Composite Sections 1

1

CVG4143 References

1

Handbook Text

Composite Design

• Chapter 6 in TextChapter 6 in Text• Clause 17 in Handbook

CVG4143 Composite Design 2

Conceptual IntroductionConcrete vs. Steel

• Plain ConcretePlain Concrete– Strong in Compression– Weak in Tension

• Structural Steel– Strong in Tension

k i i


– Weak in Compression• Local Buckling• Overall Buckling• Lateral Torsional Buckling

Composite Sections

• Main Idea:Main Idea:

– Use Concrete to carry Compression– Use Structural Steel to carry Tension

This is the philosophy of a “composite section”


of a composite section

2

Composite Beam

concreteA

structural Steel

AComposite Beam

concrete


structural Steel

Section A-A

Composite Action

Introductory Examples:y pExample 1:Given:

A simply supported beam has a 10 m span is subject to a factored load P at mid-span. Cross-section is 100 mm wide x 300 mm deep. Material is Fy=300 MPa


Required:– Calculate maximum load P based on

Elastic Flexural Resistance– Sketch the deformed configuration of

the beam under load P

Composite ActionExample 1-Solution:

1 Based on Elastic Analysis (only1. Based on Elastic Analysis (only outermost fibres are assumed to yield) , Elastic Flexural Resistance

2. External Moment =

2

26

6100 3000.90 300 10 405

6

rx x y ybhM S F F

kNm

φ φ

−

= =

×= × × =

10PL P


3. Equate external moment to flexural resistance

( )10 2.54 4fx

PL PM P kNm×= = =

2.5 405 162rfx x P kNm PM kNM= ⇒ = ⇒ =

Composite Action

Deformed Configurationg

P

P/2 P/2


3

Composite Action

Example 2:A h i t ll li th bAssume we horizontally slice the beam cross section into two pieces (along the NA)

1. What would be the maximum factored load P that can be carried by a single beam?

2. What would be the maximum factored load P that can be carried by the system?


Sketch the deformed shape in both cases

Composite Action

Example 2-Solution for Single Beamp gFor a single piece:

Resistance dropped to 25% of that

( )( )2

1/2

26

26

100 1500.90 300 10 101.256

x y yrx

b hM S F F

kNm

φ φ

−

= =

×= × × =

( )1/2 (1/2) (1/2)2.5 101.25 40.5fx rx P kNm PM M kN⇒ == ⇒ =


Resistance dropped to 25% of that determined in Example 1

P

P/2 P/2

Composite Action

Example 2-Solution for the SystemFor both pieces (system), deformed configuration is

Notes1. Slippage takes place between the underside of the top beam and the top of the bottom beam2 Since there are no gaps between both beams they

P

P/2P/2


2. Since there are no gaps between both beams, they have the same deflection curve , i.e., Also,

3. Since both beams have identical cross-sections, their moment of inertia is identical. Also they are made of the same material, thus they have the same moment Young Modulus. We have

t bv v='' ''t bv v=

'' ''t t t b b bE I v E I v=

Composite Action

We have just demonstrated, that for j ,the given problem, as the system deforms, the internal moment carried by both beams is equal

4. Thus, when the top beam attains its elastic flexural resistance, the bottom beam will also attain the same

rt rbM M=


beam will also attain the same flexural resistance. (This would not be the case if we had sliced the original beam into two unequal cross sections)

4

Composite Action

5. The resistance of the system is5. e es sta ce o t e syste sthe sum of the resistances of the two beams

22 101.25 202.5

rs rt rb rtM M M MkNm

= + =

= × =

2(1/2) 2(1/2)2.5 202.5 81fx rsM M P kNm P kN= ⇒ = ⇒ =


The capacity of the system of sliced beams is twice that of a single beam, but half that of the unsliced beam in Example 1

Composite Action

Q: How can we recover the full Qresistance attained in Example 1?

