Theoretical Mechanics
3
In the first part,statics,we started from the axioms of statics,then
obtained the equilibrium conditions of a rigid body by
simplification of the system of forces,They are used to investigate
equilibrium problems of rigid bodies and of system of rigid
bodies,In this chapter,we shall introduce a theorem which is
suitable to study equilibrium problems of any system of particles,
in general,Using the concepts of displacements and work,the
theorem defines the equilibrium conditions for any system of
particles,It is called the theorem of virtual displacements,It is
the most general theorem to study equilibrium problems,In
addition,combining it with D‘Alembert’s principle,we can obtain
a general equation of dynamics which can be used to solve the
problems of dynamics,
4
在第一篇静力学中，我们从静力学公理出发，通过力系
的简化，得出刚体的平衡条件，用来研究刚体及刚体系统的
平衡问题。在这一章里，我们将介绍普遍适用于研究任意质
点系的平衡问题的一个原理，它从位移和功的概念出发，得
出任意质点系的平衡条件。该原理叫做 虚位移原理 。它是研
究平衡问题的最一般的原理，不仅如此，将它与达朗伯原理
相结合，就可得到一个解答动力学问题的动力学普遍方程。
5
§ 16–1 Constraints and their classification
§ 16–2 Degrees of freedom and generalized
coordinates
§ 16–3 Virtual displacements and virtual work
§ 16–4 Ideal constraints
§ 16–5 Theorem of virtual displacements
Chapter 16,Theorem of virtual displacements
6
§ 16–1 约束及其分类
§ 16–2 自由度 广义坐标
§ 16–3 虚位移和虚功
§ 16–4 理想约束
§ 16–5 虚位移原理
第十六章 虚位移原理
7
§ 16-1 Constraint and their classification
1,Constraints and the equations of constraints
All kinds of conditions which limit the motion of a particle or a
system of particles are called constraints,
The equations which express these limiting conditions are
called the equations of constraints,
A single pendulum in a plane
.222 lyx ??
A crankguide
,222 ryx AA ??
.0,)()( 222 ????? BABAB ylyyxx
For example,
8
§ 16-1 约束及其分类
一、约束及约束方程
限制质点或质点系运动的各种条件称为约束。
将约束的限制条件以数学方程来表示，则称为约束方程。
平面单摆
222 lyx ??
例如,
曲柄连杆机构
222 ryx AA ??
0,)()( 222 ????? BABAB ylyyxx
9
By their forms and characters constraints can be classified to into
different types as follows
2,Classification of constraints
(1) Geometrical constraints and constraints of motion
Conditions which limit the geometric position in space of a particle
or of a system of particles are called geometrical constraints,
Examples are,the limiting conditions shown in the cases given
above,
Conditions which limit the motion of a particle or of a system of
particles are called constraints of motion,
An example is,the pure rolling of wheels along a tangent track.,
10
根据约束的形式和性质，可将约束划分为不同的类型，通
常按如下分类,
二、约束的分类
1、几何约束和运动约束
限制质点或质点系在空间几何位置的条件称为 几何约束 。
如前述的平面单摆和曲柄连杆机构例子中的限制条件都是几
何约束。
当约束对质点或质点系的运动情况进行限制时，这种约
束条件称为 运动约束 。
例如,车轮沿直线轨道作纯滚动时。
11
Geometrical constraint,
Constraint of motion,
).0(
,0
,
??
??
?
?
?
?? rx
rv
ry
A
A
A
A constraint which is time-dependent is called an unsteady
constraint,A constraint which does not depend on time is called a
steady constraint.In the examples above,the constraints do not
change with the time,so they are all steady constraints,
(2) Steady constraints and unsteady
constraints
For example,the object M is made of a
fixed ring and a rope tied to the ring,At the
beginning,the length of the single pendulum
is l0,drag the rope with uniform speed v,In
the equation of the constraint,x2+y2=( l0 -
vt )2,the time t appears directly,
12
几何约束,
运动约束,
)0(
0
??
??
?
?
?
?? rx
rv
ry
A
A
A
当约束条件与时间有关，并随时间变化时称为 非定常约束 。
约束条件不随时间改变的约束为 定常约束 。
前面的例子中约束条件皆不随时间变化，它们都是定常约束。
2、定常约束和非定常约束
例如,重物 M由一条穿过固定圆环的细绳
系住。初始时摆长 l0,匀速 v拉动绳子。
x2+y2=( l0 -vt )2 约束方程中显含时间 t
13
If there appear time-derivatives of coordinates in an equation of a
constraint (as in the case of a constraint of motion and if),moreover,
these derivatives can not be removed by the infinitesimal calculus,
(hence,the coordinate derivative contained in the equation of the
constraint is not a total differential of a certain function and the
equation of the constraint can not be changed into a finite form by
integration) this constraint is called a nonholonomic constraint,
Generally,the equations of the nonholonomic constraints can not be
expressed in differential form,
If there are on time-derivatives of coordinates in the equation of a
constraint,or if such derivatives can be transformed into a finite
form by infinitesimal calculus,then this kind of constraint is called a
holonomic constraint,
(3) Holonomic and nonholonomic constraints
14
如果在约束方程中含有坐标对时间的导数（例如运动约束）
而且方程中的这些导数不能经过积分运算消除，即约束方程中
含有的坐标导数项不是某一函数全微分，从而不能将约束方程
积分为有限形式，这类约束称为 非完整约束 。一般地，非完整
约束方程只能以微分形式表达。
3、完整约束和非完整约束
如果约束方程中不含有坐标对时间的导数，或者约束方程
中虽有坐标对时间的导数，但这些导数可以经过积分运算化为
有限形式，则这类约束称为 完整约束 。
15
A constraint which limits the
movement of a particle of a
system in two opposite
directions at the same time is
called a double face constraint,
A constraint which limits the
movement of a particle of a
system in a single direction is
called a single face constraint,
For example,if a wheel is purely rolling along a linear rail,
is a differential equation,but after integration we get
, This constraint is a holonomic one,0?? ??? rx A Crx
A ?? ?
(4) Single and double face constraints
A geometrical constraint must be a holonomic one,but a holonomic
one is not necessarily a geometrical one,A nonholonomic constraint
must be a constraint of motion,but a constraint of motion one is not
necessarily a nonholonomic one,
刚杆
x2+y2=l2
绳
x2+y2? l2
16
在两个相对的方向上同时
对质点或质点系进行运动限制
的约束称为 双面约束 。只能限
制质点或质点系单一方向运动
的约束称为 单面约束 。
例如,车轮沿直线轨道作纯滚动,是微分方程，但
经过积分可得到 （常数），该约束仍为完整约束。
0?? ??? rx A
Crx A ?? ?
4、单面约束和双面约束
几何约束必定是完整约束，但完整约束未必是几何约束。
非完整约束一定是运动约束，但运动约束未必是非完整约束。
刚杆
x2+y2=l2
绳
x2+y2? l2
17
The equation of constraint of a double face constraint is an equality,
the equation of constraint of a single face constraint is an inequality,
We will discuss on the following only particle or a system of
particles which is subjected to steady,double face and holonomic
constraints,the general form of their equations is (s is the number of
the constraints,n is the number of the particles of the system),
),,2,1( 0),,;;,,( 111 sjzyxzyxf nnnj ???? ??
18
双面约束的约束方程为等式，单面约束的约束方程为不等式。
我们只讨论质点或质点系受定常、双面、完整约束的情况，
其约束方程的一般形式为（ s为质点系所受的约束数目,n为质
点系的质点个数）
),,2,1( 0),,;;,,( 111 sjzyxzyxf nnnj ???? ??
