Limit state of Serviceability

• Limit State of Serviceability-

– limit state of deflection-
– short term and long term deflection-
– IS code recommendations-

• limit state of cracking-

– estimation of crack width-
– numerical examples
Need for Limit State of Serviceability

1. Structures designed by limit state of collapse are

found to be of comparatively smaller sections.
2. They must be checked for deflection and width of
cracks.
3. Excessive deflection of a structure or part thereof
adversely affects the appearance and efficiency of the
structure, finishes or partitions.
4. Excessive cracking of concrete also seriously affects
the appearance and durability of the structure.
Short- and Long-term Deflections
Short-term deflection refers to the immediate deflection
after casting and application of partial or full service
loads.

The following factors influence the short-term deflection

1. magnitude and distribution of live loads,

2. span and type of end supports,
3. cross-sectional area of the members,
4. amount of steel reinforcement and the stress
5. developed in the reinforcement,
6. characteristic strengths of concrete and steel,
7. amount and extent of cracking. ,
• Long-term deflection occurs over a long period of
time largely due to shrinkage and creep of the
materials.
• long-term deflection is almost two to three times of
the short-term deflection.
•
• factors influencing the long-term deflection
• humidity and temperature ranges during curing,
• age of concrete at the time of loading
• type and size of aggregates,
• water-cement ratio,
• amount of compression reinforcement,
• size of members etc.
Code Provisions

Clause 35.3 of IS 456 refers to the limit state

of serviceability comprising
1.Deflection in CL. 35.3.1 and
2.Cracking in CL. 35.3.2.
Control of Deflection CL 23.2

both the requirements are to be fulfilled

 Design Procedure to Limit Deflections within acceptable
limits as per IS 456-2000

1. Empirical Method
based on Span to effective depth ratios
2. Deflection Computations
Computing actual deflections due to short term
and long term effects as per Annex C
 Span to effective depth ratios CL 23.2.1

Concept recommended by IS 456 to be used as the

guidelines to assume the initial dimensions of the members
which will generally satisfy the deflection limits.
CL 23.2.1 stipulates different span to effective depth ratios
Basic values of Span to Effective Depth ratios
Modification Factor (ks) for Span > 10m

ks = 10/ Span in m
L/d = Basic Value x ks
Modification factor for Tension Reinforcement (kt)

L/d = Basic Value x ks x kt

Modification factor for Compression Reinforcement (kc)

L/d = Basic Value x ks x kt x kc

Reduction factor for Flanged Beams (kf)

L/d = Basic Value x ks x kt x kc xkf

Example 1

Check for the limit state of deflection using empirical

method for the rectangular beam with following data.
Ast = 4- #20; Asc = 2-#16, Beam Size = 300x650 mm
Effective cover = 50mm. Fe415. Span= 12m and simply
supported.
Step1

d =650-50 = 600mm

Ast = 1256 mm2 ; Asc = 402 mm2

pt = 100 x1256/(300x600) = 0.7%

pc = 100 x402/(300x600) = 0.22%

For Spans up to 10m :

L/d (Basic Value) = 20 CL 23.2.1 (a)

Step 2: Modification factors

Span > 10 mm

• ks = 10/12 =0.833 CL 23.2.1 (b)

pt = 0.7% CL 23.2.1 (c)

• fs = 0.58 fy = 240 N/mm2

• From Fig 4; kt = 1.1

pc = 0.22% CL 23.2.1 (d)

• From Fig 5; kc = 1.07

 Rectangular Beam

• bw/bf =1 ; From Fig6

• kf = 1 CL 23.2.1 (e)
Step 3: Check (L /d) ratio

Permissible Value

L/d = 20 x 0.833 x 1.1 x 1.07 x 1 = 19.6

Actual Value

L/d = 12x1000/600 = 20 > 19.6 NOT OK

Example 2

Check for the limit state of deflection using empirical

method for the T beam with following data.
Ast = 1600 mm2; Asc = 900 mm2, bw= 300mm, d=400mm,
bf= 900mm; Fe415; Span= 8m and continuous
Step1

d =400mm

pt = 100 x1600/(900x400) = 0.44%

pc = 100 x900/(900x400) = 0.25%

For Spans up to 10m :

