1
M e c h a n i c s o f M a t e r i a l s
2
3
CHAPTER 10 STABILIZATION OF
COMPRESSIVE COLUMNS
§ 10–1 CONCEPTS OF STABILITY OF COMPRESSED COLUMNS
§ 10–2 EULER’S FORMULA OF THE CRITICAL FORCE OF SLENDER
COMPRESSED COLUMNS
§ 10–3 CRITICAL STRESS OF COMPRESSED COLUMNS AS STRESS
EXCEEDS PROPORTIONAL LIMIT
§ 10-4 STABILITY CHECK AND REASONABLE SECTION OF
COMPRESSED COLUMNS
4
第十章 压杆稳定
§ 10–1 压杆稳定性的概念
§ 10–2 细长压杆临界力的欧拉公式
§ 10–3 超过比例极限时压杆的临界应力
§ 10-4 压杆的稳定校核及其合理截面
5
§ 10–1 CONCEPTS OF STABILITY OF COMPRESSED COLUMNS
The load-carrying
capacity of structure
members,
① Strength
② Rigidity
③ Stability
Some structure members
in engineering have enough
strength and rigidity but
they are unable to work
safely and reliably,
6
§ 10–1 压杆稳定性的概念
构件的承载能力,① 强度
②刚度
③稳定性
工程中有些构
件具有足够的强度、
刚度，却不一定能
安全可靠地工作。
7
P
8
P
9
1,Stable and instable equilibrium,
1),instable equilibrium
10
一、稳定平衡与不稳定平衡,
1,不稳定平衡
11
2),Stable equilibrium
12
2,稳定平衡
13
3),Stable and instable equilibrium
14
3,稳定平衡和不稳定平衡
15
2,Loss of stability and the critical pressure of the column,
1).Ideal compressive columns,the material is absolutely ideal； the axis is absolutely
straight； the compressive force is absolutely along the axis of the column,
2),Stable and instable equilibrium of compressive columns,
Instable
equilibrium
Stable
equilibrium
16
二、压杆失稳与临界压力,
1.理想压杆：材料绝对理想；轴线绝对直；压力绝对沿轴线作用。
2.压杆的稳定平衡与不稳定平衡,
稳
定
平
衡
不
稳
定
平
衡
17
3).loss of stability of compressed
column,
4).Critical pressure of compressed columns
Stable
equilibrium
Instable
equilibrium
Critical state
Critical pressure,Pcr
corresponding
pressure
intermediate state
18
3.压杆失稳,4.压杆的临界压力
稳
定
平
衡
不
稳
定
平
衡
临界状态
临界压力, Pcr
过 度
对 应 的
压力
19
§ 10–2 EULER’S FORMULA OF THE CRITICAL PRESSURE OF
SLENDER COMPRESSED COLUMNS
1,Critical pressure for the column with two hinged ends,
PyyxM ?),(
y
EI
P
EI
My ??????
① bending moment,
② Approximate differential
equation of the deflection curve
,02 ???????? ykyy
EI
Py
EI
Pk ?2
:w h e r e
P
P
x
y
P
M
Suppose the pressure has reached the critical value and the column has been in
tiny bending state as shown in the figure,Start to determine the critical force with
the deflective curve,
P
x L
20
§ 10–2 细长压杆临界力的欧拉公式
一、两端铰支压杆的临界力,
PyyxM ?),(
假定压力已达到临界值，杆已经处于微弯状态，如图,
从挠曲线入手，求临界力。
y
EI
P
EI
My ??????
① 弯矩,
② 挠曲线近似微分方程,
02 ???????? ykyy
EI
Py
EI
Pk ?2
:其中
P
x L
P
x
y
P
M
21
③ Solution of the differential equation,
④ Determine the integral constants,
xBxAy c o ss i n ??
0)()0( ?? Lyy
?
?
?
??
???
0c o ss i n
00
kLBkLA
BA 0
c o s s i n
1 0
?
kLkL
0s in ?kL
EI
P
L
nk ?? ? so
The critical force Pcr is the smallest pressure under tiny bending,therefore
we only take n=1 and the column will bend about the axis with the smallest
moment of inertia,
2
m i n
2
L
EIP
cr
??
That is
22
③ 微分方程的解,
④ 确定积分常数,
xBxAy c o ss i n ??
0)()0( ?? Lyy
?
?
?
??
???
0c o ss i n
00
:
kLBkLA
BA
即 0c o s s i n
1 0
??
kLkL
0s in ?? kL
EI
P
L
nk ??? ?
临界力 Pcr 是微弯下的最小压力,故，只能取 n=1 ；且
杆将绕惯性矩最小的轴弯曲。
2
m i n
2
L
EIP
cr
???
23
2,Application range of the formula,
3,Euler’s formula of the critical pressure for the column with
other end conditions,
1).Ideal compressive columns； 2).In linear elastic range；
3).The ends of the column are supported by hinges,
?—Length coefficient（ or constraint coefficient）
Euler’s formula of the critical pressure for
the compressive column with two hinged ends
General form of Euler’s formula of
the critical pressure
2
m i n
2
L
EIP
cr
??
2
m i n
2
)( L
EI
P cr
?
?
?
24
二、此公式的应用条件,
三、其它支承情况下，压杆临界力的欧拉公式
1.理想压杆； 2.线弹性范围内； 3.两端为球铰支座。
?—长度系数 （ 或约束系数 ） 。
两端铰支压杆临界力的欧拉公式
压杆临界力欧拉公式的一般形式
2
2
L
EIP
cr
m i n??
2
2
)(
m i n
L
EI
P cr
?
?
?
25
0.5
l
Table10–1 Euler’s formula of the slender compressive column under various constraint conditions
Supports Two hinged
ends
One free end
and one
hinged end
Two fixed
ends
One fixed
end and one
free end
Two fixed ends
but one of them is
movable laterally,
The shape of
the deflective
curve in lost
of stability
Pcr
A
B
l
Euler’s
formula of
the critical
forcePcr
Length
coefficientμ
2
2
l
EIP
cr
??
2
2
)7.0( l
EIP
cr
??
2
2
)5.0( l
EIP
cr
??
2
2
)2( l
EIP
cr
??
2
2
l
EIP
cr
??
