1
Mechanics of Materials
2
3
§ 5–1 FORWORD
§ 5–2 NORMAL STRESS ON THE CROSS SECTION OF THE BEAM IN
PLANAR BENDING
§ 5–3 SHEARING STRESS ON THE CROSS SECTION OF THE BEAM IN PLANAR
BENDING
§ 5–4 STRENGTH CONDITIONS OF THE NORMAL STRESS AND SHEARING
STRESS OF THE BEAM ? REASONABLE SECTION OF THE BEAM
§ 5–5 PLANAR BENDING OF UNSIMMETRIC BEAM?BENDING CENTER OF THE
OPENED THIN-WALLED SECTION
§ 5–6 LIMIT BENDING MOMENT OF A MATERIAL WITH PLASTICITY
CONSIDERED
CHAPTER 5 STRESSES IN BENDING
4
§ 5–1 引言
§ 5–2 平面 弯曲时梁横截面上的正应力
§ 5–3 梁横截面上的剪应力
§ 5–4 梁的正应力和剪应力强度条件 ? 梁的合理截面
§ 5–5 非对称截面梁的平面弯曲 ?开口薄壁截面的弯曲中心
§ 5–6 考虑材料塑性时的极限弯矩
第五章 弯曲应力
§ 5－１ FORWARD
1,(Internal forces) stresses on the cross section of the bending
member
Internal
forces
Shearing force Q Shearing stress t
Bending moment M Normal stress s
§ 5－１ 引言
1、弯曲构件横截面上的（内力）应力
内力
剪力 Q 剪应力 t
弯矩 M 正应力 s
There are only normal stresses in the section under plane bending beam of
pure bending (There is only M and no Q in the cross section),
There are shearing stresses in the section under plane bending beam of
transverse bending,(There are Q and M in the cross section),
2,Study method
Longitudinal plane
of symmetry
P1 P2 Such as,
平面弯曲时横截面 s 纯弯曲梁 (横截面上只有 M而无 Q的情况 )
平面弯曲时横截面 t 剪切弯曲 (横截面上既有 Q又有 M的情况 )
2、研究方法
纵向对称面
P1 P2 例如,
Deformation of some portion of the
beam in which there are only bending
moment and no shearing stress is called
pure bending,Such as portion AB,
P P a a
A B
Q
M
x
x
Pure Bending,
P
P
Pa
某段梁的内力只有弯矩
没有剪力时，该段梁的变
形称为纯弯曲。如 AB段。
P P a a
A B
Q
M
x
x
纯弯曲 (Pure Bending),
P
P
Pa
§ 5－ 2 NORMAL STRESS ON THE CROSS SECTION OF THE BEAM IN
PLANAR BENDING
1.Experiment of pure
bending of the beam
Lateral lines(a b,c d） keep l
straight lines and rotate through some
angles after deformation,longitudinal
straight lines change into curves with
upper fibers constructed and lower
fibers elongated,lateral lines are still
normal to longitudinals lines after
deformation,
（ 1） Geometric law of the
deformation,
1,NORMAL STRESS ON THE
CROSS SECTION OF THE BEAM IN
PLANAR BENDING
Neutral
layer
Longitudinal
plane of
symmetry
b d
a c
a
b
c
d
M M
§ 5－ 2 平面 弯曲时梁横截面上的正应力
1.梁的纯弯曲实验
横向线 (a b,c d）变
形后仍为直线，但有转动；
纵向线变为曲线，且上缩
下伸；横向线与纵向线变
形后仍正交。
（一）变形几何规律,
一,纯弯曲时梁横截面
上的正应力
中性层
纵向对称面
中性轴
b d
a c
a
b
c
d
M M
?There are only normal stresses on cross sections,
?Hypothesis of plane section,The cross sections remain still planes and only
rotate through some angles around their neutral axes after deformation,
（ May be proved by symmetry and the method of infinite division）
?Neutral layer,A layer at a certain height inside the beam in which the
longitudinal fibers are neither to be elongated nor to be shortened and
they are neither subject to tension nor compression.This layer is called
the neutral layer
?Neutral axis,The intersection of the neutral layer with any cross section,
2,Two concepts,
3,Deduction,
?横截面上只有正应力。
?平面假设：横截面变形后仍为平面，只是绕中性轴发生转动，
距中性轴等高处，变形相等。
（可由对称性及无限分割法证明）
3.推论
2.两个概念
?中性层：梁内一层纤维既不伸长也不缩短，因而纤维不
受拉应力和压应力，此层纤维称中性层。
?中性轴：中性层与横截面的交线。
A1 B1 O1 O
4,Geometric equation,
( 1 ),,,,,,
?
? yx ?
a b
c d A B
dq ?
x
y
11111 OOBA
AB
ABBA
x
?????
) ) )
OO1 )
?q?
q?q? yy ??+?
d
dd)(
Neutral axis
Y
x
z
A1 B1 O1 O
4,几何方程,
( 1 ),,,,,,
?
? yx ?
a b
c d A B
dq ?
x
y
11111 OOBA
AB
ABBA
x
?????
) ) )
OO1 )
?q?
q?q? yy ??+?
d
dd)(
（ 2） Physical relation,
Assume there is no extrusion between the longitudinal fibers,therefore，
an arbitrary point in the beam is at an uniaxial stressed state,
( 2 ),,,,,, ??s EyE xx ??
sx sx
（ 3） Static relations,
0ddd ????? ???? ???s zAAAx ESAyEAEyAN
s e c t i o n, t h eofc e n t e r e t h
t h r o u g his a x i s e u t r al )( so 0 nzS z ?
Neutral axis
Y
x
z
（二）物理关系,
假设：纵向纤维互不挤压。于是，任意一点均处于单项应
力状态。
( 2 ),,,,,, ??s EyE xx ??
sx sx
（三）静力学关系,
0ddd ????? ???? ???s zAAAx ESAyEAEyAN
轴过形心中性 )( 0 zS z ??
0dd)d( ????? ???? ???s yz
AAAy
EIAyzEAE y zzAM
（ Symmetric plane）
MEIAyEAEyyAM z
AAAz
????? ???? ???s dd)d( 2
2
z
z
EI
M?
?
1
… …(3) EIz flexural rigidity
of the beam
(4),....,
z
x I
M ? s Neutral axis
Y
x
z
0dd)d( ????? ???? ???s yz
AAAy
EIAyzEAE y zzAM
（对称面）
MEIAyEAEyyAM z
AAAz
????? ???? ???s dd)d( 2
2
z
z
EI
M?
?
1
… …(3) EIz 杆的抗弯刚度。
( 4 ),....,
z
x I
M y?s
D
d
D
d ? a
b
B
)1(6 s e c 3
32
m a x BH
bhBH
y
IWt i o n z
z ?????
)1(32 4
3
m a x
a? ????? Dy IW zz
Circular section
（ 4） Maximum normal stress,
section modulus
of bending。
m a x
y
IW z
z ?
)5.,,,,,,,, ( m a x
zW
M
?s
Neutral axis
Y
x
z
（四）最大正应力,
zW
M?
m a xs
… …(5)
D
d
D
d ? a
)1(32 4
3
m a x
a? ????? Dy IW zz圆环
b
B
)1(6 3
32
ma x BH
bhBH
y
IW z
z ?????回字框
m a xy
I W z
z ? 抗弯截面模量。?
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
z
y
Solution,?Plot the diagram of M
and determine the bending moment of
the section
k N m60)22( 1
2
1 ??? ?x
qxq LxM
30
Example 1 The simply supported beam
subjected to uniformly distributed load is
shown in the figure,Try to determine：
(1） the normal stresses at points 1,2
on the section 1—1；
（ 2） the maximum normal stress on this
section；
(3) knowing maximum normal stress in
the whole beam；
（ 4） as E=200GPa,determine the
radius of curvature of section 1—1,
例 1 受均布载荷作用的简支梁
如图所示，试求,
（ 1） 1——1截面上 1,2两点
的正应力；
（ 2）此截面上的最大正应力；
（ 3）全梁的最大正应力；
（ 4）已知 E=200GPa，求 1—1
截面的曲率半径。
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
z
y
解,?画 M图求截面弯矩
k N m60)22( 1
2
1 ??? ?x
qxq LxM
30
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
z
y
k N m5.678/3608/ 22m ax ???? qLM
4512
33
m108 3 2.51012 1 8 01 2 012 ?? ?????? bhI z
34 m1048.6
2/
???? hIW
zz
M P a7.6110
8 3 2.5
6060
5
1
21
??
??
?
??
zI
yM
ss
?Determine the stress
180
30
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
z
y
k N m5.678/3608/ 22m ax ???? qLM
4512
33
m108 3 2.51012 1 8 01 2 012 ?? ?????? bhI z
34 m1048.6
2/
???? hIW
zz
M P a7.6110
8 3 2.5
6060
5
1
21
??
??
?
