COMPOSITE BEAMS AND SLABS

frontdotardUrban and Civil

Nov 15, 2013 (3 years and 11 months ago)

82 views

COMPOSITE BEAMS AND SLABS

Introduction

The term composite can be used of any struct
ural medium in which two or more materials
interact
to provide the required
strength and stiffness. In steel
work construction, the term
refers to cross
-
section which
combine steel section with concrete in such a way that the two
act together as one unit. Examples are shown below.



In situ concrete


in situ concrete

precast unit









Headed stud

connector











The performance of composite bea
ms

is similar to that of reinforced concrete (rc)
beams, but
there are two main differences.

Firstly, the steel section has a significant depth and its second moment o
f area may not be
ignored, unlike
that

of steel bar reinforcement.


Secondly, the concrete

to reinforcement
bond
whic
h essentially for rc action is absent, in
composite
beams

generally must be provided by shear connection.

Design
method for composit
e

beams there
fore follows those methods for r
c

with modification
as indicated. Due to the presence of the concrete slab, problem
s

of steel compression flange
instability and local bucking of the steel member
s

are not usually relevant in simply
supported member except doing erection.

Recommendation
s

f
or design in composite construction
s are not included in part 1 of BS

5950 but are include
d in part 3 and 4 of BS

5950.

Advantages

of Composite
Beam (CB) Construction


The advantages of C
B over normal steelwork beams are

1.

The

increased moment capacity an
d

2.


Stiffness
, or alternatively the reduced steel sizes for the same moment capacity.


Disadvantages of CB

The disadvantage of composite construction is the need to provide shear

connectors to ensure
interaction

of the parts.

The following are considered
when dealing with composite structure
s

a)


Shear

and moment capacity


Essentially composite beams are T
-

beams with wide concrete flanges
.

Effective

breadth
(bs)

may be ta
ken as one
-

fifth of the span for
simply supported. While continuous

and
cantilev
er beams they are
treated
sep
a
rately

(see BS

5950 part 3).


Shear capacity is based on the resistance of the web of the steel section alone.


P
v
= 0.6P
y

A
v

Moment capacity
(Mc)
is based
on
the
as
sumed ultimate steel conditions as shown in Figs 1
and 2

i.

When NA
(


)

is
within the concrete slab


depth (




Mc=A
b
P
y

(



+




-



)


ii.

When NA
(


)

is within the steel beam


Mc =
A
b
P
y

(





+



)
-

2





(








)

















=
0.4

















+









D















Where,


=
A
b
P
y





0.4









A
b

= steel area

Fig 1; NA is within the concrete slab



















0.4














































Where
,



=



-

0.2





























A
b

= steel area


Fig. 2
;

NA

within the steel beam





Shear Connectors

There are various forms of it but the preferred type is the headed stud.


Shear connectors must
perform

the primary function of
:

a)

Transferring
of
shear at the
steal /concrete int
erface (equivalent to bond),

h
ence
control
ling

slip between the two parts
.

b)

Secondly,

carries

the tensile force between the part
s, hence controlling

separation.

The performance of all shear connector
s

is affected by

a)

L
ateral restra
int of

the surrounding concrete


b)

Presence

of tension in the concrete, and

c)


Type

of concret
e used (normal or light weight)


Shear force (


)

=








Where,


=













(where
NA
is
in concrete)













=













(where

NA
is
in
steel)


And N
sc

=
No of studs required

The connector force (












Shear strength





of headed studs


Diameter

(
mm
)

Height

(
mm
)

Shear

strength




in (K
N)
for


in
(N/mm
2
)

20

30

40

50

22

100

112

126

139

153

19

100

90

100

109

119

16

75

6

74

82

90


Local shear in Concrete


The total shear connection
depends
on

a)


The

shear connector
itself and

b)

The

abilit
y of
the surrounding concrete to transmit

the shear stresses.

Therefore,

longitudinal

shear failure is possible as such transverse

r
einforcement

must

be
provided with strength greater than the applied shear per unit length

(q)
:


q


0.15







And

q


0.9



+
0.7





Where
,




is either (


+


) o
r 2



depending on the shear path






=

the
design strength of the
r
einforcement







= the concrete cube

strength






is

either (2 (
connector width

+
stud height)
)


Or 2
(
slab

depth)


Deflection



As in steel beam design, defection ought
to be checked

for at the serviceability limit state
(un
-

factored loads)

The values of Neutral Axis (NA)

depth





and
the Equivalent S
ec
ond M
oment

area





allows defection to be calcul
ated using normal elastic formulae
with

a value of





= 205 K
N/mm
2
.


