M e c h a n i c s o f M a t e r i a l s
CHAPTER 11 ENERGY METHOD
§ 11–1 GENERAL EXPRESSIONS OF THE STRAIN ENERGY
§ 11–2 MOHR’S THEOREM(METHOD OF UNIT FORCE)
§ 11–3 CATIGLIANO’S THEOREM
第十一章 能量方法
§ 11–1 变形能的普遍表达式
§ 11–2 莫尔定理 (单位力法 )
§ 11–3 卡氏定理
§ 11–1 GENERAL EXPRESSIONS OF THE STRAIN ENERGY
1,Principle of energy,
2,Calculation of the strain energy of rods,
1),Calculation of the strain energy of rods in tension or compression,
?? L xEA xNU d2 )(
2
?
?
?
n
i ii
ii
AE
LNU
1
2
2
??21?u
Strain energy stored in the elastic body is equal to the work done by external
forces,that is,
WU ?
Method to analyze and calculate displacements, deformations and internal
forces of deformable bodies by this kind of relation is called energy method,
or Density of the strain energy,
§ 11–1 变形能的普遍表达式
一、能量原理,
二、杆件变形能的计算,
1.轴向拉压杆的变形能计算,
?? L xEA xNU d2 )(
2 ?
?
?
n
i ii
ii
AE
LNU
1
2
2
或 ??
2
1,?u比能
弹性体内部所贮存的变形能,在数值上等于外力所作
的功,即
WU ?
利用这种功能关系分析计算可变形固体的位移、变形
和内力的方法称为能量方法。
2,Calculation of the strain energy of rods in torsion,
?? L
P
n x
GI
xM
U d
2
)(
2
?
?
?
n
i Pii
ini
IG
LMU
1
2
2
??21?u
3,Calculation of strain energy of rods in bending,
?? L xEI
xM
U d
2
)(
2
?
?
?
n
i ii
ii
IE
LMU
1
2
2
??21?u
or
Density of the strain energy,
or
Density of the strain energy,
2.扭转杆的变形能计算,
?? L
P
n x
GI
xM
U d
2
)(
2
?
?
?
n
i Pii
ini
IG
LMU
1
2
2

??
2
1,?u比能
3.弯曲杆的变形能计算,
?? L xEI
xM
U d
2
)(
2
?
?
?
n
i ii
ii
IE
LMU
1
2
2

??
2
1,?u比能
3,General expressions of the strain energy,
Strain energy is independent of the order of loading,Deformations due to
mutually independent load may be summed up each other,
For slender columns,the strain energy due to shearing forces may be neglected,
x
EI
xM
x
GI
xM
x
EA
xN
U
LL
P
n
L
d
2
)(
d
2
)(
d
2
)( 222
??? ???
?? L xEA xQ d2 )(
2
S? ?S?
x
EI
xM
x
GI
xM
x
EA
xN
U
LL
P
n
L
d
2
)(
d
2
)(
d
2
)( 222
??? ???
Deflection factor of shear
三、变形能的普遍表达式,
变形能与加载次序无关;相互独立的力(矢)引起的变形能
可以相互叠加。
细长杆,剪力引起的变形能可忽略不计。
?? L xEA xQ d2 )(
2
S? 剪切挠度因子?S?
x
EI
xM
x
GI
xM
x
EA
xN
U
LL
P
n
L
d
2
)(
d
2
)(
d
2
)( 222
??? ???
x
EI
xM
x
GI
xM
x
EA
xN
U
LL
P
n
L
d
2
)(
d
2
)(
d
2
)( 222
??? ???
Solution,In energy method( work done by
external forces is equal to the strain energy)
① Determine internal forces
?? s i n)( PRM T ?
)c o s1()( ?? ?? PRM N
A
Bending moment,
Torque,
Example 1 A semicircle rod as shown in the figure is lie in horizontal plane,
A vertical force P act at its point A,Determine the displacement of point A in
vertical direction,
P R
O
Q
MN
MT
A
A
P N
B
?
