1
Mechanics of Materials
2
3
CHAPTER 12 SATATICALLY INDETERMINATE
PROBLEMS
§ 12–1 SUMMARY OF STATICALLY INDETERMINATE
STRUCTURES
§ 12–2 SOLVE STATICALLY INDETERMINATE STRUCTURES BY
THE FORCE METHOD
§ 12–3 SYMMETRY AND THE APPLICATION OF THE
PROPERTIES OF SYMMETRY
§ 12-4 CONTINUOUS BEAMS AND THREE-MOMENT
EQUATIONS
4
第十二章 静不定结构
§ 12–1 静不定结构 概述
§ 12–2 用力法解静不定结构
§ 12–3 对称及对称性质的应用
§ 12-4 连续梁与三弯矩方程
5
Structures that the whole constraint reactions and the internal forces in such
structures can not be determined only by the static equilibrium equations are all
called statically indeterminate structures or systems,
§ 12–1 SUMMARY OF STATICALLY INDETERMINATE
STRUCTURES
In the statically indeterminate structure the constraints in excess of the number
needed to support the structure in a statically determinate manner are called static
redundant constraints,The reactions corresponding to the redundant constraints
are called redundant reactions,The number of redundant reactions is the degree
of static indeterminacy,
6
用静力学平衡方程无法确定全部约束力和内力的结构，统
称为 静不定结构或系统,也称为 超静定结构或系统 。
§ 12–1 静不定结构 概述
在静不定结构中，超过维持静力学平衡所必须的约束称为 多
余约束,多余约束相对应的反力称为 多余约束反力,多余约束的
数目为结构的 静不定次数 。
7
Classification of the statically indeterminate problems
Analytical methods
First class,There are redundant constraints only outside the structure，
that is the constraint reactions are statically indeterminate,such a structure
is called the statically indeterminate system in constraint,
Second class,There are redundant constraints only inside the structure，
that is the internal forces are statically indeterminate,such a structure is
called the statically indeterminate system in internal forces,
Third class,There are redundant constraints inside and outside the
structure,that is the constraint reactions and internal forces are all
statically indeterminate,
1).Force method,In this method unknown forces are basic unknown
quantities,
2).Displacement method,In this method unknown displacements are
basic unknown quantities,
8
静
不
定
问
题
分
类
第一类：仅在结构外部存在多余约束，即支反力是静
不定的，可称为外力静不定系统。
第二类：仅在结构内部存在多余约束，即内力是静不
定的，可称为内力静不定系统。
第三类：在结构外部和内部均存在多余约束，即支反
力和内力是静不定的。
分析方法
1.力法：以未知力为基本未知量的求解方法。
2.位移法：以未知位移为基本未知量的求解方法。
9 First class Second class Third class
10 第一类 第二类 第三类
11
§ 12–2 SOLVE STATICALLY INDETERMINATE STRUCTURE BY
THE FORCE METHOD
1,Thoughts of the force method（ explain by examples）
Solution,① Determine the number of
redundant constraint reactions.(one)
② Select the redundant constraint and take
it out,substitute it by its reaction,Write out the
compatibility equation of deformation,See Fig.(b)
C
2l
P
A B
2l
(a)
P
A B C X
1
(b)
Example 1 EI is a constant in the
beam shown in the figure.Try to
determine its constraint reactions,plot the
bending-moment diagram and determine
the deflection of the middle point C on the
beam,
12
§ 12–2 用力法解静不定结构
一、力法的基本思路（举例说明）
解,① 判定多余约束反力的数目
（一个）
② 选取并去除多余约束，代
以多余约束反力，列出变形
协调方程，见图 (b)。
C
2l
[例 1 ] 如图所示，梁 EI为常数。
试求支座反力，作弯矩图，并
求梁中点的挠度。
P
A B
2l
(a)
P
A B C X
1
(b)
13
011 1 ?????? PXB
Compatibility equation of deformation
③ Calculate and by the
energy method P1? 11X?
P
A B C (c) x
(d)
x A
B
X1
A B 1
x
(e)
From Mohr’s theorem we get,(Fig.c、
d,e)
EI
Pl
xx
l
xP
EI
l
lP
48
5
d)
2
(
1
3
2
1
??
?????? ?
EI
lXxxxX
EI
l
X 3d
1 31
0 11 1
?????? ?
14
011 1 ?????? PXB
变形协调方程
③ 用能量法计算 和
P1? 11X?
P
A B C (c) x
(d)
x A
B
X1
A B 1
x
(e)
由莫尔定理可得 (图 c,d,e)
EI
Pl
xx
l
xP
EI
l
lP
48
5
d)
2
(
1
3
2
1
??
?????? ?
EI
lXxxxX
EI
l
X 3d
1 31
0 11 1
?????? ?
15
④ Determine redundant reactions
Substituting the above result into the
compatibility equation of deformation we
get
0
48
5
3
33
1 ??
EI
Pl
EI
lX PX
16
5
1 ?
⑤ Determine other constraint reactions,
C
P
A B ( f )
16
5P
16
11P
16
3Pl
⑥ Plot the bending-moment diagram as
shown in Fig.(g),
(g) + –
16
3Pl
32
5Pl
⑦ Determine the deflection at the middle
point of the beam,
The reactions at the end A may be found
out by the static equilibrium equations,
Their magnitudes and directions are
shown in Fig.(f),
16
④ 求多余约束反力
将上述结果代入变形协调方程得
0
48
5
3
33
1 ??
EI
Pl
EI
lX
PX
16
5
1 ?
⑤ 求其它约束反力
由平衡方程可求得 A端反
力，其大小和方向见图 (f)。
C
P
A B ( f )
16
5P
16
11P
16
3Pl
⑥ 作弯矩图，见图 (g)。
(g) + –
16
3Pl
32
5Pl
⑦ 求梁中点的挠度
17
Select the primary structure (see Fig.( b)) as our study object,The structure
subjected to the unit load is shown in Fig.(h),P
A B C X
1
(b)
x
1
A B C
(h)
From Mohr’s theorem we get
)(
768
7
d)(])
2
(
16
5
[
1
3
2
0
??
?????? ?
EI
Pl
xxPxx
l
P
EI
y
l
C
Attention,For the same statically
indeterminate structure,if we select different
redundant constraints,the primary structures
are different too,In the above example if we
select the rotational constraint at the fixed as
the redundant one,the primary structure is the
simplely supported beam as shown in Fig.(i)
C
P
A
B (i)
X1
18
选取基本静定系 ( 见图 ( b)) 作为计算对象。单位载荷如图 (h) 。
P
A B C X
1
(b)
x
1
A B C
(h)
用莫尔定理可得
)(
768
7
d)(])
2
(
16
5
[
1
3
2
0
??
?????? ?
EI
Pl
xxPxx
l
P
EI
y
l
C
注意,对于同一静不定结构，若选
取不同的多余约束，则基本静定系
也不同。本题中若选固定端处的转
动约束为多余约束，基本静定系是
如图 (i)所示的简支梁。
C
P
A
B (i)
X1
19
2,Canonical equation of the force method
Rewrite the compatibility equation of deformation with the unknown force as
the unknown quantity in above example into the following expression,
01111 ??? PX?
It is the standard form of the compatibility equation of deformation,That is
the so-called canonical equation of the force method,
X1——Redundant unknown quantity；
11——the displacement at the acting point of X1 that is induced by the unit force
X1 and is along the direction of X1 on the primary structure;
1P——the displacement at the acting point of X1 that is induced by the original
load and is along the direction of X1 on the primary struture,
?
