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.
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