Module
7
Limit State of
Serviceability
Version 2 CE IIT, Kharagpur

Lesson
17
Limit State of
Serviceability
Version 2 CE IIT, Kharagpur

Instruction Objectives:
At the end of this lesson, the student should be able to:
•

explain the need to check for the limit state of serviceability after designing
the structures by limit state of collapse,

•

differentiate between short- and long-term deflections,

•

state the influencing factors to both short- and long-term deflections,

•

select the preliminary dimensions of structures to satisfy the requirements
as per IS 456,

•

calculate the short- and long-term deflections of designed beams.

7.17.1 Introduction
Structures designed by limit state of collapse are of comparatively smaller
sections than those designed employing working stress method. They, therefore,
must be checked for deflection and width of cracks. Excessive deflection of a
structure or part thereof adversely affects the appearance and efficiency of the
structure, finishes or partitions. Excessive cracking of concrete also seriously
affects the appearance and durability of the structure. Accordingly, cl. 35.1.1 of
IS 456 stipulates that the designer should consider all relevant limit states to
ensure an adequate degree of safety and serviceability. Clause 35.3 of IS 456
refers to the limit state of serviceability comprising deflection in cl. 35.3.1 and
cracking in cl. 35.3.2. Concrete is said to be durable when it performs
satisfactorily in the working environment during its anticipated exposure
conditions during service. Clause 8 of IS 456 refers to the durability aspects of
concrete. Stability of the structure against overturning and sliding (cl. 20 of IS
456), and fire resistance (cl. 21 of IS 456) are some of the other importance
issues to be kept in mind while designing reinforced concrete structures.
This lesson discusses about the different aspects of deflection of beams
and the requirements as per IS 456. In addition, lateral stability of beams is also
taken up while selecting the preliminary dimensions of beams. Other
requirements, however, are beyond the scope of this lesson.

7.17.2 Short- and Long-term Deflections
As evident from the names, short-term deflection refers to the immediate
deflection after casting and application of partial or full service loads, while the
long-term deflection occurs over a long period of time largely due to shrinkage

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and creep of the materials. The following factors influence the short-term
deflection of structures:
(a) magnitude and distribution of live loads,
(b) span and type of end supports,
(c) cross-sectional area of the members,
(d) amount of steel reinforcement and the stress developed in the
reinforcement,
(e) characteristic strengths of concrete and steel, and
(f) amount and extent of cracking.
The long-term deflection is almost two to three times of the short-term
deflection. The following are the major factors influencing the long-term deflection
of the structures.
(a) humidity and temperature ranges during curing,
(b) age of concrete at the time of loading, and
(c) type and size of aggregates,
water-cement ratio, amount of
compression reinforcement, size of members etc., which influence the
creep and shrinkage of concrete.

7.17.3 Control of Deflection
Clause 23.2 of IS 456 stipulates the limiting deflections under two heads
as given below:
(a) The maximum final deflection should not normally exceed span/250
due to all loads including the effects of temperatures, creep and shrinkage and
measured from the as-cast level of the supports of floors, roof and all other
horizontal members.
(b) The maximum deflection should not normally exceed the lesser of
span/350 or 20 mm including the effects of temperature, creep and shrinkage
occurring after erection of partitions and the application of finishes.
It is essential that both the requirements are to be fulfilled for every
structure.

7.17.4 Selection of Preliminary Dimensions
The two requirements of the deflection are checked after designing the
members. However, the structural design has to be revised if it fails to satisfy any
one of the two or both the requirements. In order to avoid this, IS 456
recommends the guidelines to assume the initial dimensions of the members
which will generally satisfy the deflection limits. Clause 23.2.1 stipulates different
span to effective depth ratios and cl. 23.3 recommends limiting slenderness of
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beams, a relation of b and d of the members, to ensure lateral stability. They
are given below:
(A) For the deflection requirements
Different basic values of span to effective depth ratios for three different
support conditions are prescribed for spans up to 10 m, which should be modified
under any or all of the four different situations: (i) for spans above 10 m, (ii)
depending on the amount and the stress of tension steel reinforcement, (iii)
depending on the amount of compression reinforcement, and (iv) for flanged
beams. These are furnished in Table 7.1.
(B) For lateral stability
The lateral stability of beams depends upon the slenderness ratio and the
support conditions. Accordingly cl. 23.3 of IS code stipulates the following:
(i) For simply supported and continuous beams, the clear distance
between the lateral restraints shall not exceed the lesser of 60b or 250b2/d,
where d is the effective depth and b is the breadth of the compression face
midway between the lateral restraints.
(ii) For cantilever beams, the clear distance from the free end of the
cantilever to the lateral restraint shall not exceed the lesser of 25b or 100b2/d.
Table 7.1 Span/depth ratios and modification factors
Sl.
No.
1
2