??
zI
yM
ss
?求应力
180
30
M P a6.921048.6 60 41m a x1 ????
zW
Ms
m4.1941060 832.5200
1
1 ??
???
M
EI z?
M P a2.1041048.6 5.67 4m a xm a x ????
zW
Ms
?Determine the radius of curvature
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
30
M P a6.921048.6 60 41m a x1 ????
zW
Ms
m4.1941060 832.5200
1
1 ??
???
M
EI z?
M P a2.1041048.6 5.67 4m a xm a x ????
zW
Ms
?求曲率半径
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
30
§ 5－ 3 SHEARING STRESS ON THE CROSS SECTION OF THE BEAM
1,Shearing stresses on the cross section of
the beam with rectangular sections
0)(112 ????? dxbNNX t
Q(x)+d Q(x)
M(x) y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
dx x
Fig,a
Fig,b
Fig,c
1),Two assumptions,
① Shearing stresses are all parallel to the
shearing force；② Shearing stresses are equal
at the same distance to the neutral axis,
2),Researching method;the free body is in
equilibrium,
① Take an infinitesimal segment of the
beam as shown in figure b；
② Take a small block from the
infinitesimal segment as shown in figure c
,the equilibrium equation is
§ 5－ 3 梁横截面上的剪应力
一,矩形截面 梁横截面上的剪应力
1、两点假设,
① 剪应力与剪力平行；
②矩中性轴等距离处，剪应力
相等。
2、研究方法：分离体平衡。
① 在梁上取微段如图 b；
②在微段上取一块如图 c，平衡
0)(112 ????? dxbNNX t
dx x
Q(x)+d Q(x) M(x)
y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
图 a
图 b
图 c
dx x
Q(x)+d Q(x) M(x)
y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
Fig.a
Fig.b
Fig.c
z
z
AzA I
MSAy
I
MAN *** ??? ??
** dd1 s
z
z
I
SMMN *+? )d(
2
z
z
z
z
bI
QS
bI
S
x
M ** ??
d
d
1t
From the theory of the conjugate
shearing stress
zbI
QSy *???
1)( ttt
)
4
(
2
)
2
(
2
2 22 yhbyhb
yh
AyS cz ???
+
?? ***
dx x
Q(x)+d Q(x) M(x)
y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
图 a
图 b
图 c
z
z
AzA I
MSAy
I
MAN *** ??? ??
** dd1 s
z
z
I
SMMN *+? )d(
2
z
z
z
z
bI
QS
bI
S
x
M ** ??
d
d
1t
由剪应力互等
zbI
QSy *???
1)( ttt
)
4
(
2
)
2
(
2
2 22 yhbyhb
yh
AyS cz ???
+
?? ***
tt 5.123m a x ?? AQ
)4(2 2
2
yhIQSo
z
??矩t
Q
Direction of t,The same as that of shearing force in the cross section；
Magnitude of t,Distribution along the width is uniform and distribution along
the height is parabolic,The maximum shearing stress is 1.5
times as much as the mean shearing stress,
2,Shearing stress in the section of the beam with other shapes in section
1) Method to determine the shearing stress is the same as that for the rectangular
section,Calculation formula of the shearing stress is also,
z
z
bI
QS *?
1
t
Where Q is shearing force in the section； Sz* is the static moment of the part
section under the point y about neutral axis,
*
tt 5.123m a x ?? AQ
)4(2 2
2
yhIQ
z
??? 矩t
Q
t方向：与横截面上剪力方向相同；
t大小：沿截面宽度均匀分布，沿高度 h分布为抛物线。
最大剪应力为平均剪应力的 1.5倍。
二、其它截面梁 横截面上的剪应力
1、研究方法与矩形截面同 ;剪应力的计算公式亦为,
z
z
bI
QS *?
1
t
其中 Q为截面剪力； Sz 为 y点以下的面积对中性轴之静矩； *
2)The maximum bending shearing stresses of several sections in common use,
Iz is the moment of inertia of the whole section about axis z； b is the width of
the section at point y,
① Ⅰ -section,
maxt
mint
b,Vertical shearing stresses act mainly in the web（ 95~97%） with tmax
≈tmin,Therefore,for Ⅰ -section steel the maximum shearing
stress is
Conclusion,
Af —Area of the web ; max A
Q t
f
≈
a,Because of the relation tmax in
flanges＜＜ tmax in the web,we
only calculate tmax in the web,
; max A Q t
f
≈
2、几种常见截面的最大弯曲剪应力
Iz为整个截面对 z轴之惯性矩； b 为 y点处截面宽度。
① 工字钢截面,
maxt
mint; ? max A Q t
f
结论,翼缘部分 tmax?腹板上的 tmax,只计算腹板上的 tmax。
铅垂剪应力主要腹板承受（ 95~97%），且 tmax≈ tmin
故工字钢最大剪应力
Af —腹板的面积。 ; ? max A
Q t
f
② Circular section,
tt 3434m a x ?? AQ
③ Thin-walled cirque,
tt 22m a x ?? AQ
④ Channel steel：,
x
y
z
P
0)d(? ? ?? ?A a r mf o r c ex dAM t
R
Hhe ?
Q
e
Q
e
h;
z
z
bI
QS *?t; 21
zI
QA *?t
On the web
On the flange Resultant H
Resultant R,R ≈ Q
② 圆截面,
tt 3434m a x ?? AQ
③ 薄壁圆环,
tt 22m a x ?? AQ
④ 槽钢,
x
y
z
P
QRR
z
z
bI
QS ?*?,合力为腹板上 ; t
。合力为翼缘上 H
zI
QA; 21
*
?t
0)d(? ? ?? Ax dAM 力臂t
R
Hhe ?
Q
e
Q
e
h
§ 5-4 STRENGTH CONDITIONS OF THE NORMAL STRESS AND SHEARING
STRESS OF THE BEAM ? REASONABLE SECTION OF THE BEAM
1),Analysis of the critical plane and the critical point,
① For a general section the maximum normal stress occurs at the upper and
lower edges of the section in which the bending moment is maximum and the
maximum shearing stress occurs at neutral axis of the section in which the
shearing stress is maximum,
Q
t s s
s
M
t
1,Strength conditions of the normal stress and the shearing stress of the beam
§ 5-4 梁的正应力和剪应力强度条件 ? 梁的合理截面
1、危险面与危险点分析,
① 一般截面，最大正应力发生在弯矩绝对值最大的截面的上
下边缘上；最大剪应力发生在剪力绝对值最大的截面的中
性轴处。
Q
t s s
s
M
t
一、梁的正应力和剪应力强度条件
2),Strength conditions of the normal stress and the shearing stress,
? ?tt ?? *
z
z
Ib
SQ m a xm a x
m a x? ?
ss ??
zW
M m a x
m a x
3),Applications of the strength conditions,can make three kinds of
calculations about the strength can be made according to these
conditions,
s
M Q
t t
s
② For the thin-walled section with webs it is the same case about the
maximum normal stress and the maximum shearing stress as case①, There is a
possible critical point that is at the intersection of the web and the flange of the
section in which Q and M are both larger.（ Explain later）
2、正应力和剪应力强度条件,
② 带翼缘的薄壁截面，最大正应力与最大剪应力的情况与上
述相同；还有一个可能危险的点，在 Q和 M均很大的截面
的腹、翼相交处。（以后讲）
? ?tt ?? *
z
z
Ib
SQ m a xm a x
m a x? ?
ss ??
zW
M m a x
m a x
3、强度条件应用,依此强度准则可进行三种强度计算,
s
M Q
t t
s
4),Some special cases about the strength of the shearing stress needs to be
checked,
① If the span of the beam is shorter,M is smaller and Q is larger we must
check the strength of the shearing stress,
③ For the anisotropic material (such as wood) the Capacity is lower,therefore,
we must check the strength of the shearing stress,
、校核强度,① Ckeck the strength,
② Design the dimension of the section,
③ Determine the loads,
][ ];[ m a xm a x ttss ??
][
m a x
s
MW
z ?
)(][ ];[ m a xm a x MfPWM z ?? s
② For the riveted or melted composite section if the ratio of the thickness of the
web and the height is smaller than the corresponding ratio of the hot rolled steel,
we must check the strength of the shearing stress,
4、需要校核剪应力的几种特殊情况,
② 铆接或焊接的组合截面，其腹板的厚度与高度比小于型钢的
相应比值时，要校核剪应力。
① 梁的跨度较短,M 较小，而 Q较大时，要校核剪应力。
③ 各向异性材料（如木材）的抗剪能力较差,要校核剪应力。
、校核强度,① 校核强度,
② 设计截面尺寸,
③ 设计载荷,
][ ];[ m a xm a x ttss ??
][
m a x
s
MW
z ?