?
20
二、力法正则方程
上例中以未知力为未知量的变形协调方程可改写成下式
01111 ??? PX?
变形协调方程的标准形式，即所谓的力法正则方程。
X1——多余未知量；
11——在基本静定系上,X1取单位值时引起的在 X1作用点沿
X1方向的位移；
1P——在基本静定系上,由原载荷引起的在 X1作用点沿
X1方向的位移；
?
?
21
For the statically indeterminate structure with more redundant unknown
reactions its canonical equation is as following,
0
0
0
2211
22222121
11212111
??????
??????
??????
nPnnnnn
Pnn
Pnn
XXX
XXX
XXX
???
???
???
?
??
?
?
From the reciprocal theorem of displacement we know,
jiij ?? ?
ij,Influence coefficient,It is the displacement at the acting point of Xi that is
induced by the unit force Xj and is along the direction of Xi on the primary
structure;
iP,Free term,It expresses the displacement at the acting point of Xi that is
induced by the original load and is along the direction of Xi on the primary
structure,
?
?
22
对于有无数多余约束反力的静不定系统的正则方程如下,
0
0
0
2211
22222121
11212111
??????
??????
??????
nPnnnnn
Pnn
Pnn
XXX
XXX
XXX
???
???
???
?
??
?
?
由位移互等定理知,
jiij ?? ?
ij,影响系数，表示在基本静定系上由 Xj取单位值时引起的
在 Xi作用点沿 Xi方向的位移；
iP,自由项，表示 在基本静定系上,由原载荷引起的在 Xi
作用点沿 Xi 方向的位移。
?
?
23
Example 2 Try to determine all constraint reactions of the rigid frame
shown in the figure,EI of the rigid frame is constant,
q
a
A
B
a
Solution,① The rigid frame has two redundant
constraints,
② Select the redundant constraints and take them
out,substitute them by their reactions,
q
A
B
X1 X2 ③ Write out the canonical equation of the force method
0
0
2222121
1212111
????
????
P
P
XX
XX
??
??
From Mohr’s theorem we have,
④ Calculate the coefficient ij and the free term iP ? ?
24
例 2 试求图示刚架的全部约束反力，刚架 EI为常数。
q
a
A
B
a
解,① 刚架有两个多余约束。
② 选取并去除多余约束，代以多
余约束反力。
q
A
B
X1 X2
③ 建立力法正则方程
0
0
2222121
1212111
????
????
P
P
XX
XX
??
??
用莫尔定理求得
④ 计算系数 ij和自由项 iP ? ?
25
q
A
B
x1 x 2
A
B
x1 x 2 1
1
A
B
x1 x 2
EI
qaxaqx
EI
a
P 6d)2
1(1 4
2
2
201 ??????? ?
EI
qaxxqx
EI
a
P 8d)2
1(1 4
22
2
202 ??????? ?
EI
axaxx
EI
aa
3
4)dd(1 3
20
2
10
2
111 ??? ???
EI
axx
EI
a
3
d1
3
20
2
222 ?? ??
EI
axax
EI
a
2
d1
3
20 22112 ??? ???
26
q
A
B
x1 x 2
A
B
x1 x 2 1
1
A
B
x1 x 2
EI
qaxaqx
EI
a
P 6d)2
1(1 4
2
2
201 ??????? ?
EI
qaxxqx
EI
a
P 8d)2
1(1 4
22
2
202 ??????? ?
EI
axaxx
EI
aa
3
4)dd(1 3
20
2
10
2
111 ??? ???
EI
axx
EI
a
3
d1
3
20
2
222 ?? ??
EI
axax
EI
a
2
d1
3
20 22112 ??? ???
27
⑤ Determine the redundant reactions,
Substituting the above results into the canonical equation of the force method we
get
0
832
0
623
4
4
2
3
1
3
4
2
3
1
3
???
???
EI
qa
X
EI
a
X
EI
a
EI
qa
X
EI
a
X
EI
a
)(
7
3
)(
28
1
2
1
??
???
qaX
qaX
⑥ Determine other reactions
Other reactions determined by
static equilibrium equations are all
shown in the figure,
q
A
B
qa73
qa281
qa74
qa281
2
28
3 qa
28
⑤ 求多余约束反力
将上述结果代入力法正则方程可得
0
832
0
623
4
4
2
3
1
3
4
2
3
1
3
???
???
EI
qa
X
EI
a
X
EI
a
EI
qa
X
EI
a
X
EI
a
)(
7
3
)(
28
1
2
1
??
???
qaX
qaX
⑥ 求其它支反力
由平衡方程得其它支反力，
全部表示于图中。
q
A
B
qa73
qa281
qa74
qa281
2
28
3 qa
29
§ 12–3 SYMMETRY AND THE APPLICATION OF
THE PROPERTIES OF SYMMETRY
1,Symmetric and antisymmetric deformations of symmetric structures
If the dimension, shape,material and constraint conditions of a structure are all
symmetric about an axis,such a structure is called a symmetric structure,As such a
structure is subjected to symmetric loads about the symmetric axis of the structure,the
structure will produce symmetric deformations,If the external loads are antisymmetric,
then the structure will produce antisymmetric deformations,
E1I1 E1I1
EI
E1I1 E1I1
EI
E1I1 E1I1
EI
30
§ 12–3 对称及对称性质的应用
一、对称结构的对称变形与反对称变形
结构几何尺寸、形状，构件材料及约束条件均对称于某一
轴，则称此结构为 对称结构 。当对称结构受力也对称于结构对
称轴，则此结构将产生 对称变形 。若外力反对称于结构对称轴，
则结构将产生 反对称变形 。
E1I1 E1I1
EI
对
称
轴
E1I1 E1I1
EI
对
称
轴
E1I1 E1I1
EI
对
称
轴
31
Sym
me
tric
axi
s
X1
X2
X2
X3
P
X1
X3
Such as,
X1
X3
P
X1
X3
P
X2
X2
P P
We can properly use the properties of symmetry and antisymmetry to
determine some unknown quantities without solving the equations and make the
calculation process much simpler,For example,antisymmetric internal forces in
the symmetric section are zero or knowing in the symmetric deformation,while
symmetric internal forces in the antisymmetric section are zero or knowing in the
antisymmetric deformation,
32
正确利用对称、反对称性质，则可推知某些未知量，可
大大简化计算过程：如对称变形对称截面上，反对称内力为
零或已知；反对称变形对称截面上，对称内力为零或已知。
对
称
轴
X1
X2
X2
X3
P
X1
X3
例如,
X1
X3
P
X1
X3
P
X2
X2
P P
33
Example 3 Try to determine the whole constraint
reactions of the rigid frame shown in the figure,EI of
the frame is constant,
A B
C
P P
a a
P P X
1 X1 01111 ??? PX?
Solution,There are three redundant constraints
in the frame shown in the figure,Because of
symmetry of the structure and antisymmetry of
loads the axial force and the bending moment in the
symmetric section are all zero,There is only one
redundant unknown force(shearing force）,
therefore,we only need to write one canonical
equation,
1P and 11 are determined by Mohr’s
theorem
??
34
[例 3 ] 试求图示刚架的全部约束反力。刚架 EI为常数。
A B
C
P P
a a
解：图示刚架有三个多余未知力。但
由于结构是对称的，而载荷反对称，
故对称轴横截面上轴力、弯矩为零，
只有一个多余未知力（剪力），只需
列出一个正则方程求解。
P P X
1 X1
01111 ??? PX?
