# Chapter 4 Axially Loaded Compression Members 钢结构基本原理

## Transcription

Chapter 4 Axially Loaded Compression Members 钢结构基本原理
```Chapter 4 Axially Loaded Compression Members

The strength and stiffness of axially loaded compression members

The overall stability of axially loaded compression members

4.4 The local stability of solid-web axially loaded compression members

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Chapter 4 Axially Loaded Compression Members

4.1 The application and section types of axially loaded compression members
4.1.1 The application of axially loaded compression members
The axially loaded compression
members can classify into two categories:
tension and compression. Both of them
must meet the requirements of
supporting capacity limit state and
serviceability limit state.
（
Serviceability limit state is decided

Supporting capacity limit state

） local stability.
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The application of axially loaded compression members
Chapter 4 Axially Loaded Compression Members

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4.1.2 Section types
Axially loaded compression members has many kinds of section types，it
can be divided into the structure steel sections and built-up sections，and the
built-up sections contains solid-web and lattice form.
（a）
（b）
（c）
(a) structure steel sections
(b) solid-web
(c) lattice form
Chapter 4 Axially Loaded Compression Members

l
l
l =l

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）

1
x
y
y
x (虚轴)
y
y
1
x (虚轴)
y
y
(实轴)
x
1
(实轴)
x
1
x
Chapter 4 Axially Loaded Compression Members

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4.2 The strength and stiffness of axially loaded compression members
4.2.1 The strength of axially loaded compression members
To undamaged section members ，we must control the gross sections’
mean stress under the materials' yield strength：
N fy
 
 f
A R
But to weaken section members, we must control the net sections' under
the materials' tensile strength. According to Code for design of steel structures
（50017－2003），partial safety factor for resistance :  uR  1.25 R
fu
fu
 R fu fy
N



 
 0.8  f
An  uR  uR f y  R
fy
Chapter 4 Axially Loaded Compression Members

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As to make the calculation more simple and convenient, Code for design
of steel structures （50017－2003）chooses the yield ratio as 0.8. So to the
axially loaded members, we can calculate the strength as:
N

 f
An
N ——design value
An ——net section area
f ——tensile strength or compressive strength
back
Chapter 4 Axially Loaded Compression Members

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4.2.2 The stiffness of axially loaded compression members
As not to make excessive deformability , the axially loaded
compression members must have enough stiffness, so we can through
controlling the slenderness ratio to assurance members’ stiffness, that is:
max
l0
   
i
max——
the maximal slenderness ratio
l 0 —— members' effective length
i —— radius of gyration
  —— allowable slenderness ratio
back
Chapter 4 Axially Loaded Compression Members

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4.3 The overall stability of axially loaded compression members
4.3.1 Introduction of Stabilities problem
In dealing with compression members, the problem of stability is of
great importance. Unlike tension members, while the load tends to hold the
members in alignment, compression members are very sensitive to factors
that may tend to cause lateral displacements or buckling. The situation is
similar in ways to the lateral buckling of beams. The bucking problem is
intensified and the load-carrying capacity is affected by such factors as
eccentric load ,imperfection of material, and initial crookedness of the
member. Also, residual stresses play a role. There are the variable stresses
that are “locked up” in the member as a result of the method of manufacture,
which involves unequal cooling rates within the cross section. The parts that
cool first will have residual compression stresses, while parts that cool that
cool last will have residual tension stresses. Residual stresses may also be
induced by non-uniform plastic deformation caused by cold working, such
as in the straightening process.
Chapter 4 Axially Loaded Compression Members

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4.3.2 Ideal axially loaded compression members and buckling forms
Ideal axially loaded compression members' basic assumption：
1）the member bar must be prismatic ideal member
2）the loads are coincident with the longitudinal centroidal axis of the members
3）without the initial stress influence
4）the material must be homogeneous、isotropic and infinite elastic ，
correspond with hooke's law .