?=1 ??0.7 ?=0.5 ?=2 ?=1
Pcr
A
B
l
Pcr
A
B
l 0.7
l
C C
D
C,Inflection
point
C,D,
Inflection
point
0.5
l
Pcr Pcr
l
2l l
C,Inflection point
26
0.5
l
表 10–1 各种支承约束条件下等截面细长压杆临界力的欧拉公式
支承情况 两端铰支 一端固定另端铰支 两端固定 一端固定另端自由 两端固定但可沿横向相对移动
失
稳
时
挠
曲
线
形
状
Pcr
A
B
l
临界力 Pcr
欧拉公式
长度系数 μ
2
2
l
EIP
cr
??
2
2
)7.0( l
EIP
cr
??
2
2
)5.0( l
EIP
cr
??
2
2
)2( l
EIP
cr
??
2
2
l
EIP
cr
??
?=1 ??0.7 ?=0.5 ?=2 ?=1
Pcr
A
B
l
Pcr
A
B
l 0.7
l
C C
D
C— 挠曲
线拐点
C,D— 挠
曲线拐点
0.5
l
Pcr Pcr
l
2l l
C— 挠曲线拐点
27
P
Mkyky 22 ????
MPyxMyEI ??????? )(
EI
Pk ?2
P
Mkxdkxcy ??? s i nc o s
0,;0,0 ???????? yyLxyyx
Solution,The deformation of the column
is shown in the figure,The approximate
differential equation of its deflection curve is,
The boundary conditions are,
Example 1 Try to deduce the critical force of the following slender column
by the approximate differential equation of the deflective curve,
P
L
x
P
M0
P
M0
P
M0
x P
M0
Let
kxckxdy s i nc o s ???
28
P
Mkyky 22 ????
MPyxMyEI ??????? )(
EI
Pk ?2:令
0,;0,0 ???????? yyLxyyx
解：变形如图，其挠曲线近似微分方程为,
边界条件为,
[例 1 ] 试由挠曲线近似微分方程，导出下述细长压杆的临界力
公式。
P
L
x
P
M0
P
M0
P
M0
x P
M0
P
Mkxdkxcy ??? s i nc o s
kxckxdy s i nc o s ???
29
2
2
2
2
)2/(
4
L
EI
L
EI
P cr
??
??
?2?kL
In order to determine the minimum critical pressure―k‖must be
the minimum value except zero,that is,
Therefore the critical pressure is,
2 ?nkL ??
? = 0.5
?? nkLnkLd
P
Mc ????? 2,0,and
30
?? nkLnkLd
P
Mc ????? 2,0,并
2
2
2
2
)2/(
4
L
EI
L
EI
P cr
??
??
?2?kL
为求最小临界力,, k” 应取除零以外的最小值, 即取,
所以，临界力为,
2 ?nkL ??
? = 0.5
31 ③ Critical pressure of the column
Example 2 Determine the critical pressure of the following slender compressive
column,
,
12
3 hb
I y ?
?=1.0,Solution,① About the axis y,two ends are hinged,
2
2
2
L
EI
P yc r y
?
?
,
12
3bh
I z ?
?=0.7,② About the axis z,the left end is fixed and the right end is hinged,
2
1
2
)7.0( L
EI
P zc r z
?
?
),m i n ( c r zc r ycr PPP ?
y
z
L1
L2
y
z h
b
x
32
③ 压杆的临界力
[例 2] 求下列细长压杆的临界力。
,
12
3 hb
I y ?
?=1.0,解, ① 绕 y 轴，两端铰支,
2
2
2
L
EI
P yc r y
?
?
,
12
3bh
I z ?
?=0.7,
② 绕 z 轴，左端固定，右端铰支,
2
1
2
)7.0( L
EI
P zc r z
?
?
),m i n ( c r zc r ycr PPP ?
y
z
L1
L2
y
z h
b
x
33
4912
3
m i n m1017.41012
1050 ?? ?????I
48m i n m1089.3 ???? zII
2
2
m in
2
)( l
EIP
cr ?
??
Example 3 Determine the critical pressure of the following slender
compressive columns.Knowing:L=0.5m and E=200GPa,
Solution,Fig.(a)
Fig.(b)
2
1
m in
2
)( l
EIP
cr ?
?? kN14.67
)5.07.0(
20017.4
2
2
?
?
??? ?
kN8.76
)5.02(
200389.0
2
2
?
?
??? ?Fig.(a) Fig.(b)
50
10
P
L
P
L
(45?45? 6)
Equal-leg angle
steel
y z
34
4912
3
m i n m1017.41012
1050 ?? ?????I
2
1
m in
2
)( l
EIP
cr ?
??
48m i n m1089.3 ???? zII
2
2
m in
2
)( l
EIP
cr ?
??
[例 3] 求下列细长压杆的临界力。已知,L=0.5m, E=200GPa,
图 (a) 图 (b)
解：图 (a)
图 (b)
kN14.67
)5.07.0(
20017.4
2
2
?
?
??? ?
kN8.76
)5.02(
200389.0
2
2
?
?
??? ?
50
10
P
L
P
L
(45?45? 6)
等边角钢
y z
35
§ 10–3 CRITICAL STRESS OF COMPRESSED COLUMNS AS STRESS
EXCEEDS THE PROPORTIONAL LIMIT
A
P cr
cr ??
1,Basic concepts
1).Critical stress,Mean stress in the cross section of the compressive column in
the critical state,
3).Flexibility,
2
2
2
2
2
2
)/()( ?
?
?
?
?
?? E
iL
E
AL
EI
A
P cr
cr ????
2).Critical stress for slender
compressed columns,
— iL?? ?
2
2
?
? ? E
cr ? That is,
— AIi ?
Radius of inertia
Flexibility (or slenderness ratio) of the column
36
§ 10–3 超过比例极限时压杆的临界应力
A
P cr
cr ??
一,基本概念
1.临界应力：压杆处于临界状态时横截面上的平均应力。
3.柔度,
2
2
2
2
2
2
)/()( ?
?
?
?
?
?? E
iL
E
AL
EI
A
P cr
cr ????
2.细长压杆的 临界应力,
—惯性半径。—
A
Ii ?
）—杆的柔度（或长细比—
i
L?? ?
2
2
?
?