A: By reconnecting both pieces at the interface, we are able to recover


a) the full original capacity (if connection is fully effective) or b) part of it (when connection is partially effective)

Composite Action

If we fully reconnect the two pieces together, the deformed shape of the composite system is

NoteThe deformed configuration of the system is identical to that in Example 1. Thus, capacity of the composite system is no different from h f l 1

P

P/2 P/2


that of Example 1

How much did we gain due to composite action in this problem? The capacity of the composite system doubled

What is a full Connection?

A full connection:A full connection: – Prevents relative slip between the

two connected pieces– Forces the plane section of the

whole system to remain plane after deformation (as for the case of a single beam-Example 1)


5

Composite Action

• To think about:To think about:

– How to modify the above solutions to calculate the plasticflexural resistance of the systems?


Types of Composite Sections

1. Solid Slabs (our focus)1. Solid Slabs (our focus)

2. Ribbed Slabs with ribs parallelto the beam

3. Ribbed Slabs with ribs


perpendicular to the beam

Effective Slab Width

• The portion of the slab width thatThe portion of the slab width that can be considered to act with the structural steel section is called the effective slab width b

Bb


Effective Slab Width

Determine Effective Slab Width “b”

Case 1: (central beam)Slabs extending on both sides of steel beam, b = lesser of (Span/4, centre to centre distance between steel beams)

Case 2: (edge beam)Slabs extending on one side of the steel beam b =


flange width of steel section bfl +the lesser of (Span/10, ½ clear distance to adjacent steel beam)

(Clause S16-01 Cl. 17.4)

6

Definitions for determining Effective Slab width “b”

For central beam,

Span

,Slab extends from both sides

Centre to centre Distance

flange width of steel section b

clear distance to adjacent beam


SupportSupport For edge beams,

Slab extends from one side

Methods of connecting solid slabs and steel

• Text Fig. 6.1 – P 156 (9th ed.)Text Fig. 6.1 P 156 (9 ed.)

1. End Welded Studs(most common-our focus)


2. Channel Connectors

Fundamental Concepts

• At any section of a composite (or y p (plain) beam:

sum of the internal axial forces

(N internal)=

internal normal force


internal normal force induced by loads

(N external)=

zero -for a beam

Fundamental Concepts

• Also,Also,

Sum of the moments of Internal forces

(M internal)=


Internal moments Induced by loads

(M external)

7

Components of Composite Sections

Cr‘

structura l steel

Shear studs

Tr

concrete slabCr

Vh

Vh


sectionof zeromoments

sectionof maximummoment

Preliminaries

• It is of interest to calculate theIt is of interest to calculate the capacity of each of the three components

– Compressive capacity of concrete slab


– Shear capacity of studs

– Tensile capacity of structural steel

Preliminaries

• Maximum compressiveMaximum compressive resistance of the effective width of the concrete slab

• Maximum tensile resistance of the structural steel section

1max ' 'r c cC f btφ α=

max T A Fφ=


• Sum of the factored resistances of shear connectors at the interface (S16-09 Cl. 17.7)rQ

max r s yT A Fφ=

Important Note

• The capacity of the concreteThe capacity of the concrete slab, studs, and structural steel is in general different from the actual internal forces induced in each of them.

• This fundamental concept is


• This fundamental concept is best explained using the following weakest link analogy

8

The weakest link analogy

concretecompressivecapac ity “Cr’max”

shear studs Qr

Struc tura l

sectionof zero

sectionof maximum

t

structura l steel

Shear studs

Tr

concrete slabCr

Vh

Vh

‘

rQmax 'rC


Struc tura l Steel SectionTensile Capacity“Tr max”

Loading

moments moment

max rT

The ‘weakest link’ analogy

• Each of the three rings has aEach of the three rings has a different capacity

• However, under any applied load, all three rings are subject to the same load


• Depending on the capacities of the rings, there are three conceivable failure scenarios

The ‘weakest link’ analogy

Possible failure ScenariosPossible failure Scenarios

Vh= max Tr

Cr’ =max Tr

Cr ‘=Qr

Tr = max Cr’