L/d (Basic Value) = 26 CL 23.2.1 (a)

Step 2: Modification factors

Span < 10 mm

• ks = 1 CL 23.2.1 (b)

pt = 0.44% CL 23.2.1 (c)

• fs = 0.58 fy = 240 N/mm2

• From Fig 4; kt = 1.3

pc = 0.25% CL 23.2.1 (d)

• From Fig 5; kc = 1.08

 Flanged Beam
• bw/bf =300/900 = 0.33 ; From Fig6
• kf = 0.8 CL 23.2.1 (e)
Step 3: Check (L /d) ratio

Permissible Value

L/d = 26 x 1 x 1.3 x 1.08 x 0.8 = 29.2

Actual Value

L/d = 8x1000/400 = 20 < 29.2 OK

Example 3 - Deflection Computation as per Annex C

A cantilever beam of span 4m and size=300x600 mm, is

subjected to a Maximum bending moment,
M =160 kNm. under service conditions due to UDL.

The beam is reinforced with 4-#20 on the tension face at an

effective cover=50mm.

M20 and Fe 415 steel. Compute deflection due to short and

long term effects.
Step 1: Properties of Uncracked Section

Igr (Gross Moment of Inertia) = bD3/12

Igr = 300x6003/12 = 54x108 mm4

yt = D/2 = 300 mm
Step 2: Properties of Cracked Section

This is determined based on the area of transformed section.

Steel is replaced by an equivalent area of concrete = m Ast

Ast = 1256 mm2 ( 4-#20)

Ec = 5000 (20)0.5 = 22361 N/mm2 CL 6.2.3.1

Modular ratio m = Es/Ec = 200000/22361 = 8.94

 50
m Ast

550-X
NA NA
550
X Comp. X

300 300
C/S Transformed Area

Position of NA.

Moment of Area in Compression about NA = 300 X2/2

Moment of equivalent Area in Tension about NA = m Ast (550-X)
300 X2/2 = 8.94 x1256 x(550-X)

X = 168.9 mm

Lever arm

z = d – (X/3) = 550 – (168.9/3) = 493.7 mm

Cracked Moment of Inertia

Ir = {300 x 168.93/3} + 8.94x1256x(550-168.9)2

=21.1 x108 mm4

Cracking Moment Mr

fcr = 0.7 (20)0.5 = 3.13 N/mm2 Cl. 6.2.2

Mr = fcr Igr /yt

= 3.13(54)(108)/300 = 56.34 kNm

Effective Moment of Inertia Ieff C-2.1

Ir
Ieff 
1.2  ( Mr / M )( Z / d )(1  X / d )(bw / b)
 Ir
Ieff 
1.2 - (56.34/160 ) (493.7/550 ) (1- (168.9/550 )) (300/300)

Ieff = 1.02 Ir

Ir ≤ Ieff

Ieff = 1.02 x 21.1 x108 = 21.52 x108 mm4 < Igr

Condition is Satisfied
 a
Step 3: Short-term deflection ( i)

ai= wL4 / 8 Ec Ieff

= (wL2/2) (L2/4 Ec Ieff )

= M L2 / (4 Ec Ieff )

= 160 x106 x 40002 / (4x22361x 21.52x108)

ai = 13.3 mm
Step 4: Deflection due to shrinkage C-3.1

acs = k3 ccs L2
k3 = 0.5 (cantilever)
ccs = k4 ecs / D

pt = 100 x 1256/ (300 x550) = 0.76%; pc=0

(pt – pc) = 0.76

For 0.25 < (pt-pc) < 1.0

k4 = 0.72 x0.76/ (0.76)0.5 = 0.63 < 1.0

K4=0.63
ecs = 0.0003 CL 6.2.4.1

ccs = 0.63 x0.0003 / 600 = 3.15 x10-7

acs = 0.5 x 3.15 x10 -7 x (4000)2 = 2.52mm

Step 5: Deflection due to Creep C-4.1

acc(perm) = aicc(perm) - ai(perm)

Compute aicc(perm)

Assuming the age of concrete at loading as 28 days, CL.

