Mechanics of Materials
Tension( Compression) Torsion Plane bending
Internal
force
Stress
Deforma
tion
N
N > 0
x— Axle of the
pole
A
Mn > 0
x— Axle of the
pole
A M
n A M
Q M > 0
Q > 0 x—Parallel to the axle of the
pole
x
s
A
xN )(?s
L
() d
()L
NxLx
E A x?? ?
O
tr
p
n
I
M rrt ?)(
z
x I
My?s
s
t x
y
z
z
y bI
QS ??t
A B
x
GI
M
ABL p
n
AB d???
q u
f
x
q? f′
u? f
EI
xMxf )()( ????
拉 （压） 扭 转 平面弯曲
内
力
应
力
变
形
N
N > 0
x—杆轴
A
Mn > 0
x—杆轴
A M
n A M
Q M > 0
Q > 0 x—平行于杆轴
x
s
A
xN )(?s
L
x
xEA
xNL
L
d
)(
)(d ??
O
tr
p
n
I
M rrt ?)(
z
x I
My?s
s
t x
y
z
z
y bI
QS ??t
A B
x
GI
M
ABL p
n
AB d???
q u
f
x
q? f′
u? f
EI
xMxf )()( ????
Tension( Compression) Torsion Plane bending
Strength
conditions
Rigidity
conditions
Strain
energy
][max ss ?
][
m a x
m i n s
NA ?
][m a x sAN ?
][max tt ?
][
|| m a x
t
n
t
MW ?
m a x []ntMW t?
][max ss ? ][max tt ?
][
m a x
s
MW
z?
][m a x szWM ?
][max qq ?
][m a x qq ?
??
?
??
??
L
f
L
f || m a x
x
EA
xNU
L
d
2
)(2??
x
GI
xMU
L
n d
2
)(2?? x
EI
xMU
L
d
2
)(2??
拉 （压） 扭 转 平面弯曲
强
度
条
件
刚
度
条
件
变
形
能
][max ss ?
][
m a x
m i n s
NA ?
][m a x sAN ?
][max tt ?
][
|| m a x
t
n
t
MW ?
m a x []ntMW t?
][max ss ? ][max tt ?
][
m a x
s
MW
z?
][m a x szWM ?
][max qq ?
][m a x qq ?
??
?
??
??
L
f
L
f || m a x
x
EA
xNU
L
d
2
)(2??
x
GI
xMU
L
n d
2
)(2?? x
EI
xMU
L
d
2
)(2??
Tension and
Compression
Torsion
Plane
Bending
The internal-force calculation
Take the left part of point A as the study object,the internal forces of point A are calculated by the
following formulas,(where, Pi,Pj” are the external forces of the left part of point A),
?? ?? ) () ( jiAn mmM
? ? ? ??? ?? )( )( jAiAA PmPm M
? ? ? ??? ???? jiA PP Q
?? ???? )()( jiA PPN
拉
压
扭
转
平
面
弯
曲
内力计算
以 A点左侧部分为对象,A点的内力由下式计算,
(其中,Pi,Pj”均为 A 点左侧部分的所有外力 )
?? ?? ) () ( jiAn mmM
? ? ? ??? ?? )( )( jAiAA PmPm M
? ? ? ??? ???? jiA PP Q
?? ???? )()( jiA PPN
Relations among the shearing force,bending moment and external forces,
? ? ? ?xq
x
xQ ?
d
d
)(d )(d xQx xM ?
)(d )(d 2
2
xqx xM ?
Applications of symmetry and antisymmetry,
For the symmetric structure under the action of symmetric loads the diagram of its shearing
stress Q is antisymmetric and the diagram of the bending moment M is symmetric,For the
symmetric structure under the action of antisymmetric loads the diagram of its shearing stress Q
is symmetric and the diagram of the bending moment M is antisymmetric,
弯曲剪力、弯矩与外力间的关系
? ? ? ?xq
x
xQ ?
d
d
)(d )(d xQx xM ?
)(d )(d 2
2
xqx xM ?
对称性与反对称性的应用,
对称结构在对称载荷作用下,Q图反对称,M图对称；对称
结构在反对称载荷作用下,Q图对称,M图反对称。
Relations among the shearing force,bending moment and external forces,
Exter
nal
forces
No- external force segment Uniform-load segment Concentrated force Concentrated couple
q=0 q>0 q<0
Charact
eristics
of Q-
diagram
Charact
eristics
of M-
digram
C
P
C
m
Horizontal straight line
x
Q
Q>0
Q
Q<0
x
Inclined straight line
Increasing function
x
Q
x
Q
Decreasing function
x
Q
C
Q1
Q2
Q1–
Q2=P
Sudden change from the
left to the right
x
Q
C
No changed
inclined straight line
x
M
x
M
Decreasing function
Curves
x
M
x
M
Dog-ear from the left to
the right
Sudden change from
the left to the right
Opposite
to M x
M M
x
M1
M2
mMM ?? 21basin- basin
tomb-like
Increasing function
剪力, 弯矩与外力间的关系
外
力
无外力段 均布载荷段 集中力 集中力偶
q=0 q>0 q<0
Q
图
特
征
M
图
特
征
C
P
C
m
水平直线
x
Q
Q>0
Q
Q<0
x
斜直线
增函数
x
Q
x
Q
降函数
x
Q
C
Q1
Q2
Q1–
Q2=P
自左向右突变
x
Q
C
无变化
斜直线
x
M
增函数
x
M
降函数
曲线
x
M
坟状
x
M
盆状
自左向右折角
自左向右突变
与
m
反
x
M
折向与 P反向
M
x
M1
M2
mMM ?? 21
Steps to solve statically indeterminate problems,
① Equilibrium equations
② Geometric equations——compatibility equations of deformation
③ Physical equations——relations between the deformation
④ Complementary equations
⑤ Solve the equilibrium equations and complementary equations
Applications of deformations,
Determine displacements and solve the
statically indeterminate problems,
Applications of the strain energy,
Determine displacements and solve the
dynamic loading problem,
j
h
d
K
?