3
4
5

Items
Basic values of span to
effective depth ratio for
spans up to 10 m
Modification factors for
spans > 10 m

Modification factors
depending on area and
stress of steel
Modification factors
depending as area of
compression steel
Modification factors for
flanged beams

Cantilever
7

Simply
supported
20

Continuous
26

Not applicable Multiply values of row 1 by
as deflection
10/span in metres.
calculations
are to be
done.
Multiply values of row 1 or 2 with the modification
factor from Fig.4 of IS 456.
Further multiply the earlier respective value with
that obtained from Fig.5 of IS 456.
(i) Modify values of row 1 or 2 as per Fig.6 of IS
456.
(ii) Further modify as per row 3 and/or 4 where
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reinforcement percentage to be used on area
of section equal to bf d.

7.17.5 Calculation of Short-Term Deflection
Clause C-2 of Annex C of IS 456 prescribes the steps of calculating the
short-term deflection. The code recommends the usual methods for elastic
deflections using the short-term modulus of elasticity of concrete Ec and
effective moment of inertia Ieff given by the following equation:
I eff

=

Ir
; but I r ≤ I eff
1.2 - (M r / M )( z / d )( 1 − x / d )( bw / b )

≤ I gr

(7.1)
where Ir = moment of inertia of the cracked section,
Mr = cracking moment equal to (fcr Igr)/yt , where fcr is the modulus of
rupture of concrete, Igr is the moment of inertia of the gross section
about the centroidal axis neglecting the reinforcement, and yt is the
distance from centroidal axis of gross section, neglecting the
reinforcement, to extreme fibre in tension,
M = maximum moment under service loads,
z = lever arm,
x = depth of neutral axis,
d = effective depth,
bw = breadth of web, and
b = breadth of compression face.
For continuous beams, however, the values of Ir, Igr and Mr are to be
modified by the following equation:
⎡X + X2 ⎤
X e = k1 ⎢ 1
⎥ + (1 - k1 ) X o
2
⎣
⎦
(7.2)
where Xe

= modified value of X,

X1, X2 = values of X at the supports,
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Xo

= value of X at mid span,

k1

= coefficient given in Table 25 of IS 456 and in Table 7.2 here, and

X

= value of Ir, Igr or Mr as appropriate.

Table 7.2 Values of coefficient k1
k1

k2

0.5
or
less
0

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

0.03

0.08

0.16

0.30

0.50

0.73

0.91

0.97

1.0

Note: k2 is given by (M1 + M2)/(MF1 + MF2), where M1 and M2
moments, and MF1 and MF2 = fixed end moments.

= support

7.17.6 Deflection due to Shrinkage
Clause C-3 of Annex C of IS 456 prescribes the method of calculating the
deflection due to shrinkage α cs from the following equation:

α cs = k 3 ψ cs l 2
(7.3)
where k3 is a constant which is 0.5 for cantilevers, 0.125 for simply supported
members, 0.086 for members continuous at one end, and 0.063 for fully
continuous members; ψ cs is shrinkage curvature equal to k4 ε cs /D where ε cs is
the ultimate shrinkage strain of concrete. For ε cs , cl. 6.2.4.1 of IS 456
recommends an approximate value of 0.0003 in the absence of test data.
k 4 = 0.72( pt - p c ) / pt ≤ 1.0, for 0.25 ≤ pt - pc < 1.0
= 0.65( pt - p c ) / pt ≤ 1.0, for pt - pc ≥ 1.0
(7.4)
where pt = 100Ast/bd and pc = 100Asc/bd, D is the total depth of the section,
and l is the length of span.

