1
Theoretical Mechanics
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3
Chapter 3,General Coplanar Force System
General coplanar force system,Several forces act on a body,the
lines of action of these forces are in a plane,but they don’t
intersect at one point and are not parallel,
[Example]
Reduction of a force system to a given point,Reduce an
unknown force system(a general coplanar force system) to a
known force system渎雁淫詫颠E渎雁淫詫颠urrent forces and
a coplanar system of force couples)
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第三章 平面任意力系
平面任意力系, 各力的作用线在同一平面内,既不汇交为一点
又不相互平行的力系叫 ~。
[例 ]
力系向一点简化, 把未知力系(平面任意力系)变成已知
力系(平面汇交力系和平面力偶系)
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Chapter 3,General Coplanar Force System
§ 3–1 Theorem of translation of a force
§ 3–2 Reduction of a general coplanar force system
to a given point
§ 3–3 Result of reduction of a general coplanar
force system ? the law of the resultant moment
§ 3–4 Conditions and equations for the equilibrium of
a general coplanar force system
§ 3–5 equilibrium equations of a coplanar system of
parallel forces
§ 3–6 The concepts of statically determinate and statically
indeterminate problems ? the equilibrium of a body
system
§ 3–7 Analysis of internal force of simple plane truss
Exercises
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第三章 平面一般力系
§ 3–1 力线平移定理
§ 3–2 平面一般力系向一点简化
§ 3–3 平面一般力系的简化结果 ? 合力矩定理
§ 3–4 平面一般力系的平衡条件和平衡方程
§ 3–5 平面平行力系的平衡方程
§ 3–6 静定与静不定问题的概念 ?物体系统的平衡
§ 3–7 平面简单桁架的内力分析
平面一般力系习题课
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§ 3-1 Theorem of translation of a force
Theorem of translation of a force,A force acting on a rigid body
can be moved parallel to its lines of action to any point of the
body,if we add a couple with a moment equal to the moment of
the force about the point to which it is translated,
[proof]
force system force force system ),( FFF ????FFF ???,,F
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§ 3-1 力线平移定理
力的平移定理, 可以把作用在刚体上点 A的力 平行移到任一
点 B,但必须同时附加一个力偶。这个力偶
的矩等于原来的力 对新作用点 B的矩。 F
F
[证 ] 力 力系 ),力偶(力 FFF ????FFF ???,,F
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① The theorem of translation of a force gives a relationship
between the force and force couple,force force+force
couple,
② The condition of translation of a force is to add a force couple
m,where m is given by m=F?d,
③ The theorem of translation of a force is the theoretical basis
of the reduction of a force system,
Explanation,
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① 力线平移定理揭示了力与力偶的关系:力 力 +力偶
(例断丝锥)
②力平移的条件是附加一个力偶 m,且 m与 d有关,m=F?d
③ 力线平移定理是力系简化的理论基础。
说明,
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§ 3-2 Reduction of a general
coplanar force system to a given point
A general coplanar
force system
A coplanar system of
concurrent forces + a
system of force couples
reduced to a
given center
(unknown force system) (known force system)
A coplanar system
of concurrent forces
Force R' (principal vector)
(acts on the given center)
A system of
force couples
Force couple MO (principal
moment) (acts in the plane)
an arbitrary
point
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§ 3-2 平面一般力系向一点简化
一般力系(任意力系) 向一点简化 汇交力系 +力偶系
(未知力系) (已知力系)
汇交力系 力, R'(主矢 ), (作用在简化中心 )
力 偶 系 力偶, MO (主矩 ), (作用在该平面上 )
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?????? iFFFFR ?321' v e c t o r p r i n c i p a l
?????
????
)()()(
m o m e n t p r i n c i p a l
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321
iOOO
O
FmFmFm
mmmM
?
?
magnitude,
Principal
vector direction,
for a given center of simplification (it is independent
on the position of the given center),
[because the principal vector equals to the resultant
of the forces]
R?
2222 )()(''' ?? ???? YXRRR
yx
?
??? ??
X
Y
R
R
x
y 11 tgtg?
(moving effect)
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大小,
主矢 方向,
简化中心 (与简化中心位置无关 )
[因主矢等于各力的矢量和 ]
R?
?????? iFFFFR ?321'主矢
?????