???
2
11
:b o d y f a l l i n g f r e e ( 1 )
2
( 2 ) H o r i z o n t a l i m p a c t,
j
v
K
d g
?
?
△ j,Static displacement of the
impact body at the falling point
超静定问题的方法步骤,
① 平衡方程
② 几何方程 ——变形协调方程
③ 物理方程 ——变形与力的关系
④ 补充方程
⑤ 解由平衡方程和补充方程组
变形的应用,
求位移和解决超静定问题
变形能的应用,
求位移和解决动载问题
j
h
d
K
?
???
2
11
,( 1 ) 自由落体
j
g
v
d
K
?
?
2
,)2( 水平冲击
△ j:冲击物落点的静位移
M a t e r i a l t e s t i n g
sp se ss
sb
s
a
b
ep
et
ee
st
f
g h
e
s?MPa?
0.05 0.10 0.15 0.20 0.25
450
400
350
300
250
200
150
100
50
0
Typical points in the s ? ? Curve of the low – carbon steel
p
e
材料试验
sp se ss
sb
s
a
b
ep
et
ee
st
f
g h
e
s?MPa?
0.05 0.10 0.15 0.20 0.25
450
400
350
300
250
200
150
100
50
0
低碳钢 s?e曲线上特征点
p
e
? ?1, A l l o w a b l e s t r e s s i s, jx
n
s
s ?
0, 22, i m i t i n g s t r e s s i s, {,,}j x s bL s s s s?
3,Safety coefficient,n
Posson’s ration (or the lateral contraction coefficient
e
?
e
?
?
Three elastic constants
?
t?G
e
s?E
)1(2 ??
? EG
? ?1, jx
n
s
s ?,许 用 应 力
},,2.0{,2 bsjx ssss ?、极限应力
3,安全系数, n
泊松比（或横向变形系数）
e
?
e
?
?
三个弹性常数
?
t?G
e
s?E
)1(2 ??
? EG
n n
（ Resultant ）
（ Resultant ）
P
P
Pc
n n
Q
h ?
b
h
t 1 T
t max
Note,b
Practical calculations of shear and bearing
? ?tt ??
A
Q
? ?c
c
c
c A
P ss ??
Restricted torsion of the rod with rectangular sections
2m a x
m a x,
n
P
M
W h b
pW
t???
3,n
P
P
M I h b
GI
q??? m a x1
?tt ?
n n
（合力）
（合力）
P
P
Pc
n n
Q
h ?
b
h
t 1 T
t max
注意, b
剪切与挤压的实用计算
? ?tt ??
A
Q
? ?c
c
c
c A
P ss ??
矩形截面杆约束扭转
2m a x
m a x,
n
P
M
W h b
pW
t??? 其 中
3,,n
P
P
M I h b
GI
q??? 其 中 m a x1
?tt ?
64
64
3
4
4
3
nR
Gd
K
K
P
Gd
nPR
?
???
Calculations for the cylindric close – coiled helical springs
m a x 3
8
E x a c t v a l u e, ; w h e r e,
4 1 0, 6 1 5; i s t h e s p r i n g i n d e x,
44
DP
k
d
CD
k C
C C d
t
?
?
?
? ? ?
?
Conditions for the non – symmetric beam to produce the planar
bending,
① The external forces must act in the principal planes of inertia ;
② The neutral axis is the principal axes of the centroid;
③ If the external forces are transverse forces,they must be through the center
of bending,
P
x
y
z
O
m a x 3
8A p p r o x i m a t e v a l u e, ( 1 )
2
d D P
Dd
t
?
? ? ?
64
,
64
3
4
4
3
nR
Gd
K
K
P
Gd
nPR
?
???
其中
圆柱形密圈螺旋弹簧的计算
m a x 3
8;
4 1 0, 6 1 5;
44
DP
k
d
CD
k C
C C d
t
?
?
?
? ? ?
?
精 确 值, 其 中,
为 弹 簧 指 数
非对称截面梁发生平面 弯曲的条件
① 外力必须作用在主惯性面内 ;
② 中性轴为形心主轴 ;
③ 若是横向力，还必须过弯曲中心。
P
x
y
z
O
3m a x
8)1
2
(
d
DP
D
d
?
t ??近似值：
Conjugate beam ——relations between the real beam and the imaginary beam
① The direction of the axis x and the origin of coordinates are
identical,
② Having same geometric shapes,
④ Set up the,force” boundary conditions of the imaginary beam according to the,displacement”
boundary conditions of the real beam,
AAAA QEIMEI f ?? q ;
EI
Q
EI
M
f xxxx ?? q ;
⑤ Determine the displacement” of the real beam according to the,internal force” of the imaginary
beam,
a, Fixed end Free end
b, Hinged support Hinged support
c, Middle hinged support Middle hinges
( ) ( ),L e t q x M x??③ According to this equation establish distributed loads on the imaginary
beam,
共轭梁法 ——实梁与虚梁的关系
① x 轴指向及 坐标原点完全相同。 ② 几何形状完全相同。
④ 依实梁的“位移”边界条件，建立虚梁的“力”边界条件。
AAAA QEIMEI f ?? q ;
EI
Q
EI
M
f xxxx ?? q ;
⑤ 依虚梁的“内力”，求实梁的“位移”。
a,固定端 自由端
b,铰支座 铰支座
c,中间铰支座 中间铰链
载荷。依此建立虚梁上的分布令,)()( xMxq ??
③
Example 1 A folding rod is shown in the figure,A bearing is at position A and the rod may
rotate freely in the bearing but can not more up and down,Knowing,E=210GPa,G=0.4E,
[σ]=160MPa,[τ]=80MPa.Try to check the strength of the rod and determine the vertical
displacement of point B,
5
10
20
A
P=60N
B
x 500
C
x1
Solution,The vertical displacement of point B
consists of two parts,the bending of rod BA and
torsion of rod CA and the rotation of section A,
P
A C
L
f
A B
q0
L EI EI
qLf
B 8
4
?