? Ecr ?即：
37
4).Division of large flexibility column,
Pcr
E ?
?
?? ??
2
2
P
P
E ?
?
?? ?? 2
2,Calculation of the critical stress for the columns with middle or small flexibility
1),Empirical formula with the linear variation
① As?P<?<?S,
,scr ba ??? ???
s
s
b
aSo ??? ???
?? bacr ??
Ps ??? ?? is one of the middle flexibility column,
The column with
Its critical stress may be determined by the empirical formula,
P?? ?
P?? ?
The columns with is called the large flexibility column(or slender clumn),
The critical pressure of this kind of column is calculated by Euler’s formula,
The columns with is called middle or small flexibility column,Its critical
pressure of this kind of column can not be calculated by Euler’s formula,
38
4.大 柔度杆的分界,
Pcr
E ?
?
?? ??
2
2
欧拉公式求。长细杆），其临界力用的杆称为大柔度杆（或满足 P?? ?
P
P
E ?
?
?? ?? 2
求。临界力不能用欧拉公式的杆为中小柔度杆，其 P?? ?
二、中小柔度杆的临界应力计算
1.直线型经验公式
① ?P<?<?S 时,
scr ba ??? ???
s
s
b
a ??? ????
界应力用经验公式求。的杆为中柔度杆，其临 Ps ??? ??
?? bacr ??
39
i
L?? ?
cr?
2
2
?
?
?
E
cr
?
③ Total figure of the critical
stress
② As ?S<?,
scr ?? ?
?? bacr ??
P?
S?
b
s
s
? a ??
?
P
P
E
?
?
?
2
?
The critical stress of this column is its yield
limit,
is one of the small flexibility
S?? ?
The column with
column,
40
i
L?? ?
cr?
界应力为屈服极限。的杆为小柔度杆，其临 S?? ?
2
2
?
?
?
E
cr
?
③ 临界应力总图
② ?S<? 时,
scr ?? ?
?? bacr ??
P?
S?
s
b
a s ? ? ?
?
P
P
E
?
?
?
2
?
41
2)Empirical formula with the parabolic variation
2
11 ?? bacr ??
S
c
E
?
???
56.0
43.0
2
??,
The common form in the
structural engineering is,
① As ?P<?<?s,
?
?
?
?
?
?
?
?
?
?
?
???
2
1
c
scr ?
?
???
② As ?s<?,
scr ?? ?
For A3steel,A5steel and 16Mn steel,
,c?? ?
As This formula is used to determine the critical stress,
42
2.抛物线型经验公式
2
11 ?? bacr ??
S
c
E
AA
?
?
??
56.0
43.016
2
53 ??,锰钢：钢和钢、对于
。时，由此式求临界应力 c?? ?
我国建筑业常用,
① ?P<?<?s 时,
?
?
?
?
?
?
?
?
?
?
?
???
2
1
c
scr ?
?
???
② ?s<? 时,
scr ?? ?
43
4121 cm63.23,cm3 6 7.8 ?? yIA
zy II ?
cm68.1
3 6 7.82
26.47m i n ?
?
??
A
Ii 1233.89
68.1
150 ?????
ci
l ???
Solution,For one piece of angle steel,
After two pieces of angle steel is
combined as shown in the figure,
41m i n cm26.4763.2322 ????? yy III
Therefore the critical pressure should be determined by the empirical formula
y
z
Example 4 A compressive column is made from two equal- leg angle steel of
56?56?8,Its two ends are both hinged,Its length is L=1.5m,The pressure is
P=150kN,The angle steel is A3 steel,Try to determine the critical pressure and the
safety coefficient of stability nst by Euler’s formula or the empirical formula with the
parabolic variation,
with the parabolic variation,
44
[例 4 ] 一压杆长 L=1.5m，由两根 56?56?8 等边角钢组成，两端
铰支，压力 P=150kN，角钢为 A3钢，试用 欧拉公式或抛物线公
式求 临界压力和稳定安全系数 nst。
4121 cm63.23,cm3 6 7.8 ?? yIA
zy II ?
cm68.1
3 6 7.82
26.47m i n ?
?
??
A
Ii 1233.89
68.1
150 ?????
ci
l ???
解：一个角钢,
两根角钢图示组合之后
41m i n cm26.4763.2322 ????? yy III
所以，应由抛物线公式求 临界压力。
y
z
45
M P a7.181])
123
3.89(43.01[235])(43.01[ 22 ?????
c
scr ?
???
kN304107.18110367.82 64 ??????? ?crcr AP ?
02.2
1 5 0
3 0 4 ???
P
Pn cr
st
Safety coefficient
46
M P a7.181])
123
3.89(43.01[235])(43.01[ 22 ?????
c
scr ?
???
kN304107.18110367.82 64 ??????? ?crcr AP ?
02.2
1 5 0
3 0 4 ???
P
Pn cr
st
安全系数
47
§ 10–4 STABILITY CHECK AND REASONABLE SECTIONS OF
COMPRESSED COLUMNS
1,Permissible stress for the compressed columns in stabilization,
1).Determine the permissible stress by the method of the safety coefficient,
? ?
st
cr
st n
?? ?
2),Determine the permissible stress by the method of the reduction
coefficient,
? ? ? ???? ?W
2,Stability condition of the compressed columns,
? ? st
A
P ?? ??
??
Reduction coefficient,It is related to material
properties and the flexibility of columns,and, 1??
48
§ 10–4 压杆的稳定校核及其合理截面
一、压杆的稳定许用应力,
1.安全系数法确定许用应力,
? ?
st
cr
st n
?? ?
2.折减系数法确定许用应力,
? ? ? ???? ?st
柔度有关。
其值与材料性能及压杆折减系数 1,,?? ??
二、压杆的稳定条件,
? ? st
A
P ?? ??
49
Example 5 The column AB in the simple crane shown in the figure is made
from circular deal,Its length is L= 6m and its diameter is d = 0.3m,[? ]
=11MPa,Try to determine the permissible pressure of the column,
803.0 461 ????? i Lxy ??
Solution,Apply the method of
reduction coefficient,
① Maximum flexibility
In plane xy,?=1.0
In plane zy,?=2.0
1 6 03.0 462 ????? i Lzy ??