19
§ 16-2 Degrees of freedom and generalized coordinates
The position of a free particle in space is given by（ x,y,z ),
There are 3 numbers,
The position of a free system of particles in space is given by ( xi,
yi,zi ) (i=1,2…… n),There are 3n numbers,
An unfree system of particles,under the action of s holonomic
constraints,has only (3n-s ) independent coordinates,
The degree of freedom of it is k=3n-s,
The number of the independent coordinates which determine the
position of a system under the action of holonomic constraints is
called the number of degrees of freedom,or shortly,the degree
of freedom,
For example,in the case of the crankguide,discussed above the
position coordinates xA,yA,xB and yB,have to satisfy three
equations of constraints,therefore,it has only one degree of
freedom,
20
§ 16-2 自由度 广义坐标
一个自由质点在空间的位置：（ x,y,z ) 3个
一个自由质点系在空间的位置,( xi,yi,zi ) (i=1,2…… n) 3n个
对一个非自由质点系，受 s个完整约束，（ 3n-s )个独立坐标。
其自由度为 k=3n-s 。
确定一个受完整约束的质点系的位置所需的独立坐标的数目，
称为该质点系的 自由度的数目,简称为 自由度 。
例如,前述 曲柄连杆机构 例子中,确定曲柄连杆机构位置的四
个坐标 xA,yA,xB,yB须满足三个约束方程,因此有 一个自由度 。
21
In general,if a system of n particles consisting of n particles is
under the action of s constraints,the number of degrees of freedom
is snk ?? 3
Generally,n and s are very large,but k is small,In order to
determine the position of the system,the parameters should be
selected properly (they have to be independent on one another),
This is far more convenient than using 3n rectangular coordinates
and s constraint equations of constraints,
The independent parameters which are used to determine the
position of the system are called generalized coordinates,
The selection of the generalized coordinate is not unique,They
could be the linear displacements (x,y,z,s etc.) or the angular
displacements (such as ?,?,?,? etc.),If the constraints are
holonomic ones,the number of generalized coordinates is equal to
the number of the degrees of freedom,
22
一般地，受到 s个约束的、由 n个质点组成的质点系，其自由度为
snk ?? 3
通常,n 与 s 很大而 k 很小。为了确定质点系的位置，用
适当选择的 k 个参数（相互独立），要比用 3n个直角坐标和 s个
约束方程方便得多。
用来确定质点系位置的独立参数，称为 广义坐标 。
广义坐标的选择不是唯一的。广义坐标可以取线位移（ x,y,
z,s 等）也可以取角位移（如 ?,?,?,? 等）。在完整约束情
况下，广义坐标的数目就等于自由度数目。
23
For example,in a crankguide,the outer corner ? of the crank OA
can be selected as the generalized coordinate,then we have
.0,s i nc o s
,s i n,c o s
222 ????
??
BB
AA
yrlrx
ryrx
??
??
After choosing generalized
coordinates,the rectangular
coordinates of any particle of
the system can be expressed
as functions of the
generalized coordinates,
24
例如,曲柄连杆机构中，可取曲柄 OA的转角 ?为广义坐标，则,
0,s i nc o s
s i n,c o s
222 ????
??
BB
AA
yrlrx
ryrx
??
??
广义坐标选定后，
质点系中每一质点的直
角坐标都可表示为广义
坐标的函数。
25
In the case of a double pendulum which can swing only in the
vertical plane we have e.g,
.)()(
,
),,(,),(
22
12
2
12
22
1
2
1
2211
byyxx
ayx
yxyx
????
??
There are two degrees of
freedom,s? and ? can be
chooses as the generalized
coordinates,
.c o sc o s,s i ns i n
,c o s,s i n
22
11
????
??
baybax
ayax
????
??
26
例如,双锤摆。设只在铅直平面内摆动。
22
12
2
12
22
1
2
1
2211
)()(
),(,),(
byyxx
ayx
yxyx
????
??
两个自由度
取广义坐标 ?,?
????
??
c o sc o s,s i ns i n
c o s,s i n
22
11
baybax
ayax
????
??
27
Generally,a system of n particles has k degree of freedoms,
Choosing q1,q2,…… qk as the generalized coordinates,the
coordinates (the position vector) of every particle of the system
can be expressed as functions of the generalized coordinates,
),,,(
),,,(
),,,(
),,,(
21
21
21
21
kii
kii
kii
kii
qqqrr
qqqzz
qqqyy
qqqxx
?
?
?
?
?
?
?
?
),,2,1( ni ??
28
一般地，设有由 n个质点组成的质点系，具有 k个自由
度，取 q1,q2,……, qk为其广义坐标，质点系内各质点的
坐标及矢径可表为广义坐标的函数。
),,,(
),,,(
),,,(
),,,(
21
21
21
21
kii
kii
kii
kii
qqqrr
qqqzz
qqqyy
qqqxx
?
?
?
?
?
?
?
?
),,2,1( ni ??
29
§ 16-3 Virtual displacements and virtual work
At a fixed instant during the motion of a system of particles,an
arbitrary infinitesimal displacement of the particles of the system,
which is allowed by the constraints is called a virtual
displacement of the system of particles (at this given instant),
Virtual displacements can be linear displacements,but they can be
al so e.g,angular displacements,They are usually expressed using
the symbol of variation ?,
M
30
§ 16-3 虚位移和虚功
在质点系运动过程的某瞬时，质点系中的质点发生的为
约束允许的任意的无限小位移，称为质点系（在该瞬时）的
虚位移 。
虚位移可以是线位移，也可以是角位移。通常用变分符
号 ? 表示虚位移。
M
31
The virtual displacement is different from the real one during
real motion,
Real displacements really happen under the action of a certain force
during a certain time,The virtual displacement is an arbitrary one
permitted by the constraints at a fixed time,
A real displacement has a well-determined direction,it is a small or
a finite one,A virtual displacements are small displacements,of
several different directions permitted by to the constraints,
A real displacement happens during a time interval,but a virtual one
is just a pure geometrical concept,it happens in an infinite small
time interval,
Under the action of the steady
constraints,a small real displacement is
identical to a virtual one,However if the
constraints are unsteady,a small real
displacement is no longer a virtual one,
32
虚位移与真正运动时发生的实位移不同 。
实位移是在一定的力作用下和给定的初条件下运动而实际
发生的；虚位移是在约束容许的条件下可能发生的。
实位移具有确定的方向，可能是微小值，也可能是有限值；
虚位移则是微小位移，视约束情况可能有几种不同的方向。
实位移是在一定的时间内发生的；虚位移只是纯几何的概
念，完全与时间无关。
在定常约束下，微小的实位移必
然是虚位移之一。而在非定常约束下，
微小实位移不再是虚位移之一。
33
There are certain relationships among the virtual displacements of
the particles of the system,In order to determine them two ways are
developed,
(1) Geometrical method
The knowledge of kinematics tells us the displacement of a particle
is proportional to its velocity,hence
Therefore,we can analyze the relationships among virtual
displacements by an analysis of velocities,
.dtvrd ??
34
质点系中各质点的虚位移之间存在着一定的关系,确定这
些关系通常有两种方法,
(一 ) 几何法 。由运动学知，质点的位移与速度成正比，即
因此可以用分析速度的方法分析各点虚位移之间的关系。
dtvrd ??
35
(2) Analytical method
The coordinates of every particle of the system can be expressed as
functions of the generalized coordinates (q1,q2,……,qk),the
variations of the generalized coordinates are,the
virtual displacements of every particle are, The projections on
rectangular coordinates can be expressed as follows,
kqqq ???,,,21 ?
ir?
k
k
iii
i
k
k
iii
i
k
k
iii
i
q
q
z
q
q
z
q
q
z
z
q
q
y
q
q
y
q
q
y
y
q
q
x
q
q
x
q
q
x
x
????
????
????
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
2
2
1
1
2
2
1
1
2
2
1
1
),2,1( ni ??
36
(二 ) 解析法 。质点系中各质点的坐标可表示为广义坐标的函数
（ q1,q2,……,qk），广义坐标分别有变分,各
质点的虚位移 在直角坐标上的投影可以表示为
kqqq ???,,,21 ?
ir?
k
k
iii
i
k
k
iii
i
k
k
iii
i
q
q
z
q
q
z
q
q
z
z
q
q
y
q
q
y
q
q
y
y
q
q
x
q
q
x
q
q
x
x
????
????
????
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
2
2
1
1
2
2
1
1
2
2
1
1
),2,1( ni ??
37
[Example 1] Analyze the virtual
displacements of the points C,A and
B when the system is at the position
shown in the diagram (suppose
OC=BC= a and OA=l)
Solution
This is a system with one degree of
Freedom,We select the angle ? between
the rod OA and the axis x as the generalized coordinate,
1,Geometrical method
,
s in2
1
s in2
,
???
?
?
?
???
?
a
a
PB
PC
r
r
l
a
r
r
B
C
A
C
38
[例 1] 分析图示机构在图示位置时，
点 C,A与 B的虚位移。
(已知 OC=BC= a,OA=l )
解,此为一个自由度系统，取
OA杆与 x 轴夹角 ?为广义坐标。
1、几何法
???
?
?
?
s i n2
1
s i n2
???
?
a
a
PB
PC
r
r
l
a
r
r
B
C
A
C
39
.0,s in2
,c o s,s in
,c o s,s in
,,
????
?????
?????
??
BB
AA
CC
AC
yax
lylx
ayax
lrar
?????
????????
????????
??????
Transforming the coordinates
of the points C,A and B into
function of the generalized
coordinate ?we get
.0,c o s2
,s in,c o s
,s in,c o s
??
??
??
BB
AA
CC
yax
lylx
ayax
?
??
??
2,Analytical method Writing the variation of the
generalized coordinate ?,as we
can obtain the projections of the
virtual displacement of every point
on the corresponding rectangular
axes as,
.0,s i n2
,c o s,s i n
,c o s,s i n
????