EI
PL
f B
3
3
?
例 1 拐杆如图,A处为一轴承，允许杆在轴承内自由转动，但不能
上下移动，已知,E=210Gpa,G=0.4E,[s]=160MPa,[t]=80MPa
,试校核此杆的强度并求 B点的垂直位移。
5
10
20
A
P=60N
B
x 500
C
x1
解,B点的垂直位移由两部
分组成，即,BA弯曲和 CA杆
扭转,A截面转动而引起。
P
A C
L
f
A B
q0
L EI EI
qLf
B 8
4
?
EI
PL
f B
3
3
?
P=60N
A
B
C ABAABBBB LEIPLfff ????? 3 321
P
ACAB
AB
AB
GI
LPL
L
EI
PL
??
3
3
??
???
??? 3
3
3
10
1052 1 03
123.060
3
4 10202104.0
325.03.0603.0 ?
??
????
?
mm22.8?
A
P
fB1
B
P
A
C
B
MA=PLAB
fB2
P=60N
A
B
C ABAABBBB LEIPLfff ????? 3 321
P
ACAB
AB
AB
GI
LPL
L
EI
PL
??
3
3
??
???
??? 3
3
3
10
1052 1 03
123.060
3
4 10202104.0
325.03.0603.0 ?
??
????
?
mm22.8?
A
P
fB1
B
P
A
B
MA=PLAB
fB2
P
n
W
Mτ m a x
m a x ?
M P a46.11
0, 0 2
1618
3 ??
??
?
m a xm a x
zW
M?s
M P a2160, 0 1005.0 618 2 ????
? ? m a x ss ? The strength is sufficient
P=60N
A
B
C
18 Nm
P
n
W
Mτ m a x
m a x ?
M P a46.11
0, 0 2
1618
3 ??
??
?
m a xm a x
zW
M?s
M P a2160, 0 1005.0 618 2 ????
? ? m a x ss ?
强度不足
P=60N
A
B
C
18 Nm
Solution,The bending moment of the real
beam is shown in the figure,
The supports and loads of the imaginary
beam are also shown in the figure,
22
2
1 PaaPaQ
B ?????
)(48109 2 ?? PaN B
Example 2 Determine the deflection and rotational
angle of point B on the straight beam with equal
sections by the conjugate –beam method ( knowing
AB=2a,BD=CD=0.5a,E,I and P.)
2Pa
8
2Pa
3
4a
NB NC
P A B C D
Pa?
4
Pa
P A B C
D
4
Pa
Pa?
x
M
－
+
+
解,实梁弯矩如图,
虚梁支座及载荷如图,
22
2
1 PaaPaQ
B ?????
)(48109 2 ?? PaN B
例 2 用共轭梁法求下列等截面直
梁 B点的挠度及转角。 (AB=2a，
BD=CD=0.5a,E,I,P均已知 )
P A B C D
Pa?
4
Pa
4
Pa
Pa?
x
M
P A B C
D
2Pa
8
2Pa
3
4a
NB NC
－
+
+
222
48
61
48
1 0 9 PaPaPaQ
B ?????
32
3
4
3
4 PaaPaM
B ???
EI
Pa
EI
Mf B
B 3
4 3??
EI
Pa
EI
Q B
B 48
61 2??? ?
?q
EI
Pa
EI
Q B
B
2
?? ??q
Determine the displacement of the real beam,
P A B C D
Pa?
4
Pa
2Pa
8
2Pa
3
4a
NB NC
222
48
61
48
1 0 9 PaPaPaQ
B ?????
32
3
4
3
4 PaaPaM
B ???
EI
Pa
EI
Mf B
B 3
4 3??
EI
Pa
EI
Q B
B 48
61 2??? ?
?q
EI
Pa
EI
Q B
B
2
?? ??q
求实梁位移
P A B C D
Pa?
4
Pa
2Pa
8
2Pa
3
4a
NB NC
Example 3 The structure is shown in the figure, E=210Gpa,ss=240MPa,LBC=1m，
ABC=1cm2,AB is a beam with rectangular sections,b=10cm,h=30cm,L=2m,q0=20 kN/m,
Determine the safety coefficient of the structure,
Solution:The beam is statically indeterminate
to the first degree,
BCBNBqB BCfff ????
BC
BCBCBC
EA
LN
EI
LN
EI
Lq ??
38
34
0
kN14.8?BCN
EI
q0
L A
B EI
q0
L NBC A
B
C
例 3 结构如图,E=210Gpa,ss=240MPa,LBC=1m,ABC=1cm2,
AB为矩形截面梁,b=10cm,h=30cm,L=2m,q0=20 kN/m,求结构的安
全系数。
解,一次静不定梁,
BCBNBqB BCfff ????
BC
BCBCBC
EA
LN
EI
LN
EI
Lq ??
38
34
0
kN14.8?BCN
q0
L A
B EI
q0
L NBC A
B
EI
C
?? m a xm a x
z
AB W
M
s
M P a8.15
0, 30, 1
62 3 7 2 0
2 ??
?
M P a4.81
10
8 1 4 0
4
??
?
?
BC
BC
BC
A
N
s
94.2
4.81
240
m a x
???
s
s sn
The bending moment is shown in the figure,q0
L NBC A
B
EI
–23.72kN·m
1.64kN·m
x
M
?? m a xm a x
z
AB W
M
s
M P a8.15
0, 30, 1
62 3 7 2 0
2 ??
?
M P a4.81
10
8 1 4 0
4
??
?
?
BC
BC
BC
A
N
s
94.2
4.81
240
m a x
???
s
s sn
弯矩如图, q0
L NBC A
B
EI
–23.72kN·m
1.64kN·m
x
M
y1 zC
yC
y2
Example 4 The beam and its section are shown in the
figure, y2=2y1,IZC,q and L are known,[sy]=3[sL]、
Try to determine the reasonable length of a ； If y2=4y1,
how much is the reasonable length of a?
