5.
2 Reinforced Co
effective depth hac
an empirical approach t atioof effective span to However, in Some
been found to satisfy the irements satisfactorily.
cases such as in case of a C Jeam of span greater than 10 m, deflection
calculation should be made. approach of effective span to
The calculation of deflection and the empirical
effective depth ratio are discussed below.
5.2.1 Calculation of Deflection
short-term and long-term deflections. TThe short-term
The total deflection consists of application
immediate deflection after the sustained of load,
deflection is instantaneous or under load. The
due to shrinkage and creep follows.
and long-term deflection is made as
calculation of short-term and long-term deflections is
1. Short-term Deflection
method for elastic deflection
calculated bythe usual
The short-term deflection may be
elasticity of concrete E, and effective moment of
of
USing the short-term modulus of elasticity is taken as secant modulus. It is
given
modulus
inertia I. The short-term
by, N/mm?) (S.1)
E = 5700/f (in
along the span of beam with variation in bendine
varies
The mOment of inertia (I) used for determining 1, of beam near
sections near
sections
moment. The effective
mid-span and supports as shown in Fig. 5.1. At sections near
the points of inflection, section may be uncracked and
stresses are low, so the entire cracks and
the point of inflection, mid-span, concrete in tension near bottom
sections near effective section. At
fully effective. At equivalent area of steel constitute
concrete and
the uncracked concrete cracks at the top resulting in effective area
sections near interior supports,
as shown in the figure.
Section near Section near
Section near
inflection point mid-span interior support
Fig. 5.1 Effective Sections of Continvous -beom
Sevi
effective moment of inertia lies between that of the gross , that
Therefore, the value of
the following
and cracked section. The IS Code has recommended
includes the effect of load level and degree of cracking. (52)
1.2
Mor Z
Md d)b,
Serviceability Requirements 5.3
subject to l,sI,sI,
where L,= Moment of inertia of the cracked section
of tension and compression
considering cquivalent area
reinforcement
L,= Moment of inertia of uncracked sectionneglecting
reinforcement
M,, =Cracking moment of the section
M = Maximum moment under service load
d=Effective depth of section
b.. = Breadth of web of section
b, = Breadth of compression flange of section
Z=Lever arm of section
X= Depth of neutral axis of section
The moment of inertia about neutral axis of the cracked rectangular section
considering the reinforcement () and uncracked rectangular section neglecting the
reinforcement () are given by,
,= bx'3 + (1.5m -1) A, (a-d +mA, (d-x² (5.3)
I, = bD³12 (5.4)
where, b=Width of rectangular section
d= Effective depth of rectangular section
D=Overall depth of rectangular section
m = Modular ratio
Ae = Area of compression reinforcement
A, = Area of tension reinforcement
d= Effective cover to compression reinforcement
x = Depth of neutral axis of section
The cracking moment of section is given by,
(5.5)
M, =14f/y.
f,=Modulus of rupture of concrete
=0.7,/f (in Nimm') (5.6)
the section
y, = Distance of extremne tension fibre from the centroid of
For continuous beam, deflection shall be calculated using the values ofI,, I, and
Me, modified by the following equation,
(5.7)
X = K(X, + X,)2 +(1 - K) Xo
where, X,= Modified value of X
X = Value of I, I, and M, as appropriate
X, X, = Value of X at supports
Xo = Value of X at mid-span
K= Coefficient given in Table 5.1
5.4 Reintorcod (onroto Dosgn
Values of coollcientK
Table 5.1
I4
13
M +M, ),9 L0 I2
0.5 0.6 0.7 08
M+ M2 0.91 ) 1 S
0,08 0,16 03 0.5 0,73
0.03
where M,, M, =Support moments
Mn, M2 = Fixcd cnd moments clastic ieflection
a9,
usual method for
Short-term deflcction is determined by (5.2)
ds = a WT'I(E, 1,)
condition of
bcarn and load
where, W= Total load support
a = Cocfficient depending on the uniforrnly distributed
distribution
beam subjected to
=5/384 for simply supported
load subjected to unifornly distributed
beam
=]/384 for fixed supported
load
2. Long-term Deflection shrinkage of concrete. They
are
creep and
long-term deflections are due to
The
determined as follows: deformation
Creep Creep is arn inelastic tine. The
Deflection due to rate with
(a) Long-term sustained Joad. It increases at a decreasing
instantancous and creep strains
with time under Jong-term strain comprisingremains unaffected as it exhíbit
instantaneous strain and tension stecl
in Fig. 5.2. The strain in increased curvature and
are shown compressive strain results in
incrcase in deflection plus long
no creep. The The combined initial short-term modulus of