????
)()()(
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321
iOOO
O
FmFmFm
mmmM
?
?主矩
2222 )()(''' ?? ???? YXRRR
yx
?
??? ??
X
Y
R
R
x
y 11 tgtg?
( 移动效应 )
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magnitude,
direction,+ -
The given center for simplification,(it is dependent
on the position of the given center),
[because the principal moment equals to the
algebraic sum of moments of the forces about the
given center]
)( iOO FmM ??
(rotating effect)
Fixed support (rigid embedding)
Often used in the engineering
awning turning tool
MO Principal moment
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大小,
主矩 MO 方向, 方向规定 + —
简化中心, (与简化中心有关 )
(因主矩等于各力对简化中心取矩的代数和)
)( iOO FmM ??
( 转动效应 )
固定端(插入端)约束 在工程中常见的
雨 搭 车 刀
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Fixed support (rigid embedding) Explanation,
① Consider the forces Fi acting
in the same plane;
② Reduce Fi to the point A,
get the force and the force
couple;
③ the direction of RA is
uncertain,but it can be
determined by its rectangular
components YA and XA;
④ YA,XA and MA are the reaction
forces of the fixed support;
⑤ YA and Xa restrict the
translation of the rigid body,
Ma restrict the rotation,
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固定端(插入端)约束 说明
① 认为 Fi这群力在同一
平面内 ;
② 将 Fi向 A点简化得一
力和一力偶 ;
③ RA方向不定可用正交
分力 YA,XA表示 ;
④ YA,XA,MA为固定端
约束反力 ;
⑤ YA,XA限制物体平动,
MA为限制转动。
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§ 3-3 Result of reduction of a general
coplanar force system ? the law of the resultant moment
② =0,MO≠0,the system can be reduced to a force couple
with the moment MO=M,Here,the forces acting on the rigid
body are equivalent to the action of a force couple,Because
the force couple can be transferred anywhere in its plane of
action,the principal moment is independent on the given
center of simplification,
R?
① =0,MO =0,the force system is in equilibrium,This case
will be further discuss in the follow section,R
?
Result of the reduction,principal vector,principal moment MO,R?
③ ≠0,MO =0,the system can be reduced to a resultant force
going through the given center of simplification,The result of
the reduction is a resultant force (the resultant force of the
force system),,(It depends on the given center of
simplification,if the given center is changed,the principal
moment is not equal to zero.)
R?
RR ??
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§ 3-3 平面一般力系的简化结果 ? 合力矩定理
简化结果,主矢,主矩 MO,下面分别讨论 。
② =0,MO≠0 即简化结果为一合力偶,MO=M 此时刚
体等效于只有一个力偶的作用,因为力偶可以在刚体平
面内任意移动,故这时,主矩与简化中心 O无关。
R?
① =0,MO =0,则力系平衡,下节专门讨论。 R?
R?
③ ≠0,MO =0,即简化为一个作用于简化中心的合力。这时,
简化结果就是合力 ( 这个力系的合力),。 (此时
与简化中心有关,换个简化中心,主矩不为零)
R?
RR ??
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R?
④ ≠0,MO ≠0,this is the most general case,It can be reduced to
a resultant force, R
R
Md O?
The magnitude of the resultant force is equal to the
magnitude of the principal vector of the original force
system,
The position of the action line of the resultant force
is given by,
R
R
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R?④ ≠0,MO ≠0,为最一般的情况。此种情况还 可以继续简
化为一个合力 。 R
合力 的大小等于原力系的主矢
合力 的作用线位置
R
Md O?
R
R
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Conclusions,
)(
1
?
?
?
n
i
iOO FmM
m o m e n t ) ( p r i n c i p a l)( OO MdRRm ???
)()(
1
?
?
?
n
i
iOO FmRM
The results of the reduction of a general coplanar force system are,
① the resultant moment MO ; ② the resultant force,
The law of the resultant moment,
Because
the moment of the resultant force about point O is
——The law of the resultant moment
Because the given center of simplification is arbitrary,this
equation has a general meaning,
The moment of the resultant of a general coplanar force system
about any center is equal to the algebraic sum of the moments
of the forces about that center,
R
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结论,
)(
1
?
?
?
n
i
iOO FmM
)()( 主矩OO MdRRm ???