Vh = Cr’max


max Tr weakestFull interactionNA lies in concrete

Qr weakestPartial interactionNA lies in steel &NA lies in concrete

Tr = Qr

max Cr’ weakestFull interactionNA lies in steel

r

Classification of Composite Sections

• Case 1: S16-09 Cl.17.9.3 (a)( )[max Tr is the weakest link]

– Full shear connection at the concrete steel interface [Qr does

max max 'r rT C≤

rr QT ≤max


[Qr not govern the design]

– Plastic Neutral axis lies in the slab

9


(cont’d)• Case 2: S16-09 Cl.17.9.3 (b)Case 2: S16 09 Cl.17.9.3 (b)

[ is the weakest link]

– Full shear connection [Q does not

max ' maxr rC T≤

max 'r rC Q≤

max 'rC


Full shear connection [Qr does not govern design]

– Plastic Neutral axis in steel section


(cont’d)• Case 3: S16-09 Cl.17.9.3 (c)( )

[Qr is the weakest link]

– Partial shear connection [shear i f d d i ]

max 'max

r r

r r

Q CQ T

≤≤


capacity of studs governs design]

– Two Plastic Neutral Axes, one in steel and the other in Concrete

Case 1- Internal Forces

Internal Force DiagramInternal Force Diagram

NA

rT

'rC

b t

'ed

ca 1β=

c

'1 cc fφα


yFφ

Composite Section Neutral Axis in Slab

Case 1-Compressive Force

Compressive force in concrete block

= specified concrete strength

1' 'r c cC f baφ α=

1 0.85 0.0015 ' 0.67cfα = − ≥

'cf

0.65cφ =


specified concrete strengthb = the effective slab widtha = depth of concrete block

cf

10

Case 1-Equilibrium of Internal Forces

• Tensile Force (structural steel)Tensile Force (structural steel)

• External applied axial force = net internal axial force=0

ysr FAT φ=

0 'r rT C= −


• Solve for a

1 's y

c c

A Fa

f bφφ α

=

Case 1Neutral Axis Depth

Note: Concrete block depth a Neutral Axis (NA) depth cthey are related through

(Cl. 10.1.7 in A23.3-04)

≠ca 1β=

1 0.97 0.0025 ' 0.67cfβ = − ≥


Fix Figures 6.2 and 6.3 in “Limit States Design in Structural Steel” to reflect this observation

Case 1 – Resisting Moment

• Internal moment armInternal moment arm

• Resisting moment

'2 2d ae t= + −

' ' 'M C e T e= =


'rc s yM A F eφ=rc r rM C e T e= =

Case 1 - Horizontal Shear Force

• A horizontal shear force needsA horizontal shear force needs to be transferred at the concrete-steel interface

• This force ensures proper interaction between the two components (concrete and steel)


components (concrete and steel)

• For case 1, the horizontal shear force ysh FAV φ=

11

Q: When to design according to the Case 1

Procedure?

• A: When steel section is the weakest link, i.e. when

max max 'r rT C≤

rr QT ≤max


When to design according to the Case 1

procedure?

Given:

Required: Determine whether or not case 1 procedure should be used

Hint: put each of the two inequalities above in a mathematical form,

rscyc QtbdAFf ,,,,,,,,' φφ


above in a mathematical form, suitable for design and/or computer code (refer to your notes to do so)

When to design according to case 1

procedure? Determine resistances

If both of the following two conditions are met

1max ' 'r c cC f btφ α=

ysr FAT φ=max

1 0.85 0.0015 ' 0.67cfα = − ≥


conditions are met

then go to procedure for case 1

max r rT Q≤max max 'r rT C≤

Case 1 Procedure -Summary

1. Calculate depth of concrete pblock

2. Calculate internal moment arm

3. Calculate resisting moment


g

4. Calculate horizontal shear force

Composite Sections 1

Documents

Transcript of Composite Sections 1