Chapter 4 Axially Loaded Compression Members

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Buckling modes of strut :
N
1
N
1
V
1-1
(a)
bending-buckling
2
N
2
3
3
¦Õ
V
2-2
(b)
torsional-buckling
3-3
(c)
bending and torsional-buckling
Chapter 4 Axially Loaded Compression Members
①elastic bending-buckling——Euler’s formula
N
y
N
z
-EIY ''
z
y1 y2
z
（

）
(1)Critical force of bending-buckling
l0

4.3.3 Crippling load of axially loaded compression members
y
N
y
N
critical state of hinged-hinged axially loaded compression bar
Chapter 4 Axially Loaded Compression Members

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Critical force：
N cr   2 EI / l 2   2 EA /(l / i) 2
  2 EA / 2
Critical stress：
N cr  2 E
 cr 
 2
A

derivation process of critical force
Chapter 4 Axially Loaded Compression Members

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Elastic bending assumption of axially loaded compression members
requires: critical stress must be less than material’s proportional limit, that is:
 2Ε
 cr  2  f p

solve：
  
E
 p
fp
 pis critical slenderness
Chapter 4 Axially Loaded Compression Members

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②the elastoplasticity critical force and stress of ideal axially loaded
compression members
As to axially loaded compression members (slenderness rationλ<λp )take
place to bending-bucking，the formula of critical force and stress is：
N cr 
 2 Et I
l2
 2 Et
 cr  2

Chapter 4 Axially Loaded Compression Members

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（2）critical force of torsional-buckling
the torsional-buckling critical force computational formula of the axially
loaded compression members is:
 π 2 EI ω
1
N z   2  GI t  2
 l
 i0
the first part of the formula is warping torsion, it is related to the length
of the members, the second part is free torsion.
E
Chapter 4 Axially Loaded Compression Members

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In order to facilitate the application, we can convent the torsionalbuckling calculation of the critical force into bending-buckling in the form of
Euler‘s formula, that is ：
  2 EI ω
 1  2E
N z   2  GI t  2  2  A
z
 l
 i0
The meaning of the formula is that we can written the torsional-buckling
calculation formula in the form of Euler’s formula, but we should change the
λ into equivalent slenderness ratio λz。
z 
Iω
Ai02

2
2
lω  GI t  E


Iω
Ai02
lω2  I t 25.7
Chapter 4 Axially Loaded Compression Members

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（3）critical force of bending and torsional-buckling
The critical force of bending and torsional-buckling calculation formula
is：
N
 N N z  N   N a0 / i0   0
2
Ey
2
As to facilitate the application ，we can write the formula in the form of
Euler’s formula, that is:
N cr   2 EA / 2yz
Chapter 4 Axially Loaded Compression Members

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Equivalent slenderness ratio is：
1
yz 
2

 
 2  2 
z
 y
2
y


2 2
z
 4(1  a02 / i02 )2y 2z 

1/ 2
a 0 ——distance from section centroid to shear center
i0 ——radius of gyration to shear center
 y ——slenderness ratio of bending-buckling to symmetry axis
z ——equivalent slenderness ratio of torsional-buckling
Chapter 4 Axially Loaded Compression Members

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4.3.4 the effect of initial defect to the axially loaded compression members
The bucking problem is intensified and the load-carrying capacity is
affected by such factors as eccentric load ,imperfection of material, and initial
crookedness of the member. Also, residual stresses play a role. There are the
variable stresses that are “locked up” in the member as a result of the method
of manufacture, which involves unequal cooling rates within the cross section.
The parts that cool first will have residual compression stresses, while parts
that cool that cool last will have residual tension stresses. Residual stresses
may also be induced by non-uniform plastic deformation caused by cold
working, such as in the straightening process.
Chapter 4 Axially Loaded Compression Members

4.3.5 Overall stability calculation of actual axially loaded compression
members
（1）Calculation analysis of stability bearing capacity
Euler’s formula is the basic formula of calculating the stability,but as
the influence of defects,we have to use the column curves which is come
from theory calculation and experiment to do the stability calculation.
（ （2）Column curves of axially loaded compression members

）
Chapter 4 Axially Loaded Compression Members

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¦Õ
= ¦fÒ
cr
y
1.0
upon-boundary
0.8
a
0.6
b
under-boundaryc
0.4
d
0.2
0
40
80
120
Column curves
160
200 λ√f y /235
Chapter 4 Axially Loaded Compression Members

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(3)Calculation formula of overall stability
According to the priciple of the critical force  crof overall stability
should be less than axial force ,at the same time considerate the subentry
coefficient  R ,that is:
N cr
 cr f y
N
 



 f
A
A R
fy  R
N
 
 f
A
or
N
 f
A
N ——design value of axial force
A ——gross section area
f ——design value of compression strength
 ——   cr / f y , overall stability coefficient
back
Chapter 4 Axially Loaded Compression Members

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We should put attention to the slenderness ratio  , when we calculate
the slenderness ration  , we can considerate in these aspects.