?
)()(
1
?
?
?
n
i
iOO FmRM
平面任意力系的简化结果,①合力偶 MO ; ②合力
合力矩定理,由于主矩
而合力对 O点的矩
———合力矩定理
由于简化中心是任意选取的,故此式有普遍意义。
即,平面任意力系的合力对作用面内任一点之矩等于力系
中各力对于同一点之矩的代数和。
R
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§ 3-4 Conditions and equations for the
equilibrium of a general coplanar force system
If =0 then forces are in equilibrium,
MO=0 then force couples are in equilibrium.
Therefore,the necessary and sufficient conditions for the
equilibrium of any general coplanar force system are,
The principal vector and the principal moment MO of the
system both are equal to zero,
I.e., 0)()(' 22 ??? ?? YXR
0)( ?? ? iOO FmM
R?
R?
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§ 3-4 平面一般力系的平衡条件与平衡方程
由于 =0 为力平衡
MO=0 为力偶也平衡
R?
所以 平面任意力系平衡的充要条件为,
力系的主矢 和主矩 MO 都等于零,即,
0)()(' 22 ??? ?? YXR
0)( ?? ? iOO FmM
R?
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0?? X
0)( ?? iA Fm
0)( ?? iB Fm
② the equations
of two moments
Condition,the axis x
is not line AB ?
0)( ?? iA Fm
0)( ?? iB Fm
0)( ?? iC Fm
③ the equations
of three moments
Condition,the points
A,B and C are not
collinear
The three independent equations above can only
determine three unknown quantities,
0?? X
0?? Y
0)( ?? iO Fm
① the equations
of one moment
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0?? X
0)( ?? iA Fm
0)( ?? iB Fm
② 二矩式
条件,x 轴不 AB
连线
?
0)( ?? iA Fm
0)( ?? iB Fm
0)( ?? iC Fm
③ 三矩式
条件,A,B,C不在
同一直线上
上式有三个独立方程,只能求出三个未知数。
0?? X
0?? Y
0)( ?? iO Fm
① 一矩式
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[Example] P and a are given,Determine the reaction forces of
the constraints at the point A and B,
Solution,
① investigate beam AB,
② draw the forces diagram
(If we write,free from constraints”
under the diagram,we can directly
draw the reaction forces of the
constraints on the original diagram
of the whole structure.)
0)( F r o m ?? iA Fm
3
2,032 PNaNaP
BB ???????
0?? X 0?AX
0?? Y
3,0
PYPNY
ABB ?????
free from constraints
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[例 ] 已知,P,a,求,A,B两点的支座反力?
解:①选 AB梁研究
②画受力图( 以后注明
解除约束,可把支反
力直接画在整体结构
的原图上 )
0)( ?? iA Fm由
3
2,032 PNaNaP
BB ???????
0?? X 0?AX
0?? Y
3,0
PYPNY
ABB ?????
解除约束
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R?
§ 3-5 Equilibrium equations of
a coplanar system of parallel forces
Coplanar system of parallel forces,If all lines of action of the
forces acting on a body are parallel and in one plane,they form a
coplanar system of parallel forces,
Assuming F1,F2 … Fn to be a
coplanar system of parallel forces,
the reduction to the point O are,
,
The position of the line of action of
the resultant force is,
?
???
F
xF
R
Mx iiO
R '
??? FRR O '
?? ?? iiiOO xFFmM )(
Principal vector
principal moment
The necessary and sufficient conditions are,
Principal vector =0,Principal moment MO =0,
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设有 F1,F2 … Fn 各平行力系,
向 O点简化得,
合力作用线的位置为,
平衡的充要条件为 主矢 =0
主矩 MO =0
?
???
F
xF
R
Mx iiO
R ' R?
