Download - AZ024 Lecture 3 (rev[1] a)
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Topic: Reinforced Concrete (II)- Design in RC Beam and slab-1
1.0 Introduction to Structural Element beams
Beams Horizontal members carrying roof and floor loads. They resist loads in
bending, shear and bond and may be simply supported or continuous. For in-situ
construction beam are often flanged, of T or L shape where part of the floor slab acts
with the beam.
2.0 General Terms in Limit state design (BS 8110(1995))
Characteristics loads:- the working or service loads, classified into dead, imposed and
wind loads. They have a low probability of being exceed during the life of the
structure.
Characteristics strengths:- the strength of materials below which not more than 5% of
test results fall. For concrete this is the cube strength at 28days and for reinforcement,
the yield stress.
Design load:- the characteristics loads multiplied by the partial factors of safety for
the load
Design strength:- the characteristics strengths divided by partial factors of safety for
materials
Limit state:- States where the structure has become unfit for use. The main limit state
are:
(1)ultimate limit state: to satisfy this the strength must be adequate to carry the
loads. Account must also be taken of stability
(2)serviceability limit state: to satisfy these both deflection and cracking must
not be excessive.
Limit State Design
The condition of a structure when it becomes unserviceable is called a Limit State
Loads and partial factor of safety
Characteristics dead load, KG
This is the weight of the structure complete with finishes, fixtures and partitions.
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Characteristics imposed load, KQ
This load depends on the use of the Buildings,
Characteristics wind load, KW
This is defined and calculated in accordance with code of practice, Ch.V, Part 2. (it is
out of scope in this course)
The design load
= characteristics Load x Partial factor of safety for loads
= KKF
Where K takes account of (1) possible overloads (2) inaccurate assessment of the
effects of loading and unknown stress distribution within the structure.
Materials and partial factors of safetyThe characteristics strength is defined as the cube strength of concrete CUf at 28 days,
and the yield of reinforcement yf , below which not more than 5% of the test results
fall.
The resistance of sections is based on the design strength
Design strength = characteristics strength / partial factors of safety for materials
=M
Kf
Table 1 Design Load
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Table 2 Design Strengths
The maximum design stress in the concrete is given by CUCU ff 67.05.1/ = where the
factor 0.67 takes account of the ratio between the characteristics cube strength and the
bending strength in a flexural member.
Assumption takes in single reinforced rectangular beams
The ultimate moment of resistance of a section is based on the assumption given in
the BS8110.
(1)The strains in the materials derived assuming that plane sections remain plane.(2)The stresses in the concrete are derived using either (a) the design stress-strain
curve given in BS8110 with 5.1=m , or (b) a uniform compressive stress of
CUf45.0 over the whole compressive zone.
(3)Depth of the stress block = 0.9 x depth of the neutral axis = x9.0
(4)The tensile strength of the concrete is ignored
(5)The stresses in the reinforcement are derived from the stress-strain curve given in
BS8110 where .15.1=m
On the basis of the above assumptions, the strain and stress diagrams for a beam
section are shown below:-
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h =overall depth of the section
d =effective depth = depth of centre line of the steel
b =breath of the beam
x =depth to the neutral axis
SA =area of steel in tension
C =strain in concrete
T =strain in steel
stf =stress in the steel in tension
The term balance design refers to a beam with the maximum ultimate moment of
resistance, for example, with sufficient steel to causes the neutral axis to be at its
maximum depth of 1/2d.
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Considering the rectangular stress diagram,
The concrete stress =CUf45.0
The steel stress = yf87.0
C(force in the concrete in compression) = CUCU bdfdbf 20.09.05.045.0 =
T(force in the steel in tension) = syAf87.0
Z = level arm = ( ) ddd 775.09.04
=
RCM (moment of resistance with respect of the concrete) = ZC
= 22156.0775.09.05.045.0 kbdfbdddbf CUCU ==
where CUfk 156.0=
RSM (moment of resistance with respect of the steel) = ZAfTZ sy87.0=
Let RSRC MM=
ZAfbdf syCU 87.0156.02=
Percentage of steel in the tension
( )%
23
775.087.0
156.0100100 2
y
CU
y
CUs
f
f
bddf
bdf
bd
Ap =
=
=
Point to notes
(1)the nominal cover should always be at least equal to the size of the bar and in the
case of bundles of 3 or more bars should be equal to the size of a single bar of
equivalent areas
(2)further recommendations regarding cover are given in BS8110. These depend on
conditions of exposure and concrete grade. For example, for grade 25 concrete for
mild exposure, for example, completely protected against the weather except
during construction, the cover given is 20mm. The cover is lower with higher
grades of concrete and greater when condition of exposure become more severe.
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Doubly reinforced beams
If the concrete alone cannot resist the applied moment, reinforcement can be added to
strengthen the beam section in compression.