用莫尔定理求 1P和 11。 ? ?
35
P x1 x 2 x1 x 2 1
EI
PaxaPx
EI
a
P 2d2)(
2 3
20 21 ??????? ?
EI
Paxaxx
EI
a a
12
7]d)
2
(d[2
3
2
22
0 01
2
111 ????? ? ??
0
212
7 3
1
3
??
EI
PaX
EI
Pa
Then
PX 761?
From static equilibrium equations we can find out,
PRR BA 76??
PHH BA ??
PaMM BA 74??
A B
P P
MB R
B
HB
MA RA
HA
36
P x1 x 2 x1 x 2 1
EI
PaxaPx
EI
a
P 2d2)(
2 3
20 21 ??????? ?
EI
Paxaxx
EI
a a
12
7]d)
2
(d[2
3
2
22
0 01
2
111 ????? ? ??
0
212
7 3
1
3
??
EI
PaX
EI
Pa则 PX
7
6
1?
由平衡方程求得,
PRR BA 76??
PHH BA ??
PaMM BA 74??
A B
P P
MB R
B
HB
MA RA
HA
37
§ 12-4 CONTINUOUS BEAMS AND
THREE-MOMENT EQUATIONS
1,Continuous beam and the degree of static indeterminacy
0 1 2 n-1 n+1 n
l1 l2 ln ln+1
Mn-1 M1 M2 Mn Mn+1
In order to reduce the stress and bending deformation of the straight beam that the
span is longer we often install several supports in the middle of the beam,This kind
of structures often seen in the civil, bridge and mechanical engineering are called
continuous beams,Suppose taking out the supports in the middle the beam becomes
into statically determinate beam with two hinged ends,Therefore the supports in the
middle are redundant constraints,the number of the supports in the middle is that of
redundant constraints,also the degree of the statically indeterminacy of the beam,
38
§ 12-4 连续梁与三弯矩方程
为减小跨度很大直梁的弯曲变形和应力，常在其中间安
置若干中间支座，在建筑、桥梁以及机械中常见的这类结构称
为 连续梁 。撤去中间支座，该梁是两端铰支的静定梁，因此中
间支座就是其多余约束，有多少个中间支座，就有多少个多余
约束，中间支座数就是连续梁的 超静定次数 。
一、连续梁与超静定次数
0 1 2 n-1 n+1 n
l1 l2 ln ln+1
M1 M2 Mn-1 Mn Mn+1
39
2,Three-moment equation
Continuous beam is a statically indeterminate structure,Choice of its primary
structure is various, If select the beam taken out the middle supports as the primary
structure,each constraint reaction can produces effects on the displacement at the
position of each middle support,therefore each equation of canonical equations will
include reduntant reactions,which makes the calculation complex,Suppose to cut
off the beam at each middle support and install the hinge,so the continuous beam is
changed into several simply supported beams and each simply supported beam is one
primary beam,It is equivalent to release internal constraint reactions at each middle
support,that is considering the internal moments M1,M2,… Mn-1,Mn as
redundant constraint reactions (see the above figure),So we need to act a couple of
moments which are equal in magnitude and opposite in direction in two sections at
the sides of a point of the beam above each support,the displacements
corresponding to the couple of moments are relative angles of rotation of the two
sections,
40
二、三弯矩方程
连续梁是超静定结构，静定基可有多种选择，如果选撤去
中间支座为静定基，则因每个支座反力将对静定梁的每个中间
支座位置上的位移有影响，因此正则方程中每个方程都将包含
多余约束反力，使计算非常繁琐。如果设想将每个中间支座上
的梁切开并装上铰链，将连续梁变成若干个简支梁，每个简支
梁都是一个静定基。这相当于把每个支座上梁的内约束解除，
即将其内力弯矩 M1,M2,… Mn-1,Mn,… 作为多余约束力 (见上
图 )，则每个支座上方的铰链两侧截面上需加上大小相等、方向
相反的一对力偶矩，与其对应的位移是两侧截面的相对转角。
41
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ???
n-1 n+1 n
ln ln+1
Mn-1 M
n+1
n-1 n n+1 n
Mn
1 1
dwn
dxn
MnP
dwn+1
dxn+1 xn xn+1
wn wn+1
an bn+1
At sides of support n we take two segments from the primary beam with spans
lnand ln+1.Suppose the relative angle of rotation of the sections of the beam at the
sides of support n is n,then ?
42
如从基本静定系中任意取出两个相邻跨度 ln,ln+1，设 n支
座上方，铰链两侧的相对转角为 n，则
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ???
n-1 n+1 n
ln ln+1
Mn-1 M
n+1
n-1 n n+1 n
Mn
1 1
dwn
dxn
MnP
dwn+1
dxn+1 xn xn+1
wn wn+1
an bn+1
?
43
1.Determine nP,
n
n
l
xM ??
1
1
?
????
n
n
l
xM
From Mohr’s theorem we get
)d
1
d
1
(
1
dd
1
1
11
1
1
11)1(
??
??
?
?
??
?
?
???
??
???
nn
nn
l
nn
n
l
nn
n
l
n
nnPn
l
n
nnnP
nP
x
l
x
lEI
E I l
xxM
E I l
xxM
??
Where,
nnnP ddxM ?? 11)1( ??? ? nnPn ddxM ?
When only the external loads are acted on the primary structure the bending
moment in the span ln and ln+1 is respectively designated by MnP and M(n+1)P,
When the unit moment is only acted on the primary structure the bending
moment in the span ln and ln+1 is respectively designated by
and
?
44
1.求 nP,
静定基上只作用外载荷时，跨度 ln上弯矩记为 MnP，跨度
ln+1上弯矩记为 M(n+1)P。当只作用单位力偶矩时，跨度 ln上和
ln+1上弯矩分别记为
n
n
l
xM ??
1
1
?
????
n
n
l
xM
则由莫尔定理得
)d
1
d
1
(
1
dd
1
1
11
1
1
11)1(
??
??
?
?
??
?
?
???
??
???
nn
nn
l
nn
n
l
nn
n
l
n
nnPn
l
n
nnnP
nP
x
l
x
lEI
E I l
xxM
E I l
xxM
??
式中,
nnnP ddxM ?? 11)1( ??? ? nnPn ddxM ?
?
45
nnl nn ax
n
?? ?? d 1111 d ???? ?? nn
l nn
bx
n
??
)(
1
1
11
?
?????
n
nn
n
nn
nP l
b
l
a
EI
??
Hence
)(
3
1
)d()(
1
)d()(
1
11
1
1
1
1
1
??
?
?
?
? ???? ??
?
nnl n
n
n
n
n
l n
n
n
n
n
nn llEIxl
x
l
x
EI
x
l
x
l
x
EI nn
?
Similarly we can find out
nnn lEI6
1
)1( ??? 1)1( 6
1
?? ? nnn lEI?
Substituting the above result into the following equation
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ???
we get
46
nnl nn ax
n
?? ?? d 1111 d ???? ?? nn
l nn
bx
n
??
)(
1
1
11
?
?????
n
nn
n
nn
nP l
b
l
a
EI
??
因此
)(
3
1
)d()(
1
)d()(
1
11
1
1
1
1
1
??
?
?
?
? ???? ??