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7.17.7 Deflection Due to Creep
Clause C-4 of Annex C of IS 456 stipulates the following method of
calculating deflection due to creep. The creep deflection due to permanent loads
α cc ( perm ) is obtained from the following equation:

α cc ( perm ) = α 1cc ( perm ) - α 1(perm)
(7.5)
where α 1cc ( perm ) = initial plus creep deflection due to permanent loads obtained
using an elastic analysis with an effective modulus of
elasticity,
E ce = E c /(1 + θ ), θ being the creep coefficient, and

α 1( perm ) = short-term deflection due to permanent loads using Ec.

7.17.8

Numerical Problems

Problem 1:
Figures 7.17.1 and 2 present the cross-section and the tensile steel of a
simply supported T-beam of 8 m span using M 20 and Fe 415 subjected to dead

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load of 9.3 kN/m and imposed loads of 10.7 kN/m at service. Calculate the shortand long-term deflections and check the requirements of IS 456.
Solution 1:
Step 1: Properties of plain concrete section
Taking moment of the area about the bottom of the beam
yt =

I gr

(300)(600)(300) + (2234 - 300)(100)(550)
= 429.48 mm
(300)(600) + (2234 - 300)(100)

300(429.48) 3
2234(170.52) 3 1934(70.52) 3
=
+
= (11.384) (10) 9 mm 4
3
3
3

This can also be computed from SP-16 as explained below:
Here, bf /bw = 7.45, Df /D = 0.17. Using these values in chart 88 of SP-16,
we get k1 = 2.10.
Igr = k1bw D3/12 = (2.10)(300)(600)3/12 = (11.384)(10)9 mm4
Step 2: Properties of the cracked section (Fig.7.17.2)

f cr = 0.7

f ck (cl. 6.2.2 of IS 456) = 3.13 N/mm2
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Mr = fcr Igr /yt = 3.13(11.384)(10)9/429.48 = 82.96 kNm
Es = 200000 N/mm2
Ec = 5000

f ck (cl. 6.2.3.1 of IS 456) = 22360.68 N/mm2

m = Es /Ec = 8.94
Taking moment of the compressive concrete and tensile steel about the neutral
axis, we have (Fig.7.17.2)
bf x2/2 = m Ast (d – x) gives
or

(2234)(x2/2) = (8.94)(1383)(550 – x)

x2 + 11.07 x – 6087.92 = 0 . Solving the equation, we get x = 72.68 mm.
z = lever arm = d – x/3 = 525.77 mm
2234(72.68) 3
Ir =
+ 8.94(1383)(550 - 72.68) 2 = 3.106(10) 9 mm 4
3
M = wl2/8 = (9.3 + 10.7)(8)(8)/8 = 160 kNm

I eff =

=

So,

Ir
M z
x b
1.2 - r (1 - ) ( w )
d b
M d

…. (Eq. 7.1)

Ir
= 0.875 I r . But I r ≤ I eff ≤ I gr
82.96 525.77
72.68 300
1.2 - (
)(
) (1 )(
)
160
550
550 2234

Ieff = Ir = 3.106(10)9 mm4.

Step 3: Short-term deflection (sec. 7.17.5)
E c = 5000

f ck (cl. 6.2.3.1 of IS 456) = 22360.68 N/mm2

Short-term deflection = (5/384) wl4/EcIeff
= (5)(20)(8)4(1012)/(384)(22360.68)(3.106)(109) = 15.358 mm
(1)

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Step 4: Deflection due to shrinkage (sec. 7.17.6)
k 4 = 0.72( pt - pc ) / pt = 0.72(0.84) 0.84 = 0.6599

ψ cs = k 4 ε cs / D = (0.6599)(0 .0003)/600 = 3.2995(10) -7
k3 = 0.125 (from sec. 7.17.6)

α cs = k 3 ψ cs l 2 (Eq. 7.3) = (0.125)(3.2995)(10)-7(64)(106) = 2.64 mm
(2)

Step 5: Deflection due to creep (sec. 7.17.7)
Equation 7.5 reveals that the deflection due to creep

α cc ( perm )

can be

obtained after calculating α 1cc ( perm ) and α 1( perm ) . We calculate α 1cc ( perm ) in the
next step.
Step 5a: Calculation of α 1cc ( perm )
Assuming the age of concrete at loading as 28 days, cl. 6.2.5.1 of IS 456
gives θ = 1.6. So, Ecc = Ec /(1 + θ ) = 22360.68/(1 + 1.6) = 8600.2615
N/mm2 and m = Es /Ecc = 200000/8600.2615 = 23.255
Step 5b: Properties of cracked section