6.2.5.1; θ = 1.6

Ece = Ec /(1 + θ) =22361/ (2.6) = 8600.4 N/mm2

m = Es /Ece = 200000/8600.4 = 23.25

Position of NA
300 X2/2 = 23.25 x1256 x(550-X)

X = 244.05 mm

Lever arm

z = d – (X/3) = 550 – (244.05 /3) = 468.65 mm

Cracked Moment of Inertia

Ir = {300 x 244.053/3} + 23.25x1256x(550 -244.05)2

=41.9 x108 mm4

Assume 50% of Moment as due to permanent
Loads
M = 0.5 x160 = 80 kNm
 Effective Moment of Inertia Ieff C-2.1

Ir
Ieff 
1.2 - (56.34/80 )(468.65/55 0 ) (1- (244.05/ 550 )) (300/300)

Ieff = 1.154Ir

Ir ≤ Ieff.

Ieff = 1.154 x 41.9 x108 = 48.37x108 mm4 ≤ Igr

Condition is Satisfied
aicc (Perm) = M L2 / (4 Ece Ieff )
= 80x106x40002/(4 x 8600.4x 48.37x108)

= 7.69 mm
Compute ai(perm)
ai (Perm) = M L / (4 E I
2
c eff )

= 80x106x40002/(4 x 22361 x 48.37x108)

= 2.96mm

acc(perm) = 7.69 – 2.96 = 4.73 mm

Step 6: Limiting Values CL 23.2

CL 23.2.(a)
Maximum allowable deflection
= 4000/250 = 16 mm

Actual deflection = 13.3 + 2.52+ 4.73

= 20.55mm > 16 mm NOT OK
CL 23.2.(b)
Maximum allowable deflection
= 4000/350 = 11.4 mm < 20 mm

Actual deflection = 2.96 + 2.52+ 4.73

= 10.21mm < 11.4 mm OK
Limit State of Serviceability - Cracking

CL 35.3.2

Maximum Crack Width

< 0.3mm (Normal )

< 0.2 mm (exposed to moisture, contact
with soil or ground water)
< 0.1 mm (Severe Table 3)
Control of Crack Width

1. As per General Detailing rules specified in CL

26 ; IS 456 with regard to minimum Ast,
maximum spacing of rebars, minimum cover
etc.

2. Crack width Computation As per Annex F

Crack width Computation As per Annex F
Example

For the beam C/S shown compute crack widths

at points A,B,C and D. The beam is simply
supported over a span of 6m and subjected to
a service load including self weight of 16 kN/m.
Ast = 3- #20, M20 and Fe 415
 X
d = 450 NA

.
h = 500
D
(2/3) (d-X)
50 100 100 50

50
. .A .C
B
b = 300
Step 1: Basic Parameters

Ast = 3 - #20 = 942 mm2

Ms = ws L2/8 = 16 x 36/8 = 72 kNm

Ec = 5000 (20)0.5 = 22361 N/mm2 CL 6.2.3.1

Es = 200000 N/mm2
m = Es/Ec = 8.94
Step 2: Position of NA

300 X2/2 = 8.94 x 942 x (450-X)

X = 133.36 mm

Step 3: Cracked Moment of Inertia (Ir)

Ir = 300 x 133.363/3 + 8.94 x 942x(450-133.36)2

Ir = 1081.52 x106 mm4
Step 4: Strain in Concrete at the level of steel
ecs = fcs/ Ec
fcs = Ms(d-x)/ Ir (M/I = f/y; f = M x y /I)