?????
?????
BB
AA
CC
yax
lylx
ayax
?????
????????
????????
??
40
0,s in2
c o s,s in
c o s,s in
,
????
?????
?????
??
BB
AA
CC
AC
yax
lylx
ayax
lrar
?????
????????
????????
??????
将 C,A,B点的坐标表示成
广义坐标 ? 的函数，得
0,c o s2
s in,c o s
s in,c o s
??
??
??
BB
AA
CC
yax
lylx
ayax
?
??
??
2、解析法
对广义坐标 ? 求变分，得各点
虚位移在相应坐标轴上的投影,
0,s i n2
c o s,s i n
c o s,s i n
????
?????
?????
BB
AA
CC
yax
lylx
ayax
?????
????????
????????
41
The work done by a force along the virtual displacement of the
particle is called the virtual work,it is written as,
F r?
W?
zZyYxXW
rFW
????
??
???
??
42
力 在质点发生的虚位移 上所作的功称为 虚功,记为 。 F r? W?
zZyYxXW
rFW
????
??
???
??
43
§ 16-4 Ideal constraints
If the sum of the virtual works of all reaction forces of constraints
of the system of particles along any virtual displacement is zero,
then these constraints are called ideal constraints,
The condition for the action on ideal constraints of the system of
particles is
.0???? ? iiN rNW ??
44
§ 16-4 理想约束
如果在质点系的任何虚位移上，质点系的所有约束反力的
虚功之和等于零，则称这种约束为 理想约束。
质点系受有理想约束的条件,
0???? ? iiN rNW ??
45
Typical examples of ideal constraints are shown in the following,
1,Smooth supporting plane 2,Smooth hinge
0' ????? ? rNrNW N ???0??? rNW N ??
3,Rigid rod without weight
4,Soft cord which can not be
stretched
5,Pure rolling of a rigid
body on a rugged surface
0)( ??? ?? CN rFNW ??
46
理想约束的典型例子如下,
1、光滑支承面 2、光滑铰链
3、无重刚杆
4、不可伸长的柔索
5、刚体在粗糙面上的纯滚动
0' ????? ? rNrNW N ???0??? rNW N ??
0)( ??? ?? CN rFNW ??
47
§ 16-5 Theorem of virtual displacement
1,Theorem of virtual displacement
The necessary and sufficient condition for the equilibrium of a
system of particles with steady and ideal constraints is that the sum
of the virtual works of all the positive forces acting on this system
along any virtual displacement is zero,hence
? ?? 0ii rF ?
or
? ??? 0)( iiiiii zZyYxX ???
48
§ 16-5 虚位移原理
一、虚位移原理
具有定常、理想约束的质点系，平衡的必要与充分条件是：
作用于质点系的所有主动力在任何虚位移上所作的虚功之和等
于零。即
? ?? 0ii rF ?
解析式,
? ??? 0)( iiiiii zZyYxX ???
49
? ?? 0ii rF ?
Proof (1) Necessity,
When the system of particles is in equilibrium then,
One to the equilibrium of the whole system,an arbitrary
particle Mi is in equilibrium,too,
0?? ii NFfor any arbitrary virtual displacement of the particle M
i,,we
have ir?
.0)( ??? iii rNF ?
Because the constraints is ideal we have
.0? ?? ii rF ?
.0? ?? ii rN ?
Therefore
,0)( ???? iii rNF ?
.0???? ?? iiii rNrF ??
For the whole system we get
50
证明, (1) 必要性：即质点系处于平衡时，必有 ? ?? 0
ii rF ?
∵ 质点系处于平衡 ∴ 选取任一质点 Mi也平衡。
0?? ii NF
对质点 Mi 的任一虚位移,有
ir?
0)( ??? iii rNF ?
0)( ?? ?? iii rNF ?
0?????? iiii rNrF ??
由于是理想约束
? ?? 0ii rF ?
? ?? 0ii rN ?
所以
对整个质点系,
51
For the whole system of particles it follows
0)( ?? ?? iii rNF ?, Under the action of ideal constraints,then we have,
? ?? 0ii rN ?
? ?? 0 T h e r e f o r e ii rF ?in conflict with the initial proposition,
Therefore,if,the system of particles must be in
equilibrium,
? ?? 0ii rF ?
iRird ii rdr ??
A real displacement happens in the direction,choosing
we get
0)( ????? iiiii rRrNF ??
(2) Sufficiency,
If the system of particles satisfies it must be in
equilibrium,If,but if the system is not in equilibrium,
then at least the ith particle is not balanced,
? ?? 0ii rF ?
? ?? 0ii rF ?
0 ??? iii RNF
52
(2) 充分性：即当质点系满足,质点系一定平衡。
若,而质点系不平衡，则至少有第 i个质点不平衡。
? ?? 0ii rF ?
? ?? 0ii rF ?
在 方向上产生实位移,取,则
iR ird ii rdr ??
0)( ????? iiiii rRrNF ??
0)( ?? ?? iii rNF ?对质点系,(理想约束下,) ? ?? 0ii rN ?
? ??? 0 ii rF ? 与前题条件矛盾
故 时质点系必处于平衡。 ? ?? 0ii rF ?
0 ??? iii RNF
53
2,Applications of the theorem of virtual displacements
(1) A system in equilibrium being given,determine the
relationships among the positive forces,
(2) The positive forces of the system being given,determine the
conditions of equilibrium,
(3) Determine the reaction forces of the constraints it the system
is in equilibrium under the action of the known positive
forces,
(4) Determine the internal force of the double force equivalent
rod of balanced truss,
54
二、虚位移原理的应用
1,系统在给定位置平衡时，求主动力之间的关系；
2,求系统在已知主动力作用下的平衡位置；
3,求系统在已知主动力作用下平衡时的约束反力；
4,求平衡构架内二力杆的内力。
55
[Example 1] An ellipsograph mechanism is shown in the fig,The
length of the rod AB is l,its weight and friction can be neglected,
the hinges are smooth,Determine the relationship between the
magnitude of the positive forces when the system is in
equilibrium as shown in the fig,
Solution,
Investigate the whole
system,The constraints of
the system are all
holonomic,ideal and steady,
A
B
56
例 1 图示椭圆规机构，连杆 AB长 l，杆重和滑道摩擦不计，
铰链为光滑的，求在图示位置平衡时，主动力大小 P和 Q之间
的关系。
解,研究整个机构。
系统的所有约束都是
完整、定常、理想的。
57
1,Geometric method
Let A and B perform the virtual
displacements and,Then,from
the theorem of virtual displacements,we
can obtain the equation for the virtual
work as following
Ar? Br?
.0)tg ???? ArQ(P ??
Because is an arbitrary quantity we get
Ar? ? t gQP ?
.0 ?? BA rQrP ??
,tg
,c o ss i n,H o w e v e r
???
????
??
????
AB
BA
rr
rr
58
1、几何法,使 A发生虚位移,
B的虚位移,则由虚位移原理，
得虚功方程,
Ar?
Br?
0)tg ???? ArQ(P ??
由 的任意性，得 Ar? ? t gQP ?
0 ?? BA rQrP ??
tg
c o ss i n
???
????
??
????
AB
BA
rr
rr而
59
2,Analytical method
Because the system has only one
degree of freedom,the angle ? can be
selected as the generalized coordinate,
,c o s,s i n
,s i n,c o s
? ? ??? ? ??
??
lylx
lylx
AB
AB
???
??
Because is an arbitrary quantity we get ??
,t g?QP ?
,0??? BA xQyP ??
.0)s i nc o s( ??? ???? lQP
60
2、解析法 由于系统为单自由度,
可取 ?为广义坐标。
? ? ??? ? ??
??
c o s,s i n
s i n,c o s
lylx
lylx
AB
AB
???
??
由于 任意，故 ?? ? t gQP ?
,0??? BA xQyP ??
0)s i nc o s( ??? ???? lQP
61
Solution
This is a system with two degree of
freedom,Choosing the angles? and ?
as generalized coordinates,then there
are two ways to solve the problem,
[Example 2] The rod OA is joined with the rod AB at point A,In
addition,there is a hinge-support at point O as shown,The
lengths of the rods are 2a and 2b,respectively,their weights are
P1and P2,Suppose a horizontal force F acts on point B which
makes the system to be equilibrium,Determine the angles and to,
y
62
解,这是一个具有两个自由度的系
统，取角 ?及 ?为广义坐标，现用两
种方法求解。
例 2 均质杆 OA及 AB在 A点用铰连接，并在 O点用铰支承，
如图所示。两杆各长 2a和 2b，各重 P1及 P2，设在 B点加水平
力 F 以维持平衡，求两杆与铅直线所成的角 ?及 ? 。
y
63
Applying the theorem of virtual
displacement we get
)(, 021 axFyPyP BDC ??? ???