)(][ ];[ m a xm a x MfPWM z ?? s
Solution,?Plot the internal-force
diagram and determine the internal force in
the critical section
Example 2 A wooden beam with the section of
rectangular sections (b?h=0.12m?0.18m） is
shown in the figure,[s] =7MPa,[t] =0,9 MPa，
Try to determine the ratio of the maximum normal
stress to the maximum shearing stress,and check
the strength of the beam,
N5 4 0 02 33 6 0 02m a x ???? qLQ
Nm4 0 5 08 33 6 0 08
22
m a x ?
??? qLM
q=3.6kN/m
x
M
+
8
2qL
A B
L=3m
Q
2
qL
2
qL
–
+
x
解,?画内力图求危面内力
例 2 矩形 (b?h=0.12m?0.18m） 截面
木梁如图,[s]=7MPa,[t]=0,9 M
Pa，试求最大 正应力和最大剪应力
之比,并校核梁的强度。
N5 4 0 02 33 6 0 02m a x ???? qLQ
Nm4 0 5 08 33 6 0 08
22
m a x ?
??? qLM
q=3.6kN/m
x
M
+
8
2qL
A B
L=3m
Q
2
qL
2
qL
–
+
x
?Determine the maximum stress and
check the strength
?The ratio of the stresses
7.1632m a x
m a x
m a x ???
h
L
Q
A
W
M
zt
s
q=3.6kN/m
x
M
+
8
2qL
Q
2
qL
2
qL
–
+
x
][7 M P a6, 2 5 M P a
18.012.0
4 0 5 066
22
m a xm a x
m a x
s
s
???
?
?
???
bh
M
W
M
z
][0, 9 M P a0, 3 7 5 M P a
18.012.0
54005.15.1 m a x
m a x
t
t
???
?
???
A
Q
?求最大应力并校核强度
?应力之比
7.1632m a x
m a x
m a x ???
h
L
Q
A
W
M
zt
s
q=3.6kN/m
x
M
+
8
2qL
Q
2
qL
2
qL
–
+
x
][7 M P a6, 2 5 M P a
18.012.0
4 0 5 066
22
m a xm a x
m a x
s
s
???
?
?
???
bh
M
W
M
z
][0, 9 M P a0, 3 7 5 M P a
18.012.0
54005.15.1 m a x
m a x
t
t
???
?
???
A
Q
y1
y2
G
A1
A2
A3
A4
Solution,?Plot the bending moment
diagram and determine the internal force on
the critical plane
Example 3 The cast-iron beam having T-shape
sections is subjected to two concentrated forces as
shown in the figure,[sL]=30MPa,[sy]=60 MPa，
Its center of section is at point C,y1=52mm,
y2=88mm,Iz=763cm4, Try to check the strength
of this beam,And explain how to place the beam
having T-shape section is more reasonable,
???? kN5.10;kN5.2 BA RR
p a r t )u p p e r o n t h en c o m p r e s s i o
p a r tl o w e r i n t h et e n s i o n (k N m5.2?CM
）
（
p a r tl o w e r i n t h en c o m p r e s s i o
p a r t,u p p e r i n t h et e n s i o n k N m4??BM
4
?Plot the distribution of the stress in the
critical plane and determine the critical point,
P1=9kN
1m 1m 1m
P2=4kN
A B C D
x
2.5kNm
-4kNm
M
y1
y2
G
A1
A2
A3
A4
解,?画弯矩图并求危面内力
例 3 T 字形截面的铸铁梁受力如
图，铸铁的 [sL]=30MPa,[sy]=60
MPa，其截面形心位于 C点，
y1=52mm,y2=88mm,
Iz=763cm4,试校核此梁的强度。
并说明 T字梁怎样放置更合理？
???? kN5.10;kN5.2 BA RR
)(k N m5.2 下拉、上压?CM
（上拉、下压）k N m4??BM
4
?画危面应力分布图，找危险点
P1=9kN
1m 1m 1m
P2=4kN
A B C D
x
2.5kNm
-4kNm
M
?Check the strength
M P a2.2810763 885.2 822 ????? ?
z
C
LA I
yMs
M P a2.27107 6 3 524 813 ????? ?
z
B
LA I
yMs
M P a2.46107 6 3 884 824 ????? ?
z
B
yA I
yMs
? ?LL ss ?? 2.28m a x
? ?yy ss ?? 2.46m a x
?It is reasonable to put the large part
in the top place,
y1
y2
G
A1
A2
A3
A4
x
2.5kNm
-4kNm
M
y1
y2
G
A3
A4
⊕
○
?校核强度
M P a2.2810763 885.2 822 ????? ?
z
C
LA I
yMs
M P a2.27107 6 3 524 813 ????? ?
z
B
LA I
yMs
M P a2.46107 6 3 884 824 ????? ?
z
B
yA I
yMs
? ?LL ss ?? 2.28m a x
? ?yy ss ?? 2.46m a x
?T字头在上面合理。
y1
y2
G
A1
A2
x
2.5kNm
-4kNm
M
y1
y2
G
A3
A4
A4
A3
⊕
○
2,Reasonable section of the beam
（ 1） Reasonable ration between the height and the width of the section of
the wooden beam with rectangular section
Li Jie of north-Song dynasty pointed out in the book
of written in the year 1100 that the reasonable ratio
of height and width (h/b) of a rectangular wooden
beam is 1.5,
R
b
h
T·Young pointed out in the book of written in 1807 that the strength
reaches the maximum when the reasonable ratio of height to width
of a rectangular woden beam is h/b= and the rigidity reaches
the maximum when h/b=,
2
3
二、梁的合理截面
（一）矩形木梁的合理高宽比
R
北宋李诫于 1100年著 ?营造法式 ?一书中指出,
矩形木梁的合理高宽比 ( h/b = ) 1.5
英 (T.Young)于 1807年著 ?自然哲学与机械技术讲义 ?一书中指出,
矩形木梁的合理高宽比 为
刚度最大。时强度最大时,3 ;,2 ?? bhbh
b
h
A
Q
3
433.1
mm a x ?? tt 32
3
1
1
DW
z
??
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 tt ?
)2/( ;,4 12
2
1 s DRRaaDA ??? ??
Strength,Normal stress,Shearing stress,
1 If areas of sections are the same,select the section with a larger modulus
in bending,
? ? ss ??
zW
M ? ?tt ??
z
z
bI
QS *
2,Reasonable section of the beam with the other material and other section
shape
z
D1
z a
a
A
Q
3
433.1
mm a x ?? tt
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 tt ?
强度：正应力,剪应力,
1、在面积相等的情况下，选择抗弯模量大的截面
? ? ss ??
zW
M ? ?tt ??
z
z
bI
QS *
（二）其它材料与其它截面形状梁的合理截面
z
D1
z a
a
32
3
1
1
DW
z
??
)2/( ;,4 12
2
1 s DRRaaDA ??? ??
mtt 2m a x ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ?
1
222
1 67.1,4 ])8.0([4 s DDDDDA ??? ??
4/2,24 1121
2
1 s DaaDA ?? ??
1
3
1
2
4 67.1 6
4
6 zz W
abhW ???
mtt 5.1m a x ?
z D
0.8
D
a1
2a
1 z
mtt 2m a x ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ?
1
222
1 67.1,
4
])8.0([
4 DD
DDD ??? 时当 ??
4/2,24 1121
2
1 DaaD ?? ?? 时当
1
3
1
2
4 67.1 6
4
6 zz W
abhW ???
mtt 5.1m a x ?
z D
0.8
D
a1
2a
1 z
)(= 3.2 mm a x
fA
Qtt ?
For theⅠ -section the method to
determine the shearing stresses is
similar to that for the 口 -section,
15 57.4 zz WW ?
122222
21 05.1,6.18.02
4 s Daaa
DA ?????
0.8a2
a2
1.6
a 2 2a 2 z
)(= 3.2 mm a x
fA
Qtt ?
工字形截面与框形截面类似。
15 57.4 zz WW ?
12
2
2
2
2
2
1 05.1,6.18.02
4 Daaa
D ???? 时当 ?
0.8a2
a2
1.6
a 2 2a 2 z
For the materials with different capacities to resist tension and compression like the
cast iron,it is better to use the T-shape section and to make the neutral axis be close
to the side of having a weaker capacity to resist deformations,that is if the capacity
for the material to resist tension is weaker and the upper side of the critical section
is in tension,we must make the neutral axis be close to the upper part,It is shown
in the following figure,
2,Select shapes of the section according to material properties
s
G
z
对于铸铁类抗拉、压能力不同的材料，最好使用 T字形类的截
面，并使中性轴偏于抗变形能力弱的一方，即：若抗拉能力弱，
而梁的危险截面处又上侧受拉，则令中性轴靠近上端。如下图,
2、根据材料特性选择截面形状
s
G
z
（ 3） select the beam with non-constant section as shown in the following figure
,
The best is to select an equal-strength beam,
that is
][)( )()(m a x ss ?? xW xMx
If it is an equal-strength rectangular section,
its height is
][
)(6)(
sb
xMxh ?
At the same time
][)(5.1m a x tt ?? xbh Q
][5.1)( tb
Qxh ??
