)()()( QxxqxMxQ
x ???? ?
000 0 d)d)(()( MxQxxxqxM
x x ??? ? ?
载荷。依此建立虚梁上的分布令,)()( xMxq ??
④
)()( xqxM ???
虚梁对应方程,
)()( xMxEI f ??
The quantities with subscripts ―0‖dentte they are in the coordinate
origin coordinate origin,
Integral of the―displacement‖ differential equation of the real beam
)()( xMxfEI ????
00 ))(()( ?? EIdxxMxfEIEI
x ????? ?
000 0 d)d))((()( E I fxEIxxxMxE I f
x x ???? ? ? ?
)()( xQxEI ??
⑥ Set up the ―force‖ boundary conditions of the imaginary beam according to
the ―displacement‖ boundary conditions of the real beam,
AAAA QEIMEI f ?? ? ;
)()( xMxEI f ??
下脚标带,0‖的量均为坐标原点的量。
实梁“位移”微分方程的积分
)()( xMxfEI ????
00 ))(()( ?? EIdxxMxfEIEI
x ????? ?
000 0 d)d))((()( E I fxEIxxxMxE I f
x x ???? ? ? ?
)()( xQxEI ??
⑥ 依实梁的“位移”边界条件建立虚梁的“力”边界条件。
AAAA QEIMEI f ?? ? ;
中间铰
支座 A
0?Af
0?A?
0?Af
0?A?
右左 AA ?? ?
右左 AA ff ? 右左 AA MM ?
右左 AA QQ ?
固定端 A
A
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0?Af
0?A?
0?? 右左 AA ??
0?Af
0?? 右左 AA QQ
0 ?AM
固定端 A
A
自由端 A A
自由端 A A
铰支端 A
A
铰支端 A
A
中间铰
支座 A
中间铰 A
中间铰 A
Imaginary beam Real beam Conjugated beam
Supports and ends Displacement boundary Supports and ends Force boundary
Fixed
end A
Free
end A
Hinged
support A
Middle
hinged
support A
Middle
hinge A
Free
end A
Fixed
end A
Hinged
support A
Middle
hinge A
Middle hinged
support A
中间铰
支座 A
虚虚 梁梁
实实 梁梁
共 轭 梁
支 承 和 端 部 情 况
位移边界 相应的支承和端部情况 力边界
0?Af
0?A?
0?Af
0?A?
右左 AA ?? ?
右左 AA ff ? 右左 AA MM ?
右左 AA QQ ?
固定端 A
A
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0 ?AM
0 ?AQ
0?Af
0?A?
0?? 右左 AA ??
0?Af
0?? 右左 AA QQ
0 ?AM
固定端 A
A
自由端 A A
自由端 A A
铰支端 A
A
铰支端 A
A
中间铰
支座 A
中间铰 A
中间铰 A
Summary,Relations between the equal-section real beam and the
imagine beam as following,
① Coordinate origin and direction
of axis x are all completely same,
② Geometric shapes are
completely same,
④ Set up the ―force‖ boundary conditions of the imaginary beam according to
the ―displacement‖ boundary conditions of the real beam,
AAAA QEIMEI f ?? ? ;
EI
Q
EI
M
f xxxx ?? ? ;
⑤ Determine the ―displacement‖ of the real beam according to the ―internal
force‖ of the imaginary beam,
a, Fixed end Free end
b,
c, middle hinged supports
middles hinge
L e t ( ) ( )
,
q x M x a c c o r d i n g t o t h i s e q u a t i o n e s t a b l i s h
u n i f o r m l o a d s o n t h e i m a g i n a r y b e a m
??：
③
Hinged
supports
Hinged
supports
总结：等截面实梁与虚梁的关系如下,
① x 轴指向及 坐标原点完全相同。 ② 几何形状完全相同。
④ 依实梁的“位移”边界条件，建立虚梁的“力”边界条件。
AAAA QEIMEI f ?? ? ;
EI
Q
EI
M
f xxxx ?? ? ;
⑤ 依虚梁的“内力”，求实梁的“位移”。
a,固定端 自由端
b,铰支座 铰支座
c,中间铰支座 中间铰链
载荷。依此建立虚梁上的分布令,)()( xMxq ??
③
Solution,? Set up coordinates and
the imaginary beam
Example 2 Determine the displacement at point B (deflection and angle of
rotation) of the following straight beams with equal sections in shape and area
2)(
2
q)( xLxM ???
2)(
2
q)( )( xLxMxq ????
q
L
A
B
f
x
2
2
0
qLq ?