Substitution into equation (a) results in
.0)co s2s i n()co s2s i n2s i n( 221 ??????? ????????? bFbPaFaPaP
Solution 1
.c o s2c o s2,s in2s in2
,s ins in2
,c o sc o s2
,s in,c o sB u t
? ? ?? ? ????
? ? ?? ? ??
??
? ? ???
baxbax
bay
bay
ayay
BB
D
D
CC
????
???
??
???
64
应用虚位移原理,
)( 021 axFyPyP BDC ??? ???
代入 (a)式，得,
0)co s2s i n()co s2s i n2s i n( 221 ??????? ????????? bFbPaFaPaP
解法一,
? ? ?? ? ????
? ? ?? ? ??
??
? ? ???
c o s2c o s2,s in2s in2
s ins in2
,c o sc o s2
s in,c o s
baxbax
bay
bay
ayay
BB
D
D
CC
????
???
??
???而
65
Because is independent on,we get ??
.0c o s2s i n
,0c o s2s i n2s i n
2
21
?????
???????
??
???
bFbP
aFaPaP
.2 t g,22tg
221 P
F
PP
F ?
?? ??
So we obtain
??
66
由于 是彼此独立的，所以,????,
0c o s2s i n
0c o s2s i n2s i n
2
21
?????
???????
??
???
bFbP
aFaPaP
221
2 t g,
2
2tg
P
F
PP
F ?
?? ??
由此解得,
0)c o s2s i n()c o s2s i n2s i n( 221 ??????? ????????? bFbPaFaPaP
67
.0s i nco s 2 ?? ???? DB rPrF
But,,2 ?????? brbr
DB ??
Substitution into the equation above
results in
.22tg
22 P
F
bP
bF ?
?
??
??
???
Solution 2
Firstly,keep ? constant,but let ? change by a variation, Then
we obtain a set of virtual displacements of the system as show in
the diagram,
??
68
0s i nco s 2 ?? ???? DB rPrF
而 ?????? brbr
DB ??,2
代入上式，得
22
22tg
P
F
bP
bF ?
?
??
??
???
解法二,
先使 ? 保持不变，而使 ? 获得变分,得到系统的一
组虚位移，如图所示。
??
69
Keep ? constant,but let ? choux of, Then we have another set of
virtual displacements of the system as shown in the,
??
,BDA rrr ??? ??
.0s i ns i nco s 21 ??? ?????? DCB rPrPrF
But
.2
,
?????
???
arrr
ar
ADB
C
???
?
Substitution into the equation above gives
,22tg
21 PP
F
???
,0)s i n2s i n2co s( 21 ?????? ????? aPaPaF
diagram
70
再使 ? 保持不变，而使 ? 获得变分,得到系统的另
一组虚位移，如图所示。
??
BDA rrr ??? ??
0s i ns i nco s 21 ??? ?????? DCB rPrPrF
而
?????
???
arrr
ar
ADB
C
2
,
???
?
代入上式后，得,
22tg
21 PP
F
???
0)s i n2s i n2co s( 21 ?????? ????? aPaPaF
图示中,
71
Solution
Relieve the hinged
support B,substitute
the corresponding
reaction force of the
constraint,
BR
,0211 ????? ????? mrPrRrP CBB
,211
BB
C
B
B rmr
rP
r
rPR
?
??
?
?
?
? ????
[Example 3] This is a
multispan statically
determinate a beam,
Determine the reaction
forces on the hinged
support B,
72
例 3 多跨静定梁，
求支座 B处反力。
解,将支座 B 除
去，代入相应的
约束反力 。
BR
0211 ????? ????? mrPrRrP CBB
BB
C
B
B rmr
rP
r
rPR
?
??
?
?
?
? ????
2
1
1
73
.
96
11
8
11
12
11
12
1
6
1
4
,
8
11
,
2
1
,M o r e o v e r 1
?????????
??
B
C
B
E
B
G
B
B
C
B
r
r
r
r
r
r
r
r
r
r
r
?
?
?
?
?
?
?
??
?
?
?
?
.961181121 21 mPPR B ????
Hence,
74
96
11
8
11
12
11
12
1
6
1
4
,
8
11,
2
1 1
?????????
??
B
C
B
E
B
G
B
B
C
B
r
r
r
r
r
r
r
r
r
r
r
?
?
?
?
?
?
?
??
?
?
?
?
而
mPPR B 961181121 21 ????
BB
C
B
B rmr
rP
r
rPR
?
??
?
?
?
? ????
2
1
1
75
[Example 4] A sliding sleeve D is slipping
on the rod AB and drives the rod CD
sliding vertical,If ?=0o,the spring is at its
largest length,the stiffness of the spring is
5(kN/m),Determine the moment M acting
on the rod AB when the system is in
equilibrium at a certain angle ?,
Solution
The problem is to determine the relationships among the positive
forces acting on the system when it is in equilibrium,The spring
force can be treated as a positive force,and the constraints are ideal
ones,Therefore,the problem can be solved by the theorem of
virtual displacements,
76
例 4 滑套 D套在光滑直杆 AB上，并带动
杆 CD在铅直滑道上滑动。已知 ?=0o时，
弹簧等于原长，弹簧刚度系数为
5(kN/m)，求在任意位置（ ? 角）平衡
时，加在 AB杆上的力偶矩 M？
解,这是一个已知系统平衡，求作用于系统上主动力之间关系
的问题。将弹簧力计入主动力，系统简化为理想约束系统，故
可以用虚位移原理求解。
77
Select the system to consist of the rods AB,CD and select
the sliding sleeve D as the object to be investigated,
,ts e c3.0s
,s e c3.0
),kN( |s e c1|5.1||
),m( |s e c1|3.0||
.
c o s
3 0 0
6 0 0,W h e n
).mm(3 0 03 0 06 0 0,0W h e n
0
0
0
? ? ???
?
?
?
?
??
?
g
s
llkFF
ll
l
l
D
?
?
??????
???
???
????
?
From the theorem of virtual displacements,we have
).mkN( c o s )c o s1(s in45.0 3 ??? ? ??M
,0?? sFM ???,0] t gs e c3.0|s e c1|5.1[ ???? ?????M
78
选择 AB杆,CD杆和滑套 D的系统为研究对象。
? ? ???
?
?
?
?
?
?
g
s
llkFF
ll
l
l
D
ts e c3.0s
s e c3.0
)kN( |s e c1|5.1||
)m( |s e c1|3.0||
c o s
3 0 0
6 0 0,
)mm(3 0 03 0 06 0 0,0
0
0
0
?
?
??????
???
??
????
角时
时
?
由虚位移原理，得,
)mkN( co s )co s1(s i n45.0 3 ??? ? ??M
0?? sFM ??? 0] t gs e c3.0|s e c1|5.1[ ???? ?????M
79
The system with ideal constraints should be selected as the object
to be investigated without relieving the constraints,The system has
at least one degree of freedom,If there are nonideal constraints in
the system,such as a spring force or a friction force etc,they can
be treated as a positive force,Then the system is one with ideal
constraint,so it can be selected as the object to be investigated,
If you want to determine the reaction forces of the constraints,you
have to relieve the constraints firstly,then add the reaction forces
of the constraints,they are positive forces,You should get rid of
the constraints one by one,after every relief,a reaction force of
each constraint should be added in the diagram as a positive force
the number of degrees of freedom increases by one at each step,
Steps and key points in solving the equilibrium problem of a
system of particles using the theorem of virtual displacements,
1,Choose the object to be investigate d correctly
80
以不解除约束的理想约束系统为研究对象，系统至少有
一个自由度。若系统存在非理想约束，如弹簧力、摩擦力等，
可把它们计入主动力，则系统又是理想约束系统，可选为研
究对象。
若要求解约束反力，需解除相应的约束，代之以约束反
力，并计入主动力。应逐步解除约束，每一次研究对象只解
除一个约束，将一个约束反力计入主动力，增加一个自由度。
应用虚位移原理求解质点系平衡问题的步骤和要点,
1、正确选取研究对象,
81
2,Analyze the forces correctly
Draw the positive forces in the force diagram,e.g,spring forces,
friction forces and the reaction forces of the constraints which are
to be determined,
3,Analyze the virtual displacements correctly,determine the
relationships between the virtual displacements
4,Use the theorem of virtual displacements to set up the
equations
5,Solve the equations of virtual work and determine the
unknown quantities
82
2、正确进行受力分析,
画出主动力的受力图，包括计入主动力的弹簧力、摩擦
力和待求的约束反力。
3、正确进行虚位移分析，确定虚位移之间的关系 。
4、应用虚位移原理建立方程。
5、解虚功方程求出未知数。
83
84
3
In the first part,statics,we started from the axioms of statics,then
obtained the equilibrium conditions of a rigid body by
simplification of the system of forces,They are used to investigate
equilibrium problems of rigid bodies and of system of rigid
bodies,In this chapter,we shall introduce a theorem which is
suitable to study equilibrium problems of any system of particles,
in general,Using the concepts of displacements and work,the
theorem defines the equilibrium conditions for any system of
particles,It is called the theorem of virtual displacements,It is
the most general theorem to study equilibrium problems,In
addition,combining it with D‘Alembert’s principle,we can obtain
a general equation of dynamics which can be used to solve the
problems of dynamics,
4
在第一篇静力学中，我们从静力学公理出发，通过力系
的简化，得出刚体的平衡条件，用来研究刚体及刚体系统的
平衡问题。在这一章里，我们将介绍普遍适用于研究任意质
点系的平衡问题的一个原理，它从位移和功的概念出发，得
出任意质点系的平衡条件。该原理叫做 虚位移原理 。它是研
究平衡问题的最一般的原理，不仅如此，将它与达朗伯原理
相结合，就可得到一个解答动力学问题的动力学普遍方程。
5
§ 16–1 Constraints and their classification
§ 16–2 Degrees of freedom and generalized
coordinates
§ 16–3 Virtual displacements and virtual work
§ 16–4 Ideal constraints
§ 16–5 Theorem of virtual displacements
Chapter 16,Theorem of virtual displacements
6
§ 16–1 约束及其分类
§ 16–2 自由度 广义坐标
§ 16–3 虚位移和虚功
§ 16–4 理想约束
§ 16–5 虚位移原理
第十六章 虚位移原理
7
§ 16-1 Constraint and their classification
1,Constraints and the equations of constraints
All kinds of conditions which limit the motion of a particle or a
system of particles are called constraints,
The equations which express these limiting conditions are
called the equations of constraints,
A single pendulum in a plane
.222 lyx ??