= 72x106 (450-133.36) / 1081.52 x106

= 21.08 N/mm2
ecs = 21.08/ 22361 = 9.43x10 - 4
Step 5: Strain e1 at A,B,C and D
e1 @ D (2/3)d-x

ecs d-x h-x

e1 @ A,B,C
At A,B,C
e1 = ecs ( h-X) /(d-X)
= 9.43x10 – 4 (500-133.36)/(450-133.36)
= 1.09x10-3
At D
e1 = ecs (2/3)( d-X) /(d-X)
= 9.43x10 – 4 (2/3)
= 6.28x10 - 4
 em at A,B,C,D
Step 6:

em = e1 – {b(h-X)(a-X) / (3 Es As(d-X)}
b(h-X) / (3 Es As (d-X) =

300(500-133.36) / (3 x 200000 x 942 x (450-133.36)

= 6.15x10-7

em = e1 – 6.15x10 -7 (a-X)
em = e1 – 6.15x10 -7 (a-X)

At A,B,C
e
a=h = 500mm, 1 = 1.09x10-3

em = e1 – 6.15x10 -7 (a-X) = 8.64x10-4

At D
e
a=X +(2/3)(d-X) = 344.45 mm, 1 = 6.28x10 - 4

em = e1 – 6.15x10 -7 (a-X) = 4.98x10-4

Step 7: Crack width wcr
At A
acr = 50-10 = 40mm
e
Cmin = 40mm; m = 8.64x10-4
wcr = 3 x 40 x 8.64x10-4 = 0.103mm
At B
acr = (502 + 502)0.5 -10 = 60.7mm
e
Cmin = 40mm ; m = 8.64x10-4
wcr
= 3 x60.71x8.64x10-4 /( 1+ 2(60.71-40)/(500 -133.36)) =
0.141mm
At C
acr = (502 + 502)0.5 -10 = 60.7mm
Cmin = 40mm
wcr = 0.141 mm

At D

acr = (502 + ((450-133.36)/3)2)0.5 -10

= 106.8mm
e
Cmin = 40mm; m = 4.98x10-4
wcr = 0.117 mm
DEFLECTIONS OF BEAMS
 Deflections
• Deflection of structure or part thereof shall not adversely affect
the appearance or efficiency of structure or finishes or partitions.

• Deflection shall generally be limited to the following:

i. Final deflection due to all loads including the effects of
temperature, creep and shrinkage and measured from as-cast level
of supports of floors, roofs and all other horizontal members
should not normally exceed span/250.

ii.Deflection including effects of temperature, creep and shrinkage

occurring after erection of partitions and application of finishes
should not normally exceed (span/350) or 20mm whichever is less.
• Factors influencing limits on deflection in flexural members

• Aesthetic/psychological discomfort

• Crack width limitation

• Effect on attached structural and non structural elements

• Ponding in (roof) slabs

 • Limits on deflections of flexural members

i. Final deflection due to all loads including temperature effect,

long term effects of creep and shrinkage after construction of
structural members and before the construction of partitions and
finishes should be less than Span
250

ii. Deflections due to various loads acting on the structural members

including temperature effect, long-term effects of creep and
shrinkage that occur after the construction of partitions and
finishes should be less than Span or 20 mm
350
 Short-Term Deflections
• Short-term deflections due to the applied service loads are
generally based on the assumptions of linear elastic behavior,
and for this purpose, reinforced concrete is treated as a
homogenous material.
WL 3
  kw
EI
Where
• W = total load on the span ML 2
 km
EI
M = maximum moment
Kw and Km – constants depend on the load distribution conditions of end
restraint and variation in the flexural rigidity EI.
• For the purpose of calculating short–term deflections in reinforced concrete
flexural members elastic theory may be used of.
• Flexural rigidity is to be considered in the calculation.
 Short-Term Deflections
• Modulus of elasticity of concrete depends on:
– Concrete quality,
– Age
– Stress level and
– Rate or duration of applied load.
• Short-term loading upto service load levels, IS:456-2000 specifies the
modulus of elasticity as Ec=5000√fck.
• Second moment of area, I, to be considered in the deflection
calculations is influenced by
– Percentage reinforcement
– Extent of flexural cracking,
• which in turn depends on the applied bending moment and the
modulus of rupture, fcr, of concrete.
• Flexural rigidity, EI is obtainable as the slope(secant modulus)
of the moment-curvature relationship.
i. EIT – Based on uncracked transformed section
ii. EIgr – Based on gross uncracked transformed section ignoring steel
iii. EIeff – Based on effective section
iv. EIcr– Based on “cracked transformed section”
 EIT - True EI for M < Mcr (EIT - constant)
 EIeff - True EI for M > Mcr (EIeff – depends on the load level)

