Continuous Beam

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Deflection
• For rectangular beam,
1. The final deflection should not exceed span/250
2. Deflection after construction of finishes and
partitions should not exceed span/500 or
20mm, whichever is the lesser, for spans up to
10 m.
BS 8110 uses an approximate method based on
permissible ratios of the span/effective depth.
Deflection (clause 3.4.6.3)
• This basic span/effective depth ratio is used in
determining the depth of the reinforced
concrete beam.
• For beam with span less than 10 m
Reinforcement details (clause 3.12, BS
8110)
• The BS 8110 spell out a few rules to follow
regarding:
1. Maximum and minimum reinforcement area
2. Spacing of reinforcement
3. Curtailment and anchorage of reinforcement
4. Lapping of reinforcement
Reinforcement areas (clause 3.12.5.3
and 3.12.6.1, BS 8110)
• Minimum area of reinforcement is provided to
control cracking of concrete.
• Too large an area of reinforcement will hinder
proper placing and compaction of concrete
around reinforcement.
• For rectangular beam with b (width) and h
(depth), the area of tensile reinforcement, As
should lie:
• 0.24% bh ≤As ≤ 4% bh for fy = 250 N/mm
2
• 0.13% bh ≤As ≤ 4% bh for fy = 500 N/mm
2



Spacing of reinforcement (clause
3.12.11.1, BS 8110)
• The minimum spacing between tensile
reinforcement is provided to achieve good
compaction. Maximum spacing is specified to
control cracking.
• For singly reinforcement simply supported beam
the clear horizontal distance between tension bars
should follow:
• h
agg
+ 5 mm or bar size≤ s
b
≤ 280 mm f
y
= 250
N/mm
2
• h
agg
+ 5 mm or bar size≤ s
b
≤ 155 mm f
y
= 500
N/mm
2
(h
agg
is the maximum aggregate size)


Curtailment (clause 3.12.9, BS 8110)
• The area tensile reinforcement is calculated
based on the maximum bending moment at mid-
span. The bending moment reduces as it
approaches to the supports. The area of tensile
reinforcement could be reduced (curtailed) to
achieve economic design.
Curtailment (clause 3.12.9, BS 8110)
Simply
supported
beam
Continuous
beam
(Chanakya Arya, 2009)
For beam subjected
to predominantly
UDL
Anchorage (clause 3.12.9, BS 8110)
• At the end support, to achieve proper anchorage
the tensile bar must extend a length equal to one
of the following:
1. 12 times the bar size beyond the centre line of
the support
2. 12 times the bar size plus d/2 from the face of
support
(Chanakya Arya, 2009)
Anchorage (clause 3.12.9, BS 8110)
• In case of space limitation, hooks
or bends in the reinforcement can
be use in anchorage.
• If the bends started after the
centre of support, the anchorage
length is at least 4f but not greater
than 12f.
• If the hook started before d/2 from
the face of support, the anchorage
length is at least 8r but not greater
than 24f.


Continuous L and T beam
• For continuous beam, various loading
arrangement need to be considered to obtain
maximum design moment and shear force.
Continuous L and T beam
• The analysis to calculate the bending moment
and shear forces can be carried out by
1. using moment distribution method
2. Provided the conditions in clause 3.4.3 of BS
8110 are satisfied, design coefficients can be
used.
Clause 3.4.3 of BS 8110: Uniformly-loaded continuous beams
with approximately equal spans: moments and
shears
Effective span – for continuous beam the effective span should
normally taken as the distance between the centres of supports
L- and T- beam
• Beam and slabs are cast monolithically, that is,
they are structurally tied.
• At mid-span, it is more economical to design
the beam as an L or T section by including the
adjacent areas of the slab. The actual width of
slab that acts together with the beam is
normally termed the effective flange.
Clause 3.4.1.5: Effective width of
flanged beam
Effective span – for continuous beam the effective span
should normally taken as the distance between the centres of
supports
L- and T- beam
• The depth of neutral axis in relation to the
depth of the flange will influence the design
process.
• The neutral axis

• When the neutral axis lies within the flange,
the breadth of the beam at mid-span(b) is
equal to the effective flange width. At the
support of a continuous beam, the breadth is
taken as the actual width of the beam.

L- and T-beam
• At the internal supports, the bending moment
is reversed and it should be noted that the
tensile reinforcement will occur in the top half
of the beam and compression reinforcement
in the bottom half of the beam.
Example 3.10 continuous beam design
(Chanakya Arya, 2009)
• A typical floor plan of a small building structure is shown in
figure below. Design B1/5 assuming the slab supports an
imposed load of 4 kN/m
2
and finishes of 1.5 kN/m
2
. The
overall sizes of the beams and slab are indicated on the
drawing. The column are 400x400 mm. The characteristic
strength of the concrete is 35 N/mm
2
and of the steel
reinforcement is 500 N/mm
2
. The cover maybe assumed to be
30 mm.
Example 3.10 Design of shear reinforcement
(Chanakya Arya, 2009)
Example 3.10 Design of shear reinforcement
(Chanakya Arya, 2009)
Example 3.10 Design of shear reinforcement
(Chanakya Arya, 2009)
• The beam B1/5 does not satisfy the conditions
in clause 3.4.3, the moment and shear force
on the beam can be estimated using moment
distribution method or stiffness method.
• Two load cases must be considered: (1)
maximum design load on all spans, (2)
maximum and minimum design loads on the
alternate spans.
Example 3.10 Design of shear reinforcement
(Chanakya Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
For case 1: maximum design load on all spans
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement
(Chanakya Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
For case 2: maximum and minimum design loads on
the alternate spans
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
B1-3 span
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
B3-5 span
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Design shear force and bending moment
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
It is design as T
beam at the
mid-span
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
It is design as
rectangular
beam at the
supports
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
• The simplified rules in clasue 3.12.10.2 do not apply.
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Curtailment of steel reinforcement
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
• From clause 3.12.9.1, the cut off point of the
bars in the tension zone is obtained by
extending the bars an anchorage length in
accordance to Table 3.7 in BS 8110.
• For f
cu
= 35 N/mm
2
and f
y
= 500 N/mm
2
,
deformed type 2 bars
• Anchorage length = 38f
• The theoretical cut off point
= 310 + 38 x 25 = 1260 mm
Hence the 2H25 bars can be stopped at, say 1.3
m from support B3

Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Design for shear
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Example 3.10 Design of shear reinforcement (Chanakya
Arya, 2009)
Deflection
OK

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