A crankguide
,222 ryx AA ??
.0,)()( 222 ????? BABAB ylyyxx
For example,
8
§ 16-1 约束及其分类
一、约束及约束方程
限制质点或质点系运动的各种条件称为约束。
将约束的限制条件以数学方程来表示，则称为约束方程。
平面单摆
222 lyx ??
例如,
曲柄连杆机构
222 ryx AA ??
0,)()( 222 ????? BABAB ylyyxx
9
By their forms and characters constraints can be classified to into
different types as follows
2,Classification of constraints
(1) Geometrical constraints and constraints of motion
Conditions which limit the geometric position in space of a particle
or of a system of particles are called geometrical constraints,
Examples are,the limiting conditions shown in the cases given
above,
Conditions which limit the motion of a particle or of a system of
particles are called constraints of motion,
An example is,the pure rolling of wheels along a tangent track.,
10
根据约束的形式和性质，可将约束划分为不同的类型，通
常按如下分类,
二、约束的分类
1、几何约束和运动约束
限制质点或质点系在空间几何位置的条件称为 几何约束 。
如前述的平面单摆和曲柄连杆机构例子中的限制条件都是几
何约束。
当约束对质点或质点系的运动情况进行限制时，这种约
束条件称为 运动约束 。
例如,车轮沿直线轨道作纯滚动时。
11
Geometrical constraint,
Constraint of motion,
).0(
,0
,
??
??
?
?
?
?? rx
rv
ry
A
A
A
A constraint which is time-dependent is called an unsteady
constraint,A constraint which does not depend on time is called a
steady constraint.In the examples above,the constraints do not
change with the time,so they are all steady constraints,
(2) Steady constraints and unsteady
constraints
For example,the object M is made of a
fixed ring and a rope tied to the ring,At the
beginning,the length of the single pendulum
is l0,drag the rope with uniform speed v,In
the equation of the constraint,x2+y2=( l0 -
vt )2,the time t appears directly,
12
几何约束,
运动约束,
)0(
0
??
??
?
?
?
?? rx
rv
ry
A
A
A
当约束条件与时间有关，并随时间变化时称为 非定常约束 。
约束条件不随时间改变的约束为 定常约束 。
前面的例子中约束条件皆不随时间变化，它们都是定常约束。
2、定常约束和非定常约束
例如,重物 M由一条穿过固定圆环的细绳
系住。初始时摆长 l0,匀速 v拉动绳子。
x2+y2=( l0 -vt )2 约束方程中显含时间 t
13
If there appear time-derivatives of coordinates in an equation of a
constraint (as in the case of a constraint of motion and if),moreover,
these derivatives can not be removed by the infinitesimal calculus,
(hence,the coordinate derivative contained in the equation of the
constraint is not a total differential of a certain function and the
equation of the constraint can not be changed into a finite form by
integration) this constraint is called a nonholonomic constraint,
Generally,the equations of the nonholonomic constraints can not be
expressed in differential form,
If there are on time-derivatives of coordinates in the equation of a
constraint,or if such derivatives can be transformed into a finite
form by infinitesimal calculus,then this kind of constraint is called a
holonomic constraint,
(3) Holonomic and nonholonomic constraints
14
如果在约束方程中含有坐标对时间的导数（例如运动约束）
而且方程中的这些导数不能经过积分运算消除，即约束方程中
含有的坐标导数项不是某一函数全微分，从而不能将约束方程
积分为有限形式，这类约束称为 非完整约束 。一般地，非完整
约束方程只能以微分形式表达。
3、完整约束和非完整约束
如果约束方程中不含有坐标对时间的导数，或者约束方程
中虽有坐标对时间的导数，但这些导数可以经过积分运算化为
有限形式，则这类约束称为 完整约束 。
15
A constraint which limits the
movement of a particle of a
system in two opposite
directions at the same time is
called a double face constraint,
A constraint which limits the
movement of a particle of a
system in a single direction is
called a single face constraint,
For example,if a wheel is purely rolling along a linear rail,
is a differential equation,but after integration we get
, This constraint is a holonomic one,0?? ??? rx A Crx
A ?? ?
(4) Single and double face constraints
A geometrical constraint must be a holonomic one,but a holonomic
one is not necessarily a geometrical one,A nonholonomic constraint
must be a constraint of motion,but a constraint of motion one is not
necessarily a nonholonomic one,
刚杆
x2+y2=l2
绳
x2+y2? l2
16
在两个相对的方向上同时
对质点或质点系进行运动限制
的约束称为 双面约束 。只能限
制质点或质点系单一方向运动
的约束称为 单面约束 。
例如,车轮沿直线轨道作纯滚动,是微分方程，但
经过积分可得到 （常数），该约束仍为完整约束。
0?? ??? rx A
Crx A ?? ?
4、单面约束和双面约束
几何约束必定是完整约束，但完整约束未必是几何约束。
非完整约束一定是运动约束，但运动约束未必是非完整约束。
刚杆
x2+y2=l2
绳
x2+y2? l2
17
The equation of constraint of a double face constraint is an equality,
the equation of constraint of a single face constraint is an inequality,
We will discuss on the following only particle or a system of
particles which is subjected to steady,double face and holonomic
constraints,the general form of their equations is (s is the number of
the constraints,n is the number of the particles of the system),
),,2,1( 0),,;;,,( 111 sjzyxzyxf nnnj ???? ??
18
双面约束的约束方程为等式，单面约束的约束方程为不等式。
我们只讨论质点或质点系受定常、双面、完整约束的情况，
其约束方程的一般形式为（ s为质点系所受的约束数目,n为质
点系的质点个数）
),,2,1( 0),,;;,,( 111 sjzyxzyxf nnnj ???? ??