??? FRR O '主矢
?? ?? iiiOO xFFmM )(主矩
§ 3-5 平面平行力系的平衡方程
平面平行力系,各力的作用线在同一平面内且相互平行的力系叫 ~。
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Therefore,the equations for the
equilibrium of the coplanar system
of parallel forces are,
Condition,line AB is not
parallel to the line of the
action of the force,
0)( ?? iA Fm
0)( ?? iB Fm
the equations
of two moments
0?? Y
0)( ?? iO Fm
the equations
of one moment
Actually,the x projections of all
the forces are zero,,
Therefore,there are only two
independent equations,they can
determine only two unknown
quantities,
0?? X
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所以 平面平行力系的平衡方程为,
0)( ?? iA Fm
0)( ?? iB Fm
二矩式
条件,AB连线不能平行
于力的作用线
0?? Y
0)( ?? iO Fm
一矩式
实质上是各力在 x 轴上的投影
恒等于零,即
恒成立,所以只有两个
独立方程,只能求解两个独立
的未知数。
0?? X
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0,0 F r o m ??? AXX
02
2; 0)(
????????
??
aPm
a
aqaR
Fm
B
A
0?? Y 0????? PqaRY
BA
)kN(122028.0162 8.02022 ??????????? PamqaR B
)kN(24128.02020 ???????? BA RqaPY
[Example] Let P=20kN,m=16kN·m,q=20kN/m,a=0.8,
Determine the reaction forces at the points A and B,
Solution,
Study the beam AB,
Solving them,we obtain,
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0,0 ??? AXX由
02
2; 0)(
????????
??
aPm
a
aqaR
Fm
B
A
0?? Y 0????? PqaRY
BA
)kN(122028.0162 8.02022 ??????????? PamqaR B
)kN(24128.02020 ???????? BA RqaPY
[例 ] 已知,P=20kN,m=16kN·m,q=20kN/m,a=0.8m
求,A,B的支反力。
解:研究 AB梁
解得,
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§ 3-6 The concepts of statically determinate and statically
indeterminate problems ? the equilibrium of a body system
1,The concepts of statically determinate and statically
indeterminate problems
We have learned,
Coplanar system of concurrent forces,two independent
equations can only determine two unknown quantities,
System of force couples has an independent equation,so it
can only determine one unknown quantity,
General coplanar force system,three independent
equations which can determine three independent quantities,
0?? X
0?? Y
? ? 0im
0?? X
0?? Y
0)( ?? iO Fm
When The number of equations≥the number of unknown quantities
it is a statically determinate problem (which can be solved),
When The number of equations<the number of unknown quantities
it is a statically indeterminate problem,
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§ 3-6 静定与静不定问题的概念 ? 物体系统的平衡
一、静定与静不定问题的概念
我们学过,
平面汇交力系 两个独立方程,只能求两个独立
未知数。
一个独立方程,只能求一个独立未知数。
三个独立方程,只能求三个独立未知数。
0?? X
0?? Y
? ? 0im
0?? X
0?? Y
0)( ?? iO Fm
力偶系
平面
任意力系
当,独立方程数目 ≥ 未知数数目时,是静定问题(可求解)
独立方程数目 <未知数数目时,是静不定问题(超静定问题)
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[Example]
Statically indeterminate problems can be solved by the
harmonious conditions of displacement in others courses of
mechanics (material mechanics,structural mechanics and elastic
mechanics),
Statically determinate Statically indeterminate
(three unknown quantities) (four unknown quantities)
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[例 ]
静不定问题在强度力学 ( 材力,结力,弹力)中用位移
谐调条件来求解 。
静定(未知数三个) 静不定(未知数四个)
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[Example]
2,Equilibrium of a body system
External forces,
forces which act on the body system from outside,
Internal forces,
forces among the bodies of the body system,
Body system (BS),A system composed of several rigid bodies
connected together by constraints,
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[例 ]
二、物体系统的平衡问题
外力,外界物体作用于系统上的力叫外力。
内力,系统内部各物体之间的相互作用力叫内力。
物体系统( 物系 ):由若干个物体通过约束所组成的系统叫 ~。
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The characteristics of the equilibrium of a body system,
① The body system is static
② Every body in the body system is also in equilibrium,
for every body there are 3 equilibrium equations,
so the whole system of the bodies has 3n equations,
(n is number of bodies in the body system)
The general method to solve the problems of the body system,
from whole parts (usual way); from parts whole (unusual way)
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物系平衡的特点,
①物系静止
②物系中每个单体也是平衡的。每个单体可列 3个
平衡方程,整个系统可列 3n个方程(设物系中
有 n个物体)
解物系问题的一般方法,
由整体 局部 (常用),由局部 整体 (用较少)
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[Example] Let OA horizontal (OA=R,AB= l) and the pressure force
to be P,Determine,① M,② the reaction forces of point O,
③ the internal forces of the rod AB,
④ the side pressure acting on the lead rail,
0 f r o m ?? X
0s i n ??? ?BSN
0?? Y
0co s ??? ?BSP
?? gPNPS B t,c o s ???
