1
Theoretical Mechanics
2
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(a coplanar system of concurrent forces and
a coplanar system of force couples)
4

[例 ]

5
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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§ 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

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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

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。

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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

（未知力系） （已知力系）

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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
21
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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[因主矢等于各力的矢量和 ]
R?
?????? iFFFFR ?321'主矢
?????
????
)()()(
21
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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（因主矩等于各力对简化中心取矩的代数和）
)( iOO FmM ??
（ 转动效应 ）

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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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① 认为 Fi这群力在同一

② 将 Fi向 A点简化得一

③ RA方向不定可用正交

④ YA,XA,MA为固定端

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

② =0,MO≠0 即简化结果为一合力偶,MO=M 此时刚

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,

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

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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)(
1
?
?
?
n
i
iOO FmM
)()( 主矩OO MdRRm ???
?
)()(
1
?
?
?
n
i
iOO FmRM

———合力矩定理

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

MO=0 为力偶也平衡
R?

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

?
0)( ?? iA Fm
0)( ?? iB Fm
0)( ?? iC Fm
③ 三矩式

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？

②画受力图（ 以后注明

0)( ?? iA Fm由
3
2,032 PNaNaP
BB ???????
0?? X 0?AX
0?? Y
3,0
PYPNY
ABB ?????

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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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?
???
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

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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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0)( ?? iA Fm
0)( ?? iB Fm

0?? Y
0)( ?? iO Fm

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

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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

0?? X
0?? Y
? ? 0im
0?? X
0?? Y
0)( ?? iO Fm

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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 ]

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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 ]

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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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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

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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,

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 ???

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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 ]

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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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§ 3-7 平面简单桁架的内力分析
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Structure of truss in engineering
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53
Structure of truss in engineering
54

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Structure of truss in engineering
56

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Structure of truss in engineering
58

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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

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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)
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③外力作用在节点上。

( 基本三角形 )

(a) (b) (c)
63
The common simplified models for
truss in engineering mechanics
64

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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

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
67
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

0?? Y 0,'3 ?? SP
,kN 10'3 ?　解得 S

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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

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

② 选截面 I-I,取左半部研究
I
I
A'
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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

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

21 SS ??且四杆节点无载荷、其中两两在

21 SS ??
43 SS ??

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,

③ 平衡;0,0' ?? OMR

)()(
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）

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

? ?
? ?
? ?
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

? ? 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

?
?
?
?
0
0
Y
X

0?? im

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

① ①
② ②
③ ③
④ ④
?

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

? ? 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？

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

045s i n,0 ???????? EDPCESm oCAE

)N(1 4 1 41707.0 11 0 0 045s i n ?????????? CEEDPS oCA

87
Solution,
Study the whole rigid structure;
Draw the force diagram;
Select a coordinate system,
write down the equations
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,??????
CD
AC ??
N48 ;N1 3 6,o b t a i n We ???? AA YX
88

?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 ??????
CD
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

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
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？

? ? 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

0?? Fm由
0512 ?????? PQY G
)kN(502 10550 ????? GY

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