P
x
（三）采用变截面梁,如下图,
最好是等强度梁，即
][)( )()(m a x ss ?? xW xMx
若为等强度矩形截面，则高为
][
)(6)(
sb
xMxh ?
同时
][)(5.1m a x tt ?? xbh Q
][5.1)( tb
Qxh ??
P
x
§ 5-5 PLANAR BENDING OF THE UNSYMMETRIC BEAM?BENDING
CENTER OF THE OPENED THIN-WALLED SECTION
0dd)d( ????? ?? ?? ???s yzAAAy EIAyzEAE y zzAM
0dd)d( ????? ?? ?? ???s zAAA ESAyEAEyAN
o 0 SI yz ?
The geometric equation and physical equation are not changed,P
x
y
z
O
z so 0?zS (neutral) axis z is through the center of the section,
Axes y and z must be the principal axes of inertia of the section
and the acting lines of external forces must coincide with the
principal axes of inertia of the section
§ 5-5 非对称截面梁的平面弯曲 ? 开口薄壁截面的弯曲中心
轴过形心中性 )( z 0 ??zS
0dd)d( ????? ?? ?? ???s yzAAAy EIAyzEAE y zzAM
0dd)d( ????? ?? ?? ???s zAAA ESAyEAEyAN
外力要与主轴共线。轴必须为截面主惯性轴、,0 zyI yz ??
几何方程与物理方程不变。 P
x
y
z
O
MEIAyEAEyyAM z
AAAz
????? ?? ?? ???s dd)d( 2
2
,0)d(? ? ?? A a r mx dAM t
Use this expression to determine the formula of the normal stress,The method to
study the shearing stress and the formula of the shearing stress are not changed,
Bending center(center of the shearing force)：
Acting point of the lateral force not to make the
rod produce torsion.（ such as the preceding origin
O of the coordinates.）
P
x
y
z
O
Use this expression to determine the distance e
from axis x to the axis of the rod,
MEIAyEAEyyAM z
AAAz
????? ?? ?? ???s dd)d( 2
2
exdAM Ax 轴到杆轴的距离依此确定力臂,0)d(? ? ?? t
依此确定正应力计算公式。 剪应力研究方法与公式形式不变。
弯曲中心 (剪力中心 )：使杆不发生扭转的横向力作用点。
（如前述坐标原点 O） P
x
y
z
O
Channel steel,
P
0)d( a r m f o r c e? ? ?? Ax dAM t RHhe ?
Q
e;
z
z
bI
QS *?t; 21
zI
QA *?t
On the web
On the flange Resultant H
Resultant R,R ≈ Q
x
y
z
P
s
M
?
Conditions for the unsymmetrical beam to produce planar
bending,Acting lines of external forces must lie in the principle plane of
inertia and the neutral axis is the centroid principal axis,Lateral forces must
be through the bending center,
槽钢,
非对称截面梁发生平面弯曲的条件：外力必须作用在主惯性面
内，中性轴为形心主轴，,若是横向力，还必须过弯曲中心。
x
y
z
P
P
s
M
QRR
z
z
bI
QS ?*?,合力为腹板上 ; t
。合力为翼缘上 H
zI
QA; 21
*
?t
0)d(? ? ?? Ax dAM 力臂t RHhe ?
Q
e
z
z
bI
QSτ *?
Determination of the bending center,
??? A a r mC dAM )d(t
(1)For the section with double symmetric axes its
cencroid is the bending center,
(2)For the section of antisymmetry the antisymmetric
center of the section is the bending center,
(3)If the section consists of two narrow and long
rectangles,the intersection of longitudinal lines
of the two rectangles is the bending center,
(4)General method to determine the bending center,
yC eQM ?
C
C
C
Qy
e
C
Determine the shearing stress at any point
Reduce to the cencroids
Determine the distance e from the
bending center to the centroid,
z
z
bI
QSτ *?,求任意一点剪应力
弯曲中心的确定,
??? AC dAM 力臂向形心简化 )d(,t
(1)双对称轴截面，弯心与形心重合。
(2)反对称截面，弯心与反对称中心重合。
(3)若截面由两个狭长矩形组成，
弯心与两矩形长中线交点重合。
(4)求弯心的普遍方法,
yC eQMe ?,求弯心到形心距离
C
C
C
Qy
e
C
ss
ss
§ 5-6 LIMIT BENDING MOMENT OF A MATERIAL
WITH PLASTICITY CONSIDERED
（ 1） Physical relation,sx ss ??
After complete yielding the plane assumption is not applied again.But the
assumption about no bearing between longitudinal fibers is still applied,
s
?
s s
s s
s?? diagram of an idea
elastic plastic material
ss
ss
Diagram of distribution
of the elastic limit
Diagram of distribution of
the plastic limit
ss
ss
§ 5-6 考虑材料塑性时的极限弯矩
（一）物理关系为,
sx ss ??
全面屈服后,平面假设不再成立 ;仍做纵向纤维互不挤压假设。
s
?
s s
s s
理想弹塑性材料的
s??图
ss
ss
弹性极限
分布图
塑性极限
分布图
（ 2） Static relations,
f o r m u l a ) b y t h i s a x i s n eu t r a l t h eofp o s i t i o n t h e( D e t e r m i n e So CS AA ?
0)(dd)(d ???+??? ??? ? CSsA sA sA AAAAAN
SC
ssss
)f o r m u l a b y t h i s a x i s ofp o s i t i o n t h e( D e t e r m i n e So ySS CySy ?
0)(dd)()d( ???+??? ??? ? cySysA sA sAy SSAzAzzAM
LC
ssss
（ 1） Physical relation,
sx ss ??
y
z
x
ss
Mjx
Cross section Distribution of
normal stresses
（二）静力学关系,
)( 依此确定中性轴的位置CS AA ??
0)(dd)(d ???+??? ??? ? CSsA sA sA AAAAAN
SC
ssss
)( 轴的位置依此确定 yCySy SS ??
0)(dd)()d( ???+??? ??? ? cySysA sA sAy SSAzAzzAM
LC
ssss
（一）物理关系为,
sx ss ??
y
z
x
ss
Mjx
横截面图 正应力分布图
??? ? ?+?? nc o m p r e s s i ot e n s i o n dd)d( A SA SAz AyAyyAM sss
jxzzS MSS ??? )( nc o m p r es s i ot e n s i o ns
y
z
x
ss
Mjx
Cross section Distribution of
normal stresses
) ( z z S jx S S M compression tension ? ? s S S W s ?
??? ? ?+?? 压拉 A SA SAz AyAyyAM dd)d( sss
jxzzS MSS ??? )( 压拉s
y
z
x
ss
Mjx
横截面图 正应力分布图 )(
zzSjx SS 压拉 ?? s SSW s?
Example 4 Try to determine the ratio of the elastic limit bending
moment M max to the plastic one Mjx of the beam with rectangular
sections,
Solution,
ss
bhWM ss
6
2
m a x ??
ssszsSjx
bhhbhSWM ssss
44222
2
t e n s i o n ??????
3
2m a x ?
jxM
M
[例 4] 试求矩形截面梁的弹性极限弯矩 M max与塑性极限弯矩 Mjx
之
比。
解,
ss
bhWM ss
6
2
m a x ??
ssszsSjx
bhhbhSWM ssss
44222
2
?????? 拉
3
2m a x ?
jxM
M
81
Chapter 5 Exercises
1,What are the two assumptions when the formula of the normal
stress of the beam in bending is derived?