Solution,The bending moment is shown in the figure,
2
2
1
qaM ? )
4(2
2
2
2 a
LqM ??
It is reasonable for the stresses of the critical sections to
reach the limit state simultaneously,
? ?2 y 3 L1, 5yLIf s s s s??? ? ???
a
q
a L
A B
D1
x
D2
D3
M
x M1
M2
y1 zC
yC
y2
例 4 梁及截面如图,y2=2y1,IZC,q
,L均已知,[sy]=3[sL]、试确定 a的
合理长度； 如果 y2=4y1,a的合理长
度又是多少？
解,弯矩如图,
2
2
1
qaM ?
)4(2 2
2
2 a
LqM ??
危险面的应力同时达到极限状态合理。
? ? ? ?L3y2 5.1 ssss ??? Ly若
a
q
a L
A B
D1
x
D2
D3
M
x M1
M2
y1 zC
yC
y2 a
q
a L
A B
D1
x
D2
D3
? ?L211 ss ??
z
L I
yM
? ?213L L
z
My
I
ss??
If y2=4y1,how much is the reasonable length of a?
,it is reasonable,
6
3 La ?When
So the proper conditions should be,
M
x M1
M2
y1 zC
yC
y2 a
q
a L
A B
D1
x
D2
D3
? ?L211 ss ??
z
L I
yM
? ?213L L
z
My
I
ss??
时，合理。
6
3 La ?
如果 y2=4y1,a的合理长度又是多少？
:合理条件应为?
M
x M1
M2
? ?L211 ss ??
z
L I
yM
? ?y222 ss ??
z
y I
yM
2
2
1
qaM ? )
4(2
2
2
2 a
LqM ??
? ?3 L 2 y1, 3 3LyIf s s s s??? ? ? ??
S o t h e p r o p e r
c o n d i t i o n s s h o u l d b e
?
?
?
D1
x
D2
D3 4
L a? It is reasonable,When
M
x M1
M2
? ?L211 ss ??
z
L I
yM
4
L a? 时，合理。
? ?y222 ss ??
z
y I
yM
2
2
1
qaM ? )
4(2
2
2
2 a
LqM ??
? ? ? ?y2L3 33.1 ssss ??? yL若
?
?
?
?,合理条件应为
D1
x
D2
D3
M
x M1
M2
Example 5 Use the conjugate beam method
to determine the deflections of points A and D
and rotational angles of points A and B of the
straight beam with equal section,
Solution,The bending moment of the real
beam obtained from the superposition
method is shown in the figure,
The supports and loads of the imaginary
beam are also shown in the figure,
P=qL/2
A B C
D
L/2 L/2 L/2
A
B
C
D
qL2/8
qL2/8 M
x
qL2/4
qL2/4
例 5 用共轭梁法求下列等截面直
梁 A,D点的挠度及 A,B点的转
角。
qL2/8
解,用叠加法求实梁弯矩如图,
虚梁支座及载荷如图,
P=qL/2
A B C
D
L/2 L/2 L/2
M
x
A
B
C
D
qL2/4
qL2/4
qL2/8
A
B
C
D
qL2/4
qL2/8
2416
33 qLqL
RQ AB ?????
3
48
5 qLR
A ??
4
24
1 qLM
A ?
Determine the reactions of the
imaginary beam and the internal
forces of the appointed points,
A
B
C
D 0
qL3/12
qL3/8 qL3/16
AR
AM
A
B
C
D
qL2/4
qL2/8
2416
33 qLqL
RQ AB ?????
3
48
5 qLR
A ??
4
24
1 qLM
A ?
求虚梁支反力和指定点内力
A
B
C
D 0
qL3/12
qL3/8 qL3/16
AR
AM
6822
1 2 LqLLM
D ?
EI
Mf i
i ?
EI
Q i
i ?q
38416
3
823
2 42 qLLqLL ???
Determine the displacements of the real
beam,
qL2/4
A
B
C
qL2/8
QD
D
qL2/8
MD
L/2
6822
1 2 LqLLM
D ?
EI
Mf i
i ?
EI
Q i
i ?q
38416
3
823
2 42 qLLqLL ???
求实梁位移
qL2/4
A
B
C
qL2/8
QD
D
qL2/8
MD
L/2
Example 6 Determine the deflections of points A,D
and E (the mid – point of BC ) of the straight beam with
equal sections by the superposition method,
EI
qaf
A 8
4
1 ?
Solution,The structure and the load are decomposed
like the figure,
EI
qa
a
EI
a
qa
f A
33
2
2
4
2
2 ?
?
?
EI
qaf
E 8
4
2 ??
EI
qaff A
D 62
4
2
2 ??
q=P/a
A
B
C D
a E a 2a
A B
q
P
A
B
C D
P qa2/2
例 6 用叠加法求下列等截面直梁 A
,D,E（ BC之中点）点的挠度。
EI
qaf
A 8
4
1 ?
解,结构和载荷分解如图。
EI
qa
a
EI
a
qa
f A
33
2
2
4
2
2 ?
?
?
EI
qaf
E 8
4
2 ??
EI
qaff A
D 62
4
2
2 ??
q=P/a
A
B
C D
a E a 2a
A B
q
P
A
B
C D
P qa2/2
4
3 3D
qaf
EI
?
EI
qaf
A 3
4
4 ?
EI
qaf
E 4
4
4 ??
EI
qaff
AD 3
22 4
44 ??
EI
qaff
i
AiA 24
19 44
1
?? ?
? EI
qaff
i
EiE 8
3 44
1
??? ?
?
EI
qaff
i
DiD 6
7 44
1
?? ?
?
C D
P
C
B A D P Pa
4
3 3D
qaf
EI
?
EI
qaf
A 3
4
4 ?
EI
qaf
E 4
4
4 ??
EI
qaff
AD 3
22 4
44 ??
EI
qaff
i
AiA 24
19 44
1
?? ?
? EI
qaff
i
EiE 8
3 44
1
??? ?
?
EI
qaff
i
DiD 6
7 44
1
?? ?
?
