Intersections V04 No03 12

8/9/2019 Intersections V04 No03 12

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Article No.12, Intersections/Intersecii, Vol.4, 2007, No.3, Structural Mechanics 129

Structural Mechanics

Composite columns with circular section.

Shear transfer mechanism

Petru Moga, tefan I. Guiu and Ctlin Moga

Faculty of Civil Engineering, Technical University, Cluj-Napoca, 400027, Romania

Summary

The mechanisms of the shear transfer over the interface between the circular steeltube and the concrete core as well as the design of the shear connection arepresented in this paper.

A numerical example for the evaluation the longitudinal shear stresses over theinterface between structural steel and concrete is also presented here.

KEYWORDS: composite columns, shear stresses, shear transfer, connection.

1. INTRODUCTION

A composite column may either be a concrete partially or completely encased

section or a concrete- filled section, Figure 1.

Figure 1

In case of the concrete-filled hollow steel sections, there are three mechanisms


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P. Moga, t. I. Guiu, C. Moga



which are often referred to as the natural bond, by which shear stresses can betransferred over the interface between the steel tube and the concrete core, theseare: adhesion, interface interlocking and friction, Figure 2.

Figure 2

If the natural bond is not enough to achieve the required shear resistance there is

the possibility of using mechanical shear connectors, the most widely used typesbeing headed studs and the shot-fired nails.

The shear stresses, which take place on the interface steel-concrete, are shown inFigure 3.

Figure 3

Figure 4

Taking into account that the structural

steel and the concrete have differentmechanical characteristics, the concrete core is

transformed into an equivalent steel sectionusing the modular ratio n.

The mechanical model is presented inFigure 4.

The equivalent in the steel area of the

concrete core will be:


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Composite columns with circular section. Shear transfer mechanism



n

AA cechiv.c = (1)

where the modular ratio can be taken as:

cm

ai

EE2n2n = (2)

2. TANGENTIAL STRESSES

The state of tangential stresses caused by the shear force T=VSd.c is presented in

Figure 5, where xz is the shear stress given by the Juravski formula:

y

y

zxxzbI

TS== (3)

where:

- Sy section modulus of the slipping portion referred to neutral axis:

( )322y zR3

2S = (4)

- b width of the section at the distance z from the neutral axis:

22zR2b = (5)

- Iy moment of inertia of the cross-section:

64D

4RI

44

y == (6)

By replacing the above parameters into relation (3) it results:

y

22

xzI3

)zR(T = (7.a)

=

2

2

xzR

z1

A

T

3

4 (7.b)


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Figure 5

3. LONGITUDINAL SHEAR BETWEEN

STEEL TUBE AND CONCRETE

The longitudinal shear stress l , equal with the normal component n , of the shear

stress xz , will be:

R

z

R

z1

A

T

3

4sin

2

2

0xzln

=== , )zR(z

R

1

A

T

3

4 223l

= (8)

The maximum value of l and n is obtained from the condition:

3

Rz0)z(f

' == .

It results:


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A

T51.0

A

T

39

8max.n ==

The longitudinal and the normal shear stresses variation are presented in Fig 6.

Figure 6

The sum of n divided by a quarter of the interior surface of the steel tube will give

the value of the longitudinal tangential stress on the interface between steel pipeand concrete core:

RR

T10689

R2

R3

24

R2

4

c

4

c

max.n

c

n

f=

=

=

(9)

where:

Rc- the interior radius of the steel tube;

R - the radius of the concrete core taking into account the modulus ratio andresults from:

n

RRR

n

RA c

22c

echiv.c === (10)


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By replacing the value R in the relation (9) it is obtained:

Td

n275.0T

R

n10689

2c

2c

4f ==

(11)

The longitudinal shear force for the entire length of the composite column andrespectively for a half of cross-section, will be given by:

lTd

n43.0ld

2

1L

c

cfl == (12)

If the natural bond is not enough to achieve the shear resistance, it will benecessary to use mechanical shear connectors and their number will result from thecondition:

Rd

l

P

L2N

(13)

In accordance with Eurocode 4, the design value of the longitudinal shear strength

Rd in case of a concrete filled circular hollow sections is Rd =0.55 N/mm2.

From the relation (12) the maximum value of the shear force when it is notnecessary to use mechanical shear connectors will result:

n

d2T

2c

max= (results T in [N] for dcin [mm] ) (14)

It is necessary to underline that the shear force T represents the part of the shearforce, taken over by the concrete component of the composite column, which canbe evaluated with the relation:

SaSd

cvav

cvSdSc VV

AA

AVV =

+= (15)

where: aav A2

A

= andn

AAA cechiv.ccv ==


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4. NUMERICAL EXAMPLE

For a composite steel-concrete column the longitudinal shear stress over theinterface between the circular steel tube and the concrete core is evaluated.

4.1. Design data

Composite column cross-section and loading (Figure 7)

Figure 7

Cross-section characteristics:

Circular pipe: 325 10 mm

Steel: S 355:

- fy=355 N/mm2

- Ea= 210 000 N/mm2

- Aa=99 cm2

Concrete: C 25/30:

- fck = 25 N/mm2

- Ecm = 30 500 N/mm2

-2

c cm731A =

4.2. Longitudinal shear

Modulus ratio n:

8.13E

E2n2n

cm

ai ===

Equivalent in steel area of the concrete core:

2

echiv.c cm538.13

731A ==

Shear force of the concrete component:

2

aav cm6399

2

A

2

A ===


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kN55.365363

5380VSc =

+=

The shear force, which can be achieved by the natural bond:

kN55.36VkN50108.13

3052n

d2T Sc32

2

cmax =>===

,

so it results that are not necessary mechanic shear connectors.

The longitudinal shear stress:

2

Rd

2

22

c

f mm/N55.0mm/N40.055036305

8.13275.0T

d

n275.0 =

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Intersections V04 No03 12