2,A force couple M is acted on the free end of a cantilever beam with
rectangular sections,The longitudinal normal strain of the top section of
the beam is 0.0008,Try to determine the radius of curvature of the axis of
the beam,
3,A simply supported beam with square sections is subjected to
uniform loads,If [σ]=6[τ],try to determine the ratio L/a when the
maximum normal stress and the maximum shearing stress of the beam
reach their allowable stresses simultaneously,
82
第五章 练习题
一、推导梁弯曲正应力公式时，采用了哪两个
假设？
二、矩形截面悬臂梁在自由端作用力偶 M。已知
梁顶面的纵向正应变为 0.0008，试求梁轴线的曲率半
径。
三、正方形截面简支梁受均布载荷作用，若
[σ]=6[τ]，试求当梁的最大正应力和最大剪应力同时
达到许用应力时，比值 L/a的大小,
83
4,The allowable stress of the wooden beam shown in the figure is
[σ]=10MPa,Try to determine the maximum diameter d of the circular
hole in case that the strength of the beam is satisfied (neglecting stress
concentrations),
84
四、图示木梁的许用应力 [σ ]=10MPa.试求在
保证梁强度的条件下圆孔的最大直径 d(不考虑应
力集中 ),
85
86
Mechanics of Materials
2
3
§ 5–1 FORWORD
§ 5–2 NORMAL STRESS ON THE CROSS SECTION OF THE BEAM IN
PLANAR BENDING
§ 5–3 SHEARING STRESS ON THE CROSS SECTION OF THE BEAM IN PLANAR
BENDING
§ 5–4 STRENGTH CONDITIONS OF THE NORMAL STRESS AND SHEARING
STRESS OF THE BEAM ? REASONABLE SECTION OF THE BEAM
§ 5–5 PLANAR BENDING OF UNSIMMETRIC BEAM?BENDING CENTER OF THE
OPENED THIN-WALLED SECTION
§ 5–6 LIMIT BENDING MOMENT OF A MATERIAL WITH PLASTICITY
CONSIDERED
CHAPTER 5 STRESSES IN BENDING
4
§ 5–1 引言
§ 5–2 平面 弯曲时梁横截面上的正应力
§ 5–3 梁横截面上的剪应力
§ 5–4 梁的正应力和剪应力强度条件 ? 梁的合理截面
§ 5–5 非对称截面梁的平面弯曲 ?开口薄壁截面的弯曲中心
§ 5–6 考虑材料塑性时的极限弯矩
第五章 弯曲应力
§ 5－１ FORWARD
1,(Internal forces) stresses on the cross section of the bending
member
Internal
forces
Shearing force Q Shearing stress t
Bending moment M Normal stress s
§ 5－１ 引言
1、弯曲构件横截面上的（内力）应力
内力
剪力 Q 剪应力 t
弯矩 M 正应力 s
There are only normal stresses in the section under plane bending beam of
pure bending (There is only M and no Q in the cross section),
There are shearing stresses in the section under plane bending beam of
transverse bending,(There are Q and M in the cross section),
2,Study method
Longitudinal plane
of symmetry
P1 P2 Such as,
平面弯曲时横截面 s 纯弯曲梁 (横截面上只有 M而无 Q的情况 )
平面弯曲时横截面 t 剪切弯曲 (横截面上既有 Q又有 M的情况 )
2、研究方法
纵向对称面
P1 P2 例如,
Deformation of some portion of the
beam in which there are only bending
moment and no shearing stress is called
pure bending,Such as portion AB,
P P a a
A B
Q
M
x
x
Pure Bending,
P
P
Pa
某段梁的内力只有弯矩
没有剪力时，该段梁的变
形称为纯弯曲。如 AB段。
P P a a
A B
Q
M
x
x
纯弯曲 (Pure Bending),
P
P
Pa
§ 5－ 2 NORMAL STRESS ON THE CROSS SECTION OF THE BEAM IN
PLANAR BENDING
1.Experiment of pure
bending of the beam
Lateral lines(a b,c d） keep l
straight lines and rotate through some
angles after deformation,longitudinal
straight lines change into curves with
upper fibers constructed and lower
fibers elongated,lateral lines are still
normal to longitudinals lines after
deformation,
（ 1） Geometric law of the
deformation,
1,NORMAL STRESS ON THE
CROSS SECTION OF THE BEAM IN
PLANAR BENDING
Neutral
layer
Longitudinal
plane of
symmetry
b d
a c
a
b
c
d
M M
§ 5－ 2 平面 弯曲时梁横截面上的正应力
1.梁的纯弯曲实验
横向线 (a b,c d）变
形后仍为直线，但有转动；
纵向线变为曲线，且上缩
下伸；横向线与纵向线变
形后仍正交。
（一）变形几何规律,
一,纯弯曲时梁横截面
上的正应力
中性层
纵向对称面
中性轴
b d
a c
a
b
c
d
M M
?There are only normal stresses on cross sections,
?Hypothesis of plane section,The cross sections remain still planes and only
rotate through some angles around their neutral axes after deformation,
（ May be proved by symmetry and the method of infinite division）
?Neutral layer,A layer at a certain height inside the beam in which the
longitudinal fibers are neither to be elongated nor to be shortened and
they are neither subject to tension nor compression.This layer is called
the neutral layer
?Neutral axis,The intersection of the neutral layer with any cross section,
2,Two concepts,
3,Deduction,
?横截面上只有正应力。
?平面假设：横截面变形后仍为平面，只是绕中性轴发生转动，
距中性轴等高处，变形相等。
（可由对称性及无限分割法证明）
3.推论
2.两个概念
?中性层：梁内一层纤维既不伸长也不缩短，因而纤维不
受拉应力和压应力，此层纤维称中性层。
?中性轴：中性层与横截面的交线。
A1 B1 O1 O
4,Geometric equation,
( 1 ),,,,,,
?
? yx ?
a b
c d A B
dq ?
x
y
11111 OOBA
AB
ABBA
x
?????
) ) )
OO1 )
?q?
q?q? yy ??+?
d
dd)(
Neutral axis
Y
x
z
A1 B1 O1 O
4,几何方程,
( 1 ),,,,,,
?
? yx ?
a b
c d A B
dq ?
x
y
11111 OOBA
AB
ABBA
x
?????
) ) )
OO1 )
?q?
q?q? yy ??+?
d
dd)(
（ 2） Physical relation,
Assume there is no extrusion between the longitudinal fibers,therefore，
an arbitrary point in the beam is at an uniaxial stressed state,
( 2 ),,,,,, ??s EyE xx ??
sx sx
（ 3） Static relations,
0ddd ????? ???? ???s zAAAx ESAyEAEyAN
s e c t i o n, t h eofc e n t e r e t h
t h r o u g his a x i s e u t r al )( so 0 nzS z ?
Neutral axis
Y
x
z
（二）物理关系,
假设：纵向纤维互不挤压。于是，任意一点均处于单项应
力状态。
( 2 ),,,,,, ??s EyE xx ??
sx sx
（三）静力学关系,
0ddd ????? ???? ???s zAAAx ESAyEAEyAN
轴过形心中性 )( 0 zS z ??
0dd)d( ????? ???? ???s yz
AAAy
EIAyzEAE y zzAM
（ Symmetric plane）
MEIAyEAEyyAM z
AAAz
????? ???? ???s dd)d( 2
2
z
z
EI
M?
?
1
… …(3) EIz flexural rigidity
of the beam
(4),....,
z
x I
M ? s Neutral axis
Y
x
z
0dd)d( ????? ???? ???s yz
AAAy
EIAyzEAE y zzAM
（对称面）
MEIAyEAEyyAM z
AAAz
????? ???? ???s dd)d( 2
2
z
z
EI
M?
?
1
… …(3) EIz 杆的抗弯刚度。
( 4 ),....,
z
x I
M y?s
D
d
D
d ? a
b
B
)1(6 s e c 3
32
m a x BH
bhBH
y
IWt i o n z
z ?????
)1(32 4
3
m a x
a? ????? Dy IW zz
Circular section
（ 4） Maximum normal stress,
section modulus
of bending。
m a x
y
IW z
z ?
)5.,,,,,,,, ( m a x
zW
M
?s
Neutral axis
Y
x
z
（四）最大正应力,
zW
M?
m a xs
… …(5)
D
d
D
d ? a
)1(32 4
3
m a x
a? ????? Dy IW zz圆环
b
B
)1(6 3
32
ma x BH
bhBH
y
IW z
z ?????回字框
m a xy
I W z
z ? 抗弯截面模量。?
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
z
y
Solution,?Plot the diagram of M
and determine the bending moment of
the section
k N m60)22( 1
2
1 ??? ?x
qxq LxM
30
Example 1 The simply supported beam
subjected to uniformly distributed load is
shown in the figure,Try to determine：
(1） the normal stresses at points 1,2
on the section 1—1；
（ 2） the maximum normal stress on this
section；
(3) knowing maximum normal stress in
the whole beam；
（ 4） as E=200GPa,determine the
radius of curvature of section 1—1,
例 1 受均布载荷作用的简支梁
如图所示，试求,
（ 1） 1——1截面上 1,2两点
的正应力；
（ 2）此截面上的最大正应力；
（ 3）全梁的最大正应力；
（ 4）已知 E=200GPa，求 1—1
截面的曲率半径。
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
z
y
解,?画 M图求截面弯矩
k N m60)22( 1
2
1 ??? ?x
qxq LxM
30
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
z
y
k N m5.678/3608/ 22m ax ???? qLM
4512
33
m108 3 2.51012 1 8 01 2 012 ?? ?????? bhI z
34 m1048.6
2/
???? hIW
zz
M P a7.6110
8 3 2.5
6060
5
1
21
??
??
?
??
zI
yM
ss
?Determine the stress
180
30
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
z
y
k N m5.678/3608/ 22m ax ???? qLM
4512
33
m108 3 2.51012 1 8 01 2 012 ?? ?????? bhI z
34 m1048.6
2/
???? hIW
zz
M P a7.6110
8 3 2.5
6060
5
1
21
??
??
?
??
zI
yM
ss
?求应力
180
30
M P a6.921048.6 60 41m a x1 ????
zW
Ms
m4.1941060 832.5200
1
1 ??
???
M
EI z?