T
O
MN
[例 1 ] 图示半圆形等截面曲杆位于水平面内,在 A点受铅垂力 P
的作用,求 A点的垂直位移。
解:用能量法(外力功等于应变能)
① 求内力
?? s i n)(,PRM T ?弯矩
)c o s1()(,?? ?? PRM N扭矩
A
P R
O
Q
MT
A
A
P N
B
?
T
O
③ Work done by external forces is equal to the strain energy
② Strain energy,
??? ??? LL
P
L
x
EI
xM
x
GI
xMx
EA
xNU d
2
)(
d
2
)( d
2
)( 22n2
?? ?
?
?
??
?
?
?
?
0
222
0
222
d
2
)( s i n
d
2
)c o s1(
R
EI
RP
R
GI
RP
P
EI
RP
GI
RP
P 44
3 3232 ??
??
,
2
UfPW A ??
Let
EI
PR
GI
PR
f
P
A 22
3 33 ??
??
then
③ 外力功等于应变能
② 变形能,
??? ??? LL
P
L
x
EI
xM
x
GI
xMx
EA
xNU d
2
)(
d
2
)( d
2
)( 22n2
?? ?
?
?
??
?
?
?
?
0
222
0
222
d
2
)( s i n
d
2
)c o s1(
R
EI
RP
R
GI
RP
P
EI
RP
GI
RP
P 44
3 3232 ??
??
UfPW A ??
2
?
EI
PR
GI
PR
f
P
A 22
3 33 ??
???
Example 2 Determine the deflection of point C by the energy
method,where the beam is of equal section and straight,
CPfW 2
1?
Solution,Work done by external
forces is equal to the strain energy
?? L xEI xMU d2 )(
2
)0(; 2)( axxPxM ???
By using symmetry we get,
EI
aPxxP
EI
U
a
12
d)
2
(
2
12 32
0
2 ?? ?
EI
Pat h e n fUW
C 6,
3
??Thinking:
For the distributed load,can
we determine the displacement of point C
by this method?
q
C a a
A
P
B
f
Let
[例 2 ] 用能量法求 C点的挠度。梁为等截面直梁。
CPfW 2
1?
解,外力功等于应变能
?? L xEI xMU d2 )(
2
)0(; 2)( axxPxM ???
应用对称性,得,
EI
aPxxP
EI
U
a
12
d)
2
(
2
12 32
0
2 ?? ?
EI
PafUW
C 6
3
????
思考:分布荷载时,可否用此法求 C点位移? q
C a a
A
P
B
f
§ 11–2 MOHR’S THEOREM(METHOD OF UNIT FORCE)
AC fUUU ???? 10
?? L xEI xMU d2 )(
2
?? L xEI xMU d2 )(
2
0
0
? ?? LC xEI xMxMU d2 )]()([
2
0
?? LA xEI xMxMf d)()( 0
Determine the displacement f A of an
arbitrary point A,
1,Provement of the theorem,
a
A
Fig
fA
q(x)
Figc
A
0 P =1 q(x)
f
A
Figb
A
=1 P0
§ 11–2 莫尔定理 (单位力法 )
AC fUUU ???? 10
?? L xEI xMU d2 )(
2
?? L xEI xMU d2 )(
2
0
0
? ?? LC xEI xMxMU d2 )]()([
2
0
?? LA xEI xMxMf d)()( 0
求任意点 A的位移 f A 。
一、定理的证明,
a
A

fA
q(x)
图 c
A
0 P =1 q(x)
f
A
图 b
A
=1 P0
Mohr’s theorem(method of unit force)
2,General form of Mohr’s theorem
x
EI
xMxMf
LA
d)()( 0??
?? ??? L
P
nn
LA
x
GI
xMxM
x
EA
xNxN d)()(d)()( 00? x
EI
xMxM
L
d)()( 0?
莫尔定理 (单位力法 )
二、普遍形式的莫尔定理
x
EI
xMxMf
LA
d)()( 0??
?? ??? L
P
nn
LA
x
GI
xMxM
x
EA
xNxN d)()(d)()( 00? x
EI
xMxM
L
d)()( 0?