)(xq
A B
L
Determine the bending moment of
the real beam in order to obtain loads
of the imaginary beam
? Determine the shearing force and
the bending moment at the point B of
the imaginary beam in order to obtain
the deflection and the rotational angle
at the same point of the real beam,
解,? 建立坐标和虚梁
[例 2] 求下列等截面直梁 B点的位移（挠度和转
角）。
?求虚梁 B点的剪力和弯矩，
以求实梁 B点的转角和挠度
求实梁的弯矩方程
以确定虚梁荷载
2)(
2
q)( xLxM ???
2)(
2
q)( )( xLxMxq ????
q
L
A
B
f
x
2
2
0
qLq ?
)(xq
A B
L
3
0 LqQEI
BB ???
84
3 4qLLAMEI f
qBB ????
EI
qL
B 6
3
?? ?
EI
qLf
B 8
4
??
)( xqQ B ?
2
2
0
qLq ?
)(xq
A B
L
( ) i n t
i n t
B B M o m e u t o f q x w i t h r e s p e c t t o P o B i n
t h e l e f t s i d e o f t h i s P o
M ?
Area of in the left side of the
point B
The moment of the area of
of left side of the point B about
point B
?求虚梁 B点的剪力和弯矩，以求实梁 B点的转角和挠度
3
0 LqQEI
BB ???
84
3 4qLLAMEI f
qBB ????
EI
qL
B 6
3
?? ?
EI
qLf
B 8
4
??
面积) 的(点左侧 xqBQ B ?
2
2
0
qLq ?
)(xq
A B
L
面积对) 的(点左侧 xqBBM ?
B点之矩
Solution,? Set up coordinates
and the imaginary beam
?Determine the shearing force and
bending moment of point B in
the imaginary beam,
Find out the bending-moment
equation to determine loads of the
imaginary beam,
) )( M ( xxq ??
???? 2 ; 2 qaRqaR DA
q qa2
qa
A B
C D
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
a a a
f
x
D
解,? 建立坐标和虚梁
?求虚梁 B点的剪力和弯矩
求实梁的弯矩方程以确定
虚梁荷载
) )( M ( xxq ??
???? 2 ; 2 qaRqaR DA
q qa2
qa
A B
C D
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
a a a
f
x
D
72
13 3 ?? qaR
A
3
23
72
5
22
1
72
13 qaaqaqa
BQ ????
?How about the displacement near
point C？
4
23
72
7
322
1
72
13 qaaaqaaqa
BM ?????
EI
qa
B 72
5 3
?? EI
qaf
B 72
7 4??
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
A B C
a a a
D
qa2/2
3qa2/8
?求虚梁 B点的剪力和弯矩
72
13 3 ?? qaR
A
3
23
72
5
22
1
72
13 qaaqaqa
BQ ????
?C点左右位移怎样？
4
23
72
7
322
1
72
13 qaaaqaaqa
BM ?????
EI
qa
B 72
5 3
?? EI
qaf
B 72
7 4??
qa2/2
x M qa2/2
qa2/2 3qa2/8
–
+
A B C
a a a
D
qa2/2
3qa2/8
① Convert the change of section into the bending moment,
② Geometric shape,the length is not change,moment of inertia
changed into I0,
③ Corresponding equation of
the real beam,
Corresponding equation of
the imaginary beam,
)()(0 xMxfEI ?????
)()( xqxM ???
4,Method of conjugate beam for non-constant beam,
00
0 )(
)(
)()(
EI
xM
I
I
xEI
xMxf ????????
)()(0 xMxfEI ?????
)(
)()( 0
xI
IxMxM ??
L e t ( ) ( )q x M x???：④ According to this establish the uniform load on the imaginary
beam,They are all the same as the equal section straight beam
in other aspects,
① 将截面的变化折算到弯矩之中去。
② 几何形状：长度不变，惯性矩变为 I0 。
③ 实梁对应方程,
虚梁对应方程,
)()(0 xMxfEI ?????
)()( xqxM ???
四、变截面直梁的共轭梁法,
00
0 )(
)(
)()(
EI
xM
I
I
xEI
xMxf ????????
)()(0 xMxfEI ?????
)(
)()( 0
xI
IxMxM ??
其它与等截面直梁完全相同。
载荷。依此建立虚梁上的分布令,)()( xMxq ???
④
Example 3 Determine the displacement
of point C on the following beam with
non-constant section,Knowing：
IDE=2IEB =2IAD,
Solution,? Set up coordinates and
the imaginary beam,
) )( (xMxq ???
ADADAD xMxI
IxMxM )(
)(
)()( 0 ???
2
)()()( DE
DE
AD
DEDE
xM
I
IxMxM ???
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
[例 3] 求下列变截面直梁 C点的
位移，已知,IDE =2IEB =2IAD 。
解,? 建立坐标和虚梁
) )( (xMxq ???
ADADAD xMxI
IxMxM )(
)(
)()( 0 ???
2
)()()( DE
DE
AD
DEDE
xM
I
IxMxM ???