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Taking moment of compressive concrete and tensile steel about the
neutral axis (assuming at a distance of x from the bottom of the flange as
shown in Fig.7.17.3):
2234(100)(50 + x ) = (23.255)(1383)(450 - x ) or
which gives x = 112.92 mm. Accordingly,
512.36 mm.

x = 12.92 mm

z = lever arm = d – x/3 =

Ir = 2234(100)3/12 + 2234(100)(62.92)2 + 23.255(1383)(550 – 112.92)2
+ 300(12.92)3/3

= 7.214(109) mm4

Mr = 82.96 kNm (see Step 2)
M = wperm l2/8 = 9.3(8)(8)/8 = 74.4 kNm.
I eff =

Ir
= 0.918 I r
82.96 512.36
112.92 300
(1.2) - (
)(
) (1 )(
)
74.4
550
550
2234

However, to satisfy Ir ≤ Ieff ≤ Igr, Ieff should be equal to Igr. So, Ieff = Igr =
11.384(109). For the value of Igr please see Step 1.
Step 5c: Calculation of α 1cc ( perm )
=
5wl4/384(Ecc)(Ieff)
α 1cc ( perm )
5(9.3)(8)4(10)12/384(8600.2615)(11.384)(109)

=

= 5.066 mm
(3)
Step 5d: Calculation of α 1( perm )
=
5wl4/384(Ec)(Ieff)
α 1( perm )
5(9.3)(8)4(10)12/384(22360.68)(11.384)(109)

=

= 1.948 mm
(4)
Step 5e: Calculation of deflection due to creep

α cc ( perm ) = α 1cc ( perm ) - α 1( perm )
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= 5.066 – 1.948 = 3.118 mm
(5)
It is important to note that the deflection due to creep α cc ( perm )

can be

obtained even without computing α 1cc ( perm ) . The relationship of α cc ( perm )

and

α 1( perm ) is given below.
α cc ( perm ) = α 1cc ( perm ) - α 1( perm )
= {5wl4/384(Ec)(Ieff)} {(Ec /Ecc) – 1} = α 1( perm ) ( θ )
Hence, the deflection due to creep, for this problem is:

α cc ( perm ) = α 1( perm ) ( θ ) = 1.948(1.6) = 3.116 mm

Step 6: Checking of the requirements of IS 456
The two requirements regarding the control of deflection are given in sec.
7.17.3. They are checked in the following:
Step 6a: Checking of the first requirement
The maximum allowable deflection = 8000/250 = 32 mm
The actual final deflection due to all loads
= 15.358 (see Eq.1 of Step 3) + 2.64 (see Eq.2 of Step 4)
+ 3.118 (see Eq.5 of Step 5e) = 21.116 mm < 32 mm. Hence,
o.k.
Step 6b: Checking of the second requirement
The maximum allowable deflection is the lesser of span/350 or 20 mm.
Here, span/350 = 22.86 mm. So, the maximum allowable deflection = 20 mm.
The actual final deflection = 1.948 (see Eq.4 of Step 5d) + 2.64 (see Eq.2 of Step
4) + 3.118 (see Eq.5 of step 5e) = 7.706 mm < 20 mm. Hence, o.k.
Thus, both the requirements of cl.23.2 of IS 456 and as given in sec.
7.17.3 are satisfied.

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7.17.9 Practice Questions and Problems with Answers
Q.1:

Why is it essential to check the structures, designed by the limit state of
collapse, by the limit state of serviceability?

A.1:

See sec. 7.17.1.

Q.2:

Explain short- and long-term deflections and the respective influencing
factors of them.

A.2:

See sec. 7.17.2.

Q.3:

State the stipulations of IS 456 regarding the control of deflection.

A.3:

See sec. 7.17.3.

Q.4: How would you select the preliminary dimensions of structures to satisfy (i)
the deflection requirements, and (ii) the lateral stability ?
A.4:

See secs. 7.17.4 A for (i) and B for (ii).