T1
A
B
W
T2
x
y
z
O
50
[例 5 ] 图示起重机,AB 杆为圆松木，长 L= 6m,[? ] =11MPa，
直径,d = 0.3m,试 求此杆的 许用 压力。
803.0 461 ????? i Lxy ??
解,折减系数法
① 最大柔度
x y面内, ?=1.0
z y面内, ?=2.0
1 6 03.0 462 ????? i Lzy ??
T1
A
B
W
T2
x
y
z
O
51
? ? ? ???? ?st
? ? ? ? kN911011117.0
4
3.0 62 ??????? ??
stBCBC AP
② Determine the reduction coefficient
③ Determine the permissible pressure
117.016030003000,80 22 ???? ??? 时
For wooden column:as
52
? ? ? ???? ?st
? ? ? ? kN911011117.0
4
3.0 62 ??????? ??
stBCBC AP
② 求折减系数
③ 求 许用 压力
117.016030003000,80,22 ???? ??? 时木杆
53
3,Reasonable section of
the column,
i
L?? ?
2
m in
2
)( L
EI
P cr
?
?
?
m in
A
I
i?
maxmin II ?
It is reasonable,
保国寺大殿的拼柱形式
Sakyamni wooden pagoda
Yingzhou wooden pagoda
It was built in 1056 with the double-sleeve
structure.The pagoda plane is octagon.It
underwent the level 8 earthquake of 1305,
54
三、压杆的合理截面,
i
L?? ?
2
m in
2
)( L
EI
P cr
?
?
?
m in
A
I
i?
maxmin II ?
合理
保国寺大殿的拼柱形式
1056年建，“双筒体”结构，塔身平面
为八角形。经历了 1305年的八级地震。
55
4
1
4
1
0
2
1
cm6.25,cm3.1 9 8
,cm52.1,cm74.12
??
??
yz II
zA
41 cm6.3 9 63.1 9 822 ???? zz II
???? ])2/([2 2011 azAII yy
])2/52.1(74.126.25[2 2a????
,)2/52.1(74.126.253.198 2a???
Solution,For each channel steel of number 10,the
centroid is at the point C1,
After two channel steels are combined,
cm32.4?a
P
L
z0
y y
1
z1 C1
a
Example 6 The upright column shown in the figure is made from two channel
steels,Its length is L=6m,Knowingthe material is A3steel,E=200GPa and
,Its lower end is fixed and its upper end is supported by a spherical
hinge,How much is the value of a when the critical pressure of the upright column
is the maximum?
That is it is
reasonable as
M P ap 2 0 0??
so
56
4
1
4
1
0
2
1
cm6.25,cm3.1 9 8
,cm52.1,cm74.12
??
??
yz II
zA
41 cm6.3 9 63.1 9 822 ???? zz II
???? ])2/([2 2011 azAII yy
])2/52.1(74.126.25[2 2a????
时合理即 2)2/52.1(74.126.253.198, a???
[例 6 ] 图示立柱,L=6m，由两根 10号槽钢组成，材料为 A3钢
E=200GPa，, 下端固定，上端为球铰支座,试问 a=？
时，立柱的 临界压力最大，值为多少？
解, 对于单个 10号槽钢, 形心在 C1点 。
两根槽钢图示组合之后,
cm32.4?a
y1
P
L
z0
y
z1 C
1
a
M P ap 2 0 0??
57
5.1 0 6
1074.122
106.3 9 6
67.0
2
67.0
4
8
1
?
??
?
?
?
?
?
??
?
?
A
Ii
L
z
?
?
??
?
?? ??
?
???? 3.99
10200
10200
6
922
P
p
E
kN8.443
)67.0(
106.396200
)( 2
22
2
2
?
?
????? ??
?
?
l
EIP
cr
Determine the critical force,
It is a large flexibility rod,The critical force is determined
by Euler’s formula,
58
5.1 0 6
1074.122
106.3 9 6
67.0
2
67.0
4
8
1
?
??
?
?
?
?
?
??
?
?
A
Ii
L
z
?
?
??
?
?? ??
?
???? 3.99
10200
10200
6
922
P
p
E
kN8.443
)67.0(
106.396200
)( 2
22
2
2
?
?
????? ??
?
?
l
EIP
cr
求临界力,
大柔度杆，由欧拉公式求临界力。
59
Chapter 10 Exercises
1,How to distinguish the stable equilibrium and the unstable
equilibrium of compressed columns?
2,A compressed rod produces the bending deformation due to the loss
of stability,A beam produces the bending deformation due to the action
of transverse forces,What are the differences of the two in nature?
3.Three circular rods with the same diameter d=16cm are shown in the
figure,Knowing their materials are A3 steel,E=200GPa
and, M P a
p 200??
Try to determine,
① Which compressed rod loses stability easily?
② The maximum critical value of the
compressive forces for the three rods,
60
第十章 练习题
一, 如何区别压杆的稳定平衡和不稳定平衡？
二, 压杆因失稳而产生弯曲变形, 与梁在横向
力作用下产生弯曲变形, 在性质上有何区别？
三, 三根直径均为 d=16cm 的圆杆如图所示,
材料均为 A3钢, E=200GPa,。 M P a
p 200??
试求,
① 哪一根压杆最容易失稳？
②三杆中最大的临界压力值。
61
Solution,①
Rod (a),
Rod (b),Rod (a) loses stability more easily,
Rod (c),
cmdi 44 ??
12545001 ???? i l??
5.1 2 24 7 0 07.0 ????
5.1 1 24 9 0 05.0 ????
?
5.1 1235.9920 0 1020 0 322 ????? ?? ?? ??? pEp
? ? ? ? KNP lEIcr 31362 4222 95.064 1610200 ??? ?? ???? ????
② The critical force of Rod (c) is maximum,
62
解：①
杆 (a),
杆 (b),杆 (a)最易失稳
杆 (c),
② 杆 (c)的临界力最大
cmdi 44 ??
12545001 ???? i l??
5.1 2 24 7 0 07.0 ????
5.1 1 24 9 0 05.0 ????
?
5.1 1235.9920 0 1020 0 322 ????? ?? ?? ??? pEp
? ? ? ? KNP lEIcr 31362 4222 95.064 1610200 ??? ?? ???? ????
