2.2020 Week 2 Steel compression PDF

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Warning: Error during font loading: XMLHttpRequest is not definedSteel design toAS 4100Design of CompressionUniversity of South AustraliaMembersTypes of compression MembersColumnsStruts / Braces◼ Ridge ◼ EavesTruss members◼ Top & bottom chord ◼ Web membersFailure ModesInelastic buckling◼ buc...

Description

Steel design to AS 4100 Design of Compression Members 受压构件设计

University of South Australia

Types of compression Members Columns Struts / Braces ◼ ◼

支柱／ 支撑

网格

屋檐

桁架

Truss members ◼ ◼

上弦和下弦

Top & bottom chord Web members 腹杆

失效模式

Failure Modes Compressive yielding ◼

粗壮的柱子会这样失败

very stocky columns fail this way (L /r 250) theoretical load is given by the Euler formula 理论载荷由欧拉公式给出 ideal columns with axial loading 轴压理想柱

2

Pc =

𝜋 ∗ 𝐸𝐼

𝑘𝑒 ∗ 𝑙

2

Failure Modes Inelastic buckling ◼

非弹性屈曲

buckling load of real columns < Euler load

实柱的屈曲荷载小于欧拉荷载

Initial camber ( L/1000 ) 初拱度 residual stress from 残余应力 ◼ ◼ ◼ ◼

rolling fabrication cold straightening hot dip galvanizing

滚压 装配 冷矫直 热浸镀锌

滚压是一种压力光整加工，是利用金属在常温状态的冷塑性特点的无切削的塑性加工方法。 冷矫直是在温度应低于２００℃对金属塑性加工产品的形状缺陷进行的矫正。 热浸镀锌是将已清洗洁净的铁件，经由助镀剂的润湿作用，浸入锌浴中，使钢铁与熔融锌反应生成一合金化的皮膜。 残余应力：消除外力或不均匀的温度场等作用后仍留在物体内的自相平衡的内应力。

轴向压缩力

Compression Design to AS4100 Section 6

Axial compressive force N* must be ≤ Section capacity Member capacity

截面承载力

构建承载力

Ns and Nc In Plane (Major Axis) Nc Out of Plane (Minor Axis) 标称值

where N and Nc & Nc are nominal values f = 0.9 for compression members (AS4100 Table 3.4) 标称值：用以标志或识别元件、器件或设备的适当近似值

Section Capacity 仅考虑屈服即材料失效

considers yielding ie material failure ONLY allows for local buckling 允许局部屈曲

• • •

k = 0 OneStee UCs (Table 16 1 ) 澳大利亚钢铁和矿业公司ＯｎｅＳｔ ｅｅｌ k  0 fo OneStee UBs (Table 14 1 ) k calculate fo buil up sections (c ) • plate elemen slenderness effective width

N = f k A fy cl 6.2.1

板单元长细比、有效宽度

Local buckling

局部屈曲

Local buckling (from The Behaviour and Design of Steel Structures to AS 4100, 3rd ed, Trahair & Bradford, Fig. 4.1)

Member Capacity

非弹性稳定极限

考虑整体屈曲

Considers overall Buckling Nc = f ac Ns cl 6.3.3 N c = f ac N s • ac & ac  1 • •

membe capacity is the critica check shows the efficiency o the section

欧拉公式（弹性稳定 极限）

depends on /r, type of section, grade of steel 钢种 from 5 curves based on experimental data 基于实验数据的５条曲线

Member Capacity Things you need to know to determine the Member Capacity: • The type of membe – how it was made • Form factor kf OneSteel • Ne Cross sectional Area, An Handbook for open • Yield strength fy sections • Radius of gyration r and ry • Effective lengths for X and Y axis, Le and Ley 净横截面积

回转半径

Slenderness reduction factor a ) 长细比折减系数

Once you have determined Le and Le , you need to calculate :  lex  nx =   k f  rx 

fy 250

&

 ley  ny = r  y

  kf  

fy 250

Member section constant a )

Section type

Method of manufacture Material Thickness

Slenderness

Slenderness reduction factor a ) Residual stress

Class Activity What is the effective length about x & y axis ? Le = ? Le = ?

Class Activity – Group #1 墙竖框

Wall Mullion (UB) Lex = ? Ley = ?

Class Activity – Group #2

Compression Strut (SHS) Lex = ? Ley = ?

Class Activity – Group #3 Truss Web Member (RHS) Lex = ? Ley = ?

Class Activity – Group #4 阳台柱

Verandah Column (CHS) Lex = ? Ley = ?

Class Activity – Group #5 门式刚架棚柱

Portal Frame Shed Column (UB) Lex = ? Ley = ?

Class Activity – Group #6 Sign Post (CHS) Lex = ? Ley = ?

路标

Class Activity – Group #7 屋顶柱

Roof Column (SHS) Lex = ? Ley = ?

Class Activity – Group #8 临时预制板支撑

Temporary Precast Panel Brace (SHS) Lex = ? Ley = ?

Lex

Example A1 UC properties Table 15 and 16 Onestee Design a column (UC grade 300Plus) for the following loads.