3,What we must pay attention to as we apply Mohr’s theorem,
④ Coordinate of M0(x) must be coincide with that of M(x),For each segment
the coordinate may be set up freely,
⑤ Mohr’s integrationmust be through the whole structure,
② M0,The internal force of the structure as we act a generalized unit force
along the direction,of the generalized displacement that is to be determined,
where the applied force is taken out,
① M(x),The internal force of the structure acted by original loads,
③ The product of the applied generalized unit force and the generalized
displacement to be determined determined must be of the dimension of work,
三、使用莫尔定理的注意事项,
④ M0(x)与 M(x)的坐标系必须一致,每段杆的坐标系可
自由建立。
⑤ 莫尔积分必须遍及整个结构 。
② M0——去掉主动力,在所求 广义位移 点,沿所求
广义位移 的 方向加 广义单位力 时,结构产生的内力。
① M(x):结构在原载荷下的内力。
③ 所加广义单位力与所求广义位移之积,必须为功的量纲。
Example 3 Determine the displacement and the angle of rotation of point C by
the energy method,
2
)(
2qx
a q xxM ??
?
?
?
??
?
?
???
??
?
)2(; )2(
2
)0(;
2
)(
0
axaxa
x
ax
x
xM
Solution,① Plot the diagram of the structure acted by the unit load
② Determine the internal force
B A
a a C
q
B A
a a C
0 P =1
x
[例 3] 用能量法求 C点的挠度和转角。梁为等截面直梁。
2
)(
2qx
a q xxM ??
?
?
?
??
?
?
???
??
?
)2(; )2(
2
)0(;
2
)(
0
axaxa
x
ax
x
xM
解,① 画单位载荷图
② 求内力
B A
a a C
q
B A
a a C
0 P =1
x
d)()(d)()(
2
0
0
0 ?? ??
a
a
a
C xEI
xMxMx
EI
xMxMf
?
a
x
EI
xMxM
0
0 d)()(2
Symmetry
EI
qaxxqxqax
EI
a
24
5d
2
)
2
(2
4
0
2
??? ?
③ Deformation
B A
a a C
0 P =1 B A
a a C
q
x
( )
d)()(d)()(
2
0
0
0 ?? ??
a
a
a
C xEI
xMxMx
EI
xMxMf
?
a
x
EI
xMxM
0
0 d)()(2
对称性
EI
qaxxqxqax
EI
a
24
5d
2
)
2
(2
4
0
2
??? ?
③ 变形
B A
a a C
0 P =1 B A
a a C
q
x
( )
④ Determine the angle of rotation,Set up the coordinate again (as shown in the figure)
?? ?????
aa
x
a
xqxqax
EI
x
a
xqxqax
EI 0 2
2
2
2
2
0
1
1
2
1
1 d2)2(
1d
2
)
2
(1
2
)(,
2
1
1
qxq a xxMAC ??
a
xxM
2)(
1
0 ??
2
)(,
2
2
2
qxq a xxMBC ??
a
xxM
2)(
2
0 ?
q
B A
a a C
x2 x1
B A
a a
C
MC0=1
d ) ( ) (
) ( ) (
) ( 0
0
) ( 0
0
?
?
?
?
a
BC
a
AB
xEI x M x M
dx EI x M x M
c?
=0
④ 求转角,重建坐标系(如图)
?? ?????
aa
x
a
xqxqax
EI
x
a
xqxqax
EI 0 2
2
2
2
2
0
1
1
2
1
1 d2)2(
1d
2
)
2
(1
2
)(,
2
1
1
qxq a xxMAC ??
a
xxM
2)(
1
0 ??
2
)(,
2
2
2
qxq a xxMBC ??
a
xxM
2)(
2
0 ?
q
B A
a a C
x2 x1
B A
a a
C
MC0=1
d ) ( ) (
) ( ) (
) ( 0
0
) ( 0
0
?
?
?
?
a
BC
a
AB
xEI x M x M
dx EI x M x M
c?
=0
PxxM AB ?)(
xxM AB ?)(0
PxM n