?
nnl n
n
n
n
n
l n
n
n
n
n
nn llEIxl
x
l
x
EI
x
l
x
l
x
EI nn
?
类似地可求出
nnn lEI6
1
)1( ??? 1)1( 6
1
?? ? nnn lEI?
将上述结果代入方程
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ??? 得
47
1
11
1111
66
)(2
?
??
????
???
???
n
nn
n
nn
nnnnnnn
l
b
l
a
lMllMlM
?? Three-moment equation
For each middle support of the continuous beam we can write
out a three-moment equation,Therefore the number of three-
moment equations written out in possibility is equal to that of the
middle supports and also equal to the degrees of the static
indeterminacy,Each equation includes only three redundant
constraint moments,This makes the calculation about the
continuous beam simpler,
48
1
11
1111
66
)(2
?
??
???? ??????
n
nn
n
nn
nnnnnnn l
b
l
a
lMllMlM
??
三弯矩方程
对于连续梁的每一个中间支座都可以列出一个三弯矩方程，
所以可能列出的方程式的数目恰好等于中间支座的数目，也就
是等于静不定的次数。而且每一个方程式中只含有三个多余约
束力偶矩，这就使得计算得以一定的简化。
49
Example 4 Try to plot the bending-moment diagram of the continuous
beam AC with on equal rigidity in bending by the three-moment equation.The
structure is shown in Fig.(a),
A B C
q P=ql
l l/2 l/2
Solution,The beam AC has
two spans,The length of the spans
is l1=l2=l 。 The number of the
middle support is 1,that is n=1,
There are no moments at the
supports 0and2,hence,;001 ??? MM n;1 Bn MMM ??
021 ??? MM n
(a)
A B C
q P=ql MB
(b)
50
[例 4 ] 试用三弯矩方程作等刚度连续梁 AC的弯矩图。见图 (a)。
A B C
q P=ql
l l/2 l/2
解,AC梁总共有二跨，跨
长 l1=l2=l 。中间支座编号应
取为 1，即 n=1。由于已知 0，
2两支座上无弯矩，故;001 ??? MM n;1 Bn MMM ??
021 ??? MM n
(a)
A B C
q P=ql MB
(b)
51
8;
12
3
21
3
1
qlql
nn ???? ? ????
2
1 ;
2
1
1
1 ??
?
?
n
n
n
n
l
b
l
a
A B C
q P=ql
w1 w2 8
2ql
4
2ql
(c) From Fig.(c) and Fig.(d) we get,
1
A B C
1
(d)
Substituting it into
three- moment equation
we get
)
2
1
82
1
12
(60)(20
33
1 ????????
qlqlllM
So we have
2
1 32
5 qlM ?? (The direction is contrary to what is indicated in
Fig.(b))
52
8;
12
3
21
3
1
qlql
nn ???? ? ????
2
1 ;
2
1
1
1 ??
?
?
n
n
n
n
l
b
l
a
A B C
q P=ql
w1 w2 8
2ql
4
2ql
(c) 由图 (c)和 (d)图得,
1
A B C
1
(d) 代入三弯矩方程可得
)
2
1
82
1
12
(60)(20
33
1 ????????
qlqlllM
解得
2
1 32
5 qlM ?? (方向与图 (b)所示相反 )
53
Multiplying the unit bending
moment diagram shown in Fig.(d)
by
And we can get the bending-
moment diagram of the
simply supported beam due to MB,
Adding it and the bending-moment diagram (Fig.(c) )due to
the loads we get the bending-moment diagram(see Fig,(e)) of
the beam AC,
2
32
5 ql? + +
–
(e)
2
32
5 ql
2
64
11 ql
54
将图 (d)中的单位弯矩图乘以
便得到 MB在简支梁上产生
的 M图，再与载荷引起的 M
图 (c)相加，就得到梁 AC的
弯矩图，见图 (e)。
2
32
5 ql? + +
–
(e)
2
32
5 ql
2
64
11 ql
55
Chapter 12 Exercises
1,Try to determine the axial force of the rod BC in tension shown in
the figure by the force method,
Solution,
We get
BCP LX ???? 1111?
EIL
L
EI xx dx 3011
3?? ? ??
? ?
EIPL
L
EI
xPx
P dx 301
3???? ? ?
EI
LX
BCL
1???
? ?231 1/ AL IPX ??
56
第十二章 练习题
一, 用力法求图示结构中拉杆 BC的轴力 。
解,
求得
BCP LX ???? 1111?
EIL
L
EI xx dx 3011
3?? ? ??
? ?
EIPL
L
EI
xPx
P dx 301
3???? ? ?
EI
LX
BCL
1???
? ?231 1/ AL IPX ??
57
2,Try to plot the bending-moment diagram of the rigid
frame (neglecting the effect of the axial force),
Solution,① Determine the redundant unknown forces,
We get
② Plot the bending-moment diagram
The bending moment of Section A is,
The bending moment of Section B is,
01111 ??? PX?
? ? ? ? ? ? ? ?
2
2
1
12
2
1
1 222
2/
0
11
1
2/
0
11
11 EI
L
EI
LL
EI
L
EI dxdx ???? ??
???????
? ? ? ? ? ? ? ?
2
21
1
2
12
2
4 11
1
12 816
2
2/
0
1
10
1
1 EI
LPL
EI
PLL
EI
L
EI
x
P dxdx
PLP ?????? ?? ???
? ?1221 121 181 ILIL ILPLX ???
1XM A ??
14 1 XM
PL
B ??
58
二、试作刚架的弯矩图（轴力的影响不计）。
解, ① 求多余未知力
求得
② 作弯矩图
截面 A的弯矩为
截面 B的弯矩为
01111 ??? PX?
? ? ? ? ? ? ? ?
2
2
1
12
2
1
1 222
2/
0
11
1
2/
0
11
11 EI
L
EI
LL
EI
L
EI dxdx ???? ??
???????
? ? ? ? ? ? ? ?
2
21
1
2
12
2
4 11
1
12 816
2
2/
0
1
10
1
1 EI
LPL
EI
PLL
EI
L
EI
x
P dxdx
PLP ?????? ?? ???
? ?1221 121 181 ILIL ILPLX ???
1XM A ??
14 1 XM
PL
B ??
59
3,Try to plot the bending-moment diagram of the beam
shown in the figure by the three-moment equation,Knowing EI
is a constant,Assume m is acted in the left of Support B,
Solution,Let,,
As n=1,,,
,,
,we get the equation
As n=2，,
we get the equation
Combining the two equations we get
AMM ?1 BMM ?2
01 ?? 2/2 mL???
3/2 Lb ?
mLLLMLM
LmL
BA ???
??
?
? ????? 3262
03 ?M 3/22 La ?
? ? mLL LLLMLM mLBA 262 322 ??????? ??????
mM A 72? mM B 73?
01 ?L 00 ?M
60
三, 试用三弯矩方程画图示梁的弯矩图 。 已知 EI
为常数 。 设 m作用在 B支座稍左处 。
解：令, 。
当 n=1时,,,
,,
,得方程
当 n=2时,,
得方程
联立两方程求得
AMM ?1 BMM ?2
01 ?? 2/2 mL???
3/2 Lb ?
mLLLMLM
LmL
BA ???
??
?
? ????? 3262
03 ?M 3/22 La ?
? ? mLL LLLMLM mLBA 262 322 ??????? ??????
mM A 72? mM B 73?