Solution,
Study the point B
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[例 ] 已知,OA=R,AB= l,当 OA水平时,冲压力为 P时,
求:① M=?② O点的约束反力?③ AB杆内力?
④冲头给导轨的侧压力?
0?? X由
0s i n ??? ?BSN
0?? Y
0co s ??? ?BSP
?? gPNPS B t,c o s ???
解,研究 B
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0)( ?? Fm O
0c o s ???? MRS A ?
0?? X
0s i n ??? ?AO SX
0?? Y
0c o s ??? OA YS ?
PRM ??
PY O ??? t gPX O ??
[the sign of minus means the direction of the
forces is opposite to the direction indicated in
the diagram]
Then study the wheel,
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0)( ?? Fm O
0c o s ???? MRS A ?
0?? X
0s i n ??? ?AO SX
0?? Y
0c o s ??? OA YS ?
PRM ??
PY O ??? t gPX O ??
[负号表示力的方向与图中所设方向相反 ]
再研究轮
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Because there are so many kinds of body systems,the systems which
only composed of straight elements are called----truss
§ 3-7 Analysis of internal force of plane truss
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由物系的多样化,引出仅由杆件组成的系统 ——桁架
§ 3-7 平面简单桁架的内力分析
51
Structure of truss in engineering
52
工程中的桁架结构
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Structure of truss in engineering
54
工程中的桁架结构
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Structure of truss in engineering
56
工程中的桁架结构
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Structure of truss in engineering
58
工程中的桁架结构
59
Truss, a rigid structure system composed of straight
elements,connected at their ends by pins,
without deformation under the action of forces,
joint
element
60
桁架,由杆组成,用铰联接,受力不变形的系统。
节点
杆件
61
The model of truss in mechanics
(basic triangle) Triangles ensure stability
The advantage of the truss,light,the capability of the material
can exert sufficiently,
The characters of the truss,① straight members,the weight of
themselves can be neglected,they are all double force equivalent
rods;
② The straight elements are connected at their ends by joints,
③ All external loads on the truss act only at the joints,
(a) (b) (c)
62
桁架的优点:轻,充分发挥材料性能。
桁架的特点:①直杆,不计自重,均为二力杆;②杆端铰接;
③外力作用在节点上。
力学中的桁架模型
( 基本三角形 )
三角形有稳定性
(a) (b) (c)
63
The common simplified models for
truss in engineering mechanics
64
工程力学中常见的桁架简化计算模型
65
1,Method of Isolation joints,
P=10kN,determine the internal forces on every elements,[Example]
,0?? X 0?BX
,0)( ?? Fm A
,0)( ?? Fm B
024 ?? PY B
042 ?? ANP kN 5,0 ???? BAB YNX
Solution,
① study the whole structure,
determine the reaction forces,
② Study the joints A,C and D one by one,determine the internal
forces of every member,
0?? X 030c o s 012 ?? SS
0?? Y 030s in 01 ?? SN A
kN10,kN66.8 12 ??? SSWe get,(denote that the rod is compressed)
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,0?? X 0?BX
,0)( ?? Fm A
,0)( ?? Fm B
024 ?? PY B
042 ?? ANP
kN 5,0 ???? BAB YNX
解, ① 研究整体,求支座反力
一、节点法 已知:如图 P=10kN,求各杆内力? [例 ]
② 依次取 A,C,D节点研究,计算各杆内力。
0?? X 030c o s 012 ?? SS
0?? Y 030s in 01 ?? SN A
)(kN10,kN66.8 12 表示杆受压解得 ??? SS
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0?? X
0?? Y
030c o s'30c o s 0104 ?? SS
030s i n30s i n' 04013 ???? SSS
1
'
1 SS ?
kN 10,kN 10 43 ??? SS
kN 66.75 ?S
0?? X 0'25 ?? SS
2'2 SS ?