（1）Section is bisymmetry or polar symmetry （bending-buckling）
x  l0x / ix
 y  l0 y / i y
l0y ——effective length
l0x 、
y
y
x
x
y
x
t
b
x
y
Chapter 4 Axially Loaded Compression Members

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（2）section is monosymmetric（bending and torsion-buckling ）
For monosymmetric section，we should considerate torsion, so in the
same condition, the critical force of bending and torsion-buckling is less than
bending-buckling, so we need use the equivalent slenderness ratio instead.
1
yz 
2

 
 2  2 
z
 y
2
y

  i A / I t / 25.7  I ω / l
2
z
2
0
i02  e02  ix2  iy2

2 2
z
2
ω



 4 1  e02 / i02 2y 2z 

1/ 2
y
x
x
y
Practical application often use simplified calculation formula （sheet 4.5.3)
Chapter 4 Axially Loaded Compression Members

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4.4 The local stability of solid-web axially loaded compression members
The solid-web axially loaded compression members depend on web
plate and flange to bear the compressive force.When web plate and flange
are thinner, under the axis compressive force, both the web plate and
flanges probably attain critical bearing capacity and then destabilizate. This
kind of destabilization usually take place partially , so be called crippling.
web-buckling
D
flange-buckling
P
C
G
E
b
1
F
B
E
D
C
F
Q
A
B
A
Crippling of axially loaded members
Chapter 4 Axially Loaded Compression Members
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（1） Rectangular flat-plate of simply supported on four sides
x
N
N
w
b

4.4.1 Stability of rectangular flat-plate in one-way uniform compression
a
y
simply supported rectangular flat-plate in one-way uniform compression
Chapter 4 Axially Loaded Compression Members

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the formula of calculate the critical force of the simple supported on four sides
rectangular plate:
N cr
 E t
 cr 
k
2  
t
12 1    b 
2

2

（2）Rectangular flat-plate of the simple supported on three sides and the
other side is free
According to theory analysis , for longer plate, the buckling coefficient can
be calculated by the bottom formular：
b2
k  0.425  2
a
Chapter 4 Axially Loaded Compression Members

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For long plate a
 b ，k  0.425
When plate is on elasto-plasticity process , we can use this formual to
calculate the critical force :
 k 2 E  t 
 cr 
2  
121     b 
2
In the formula： 
Et
E
Chapter 4 Axially Loaded Compression Members

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4.4.2 the flakiness ratio limit value of flange
the flakiness ratio limit value of flange should meet the formula’s demand：
b1
235
 (10  0.1 )
t
fy
λ——the maximum slenderness ratio ，whenλ<30，takeλ＝30；
whenλ>100，takeλ＝100
4.4.3 the ratio of height to thickness limit value of web
h0
235
 (25  0.5 ) 
tw
fy
h、 t w——width and thickness of web
λ——the maximum slenderness ratio ，whenλ<30，takeλ＝30；
whenλ>100，takeλ＝100
Chapter 4 Axially Loaded Compression Members

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4.4.3 sectional dimension limit value of axially loaded compression steel
tube
 235 
D

 100  
 f 
t
 y 
D ——outer diameter of steel tube
t ——wall thickness of steel tube
i y  l0 y / 
Chapter 4 Axially Loaded Compression Members

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4.5 The design of solid-web axially loaded compression members
4.5.1 Design principles
1）Equistability principle
2）Width limb and thin wall
3）Manufacture saving of labor
4） Connect convenient
4.5.2 design section
(1) Calculate the parameter
1)Assume the slenderness ratio 
2)According to the section type select  x and  y , calculate the area A 
3)Calculate the radius of gyration ix  l0 x /  and iy  l0 y / 
N
 min f
Chapter 4 Axially Loaded Compression Members

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(2) Select profile steel or initial determine the dimension of compound section
if we use profile steel, we can select it according to A 、i y、ix.
but if we use compound section, we should according to the appendix 4 to
determine b  iy /  2 、h  ix / 1 , and the same time we should accord with the
design principle .