C D
P
C
B A D P Pa
Tension( Compression) Torsion Plane bending
Internal
force
Stress
Deforma
tion
N
N > 0
x— Axle of the
pole
A
Mn > 0
x— Axle of the
pole
A M
n A M
Q M > 0
Q > 0 x—Parallel to the axle of the
pole
x
s
A
xN )(?s
L
() d
()L
NxLx
E A x?? ?
O
tr
p
n
I
M rrt ?)(
z
x I
My?s
s
t x
y
z
z
y bI
QS ??t
A B
x
GI
M
ABL p
n
AB d???
q u
f
x
q? f′
u? f
EI
xMxf )()( ????
拉 （压） 扭 转 平面弯曲
内
力
应
力
变
形
N
N > 0
x—杆轴
A
Mn > 0
x—杆轴
A M
n A M
Q M > 0
Q > 0 x—平行于杆轴
x
s
A
xN )(?s
L
x
xEA
xNL
L
d
)(
)(d ??
O
tr
p
n
I
M rrt ?)(
z
x I
My?s
s
t x
y
z
z
y bI
QS ??t
A B
x
GI
M
ABL p
n
AB d???
q u
f
x
q? f′
u? f
EI
xMxf )()( ????
Tension( Compression) Torsion Plane bending
Strength
conditions
Rigidity
conditions
Strain
energy
][max ss ?
][
m a x
m i n s
NA ?
][m a x sAN ?
][max tt ?
][
|| m a x
t
n
t
MW ?
m a x []ntMW t?
][max ss ? ][max tt ?
][
m a x
s
MW
z?
][m a x szWM ?
][max qq ?
][m a x qq ?
??
?
??
??
L
f
L
f || m a x
x
EA
xNU
L
d
2
)(2??
x
GI
xMU
L
n d
2
)(2?? x
EI
xMU
L
d
2
)(2??
拉 （压） 扭 转 平面弯曲
强
度
条
件
刚
度
条
件
变
形
能
][max ss ?
][
m a x
m i n s
NA ?
][m a x sAN ?
][max tt ?
][
|| m a x
t
n
t
MW ?
m a x []ntMW t?
][max ss ? ][max tt ?
][
m a x
s
MW
z?
][m a x szWM ?
][max qq ?
][m a x qq ?
??
?
??
??
L
f
L
f || m a x
x
EA
xNU
L
d
2
)(2??
x
GI
xMU
L
n d
2
)(2?? x
EI
xMU
L
d
2
)(2??
Tension and
Compression
Torsion
Plane
Bending
The internal-force calculation
Take the left part of point A as the study object,the internal forces of point A are calculated by the
following formulas,(where, Pi,Pj” are the external forces of the left part of point A),
?? ?? ) () ( jiAn mmM
? ? ? ??? ?? )( )( jAiAA PmPm M
? ? ? ??? ???? jiA PP Q
?? ???? )()( jiA PPN
拉
压
扭
转
平
面
弯
曲
内力计算
以 A点左侧部分为对象,A点的内力由下式计算,
(其中,Pi,Pj”均为 A 点左侧部分的所有外力 )
?? ?? ) () ( jiAn mmM
? ? ? ??? ?? )( )( jAiAA PmPm M
? ? ? ??? ???? jiA PP Q
?? ???? )()( jiA PPN
Relations among the shearing force,bending moment and external forces,
? ? ? ?xq
x
xQ ?
d
d
)(d )(d xQx xM ?
)(d )(d 2
2
xqx xM ?
Applications of symmetry and antisymmetry,
For the symmetric structure under the action of symmetric loads the diagram of its shearing
stress Q is antisymmetric and the diagram of the bending moment M is symmetric,For the
symmetric structure under the action of antisymmetric loads the diagram of its shearing stress Q
is symmetric and the diagram of the bending moment M is antisymmetric,
弯曲剪力、弯矩与外力间的关系
? ? ? ?xq
x
xQ ?
d
d
)(d )(d xQx xM ?
)(d )(d 2
2
xqx xM ?
对称性与反对称性的应用,
对称结构在对称载荷作用下,Q图反对称,M图对称；对称
结构在反对称载荷作用下,Q图对称,M图反对称。
Relations among the shearing force,bending moment and external forces,
Exter
nal
forces
No- external force segment Uniform-load segment Concentrated force Concentrated couple
q=0 q>0 q<0
Charact
eristics
of Q-
diagram
Charact
eristics
of M-
digram
C
P
C
m
Horizontal straight line
x
Q
Q>0
Q
Q<0
x
Inclined straight line
Increasing function
x
Q
x
Q
Decreasing function
x
Q
C
Q1
Q2
Q1–
Q2=P
Sudden change from the
left to the right
x
Q
C
No changed
inclined straight line
x
M
x
M
Decreasing function
Curves
x
M
x
M
Dog-ear from the left to
the right
Sudden change from
the left to the right
Opposite
to M x
M M
x
M1
M2
mMM ?? 21basin- basin
tomb-like
Increasing function
剪力, 弯矩与外力间的关系
外
力
无外力段 均布载荷段 集中力 集中力偶
q=0 q>0 q<0
Q
图
特
征
M
图
特
征
C
P
C
m
水平直线
x
Q
Q>0
Q
Q<0
x
斜直线
增函数
x
Q
x
Q
降函数
x
Q
C
Q1
Q2
Q1–
Q2=P
自左向右突变
x
Q
C
无变化
斜直线
x
M
增函数
x
M
降函数
曲线
x
M
坟状
x
M
盆状
自左向右折角
自左向右突变
与
m
反
x
M
折向与 P反向
M
x
M1
M2
mMM ?? 21
Steps to solve statically indeterminate problems,
① Equilibrium equations
② Geometric equations——compatibility equations of deformation
③ Physical equations——relations between the deformation
④ Complementary equations
⑤ Solve the equilibrium equations and complementary equations
Applications of deformations,
Determine displacements and solve the
statically indeterminate problems,
Applications of the strain energy,
Determine displacements and solve the
dynamic loading problem,
j
h
d
K
?