Design formulae for doubly reinforced beams are given in BS8110. These are based
on the followings:-
(1)rectangular stress block with the depth to the neutral axis, dx2
1=
(2)stress in concrete in compression = CUf45.0
(3)stress in reinforcement in compression = yf87.0
(4)stress in reinforcement in tension = yf87.0
Note: SC = force in steel in compression
T = force in steel in tension
BS 8110 states that the formulae should not be used whend
d' is >0.2. If this
requirement is not met, the stress in the compression steel will not reachyf87.0 . The
moment of resistant of the concrete 2156.0 bdfM CURC =
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However, if this is less than M (ultimate moment), the compression steel resists
( )RCMM
The force in steel in compression
( )( )'dd
MMC RCs
=
Area of compression steel =y
s
sf
CA
87.0
'=
For equilibrium,
SC CCT +=
( ) ( )'
'
87.0225.0
87.05.045.0
SyCU
SyCU
Afbdf
Afbdf
+=
+=
Area of Tension Steel =As = Force / Stress =y
f
T
87.0
Example 1:-
A simply support rectangular beam of 7m span carried a uniformly distributed loadwhich includes a self-weight of 4kN/m and imposed load of 3kN/m. the breath of the
beam is 300mm. Find the depth of the beam and the steel area required for balanced
using Grade 30 concrete and mild steel reinforcement (yield stress,
2/250 mmNfy = )
Solution:
Design Load = mkNQG KK /4.10)3(6.1)4(4.16.14.1 =+=+
Ultimate Moment, mkNwLMU /7.638)7(4.10
8
22
===
Considering the rectangular stress diagram,
The stress of concrete =CUf45.0
The stress of steel = yf87.0
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C(force in the concrete in compression)
= CUCU bdfdbf 20.0)9.0)(5.0)((45.0 =
T= Force in the steel in tension = syAf87.0
Z= ( ) ddd 775.022
9.0 =
MRC(Moment of resistance with respect to concrete)
where CUfk 156.0= .
MRS (moment of resistance with respect to steel)
zAfTM syZRS 87.0==
But RSRC MM =
2156.087.0 bdfzAf CUsy =
Percentage of steel in the tension
( )%
23
775.087.0
156.0100100 2
y
CU
y
CUs
f
f
bddf
bdf
bd
Ap =
=
=
For balance design, (when the depth x to the neutral axis is dx 5.0= ,
Moment of resistance of concrete,
2
156.0)775.0)(9.0)(5.0(45.0 bdfddbfM CUCURC ==
Moment of resistance of concrete RCM Ultimate moment (Mu)
mmd
d
bdfCU
212
.107.63)300)(30(156.0
107.63156.0
62
62
Moment of resistance of steel RSM Ultimate moment (Mu)
22156.0)775.0)()(9.0)(5.0(45.0 kbdbdfddbf
zc
cuCU ===
=
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2
6
6
1783
107.63)212)(775.0)(250)(87.0(
107.63)87.0(
mmA
A
fA
s
s
yS
=
From steel area table,
Combination Steel Area (mm2) Comment
4 nos. bar 25 dia. 1960 ok
6 nos. bar 20 dia. 1890 ok
9 nos. bar 16 dia. 1801 economical, but need to put in several
positions.
1 nos. bar 50 dia 1960 Min 2 nos. of steel bar for shear stress
Then from steel area table,
6 nos. 20mm dia. Steel bar given 22 17831890 mmmmAs >= (O.K.)
Alternatives
n x area of one bar > 1783mm2
( ) 178342>Dn whereD = Diameter of bar (in mm)
If the diameter of the bar is choose, say 20mm diameter, then the nos. of the bar
required can be easily calculated as follows:-
( )
6
67.5
1783204
2
=
>
>
n
n
n
Example 2:-
A rectangular beam is simply supported over a span of 6m and carries a dead load
including self-weight of 9kN/m and an imposed load of 6kn/m. the beam is 210mm
wide and 310mm effective depth and the inset of the compression steel is 40mm.
Design the steel for mid span of the beam for grade 25 concrete and high yield stress
reinforcement ( )/2502
mmNfy = and mild steel links or stirrups ( )/2502
mmNfy = .
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Solution:
Design Load = mkNQG KK /2.22)6(6.1)9(4.16.14.1 =+=+
kNmwL
MU 9.998)6(2.22
8
22
===
Design of reinforcement for bending moment
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For balance design, (when the depth x to the neutral axis is 0.5d),
kNmbdfddbfM CUCURC 7.78)10(7.78156.0)775.0)(9.0)(5.0(45.0
62====
Since RCU MM > , compression reinforcement is required.
Thus, doubly reinforced design is required.
Cs= Force in steel in compression
( )( )
( )( )
kNdd
MM RCU 6.891040310
7.789.99' 3
=
=
=
Area of steel in compression,
23 412)250(87.0/)10(6.8987.0
mmf
CA
y
SS ===
For equilibrium, Tensile force = Compressive Force
kNfbdfCCT yCUSC 450)412(87.09.0)5.0(45.0 =+=+=
Area of tension steel,
yS fTA 87.0/= =23 1863)250(87.0/)10(450 mm=
Proposed number of Tension Steel
6 No. 20mm dia. Steel bars given 21890mmAS =
Proposed Number of Compression Steel
2 No. 20mm dia. Steel bars given 2628mmAS =
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Appendix: Steel Area Table