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
?Determine the shearing force and
the bending moment at point C
of the imaginary beam,
32
5 2 ?? PaR
A
0?CQ
3
32
3
6282
1
428 Pa
aaPaaaPa ??
0?C?
AD
C EI
Pa
f
32
3 3
?
???
3
2
242
1
32
5 2 aaPaaPa
CM
a a
P 0.5a
A B
C D E x
f
x
M
2
Pa
4
Pa
4
Pa4Pa
?M
)(xq
4
Pa
4
Pa4Pa
?求虚梁 C点的剪力和弯矩
32
5 2 ?? PaR
A
0?CQ
3
32
3
6282
1
428 Pa
aaPaaaPa ??
0?C?
AD
C EI
Pa
f
32
3 3
?
???
3
2
242
1
32
5 2 aaPaaPa
CM
§ 6-4 DETERMINE DEFLECTIONS AND ROTATIONAL ANGLES
OF THE BEAM BY THE PRINCIPLE OF SUPERPOSITION
1,Superposition of loads,The deformation of a structure due
to the action of multi-loads is equal to the algebraic sum of
deformation resulting from the separate action of each load,
1 2 1 1 2 2( ) ( ) ( ) ( )n n nP P P P P P? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
1 2 1 1 2 2( ) ( ) ( ) ( )n n nf P P P f P f P f P? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
2,Superposition of structural forms（ rigidization method of
segment by segment）,
§ 6-4 按叠加原理求梁的 挠度与转角
一、载荷叠加,多个载荷同时作用于结构而引起的变形
等于每个载荷单独作用于结构而引起的变形的代数和。
1 2 1 1 2 2( ) ( ) ( ) ( )n n nP P P P P P? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
1 2 1 1 2 2( ) ( ) ( ) ( )n n nf P P P f P f P f P? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?、,
二、结构形式叠加（逐段刚化法）,
Solution:① Decompose the loads as
shown in the figure
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
q L af
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
② Determine the deformations of the
beam under the action of simple loads
by looking up the table,
Example 4 Determine the angle
of rotation of point A and the
deflection of point C by the principle of
superposition,
[例 4] 按叠加原理求 A点转角和 C
点
挠度。
解,① 载荷分解如图
② 由梁的简单载荷变形表,
查简单载荷引起的变形。
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
qaf
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
qaf
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
③ Sum up
qAPAA ??? ??
)43(
12
2
qaP
EI
a ??
EI
Pa
EI
qaf
C 624
5 34 ??
EI
Paf
PC 6
3
?
EI
Pa
PA 4
2
??
45
24qC
qaf
EI
?EI
qa
qA 3
3
??
q
q
P
P =
+
A
A
A
B
B
B
C
a a
③ 叠加
qAPAA ??? ??
)43(
12
2
qaP
EI
a ??
EI
Pa
EI
qaf
C 624
5 34 ??
Example 5 Determine the deflection of point C by the principle of
superposition,
Solution,① Divide the load infinitely
as shown in the figure,
② Determine the deformations of the
beam under the action of simple loads by looking up the table,
③ Sum up
22( d ) ( 3 4 )
48d P C
P b L bf
EI
??
bLbqxxqP d2d)(d 0??
2 2 2
0 ( 3 4 ) d
24
q b L b b
E I L
??
?? ? d P CqC ff
2 2 2 40, 5
00
0
( 3 4 ) d
2 4 2 4 0
L q b L b q L
b
E I L E I
? ??
q0
0.5L 0.5L
x dx x
f
C
b
[例 5] 按叠加原理求 C点挠度。 解,?载荷无限分解如图
?由梁的简单载荷变形表,
查简单载荷引起的变形。
?叠加
22( d ) ( 3 4 )
48d P C
P b L bf
EI
??
bLbqxxqP d2d)(d 0??
2 2 2
0 ( 3 4 ) d
24
q b L b b
E I L
??
?? ? d P CqC ff
2 2 2 40, 5
00
0
( 3 4 ) d
2 4 2 4 0
L q b L b q L
b
E I L E I
? ??
q0
0.5L 0.5L
x dx
b
x
f
C
Example 6 Explanation of the superposition of structural forms (digitization
method of segment by segment ),
=
+
P L1 L2
A B C
B C
P L2
f1
f2
Equivalence
x
f
x
f
21 fff ??
f
P L1 L2
A B C Rigidize the
segment AC
P L1 L2
A B C
Rigidize the
segment BC
P L1 L2
A B C M x
f
Equivalence
[例 6] 结构形式叠加（逐段刚化法 ) 原理说明。
=
+
P L1 L2
A B C
B C
P L2
f1
f2
等价
等价
x
f
x
f
21 fff ??f
P L1 L2
A B C 刚化 AC段
P L1 L2
A B C
刚化 BC段
P L1 L2
A B C M x
f
§ 6-5 RIGIDITY CHECK OF THE BEAM
))1000 1~250 1( ( m a x ??