EIT > EIgr > EIeff > EIcr

• IS: 456-2000 specifies, Icr
Ieff 
 M  z  x  b 
1.2  cr  1  w 
• Icr ≤ Ieff ≤ Igr  M  d  d  b 
• Icr = Moment of inertia of the cracked section
• Mcr = Cracking moment I gr
Mcr  fcr
yt
• fcr = Modulus of rupture of concrete
• Igr = Moment of inertia of the gross section about the centroidal axis, neglecting the
reinforcement.
• yt = Distance from centroidal axis of gross section, neglecting reinforcement to
extreme fiber in tension.
• M = Maximum moment under service load
• Z = lever arm
• x = Depth of neutral axis
• D = Effective depth
• bw= Breadth of web and,
• b = Breadth of compression face
• For continuous beams, deflections shall be calculated using the
values of Icr, Igr and Mcr, modified by the following equation:

X e  k1
 X1  X 2   1  k X
1 0
2
k1 = coefficient
X= value of Icr, Igr, or Mcr as appropriate
Xe= modified value of X
X1 and X2= values of “X” at the supports 1 and 2
Xo= value of “X” at mid-span.

Values of k1

k2 0.5 or less 1.4 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3
k1 0 1.0 0.03 0.08 0.16 0.30 0.50 0.73 0.91 0.97

M1 ,M2 = support moments  M 1  M 2  

k 2   
MF1, MF2 = fixed end moment  M F1  M F 2 
 Deflections due to shrinkage
• Deflections due to shrinkage Δcs may be computed from the equation:
 cs  k 3 cs l 2

k3 = constant depending upon the support conditions

= 0.5, for cantilever
= 0.125, for simply supported members
= 0.086, for members continuous at one end
= 0.063, for fully continues members
Ψcs= shrinkage curvature K 4  cs

D
(Pt  Pc )
K4  0.72 for
1.0 0.25 (Pt  Pc ) 1.0
Pt
100Ast 100Asc
( Pt  Pc ) Pt  ; Pc 
 0.65  1.0 for ( Pt  Pc )  1.0 bd bd
Pt
D = total depth of the section
l= length of the span.
 Deflections due to creep

• Creep deflections due to permanent loads is given by,

Δ cc(perm)  Δ i,cc(perm)  Δ i,(perm)

where,
• Δi,cc(perm)= initial plus creep deflections due to permanent loads obtained
using an elastic analysis with effective modulus of elasticity.
• Δi,(perm)= Short-term deflections due to permanent load using Ec=5000√fck
• Effective modulus of elasticity is, EC
Ece 
[1  ]

Age at loading Creep coefficient, θ

7 days 2.2
28 days 1.6
1 year 1.1
 Control of Deflections
• For control of deflection two methods are usually described in
codes of practice
I. As per the empirical method, the span-to-effective depth ratios of
members should not be more than those specified in the codes

II. In theoretical method, calculating the actual deflection and

checking it with the allowable deflection in the codes of practice
 Control of Deflections
• Empirical procedure for control of deflection is to control the span-
to-effective depth ratio.
• Deflection of beams or slabs depend on
1. Span-to-effective depth ratio
2. Type of support
3. Percentage of tension reinforcement or the stress level at service loads
if more than the necessary steel is provided at the section.
4. Percentage of compression reinforcement
5. Type of beam (whether it is flanged or rectangular)
• Vertical deflection limits may generally be assumed to be satisfied
provided that the span-to-depth ratios are not greater than the
values obtained
 Basic Span-to-depth Ratios
a. Basic values of span-to-effective depth ratios for spans up to or
less than 10m are