???
2
11
:b o d y f a l l i n g f r e e ( 1 )
2
( 2 ) H o r i z o n t a l i m p a c t,
j
v
K
d g
?
?
△ j,Static displacement of the
impact body at the falling point
超静定问题的方法步骤,
① 平衡方程
② 几何方程 ——变形协调方程
③ 物理方程 ——变形与力的关系
④ 补充方程
⑤ 解由平衡方程和补充方程组
变形的应用,
求位移和解决超静定问题
变形能的应用,
求位移和解决动载问题
j
h
d
K
?
???
2
11
,( 1 ) 自由落体
j
g
v
d
K
?
?
2
,)2( 水平冲击
△ j:冲击物落点的静位移
M a t e r i a l t e s t i n g
sp se ss
sb
s
a
b
ep
et
ee
st
f
g h
e
s?MPa?
0.05 0.10 0.15 0.20 0.25
450
400
350
300
250
200
150
100
50
0
Typical points in the s ? ? Curve of the low – carbon steel
p
e
材料试验
sp se ss
sb
s
a
b
ep
et
ee
st
f
g h
e
s?MPa?
0.05 0.10 0.15 0.20 0.25
450
400
350
300
250
200
150
100
50
0
低碳钢 s?e曲线上特征点
p
e
? ?1, A l l o w a b l e s t r e s s i s, jx
n
s
s ?
0, 22, i m i t i n g s t r e s s i s, {,,}j x s bL s s s s?
3,Safety coefficient,n
Posson’s ration (or the lateral contraction coefficient
e
?
e
?
?
Three elastic constants
?
t?G
e
s?E
)1(2 ??
? EG
? ?1, jx
n
s
s ?,许 用 应 力
},,2.0{,2 bsjx ssss ?、极限应力
3,安全系数, n
泊松比（或横向变形系数）
e
?
e
?
?
三个弹性常数
?
t?G
e
s?E
)1(2 ??
? EG
n n
（ Resultant ）
（ Resultant ）
P
P
Pc
n n
Q
h ?
b
h
t 1 T
t max
Note,b
Practical calculations of shear and bearing
? ?tt ??
A
Q
? ?c
c
c
c A
P ss ??
Restricted torsion of the rod with rectangular sections
2m a x
m a x,
n
P
M
W h b
pW
t???
3,n
P
P
M I h b
GI
q??? m a x1
?tt ?
n n
（合力）
（合力）
P
P
Pc
n n
Q
h ?
b
h
t 1 T
t max
注意, b
剪切与挤压的实用计算
? ?tt ??
A
Q
? ?c
c
c
c A
P ss ??
矩形截面杆约束扭转
2m a x
m a x,
n
P
M
W h b
pW
t??? 其 中
3,,n
P
P
M I h b
GI
q??? 其 中 m a x1
?tt ?
64
64
3
4
4
3
nR
Gd
K
K
P
Gd
nPR
?
???
Calculations for the cylindric close – coiled helical springs
m a x 3
8
E x a c t v a l u e, ; w h e r e,
4 1 0, 6 1 5; i s t h e s p r i n g i n d e x,
44
DP
k
d
CD
k C
C C d
t
?
?
?
? ? ?
?
Conditions for the non – symmetric beam to produce the planar
bending,
① The external forces must act in the principal planes of inertia ;
② The neutral axis is the principal axes of the centroid;
③ If the external forces are transverse forces,they must be through the center
of bending,
P
x
y
z
O
m a x 3
8A p p r o x i m a t e v a l u e, ( 1 )
2
d D P
Dd
t
?
? ? ?
64
,
64
3
4
4
3
nR
Gd
K
K
P
Gd
nPR
?
???
其中
圆柱形密圈螺旋弹簧的计算
m a x 3
8;
4 1 0, 6 1 5;
44
DP
k
d
CD
k C
C C d
t
?
?
?
? ? ?
?
精 确 值, 其 中,
为 弹 簧 指 数
非对称截面梁发生平面 弯曲的条件
① 外力必须作用在主惯性面内 ;
② 中性轴为形心主轴 ;
③ 若是横向力，还必须过弯曲中心。
P
x
y
z
O
3m a x
8)1
2
(
d
DP
D
d
?
t ??近似值：
Conjugate beam ——relations between the real beam and the imaginary beam
① The direction of the axis x and the origin of coordinates are
identical,
② Having same geometric shapes,
④ Set up the,force” boundary conditions of the imaginary beam according to the,displacement”
boundary conditions of the real beam,
AAAA QEIMEI f ?? q ;
EI
Q
EI
M
f xxxx ?? q ;
⑤ Determine the displacement” of the real beam according to the,internal force” of the imaginary
beam,
a, Fixed end Free end
b, Hinged support Hinged support
c, Middle hinged support Middle hinges
( ) ( ),L e t q x M x??③ According to this equation establish distributed loads on the imaginary
beam,
共轭梁法 ——实梁与虚梁的关系
① x 轴指向及 坐标原点完全相同。 ② 几何形状完全相同。
④ 依实梁的“位移”边界条件，建立虚梁的“力”边界条件。
AAAA QEIMEI f ?? q ;
EI
Q
EI
M
f xxxx ?? q ;
⑤ 依虚梁的“内力”，求实梁的“位移”。
a,固定端 自由端
b,铰支座 铰支座
c,中间铰支座 中间铰链
载荷。依此建立虚梁上的分布令,)()( xMxq ??
③
Example 1 A folding rod is shown in the figure,A bearing is at position A and the rod may
rotate freely in the bearing but can not more up and down,Knowing,E=210GPa,G=0.4E,
[σ]=160MPa,[τ]=80MPa.Try to check the strength of the rod and determine the vertical
displacement of point B,
5
10
20
A
P=60N
B
x 500
C
x1
Solution,The vertical displacement of point B
consists of two parts,the bending of rod BA and
torsion of rod CA and the rotation of section A,
P
A C
L
f
A B
q0
L EI EI
qLf
B 8
4
?