M P a2.1041048.6 5.67 4m a xm a x ????
zW
Ms
?Determine the radius of curvature
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
30
M P a6.921048.6 60 41m a x1 ????
zW
Ms
m4.1941060 832.5200
1
1 ??
???
M
EI z?
M P a2.1041048.6 5.67 4m a xm a x ????
zW
Ms
?求曲率半径
Q=60kN/m
A B
1m 2m
1
1
x
M
+
8
2qL
M1 Mmax
1 2
120
180
30
§ 5－ 3 SHEARING STRESS ON THE CROSS SECTION OF THE BEAM
1,Shearing stresses on the cross section of
the beam with rectangular sections
0)(112 ????? dxbNNX t
Q(x)+d Q(x)
M(x) y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
dx x
Fig,a
Fig,b
Fig,c
1),Two assumptions,
① Shearing stresses are all parallel to the
shearing force；② Shearing stresses are equal
at the same distance to the neutral axis,
2),Researching method;the free body is in
equilibrium,
① Take an infinitesimal segment of the
beam as shown in figure b；
② Take a small block from the
infinitesimal segment as shown in figure c
,the equilibrium equation is
§ 5－ 3 梁横截面上的剪应力
一,矩形截面 梁横截面上的剪应力
1、两点假设,
① 剪应力与剪力平行；
②矩中性轴等距离处，剪应力
相等。
2、研究方法：分离体平衡。
① 在梁上取微段如图 b；
②在微段上取一块如图 c，平衡
0)(112 ????? dxbNNX t
dx x
Q(x)+d Q(x) M(x)
y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
图 a
图 b
图 c
dx x
Q(x)+d Q(x) M(x)
y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
Fig.a
Fig.b
Fig.c
z
z
AzA I
MSAy
I
MAN *** ??? ??
** dd1 s
z
z
I
SMMN *+? )d(
2
z
z
z
z
bI
QS
bI
S
x
M ** ??
d
d
1t
From the theory of the conjugate
shearing stress
zbI
QSy *???
1)( ttt
)
4
(
2
)
2
(
2
2 22 yhbyhb
yh
AyS cz ???
+
?? ***
dx x
Q(x)+d Q(x) M(x)
y
M(x)+d M(x) Q(x) dx
s
x
y
z
s1
t1
t
图 a
图 b
图 c
z
z
AzA I
MSAy
I
MAN *** ??? ??
** dd1 s
z
z
I
SMMN *+? )d(
2
z
z
z
z
bI
QS
bI
S
x
M ** ??
d
d
1t
由剪应力互等
zbI
QSy *???
1)( ttt
)
4
(
2
)
2
(
2
2 22 yhbyhb
yh
AyS cz ???
+
?? ***
tt 5.123m a x ?? AQ
)4(2 2
2
yhIQSo
z
??矩t
Q
Direction of t,The same as that of shearing force in the cross section；
Magnitude of t,Distribution along the width is uniform and distribution along
the height is parabolic,The maximum shearing stress is 1.5
times as much as the mean shearing stress,
2,Shearing stress in the section of the beam with other shapes in section
1) Method to determine the shearing stress is the same as that for the rectangular
section,Calculation formula of the shearing stress is also,
z
z
bI
QS *?
1
t
Where Q is shearing force in the section； Sz* is the static moment of the part
section under the point y about neutral axis,
*
tt 5.123m a x ?? AQ
)4(2 2
2
yhIQ
z
??? 矩t
Q
t方向：与横截面上剪力方向相同；
t大小：沿截面宽度均匀分布，沿高度 h分布为抛物线。
最大剪应力为平均剪应力的 1.5倍。
二、其它截面梁 横截面上的剪应力
1、研究方法与矩形截面同 ;剪应力的计算公式亦为,
z
z
bI
QS *?
1
t
其中 Q为截面剪力； Sz 为 y点以下的面积对中性轴之静矩； *
2)The maximum bending shearing stresses of several sections in common use,
Iz is the moment of inertia of the whole section about axis z； b is the width of
the section at point y,
① Ⅰ -section,
maxt
mint
b,Vertical shearing stresses act mainly in the web（ 95~97%） with tmax
≈tmin,Therefore,for Ⅰ -section steel the maximum shearing
stress is
Conclusion,
Af —Area of the web ; max A
Q t
f
≈
a,Because of the relation tmax in
flanges＜＜ tmax in the web,we
only calculate tmax in the web,
; max A Q t
f
≈
2、几种常见截面的最大弯曲剪应力
Iz为整个截面对 z轴之惯性矩； b 为 y点处截面宽度。
① 工字钢截面,
maxt
mint; ? max A Q t
f
结论,翼缘部分 tmax?腹板上的 tmax,只计算腹板上的 tmax。
铅垂剪应力主要腹板承受（ 95~97%），且 tmax≈ tmin
故工字钢最大剪应力
Af —腹板的面积。 ; ? max A
Q t
f
② Circular section,
tt 3434m a x ?? AQ
③ Thin-walled cirque,
tt 22m a x ?? AQ
④ Channel steel：,
x
y
z
P
0)d(? ? ?? ?A a r mf o r c ex dAM t
R
Hhe ?
Q
e
Q
e
h;
z
z
bI
QS *?t; 21
zI
QA *?t
On the web
On the flange Resultant H
Resultant R,R ≈ Q
② 圆截面,
tt 3434m a x ?? AQ
③ 薄壁圆环,
tt 22m a x ?? AQ
④ 槽钢,
x
y
z
P
QRR
z
z
bI
QS ?*?,合力为腹板上 ; t
。合力为翼缘上 H
zI
QA; 21
*
?t
0)d(? ? ?? Ax dAM 力臂t
R
Hhe ?
Q
e
Q
e
h
§ 5-4 STRENGTH CONDITIONS OF THE NORMAL STRESS AND SHEARING
STRESS OF THE BEAM ? REASONABLE SECTION OF THE BEAM
1),Analysis of the critical plane and the critical point,
① For a general section the maximum normal stress occurs at the upper and
lower edges of the section in which the bending moment is maximum and the
maximum shearing stress occurs at neutral axis of the section in which the
shearing stress is maximum,
Q
t s s
s
M
t
1,Strength conditions of the normal stress and the shearing stress of the beam
§ 5-4 梁的正应力和剪应力强度条件 ? 梁的合理截面
1、危险面与危险点分析,
① 一般截面，最大正应力发生在弯矩绝对值最大的截面的上
下边缘上；最大剪应力发生在剪力绝对值最大的截面的中
性轴处。
Q
t s s
s
M
t
一、梁的正应力和剪应力强度条件
2),Strength conditions of the normal stress and the shearing stress,
? ?tt ?? *
z
z
Ib
SQ m a xm a x
m a x? ?
ss ??
zW
M m a x
m a x
3),Applications of the strength conditions,can make three kinds of
calculations about the strength can be made according to these
conditions,
s
M Q
t t
s
② For the thin-walled section with webs it is the same case about the
maximum normal stress and the maximum shearing stress as case①, There is a
possible critical point that is at the intersection of the web and the flange of the
section in which Q and M are both larger.（ Explain later）
2、正应力和剪应力强度条件,
② 带翼缘的薄壁截面，最大正应力与最大剪应力的情况与上
述相同；还有一个可能危险的点，在 Q和 M均很大的截面
的腹、翼相交处。（以后讲）
? ?tt ?? *
z
z
Ib
SQ m a xm a x
m a x? ?
ss ??
zW
M m a x
m a x
3、强度条件应用,依此强度准则可进行三种强度计算,
s
M Q
t t
s
4),Some special cases about the strength of the shearing stress needs to be
checked,
① If the span of the beam is shorter,M is smaller and Q is larger we must
check the strength of the shearing stress,
③ For the anisotropic material (such as wood) the Capacity is lower,therefore,
we must check the strength of the shearing stress,
、校核强度,① Ckeck the strength,
② Design the dimension of the section,
③ Determine the loads,
][ ];[ m a xm a x ttss ??
][
m a x
s
MW
z ?
)(][ ];[ m a xm a x MfPWM z ?? s
② For the riveted or melted composite section if the ratio of the thickness of the
web and the height is smaller than the corresponding ratio of the hot rolled steel,
we must check the strength of the shearing stress,
4、需要校核剪应力的几种特殊情况,
② 铆接或焊接的组合截面，其腹板的厚度与高度比小于型钢的
相应比值时，要校核剪应力。
① 梁的跨度较短,M 较小，而 Q较大时，要校核剪应力。
③ 各向异性材料（如木材）的抗剪能力较差,要校核剪应力。
、校核强度,① 校核强度,
② 设计截面尺寸,
③ 设计载荷,
][ ];[ m a xm a x ttss ??
][
m a x
s
MW
z ?