(3)Checking calculation
1) Checking calculation of the strength :
2) Checking calculation of the stiffness :
Chapter 4 Axially Loaded Compression Members

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3) Checking calculation of overall stability :
4) Checking calculation of local stability:
(4) The relevant structure request
Program 4.1
Chapter 4 Axially Loaded Compression Members
l01
l1
l01
1 3
2
a
1
h
y
x
（

）
¦Á
x

4.6 The design of lattice form axially loaded compression members
4.6.1 The overview of lattice form axially loaded compression members
y
b
(a) Bulit-up sections(b)
For large loads it is common to use a lattice
form built-up cross section.In addition to
provideing increased cross-sectional area, the
bulit-up sections allow a designer to tailor to
providing increased cross-sectional area, the
built-up sections allow a designer to tailor to
specific needs the radius of gyration values
about the x-x and y-y axes. The dashed lines
shown on the cross sections of figure bulit-uo
sections and represent tie plates, lacing bars, or
perforated cover plates and do not contribute to
the cross-sectional properties. The functions are
to hold the main longitudinal components of the
cross-section in properties. Their functions are
to hold the main longitudinal components of the
cross section in proper relative position and to
make the built-up section act as a single unit.
Chapter 4 Axially Loaded Compression Members

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4.6.2 Overall stability of lattice form built-up sections
(1) Checking calculation of real axis overall stability
We can use formula(4.31) to check calculate the overall stability
(2) Checking calculation of virtual axis overall stability
Considerated the effect of shearing deformation, we can use this formula
calculate the overall stability.
 2 EA
N cr  2 
y
1
1
 2 EA

2
y

Chapter 4 Axially Loaded Compression Members
2
2
2
assume：





EA

oy
y

2
2





EA

oy
y

 2 EA

oy

（

）
oy——equivalent slenderness ratio of virtual axis
 ——angle of shear
Chapter 4 Axially Loaded Compression Members

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because of different columns has different  , the equivalent slenderness
ratio oy is different.
1）Double lacing bar column
2
A
oy    2

sin  cos  A1
2
y


As：   40 ~ 70
  2y  27
oy
A
A1
Derivation of double lacing bar column’s equivalent slenderness ratio oy
Chapter 4 Axially Loaded Compression Members

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2) The calculation formula of battened plate column’s equivalent slenderness
oy  2y  12
3) The calculation formula of polymelia column’s equivalent slenderness
sheet 4.7
Chapter 4 Axially Loaded Compression Members

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4.6.3 Overall stability of limbs
Lacing bar：
1  0.7max
Batten plate：
1  0.5max
and equal or lesser than 40
In the formula：
 max——the maximum slenderness ratio, when max  50，take max  50 ；
4.6.4 the calculation of lacing and batten elements
(1) Transverse shearing force
V  Vmax
Af

85
fy
235
back
Chapter 4 Axially Loaded Compression Members
（

）
V
V
N1
z
l
V
¦È

V
y
V
V
N
y
V1
x
x
y
(a)
(b)
shearing force scattergram
(c)
(a)

We often use singular set lacing bar，in order to reduce the limb’s
calculation length, we can add transverse lacing bar. And when the limbs’
space or the load are bigger, we can use cross lacing bar.
V
V
N1
N1
¦È
¦È
（

）
(2) Design of lacing bar
¦È
y
Chapter 4 Axially Loaded Compression Members
V
V1
V1
V 1=V/2
V 1=V/2
Chapter 4 Axially Loaded Compression Members

（

）
Under the transverse shearing force, the axially compression of one
lacing bar beared：
N1 
V1
n  cos 
V1 ——shearing force of one lacing bar distributed
n ——singular set lacing bar n =1 ; cross lacing bar n =2
 ——angle of inclination
Chapter 4 Axially Loaded Compression Members
1
v /2
v /2
1
l/2
1
T
1
T
d
M
l/2
（

）
v /2
l

(3) Design of battened plate
v /2
1
a/2
a
(a)
(b)
Calculation diagram of battened plate
(c)
M T 
a V1l1

2
2
Chapter 4 Axially Loaded Compression Members

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）
Shearing force：
moment ：
V1l1
T
a
M T 
a V1l1

2
2
Chapter 4 Axially Loaded Compression Members

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(4) Connecting nodes and structure request
The splicing length of the batten plate and limb usual take2mm～30mm，
in order to cut down tge splicing length, we can use weld all around of three
sides , and the same time add gusset plate.