Ley

Ley

Design examples

N* = 1875kN

Lex = 8m (major axis) Ley = 4m (minor axis) Need to find fNu ≥ N* f = 0.9 for axial compression

First step is to try to get a feel for the size of section. We can use Euler’s Buckling formula to find appropriate Ix and Iy values…… 欧拉屈曲公式

Example A1 (continued)

Try a section based on Euler buckling load => Set Pcr = N* f For Ix → Pcr = p2E Ix /Lex2 Ix  Pcr Lex2/( p2E) Ix  1875x103x80002 / (0.9xp2x2x105) = 67.6 x106 mm4

Don’t forge f For Iy → Pcr = p2E Iy /Ley2

Iy  Pcr Ley2/(p E)

Iy  1875x103x40002 / (0.9xp2x2x105) = 16.9 x106 mm4

Now we look for a column which has at least these values of Ix and Iy ……………..

Example A1 (continued)

Look at UC Section Property Table 16….. Try a 250 UC 72.9→ Ix = 114 x 106 mm4, Iy = 38.8 x106 mm4 (Table 16) kf = 1.0 An = Ag = 9320 mm2

N = k A fy 标称截面承载力

Nominal Section Capacity f Ns = 0.9 x 1 x 9320 x 300 N = 2516kN

2516kN > 1875kN needed, so fNs is OK but……….

Example A1 (continued)

Remember that ………..

Member capacity N is Critical

N = f a Ns ….we need an alph c value! For standard member desig ac depends on the modified member slenderness n Cl 6.3.3 修正构件长细比

 le  n =   k f r

fy 250

Example A1 (continued)

Next Step is to check which value of the modified slenderness ratio (x or y axis) has the largest number as this will be the critical case.

 lex  nx =   k f  rx 

fy 250

300  8000  = 79 =  1 250  111 

 ley  ny =   ry

  kf  

fy 250

300  4000  = 68 =  1 250  64.5  长细比

Use the maximum of these two slenderness ratios ie The critical modified member slenderness nx is 79 (x axis)

αc alph c values p. 8 87

ab = 0 (Table 6.3.3(1)) and kf = 1 (Onesteel)

Interpolate value of alph c

We want value for 79 1. subtract lower value from higher value = 0.715-0.681 = 0.034 2. how many divisions are there = (75→80) = 5 3. divide answer by 5 = 0.034/5 = 0.0068 per division

4. Which number is closer to what we need? → 80 (79) 5. how many divisions do we need? = 80 -1→ 1 division 6. is the number in the increasing or decreasing direction? → increasing (+)

7. calculate 0.681 (+) 0.0068 = 0.6878 ~ 0.688 = αc

Design examples to AS4100 Example A1 (continued)

acx = 0.688 fNc= facNs = 0.9 x 0.688 x 2796 / 1000 = 1731 kN  N* 不充分

Member capacity is inadequate, try a larger 250 UC 89.5 section nx = 75.6 acx = 0.71

fNc= 0.9 x 0.71 x 11400 x 280 / 1000 = 2040 kN >N* Hence satisfactory => Adopt 250UC89.5 column 令人满意的

Design examples to AS 4100 Example A2

E

N*t = 30kN

N*c = 40kN

For members in a triangulated structure 三角结构 Le = L A

AB and BD designed for compression Members are weaker about the y axis

D

N*c = 60kN

B

4m

N*c = 30kN

2@ 3m

C

P* = 40 kN

More than adequate capacity in tension 超过足够的张力能力

Example A2 continued

Try a 90x90x8 EA (Grade 300) section Onestee Table 23 p19 A = A = 1350 mm , r = 3 8 mm r = 1 6 mm, k = f = 320 MPa Onestee Table 24 20

Membe AB ◼

N = 6 kN Compression)

Slenderness ratio Le ry

= 3000/17.6 = 170

320  3000  1 = 193  250  17.6 

ny =  ◼

n = 193

ac = 0.179

(don’t forge α is different here = 0.5 (column 4))

f Nc = f ac k A f = 0.9 x 0.179 x 1 x 1350 x 320 / 1000 = 69.6 kN > 60kN

Design examples to AS4100 Example A2 continued

Member BD

N

= 40kN Compression)

Slenderness ratio le ry = 4000/17.6 = 227 ny = 257 ac =0.107 f N c = f a c k A fy = 41.6 kN > 40 kN

Capacity still OK assuming 18mm bolt holes – see Cl. 6.2.1

Key points 1 – Compression members Section capacity N = “squash load ◼

maximum load capacity

Member capacity Nc & Nc ) accounts for buckling behaviour k form factor, allows for local buckling (tabulated in Onestee tables for standard sections) αc & αc accounts for slenderness

Key points 2 – Compression members Slenderness depends on: ◼

◼ ◼ ◼ ◼

Effective length about each axis, which depends on end connections and intermediate restraint conditions Radius of gyration about each axis (tabulated in Onestee ) Form factor kf Residual stresses during manufacture, accounted for with αb Yield stress of the section fy

The critical compression member for a given load will be the one with the highest slenderness, i.e. the most slender and hence most prone to buckling....