1. Cantilever 7
2. Simply supported 20
3. Fixed or Continuous 26
b. For spans above 10m, values in (a) may be multiplied by
(10/span in meters), except for cantilever.
– For cantilever beams, the actual deflections should be
calculated and the requirement for limit state of deflection be
checked.
1. Depending on the area and the stress in steel reinforcement in tension,
the values in (a) or (b) shall be modified by multiplying with the
modification factor obtained as per Fig. 4.
Factor F1 can be calculated

1
F1   2 .0
0.225  0.00322 f s  0.625 log 10 pt 
2. Depending on the area of compression reinforcement, the value of
span-to-depth ratio is further modified by multiplying with the
modification factor obtained as per Fig. 5.
Factor F2 can be calculated

1 .6 p c
F2   1 .5
 pc  0.275
3. For flanged beam, values of (a) or (b) be modified as per Fig. 6 and the
reinforcement percentage for use in Fig.4 and 5 should be based on area
of section equal to bfd.
Factor F3 can be calculated

2  bw 
F3  0.8   0.3   0.8
7  b f 

 Final Span-to-Effective Depth Ratio

• The final allowable span-to-effective depth ratio

L
 Basic Ratio F1 F2 F3 
d

• The ratio (L/d) obtained should be as follows

L L
( obtained )    F1 F 2 F 3 
d  d  Basic
 Slenderness Limits for Beam to Ensure Lateral Stability
• A simply supported or continuous beam shall be so proportioned that
clear distance between the lateral restraints does not exceed 60b or
250b 2
d whichever is less.
Where

d = effective depth of the beam and

b = breadth of compression face midway between the lateral restraints.

• For a cantilever, the clear distance from the free end of the cantilever
100b 2

to the lateral restraint shall not exceed 25b or

d
• whichever is less.
1.A simply supported beam of cross section of width 200mm and overall depth
400mm is provided with 3- 16 mm diameter HYSD bars in tension. Cover to
the reinforcement is 40mm. The span of the beam is 5.0m. The beam is
subjected to a uniformly distributed dead load of 10 kN/m and a live load of 15
kN/m. Half of the imposed load is permanent. Calculate the total long-term
deflection at the mid-span. fck=40 N/mm2, fy=415 N/mm2, θ=2.5=creep
coefficient, εcs=0.0003
2.A simply supported beam rectangular in cross section, 450mmX750mm, spanning 10m
is subjected to a dead load of 24 kN/m and an imposed load of 34.5 KN/m. The
characteristic concrete and steel strengths are fck=40 N/mm2 and fy=460 N/mm2
respectively, Es= 200 kN/m2 and Ec=5000√fck. (Ast= (3)- 40mm dia bars).
i. Determine the mid-span service-load concrete strains at the level of the tension
reinforcements at the tension face (i.e. the soffit) of the beam, and at 250 mm below the
neutral axis.
ii.If, due to creep of concrete, the value of Ec becomes half of the short-term value,
calculate the strain due to the long-term service loads.
1. A beam 200mm wide and 400mm overall depth is reinforced with 2 nos of
20mm diameter bars is acted on by a load, part of which is permanent. The
bending moment due to the total loading is50 kN-m and the bending
moment, Mp due to the permanent load is 36 kN-m. Assuming the section as
a partially cracked section, determine, the long-term curvature of the beam
under permanent load, if fck=0.55 N/mm2 appropriate to long-term loading,
the instantaneous curvature under the total load and the permanent loads, if
fck=1.0 N/mm2 for short-term loading and the difference between the
instantaneous curvature under the total and permanent load. Given that
Ec=5000√fck, Es=200 kN/m2, θ=2.5, fck=40 N/mm2 and fy=460 N/mm2.