01 ?L 00 ?M
61
62
Mechanics of Materials
2
3
CHAPTER 12 SATATICALLY INDETERMINATE
PROBLEMS
§ 12–1 SUMMARY OF STATICALLY INDETERMINATE
STRUCTURES
§ 12–2 SOLVE STATICALLY INDETERMINATE STRUCTURES BY
THE FORCE METHOD
§ 12–3 SYMMETRY AND THE APPLICATION OF THE
PROPERTIES OF SYMMETRY
§ 12-4 CONTINUOUS BEAMS AND THREE-MOMENT
EQUATIONS
4
第十二章 静不定结构
§ 12–1 静不定结构 概述
§ 12–2 用力法解静不定结构
§ 12–3 对称及对称性质的应用
§ 12-4 连续梁与三弯矩方程
5
Structures that the whole constraint reactions and the internal forces in such
structures can not be determined only by the static equilibrium equations are all
called statically indeterminate structures or systems,
§ 12–1 SUMMARY OF STATICALLY INDETERMINATE
STRUCTURES
In the statically indeterminate structure the constraints in excess of the number
needed to support the structure in a statically determinate manner are called static
redundant constraints,The reactions corresponding to the redundant constraints
are called redundant reactions,The number of redundant reactions is the degree
of static indeterminacy,
6
用静力学平衡方程无法确定全部约束力和内力的结构，统
称为 静不定结构或系统,也称为 超静定结构或系统 。
§ 12–1 静不定结构 概述
在静不定结构中，超过维持静力学平衡所必须的约束称为 多
余约束,多余约束相对应的反力称为 多余约束反力,多余约束的
数目为结构的 静不定次数 。
7
Classification of the statically indeterminate problems
Analytical methods
First class,There are redundant constraints only outside the structure，
that is the constraint reactions are statically indeterminate,such a structure
is called the statically indeterminate system in constraint,
Second class,There are redundant constraints only inside the structure，
that is the internal forces are statically indeterminate,such a structure is
called the statically indeterminate system in internal forces,
Third class,There are redundant constraints inside and outside the
structure,that is the constraint reactions and internal forces are all
statically indeterminate,
1).Force method,In this method unknown forces are basic unknown
quantities,
2).Displacement method,In this method unknown displacements are
basic unknown quantities,
8
静
不
定
问
题
分
类
第一类：仅在结构外部存在多余约束，即支反力是静
不定的，可称为外力静不定系统。
第二类：仅在结构内部存在多余约束，即内力是静不
定的，可称为内力静不定系统。
第三类：在结构外部和内部均存在多余约束，即支反
力和内力是静不定的。
分析方法
1.力法：以未知力为基本未知量的求解方法。
2.位移法：以未知位移为基本未知量的求解方法。
9 First class Second class Third class
10 第一类 第二类 第三类
11
§ 12–2 SOLVE STATICALLY INDETERMINATE STRUCTURE BY
THE FORCE METHOD
1,Thoughts of the force method（ explain by examples）
Solution,① Determine the number of
redundant constraint reactions.(one)
② Select the redundant constraint and take
it out,substitute it by its reaction,Write out the
compatibility equation of deformation,See Fig.(b)
C
2l
P
A B
2l
(a)
P
A B C X
1
(b)
Example 1 EI is a constant in the
beam shown in the figure.Try to
determine its constraint reactions,plot the
bending-moment diagram and determine
the deflection of the middle point C on the
beam,
12
§ 12–2 用力法解静不定结构
一、力法的基本思路（举例说明）
解,① 判定多余约束反力的数目
（一个）
② 选取并去除多余约束，代
以多余约束反力，列出变形
协调方程，见图 (b)。
C
2l
[例 1 ] 如图所示，梁 EI为常数。
试求支座反力，作弯矩图，并
求梁中点的挠度。
P
A B
2l
(a)
P
A B C X
1
(b)
13
011 1 ?????? PXB
Compatibility equation of deformation
③ Calculate and by the
energy method P1? 11X?
P
A B C (c) x
(d)
x A
B
X1
A B 1
x
(e)
From Mohr’s theorem we get,(Fig.c、
d,e)
EI
Pl
xx
l
xP
EI
l
lP
48
5
d)
2
(
1
3
2
1
??
?????? ?
EI
lXxxxX
EI
l
X 3d
1 31
0 11 1
?????? ?
14
011 1 ?????? PXB
变形协调方程
③ 用能量法计算 和
P1? 11X?
P
A B C (c) x
(d)
x A
B
X1
A B 1
x
(e)
由莫尔定理可得 (图 c,d,e)
EI
Pl
xx
l
xP
EI
l
lP
48
5
d)
2
(
1
3
2
1
??
?????? ?
EI
lXxxxX
EI
l
X 3d
1 31
0 11 1
?????? ?
15
④ Determine redundant reactions
Substituting the above result into the
compatibility equation of deformation we
get
0
48
5
3
33
1 ??
EI
Pl
EI
lX PX
16
5
1 ?
⑤ Determine other constraint reactions,
C
P
A B ( f )
16
5P
16
11P
16
3Pl
⑥ Plot the bending-moment diagram as
shown in Fig.(g),
(g) + –
16
3Pl
32
5Pl
⑦ Determine the deflection at the middle
point of the beam,
The reactions at the end A may be found
out by the static equilibrium equations,
Their magnitudes and directions are
shown in Fig.(f),
16
④ 求多余约束反力
将上述结果代入变形协调方程得
0
48
5
3
33
1 ??
EI
Pl
EI
lX
PX
16
5
1 ?
⑤ 求其它约束反力
由平衡方程可求得 A端反
力，其大小和方向见图 (f)。
C
P
A B ( f )
16
5P
16
11P
16
3Pl
⑥ 作弯矩图，见图 (g)。
(g) + –
16
3Pl
32
5Pl
⑦ 求梁中点的挠度
17
Select the primary structure (see Fig.( b)) as our study object,The structure
subjected to the unit load is shown in Fig.(h),P
A B C X
1
(b)
x
1
A B C
(h)
From Mohr’s theorem we get
)(
768
7
d)(])
2
(
16
5
[
1
3
2
0
??
?????? ?
EI
Pl
xxPxx
l
P
EI
y
l
C
Attention,For the same statically
indeterminate structure,if we select different
redundant constraints,the primary structures
are different too,In the above example if we
select the rotational constraint at the fixed as
the redundant one,the primary structure is the
simplely supported beam as shown in Fig.(i)
C
P
A
B (i)
X1
18
选取基本静定系 ( 见图 ( b)) 作为计算对象。单位载荷如图 (h) 。
P
A B C X
1
(b)
x
1
A B C
(h)
用莫尔定理可得
)(
768
7
d)(])
2
(
16
5
[
1
3
2
0
??
?????? ?
EI
Pl
xxPxx
l
P
EI
y
l
C
注意,对于同一静不定结构，若选
取不同的多余约束，则基本静定系
也不同。本题中若选固定端处的转
动约束为多余约束，基本静定系是
如图 (i)所示的简支梁。
C
P
A
B (i)
X1
19
2,Canonical equation of the force method
Rewrite the compatibility equation of deformation with the unknown force as
the unknown quantity in above example into the following expression,
01111 ??? PX?
It is the standard form of the compatibility equation of deformation,That is
the so-called canonical equation of the force method,
X1——Redundant unknown quantity；
11——the displacement at the acting point of X1 that is induced by the unit force
X1 and is along the direction of X1 on the primary structure;
1P——the displacement at the acting point of X1 that is induced by the original
load and is along the direction of X1 on the primary struture,
?
