1. Summary：
4.1 The application and section types of axially loaded compression members
4.2 The strength and stiffness of axially loaded compression members
4.3 The overall stability of axially loaded compression members
4.4 The local stability of solid-web axially loaded compression members
4.5 The design of solid-web axially loaded compression members
4.6 The design of lattice form axially loaded compression members
4.7 The structure and calculation of axially loaded compression column head and heel

2. 4.1.1 The application of axially loaded compression members
4.1 The application and section types of axially loaded compression members
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 by slenderness ratio.
Supporting capacity limit state requires for strength, Overall stability, local stability. + +
The application of axially loaded compression members

3. (a) structure steel sections (b) solid-web (c) lattice form
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) structure steel sections (b) solid-web (c) lattice form

5. 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：
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 :

6. ——tensile strength or compressive strength
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:
——design value
——net section area
——tensile strength or compressive strength
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7. 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:
—— the maximal slenderness ratio
—— members' effective length
—— radius of gyration
—— allowable slenderness ratio
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8. 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.

9. 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 .

10. Buckling modes of strut :
bending-buckling torsional-buckling bending and torsional-buckling

11. 4.3.3 Crippling load of axially loaded compression members
(1)Critical force of bending-buckling
①elastic bending-buckling——Euler’s formula
critical state of hinged-hinged axially loaded compression bar

12. derivation process of critical force
Critical stress：
derivation process of critical force

13. is critical slenderness
Elastic bending assumption of axially loaded compression members requires: critical stress must be less than material’s proportional limit, that is:
solve：
is critical slenderness

14. 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：

15. （2）critical force of torsional-buckling
the torsional-buckling critical force computational formula of the axially loaded compression members is:
the first part of the formula is warping torsion, it is related to the length of the members, the second part is free torsion.

16. In order to facilitate the application, we can convent the torsional-buckling calculation of the critical force into bending-buckling in the form of Euler‘s formula, that is ： 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。

17. （3）critical force of bending and torsional-buckling
The critical force of bending and torsional-buckling calculation formula is：
As to facilitate the application ，we can write the formula in the form of Euler’s formula, that is:

18. Equivalent slenderness ratio is：
——distance from section centroid to shear center
——radius of gyration to shear center
——slenderness ratio of bending-buckling to symmetry axis
——equivalent slenderness ratio of torsional-buckling

19. 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.

20. 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

21. Column curves

22. (3)Calculation formula of overall stability
According to the priciple of the critical force of overall stability should be less than axial force ,at the same time considerate the subentry coefficient ,that is: or
back
——design value of axial force
——gross section area
——design value of compression strength
—— , overall stability coefficient

23. b t 、 ——effective length
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 y b t x y
、 ——effective length

24. （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. x y
Practical application often use simplified calculation formula （sheet 4.5.3)

25. 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
flange-buckling
Crippling of axially loaded members

26. （1） Rectangular flat-plate of simply supported on four sides
Stability of rectangular flat-plate in one-way uniform compression
（1） Rectangular flat-plate of simply supported on four sides
simply supported rectangular flat-plate in one-way uniform compression

27. the formula of calculate the critical force of the simple supported on four sides rectangular plate:
（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：

28. For long plate ，
When plate is on elasto-plasticity process , we can use this formual to calculate the critical force :
In the formula：

29. 4.4.2 the flakiness ratio limit value of flange
the flakiness ratio limit value of flange should meet the formula’s demand：
λ——the maximum slenderness ratio ，whenλ<30，takeλ＝30；whenλ>100，takeλ＝100
the ratio of height to thickness limit value of web
、 ——width and thickness of web
λ——the maximum slenderness ratio ，whenλ<30，takeλ＝30；whenλ>100，takeλ＝100

30. 4.4.3 sectional dimension limit value of axially loaded compression steel tube
D ——outer diameter of steel tube
t ——wall thickness of steel tube

31. 4.5 The design of solid-web axially loaded compression members
Design principles
1）Equistability principle
2）Width limb and thin wall
3）Manufacture saving of labor
4） Connect convenient
design section
(1) Calculate the parameter
1)Assume the slenderness ratio
2)According to the section type select and , calculate the area
3)Calculate the radius of gyration and

32. (2) Select profile steel or initial determine the dimension of compound section
if we use profile steel, we can select it according to 、 、 .
but if we use compound section, we should according to the appendix 4 to determine 、 , 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 :

33. Program 4.1 3) Checking calculation of overall stability :
4) Checking calculation of local stability:
(4) The relevant structure request
Program 4.1

34. 4.6 The design of lattice form axially loaded compression members
4.6.1 The overview of lattice form axially loaded compression members
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.
Bulit-up sections

35. 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.

36. assume： then： ——equivalent slenderness ratio of virtual axis
——angle of shear

37. because of different columns has different , the equivalent slenderness ratio is different.
1）Double lacing bar column
As：
Derivation of double lacing bar column’s equivalent slenderness ratio

38. 2) The calculation formula of battened plate column’s equivalent slenderness
3) The calculation formula of polymelia column’s equivalent slenderness
sheet 4.7

39. back 4.6.3 Overall stability of limbs Lacing bar： Batten plate：
and equal or lesser than 40
In the formula：
——the maximum slenderness ratio, when ，take ；
the calculation of lacing and batten elements
(1) Transverse shearing force
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40. shearing force scattergram

41. (2) Design of lacing bar
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.

42. Under the transverse shearing force, the axially compression of one lacing bar beared：
——shearing force of one lacing bar distributed
——singular set lacing bar =1 ; cross lacing bar =2
——angle of inclination

43. (3) Design of battened plate
Calculation diagram of battened plate

44. Shearing force：
moment ：

45. (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.
Tabula plate of bulit-up colunmn

46. 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 , calculate 。
According to primary select radius of grration ,and so on
2）calculate

47. （2）Determine the limbs’ space
According to equistability princple, assume：
For lacing bar colunmn：
For batten plate column：
Calculate:

48. Program 4.2 (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
③Checking calculation of overall stability；for virtual axis use equivalent slenderness ratio
④checking calculation of limbs’ stability
(4)Design of nodes and structure
Program 4.2

49. 4.7 The structure and calculation of axially loaded compression column head and heel
(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

50. 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 ）

51. 2) Hinged joint of beam supported by column side ①Ssolid-web column’s column side hinged joint
②Lattice form built-up column’s column side hinged joint

52. 4.7.2 Column heel’s form , structure and calculation
(1) Hinged connection column heel
1) Form and structure

53. (Flash of assemble column heel)

54. 2) The calculation of base plate ①Base plate area
assume the compression distributed uniform , the area can be calculated by this formula:
In the formula: Ｌ、Ｂ——length and width of the base plate
Ｎ ——design value of axis force
——design value of concrete’s compression strength
——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

55. In the formula: ② Base plate thickness
The thickness of the base plate is determined by the moment of reaction which is supported by the base plate Plate supported at four sides ：
Plate supported at three sides or two limbs：
Jettied plate：
In the formula:
——force of reaction supported by the base plate
——short side length of the plate supported at four sides ——diagonal line length of the plate supported at three sides or two limbs

56. Program 4.3 Ｃ——the length of jettied
——coefficient，according to seach sheet 4.8
——coefficient，according to seach sheet 4.9, when ,
calculate the moment according to jettied
③Calculation of weld
(2) Design of the ribbed plate, dummy plate and boot-beam
Program 4.3