63
64
M e c h a n i c s o f M a t e r i a l s
2
3
CHAPTER 10 STABILIZATION OF
COMPRESSIVE COLUMNS
§ 10–1 CONCEPTS OF STABILITY OF COMPRESSED COLUMNS
§ 10–2 EULER’S FORMULA OF THE CRITICAL FORCE OF SLENDER
COMPRESSED COLUMNS
§ 10–3 CRITICAL STRESS OF COMPRESSED COLUMNS AS STRESS
EXCEEDS PROPORTIONAL LIMIT
§ 10-4 STABILITY CHECK AND REASONABLE SECTION OF
COMPRESSED COLUMNS
4
第十章 压杆稳定
§ 10–1 压杆稳定性的概念
§ 10–2 细长压杆临界力的欧拉公式
§ 10–3 超过比例极限时压杆的临界应力
§ 10-4 压杆的稳定校核及其合理截面
5
§ 10–1 CONCEPTS OF STABILITY OF COMPRESSED COLUMNS
The load-carrying
capacity of structure
members,
① Strength
② Rigidity
③ Stability
Some structure members
in engineering have enough
strength and rigidity but
they are unable to work
safely and reliably,
6
§ 10–1 压杆稳定性的概念
构件的承载能力,① 强度
②刚度
③稳定性
工程中有些构
件具有足够的强度、
刚度，却不一定能
安全可靠地工作。
7
P
8
P
9
1,Stable and instable equilibrium,
1),instable equilibrium
10
一、稳定平衡与不稳定平衡,
1,不稳定平衡
11
2),Stable equilibrium
12
2,稳定平衡
13
3),Stable and instable equilibrium
14
3,稳定平衡和不稳定平衡
15
2,Loss of stability and the critical pressure of the column,
1).Ideal compressive columns,the material is absolutely ideal； the axis is absolutely
straight； the compressive force is absolutely along the axis of the column,
2),Stable and instable equilibrium of compressive columns,
Instable
equilibrium
Stable
equilibrium
16
二、压杆失稳与临界压力,
1.理想压杆：材料绝对理想；轴线绝对直；压力绝对沿轴线作用。
2.压杆的稳定平衡与不稳定平衡,
稳
定
平
衡
不
稳
定
平
衡
17
3).loss of stability of compressed
column,
4).Critical pressure of compressed columns
Stable
equilibrium
Instable
equilibrium
Critical state
Critical pressure,Pcr
corresponding
pressure
intermediate state
18
3.压杆失稳,4.压杆的临界压力
稳
定
平
衡
不
稳
定
平
衡
临界状态
临界压力, Pcr
过 度
对 应 的
压力
19
§ 10–2 EULER’S FORMULA OF THE CRITICAL PRESSURE OF
SLENDER COMPRESSED COLUMNS
1,Critical pressure for the column with two hinged ends,
PyyxM ?),(
y
EI
P
EI
My ??????
① bending moment,
② Approximate differential
equation of the deflection curve
,02 ???????? ykyy
EI
Py
EI
Pk ?2
:w h e r e
P
P
x
y
P
M
Suppose the pressure has reached the critical value and the column has been in
tiny bending state as shown in the figure,Start to determine the critical force with
the deflective curve,
P
x L
20
§ 10–2 细长压杆临界力的欧拉公式
一、两端铰支压杆的临界力,
PyyxM ?),(
假定压力已达到临界值，杆已经处于微弯状态，如图,
从挠曲线入手，求临界力。
y
EI
P
EI
My ??????
① 弯矩,
② 挠曲线近似微分方程,
02 ???????? ykyy
EI
Py
EI
Pk ?2
:其中
P
x L
P
x
y
P
M
21
③ Solution of the differential equation,
④ Determine the integral constants,
xBxAy c o ss i n ??
0)()0( ?? Lyy
?
?
?
??
???
0c o ss i n
00
kLBkLA
BA 0
c o s s i n
1 0
?
kLkL
0s in ?kL
EI
P
L
nk ?? ? so
The critical force Pcr is the smallest pressure under tiny bending,therefore
we only take n=1 and the column will bend about the axis with the smallest
moment of inertia,
2
m i n
2
L
EIP
cr
??
That is
22
③ 微分方程的解,
④ 确定积分常数,
xBxAy c o ss i n ??
0)()0( ?? Lyy
?
?
?
??
???
0c o ss i n
00
:
kLBkLA
BA
即 0c o s s i n
1 0
??
kLkL
0s in ?? kL
EI
P
L
nk ??? ?
临界力 Pcr 是微弯下的最小压力,故，只能取 n=1 ；且
杆将绕惯性矩最小的轴弯曲。
2
m i n
2
L
EIP
cr
???
23
2,Application range of the formula,
3,Euler’s formula of the critical pressure for the column with
other end conditions,
1).Ideal compressive columns； 2).In linear elastic range；
3).The ends of the column are supported by hinges,
?—Length coefficient（ or constraint coefficient）
Euler’s formula of the critical pressure for
the compressive column with two hinged ends
General form of Euler’s formula of
the critical pressure
2
m i n
2
L
EIP
cr
??
2
m i n
2
)( L
EI
P cr
?
?
?
24
二、此公式的应用条件,
三、其它支承情况下，压杆临界力的欧拉公式
1.理想压杆； 2.线弹性范围内； 3.两端为球铰支座。
?—长度系数 （ 或约束系数 ） 。
两端铰支压杆临界力的欧拉公式
压杆临界力欧拉公式的一般形式
2
2
L
EIP
cr
m i n??
2
2
)(
m i n
L
EI
P cr
?
?
?
25
0.5
l
Table10–1 Euler’s formula of the slender compressive column under various constraint conditions
Supports Two hinged
ends
One free end
and one
hinged end
Two fixed
ends
One fixed
end and one
free end
Two fixed ends
but one of them is
movable laterally,
The shape of
the deflective
curve in lost
of stability
Pcr
A
B
l
Euler’s
formula of
the critical
forcePcr
Length
coefficientμ
2
2
l
EIP
cr
??
2
2
)7.0( l
EIP
cr
??
2
2
)5.0( l
EIP
cr
??
2
2
)2( l
EIP
cr
??
2
2
l
EIP
cr
??