)(][ ];[ m a xm a x MfPWM z ?? s
Solution,?Plot the internal-force
diagram and determine the internal force in
the critical section
Example 2 A wooden beam with the section of
rectangular sections (b?h=0.12m?0.18m） is
shown in the figure,[s] =7MPa,[t] =0,9 MPa，
Try to determine the ratio of the maximum normal
stress to the maximum shearing stress,and check
the strength of the beam,
N5 4 0 02 33 6 0 02m a x ???? qLQ
Nm4 0 5 08 33 6 0 08
22
m a x ?
??? qLM
q=3.6kN/m
x
M
+
8
2qL
A B
L=3m
Q
2
qL
2
qL
–
+
x
解,?画内力图求危面内力
例 2 矩形 (b?h=0.12m?0.18m） 截面
木梁如图,[s]=7MPa,[t]=0,9 M
Pa，试求最大 正应力和最大剪应力
之比,并校核梁的强度。
N5 4 0 02 33 6 0 02m a x ???? qLQ
Nm4 0 5 08 33 6 0 08
22
m a x ?
??? qLM
q=3.6kN/m
x
M
+
8
2qL
A B
L=3m
Q
2
qL
2
qL
–
+
x
?Determine the maximum stress and
check the strength
?The ratio of the stresses
7.1632m a x
m a x
m a x ???
h
L
Q
A
W
M
zt
s
q=3.6kN/m
x
M
+
8
2qL
Q
2
qL
2
qL
–
+
x
][7 M P a6, 2 5 M P a
18.012.0
4 0 5 066
22
m a xm a x
m a x
s
s
???
?
?
???
bh
M
W
M
z
][0, 9 M P a0, 3 7 5 M P a
18.012.0
54005.15.1 m a x
m a x
t
t
???
?
???
A
Q
?求最大应力并校核强度
?应力之比
7.1632m a x
m a x
m a x ???
h
L
Q
A
W
M
zt
s
q=3.6kN/m
x
M
+
8
2qL
Q
2
qL
2
qL
–
+
x
][7 M P a6, 2 5 M P a
18.012.0
4 0 5 066
22
m a xm a x
m a x
s
s
???
?
?
???
bh
M
W
M
z
][0, 9 M P a0, 3 7 5 M P a
18.012.0
54005.15.1 m a x
m a x
t
t
???
?
???
A
Q
y1
y2
G
A1
A2
A3
A4
Solution,?Plot the bending moment
diagram and determine the internal force on
the critical plane
Example 3 The cast-iron beam having T-shape
sections is subjected to two concentrated forces as
shown in the figure,[sL]=30MPa,[sy]=60 MPa，
Its center of section is at point C,y1=52mm,
y2=88mm,Iz=763cm4, Try to check the strength
of this beam,And explain how to place the beam
having T-shape section is more reasonable,
???? kN5.10;kN5.2 BA RR
p a r t )u p p e r o n t h en c o m p r e s s i o
p a r tl o w e r i n t h et e n s i o n (k N m5.2?CM
）
（
p a r tl o w e r i n t h en c o m p r e s s i o
p a r t,u p p e r i n t h et e n s i o n k N m4??BM
4
?Plot the distribution of the stress in the
critical plane and determine the critical point,
P1=9kN
1m 1m 1m
P2=4kN
A B C D
x
2.5kNm
-4kNm
M
y1
y2
G
A1
A2
A3
A4
解,?画弯矩图并求危面内力
例 3 T 字形截面的铸铁梁受力如
图，铸铁的 [sL]=30MPa,[sy]=60
MPa，其截面形心位于 C点，
y1=52mm,y2=88mm,
Iz=763cm4,试校核此梁的强度。
并说明 T字梁怎样放置更合理？
???? kN5.10;kN5.2 BA RR
)(k N m5.2 下拉、上压?CM
（上拉、下压）k N m4??BM
4
?画危面应力分布图，找危险点
P1=9kN
1m 1m 1m
P2=4kN
A B C D
x
2.5kNm
-4kNm
M
?Check the strength
M P a2.2810763 885.2 822 ????? ?
z
C
LA I
yMs
M P a2.27107 6 3 524 813 ????? ?
z
B
LA I
yMs
M P a2.46107 6 3 884 824 ????? ?
z
B
yA I
yMs
? ?LL ss ?? 2.28m a x
? ?yy ss ?? 2.46m a x
?It is reasonable to put the large part
in the top place,
y1
y2
G
A1
A2
A3
A4
x
2.5kNm
-4kNm
M
y1
y2
G
A3
A4
⊕
○
?校核强度
M P a2.2810763 885.2 822 ????? ?
z
C
LA I
yMs
M P a2.27107 6 3 524 813 ????? ?
z
B
LA I
yMs
M P a2.46107 6 3 884 824 ????? ?
z
B
yA I
yMs
? ?LL ss ?? 2.28m a x
? ?yy ss ?? 2.46m a x
?T字头在上面合理。
y1
y2
G
A1
A2
x
2.5kNm
-4kNm
M
y1
y2
G
A3
A4
A4
A3
⊕
○
2,Reasonable section of the beam
（ 1） Reasonable ration between the height and the width of the section of
the wooden beam with rectangular section
Li Jie of north-Song dynasty pointed out in the book
of written in the year 1100 that the reasonable ratio
of height and width (h/b) of a rectangular wooden
beam is 1.5,
R
b
h
T·Young pointed out in the book of written in 1807 that the strength
reaches the maximum when the reasonable ratio of height to width
of a rectangular woden beam is h/b= and the rigidity reaches
the maximum when h/b=,
2
3
二、梁的合理截面
（一）矩形木梁的合理高宽比
R
北宋李诫于 1100年著 ?营造法式 ?一书中指出,
矩形木梁的合理高宽比 ( h/b = ) 1.5
英 (T.Young)于 1807年著 ?自然哲学与机械技术讲义 ?一书中指出,
矩形木梁的合理高宽比 为
刚度最大。时强度最大时,3 ;,2 ?? bhbh
b
h
A
Q
3
433.1
mm a x ?? tt 32
3
1
1
DW
z
??
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 tt ?
)2/( ;,4 12
2
1 s DRRaaDA ??? ??
Strength,Normal stress,Shearing stress,
1 If areas of sections are the same,select the section with a larger modulus
in bending,
? ? ss ??
zW
M ? ?tt ??
z
z
bI
QS *
2,Reasonable section of the beam with the other material and other section
shape
z
D1
z a
a
A
Q
3
433.1
mm a x ?? tt
1
32
2 1, 1 8 6
)(
6 zz W
RbhW ??? ?
mm a x 5.1 tt ?
强度：正应力,剪应力,
1、在面积相等的情况下，选择抗弯模量大的截面
? ? ss ??
zW
M ? ?tt ??
z
z
bI
QS *
（二）其它材料与其它截面形状梁的合理截面
z
D1
z a
a
32
3
1
1
DW
z
??
)2/( ;,4 12
2
1 s DRRaaDA ??? ??
mtt 2m a x ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ?
1
222
1 67.1,4 ])8.0([4 s DDDDDA ??? ??
4/2,24 1121
2
1 s DaaDA ?? ??
1
3
1
2
4 67.1 6
4
6 zz W
abhW ???
mtt 5.1m a x ?
z D
0.8
D
a1
2a
1 z
mtt 2m a x ?
1
4
3
3 75.2 )0, 8-(132 zz W
DW ?? ?
1
222
1 67.1,
4
])8.0([
4 DD
DDD ??? 时当 ??
4/2,24 1121
2
1 DaaD ?? ?? 时当
1
3
1
2
4 67.1 6
4
6 zz W
abhW ???
mtt 5.1m a x ?
z D
0.8
D
a1
2a
1 z
)(= 3.2 mm a x
fA
Qtt ?
For theⅠ -section the method to
determine the shearing stresses is
similar to that for the 口 -section,
15 57.4 zz WW ?
122222
21 05.1,6.18.02
4 s Daaa
DA ?????
0.8a2
a2
1.6
a 2 2a 2 z
)(= 3.2 mm a x
fA
Qtt ?
工字形截面与框形截面类似。
15 57.4 zz WW ?
12
2
2
2
2
2
1 05.1,6.18.02
4 Daaa
D ???? 时当 ?
0.8a2
a2
1.6
a 2 2a 2 z
For the materials with different capacities to resist tension and compression like the
cast iron,it is better to use the T-shape section and to make the neutral axis be close
to the side of having a weaker capacity to resist deformations,that is if the capacity
for the material to resist tension is weaker and the upper side of the critical section
is in tension,we must make the neutral axis be close to the upper part,It is shown
in the following figure,
2,Select shapes of the section according to material properties
s
G
z
对于铸铁类抗拉、压能力不同的材料，最好使用 T字形类的截
面，并使中性轴偏于抗变形能力弱的一方，即：若抗拉能力弱，
而梁的危险截面处又上侧受拉，则令中性轴靠近上端。如下图,
2、根据材料特性选择截面形状
s
G
z
（ 3） select the beam with non-constant section as shown in the following figure
,
The best is to select an equal-strength beam,
that is
][)( )()(m a x ss ?? xW xMx
If it is an equal-strength rectangular section,
its height is
][
)(6)(
sb
xMxh ?