The another equation of the joint D can be
used to verify the result of calculation,
0?? Y 0,'3 ?? SP
It equals to,the calculation is
accurate,
3S
Substituting into
We obtain,
Substituting into we get,
,kN 10'3 ? SThe solution of which is
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0?? X
0?? Y
030c o s'30c o s 0104 ?? SS
030s i n30s i n' 04013 ???? SSS
1'1 SS ?代入
kN 10,kN 10, 43 ??? SS解得
kN 66.75 ?S解得 
0?? X 0'25 ?? SS
后代入 2'2 SS ?
节点 D的另一个方程可用来校核计算结果
0?? Y 0,'3 ?? SP
,kN 10'3 ? 解得 S
恰与 相等,计算准确无误。 3S
69
Solution,Study the whole
structure,determine the reaction
forces,
0?? X 0?AX 0?? BM
023 ??????? aPaPaY
PY A ??

0?? ?Am 04 ????? aYhS A
h
PaS ???
40?? Y 0s in5 ???? PSY A ? 05 ?S
0?? X 0co s 456 ????? AXSSS ?
h
PaS ?
6
2,Method of cuts [Example] h,a and P are given,determine the internal forces on the
element 4,5 and 6,
② Draw a cut I-I,study the left-hand
section of the truss,
I
I
A' From
70
解, 研究整体求支反力
0?? X 0?AX
0?? BM
023 ??????? aPaPaY
PY A ??

0?? ?Am由 04 ????? aYhS A
h
PaS ???
4
0?? Y 0s in
5 ???? PSY A ?
05 ?S
0?? X 0co s 456 ????? AXSSS ?
h
PaS ?
6
二、截面法 [例 ] 已知:如图,h,a,P
求,4,5,6杆的内力。
② 选截面 I-I,取左半部研究
I
I
A'
71
Instruction,
Method of Isolation joints,This is convenient when the stresses in
all elements of the truss have to be determined,
Method of cuts, This is convenient in determining the stresses in
individual members,notably for verifying results,
Before the calculation,assume first all elements to be under tension,
if the result is negative it means the element concerned is
compressed,opposite to the direction assumed,
72
说明, 节点法:用于设计,计算全部杆内力
截面法:用于校核,计算部分杆内力
先把杆都设为拉力,计算结果为负时,说明是压力,
与所设方向相反。
73
If no load acts on a three-element
joint and two elements of them are
collinear,the third is a zero element,21 SS ??
If no load acts on the four-element
joint and two pairs of them are
collinear,the internal forces of each
pair are equal in magnitude and
of the same character,
If no load acts on a two-element
joint and the two elements are
not collinear,the internal forces of
them are zero,they are called
zero elements,
3,The judgments of the internal
forces of special elements
021 ?? SS



21 SS ??
43 SS ??
74
三杆节点无载荷、其中两杆在
一条直线上,另一杆必为零杆
21 SS ??且四杆节点无载荷、其中两两在
一条直线上,同一直线上两杆
内力等值、同性。
21 SS ??
43 SS ??
两杆节点无载荷、且两杆不在
一条直线上时,该两杆是零杆。
三、特殊杆件的内力判断
021 ?? SS



75
Exercises
1,Theorem of translation of a force is the theoretical basis of
the reduction of a force system force force + force couple
③ Equilibrium,;0,0' ?? OMR
The law of the resultant moment,,
)()(
1
i
n
i
OO FmRm ?
?
?

② 0,0o r 0,0
'' ????
OO MRMR
Resultant force (principal vector);0,0' ?? OMR
Resultant force couple (principal moment),
2,The result of the general coplanar force system,
Summary,
76
,平面一般力系习题课,
一、力线平移定理是力系简化的理论基础
力 力 +力偶
③ 平衡;0,0' ?? OMR
合力矩定理
)()(
1
i
n
i
OO FmRm ?
?
?;0,0;0,0 '' ???? OO MRMR 或
① 合力(主矢);0,0' ?? OMR
② 合力偶(主矩)
二、平面一般力系的合成结果
本章小结,
77
The equilibrium equations of the coplanar system of parallel
forces ? ? 0X
0)(
0
??
??
Fm
Y
A 0)(
0)(
?
?
?
?
Fm
Fm
B
A
line AB is not parallel to
the line of action of the
force,
② the equations
of two moments
① the equations
of one moment
? ?