19
§ 16-2 Degrees of freedom and generalized coordinates
The position of a free particle in space is given by（ x,y,z ),
There are 3 numbers,
The position of a free system of particles in space is given by ( xi,
yi,zi ) (i=1,2…… n),There are 3n numbers,
An unfree system of particles,under the action of s holonomic
constraints,has only (3n-s ) independent coordinates,
The degree of freedom of it is k=3n-s,
The number of the independent coordinates which determine the
position of a system under the action of holonomic constraints is
called the number of degrees of freedom,or shortly,the degree
of freedom,
For example,in the case of the crankguide,discussed above the
position coordinates xA,yA,xB and yB,have to satisfy three
equations of constraints,therefore,it has only one degree of
freedom,
20
§ 16-2 自由度 广义坐标
一个自由质点在空间的位置：（ x,y,z ) 3个
一个自由质点系在空间的位置,( xi,yi,zi ) (i=1,2…… n) 3n个
对一个非自由质点系，受 s个完整约束，（ 3n-s )个独立坐标。
其自由度为 k=3n-s 。
确定一个受完整约束的质点系的位置所需的独立坐标的数目，
称为该质点系的 自由度的数目,简称为 自由度 。
例如,前述 曲柄连杆机构 例子中,确定曲柄连杆机构位置的四
个坐标 xA,yA,xB,yB须满足三个约束方程,因此有 一个自由度 。
21
In general,if a system of n particles consisting of n particles is
under the action of s constraints,the number of degrees of freedom
is snk ?? 3
Generally,n and s are very large,but k is small,In order to
determine the position of the system,the parameters should be
selected properly (they have to be independent on one another),
This is far more convenient than using 3n rectangular coordinates
and s constraint equations of constraints,
The independent parameters which are used to determine the
position of the system are called generalized coordinates,
The selection of the generalized coordinate is not unique,They
could be the linear displacements (x,y,z,s etc.) or the angular
displacements (such as ?,?,?,? etc.),If the constraints are
holonomic ones,the number of generalized coordinates is equal to
the number of the degrees of freedom,
22
一般地，受到 s个约束的、由 n个质点组成的质点系，其自由度为
snk ?? 3
通常,n 与 s 很大而 k 很小。为了确定质点系的位置，用
适当选择的 k 个参数（相互独立），要比用 3n个直角坐标和 s个
约束方程方便得多。
用来确定质点系位置的独立参数，称为 广义坐标 。
广义坐标的选择不是唯一的。广义坐标可以取线位移（ x,y,
z,s 等）也可以取角位移（如 ?,?,?,? 等）。在完整约束情
况下，广义坐标的数目就等于自由度数目。
23
For example,in a crankguide,the outer corner ? of the crank OA
can be selected as the generalized coordinate,then we have
.0,s i nc o s
,s i n,c o s
222 ????
??
BB
AA
yrlrx
ryrx
??
??
After choosing generalized
coordinates,the rectangular
coordinates of any particle of
the system can be expressed
as functions of the
generalized coordinates,
24
例如,曲柄连杆机构中，可取曲柄 OA的转角 ?为广义坐标，则,
0,s i nc o s
s i n,c o s
222 ????
??
BB
AA
yrlrx
ryrx
??
??
广义坐标选定后，
质点系中每一质点的直
角坐标都可表示为广义
坐标的函数。
25
In the case of a double pendulum which can swing only in the
vertical plane we have e.g,
.)()(
,
),,(,),(
22
12
2
12
22
1
2
1
2211
byyxx
ayx
yxyx
????
??
There are two degrees of
freedom,s? and ? can be
chooses as the generalized
coordinates,
.c o sc o s,s i ns i n
,c o s,s i n
22
11
????
??
baybax
ayax
????
??
26
例如,双锤摆。设只在铅直平面内摆动。
22
12
2
12
22
1
2
1
2211
)()(
),(,),(
byyxx
ayx
yxyx
????
??
两个自由度
取广义坐标 ?,?
????
??
c o sc o s,s i ns i n
c o s,s i n
22
11
baybax
ayax
????
??
27
Generally,a system of n particles has k degree of freedoms,
Choosing q1,q2,…… qk as the generalized coordinates,the
coordinates (the position vector) of every particle of the system
can be expressed as functions of the generalized coordinates,
),,,(
),,,(
),,,(
),,,(
21
21
21
21
kii
kii
kii
kii
qqqrr
qqqzz
qqqyy
qqqxx
?
?
?
?
?
?
?
?
),,2,1( ni ??
28
一般地，设有由 n个质点组成的质点系，具有 k个自由
度，取 q1,q2,……, qk为其广义坐标，质点系内各质点的
坐标及矢径可表为广义坐标的函数。
),,,(
),,,(
),,,(
),,,(
21
21
21
21
kii
kii
kii
kii
qqqrr
qqqzz
qqqyy
qqqxx
?
?
?
?
?
?
?
?
),,2,1( ni ??
29
§ 16-3 Virtual displacements and virtual work
At a fixed instant during the motion of a system of particles,an
arbitrary infinitesimal displacement of the particles of the system,
which is allowed by the constraints is called a virtual
displacement of the system of particles (at this given instant),
Virtual displacements can be linear displacements,but they can be
al so e.g,angular displacements,They are usually expressed using
the symbol of variation ?,
M
30
§ 16-3 虚位移和虚功
在质点系运动过程的某瞬时，质点系中的质点发生的为
约束允许的任意的无限小位移，称为质点系（在该瞬时）的
虚位移 。
虚位移可以是线位移，也可以是角位移。通常用变分符
号 ? 表示虚位移。
M
31
The virtual displacement is different from the real one during
real motion,
Real displacements really happen under the action of a certain force
during a certain time,The virtual displacement is an arbitrary one
permitted by the constraints at a fixed time,
A real displacement has a well-determined direction,it is a small or
a finite one,A virtual displacements are small displacements,of
several different directions permitted by to the constraints,
A real displacement happens during a time interval,but a virtual one
is just a pure geometrical concept,it happens in an infinite small
time interval,
Under the action of the steady
constraints,a small real displacement is
identical to a virtual one,However if the
constraints are unsteady,a small real
displacement is no longer a virtual one,
32
虚位移与真正运动时发生的实位移不同 。
实位移是在一定的力作用下和给定的初条件下运动而实际
发生的；虚位移是在约束容许的条件下可能发生的。
实位移具有确定的方向，可能是微小值，也可能是有限值；
虚位移则是微小位移，视约束情况可能有几种不同的方向。
实位移是在一定的时间内发生的；虚位移只是纯几何的概
念，完全与时间无关。
在定常约束下，微小的实位移必
然是虚位移之一。而在非定常约束下，
微小实位移不再是虚位移之一。
33
There are certain relationships among the virtual displacements of
the particles of the system,In order to determine them two ways are
developed,
(1) Geometrical method
The knowledge of kinematics tells us the displacement of a particle
is proportional to its velocity,hence
Therefore,we can analyze the relationships among virtual
displacements by an analysis of velocities,
.dtvrd ??
34
质点系中各质点的虚位移之间存在着一定的关系,确定这
些关系通常有两种方法,
(一 ) 几何法 。由运动学知，质点的位移与速度成正比，即
因此可以用分析速度的方法分析各点虚位移之间的关系。
dtvrd ??
35
(2) Analytical method
The coordinates of every particle of the system can be expressed as
functions of the generalized coordinates (q1,q2,……,qk),the
variations of the generalized coordinates are,the
virtual displacements of every particle are, The projections on
rectangular coordinates can be expressed as follows,
kqqq ???,,,21 ?
ir?
k
k
iii
i
k
k
iii
i
k
k
iii
i
q
q
z
q
q
z
q
q
z
z
q
q
y
q
q
y
q
q
y
y
q
q
x
q
q
x
q
q
x
x
????
????
????
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
2
2
1
1
2
2
1
1
2
2
1
1
),2,1( ni ??
36
(二 ) 解析法 。质点系中各质点的坐标可表示为广义坐标的函数
（ q1,q2,……,qk），广义坐标分别有变分,各
质点的虚位移 在直角坐标上的投影可以表示为
kqqq ???,,,21 ?
ir?
k
k
iii
i
k
k
iii
i
k
k
iii
i
q
q
z
q
q
z
q
q
z
z
q
q
y
q
q
y
q
q
y
y
q
q
x
q
q
x
q
q
x
x
????
????
????
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
2
2
1
1
2
2
1
1
2
2
1
1
),2,1( ni ??
37
[Example 1] Analyze the virtual
displacements of the points C,A and
B when the system is at the position
shown in the diagram (suppose
OC=BC= a and OA=l)
Solution
This is a system with one degree of
Freedom,We select the angle ? between
the rod OA and the axis x as the generalized coordinate,
1,Geometrical method
,
s in2
1
s in2
,
???
?
?
?
???
?
a
a
PB
PC
r
r
l
a
r
r
B
C
A
C
38
[例 1] 分析图示机构在图示位置时，
点 C,A与 B的虚位移。
(已知 OC=BC= a,OA=l )
解,此为一个自由度系统，取
OA杆与 x 轴夹角 ?为广义坐标。
1、几何法
???
?
?
?
s i n2
1
s i n2
???
?
a
a
PB
PC
r
r
l
a
r
r
B
C
A
C
39
.0,s in2
,c o s,s in
,c o s,s in
,,
????
?????
?????
??
BB
AA
CC
AC
yax
lylx
ayax
lrar
?????
????????
????????
??????
Transforming the coordinates
of the points C,A and B into
function of the generalized
coordinate ?we get
.0,c o s2
,s in,c o s
,s in,c o s
??
??
??
BB
AA
CC
yax
lylx
ayax
?
??
??
2,Analytical method Writing the variation of the
generalized coordinate ?,as we
can obtain the projections of the
virtual displacement of every point
on the corresponding rectangular
axes as,
.0,s i n2
,c o s,s i n
,c o s,s i n
????