?
?
??
?
??
?
??
??
L
f
L
f
L
f
? ??? ?m a x
1,Rigidity conditions of the beam
Where [?] is called the permissible angle of rotation； [f/L] is called the
permissible ratio of the deflection and the span,In general we can do three
kinds of calculations about the rigidity by theses conditions,
?,Check the rigidity,
??
?
??
??
L
f
L
f m a x
? ??? ?m a x
(In civil engineering,the strength often-plays an
important role and the rigidity plays a secondary role
except special cases.）
For the civil engineering,
?,Design the dimension
of the section；
?,Determine the
permissible load。
§ 6-5 梁的刚度校核
))
1000
1~
250
1(,对土建工程( m a x ?
??
?
??
?
??
?
??
??
L
f
L
f
L
f
一、梁的刚度条件
其中 [?]称为许用转角； [f/L]称为许用挠跨比。通常 依此条件
进行如下三种刚度计算,
?,校核刚度,
?,设计截面尺寸；
?,设计载荷。
(但：对于土建工程，强度常处于主要地位，
刚度常处于从属地位。特殊构件例外）
? ??? ?m a x
??
?
??
??
L
f
L
f m a x
? ??? ?m a x
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
Example 7 A hollow circular-section beam is shown in the following figure,Its
inside and outside diameter is respectively,d=40mm,D=80mm,Knowing
E=210GPa,[f/L]=0.00001 at point C,[?]=0.001rad at point B,Try to check the
rigidity of the beam,
=
+ +
=
P1=1kN
A B D C
P2
B C D A
P2=2kN
B C D A
P2
B C
a
P2
B C D A
M
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
[例 7] 下图为一空心圆截面梁，内外径分别为,d=40mm、
D=80mm，梁的 E=210GPa，工程规定 C点的 [f/L]=0.00001,B点
的 ??]=0.001弧度,试校 核此梁的刚度。
=
+ +
=
P1=1kN
A B D C
P2
B C D A
P2=2kN
B C D A
P2
B C
a
P2
B C D A
M
P2
B C
a
=
+
+
图 1
图 2
图 3
EI
aLPaf
BC 16
2
1
11 ?? ?
EI
LP
B 16
2
1
1 ??
EI
L a P
EI
ML
B 33
2
3 ?????
EI
LaPaf
BC 3
2
2
33 ??? ?
Solution,?Transform the structure
and determine the deformations under
the actions of each simple load by
looking up the table,
02 ?B?
EI
aPf
C 3
3
2
2 ??
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
P2
B C
a
=
+
+
图 1
图 2
图 3
EI
aLPaf
BC 16
2
1
11 ?? ?EI
LP
B 16
2
1
1 ??
EI
L a P
EI
ML
B 33
2
3 ?????
EI
LaPaf
BC 3
2
2
33 ??? ?
解,?结构变换，查表求简单
载荷变形。
02 ?B? EI
aPf
C 3
3
2
2 ??
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
P2
B C
a
=
+
+
图 1
图 2
图 3
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
EI
LaP
EI
aP
EI
aLPf
C 3316
2
2
3
2
2
1 ???
EI
LaP
EI
LP
B 316
2
2
1 ???
?Determine the deformation under the action of complex loads by superposition
48
1244
44
m10188
10)4080(
64
14.3
)(
64
?
?
??
???
?? dDI
?
P2
B C
a
=
+
+
图 1
图 2
图 3
P L=400mm
P2=2kN
A C
a=0.1m
200mm
D
P1=1kN
B
P1=1kN
A B D C
P2
B C D A
M
x
f
EI
LaP
EI
aP
EI
aLPf
C 3316
2
2
3
2
2
1 ???
EI
LaP
EI
LP
B 316
2
2
1 ???
?叠加求复杂载荷下的变形
48
1244
44
m10188
10)4080(
64
14.3
)(
64
?
?
??
???
?? dDI
?
m1019.5
3316
6
2
2
3
2
2
1 ???????
EI
LaP
EI
aP
EI
aLPf
C
42
2
1 10423.0)
3
200
16
400(
1880210
4.0
316
?????
???? EI
LaP
EI
LP
B?
? ? 001.010423.0 4m a x ???? ? ??
??
?
??
??
L
f
L
f m a x
? ? m101m1019.5 56m a x ?? ????? ff
?Check the rigidity
(rad)
m1019.5
3316
6
2
2
3
2
2
1 ???????
EI
LaP
EI
aP
EI
aLPf
C
)(10423.0)
3
200
16
400(
1880210
4.0
316
42
2
1 弧度?????
?
???
EI
LaP
EI
LP
B?