? ?
? ?
0)(
0
0
Fm
Y
X
O ? ?
? ?
? ?
0)(
0)(
0
Fm
Fm
X
B
A
? ?
? ?
? ?
0)(
0)(
0)(
Fm
Fm
Fm
C
B
A
3,The equilibrium equations of the general coplanar system
② the equations
of two moments
③ the equations
of there moments
① the equations
of one moment
Point A,B and C
are not collinear
Axis x is
not line AB ?
78
一矩式 二矩式 三矩式
三,
? ?
? ?
? ?
0)(
0
0
Fm
Y
X
O ?
?
? ?
? ?
0)(
0)(
0
Fm
Fm
X
B
A
A,B连线不 x轴 ?
? ?
? ?
? ?
0)(
0)(
0)(
Fm
Fm
Fm
C
B
A
A,B,C不共线
平面一般力系的平衡方程
平面平行力系的平衡方程
成为恒等式
一矩式 二矩式
? ? 0X?
0)(
0
??
??
Fm
Y
A 0)(
0)(
?
?
?
?
Fm
Fm
B
A
BA 连线不平行于力线
79
The equilibrium equations
of the coplanar system of
concurrent forces,0)( ?? Fm A? ?
?
?
?
?
0
0
Y
X
The equilibrium equations
of the coplanar system of
force couples,0?? im
4,Statically determinate and statically indeterminate systems,
If the number of the equations≥the number of the unknown quantities
— Statically determinate system
if the number of the equation < the number of the unknown quantities
— Statically indeterminate system
5,The equilibrium of a body system,
When the body system is in equilibrium,every body in the body
system is also in equilibrium,The method to solve the problem of
the body system usually is,from whole parts elements,
80
平面汇交力系的平衡方程
成为恒等式 0)( ?? Fm A? ?
?
?
?
?
0
0
Y
X
平面力偶系的平衡方程
0?? im
四、静定与静不定
独立方程数 ≥ 未知力数目 —为静定
独立方程数 <未知力数目 —为静不定
五、物系平衡
物系平衡时,物系中每个构件都平衡,
解物系问题的方法常是,由整体 局部 单体
81
7,Noticeable problems,
The force couple has no projection on coordinate axes;
the moment of the force couple M is constant,it is
independent on the coordinate axes and the given center,
① ①






6,The steps and skills of solving problems,
STEPS
Select a body to study;
Draw the force diagram
(force analysis);
Select the coordinates,
a center,and write down
equilibrium equations;
Solve the equations,
determine the unknown
quantities,
SKILLS
Choose the coordinate
axes unknown quantities;
Choose the given center of the
forces at the point of crossing;
We’d better use the characters
of the double-force equivalent
rod sufficiently;
Use the law of the resultant
moment smartly,
?
82
六、解题步骤与技巧
解题步骤 解题技巧
选研究对象 选坐标轴最好是未知力 投影轴;
画受力图(受力分析) 取矩点最好选在未知力的交叉点上;
选坐标、取矩点、列 充分发挥二力杆的直观性;
平衡方程。
解方程求出未知数 灵活使用合力矩定理。
① ①
② ②
③ ③
④ ④
?
七、注意问题
力偶在坐标轴上投影不存在;
力偶矩 M =常数,它与坐标轴与取矩点的选择无关。
83

[Example 1] All elements are connected with each other by
joints,the bottom B is inserted into the earth,P=1000N,
AE=BE=CE=DE=1m,the weight of the elements are negligible,
Determine the internal forces of the member AC and the reaction
forces of the point B,
8,Analysis of the examples
Solution,
Study the whole rigid structure;
Draw the force diagram;
Select the coordinate system Bxy
with the point B as the given centre;
Write down the equations



? ? 0X ;0?BX
0?? Bm 0??? DEPM B
)mN(100011000 ????BM
? ? 0Y ;0?? PY B PYB ?
Solving them,we obtain,
84
解, 选整体研究
受力如图
选坐标、取矩点,Bxy,B点
列方程为,
解方程得




? ? 0X ;0?BX
0?? Bm 0??? DEPM B
)mN(100011000 ????BM
? ? 0Y ;0?? PY B PYB ?