Q.5: Check the preliminary cross-sectional dimensions of Problem 1 of sec.
7.17.8 (Fig.7.17.1) if they satisfy the requirements of control of deflection.
The spacing of the beam is 3.5 m c/c. Other data are the same as those of
Problem 1 of sec. 7.17.8.
A.5:
Step 1: Check for the effective width
bf = lo /6 + bw + 6Df or spacing of the beam, whichever is less.
Here, bf = (8000/6) + 300 + 6(100) = 2234 < 3500. Hence, bf = 2234 mm
is o.k.
Step 2: Check for span to effective depth ratio
(i) As per row 1 of Table 7.1, the basic value of span to effective depth ratio is
20.
(ii) As per row 2 of Table 7.1, the modification factor is 1 since the span 8 m <
10 m.
(iii) As per row 5 of Table 7.1, the modification factor for the flanged beam is to
be obtained from Fig. 6 of IS 456 for which the ratio of web width to flange width

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= 300/2234 = 0.134. Figure 6 of IS 456 gives the modification factor as 0.8. So,
the revised span to effective depth ratio = 20(0.8) = 16.
(iv) Row 3 of Table 7.1 deals with the area and stress of tensile steel. At the
preliminary stage these values are to be assumed. However, for this problem the
area of steel is given as 1383 mm2 (2-25T + 2-16T), for which pt = Ast(100)/bf d =
1383(100)/(2234)(550) = 0.112.
fs = 0.58 fy (area of cross-section of steel required)/(area of cross-section of steel
provided)
= 0.58(415)(1) = 240.7 (assuming that the provided steel is the
same as required, which is a rare case). Figure 4 of IS 456 gives the modification
factor as 1.8. So, the revised span to effective depth ratio = 16(1.8) = 28.8.
(v) Row 4 is concerning the amount of compression steel. Here, compression
steel is not there. So, the modification factor = 1.
Therefore, the final span to effective depth ratio = 28.8.
Accordingly, effective depth of the beam = 8000/28.8 = 277.8 mm < 550
mm.
Hence,
requirements.

7.17.9

the

dimensions

of

the

cross-section

are

satisfying

the

References

1. Reinforced Concrete Limit State Design, 6th Edition, by Ashok K. Jain,
Nem Chand & Bros, Roorkee, 2002.
2. Limit State Design of Reinforced Concrete, 2nd Edition, by P.C.Varghese,
Prentice-Hall of India Pvt. Ltd., New Delhi, 2002.
3. Advanced Reinforced Concrete Design, by P.C.Varghese, Prentice-Hall of
India Pvt. Ltd., New Delhi, 2001.
4. Reinforced Concrete Design, 2nd Edition, by S.Unnikrishna Pillai and
Devdas Menon, Tata McGraw-Hill Publishing Company Limited, New
Delhi, 2003.
5. Limit State Design of Reinforced Concrete Structures, by P.Dayaratnam,
Oxford & I.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004.
6. Reinforced Concrete Design, 1st Revised Edition, by S.N.Sinha, Tata
McGraw-Hill Publishing Company. New Delhi, 1990.
7. Reinforced Concrete, 6th Edition, by S.K.Mallick and A.P.Gupta, Oxford &
IBH Publishing Co. Pvt. Ltd. New Delhi, 1996.
8. Behaviour, Analysis & Design of Reinforced Concrete Structural Elements,
by I.C.Syal and R.K.Ummat, A.H.Wheeler & Co. Ltd., Allahabad, 1989.
9. Reinforced Concrete Structures, 3rd Edition, by I.C.Syal and A.K.Goel,
A.H.Wheeler & Co. Ltd., Allahabad, 1992.
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10. Textbook of R.C.C, by G.S.Birdie and J.S.Birdie, Wiley Eastern Limited,
New Delhi, 1993.
11. Design of Concrete Structures, 13th Edition, by Arthur H. Nilson, David
Darwin and Charles W. Dolan, Tata McGraw-Hill Publishing Company
Limited, New Delhi, 2004.
12. Concrete Technology, by A.M.Neville and J.J.Brooks, ELBS with
Longman, 1994.
13. Properties of Concrete, 4th Edition, 1st Indian reprint, by A.M.Neville,
Longman, 2000.
14. Reinforced Concrete Designer’s Handbook, 10th Edition, by C.E.Reynolds
and J.C.Steedman, E & FN SPON, London, 1997.
15. Indian Standard Plain and Reinforced Concrete – Code of Practice (4th
Revision), IS 456: 2000, BIS, New Delhi.
16. Design Aids for Reinforced Concrete to IS: 456 – 1978, BIS, New Delhi.