And the lacing bar should not be smaller than ∟45×4 or ∟56×36×4. The
thickness should be thicker than 6mm. In order to add the torsional stiffness,
we should set the tabula plate.
y
x
transverse lacing bar
x
y
tabula plate
Tabula plate of bulit-up colunmn
transverse lacing bar
A
Chapter 4 Axially Loaded Compression Members

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4.6.5 Design of lattice form axilly loaded members
According to the request for utilization 、the size of axis towing force、
and the calculation length to determine the shape and the type of steels.
（1）Primary select sections
1)Assume slenderness ratio →search  x , calculate 。
A
N
x f  n
According to A primary select radius of grration i ,and so on
x
2）calculate
x  lox / ix
Chapter 4 Axially Loaded Compression Members

（

）
（2）Determine the limbs’ space
According to equistability princple, assume：oy  x
2



For lacing bar colunmn： y
oy  27
A
A1
For batten plate column：   2  2
y
oy
1
Calculate:
iy  loy / y
b  iy / 
Chapter 4 Axially Loaded Compression Members

（

）
(3) Checking calculation of sections
Checking calculation of primary sections：
①Checking calculation of strength
②Checking calculation of stiffness, for virtual axis use equivalent
slenderness ratio oy  x
③Checking calculation of overall stability；for virtual axis use
equivalent slenderness
oy  ratio
x
④checking calculation of limbs’ stability
(4)Design of nodes and structure
Program 4.2
Chapter 4 Axially Loaded Compression Members

（

）
4.7 The structure and calculation of axially loaded compression column
(1) Design princple
1) Spread the towing force accurate
2) Convenient to creation and installation
3) Economic and reasonable
(2) Connection of the beams and columns
Chapter 4 Axially Loaded Compression Members

（

）
1) Hinged joint of beam supported by column top
Cover-plate located on the column top, connected with the column by the
weld. And it should have enough stiffness，so the thickness must be between
12mm～24mm. According to the load and web, we can set stiffener under the
cover-plate.（Flash of assemble the column head ）
log-stiffener filler-plate
cover-plate
cover-plate
(a)
support-stiffener
stiffener
(b)
(c)
log-stiffener
cover-plate

Chapter 4 Axially Loaded Compression Members
supported by
column side

log-stiffener

（

）
support
(d)
(e)
(c)
Chapter 4 Axially Loaded Compression Members
dummy plate
a
1
tc
base-plate
a
（

）
boot-beam
ct

4.7.2 Column heel’s form , structure and calculation
(1) Hinged connection column heel
1) Form and structure
(a)
(b)
Chapter 4 Axially Loaded Compression Members

boot-beam
shear key
base plate
ribbed plate
boot-beam
1
a
（

）
boot-beam
b
1
sub-base
(c)
(Flash of assemble column heel) (d)
Chapter 4 Axially Loaded Compression Members

（

）
2) The calculation of base plate
①Base plate area
assume the compression distributed uniform , the area can be calculated by
this formula:
A  LB 
N
 Ao
fc
In the formula: Ｌ、Ｂ——length and width of the base plate
Ｎ ——design value of axis force
f c ——design value of concrete’s compression strength
Ao ——open area of the base plate
According to column’s dimension adjusting the length and width of the
base plate, we should make it into square or rectangle L / B  2 .
Chapter 4 Axially Loaded Compression Members

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）
② Base plate thickness
The thickness of the base plate is determined by the moment of reaction
which is supported by the base plate.
M 4    p  a2
Plate supported at four sides ：
Plate supported at three sides or two limbs： M 2(3)    p  a12
1
Jettied plate：
M  p  c2
2
In the formula:
p  N /( A  Ao ) ——force of reaction supported by the base plate
a
——short side length of the plate supported at four sides
a1 ——diagonal line length of the plate supported at three sides or two limbs
Chapter 4 Axially Loaded Compression Members

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）
Ｃ——the length of jettied
 ——coefficient，according to b / a seach sheet 4.8
 ——coefficient，according to b / a seach sheet 4.9, when b1 / a1  0.3 ,
calculate the moment according to jettied
③Calculation of weld
(2) Design of the ribbed plate, dummy plate and boot-beam
Program 4.3
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