[例 1] 已知各杆均铰接,B端插入地内,P=1000N,
AE=BE=CE=DE=1m,杆重不计。 求 AC 杆内力? B点的反力?
八、例题分析
85
Select point E as the given center,write down the equation,
Solving them,we find,045s i n,0 ???????? EDPCESm
oCAE
① ②


)N(1 4 1 41707.0 11 0 0 045s i n ?????????? CEEDPS oCA
Study the element CD; Draw the force diagram;
86
受力如图
取 E为矩心,列方程
解方程求未知数
045s i n,0 ???????? EDPCESm oCAE




)N(1 4 1 41707.0 11 0 0 045s i n ?????????? CEEDPS oCA
再研究 CD杆
87
Solution,
Study the whole rigid structure;
Draw the force diagram;
Select a coordinate system,
write down the equations
[Example 2]Let P=100N,AC=1.6m,BC=0.9m,CD=EC=1.2m,AD=2m
AB is horizontal,ED is vertical,BD is
perpendicular to the incline,
Determine and the reaction forces
of the constraints,
BDS
02.15.2,0 ??????? PYm AB
0s i nc o ss i n,0' ???????? ??? PYXX AA
5
3
2
2.1 c o s ;
5
4
2
6.1 s i n b e c a u s e,??????
AD
CD
AD
AC ??
N48 ;N1 3 6,o b t a i n We ???? AA YX
88
[例 2] 已知,P=100N,AC=1.6m,BC=0.9m,CD=EC=1.2m,AD=2m
且 AB水平,ED铅垂,BD垂直于
斜面; 求?和支座反力?
解, 研究整体
画受力图
选坐标列方程
?BDS
02.15.2,0 ??????? PYm AB
0s i nc o ss i n,0' ???????? ??? PYXX AA
5
3
2
2.1 co s ;
5
4
2
6.1 s i n ??????
AD
CD
AD
AC ??而
N48 ;N1 3 6, ???? AA YX解得
89
Again study the element AB,
draw the force diagram as above;
0s i n,0 ???????? ACYCBSm ABC ??
N7.106
5
4
9.0
6.1)48(
s i n
:f i n d w e t h e m,S ol v i n g
?
?
??
?
?
?
?
?BC
ACY
S
A
B
90
再研究 AB杆,受力如图
0s i n,0 ???????? ACYCBSm ABC ?由
N7.106
5
49.0
6.1)48(
s i n
,?
?
??
?
?
?
?
?BC
ACY
S AB解得
91
[Example 3] P d are known,determine the internal forces of the
rods a,b,c and d,Solution,from the judgments
of the zero element
0??? adc SSS
Study the point A,
? ? 0Y?
045c o s ??? PS ob
PS b 2?
Again
92
[例 3] 已知 P d,求,a.b.c.d四杆的内力?
解,由零杆判式
0??? adc SSS
研究 A点,
? ? 0Y由
045c o s ??? PS ob
PS b 2?
93
[Example 4] Consider a continuous beam,P=10kN,Q=50kN,CE is
vertical,neglect the weight of the beam,Determine the reaction
forces of the points A,B and D,(we find the number of the
unknown quantities is larger than three,so we cannot
solve it by studying
the whole rigid structure,
it must be divided into
several parts.)
0?? Fm
0512 ?????? PQY G
)kN(502 10550 ????? GY
Solution,
① study the crane
94
[例 4] 已知:连续梁上,P=10kN,Q=50kN,CE 铅垂,不计梁重
求,A,B和 D点的反力 (看出未知数多余三个,不能先整
体求出,要拆开)
0?? Fm由
0512 ?????? PQY G
)kN(502 10550 ????? GY
解, ① 研究起重机
95
? ? 0Cm 016 ' ???? GD YY
)kN(33.8650 ??? DY
0610123,0 ?????????? QPYYm DBA )kN(100?? BY
0,0 ??????? PQYYYY DBA )kN(33.48??? AY
③ Study the
whole rigid
structure,
② Then study the beam CD
96
? ? 0Cm由
016 ' ???? GD YY
)kN(33.8650 ??? DY
0610123,0 ?????????? QPYYm DBA )kN(100?? BY
0,0 ??????? PQYYYY DBA )kN(33.48??? AY






② 再研究梁 CD
97
98