?=1 ??0.7 ?=0.5 ?=2 ?=1
Pcr
A
B
l
Pcr
A
B
l 0.7
l
C C
D
C,Inflection
point
C,D,
Inflection
point
0.5
l
Pcr Pcr
l
2l l
C,Inflection point
26
0.5
l
表 10–1 各种支承约束条件下等截面细长压杆临界力的欧拉公式
支承情况 两端铰支 一端固定另端铰支 两端固定 一端固定另端自由 两端固定但可沿横向相对移动
失
稳
时
挠
曲
线
形
状
Pcr
A
B
l
临界力 Pcr
欧拉公式
长度系数 μ
2
2
l
EIP
cr
??
2
2
)7.0( l
EIP
cr
??
2
2
)5.0( l
EIP
cr
??
2
2
)2( l
EIP
cr
??
2
2
l
EIP
cr
??
?=1 ??0.7 ?=0.5 ?=2 ?=1
Pcr
A
B
l
Pcr
A
B
l 0.7
l
C C
D
C— 挠曲
线拐点
C,D— 挠
曲线拐点
0.5
l
Pcr Pcr
l
2l l
C— 挠曲线拐点
27
P
Mkyky 22 ????
MPyxMyEI ??????? )(
EI
Pk ?2
P
Mkxdkxcy ??? s i nc o s
0,;0,0 ???????? yyLxyyx
Solution,The deformation of the column
is shown in the figure,The approximate
differential equation of its deflection curve is,
The boundary conditions are,
Example 1 Try to deduce the critical force of the following slender column
by the approximate differential equation of the deflective curve,
P
L
x
P
M0
P
M0
P
M0
x P
M0
Let
kxckxdy s i nc o s ???
28
P
Mkyky 22 ????
MPyxMyEI ??????? )(
EI
Pk ?2:令
0,;0,0 ???????? yyLxyyx
解：变形如图，其挠曲线近似微分方程为,
边界条件为,
[例 1 ] 试由挠曲线近似微分方程，导出下述细长压杆的临界力
公式。
P
L
x
P
M0
P
M0
P
M0
x P
M0
P
Mkxdkxcy ??? s i nc o s
kxckxdy s i nc o s ???
29
2
2
2
2
)2/(
4
L
EI
L
EI
P cr
??
??
?2?kL
In order to determine the minimum critical pressure―k‖must be
the minimum value except zero,that is,
Therefore the critical pressure is,
2 ?nkL ??
? = 0.5
?? nkLnkLd
P
Mc ????? 2,0,and
30
?? nkLnkLd
P
Mc ????? 2,0,并
2
2
2
2
)2/(
4
L
EI
L
EI
P cr
??
??
?2?kL
为求最小临界力,, k” 应取除零以外的最小值, 即取,
所以，临界力为,
2 ?nkL ??
? = 0.5
31 ③ Critical pressure of the column
Example 2 Determine the critical pressure of the following slender compressive
column,
,
12
3 hb
I y ?
?=1.0,Solution,① About the axis y,two ends are hinged,
2
2
2
L
EI
P yc r y
?
?
,
12
3bh
I z ?
?=0.7,② About the axis z,the left end is fixed and the right end is hinged,
2
1
2
)7.0( L
EI
P zc r z
?
?
),m i n ( c r zc r ycr PPP ?
y
z
L1
L2
y
z h
b
x
32
③ 压杆的临界力
[例 2] 求下列细长压杆的临界力。
,
12
3 hb
I y ?
?=1.0,解, ① 绕 y 轴，两端铰支,
2
2
2
L
EI
P yc r y
?
?
,
12
3bh
I z ?
?=0.7,
② 绕 z 轴，左端固定，右端铰支,
2
1
2
)7.0( L
EI
P zc r z
?
?
),m i n ( c r zc r ycr PPP ?
y
z
L1
L2
y
z h
b
x
33
4912
3
m i n m1017.41012
1050 ?? ?????I
48m i n m1089.3 ???? zII
2
2
m in
2
)( l
EIP
cr ?
??
Example 3 Determine the critical pressure of the following slender
compressive columns.Knowing:L=0.5m and E=200GPa,
Solution,Fig.(a)
Fig.(b)
2
1
m in
2
)( l
EIP
cr ?
?? kN14.67
)5.07.0(
20017.4
2
2
?
?
??? ?
kN8.76
)5.02(
200389.0
2
2
?
?
??? ?Fig.(a) Fig.(b)
50
10
P
L
P
L
(45?45? 6)
Equal-leg angle
steel
y z
34
4912
3
m i n m1017.41012
1050 ?? ?????I
2
1
m in
2
)( l
EIP
cr ?
??
48m i n m1089.3 ???? zII
2
2
m in
2
)( l
EIP
cr ?
??
[例 3] 求下列细长压杆的临界力。已知,L=0.5m, E=200GPa,
图 (a) 图 (b)
解：图 (a)
图 (b)
kN14.67
)5.07.0(
20017.4
2
2
?
?
??? ?
kN8.76
)5.02(
200389.0
2
2
?
?
??? ?
50
10
P
L
P
L
(45?45? 6)
等边角钢
y z
35
§ 10–3 CRITICAL STRESS OF COMPRESSED COLUMNS AS STRESS
EXCEEDS THE PROPORTIONAL LIMIT
A
P cr
cr ??
1,Basic concepts
1).Critical stress,Mean stress in the cross section of the compressive column in
the critical state,
3).Flexibility,
2
2
2
2
2
2
)/()( ?
?
?
?
?
?? E
iL
E
AL
EI
A
P cr
cr ????
2).Critical stress for slender
compressed columns,
— iL?? ?
2
2
?
? ? E
cr ? That is,
— AIi ?
Radius of inertia
Flexibility (or slenderness ratio) of the column
36
§ 10–3 超过比例极限时压杆的临界应力
A
P cr
cr ??
一,基本概念
1.临界应力：压杆处于临界状态时横截面上的平均应力。
3.柔度,
2
2
2
2
2
2
)/()( ?
?
?
?
?
?? E
iL
E
AL
EI
A
P cr
cr ????
2.细长压杆的 临界应力,
—惯性半径。—
A
Ii ?
）—杆的柔度（或长细比—
i
L?? ?
2
2
?
?