?
20
二、力法正则方程
上例中以未知力为未知量的变形协调方程可改写成下式
01111 ??? PX?
变形协调方程的标准形式，即所谓的力法正则方程。
X1——多余未知量；
11——在基本静定系上,X1取单位值时引起的在 X1作用点沿
X1方向的位移；
1P——在基本静定系上,由原载荷引起的在 X1作用点沿
X1方向的位移；
?
?
21
For the statically indeterminate structure with more redundant unknown
reactions its canonical equation is as following,
0
0
0
2211
22222121
11212111
??????
??????
??????
nPnnnnn
Pnn
Pnn
XXX
XXX
XXX
???
???
???
?
??
?
?
From the reciprocal theorem of displacement we know,
jiij ?? ?
ij,Influence coefficient,It is the displacement at the acting point of Xi that is
induced by the unit force Xj and is along the direction of Xi on the primary
structure;
iP,Free term,It expresses the displacement at the acting point of Xi that is
induced by the original load and is along the direction of Xi on the primary
structure,
?
?
22
对于有无数多余约束反力的静不定系统的正则方程如下,
0
0
0
2211
22222121
11212111
??????
??????
??????
nPnnnnn
Pnn
Pnn
XXX
XXX
XXX
???
???
???
?
??
?
?
由位移互等定理知,
jiij ?? ?
ij,影响系数，表示在基本静定系上由 Xj取单位值时引起的
在 Xi作用点沿 Xi方向的位移；
iP,自由项，表示 在基本静定系上,由原载荷引起的在 Xi
作用点沿 Xi 方向的位移。
?
?
23
Example 2 Try to determine all constraint reactions of the rigid frame
shown in the figure,EI of the rigid frame is constant,
q
a
A
B
a
Solution,① The rigid frame has two redundant
constraints,
② Select the redundant constraints and take them
out,substitute them by their reactions,
q
A
B
X1 X2 ③ Write out the canonical equation of the force method
0
0
2222121
1212111
????
????
P
P
XX
XX
??
??
From Mohr’s theorem we have,
④ Calculate the coefficient ij and the free term iP ? ?
24
例 2 试求图示刚架的全部约束反力，刚架 EI为常数。
q
a
A
B
a
解,① 刚架有两个多余约束。
② 选取并去除多余约束，代以多
余约束反力。
q
A
B
X1 X2
③ 建立力法正则方程
0
0
2222121
1212111
????
????
P
P
XX
XX
??
??
用莫尔定理求得
④ 计算系数 ij和自由项 iP ? ?
25
q
A
B
x1 x 2
A
B
x1 x 2 1
1
A
B
x1 x 2
EI
qaxaqx
EI
a
P 6d)2
1(1 4
2
2
201 ??????? ?
EI
qaxxqx
EI
a
P 8d)2
1(1 4
22
2
202 ??????? ?
EI
axaxx
EI
aa
3
4)dd(1 3
20
2
10
2
111 ??? ???
EI
axx
EI
a
3
d1
3
20
2
222 ?? ??
EI
axax
EI
a
2
d1
3
20 22112 ??? ???
26
q
A
B
x1 x 2
A
B
x1 x 2 1
1
A
B
x1 x 2
EI
qaxaqx
EI
a
P 6d)2
1(1 4
2
2
201 ??????? ?
EI
qaxxqx
EI
a
P 8d)2
1(1 4
22
2
202 ??????? ?
EI
axaxx
EI
aa
3
4)dd(1 3
20
2
10
2
111 ??? ???
EI
axx
EI
a
3
d1
3
20
2
222 ?? ??
EI
axax
EI
a
2
d1
3
20 22112 ??? ???
27
⑤ Determine the redundant reactions,
Substituting the above results into the canonical equation of the force method we
get
0
832
0
623
4
4
2
3
1
3
4
2
3
1
3
???
???
EI
qa
X
EI
a
X
EI
a
EI
qa
X
EI
a
X
EI
a
)(
7
3
)(
28
1
2
1
??
???
qaX
qaX
⑥ Determine other reactions
Other reactions determined by
static equilibrium equations are all
shown in the figure,
q
A
B
qa73
qa281
qa74
qa281
2
28
3 qa
28
⑤ 求多余约束反力
将上述结果代入力法正则方程可得
0
832
0
623
4
4
2
3
1
3
4
2
3
1
3
???
???
EI
qa
X
EI
a
X
EI
a
EI
qa
X
EI
a
X
EI
a
)(
7
3
)(
28
1
2
1
??
???
qaX
qaX
⑥ 求其它支反力
由平衡方程得其它支反力，
全部表示于图中。
q
A
B
qa73
qa281
qa74
qa281
2
28
3 qa
29
§ 12–3 SYMMETRY AND THE APPLICATION OF
THE PROPERTIES OF SYMMETRY
1,Symmetric and antisymmetric deformations of symmetric structures
If the dimension, shape,material and constraint conditions of a structure are all
symmetric about an axis,such a structure is called a symmetric structure,As such a
structure is subjected to symmetric loads about the symmetric axis of the structure,the
structure will produce symmetric deformations,If the external loads are antisymmetric,
then the structure will produce antisymmetric deformations,
E1I1 E1I1
EI
E1I1 E1I1
EI
E1I1 E1I1
EI
30
§ 12–3 对称及对称性质的应用
一、对称结构的对称变形与反对称变形
结构几何尺寸、形状，构件材料及约束条件均对称于某一
轴，则称此结构为 对称结构 。当对称结构受力也对称于结构对
称轴，则此结构将产生 对称变形 。若外力反对称于结构对称轴，
则结构将产生 反对称变形 。
E1I1 E1I1
EI
对
称
轴
E1I1 E1I1
EI
对
称
轴
E1I1 E1I1
EI
对
称
轴
31
Sym
me
tric
axi
s
X1
X2
X2
X3
P
X1
X3
Such as,
X1
X3
P
X1
X3
P
X2
X2
P P
We can properly use the properties of symmetry and antisymmetry to
determine some unknown quantities without solving the equations and make the
calculation process much simpler,For example,antisymmetric internal forces in
the symmetric section are zero or knowing in the symmetric deformation,while
symmetric internal forces in the antisymmetric section are zero or knowing in the
antisymmetric deformation,
32
正确利用对称、反对称性质，则可推知某些未知量，可
大大简化计算过程：如对称变形对称截面上，反对称内力为
零或已知；反对称变形对称截面上，对称内力为零或已知。
对
称
轴
X1
X2
X2
X3
P
X1
X3
例如,
X1
X3
P
X1
X3
P
X2
X2
P P
33
Example 3 Try to determine the whole constraint
reactions of the rigid frame shown in the figure,EI of
the frame is constant,
A B
C
P P
a a
P P X
1 X1 01111 ??? PX?
Solution,There are three redundant constraints
in the frame shown in the figure,Because of
symmetry of the structure and antisymmetry of
loads the axial force and the bending moment in the
symmetric section are all zero,There is only one
redundant unknown force(shearing force）,
therefore,we only need to write one canonical
equation,
1P and 11 are determined by Mohr’s
theorem
??
34
[例 3 ] 试求图示刚架的全部约束反力。刚架 EI为常数。
A B
C
P P
a a
解：图示刚架有三个多余未知力。但
由于结构是对称的，而载荷反对称，
故对称轴横截面上轴力、弯矩为零，
只有一个多余未知力（剪力），只需
列出一个正则方程求解。
P P X
1 X1
01111 ??? PX?