?????
?????
BB
AA
CC
yax
lylx
ayax
?????
????????
????????
??
40
0,s in2
c o s,s in
c o s,s in
,
????
?????
?????
??
BB
AA
CC
AC
yax
lylx
ayax
lrar
?????
????????
????????
??????
将 C,A,B点的坐标表示成
广义坐标 ? 的函数，得
0,c o s2
s in,c o s
s in,c o s
??
??
??
BB
AA
CC
yax
lylx
ayax
?
??
??
2、解析法
对广义坐标 ? 求变分，得各点
虚位移在相应坐标轴上的投影,
0,s i n2
c o s,s i n
c o s,s i n
????
?????
?????
BB
AA
CC
yax
lylx
ayax
?????
????????
????????
41
The work done by a force along the virtual displacement of the
particle is called the virtual work,it is written as,
F r?
W?
zZyYxXW
rFW
????
??
???
??
42
力 在质点发生的虚位移 上所作的功称为 虚功,记为 。 F r? W?
zZyYxXW
rFW
????
??
???
??
43
§ 16-4 Ideal constraints
If the sum of the virtual works of all reaction forces of constraints
of the system of particles along any virtual displacement is zero,
then these constraints are called ideal constraints,
The condition for the action on ideal constraints of the system of
particles is
.0???? ? iiN rNW ??
44
§ 16-4 理想约束
如果在质点系的任何虚位移上，质点系的所有约束反力的
虚功之和等于零，则称这种约束为 理想约束。
质点系受有理想约束的条件,
0???? ? iiN rNW ??
45
Typical examples of ideal constraints are shown in the following,
1,Smooth supporting plane 2,Smooth hinge
0' ????? ? rNrNW N ???0??? rNW N ??
3,Rigid rod without weight
4,Soft cord which can not be
stretched
5,Pure rolling of a rigid
body on a rugged surface
0)( ??? ?? CN rFNW ??
46
理想约束的典型例子如下,
1、光滑支承面 2、光滑铰链
3、无重刚杆
4、不可伸长的柔索
5、刚体在粗糙面上的纯滚动
0' ????? ? rNrNW N ???0??? rNW N ??
0)( ??? ?? CN rFNW ??
47
§ 16-5 Theorem of virtual displacement
1,Theorem of virtual displacement
The necessary and sufficient condition for the equilibrium of a
system of particles with steady and ideal constraints is that the sum
of the virtual works of all the positive forces acting on this system
along any virtual displacement is zero,hence
? ?? 0ii rF ?
or
? ??? 0)( iiiiii zZyYxX ???
48
§ 16-5 虚位移原理
一、虚位移原理
具有定常、理想约束的质点系，平衡的必要与充分条件是：
作用于质点系的所有主动力在任何虚位移上所作的虚功之和等
于零。即
? ?? 0ii rF ?
解析式,
? ??? 0)( iiiiii zZyYxX ???
49
? ?? 0ii rF ?
Proof (1) Necessity,
When the system of particles is in equilibrium then,
One to the equilibrium of the whole system,an arbitrary
particle Mi is in equilibrium,too,
0?? ii NFfor any arbitrary virtual displacement of the particle M
i,,we
have ir?
.0)( ??? iii rNF ?
Because the constraints is ideal we have
.0? ?? ii rF ?
.0? ?? ii rN ?
Therefore
,0)( ???? iii rNF ?
.0???? ?? iiii rNrF ??
For the whole system we get
50
证明, (1) 必要性：即质点系处于平衡时，必有 ? ?? 0
ii rF ?
∵ 质点系处于平衡 ∴ 选取任一质点 Mi也平衡。
0?? ii NF
对质点 Mi 的任一虚位移,有
ir?
0)( ??? iii rNF ?
0)( ?? ?? iii rNF ?
0?????? iiii rNrF ??
由于是理想约束
? ?? 0ii rF ?
? ?? 0ii rN ?
所以
对整个质点系,
51
For the whole system of particles it follows
0)( ?? ?? iii rNF ?, Under the action of ideal constraints,then we have,
? ?? 0ii rN ?
? ?? 0 T h e r e f o r e ii rF ?in conflict with the initial proposition,
Therefore,if,the system of particles must be in
equilibrium,
? ?? 0ii rF ?
iRird ii rdr ??
A real displacement happens in the direction,choosing
we get
0)( ????? iiiii rRrNF ??
(2) Sufficiency,
If the system of particles satisfies it must be in
equilibrium,If,but if the system is not in equilibrium,
then at least the ith particle is not balanced,
? ?? 0ii rF ?
? ?? 0ii rF ?
0 ??? iii RNF
52
(2) 充分性：即当质点系满足,质点系一定平衡。
若,而质点系不平衡，则至少有第 i个质点不平衡。
? ?? 0ii rF ?
? ?? 0ii rF ?
在 方向上产生实位移,取,则
iR ird ii rdr ??
0)( ????? iiiii rRrNF ??
0)( ?? ?? iii rNF ?对质点系,(理想约束下,) ? ?? 0ii rN ?
? ??? 0 ii rF ? 与前题条件矛盾
故 时质点系必处于平衡。 ? ?? 0ii rF ?
0 ??? iii RNF
53
2,Applications of the theorem of virtual displacements
(1) A system in equilibrium being given,determine the
relationships among the positive forces,
(2) The positive forces of the system being given,determine the
conditions of equilibrium,
(3) Determine the reaction forces of the constraints it the system
is in equilibrium under the action of the known positive
forces,
(4) Determine the internal force of the double force equivalent
rod of balanced truss,
54
二、虚位移原理的应用
1,系统在给定位置平衡时，求主动力之间的关系；
2,求系统在已知主动力作用下的平衡位置；
3,求系统在已知主动力作用下平衡时的约束反力；
4,求平衡构架内二力杆的内力。
55
[Example 1] An ellipsograph mechanism is shown in the fig,The
length of the rod AB is l,its weight and friction can be neglected,
the hinges are smooth,Determine the relationship between the
magnitude of the positive forces when the system is in
equilibrium as shown in the fig,
Solution,
Investigate the whole
system,The constraints of
the system are all
holonomic,ideal and steady,
A
B
56
例 1 图示椭圆规机构，连杆 AB长 l，杆重和滑道摩擦不计，
铰链为光滑的，求在图示位置平衡时，主动力大小 P和 Q之间
的关系。
解,研究整个机构。
系统的所有约束都是
完整、定常、理想的。
57
1,Geometric method
Let A and B perform the virtual
displacements and,Then,from
the theorem of virtual displacements,we
can obtain the equation for the virtual
work as following
Ar? Br?
.0)tg ???? ArQ(P ??
Because is an arbitrary quantity we get
Ar? ? t gQP ?
.0 ?? BA rQrP ??
,tg
,c o ss i n,H o w e v e r
???
????
??
????
AB
BA
rr
rr
58
1、几何法,使 A发生虚位移,
B的虚位移,则由虚位移原理，
得虚功方程,
Ar?
Br?
0)tg ???? ArQ(P ??
由 的任意性，得 Ar? ? t gQP ?
0 ?? BA rQrP ??
tg
c o ss i n
???
????
??
????
AB
BA
rr
rr而
59
2,Analytical method
Because the system has only one
degree of freedom,the angle ? can be
selected as the generalized coordinate,
,c o s,s i n
,s i n,c o s
? ? ??? ? ??
??
lylx
lylx
AB
AB
???
??
Because is an arbitrary quantity we get ??
,t g?QP ?
,0??? BA xQyP ??
.0)s i nc o s( ??? ???? lQP
60
2、解析法 由于系统为单自由度,
可取 ?为广义坐标。
? ? ??? ? ??
??
c o s,s i n
s i n,c o s
lylx
lylx
AB
AB
???
??
由于 任意，故 ?? ? t gQP ?
,0??? BA xQyP ??
0)s i nc o s( ??? ???? lQP
61
Solution
This is a system with two degree of
freedom,Choosing the angles? and ?
as generalized coordinates,then there
are two ways to solve the problem,
[Example 2] The rod OA is joined with the rod AB at point A,In
addition,there is a hinge-support at point O as shown,The
lengths of the rods are 2a and 2b,respectively,their weights are
P1and P2,Suppose a horizontal force F acts on point B which
makes the system to be equilibrium,Determine the angles and to,
y
62
解,这是一个具有两个自由度的系
统，取角 ?及 ?为广义坐标，现用两
种方法求解。
例 2 均质杆 OA及 AB在 A点用铰连接，并在 O点用铰支承，
如图所示。两杆各长 2a和 2b，各重 P1及 P2，设在 B点加水平
力 F 以维持平衡，求两杆与铅直线所成的角 ?及 ? 。
y
63
Applying the theorem of virtual
displacement we get
)(, 021 axFyPyP BDC ??? ???