7.17.11 Test 17 with Solutions
Maximum Marks = 50,

Maximum Time = 30 minutes

Answer all questions.
TQ.1: Explain short- and long-term deflections and the respective influencing
factors of them.
(10 marks)
A.TQ.1: See sec. 7.17.2.

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TQ.2: Check the preliminary dimensions of a singly reinforced rectangular
cantilever beam of span 4 m (Fig.7.17.4) using M 20 and Fe 415.
(15 marks)
A.TQ.2: (i) From row 1 of Table 7.1, the basic value of span to effective depth
ratio is 7.
(ii) Modification factor for row 2 is 1 as this is a singly reinforced beam.
(iii) Assuming pt as 0.6 and area of steel to be provided is the same as
area of steel required, fs = 0.58(415(1) = 240.7 N/mm2. From Fig. 4 of IS 456,
the modification factor = 1.18. Hence, the revised span to effective depth ratio is
7(1.18) = 8.26.
(iv) Modification factors for rows 4 and 5 are 1 as there is no
compression steel and this being a rectangular beam. Hence, the preliminary
effective depth needed = 4000/8.26 = 484.26 mm < 550 mm. Hence, o.k.
TQ.3: Determine the tensile steel of the cantilever beam of TQ 2 (Fig. 7.17.4)
subjected to service imposed load of 11.5 kN/m using M 20 and Fe 415.
Use Sp-16 for the design. Calculate short- and long-term deflections and
check the requirements of IS 456 regarding the deflection.
(25 marks)
A.TQ.3: Determination of tensile steel of the beam using SP-16:
Dead load of the beam = 0.3(0.6)(25) kN/m = 4.5 kN/m
Service imposed loads = 11.5 kN/m
Total service load = 16.0 kN/m
Factored load = 16(1.5) = 24 kN/m
Mu = 24(4)(4)/2 = 192 kNm
For this beam of total depth 600 mm, let us assume d = 550 mm.
Mu /bd2 = 192/(0.3)(0.55)(0.55) = 2115.70 kN/m2
Table 2 of SP-16 gives the corresponding pt = 0.678 + 0.007(0.015)/0.02
= 0.683
Again, for Mu per metre run as 192/0.3 = 640 kNm/m, chart 15 of SP-16
gives pt = 0.68 when d = 550 mm.

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With pt = 0.683, Ast = 0.683(300)(500)/100 = 1126.95 mm2. Provide 420T to have 1256 mm2. This gives provided pt = 0.761%.
Calculation of deflection
Step 1: Properties of concrete section
yt = D/2 = 300 mm, Igr = bD3/12 = 300(600)3/12 = 5.4(109) mm4
Step 2: Properties of cracked section

fcr = 0.7 20 (cl. 6.2.2 of IS 456) = 3.13 N/mm2
yt = 300 mm
Mr = fcr Igr /yt = 3.13(5.4)(109)/300 = 5.634(107) Nmm
Es = 200000 N/mm2
Ec = 5000

f ck (cl. 6.2.3.1 of IS 456) = 22360.68 N/mm2

m = Es /Ec = 8.94
Taking moment of the compressive concrete and tensile steel about the neutral
axis (Fig.7.17.5):
300 x2/2 = (8.94)(1256)(550 – x) or

x2 + 74.86 x – 41171.68 = 0
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This gives x = 168.88 mm and z = d – x/3 = 550 – 168.88/3 = 493.71 mm.
Ir = 300(168.88)3/3 + 8.94(1256)(550 – 168.88)2 = 2.1126(109) mm4
M = wl2/2 = 20(4)(4)/2 = 160 kNm
I eff =

Ir
= 1.02 I r = 2.1548 (10 9 ) mm 4
5.634 493.71
168.88
(1.2) - (
)(
) (1 ) (1)
16
550
550