hence, increased deflcction. is determined by considering the modified
term deflectiondue to creep
elasticity. It is given by, (5.9)
Ec
e, t ep
At instant
Joading
X Atlong term
loading
D
Stress before
Short-term strain Long-term srain
under sustained and after creep
Cross-section under intanteneous
loading loading
Fig. 5.2 Creep strain and cur voture
Serviceability Requirements 5.5
where G=Stress in extreme compression fibre of concrete
e.= Elastic strain in extreme compression fibre of concrete
=Creep strain in extreme compression fibre of concrete
=e,Cc
C, = Creep coefficient which is primarily a function of age of loading
=2.2 for loading after 7 days
= 1.6for loading after 28 days
= 1.1for loading after 1 year
Substituting the value of ep in terms of e, in the expression for Ere
E, (5.10)
Ece= e, (1 + Cç) (1+C)
The combined initial short-term deflection plus long-term deflection due to creep is
obtained by elastic analysis using effective modulus of elasticity, Ece. So, the long
term deflection due to creep is obtained as,
(5.11)
where &,= Long-term deflection due to creep
8,. = Initial short-term plus long-term deflection
O, = Initial short-term deflection
The presence of compression reinforcement decreases long-term deflection due to
creep. This is because reinforcement in compression zone restrains creep strain due
to
to the bond between concrete and steel. Therefore, creep deflection is modified
account for the effect of compression reinforcement as,
(5.12)
where k, = 0.8S 0.45 A IA, 0.4
(b) Long-term Deflection due to Shrinkage The shrinkage of concrete in
reinforced concrete beam has a similar effect as that of creep. In plain concrete,
bond
Shrinkage merely shortens it. When reinforcement is embedded in it, the
between concrete and steel restrains shrinkage. Thus, in a singly reinforced beam,
Testrained shrinkage on reinforcement face and unrestrained shrinkage at the
unreinforced face causes curvature and hence deflection. The shrinkage curvature
for a
singly reinforced beam is shown in Fig. 5.3.
+|Csh
D d
sh
Fig. 5.3 Curvature of Beam due to Shrinkage
5.6 Reintorced (oncrete Design
|
Shrinkage curvature is given by.
'sh
(5.13)
- Beald which may
be taken
face
strain at the unreinforcedpurposes
where C =Free shrinkage 0.0003fordesign
approxinately equalto tension steel
strain at the level of
, = Freeshrinkage
D =Overall depth of beam
d= effective depth of beam
B=1-,lesn reinforcement. The compression
the area of tension considering modified value
The value of B depends on It is accounted by
reinforcement also influence its value.
of Bas follows, (5.14)
1.0
B= 0.72PPc
, < 1.0 for 0.25<p,-p, S
P-P.> 1.0 (5.15)
= 0.65 ks1.0 for
VP,
where. P,= 100 A,Jbd
P= 100A,Jbd
Area of tension and compression steel respectively
A,f A =
b.d = Width and effective depth of beam respectively
to shrinkage can be determined by moment-area method. The
The deflection due shrinkage of concrete as shown in
curvature diagram due to
M/EIdiagram is the due to shrinkage is
conditions. The deflection
Fig.5.4 for a beam with different end described below.
computed by considering the curvature diagram as
diagram
(i) Deflection for Contilever Beam It is determined by taking moment of o.,
betweenA and B about Aas,
(W) Deflection for Simply Supported Beam It is determined by taking moment ofta
diagram between A and Cabout Aas.
(i) Deflection for Beam with One End Fixed and Other End Simply SupportedpoitS
deterimined by taking moment of , diagram between the fixed end Aandthe
of maximum deflection D about the point D as,
O,, = ,ACAC + CDI2) - 0,CD-2 obtainedby
where distance AC of the point of inflection C end Ais
from the fixed
the deflected shape
equating deflecion of the point B from the
he
cqual to zero as follows. tangent of
Moment of ., diagram between Aand B about B
=0
Curature diagram
(a) Contilever beam
D
B
023
0.257
Curvature diagram
and simply
Beanm îxed at one end
supported at the other end opt
Fig. 5.4 Deflection of beam due to shrint
4CV2= 0
4C2) + o., ( - AC)( -
-o AC(l-AC+
Solution of the above equation gives.
AC=l(1 - 1/N2) maximum defecion Dis
point of
of inflection Cand follows
The distance betwveen the point
at Dequal to zero as
the tangent
obtained by equating the slope of
OAC- 0 CD=0
1/\2)
AC =CD =l1 - +0.5(1 - 1/v2))
/N2)1(1- /N2
Ooa = O l(| -
= 0.08ó 0., moment of
by taking the
It is determined
iv) Fixed at Both Ends
Deflection for Beam andCabout
diagram between A
Cas.
5.8 Reinforced Concrete Design
where / is the point of inflection. As the tangent of deflection curve at
mid-span
horizontal from symmetry, therefore the point of inflection /is at a quarter span from
the fixed end.