Modular ratio

(m)



(N/mm
2
)

Short term

Sustained

20

8.2

16.4

30

7.3

14.6

40

6.6

13.2

50

6.0

12.0




























Steel







Strain diagram


Area





and r =













Fig 3
Transformed section






= (




+
m
r (



+




)
)
/ (
1

+mr)







=




(D +



)

2

/

4 (1 +
mr) +











+







Actual deflection

(ℓ
) =













Example 1

Yodebees Consult Limited based in Jos was contracted to design a steel structure. The plan,


section

and other details are shown in Fig. Q1

Dimension:




Span of bea
m

= 10

m


Beam centers


= 6 m


Concrete slab thick
ness (dc) = 20
0

mm spanning in two ways


Screed th
ickness (ts) = 4
0 mm

Loading:



Concrete slab unit weight (



) = 23.8

KN/m
3


Screed unit weight (



= 22 KN/m
3


Imposed load

= 5
.0 KN/m
2


Characteristics cube strength (



)

= 30

N/mm
2


Self weight of beam

= 6

KN (Assumed)


Young modulus of steel = 205 KN/m
2


Characteristic strength of steel (


) = 46
0 N/mm
2

Others:



Area of reinforcement (Ae) = 0.800 mm
2
/m

(
ϴ
10mm@200mm






Modular ratio (m) = 13.2 sustained


Length of shear path

(Ls) = 380

mm


Use 22mm diameter by 100 mm high headed stud, (shear strength, Pk = 119 Kn)




Question

a)

Design the most economical composite I
-

section beam to BS 5950 Part 3
, given

that
Zx
(calculated)

be reduced by 59

%

b)

Check the suitability of the connectors and

c)

Check for deflection.




10
m





6m

Screed


Beam

Slab


6m


6m



Fig

Q1

Solution




3m



3m




3m

4
m





3m


Load computation


Dead load
due to
slab =


(dc)

= 4.76 K
N/m
2


Dead load

due to screed
=






=
0.88 K
N/m
2



Total

(gk)

= 5.64KN/m
2

Area Calculation





= 2
(bh) =

24
m
2


= (½bh) 4 = 18.
m
2


Dead load

(






On



=
gk(A)

=
135.4KN



On


= gk(A) =101.5KN




Imposed load






On


=
qk(A)

= 1
20K
N


On

=qk(A) = 90KN



Ultimate load
(w)


Uniform

d
ead load

= 1.4x6 = 8.4KN


On

= 1.4
w
d

+1.6
w
i


= 382KN


On

=1.4
w
d

+1.6
w
i


= 286KN

191

191



143

143



8.4



4.2

4.2

334 334


10m


3m

2m

2m



3m









=
1061KNm








= 338KN



A)

Design Aspect


Assume



= 275 N/mm
2

(Table 6)































=





= 3858 cm
3



But




should be reduced by 59%









1582
cm
3


Use 457 x 191 x 82 Kg/m UB (



= 1612 cm
3
)


Other parameters are:




= 104.5cm
2
; D= 460.2mm; t = 9.9mm and T= 16mm


C
heck the following


a)

Shear capacity (



) =
0.6




= 752 KN


But







= 0.45





(Section adequate)


b)


Moment capacity (



)



Assume that



is within
the concrete slab as shown




















457 x 191 x 82 (UB)










Calculate,


=
A
b
P
y




= 119mm


dc


0.4










NA is within the slab.


Moment capacity (Mc) =A
b
P
y

(



+




-



) = 1063KNm



M
x












= 0.99



1.0


(Section adequate)


B)

Shear connectors



Force in
the
concrete @ Mid

span





) = 0.4










= 2880KN



But Shear strength





= 126 KN (given).



No of studs required
(


)

=






=
30

studs


But the connector force (










= 94.5KN


The studs are to be
evenly dist
ri
buted in each half span



Spacing =






= 167mm


Shear per unit length (q) =






= 576N/mm



But
q


0.15






And q


0.9



+
0.7






But (
Ls)

=
380mm (given)


0.15





= 1710N/mm


And


0.9



+
0.7





= 600N/mm




Shear

connector is adequate




C)

Defection

Use the un
-
fact
ored
imposed
load
(
W
)
=
210KN



But r =









= 0.026



M = 13.2 (given)


































= (




+ mr (



+




)
)
/ (1

+mr)

= 184mm









Use the transformed formula to obtain moment of
inertia



)







=




(D +



)

2

/

4 (1 +
mr) +











+




= 514185 cm
4





The actual deflection is given by the formula
(ℓ
) =









= 3.3mm






But max defection =




= 27.7mm





Deflection is adequate


The section chosen is adequate to sustain the loads.