? ? 001.010423.0 4m a x ???? ? ??
??
?
??
??
L
f
L
f m a x
?校核刚度
? ? m101m1019.5 56m a x ?? ????? ff
dx
x
Q Q+dQ
M M+dM
1,Calculation of strain energy in bending,
§ 6–6 STRAIN ENERGY OF THE BEAM IN BENDING
EI
xM )(1 ?
?
?d)(
2
1dd ??? xMWU
x
EI
xMU d
2
)(d 2?
?? L xEI xMU d2 )(
2
?
?
xd
d ?
Strain energy is equal to the work of external forces.Neglect the shearing
strain energy an omit
?dd ?M
d?
M(x)
P1
M x
f
P2 dx
d? ?
dx
x
Q Q+dQ
M M+dM
一、弯曲应变能的计算,
§ 6–6 梁内的弯曲应变能
EI
xM )(1 ?
?
?d)(
2
1dd ??? xMWU
x
EI
xMU d
2
)(d 2?
?? L xEI xMU d2 )(
2
?
?
xd
d ?
应变能等于外力功。不计剪切应变能并略去 ?dd ?M
d?
M(x)
P1
M x
f
P2 dx
d? ?
Example 8 Determine the deflection of point C on the beam by the energy
method,The beam is of equal section and straight,
CPfW 2
1?
Solution,Strain energy is equal to
the work of external forces,
?? L xEI xMU d2 )(
2
)0(;
2
)( axxPxM ???
Using the symmetry we get,
EI
aPxxP
EI
U a
12
d)
2
(
2
12 32
0
2 ?? ?
EI
PafUW
C 6 h a v e W eL o t
3
??Thinking,Can we use this method
to determine the displacement of point
C when distributed loads are applied?
P
a a
q
x
f
A
C
B
[例 8] 用能量法求 C点的挠度。梁为等截面直梁。
CPfW 2
1?
解：外力功等于应变能
?? L xEI xMU d2 )(
2
)0(;
2
)( axxPxM ???
在应用对称性，得,EIaPxxPEIU a 12d)2(2 12 32
0
2 ?? ?
EI
PafUW
C 6
3
????
思考：分布荷载时，可否用此法求 C点位移？
q
x
f
P
a a
A
C
B
2,Impact problems of the beam
1).Assumptions,
?Impact bodies are rigid bodies；
?Neglecting the potential energy of weight
and the dynamic energy of the impacted
bodies；
?Impact bodies do not rebound；
?Neglecting the loss of energy of sound、
light,heat and so on(energy is conservative),
0)(21 2111 ?????? dfhmgmvUVT
mg
L
h A B C
A B C x
f
fd Before impact,
二,梁的冲击问题 1.假设,
?冲击物为刚体；
?不计被冲击物的重力势能和动能；
?冲击物不反弹；
?不计声、光、热等能量损耗（能
量守恒）。
0)(
2
1
冲击前
2
111
???
???
dfhmgmv
UVT
mg
L
h A B C
A B C x
f
fd
A B C x
f
fd
22
2
)(
2
1
)(
2
1
)(
2
1
2
1
00
d
j
d
j
j
ddd
f
f
mg
f
f
P
fkfP
??
????
After impact
222 UVT ??
The energy before and after impact
is conservative,therefore,
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgvf ????? )2)(11( 2
d
jj
d
d f
hgv
f
fK 2)2(11 ?????
jf
h
dK
211 ??? 2?dK
Coefficient of
dynamic load
(2)Sudden load,(1)Freely falling body
22
2
222
)(
2
1
)(
2
1
)(
2
1
2
1
00
冲击后
d
j
d
j
j
ddd
f
f
mg
f
f
P
fkfP
UVT
??
????
??
冲击前、后，能量守恒，所以,
A B C x
f
fd
22 )(
2
)(
2
1
d
j
d ff
mgfhmgmv ???
jdj
j
fKf
f
hgvf ????? )2)(11( 2
d
jj
d
d f
hgv
f
f
K
2)2(
11:
?
????动荷系数
jf
h
dK
211:)1( ???自由落体 2:)2( ?
dK突然荷载
h
B A C
mg E =P
3,Calculation of dynamic response,
Solution,?Determine the
static deflection of point C,
2;
2 21
1 PRCCAAf
ACj ???
The dynamic response is equal to the product of the static response and the
coefficient of dynamic load,
ABDE
A
EI
PL
EI
LR
4848
33
?? EI
PL
32
3
?
C1
A1
D
EIEIEI DEAB ??
L
C2
Example 9 A structure is shown
in the figure,AB=DE=L,A,C are
respectively the middle points of AB
and DE, Determine the dynamic
`stress of section C under
the impact of weight mg,
h
B A C
mg E =P
三、动响应计算,
解,?求 C点静挠度
2;2 21
1 PRCCAAf
ACj ???
