3 compression mem

8/13/2019 3 Compression Mem

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Chapter 2: Member Design

Section 2 : Compression Members

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Failure modes

• Local buckling

• Overall yielding

• Overall buckling



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Local buckling – Slender or Not?

I and H Sections

Fig 5 and Table11

b

B

D td

T

r

Outstand

Web

y p275=ε

ε< 15T/ b

2r 0.21

120t/d

+

ε<

or conservatively

ε< 40t/d

yg

c2

pA

Fr =

Not slender if:

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Local buckling – Slender or Not?

b

Dd

B

tWeb

Flange

HF Rectangular Hollow Sections

Fig 5 and Table 12

y p

275=ε

ε< 40t/ b

2r 0.21

120t/d

+

ε<

or conservatively

ε< 40t/d

yg

c2

pA

Fr =

Not slender if:

t3B b −= t3Dd −=



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Overall yielding

Satisfied by overall buckling check

Pc = ρy Ag

Pc =ρ

cA

g

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Elastic Buckling of Columns

2

2

EI

L

π

Euler Buckling Load

Pcr =L

2

cr 2

EA p

(L/r)

π=

Buckling stress



= L / 1000

Out of straightness

δ

+260 N/mm

-125 N/mm

+55 N/mm

Rolled Section

C

T

C C

C C

T

2

2

2

CC

T

CC

T

C

T

TWeb Distribution

Welded section

Factors that affect overall buckling of columns

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Overall buckling (1)

Buckling resistance Pc

• Class 1 2 or 3

Pc

= Ag p

c

• Class 4

Pc

= Aeff

pcs

Fc < Pc



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Factor Influences Overall buckling

Practical Region Elastic Failure

Elastic Critical

L/ry

p

Euler curve

p = E

2

2

E

Plateau dueto yielding

p y

c

1. Effective Length of Column

2. Residual Stresses

3. Member initial out-of-straightness

4. Types of cross section5. Local buckling of component plate

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Buckling Resistance

Buckling resistance Pc

• Class 1 2 or 3

Pc

= Ag p

c

• Class 4

Pc

= Aeff

pcs



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Effective Area, Aeff

tt

20tε

Rolled I Section Hot finished RHS

20t ε

20t ε

20t ε

20t ε

20t ε

For sections which are slender

under axial load (see Fig. 8 in code)

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Overall Buckling (2)

Slenderness

• Slenderness = (λ)

• Class1, 2 and 3 cross sections

λ = LE/r

• Class 4 (slender) cross sectionsλ = LE/r (Aeff /Ag)

0.5

• The value of the compressive strength pc isobtained from table 24



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Strut Curves

• Four TABLES 24a, 24b, 24c, 24c

• Corresponding to 4 strut curves

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UK Strut Curves for S275 Steel

0204060

80100

200

260

280

10 30 50 70 90 110 130 150

pc

λUK a=2.0 UK a=3.5

UK a=5.5 UK a=8.0

HF SHS, I (x-x)

H (x-x), I (y-y), welded box,

Welded I and H about x-x

H (y-y),

welded H or I about (y-y)

Colded formed SHSThick sections



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Part of Table 24

1) Values of pc in N/mm2 for strut curve a

25824924023122242

26025124223322340

26725824823922930

27026125124123225

27326425424423420

27526525524523515

275265255245235

S275

Steel Gradeλ

Table 24 Compressive strength pc

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Table 23 Allocation of Strut Curve

(d)(c)>40mmH Section

(c)(c)Angles, channels and T’s

(c)(b)<40mmH Section

(c)(b)>40mmI Section

(b)(a)<40mmI Section

(c)(c)CF SHS

(a)(a)HF SHS

y-yx-xmm

Axis of BucklingMaximum

ThicknessType of section



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Table 22 Effective Length

PositionPosition Position

Position Position PositionPosition

Direction DirectionDirection

Direction Direction Direction

1.0 L 0.85 L 0.7 L 2.0 L 1.2 L

Position

Restraint

Restraint

PracticalL E



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Table 22 Effective length

1.0L Not restrained in direction at

either end

0.85LEffectively restrained in directionat one ends

0.85LPartially restrained in direction at

both ends

0.7LEffectively restrained in direction

at both endsEffectively held in

position at both

ends

LERestraint by the other parts of the structure



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Typical effective length

factor for use in

column design

=KL

K

L

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Effective length factor for continuous columns based on

stability analysis



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Effective length factor for continuous columns based on

stability analysis

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Special Considerations

• Members in lattice frames and trusses Table 25

(angle-, tee-, channel-sections)

• Members in continuous construction Appendix E



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Summary of design procedure (2)

6. For members in lattice frames and trussesdetermine λ : Table 25

7. For class 4 (slender sections) calculateλ = LE/r (Aeff /Ag)

0.5: Clause 4.7.4

8. Select appropriate strut curve according to sectionshape and axis of buckling: Table 23

9. Obtain the compressive strength from theappropriate strut table and the appropriate valueof design strength from Table24

10.Calculate the compressive resistance from the

product of the area (effective area for slendersections) and the compressive strength: Clause4.7.4.

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Workshop Example:356 x 368 x 129 UC, Grade S275

T = 17.5 mm, Ag

= 164 cm2, r x

= 15.6 cm, r y

= 9.43 cm

Flange thickness T > 16 , py= 265 N/mm2

The section is NOT slender, therefore Pc= A

g p

c

For buckling about the x-x axis, use strut curve (b)

For buckling about the y-y axis, use strut curve (c)

Slenderness λx=38.5

λy= 63.6

For λx

= 38.5 and py

= 265 pcx

= 243 N/mm2

For λ y = 63.6 and py = 265 pcy = 188 N/mm2

pc

is the lesser value of pcx

or pcy

Pc

=164 x 102 x 188/103=3083 kN

Since Fc < Pc

i.e 2500 < 3083

Therefore, section OK

6 m

3 5 6 x 3 6 0 x 1 2 9 U C

2500 kN



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Pin-connected space frame

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Reading assignmentsBS 5950:Part 1Code:

Clause 4.7

Reference : Chapter 2: Section 2

3 compression mem

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