O,h = 0.25 7(0. 125 7+ 0.25 /) -., 0.25 /x 0.125 /
= 0.0625 ., I
5.2.2 Effective Span/Effective Depth Ratio for Deflection Control
As the determination of deflection, as described above, involves lengthy caleulation,
therefore theserviceability limit state of deflection is often satisfied by observing
empirical rules. It is achieved by limiting the ratio of effective spanto effective denh
of beam or slab. The basic principle involved in arriving at the value of effective
span to effective depth ratios may be studied by considering a simply supported
rectangular beam of effective span I, which is subjected to auniformly distributed
load w per unit length at service state. Consider that the permissible bending stress
is a-he and the beam behaves elastically: then the moment of resistance of section is
given by,
M= 0,h2
Or ol/8 = a, bD`6
or )= 40,b bDIlI3
where Z= Section modulus = bDI6
D= Overall depth of beam
b= Width of beam
Then the deflection of the beam is given by,
5
X
384 EI
60
X
384 EbD
Substituting for o = 4 G,be b (DII,)'I3 in the above equation,
S 5 x abc
I, 24 E D
(5.16)
D
loads.
where k= Constant whose value depends on end conditions and type of needs
it and
Therefore, 8/1, can belimited by setting alimit tothe ,/D ratio. However,
beams
to be corrected to account for the actual behaviour of reinforced concrete L/Dratio
slabs. This is achieved by using effective depth instead of overall depth and stressin
and by multiplying with appropriate factors for tension reinforcementspan/efective
it and for compression reinforcement. The basic values of effective
depth ratio for beams and slabs are given in Table 5.2.
Serviceability Requirements 5.9
Bosic valves of
Table 5.2 effective span to effective depth ratios for
Support
rectangular beoms and slabs
Conditions Effective span effective depth
Span < 10 m
span > 10 m
Cantilever 7
Simply supported 20 Deflection calculation should be made
Continuous 26
20 x 10/span
26 x 10/span
The effective span to effective depth
ratio for the beams and slabs
different area of tension steel and stress in it reinforced with
are
value of effective span/effective depth ratio with obtained
by multiplying the basic
multiplication factor given by,
Multiplication factor =0.225 + 0.00322 f, bd
-0.625 log1o (5.17)
100 A,
The area of tension steel and stress in it are
considered in the region of mid-span
for simply supported and continuous beam or slab. For
it is considered at support. Figure 5.5 shows the cantilever beam or slab,
reinforcements.
modification factors for tension
2.0
d
1.6
factor
Modification lN/mm
1.2 120)
1451
190
0.8 240
2901
0.4
0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.0
100 A, /bd
Steel stress at service load, f, = 0.58f, XA,, required/A,, provided
Fig. 5.5 Modification foctor for tension steel
S observed that a lower value of effective span to effective depth ratio is required
when higher percentage of steel is used. It is because the stiffness of beam does not
increase as much as the strength of the beam on increasing the area of steel. It is
because,
() the
higher percentage of steel leads to higher depth of neutral axis and hence
larger curvature for given stress in steel,
5.10 Reintorrer fonrrete Dr sign
(1i) Iarget compression zone with higher percentage of steel leads to
deflection, and greater creen
(iii) conerete in tension z0ne, which adds very significantly to the
the heam with low percentage of steel, is less effective for higher stiffness
of steel I is also observed that lower effective span to effective percentage
of
is required for higher stress in steel. It is because the higher depthin ratin
stress
induces higher strain in steel, and hence higher curvature, resulting steel
deformation.
in higher
For a doubly reinforced beam, effective span to effective depth ratio is furth.
modified forcompression steel by multiplying with a factor given by.
16p,
Multiplying factor = (3.18)
P +0.275
where D, = Percentage of compression reinforcement
= 100 A., Jbd
Figure 5.6 shows the modification factors for compression reinforcements. It is
observed that the effective span to effective depth ratio increases with increase in
arca of compression steel. It is because,
(i) the increase in stiffness of beam is more than the increase in its
moment
capacity with increase in the area of compression steel, and
(ii) the deflection of beam reduces due to increased area of
creep andshrinkage reduces.
compression steel as
I.5
factor
Modifcation
I.4
1.3
1.J
0.5 1.0 1.5 2.0 2.5 3.0
I00 A/bd
Fig. 5.6 Modificotion foctor for
compression steel
Iot cdge supported slabs spanning in iwo directions, the shorter of the (WOsppans
is considered for
calculainy
blabs of shurler spans up lo the
3.5 etfecive spanto eftective
mand renforced with milddepthsteel,ratios. Fortw0-way
the effective span
io overall depth latio given below lay generally be assumed to salisty the detlection
inits Iut loading class up lo 3 kN/n.
For sapiy supporled slabs, ,ld < 35
Fur coninuous slabs, id 40
For high strengh delorned bars of grade Fe 415, the values given aboveshhallbe
ultipliedby 0.8.