用莫尔定理求 1P和 11。 ? ?
35
P x1 x 2 x1 x 2 1
EI
PaxaPx
EI
a
P 2d2)(
2 3
20 21 ??????? ?
EI
Paxaxx
EI
a a
12
7]d)
2
(d[2
3
2
22
0 01
2
111 ????? ? ??
0
212
7 3
1
3
??
EI
PaX
EI
Pa
Then
PX 761?
From static equilibrium equations we can find out,
PRR BA 76??
PHH BA ??
PaMM BA 74??
A B
P P
MB R
B
HB
MA RA
HA
36
P x1 x 2 x1 x 2 1
EI
PaxaPx
EI
a
P 2d2)(
2 3
20 21 ??????? ?
EI
Paxaxx
EI
a a
12
7]d)
2
(d[2
3
2
22
0 01
2
111 ????? ? ??
0
212
7 3
1
3
??
EI
PaX
EI
Pa则 PX
7
6
1?
由平衡方程求得,
PRR BA 76??
PHH BA ??
PaMM BA 74??
A B
P P
MB R
B
HB
MA RA
HA
37
§ 12-4 CONTINUOUS BEAMS AND
THREE-MOMENT EQUATIONS
1,Continuous beam and the degree of static indeterminacy
0 1 2 n-1 n+1 n
l1 l2 ln ln+1
Mn-1 M1 M2 Mn Mn+1
In order to reduce the stress and bending deformation of the straight beam that the
span is longer we often install several supports in the middle of the beam,This kind
of structures often seen in the civil, bridge and mechanical engineering are called
continuous beams,Suppose taking out the supports in the middle the beam becomes
into statically determinate beam with two hinged ends,Therefore the supports in the
middle are redundant constraints,the number of the supports in the middle is that of
redundant constraints,also the degree of the statically indeterminacy of the beam,
38
§ 12-4 连续梁与三弯矩方程
为减小跨度很大直梁的弯曲变形和应力，常在其中间安
置若干中间支座，在建筑、桥梁以及机械中常见的这类结构称
为 连续梁 。撤去中间支座，该梁是两端铰支的静定梁，因此中
间支座就是其多余约束，有多少个中间支座，就有多少个多余
约束，中间支座数就是连续梁的 超静定次数 。
一、连续梁与超静定次数
0 1 2 n-1 n+1 n
l1 l2 ln ln+1
M1 M2 Mn-1 Mn Mn+1
39
2,Three-moment equation
Continuous beam is a statically indeterminate structure,Choice of its primary
structure is various, If select the beam taken out the middle supports as the primary
structure,each constraint reaction can produces effects on the displacement at the
position of each middle support,therefore each equation of canonical equations will
include reduntant reactions,which makes the calculation complex,Suppose to cut
off the beam at each middle support and install the hinge,so the continuous beam is
changed into several simply supported beams and each simply supported beam is one
primary beam,It is equivalent to release internal constraint reactions at each middle
support,that is considering the internal moments M1,M2,… Mn-1,Mn as
redundant constraint reactions (see the above figure),So we need to act a couple of
moments which are equal in magnitude and opposite in direction in two sections at
the sides of a point of the beam above each support,the displacements
corresponding to the couple of moments are relative angles of rotation of the two
sections,
40
二、三弯矩方程
连续梁是超静定结构，静定基可有多种选择，如果选撤去
中间支座为静定基，则因每个支座反力将对静定梁的每个中间
支座位置上的位移有影响，因此正则方程中每个方程都将包含
多余约束反力，使计算非常繁琐。如果设想将每个中间支座上
的梁切开并装上铰链，将连续梁变成若干个简支梁，每个简支
梁都是一个静定基。这相当于把每个支座上梁的内约束解除，
即将其内力弯矩 M1,M2,… Mn-1,Mn,… 作为多余约束力 (见上
图 )，则每个支座上方的铰链两侧截面上需加上大小相等、方向
相反的一对力偶矩，与其对应的位移是两侧截面的相对转角。
41
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ???
n-1 n+1 n
ln ln+1
Mn-1 M
n+1
n-1 n n+1 n
Mn
1 1
dwn
dxn
MnP
dwn+1
dxn+1 xn xn+1
wn wn+1
an bn+1
At sides of support n we take two segments from the primary beam with spans
lnand ln+1.Suppose the relative angle of rotation of the sections of the beam at the
sides of support n is n,then ?
42
如从基本静定系中任意取出两个相邻跨度 ln,ln+1，设 n支
座上方，铰链两侧的相对转角为 n，则
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ???
n-1 n+1 n
ln ln+1
Mn-1 M
n+1
n-1 n n+1 n
Mn
1 1
dwn
dxn
MnP
dwn+1
dxn+1 xn xn+1
wn wn+1
an bn+1
?
43
1.Determine nP,
n
n
l
xM ??
1
1
?
????
n
n
l
xM
From Mohr’s theorem we get
)d
1
d
1
(
1
dd
1
1
11
1
1
11)1(
??
??
?
?
??
?
?
???
??
???
nn
nn
l
nn
n
l
nn
n
l
n
nnPn
l
n
nnnP
nP
x
l
x
lEI
E I l
xxM
E I l
xxM
??
Where,
nnnP ddxM ?? 11)1( ??? ? nnPn ddxM ?
When only the external loads are acted on the primary structure the bending
moment in the span ln and ln+1 is respectively designated by MnP and M(n+1)P,
When the unit moment is only acted on the primary structure the bending
moment in the span ln and ln+1 is respectively designated by
and
?
44
1.求 nP,
静定基上只作用外载荷时，跨度 ln上弯矩记为 MnP，跨度
ln+1上弯矩记为 M(n+1)P。当只作用单位力偶矩时，跨度 ln上和
ln+1上弯矩分别记为
n
n
l
xM ??
1
1
?
????
n
n
l
xM
则由莫尔定理得
)d
1
d
1
(
1
dd
1
1
11
1
1
11)1(
??
??
?
?
??
?
?
???
??
???
nn
nn
l
nn
n
l
nn
n
l
n
nnPn
l
n
nnnP
nP
x
l
x
lEI
E I l
xxM
E I l
xxM
??
式中,
nnnP ddxM ?? 11)1( ??? ? nnPn ddxM ?
?
45
nnl nn ax
n
?? ?? d 1111 d ???? ?? nn
l nn
bx
n
??
)(
1
1
11
?
?????
n
nn
n
nn
nP l
b
l
a
EI
??
Hence
)(
3
1
)d()(
1
)d()(
1
11
1
1
1
1
1
??
?
?
?
? ???? ??
?
nnl n
n
n
n
n
l n
n
n
n
n
nn llEIxl
x
l
x
EI
x
l
x
l
x
EI nn
?
Similarly we can find out
nnn lEI6
1
)1( ??? 1)1( 6
1
?? ? nnn lEI?
Substituting the above result into the following equation
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ???
we get
46
nnl nn ax
n
?? ?? d 1111 d ???? ?? nn
l nn
bx
n
??
)(
1
1
11
?
?????
n
nn
n
nn
nP l
b
l
a
EI
??
因此
)(
3
1
)d()(
1
)d()(
1
11
1
1
1
1
1
??
?
?
?
? ???? ??
?
nnl n
n
n
n
n
l n
n
n
n
n
nn llEIxl
x
l
x
EI
x
l
x
l
x
EI nn
?