EI
PL
f B
3
3
?
例 1 拐杆如图,A处为一轴承，允许杆在轴承内自由转动，但不能
上下移动，已知,E=210Gpa,G=0.4E,[s]=160MPa,[t]=80MPa
,试校核此杆的强度并求 B点的垂直位移。
5
10
20
A
P=60N
B
x 500
C
x1
解,B点的垂直位移由两部
分组成，即,BA弯曲和 CA杆
扭转,A截面转动而引起。
P
A C
L
f
A B
q0
L EI EI
qLf
B 8
4
?
EI
PL
f B
3
3
?
P=60N
A
B
C ABAABBBB LEIPLfff ????? 3 321
P
ACAB
AB
AB
GI
LPL
L
EI
PL
??
3
3
??
???
??? 3
3
3
10
1052 1 03
123.060
3
4 10202104.0
325.03.0603.0 ?
??
????
?
mm22.8?
A
P
fB1
B
P
A
C
B
MA=PLAB
fB2
P=60N
A
B
C ABAABBBB LEIPLfff ????? 3 321
P
ACAB
AB
AB
GI
LPL
L
EI
PL
??
3
3
??
???
??? 3
3
3
10
1052 1 03
123.060
3
4 10202104.0
325.03.0603.0 ?
??
????
?
mm22.8?
A
P
fB1
B
P
A
B
MA=PLAB
fB2
P
n
W
Mτ m a x
m a x ?
M P a46.11
0, 0 2
1618
3 ??
??
?
m a xm a x
zW
M?s
M P a2160, 0 1005.0 618 2 ????
? ? m a x ss ? The strength is sufficient
P=60N
A
B
C
18 Nm
P
n
W
Mτ m a x
m a x ?
M P a46.11
0, 0 2
1618
3 ??
??
?
m a xm a x
zW
M?s
M P a2160, 0 1005.0 618 2 ????
? ? m a x ss ?
强度不足
P=60N
A
B
C
18 Nm
Solution,The bending moment of the real
beam is shown in the figure,
The supports and loads of the imaginary
beam are also shown in the figure,
22
2
1 PaaPaQ
B ?????
)(48109 2 ?? PaN B
Example 2 Determine the deflection and rotational
angle of point B on the straight beam with equal
sections by the conjugate –beam method ( knowing
AB=2a,BD=CD=0.5a,E,I and P.)
2Pa
8
2Pa
3
4a
NB NC
P A B C D
Pa?
4
Pa
P A B C
D
4
Pa
Pa?
x
M
－
+
+
解,实梁弯矩如图,
虚梁支座及载荷如图,
22
2
1 PaaPaQ
B ?????
)(48109 2 ?? PaN B
例 2 用共轭梁法求下列等截面直
梁 B点的挠度及转角。 (AB=2a，
BD=CD=0.5a,E,I,P均已知 )
P A B C D
Pa?
4
Pa
4
Pa
Pa?
x
M
P A B C
D
2Pa
8
2Pa
3
4a
NB NC
－
+
+
222
48
61
48
1 0 9 PaPaPaQ
B ?????
32
3
4
3
4 PaaPaM
B ???
EI
Pa
EI
Mf B
B 3
4 3??
EI
Pa
EI
Q B
B 48
61 2??? ?
?q
EI
Pa
EI
Q B
B
2
?? ??q
Determine the displacement of the real beam,
P A B C D
Pa?
4
Pa
2Pa
8
2Pa
3
4a
NB NC
222
48
61
48
1 0 9 PaPaPaQ
B ?????
32
3
4
3
4 PaaPaM
B ???
EI
Pa
EI
Mf B
B 3
4 3??
EI
Pa
EI
Q B
B 48
61 2??? ?
?q
EI
Pa
EI
Q B
B
2
?? ??q
求实梁位移
P A B C D
Pa?
4
Pa
2Pa
8
2Pa
3
4a
NB NC
Example 3 The structure is shown in the figure, E=210Gpa,ss=240MPa,LBC=1m，
ABC=1cm2,AB is a beam with rectangular sections,b=10cm,h=30cm,L=2m,q0=20 kN/m,
Determine the safety coefficient of the structure,
Solution:The beam is statically indeterminate
to the first degree,
BCBNBqB BCfff ????
BC
BCBCBC
EA
LN
EI
LN
EI
Lq ??
38
34
0
kN14.8?BCN
EI
q0
L A
B EI
q0
L NBC A
B
C
例 3 结构如图,E=210Gpa,ss=240MPa,LBC=1m,ABC=1cm2,
AB为矩形截面梁,b=10cm,h=30cm,L=2m,q0=20 kN/m,求结构的安
全系数。
解,一次静不定梁,
BCBNBqB BCfff ????
BC
BCBCBC
EA
LN
EI
LN
EI
Lq ??
38
34
0
kN14.8?BCN
q0
L A
B EI
q0
L NBC A
B
EI
C
?? m a xm a x
z
AB W
M
s
M P a8.15
0, 30, 1
62 3 7 2 0
2 ??
?
M P a4.81
10
8 1 4 0
4
??
?
?
BC
BC
BC
A
N
s
94.2
4.81
240
m a x
???
s
s sn
The bending moment is shown in the figure,q0
L NBC A
B
EI
–23.72kN·m
1.64kN·m
x
M
?? m a xm a x
z
AB W
M
s
M P a8.15
0, 30, 1
62 3 7 2 0
2 ??
?
M P a4.81
10
8 1 4 0
4
??
?
?
BC
BC
BC
A
N
s
94.2
4.81
240
m a x
???
s
s sn
弯矩如图, q0
L NBC A
B
EI
–23.72kN·m
1.64kN·m
x
M
y1 zC
yC
y2
Example 4 The beam and its section are shown in the
figure, y2=2y1,IZC,q and L are known,[sy]=3[sL]、
Try to determine the reasonable length of a ； If y2=4y1,
how much is the reasonable length of a?
Solution,The bending moment is shown in the figure,
2
2
1
qaM ? )
4(2
2
2
2 a
LqM ??