? Ecr ?即：
37
4).Division of large flexibility column,
Pcr
E ?
?
?? ??
2
2
P
P
E ?
?
?? ?? 2
2,Calculation of the critical stress for the columns with middle or small flexibility
1),Empirical formula with the linear variation
① As?P<?<?S,
,scr ba ??? ???
s
s
b
aSo ??? ???
?? bacr ??
Ps ??? ?? is one of the middle flexibility column,
The column with
Its critical stress may be determined by the empirical formula,
P?? ?
P?? ?
The columns with is called the large flexibility column(or slender clumn),
The critical pressure of this kind of column is calculated by Euler’s formula,
The columns with is called middle or small flexibility column,Its critical
pressure of this kind of column can not be calculated by Euler’s formula,
38
4.大 柔度杆的分界,
Pcr
E ?
?
?? ??
2
2
欧拉公式求。长细杆），其临界力用的杆称为大柔度杆（或满足 P?? ?
P
P
E ?
?
?? ?? 2
求。临界力不能用欧拉公式的杆为中小柔度杆，其 P?? ?
二、中小柔度杆的临界应力计算
1.直线型经验公式
① ?P<?<?S 时,
scr ba ??? ???
s
s
b
a ??? ????
界应力用经验公式求。的杆为中柔度杆，其临 Ps ??? ??
?? bacr ??
39
i
L?? ?
cr?
2
2
?
?
?
E
cr
?
③ Total figure of the critical
stress
② As ?S<?,
scr ?? ?
?? bacr ??
P?
S?
b
s
s
? a ??
?
P
P
E
?
?
?
2
?
The critical stress of this column is its yield
limit,
is one of the small flexibility
S?? ?
The column with
column,
40
i
L?? ?
cr?
界应力为屈服极限。的杆为小柔度杆，其临 S?? ?
2
2
?
?
?
E
cr
?
③ 临界应力总图
② ?S<? 时,
scr ?? ?
?? bacr ??
P?
S?
s
b
a s ? ? ?
?
P
P
E
?
?
?
2
?
41
2)Empirical formula with the parabolic variation
2
11 ?? bacr ??
S
c
E
?
???
56.0
43.0
2
??,
The common form in the
structural engineering is,
① As ?P<?<?s,
?
?
?
?
?
?
?
?
?
?
?
???
2
1
c
scr ?
?
???
② As ?s<?,
scr ?? ?
For A3steel,A5steel and 16Mn steel,
,c?? ?
As This formula is used to determine the critical stress,
42
2.抛物线型经验公式
2
11 ?? bacr ??
S
c
E
AA
?
?
??
56.0
43.016
2
53 ??,锰钢：钢和钢、对于
。时，由此式求临界应力 c?? ?
我国建筑业常用,
① ?P<?<?s 时,
?
?
?
?
?
?
?
?
?
?
?
???
2
1
c
scr ?
?
???
② ?s<? 时,
scr ?? ?
43
4121 cm63.23,cm3 6 7.8 ?? yIA
zy II ?
cm68.1
3 6 7.82
26.47m i n ?
?
??
A
Ii 1233.89
68.1
150 ?????
ci
l ???
Solution,For one piece of angle steel,
After two pieces of angle steel is
combined as shown in the figure,
41m i n cm26.4763.2322 ????? yy III
Therefore the critical pressure should be determined by the empirical formula
y
z
Example 4 A compressive column is made from two equal- leg angle steel of
56?56?8,Its two ends are both hinged,Its length is L=1.5m,The pressure is
P=150kN,The angle steel is A3 steel,Try to determine the critical pressure and the
safety coefficient of stability nst by Euler’s formula or the empirical formula with the
parabolic variation,
with the parabolic variation,
44
[例 4 ] 一压杆长 L=1.5m，由两根 56?56?8 等边角钢组成，两端
铰支，压力 P=150kN，角钢为 A3钢，试用 欧拉公式或抛物线公
式求 临界压力和稳定安全系数 nst。
4121 cm63.23,cm3 6 7.8 ?? yIA
zy II ?
cm68.1
3 6 7.82
26.47m i n ?
?
??
A
Ii 1233.89
68.1
150 ?????
ci
l ???
解：一个角钢,
两根角钢图示组合之后
41m i n cm26.4763.2322 ????? yy III
所以，应由抛物线公式求 临界压力。
y
z
45
M P a7.181])
123
3.89(43.01[235])(43.01[ 22 ?????
c
scr ?
???
kN304107.18110367.82 64 ??????? ?crcr AP ?
02.2
1 5 0
3 0 4 ???
P
Pn cr
st
Safety coefficient
46
M P a7.181])
123
3.89(43.01[235])(43.01[ 22 ?????
c
scr ?
???
kN304107.18110367.82 64 ??????? ?crcr AP ?
02.2
1 5 0
3 0 4 ???
P
Pn cr
st
安全系数
47
§ 10–4 STABILITY CHECK AND REASONABLE SECTIONS OF
COMPRESSED COLUMNS
1,Permissible stress for the compressed columns in stabilization,
1).Determine the permissible stress by the method of the safety coefficient,
? ?
st
cr
st n
?? ?
2),Determine the permissible stress by the method of the reduction
coefficient,
? ? ? ???? ?W
2,Stability condition of the compressed columns,
? ? st
A
P ?? ??
??
Reduction coefficient,It is related to material
properties and the flexibility of columns,and, 1??
48
§ 10–4 压杆的稳定校核及其合理截面
一、压杆的稳定许用应力,
1.安全系数法确定许用应力,
? ?
st
cr
st n
?? ?
2.折减系数法确定许用应力,
? ? ? ???? ?st
柔度有关。
其值与材料性能及压杆折减系数 1,,?? ??
二、压杆的稳定条件,
? ? st
A
P ?? ??
49
Example 5 The column AB in the simple crane shown in the figure is made
from circular deal,Its length is L= 6m and its diameter is d = 0.3m,[? ]
=11MPa,Try to determine the permissible pressure of the column,
803.0 461 ????? i Lxy ??
Solution,Apply the method of
reduction coefficient,
① Maximum flexibility
In plane xy,?=1.0
In plane zy,?=2.0
1 6 03.0 462 ????? i Lzy ??