动响应计算等于静响应计算与动荷系数之积,
[例 9] 结构如图,AB=DE=L,A,C 分别为 AB 和 DE 的中点，
求梁在重物 mg 的冲击下,C 面的动应力。
ABDE
A
EI
PL
EI
LR
4848
33
??
EI
PL
32
3
?
C1
A1
D
EIEIEI DEAB ??
L
C2
?Coefficient of dynamic load
3
64
11
2
11
PL
E I h
f
h
d
K
Cj
???
???
?Determine the dynamic stress in the section C
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
64
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
?动荷系数
3
64
11
2
11
PL
E I h
f
h
d
K
Cj
???
???
?求 C面的动应力
zz
C
dCjdCd W
PL
PL
E I h
W
M
KK
4
)
64
11( 3m a xm a x ????? ??
h
B A C
mg E =P
C1
A1
D
EIEIEI DEAB ??
L
C2
§ 6-7 METHOD TO SOLVE SIMPLE
STATICALLY INDETERMINATE
PROBLEMS OF THE BEAM
1,Treatment method,
Combining the compatibility equation of
deformation,physical equation with
equilibrium equations to determine the
whole unknown forces,
Solution,? Set up the primary beam
Determine the degree of statically
indeterminacy,The structure in which
redundant constraints are substituted by
reactions—primary structure。
=
EI
q0
L A
B
L
q0 MA
B A
q0
L RB A
B
x
f
§ 6-7 简单超静定 梁的求解方法
1、处理方法：变形协调方程、物理
方程与平衡方程相结合，求全部未
知力。
解,?建立静定基
确定超静定次数，用反力
代替多余约束所得到的结构 —
—静定基。
=
EI
q0
L A
B
L
q0 MA
B A
q0
L RB A
B
x
f
?Geometric equation—compatibility
equation of deformation
0??? BBRBqB fff
+
q0
L RB A
B
=
RB
A B
q0
A B
?Physical equation—relation between
the deformation and forces
?Complementary equation
EI
LRf
EI
qLf B
BRBq B 3;8
34
???
0
38
34
??
EI
LR
EI
qL B
8
3 qLR
B ??
?Solve other problems（ reaction、
stress,deformation etc.）
?几何方程 ——变形协调方程
0??? BBRBqB fff
+
q0
L RB A
B
=
RB
A B
q0
A B
?物理方程 ——变形与力的关系
?补充方程
EI
LRf
EI
qLf B
BRBq B 3;8
34
???
0
38
34
??
EI
LR
EI
qL B
8
3 qLR
B ??
?求解其它问题（反力、应力,
变形等）
?Geometric equation—compatibility
equation of deformation
Solution,? Set up the primary beam
BCBRBqB Lfff B ????
=
Example 10 Determine the
reaction at end B in the structure as
shown in the figure,
LBC EA
x
f
q0
L RB A
B
C
q0
L RB A
B
EI =
RB
A B
+ q0
A B
EI
?几何方程
——变形协调方程,
解,?建立静定基
BCBRBqB Lfff B ????
=
[例 10] 结构如图，求 B点反
力。 LBC EA
x
f
q0
L RB A
B
C
q0
L RB A
B
EI =
RB
A B
+ q0
A B
EI
=
LBC EA
x
f
q0
L RB A
B
C
RB
A B +
q0
A B
?Physical equation—relation between the
deformation and forces
?Complementary equation
? Solve other problems（ reaction、
stress,deformation etc.）
EI
LRf
EI
qLf B
BRBq B 3; 8
34
???
EA
LR
EI
LR
EI
qL BCBB ??
38
34
)
3
(8
3
4
I
L
A
L
I
qL
R
BC
B
?
??
EA
LRL BCB
BC ??
=
LBC EA
x
f
q0
L RB A
B
C
RB
A B +
q0
A B
?物理方程 ——变形与力的关系
?补充方程
?求解其它问题（反力、应力,
变形等）
EI
LRf
EI
qLf B
BRBq B 3; 8
34
???
EA
LR
EI
LR
EI
qL BCBB ??
38
34
)
3
(8
3
4
I
L
A
L
I
qL
R
BC
B
?
??
EA
LRL BCB
BC ??
§ 6-8 HOW TO INCREASE THE LOAD-CARRYING
CAPACITY OF THE BEAM
Strength,
Normal stress,
Shearing stress,
? ? m a x ?? ??
zW
M
? ??? ??
z
z
bI
QS *
zEI
xMf )(????Rigidity,
Stability,
Those parameters are all relative to the internal forces and the
properties of the section,
§ 6-8 如何提高梁的承载能力
强度：正应力,
剪应力,
? ? m a x ?? ??
zW
M
? ??? ??
z
z
bI
QS *
zEI
xMf )(????