This satisfies Ir ≤ Ieff ≤ Igr. So, Ieff = 2.1548(109) mm4.
Step 3: Short-term deflection (sec. 7.17.5)
Ec = 22360.68 N/mm2 (cl. 6.2.3.1 of IS 456)
Short-term deflection = wl4/8EcIeff
= 20(44)(1012)/8(22360.68)(2.1548)(109) = 13.283 mm
So, short-term deflection = 13.283 mm
(1)
Step 4: Deflection due to shrinkage (sec. 7.17.6)
k 4 = 0.72( 0.761) / 0.761 = 0.664

ψ cs = k 4 ε cs / D = (0.664)(0. 0003)/600 = 3.32(10) -7
k3 = 0.5 (from sec. 7.17.6)

α cs = k 3 ψ cs l 2 = (0.5)(3.32)(10)-7(16)(106) = 2.656 mm
(2)
Step 5: Deflection due to creep (sec. 7.17.7)
Step 5a: Calculation of α 1cc ( perm )
Assuming the age of concrete at loading as 28 days, cl. 6.2.5.1 of IS 456
gives
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θ = 1.6
So,

Ecc = Ec /(1 + θ ) = 8600.2615 N/mm2
m = Es /Ecc = 200000/8600.2615 = 23.255

Step 5b: Properties of cracked section

From Fig.7.17.6, taking moment of compressive concrete and tensile steel
about the neutral axis, we have:
300 x2/2 = (23.255)(1256)(550 - x)
or

x2 + 194.72 x – 107097.03 = 0

solving we get x = 244.072 mm
z = d – x/3 = 468.643 mm
Ir = 300(244.072)3/3 + (23.255)(1256)(550 – 468.643)2
= 1.6473(10)9 mm4
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Mr = 5.634( 107) Nmm (see Step 2)
M = wperm l2/2 = 4.5(42)/2 = 36 kNm

I eff =

Ir
= 2.1786 I r = 3.5888(10 9 ) mm 4
5.634 468.643
244.072
1.2 - (
)(
) (1 ) (1)
3.6
550
550

Since this satisfies Ir ≤ Ieff ≤ Igr, we have, Ieff = 3.5888(109) mm4. For the value
of Igr please see Step 1.
Step 5c: Calculation of α 1cc ( perm )

α 1cc ( perm ) = (wperm)( l4)/(8Ecc Ieff) = 4.5(4)4(10)12/8(8600.2615)(3.5888)(109)
= 4.665 mm
(3)
Step 5d: Calculation of α 1( perm )

α 1( perm ) = (wperm)( l4)/(8Ec Ieff) = 4.5(4)4(10)12/8(22360.68)(3.5888)(109)
= 1.794 mm
(4)
Step 5e: Calculation of deflection due to creep

α cc ( perm ) = α 1cc ( perm ) - α 1( perm )
= 4.665 – 1.794 = 2.871 mm
(5)
Moreover:

α cc ( perm ) = α 1cc ( perm ) ( θ ) gives α cc ( perm ) = 1.794(1.6) = 2.874 mm.

Step 6: Checking of the two requirements of IS 456
Step 6a: First requirement
Maximum allowable deflection = 4000/250 = 16 mm
The actual deflection = 13.283 (Eq.1 of Step 3) + 2.656 (Eq.2 of Step 4)
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+ 2.871 (Eq.5 of Step 5e) = 18.81 > Allowable 16 mm.
Step 6b: Second requirement
The allowable deflection is lesser of span/350 or 20 mm. Here, span/350 =
11.428 mm is the allowable deflection. The actual deflection = 1.794 (Eq.4 of
Step 5d) + 2.656 (Eq.2 of Step 4) + 2.871 (Eq.5 of step 5e) = 7.321 mm <
11.428 mm.

Remarks:
Though the second requirement is satisfying, the first requirement is not
satisfying. However the extra deflection is only 2.81 mm, which can be made up
by giving camber instead of revising the section.

7.17.12 Summary of this Lesson
This lesson illustrates the importance of checking the structures for the
limit state of serviceability after designing by the limit state of collapse. The shortand long-term deflections along with their respective influencing factors are
explained. The code requirements for the control of deflection and the necessary
guidelines for the selection of dimensions of cross-section are stated. Numerical
examples, solved as illustrative example and given in the practice problem and
test will help the students in understanding the calculations clearly for their
application in the design problems.

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