Substitution into equation (a) results in
.0)co s2s i n()co s2s i n2s i n( 221 ??????? ????????? bFbPaFaPaP
Solution 1
.c o s2c o s2,s in2s in2
,s ins in2
,c o sc o s2
,s in,c o sB u t
? ? ?? ? ????
? ? ?? ? ??
??
? ? ???
baxbax
bay
bay
ayay
BB
D
D
CC
????
???
??
???
64
应用虚位移原理,
)( 021 axFyPyP BDC ??? ???
代入 (a)式，得,
0)co s2s i n()co s2s i n2s i n( 221 ??????? ????????? bFbPaFaPaP
解法一,
? ? ?? ? ????
? ? ?? ? ??
??
? ? ???
c o s2c o s2,s in2s in2
s ins in2
,c o sc o s2
s in,c o s
baxbax
bay
bay
ayay
BB
D
D
CC
????
???
??
???而
65
Because is independent on,we get ??
.0c o s2s i n
,0c o s2s i n2s i n
2
21
?????
???????
??
???
bFbP
aFaPaP
.2 t g,22tg
221 P
F
PP
F ?
?? ??
So we obtain
??
66
由于 是彼此独立的，所以,????,
0c o s2s i n
0c o s2s i n2s i n
2
21
?????
???????
??
???
bFbP
aFaPaP
221
2 t g,
2
2tg
P
F
PP
F ?
?? ??
由此解得,
0)c o s2s i n()c o s2s i n2s i n( 221 ??????? ????????? bFbPaFaPaP
67
.0s i nco s 2 ?? ???? DB rPrF
But,,2 ?????? brbr
DB ??
Substitution into the equation above
results in
.22tg
22 P
F
bP
bF ?
?
??
??
???
Solution 2
Firstly,keep ? constant,but let ? change by a variation, Then
we obtain a set of virtual displacements of the system as show in
the diagram,
??
68
0s i nco s 2 ?? ???? DB rPrF
而 ?????? brbr
DB ??,2
代入上式，得
22
22tg
P
F
bP
bF ?
?
??
??
???
解法二,
先使 ? 保持不变，而使 ? 获得变分,得到系统的一
组虚位移，如图所示。
??
69
Keep ? constant,but let ? choux of, Then we have another set of
virtual displacements of the system as shown in the,
??
,BDA rrr ??? ??
.0s i ns i nco s 21 ??? ?????? DCB rPrPrF
But
.2
,
?????
???
arrr
ar
ADB
C
???
?
Substitution into the equation above gives
,22tg
21 PP
F
???
,0)s i n2s i n2co s( 21 ?????? ????? aPaPaF
diagram
70
再使 ? 保持不变，而使 ? 获得变分,得到系统的另
一组虚位移，如图所示。
??
BDA rrr ??? ??
0s i ns i nco s 21 ??? ?????? DCB rPrPrF
而
?????
???
arrr
ar
ADB
C
2
,
???
?
代入上式后，得,
22tg
21 PP
F
???
0)s i n2s i n2co s( 21 ?????? ????? aPaPaF
图示中,
71
Solution
Relieve the hinged
support B,substitute
the corresponding
reaction force of the
constraint,
BR
,0211 ????? ????? mrPrRrP CBB
,211
BB
C
B
B rmr
rP
r
rPR
?
??
?
?
?
? ????
[Example 3] This is a
multispan statically
determinate a beam,
Determine the reaction
forces on the hinged
support B,
72
例 3 多跨静定梁，
求支座 B处反力。
解,将支座 B 除
去，代入相应的
约束反力 。
BR
0211 ????? ????? mrPrRrP CBB
BB
C
B
B rmr
rP
r
rPR
?
??
?
?
?
? ????
2
1
1
73
.
96
11
8
11
12
11
12
1
6
1
4
,
8
11
,
2
1
,M o r e o v e r 1
?????????
??
B
C
B
E
B
G
B
B
C
B
r
r
r
r
r
r
r
r
r
r
r
?
?
?
?
?
?
?
??
?
?
?
?
.961181121 21 mPPR B ????
Hence,
74
96
11
8
11
12
11
12
1
6
1
4
,
8
11,
2
1 1
?????????
??
B
C
B
E
B
G
B
B
C
B
r
r
r
r
r
r
r
r
r
r
r
?
?
?
?
?
?
?
??
?
?
?
?
而
mPPR B 961181121 21 ????
BB
C
B
B rmr
rP
r
rPR
?
??
?
?
?
? ????
2
1
1
75
[Example 4] A sliding sleeve D is slipping
on the rod AB and drives the rod CD
sliding vertical,If ?=0o,the spring is at its
largest length,the stiffness of the spring is
5(kN/m),Determine the moment M acting
on the rod AB when the system is in
equilibrium at a certain angle ?,
Solution
The problem is to determine the relationships among the positive
forces acting on the system when it is in equilibrium,The spring
force can be treated as a positive force,and the constraints are ideal
ones,Therefore,the problem can be solved by the theorem of
virtual displacements,
76
例 4 滑套 D套在光滑直杆 AB上，并带动
杆 CD在铅直滑道上滑动。已知 ?=0o时，
弹簧等于原长，弹簧刚度系数为
5(kN/m)，求在任意位置（ ? 角）平衡
时，加在 AB杆上的力偶矩 M？
解,这是一个已知系统平衡，求作用于系统上主动力之间关系
的问题。将弹簧力计入主动力，系统简化为理想约束系统，故
可以用虚位移原理求解。
77
Select the system to consist of the rods AB,CD and select
the sliding sleeve D as the object to be investigated,
,ts e c3.0s
,s e c3.0
),kN( |s e c1|5.1||
),m( |s e c1|3.0||
.
c o s
3 0 0
6 0 0,W h e n
).mm(3 0 03 0 06 0 0,0W h e n
0
0
0
? ? ???
?
?
?
?
??
?
g
s
llkFF
ll
l
l
D
?
?
??????
???
???
????
?
From the theorem of virtual displacements,we have
).mkN( c o s )c o s1(s in45.0 3 ??? ? ??M
,0?? sFM ???,0] t gs e c3.0|s e c1|5.1[ ???? ?????M
78
选择 AB杆,CD杆和滑套 D的系统为研究对象。
? ? ???
?
?
?
?
?
?
g
s
llkFF
ll
l
l
D
ts e c3.0s
s e c3.0
)kN( |s e c1|5.1||
)m( |s e c1|3.0||
c o s
3 0 0
6 0 0,
)mm(3 0 03 0 06 0 0,0
0
0
0
?
?
??????
???
??
????
角时
时
?
由虚位移原理，得,
)mkN( co s )co s1(s i n45.0 3 ??? ? ??M
0?? sFM ??? 0] t gs e c3.0|s e c1|5.1[ ???? ?????M
79
The system with ideal constraints should be selected as the object
to be investigated without relieving the constraints,The system has
at least one degree of freedom,If there are nonideal constraints in
the system,such as a spring force or a friction force etc,they can
be treated as a positive force,Then the system is one with ideal
constraint,so it can be selected as the object to be investigated,
If you want to determine the reaction forces of the constraints,you
have to relieve the constraints firstly,then add the reaction forces
of the constraints,they are positive forces,You should get rid of
the constraints one by one,after every relief,a reaction force of
each constraint should be added in the diagram as a positive force
the number of degrees of freedom increases by one at each step,
Steps and key points in solving the equilibrium problem of a
system of particles using the theorem of virtual displacements,
1,Choose the object to be investigate d correctly
80
以不解除约束的理想约束系统为研究对象，系统至少有
一个自由度。若系统存在非理想约束，如弹簧力、摩擦力等，
可把它们计入主动力，则系统又是理想约束系统，可选为研
究对象。
若要求解约束反力，需解除相应的约束，代之以约束反
力，并计入主动力。应逐步解除约束，每一次研究对象只解
除一个约束，将一个约束反力计入主动力，增加一个自由度。
应用虚位移原理求解质点系平衡问题的步骤和要点,
1、正确选取研究对象,
81
2,Analyze the forces correctly
Draw the positive forces in the force diagram,e.g,spring forces,
friction forces and the reaction forces of the constraints which are
to be determined,
3,Analyze the virtual displacements correctly,determine the
relationships between the virtual displacements
4,Use the theorem of virtual displacements to set up the
equations
5,Solve the equations of virtual work and determine the
unknown quantities
82
2、正确进行受力分析,
画出主动力的受力图，包括计入主动力的弹簧力、摩擦
力和待求的约束反力。
3、正确进行虚位移分析，确定虚位移之间的关系 。
4、应用虚位移原理建立方程。
5、解虚功方程求出未知数。
83
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