刚度,
稳定性,
都与内力和截面性质有关 。
R
b
h
（ 1） Reasonable ration between the height and the width of the section of
the wooden beam with rectangular section
Li Jie of north-Song dynasty pointed out in the book
of written in the year 1100 that the reasonable ratio
of height and width (h/b) of a rectangular wooden
beam is 1.5,
T·Young pointed out in the book of written in 1807 that the strength
reaches the maximum when the reasonable ratio of height to width
of a rectangular woden beam is h/b= and the rigidity reaches
the maximum when h/b=,
2
3
一、选择梁的合理截面
矩形木梁的合理高宽比
北宋李诫于 1100年著 ?营造法式 ?一书中指出,
矩形木梁的合理高宽比 ( h/b)为 1.5
英 (T.Young)于 1807年著 ?自然哲学与机械技术讲义 ?一书中指出,
矩形木梁的合理高宽比 为
刚度最大。时强度最大时,3 ;,2 ?? bhbh
R
b
h
Reasonable section in general case
A
Q
3
433.1
mm a x ?? ?? 32
3
1
1
DW
z
??
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 ?? ?
)2/( ;,
4
w h e n 12
2
1 DRRaaD ??? ??
1,If areas of sections are the same,select the section with a larger
modulus in bending,
z
D1
z a
a
1, 0 5
12
1
3
2 zz I
bhI ??
一般的合理截面
A
Q
3
433.1
mm a x ?? ??
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 ?? ?
)2/( ;,
4 1
2
2
1 DRRaaD ??? ?? 时当
1、在面积相等的情况下，选择抗弯模量大的截面
z
D1
z a
a
1, 0 5
12
1
3
2 zz I
bhI ??
32
3
1
1
DW
z
??
mm ax 2 ?? ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ? 1
222
1 67.1,
4
])8.0([
4w h e n DD
DDD ??? ??
4/2,2
4 11
2
1
2
1w h e n DaaD ?? ??
1
3
1
2
4 67.1 6
4
6 zz
WabhW ???
mm ax 5.1 ?? ?
z D
0.8
D
a1
2a
1 z
59.4)8.01(
64
14
4
3 zz I
DI ??? ?
2, 0 912812 z1
4
1
3
4 I
abh I
z ???
mm ax 2 ?? ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ? 1
222
1 67.1,
4
])8.0([
4 DD
DDD ??? 时当 ??
4/2,2
4 11
2
1
2
1 DaaD ?? ?? 时当
1
3
1
2
4 67.1 6
4
6 zz
WabhW ???
mm ax 5.1 ?? ?
z D
0.8
D
a1
2a
1 z
59.4)8.01(
64
14
4
3 zz I
DI ??? ?
2, 0 912812 z1
4
1
3
4 I
abh I
z ???
55.9 15 zz II ?
)(= 3.2 mm a x
fA
Q?? ?
For the I-shape section the method
to determine the shearing stress is
similar to that for the –shape
section,
15 57.4 zz WW ?
,6.18.024 2222
2
1 aaD ????
0.8a2
a2
1.6
a 2 2a 2 z
when 12 05.1 Da ?
55.9 15 zz II ?
)(= 3.2 mm a x
fA
Q?? ?
工字形截面与框形截面类似。
15 57.4 zz WW ?
12
2
2
2
2
2
1 05.1,6.18.02
4
DaaaD ???? 时当
?
0.8a2
a2
1.6
a 2 2a 2 z
2,Select shapes of the section according to properties material
?
G
z
For example,for the material of cast iron,the T-shape section is often
used as shown in the following figure,
2,Select the beam with non-constant sections,
The best is to select equal strength beam,that is
][
)(
)()(
m a x ?? ?? xW
xMx
If it is an equal-strength rectangular section,its height is
][
)(6)(
?b
xMxh ?
At the same time
][
)(
5.1m a x ?? ??
xbh
Q
][
5.1)(
?b
Qxh ?
P
x
)(xh
2、根据材料特性选择截面形状
?
G
z
如铸铁类材料，常用 T字形类的截面，如下图,
二、采用变截面梁
最好是等强度梁，即
][
)(
)()(
m a x ?? ?? xW
xMx
若为等强度矩形截面，则高为
][
)(6)(
?b
xMxh ?
同时
][
)(
5.1m a x ?? ??
xbh
Q
][
5.1)(
?b
Qxh ??
P
x
)(xh
EI
PLy 3
m a x 0 2 1.0?
EI
PLy 3
m a x 0 1 4.0?
EI
PLy 3
m a x 0 0 7 3.0?
3,Arrange reasonably external forces（ include reactions of supports）
and make the maximum bending moment M max as small as possible,P
L/2 L/2
x
M
+
PL/4
P
L/4 3L/4
P=qL
L/5 4L/5
Symmetry
M
x
3PL/16
+
M
x
qL2/10
+
EI
PLy 3
m a x 0 2 1.0?