T1
A
B
W
T2
x
y
z
O
50
[例 5 ] 图示起重机,AB 杆为圆松木，长 L= 6m,[? ] =11MPa，
直径,d = 0.3m,试 求此杆的 许用 压力。
803.0 461 ????? i Lxy ??
解,折减系数法
① 最大柔度
x y面内, ?=1.0
z y面内, ?=2.0
1 6 03.0 462 ????? i Lzy ??
T1
A
B
W
T2
x
y
z
O
51
? ? ? ???? ?st
? ? ? ? kN911011117.0
4
3.0 62 ??????? ??
stBCBC AP
② Determine the reduction coefficient
③ Determine the permissible pressure
117.016030003000,80 22 ???? ??? 时
For wooden column:as
52
? ? ? ???? ?st
? ? ? ? kN911011117.0
4
3.0 62 ??????? ??
stBCBC AP
② 求折减系数
③ 求 许用 压力
117.016030003000,80,22 ???? ??? 时木杆
53
3,Reasonable section of
the column,
i
L?? ?
2
m in
2
)( L
EI
P cr
?
?
?
m in
A
I
i?
maxmin II ?
It is reasonable,
保国寺大殿的拼柱形式
Sakyamni wooden pagoda
Yingzhou wooden pagoda
It was built in 1056 with the double-sleeve
structure.The pagoda plane is octagon.It
underwent the level 8 earthquake of 1305,
54
三、压杆的合理截面,
i
L?? ?
2
m in
2
)( L
EI
P cr
?
?
?
m in
A
I
i?
maxmin II ?
合理
保国寺大殿的拼柱形式
1056年建，“双筒体”结构，塔身平面
为八角形。经历了 1305年的八级地震。
55
4
1
4
1
0
2
1
cm6.25,cm3.1 9 8
,cm52.1,cm74.12
??
??
yz II
zA
41 cm6.3 9 63.1 9 822 ???? zz II
???? ])2/([2 2011 azAII yy
])2/52.1(74.126.25[2 2a????
,)2/52.1(74.126.253.198 2a???
Solution,For each channel steel of number 10,the
centroid is at the point C1,
After two channel steels are combined,
cm32.4?a
P
L
z0
y y
1
z1 C1
a
Example 6 The upright column shown in the figure is made from two channel
steels,Its length is L=6m,Knowingthe material is A3steel,E=200GPa and
,Its lower end is fixed and its upper end is supported by a spherical
hinge,How much is the value of a when the critical pressure of the upright column
is the maximum?
That is it is
reasonable as
M P ap 2 0 0??
so
56
4
1
4
1
0
2
1
cm6.25,cm3.1 9 8
,cm52.1,cm74.12
??
??
yz II
zA
41 cm6.3 9 63.1 9 822 ???? zz II
???? ])2/([2 2011 azAII yy
])2/52.1(74.126.25[2 2a????
时合理即 2)2/52.1(74.126.253.198, a???
[例 6 ] 图示立柱,L=6m，由两根 10号槽钢组成，材料为 A3钢
E=200GPa，, 下端固定，上端为球铰支座,试问 a=？
时，立柱的 临界压力最大，值为多少？
解, 对于单个 10号槽钢, 形心在 C1点 。
两根槽钢图示组合之后,
cm32.4?a
y1
P
L
z0
y
z1 C
1
a
M P ap 2 0 0??
57
5.1 0 6
1074.122
106.3 9 6
67.0
2
67.0
4
8
1
?
??
?
?
?
?
?
??
?
?
A
Ii
L
z
?
?
??
?
?? ??
?
???? 3.99
10200
10200
6
922
P
p
E
kN8.443
)67.0(
106.396200
)( 2
22
2
2
?
?
????? ??
?
?
l
EIP
cr
Determine the critical force,
It is a large flexibility rod,The critical force is determined
by Euler’s formula,
58
5.1 0 6
1074.122
106.3 9 6
67.0
2
67.0
4
8
1
?
??
?
?
?
?
?
??
?
?
A
Ii
L
z
?
?
??
?
?? ??
?
???? 3.99
10200
10200
6
922
P
p
E
kN8.443
)67.0(
106.396200
)( 2
22
2
2
?
?
????? ??
?
?
l
EIP
cr
求临界力,
大柔度杆，由欧拉公式求临界力。
59
Chapter 10 Exercises
1,How to distinguish the stable equilibrium and the unstable
equilibrium of compressed columns?
2,A compressed rod produces the bending deformation due to the loss
of stability,A beam produces the bending deformation due to the action
of transverse forces,What are the differences of the two in nature?
3.Three circular rods with the same diameter d=16cm are shown in the
figure,Knowing their materials are A3 steel,E=200GPa
and, M P a
p 200??
Try to determine,
① Which compressed rod loses stability easily?
② The maximum critical value of the
compressive forces for the three rods,
60
第十章 练习题
一, 如何区别压杆的稳定平衡和不稳定平衡？
二, 压杆因失稳而产生弯曲变形, 与梁在横向
力作用下产生弯曲变形, 在性质上有何区别？
三, 三根直径均为 d=16cm 的圆杆如图所示,
材料均为 A3钢, E=200GPa,。 M P a
p 200??
试求,
① 哪一根压杆最容易失稳？
②三杆中最大的临界压力值。
61
Solution,①
Rod (a),
Rod (b),Rod (a) loses stability more easily,
Rod (c),
cmdi 44 ??
12545001 ???? i l??
5.1 2 24 7 0 07.0 ????
5.1 1 24 9 0 05.0 ????
?
5.1 1235.9920 0 1020 0 322 ????? ?? ?? ??? pEp
? ? ? ? KNP lEIcr 31362 4222 95.064 1610200 ??? ?? ???? ????
② The critical force of Rod (c) is maximum,
62
解：①
杆 (a),
杆 (b),杆 (a)最易失稳
杆 (c),
② 杆 (c)的临界力最大
cmdi 44 ??
12545001 ???? i l??
5.1 2 24 7 0 07.0 ????
5.1 1 24 9 0 05.0 ????
?
5.1 1235.9920 0 1020 0 322 ????? ?? ?? ??? pEp
? ? ? ? KNP lEIcr 31362 4222 95.064 1610200 ??? ?? ???? ????
63
64