类似地可求出
nnn lEI6
1
)1( ??? 1)1( 6
1
?? ? nnn lEI?
将上述结果代入方程
01)1(1)1( ??????? ???? nPnnnnnnnnnn MMM ??? 得
47
1
11
1111
66
)(2
?
??
????
???
???
n
nn
n
nn
nnnnnnn
l
b
l
a
lMllMlM
?? Three-moment equation
For each middle support of the continuous beam we can write
out a three-moment equation,Therefore the number of three-
moment equations written out in possibility is equal to that of the
middle supports and also equal to the degrees of the static
indeterminacy,Each equation includes only three redundant
constraint moments,This makes the calculation about the
continuous beam simpler,
48
1
11
1111
66
)(2
?
??
???? ??????
n
nn
n
nn
nnnnnnn l
b
l
a
lMllMlM
??
三弯矩方程
对于连续梁的每一个中间支座都可以列出一个三弯矩方程，
所以可能列出的方程式的数目恰好等于中间支座的数目，也就
是等于静不定的次数。而且每一个方程式中只含有三个多余约
束力偶矩，这就使得计算得以一定的简化。
49
Example 4 Try to plot the bending-moment diagram of the continuous
beam AC with on equal rigidity in bending by the three-moment equation.The
structure is shown in Fig.(a),
A B C
q P=ql
l l/2 l/2
Solution,The beam AC has
two spans,The length of the spans
is l1=l2=l 。 The number of the
middle support is 1,that is n=1,
There are no moments at the
supports 0and2,hence,;001 ??? MM n;1 Bn MMM ??
021 ??? MM n
(a)
A B C
q P=ql MB
(b)
50
[例 4 ] 试用三弯矩方程作等刚度连续梁 AC的弯矩图。见图 (a)。
A B C
q P=ql
l l/2 l/2
解,AC梁总共有二跨，跨
长 l1=l2=l 。中间支座编号应
取为 1，即 n=1。由于已知 0，
2两支座上无弯矩，故;001 ??? MM n;1 Bn MMM ??
021 ??? MM n
(a)
A B C
q P=ql MB
(b)
51
8;
12
3
21
3
1
qlql
nn ???? ? ????
2
1 ;
2
1
1
1 ??
?
?
n
n
n
n
l
b
l
a
A B C
q P=ql
w1 w2 8
2ql
4
2ql
(c) From Fig.(c) and Fig.(d) we get,
1
A B C
1
(d)
Substituting it into
three- moment equation
we get
)
2
1
82
1
12
(60)(20
33
1 ????????
qlqlllM
So we have
2
1 32
5 qlM ?? (The direction is contrary to what is indicated in
Fig.(b))
52
8;
12
3
21
3
1
qlql
nn ???? ? ????
2
1 ;
2
1
1
1 ??
?
?
n
n
n
n
l
b
l
a
A B C
q P=ql
w1 w2 8
2ql
4
2ql
(c) 由图 (c)和 (d)图得,
1
A B C
1
(d) 代入三弯矩方程可得
)
2
1
82
1
12
(60)(20
33
1 ????????
qlqlllM
解得
2
1 32
5 qlM ?? (方向与图 (b)所示相反 )
53
Multiplying the unit bending
moment diagram shown in Fig.(d)
by
And we can get the bending-
moment diagram of the
simply supported beam due to MB,
Adding it and the bending-moment diagram (Fig.(c) )due to
the loads we get the bending-moment diagram(see Fig,(e)) of
the beam AC,
2
32
5 ql? + +
–
(e)
2
32
5 ql
2
64
11 ql
54
将图 (d)中的单位弯矩图乘以
便得到 MB在简支梁上产生
的 M图，再与载荷引起的 M
图 (c)相加，就得到梁 AC的
弯矩图，见图 (e)。
2
32
5 ql? + +
–
(e)
2
32
5 ql
2
64
11 ql
55
Chapter 12 Exercises
1,Try to determine the axial force of the rod BC in tension shown in
the figure by the force method,
Solution,
We get
BCP LX ???? 1111?
EIL
L
EI xx dx 3011
3?? ? ??
? ?
EIPL
L
EI
xPx
P dx 301
3???? ? ?
EI
LX
BCL
1???
? ?231 1/ AL IPX ??
56
第十二章 练习题
一, 用力法求图示结构中拉杆 BC的轴力 。
解,
求得
BCP LX ???? 1111?
EIL
L
EI xx dx 3011
3?? ? ??
? ?
EIPL
L
EI
xPx
P dx 301
3???? ? ?
EI
LX
BCL
1???
? ?231 1/ AL IPX ??
57
2,Try to plot the bending-moment diagram of the rigid
frame (neglecting the effect of the axial force),
Solution,① Determine the redundant unknown forces,
We get
② Plot the bending-moment diagram
The bending moment of Section A is,
The bending moment of Section B is,
01111 ??? PX?
? ? ? ? ? ? ? ?
2
2
1
12
2
1
1 222
2/
0
11
1
2/
0
11
11 EI
L
EI
LL
EI
L
EI dxdx ???? ??
???????
? ? ? ? ? ? ? ?
2
21
1
2
12
2
4 11
1
12 816
2
2/
0
1
10
1
1 EI
LPL
EI
PLL
EI
L
EI
x
P dxdx
PLP ?????? ?? ???
? ?1221 121 181 ILIL ILPLX ???
1XM A ??
14 1 XM
PL
B ??
58
二、试作刚架的弯矩图（轴力的影响不计）。
解, ① 求多余未知力
求得
② 作弯矩图
截面 A的弯矩为
截面 B的弯矩为
01111 ??? PX?
? ? ? ? ? ? ? ?
2
2
1
12
2
1
1 222
2/
0
11
1
2/
0
11
11 EI
L
EI
LL
EI
L
EI dxdx ???? ??
???????
? ? ? ? ? ? ? ?
2
21
1
2
12
2
4 11
1
12 816
2
2/
0
1
10
1
1 EI
LPL
EI
PLL
EI
L
EI
x
P dxdx
PLP ?????? ?? ???
? ?1221 121 181 ILIL ILPLX ???
1XM A ??
14 1 XM
PL
B ??
59
3,Try to plot the bending-moment diagram of the beam
shown in the figure by the three-moment equation,Knowing EI
is a constant,Assume m is acted in the left of Support B,
Solution,Let,,
As n=1,,,
,,
,we get the equation
As n=2，,
we get the equation
Combining the two equations we get
AMM ?1 BMM ?2
01 ?? 2/2 mL???
3/2 Lb ?
mLLLMLM
LmL
BA ???
??
?
? ????? 3262
03 ?M 3/22 La ?
? ? mLL LLLMLM mLBA 262 322 ??????? ??????
mM A 72? mM B 73?
01 ?L 00 ?M
60
三, 试用三弯矩方程画图示梁的弯矩图 。 已知 EI
为常数 。 设 m作用在 B支座稍左处 。
解：令, 。
当 n=1时,,,
,,
,得方程
当 n=2时,,
得方程
联立两方程求得
AMM ?1 BMM ?2
01 ?? 2/2 mL???
3/2 Lb ?
mLLLMLM
LmL
BA ???
??
?
? ????? 3262
03 ?M 3/22 La ?
? ? mLL LLLMLM mLBA 262 322 ??????? ??????
mM A 72? mM B 73?
01 ?L 00 ?M
61
62