EI
PLy 3
m a x 0 1 4.0?
EI
PLy 3
m a x 0 0 7 3.0?
三、合理布置外力（包括支座），使 M max 尽可能小。
P
L/2 L/2 M
x
+
PL/4
P=qL
L/5 4L/5
对称
P
L/4 3L/4 M
x
3PL/16
+
M
x
qL2/10
+
EI
qLy 4
m a x 013.0?
EI
qLy 43
m a x 107 8 7 5.0
???
EI
qLy 43
m a x 10326.0
???
q
L
L/5
q
L/5
q
L/2 L/2
M
x
8
2qL
+
40 2 qL
50 2 qL ?
M
x -
+ -
32 2 qL ?
M
x
32
9
16
9 2qL?
32
9
16
9 2qL?
+ +
-
EI
qLy 4
m a x 013.0?
EI
qLy 43
m a x 107 8 7 5.0
???
EI
qLy 43
m a x 10326.0
???
q
L
L/5
q
L/5
q
L/2 L/2
M
x
8
2qL
+
40 2 qL
50 2 qL ?
M
x -
+ -
32 2 qL ?
M
x
32
9
16
9 2qL?
32
9
16
9 2qL?
+ +
-
Z
Y
cr
I
I
L
GEb ?
? ?
4,Lateral buckling of the beam
1),Critical load of the beam with rectangular section in pure bending
M
M
x
y
z
Z
Y
cr
I
I
L
GEb ?
? ?
四、梁的侧向屈曲
1.矩形纯弯梁的临界载荷
M
M
x
y
z
2,Critical load of the beam with the I-shape section in pure bending
M
M
x
y
z
h
??
?
?
??
?
?
???
?
?
?
?
??
Z
Y
Z
Y
Z
Y
cr I
I
I
I
EG
I
I
L
E
L 2
2
2
2
)(
2
h ??
?
From the above analysis we know if I y is too small,the strength and rigidity
of the structure are higher,but the possibility of lateral loss of stability is also
increasing,About this point we must pay enough attention,
2.工字钢形截面纯弯梁的临界载荷
M
M
x
y
z
h
??
?
?
??
?
?
???
?
?
?
?
??
Z
Y
Z
Y
Z
Y
cr I
I
I
I
EG
I
I
L
E
L 2
2
2
2
)(
2
h ??
?
由上可见,I y过小时，虽然强度和刚度较高，但侧向失稳
的可能性却增大了，这点应引起注意。
5,Select the material of high strength to increase the
permissible stress
For the same kind of materials,differences of their“E” are small
,but those of their,?jx” are larger,So the strength can be improved by
changing another same kind of material,but the rigidity and stability
can not be improved,
For the different kind of materials,differences of their E and G
are very large（ steel, E=200GPa,copper, E=100GPa）,So the
rigidity and stability can be improved by selecting different Rind of
materials, But This will cause an obvious change in the cost of
materials !
五、选用高强度材料，提高许用应力值
同类 材料,, E”值相差不多,, ?jx”相差较大, 故换
用同类材料只能提高强度，不能提高刚度和稳定性 。
不同类材料,E和 G都相差很多（钢 E=200GPa,铜
E=100GPa），故可选用不同的材料以达到提高刚度和稳
定性的目的。但是，改换材料，其 原料费用 也会随之发生
很大的改变！
111
Chapter 6 Exercises
1,Which aspects can reflect the approximation of the
approximate differential equation of the deflective
curve?
2,What are the support conditions and the continuity
conditions when the integration method is used to determine
the equation of the deflective curve of the composite beam
shown in the figure?
3,An equal-section beam with the length L and the
weight P is put on the horizontal rigid plane,If one end of
the beam is lifted by the force P/3,but the other part still
stick on the plane,Try to determine the length of the lifted
part,
EI
xM
dx
vd )(
2
2
?
112
第六章 练习题
一、挠曲线近似微分方程 的近似性
反映在哪几方面?
二、用积分法求图示组合梁的挠曲线方程时,
需应用的支承条件和连续条件是什么?
三、长度为 L,重量为 P的等截面直梁,放置在水平
刚性平面上。若在端点施力 P/3上提,未提起部分仍
保持与平面密合,试求提起部分的长度。
EI
xM
dx
vd )(
2
2
?
113
Solution,The radius of curvature at point A of the
beam is,that is
?
01 ?? EIM A
A?
? 0?AM
0213 2 ?????? aLPaPM A
La 32?
114
解,A点处梁的曲率半径为,即
?
01 ?? EIM A
A?
? 0?AM
0213 2 ?????? aLPaPM A
La 32?
115
116