At the same time
][)(5.1m a x tt ?? xbh Q
][5.1)( tb
Qxh ??
P
x
（三）采用变截面梁,如下图,
最好是等强度梁，即
][)( )()(m a x ss ?? xW xMx
若为等强度矩形截面，则高为
][
)(6)(
sb
xMxh ?
同时
][)(5.1m a x tt ?? xbh Q
][5.1)( tb
Qxh ??
P
x
§ 5-5 PLANAR BENDING OF THE UNSYMMETRIC BEAM?BENDING
CENTER OF THE OPENED THIN-WALLED SECTION
0dd)d( ????? ?? ?? ???s yzAAAy EIAyzEAE y zzAM
0dd)d( ????? ?? ?? ???s zAAA ESAyEAEyAN
o 0 SI yz ?
The geometric equation and physical equation are not changed,P
x
y
z
O
z so 0?zS (neutral) axis z is through the center of the section,
Axes y and z must be the principal axes of inertia of the section
and the acting lines of external forces must coincide with the
principal axes of inertia of the section
§ 5-5 非对称截面梁的平面弯曲 ? 开口薄壁截面的弯曲中心
轴过形心中性 )( z 0 ??zS
0dd)d( ????? ?? ?? ???s yzAAAy EIAyzEAE y zzAM
0dd)d( ????? ?? ?? ???s zAAA ESAyEAEyAN
外力要与主轴共线。轴必须为截面主惯性轴、,0 zyI yz ??
几何方程与物理方程不变。 P
x
y
z
O
MEIAyEAEyyAM z
AAAz
????? ?? ?? ???s dd)d( 2
2
,0)d(? ? ?? A a r mx dAM t
Use this expression to determine the formula of the normal stress,The method to
study the shearing stress and the formula of the shearing stress are not changed,
Bending center(center of the shearing force)：
Acting point of the lateral force not to make the
rod produce torsion.（ such as the preceding origin
O of the coordinates.）
P
x
y
z
O
Use this expression to determine the distance e
from axis x to the axis of the rod,
MEIAyEAEyyAM z
AAAz
????? ?? ?? ???s dd)d( 2
2
exdAM Ax 轴到杆轴的距离依此确定力臂,0)d(? ? ?? t
依此确定正应力计算公式。 剪应力研究方法与公式形式不变。
弯曲中心 (剪力中心 )：使杆不发生扭转的横向力作用点。
（如前述坐标原点 O） P
x
y
z
O
Channel steel,
P
0)d( a r m f o r c e? ? ?? Ax dAM t RHhe ?
Q
e;
z
z
bI
QS *?t; 21
zI
QA *?t
On the web
On the flange Resultant H
Resultant R,R ≈ Q
x
y
z
P
s
M
?
Conditions for the unsymmetrical beam to produce planar
bending,Acting lines of external forces must lie in the principle plane of
inertia and the neutral axis is the centroid principal axis,Lateral forces must
be through the bending center,
槽钢,
非对称截面梁发生平面弯曲的条件：外力必须作用在主惯性面
内，中性轴为形心主轴，,若是横向力，还必须过弯曲中心。
x
y
z
P
P
s
M
QRR
z
z
bI
QS ?*?,合力为腹板上 ; t
。合力为翼缘上 H
zI
QA; 21
*
?t
0)d(? ? ?? Ax dAM 力臂t RHhe ?
Q
e
z
z
bI
QSτ *?
Determination of the bending center,
??? A a r mC dAM )d(t
(1)For the section with double symmetric axes its
cencroid is the bending center,
(2)For the section of antisymmetry the antisymmetric
center of the section is the bending center,
(3)If the section consists of two narrow and long
rectangles,the intersection of longitudinal lines
of the two rectangles is the bending center,
(4)General method to determine the bending center,
yC eQM ?
C
C
C
Qy
e
C
Determine the shearing stress at any point
Reduce to the cencroids
Determine the distance e from the
bending center to the centroid,
z
z
bI
QSτ *?,求任意一点剪应力
弯曲中心的确定,
??? AC dAM 力臂向形心简化 )d(,t
(1)双对称轴截面，弯心与形心重合。
(2)反对称截面，弯心与反对称中心重合。
(3)若截面由两个狭长矩形组成，
弯心与两矩形长中线交点重合。
(4)求弯心的普遍方法,
yC eQMe ?,求弯心到形心距离
C
C
C
Qy
e
C
ss
ss
§ 5-6 LIMIT BENDING MOMENT OF A MATERIAL
WITH PLASTICITY CONSIDERED
（ 1） Physical relation,sx ss ??
After complete yielding the plane assumption is not applied again.But the
assumption about no bearing between longitudinal fibers is still applied,
s
?
s s
s s
s?? diagram of an idea
elastic plastic material
ss
ss
Diagram of distribution
of the elastic limit
Diagram of distribution of
the plastic limit
ss
ss
§ 5-6 考虑材料塑性时的极限弯矩
（一）物理关系为,
sx ss ??
全面屈服后,平面假设不再成立 ;仍做纵向纤维互不挤压假设。
s
?
s s
s s
理想弹塑性材料的
s??图
ss
ss
弹性极限
分布图
塑性极限
分布图
（ 2） Static relations,
f o r m u l a ) b y t h i s a x i s n eu t r a l t h eofp o s i t i o n t h e( D e t e r m i n e So CS AA ?
0)(dd)(d ???+??? ??? ? CSsA sA sA AAAAAN
SC
ssss
)f o r m u l a b y t h i s a x i s ofp o s i t i o n t h e( D e t e r m i n e So ySS CySy ?
0)(dd)()d( ???+??? ??? ? cySysA sA sAy SSAzAzzAM
LC
ssss
（ 1） Physical relation,
sx ss ??
y
z
x
ss
Mjx
Cross section Distribution of
normal stresses
（二）静力学关系,
)( 依此确定中性轴的位置CS AA ??
0)(dd)(d ???+??? ??? ? CSsA sA sA AAAAAN
SC
ssss
)( 轴的位置依此确定 yCySy SS ??
0)(dd)()d( ???+??? ??? ? cySysA sA sAy SSAzAzzAM
LC
ssss
（一）物理关系为,
sx ss ??
y
z
x
ss
Mjx
横截面图 正应力分布图
??? ? ?+?? nc o m p r e s s i ot e n s i o n dd)d( A SA SAz AyAyyAM sss
jxzzS MSS ??? )( nc o m p r es s i ot e n s i o ns
y
z
x
ss
Mjx
Cross section Distribution of
normal stresses
) ( z z S jx S S M compression tension ? ? s S S W s ?
??? ? ?+?? 压拉 A SA SAz AyAyyAM dd)d( sss
jxzzS MSS ??? )( 压拉s
y
z
x
ss
Mjx
横截面图 正应力分布图 )(
zzSjx SS 压拉 ?? s SSW s?
Example 4 Try to determine the ratio of the elastic limit bending
moment M max to the plastic one Mjx of the beam with rectangular
sections,
Solution,
ss
bhWM ss
6
2
m a x ??
ssszsSjx
bhhbhSWM ssss
44222
2
t e n s i o n ??????
3
2m a x ?
jxM
M
[例 4] 试求矩形截面梁的弹性极限弯矩 M max与塑性极限弯矩 Mjx
之
比。
解,
ss
bhWM ss
6
2
m a x ??
ssszsSjx
bhhbhSWM ssss
44222
2
?????? 拉
3
2m a x ?
jxM
M
81
Chapter 5 Exercises
1,What are the two assumptions when the formula of the normal
stress of the beam in bending is derived?
2,A force couple M is acted on the free end of a cantilever beam with
rectangular sections,The longitudinal normal strain of the top section of
the beam is 0.0008,Try to determine the radius of curvature of the axis of
the beam,
3,A simply supported beam with square sections is subjected to
uniform loads,If [σ]=6[τ],try to determine the ratio L/a when the
maximum normal stress and the maximum shearing stress of the beam
reach their allowable stresses simultaneously,
82
第五章 练习题
一、推导梁弯曲正应力公式时，采用了哪两个
假设？
二、矩形截面悬臂梁在自由端作用力偶 M。已知
梁顶面的纵向正应变为 0.0008，试求梁轴线的曲率半
径。
三、正方形截面简支梁受均布载荷作用，若
[σ]=6[τ]，试求当梁的最大正应力和最大剪应力同时
达到许用应力时，比值 L/a的大小,
83
4,The allowable stress of the wooden beam shown in the figure is
[σ]=10MPa,Try to determine the maximum diameter d of the circular
hole in case that the strength of the beam is satisfied (neglecting stress
concentrations),
84
四、图示木梁的许用应力 [σ ]=10MPa.试求在
保证梁强度的条件下圆孔的最大直径 d(不考虑应
力集中 ),
85
86