It is reasonable for the stresses of the critical sections to
reach the limit state simultaneously,
? ?2 y 3 L1, 5yLIf s s s s??? ? ???
a
q
a L
A B
D1
x
D2
D3
M
x M1
M2
y1 zC
yC
y2
例 4 梁及截面如图,y2=2y1,IZC,q
,L均已知,[sy]=3[sL]、试确定 a的
合理长度； 如果 y2=4y1,a的合理长
度又是多少？
解,弯矩如图,
2
2
1
qaM ?
)4(2 2
2
2 a
LqM ??
危险面的应力同时达到极限状态合理。
? ? ? ?L3y2 5.1 ssss ??? Ly若
a
q
a L
A B
D1
x
D2
D3
M
x M1
M2
y1 zC
yC
y2 a
q
a L
A B
D1
x
D2
D3
? ?L211 ss ??
z
L I
yM
? ?213L L
z
My
I
ss??
If y2=4y1,how much is the reasonable length of a?
,it is reasonable,
6
3 La ?When
So the proper conditions should be,
M
x M1
M2
y1 zC
yC
y2 a
q
a L
A B
D1
x
D2
D3
? ?L211 ss ??
z
L I
yM
? ?213L L
z
My
I
ss??
时，合理。
6
3 La ?
如果 y2=4y1,a的合理长度又是多少？
:合理条件应为?
M
x M1
M2
? ?L211 ss ??
z
L I
yM
? ?y222 ss ??
z
y I
yM
2
2
1
qaM ? )
4(2
2
2
2 a
LqM ??
? ?3 L 2 y1, 3 3LyIf s s s s??? ? ? ??
S o t h e p r o p e r
c o n d i t i o n s s h o u l d b e
?
?
?
D1
x
D2
D3 4
L a? It is reasonable,When
M
x M1
M2
? ?L211 ss ??
z
L I
yM
4
L a? 时，合理。
? ?y222 ss ??
z
y I
yM
2
2
1
qaM ? )
4(2
2
2
2 a
LqM ??
? ? ? ?y2L3 33.1 ssss ??? yL若
?
?
?
?,合理条件应为
D1
x
D2
D3
M
x M1
M2
Example 5 Use the conjugate beam method
to determine the deflections of points A and D
and rotational angles of points A and B of the
straight beam with equal section,
Solution,The bending moment of the real
beam obtained from the superposition
method is shown in the figure,
The supports and loads of the imaginary
beam are also shown in the figure,
P=qL/2
A B C
D
L/2 L/2 L/2
A
B
C
D
qL2/8
qL2/8 M
x
qL2/4
qL2/4
例 5 用共轭梁法求下列等截面直
梁 A,D点的挠度及 A,B点的转
角。
qL2/8
解,用叠加法求实梁弯矩如图,
虚梁支座及载荷如图,
P=qL/2
A B C
D
L/2 L/2 L/2
M
x
A
B
C
D
qL2/4
qL2/4
qL2/8
A
B
C
D
qL2/4
qL2/8
2416
33 qLqL
RQ AB ?????
3
48
5 qLR
A ??
4
24
1 qLM
A ?
Determine the reactions of the
imaginary beam and the internal
forces of the appointed points,
A
B
C
D 0
qL3/12
qL3/8 qL3/16
AR
AM
A
B
C
D
qL2/4
qL2/8
2416
33 qLqL
RQ AB ?????
3
48
5 qLR
A ??
4
24
1 qLM
A ?
求虚梁支反力和指定点内力
A
B
C
D 0
qL3/12
qL3/8 qL3/16
AR
AM
6822
1 2 LqLLM
D ?
EI
Mf i
i ?
EI
Q i
i ?q
38416
3
823
2 42 qLLqLL ???
Determine the displacements of the real
beam,
qL2/4
A
B
C
qL2/8
QD
D
qL2/8
MD
L/2
6822
1 2 LqLLM
D ?
EI
Mf i
i ?
EI
Q i
i ?q
38416
3
823
2 42 qLLqLL ???
求实梁位移
qL2/4
A
B
C
qL2/8
QD
D
qL2/8
MD
L/2
Example 6 Determine the deflections of points A,D
and E (the mid – point of BC ) of the straight beam with
equal sections by the superposition method,
EI
qaf
A 8
4
1 ?
Solution,The structure and the load are decomposed
like the figure,
EI
qa
a
EI
a
qa
f A
33
2
2
4
2
2 ?
?
?
EI
qaf
E 8
4
2 ??
EI
qaff A
D 62
4
2
2 ??
q=P/a
A
B
C D
a E a 2a
A B
q
P
A
B
C D
P qa2/2
例 6 用叠加法求下列等截面直梁 A
,D,E（ BC之中点）点的挠度。
EI
qaf
A 8
4
1 ?
解,结构和载荷分解如图。
EI
qa
a
EI
a
qa
f A
33
2
2
4
2
2 ?
?
?
EI
qaf
E 8
4
2 ??
EI
qaff A
D 62
4
2
2 ??
q=P/a
A
B
C D
a E a 2a
A B
q
P
A
B
C D
P qa2/2
4
3 3D
qaf
EI
?
EI
qaf
A 3
4
4 ?
EI
qaf
E 4
4
4 ??
EI
qaff
AD 3
22 4
44 ??
EI
qaff
i
AiA 24
19 44
1
?? ?
? EI
qaff
i
EiE 8
3 44
1
??? ?
?
EI
qaff
i
DiD 6
7 44
1
?? ?
?
C D
P
C
B A D P Pa
4
3 3D
qaf
EI
?
EI
qaf
A 3
4
4 ?
EI
qaf
E 4
4
4 ??
EI
qaff
AD 3
22 4
44 ??
EI
qaff
i
AiA 24
19 44
1
?? ?
? EI
qaff
i
EiE 8
3 44
1
??? ?
?
EI
qaff
i
DiD 6
7 44
1
?? ?
?
C D
P
C
B A D P Pa
