limit state of serviceability in rcc

Prepared by Dr. Prashanth J. Design of RCC Structures 1

UNIT 3: FLEXURE AND SERVICEABILITY LIMIT STATES

3.1 GENERAL

In the previous chapter, the behaviour of reinforced concrete beams was explained, and

procedures given for the analysis of sections. It may be noted that the results of the analysis of a

given beam section is unique, being dictated solely by the conditions of equilibrium of forces and

compatibility of strains. On the basis of these computations, it is possible to decide whether or

not the beam is „safe‟ under known moments.

The design problem is somewhat the reverse of the analysis problem. The external loads (or load

effects), material properties and the skeletal dimensions of the beam are given, and it is required

to arrive at suitable cross-sectional dimensions and details of the reinforcing steel, which would

give adequate safety and serviceability. In designing for flexure, the distribution of bending

moments along the length of the beam must be known from structural analysis. For this, the

initial cross-sectional dimensions have to be assumed in order to estimate dead loads; this is also

required for the analysis of indeterminate structures (such as continuous beams). The adequacy

of the assumed dimensions should be verified and suitable changes made, if required.

NOTE: Unlike the analysis problem, the design problem does not have a unique solution because the flexural

strength of a section is dependent on its width and effective depth, and on the area of reinforcement; and there

are several combinations of these which would give the required strength. Different designers may come up with

different solutions, all of which may meet the desired requirements.

A complete design of a beam involves considerations of safety under the ultimate limit states in

flexure, shear, torsion and bond, as well as considerations of the serviceability limit states of

deflection, crack-width, durability etc.

Prior to taking up problems related to design in flexure, it is necessary to have first an

understanding of the requirements related to the placing of flexural reinforcement, control of

deflection, as well as other guidelines for the selection of member sizes. These are discussed in

the following sections.

3.2 REQUIREMENTS OF FLEXURAL REINFORCEMENT

3.2.1 Concrete Cover

Clear cover is the distance measured from the exposed concrete surface (without plaster and

other finishes) to the nearest surface of the reinforcing bar. The Code (Cl. 26.4.1) defines the

term nominal cover as “the design depth of concrete cover to all steel reinforcements, including

links”. This cover is required to protect the reinforcing bars from corrosion and fire, and also to

give the reinforcing bars sufficient embedment to enable them to be stressed without „slipping‟


(losing bond with the concrete). As mentioned earlier, the recent revision in the Code with its

emphasis on increased durability, has incorporated increased cover requirements, based on the

severity of the environmental exposure conditions.

The „nominal cover‟ to meet durability requirements, depending on exposure condition, are

summarized in Table 3.1. As corrosion of reinforcing bars is a common and serious occurrence,

it is advisable to specify liberal clear cover in general (particularly in excessively wet and humid

environments, and in coastal areas). It may be noted that in actual construction, the clear cover

obtained may be (and often is) less than the specified clear cover; however, this should be within

the tolerance allowed and appropriate allowance should be made for such errors in construction.

In this context, it is important to note the revised tolerance specified in IS 456 (2000), according

to which the maximum deviation in clear cover from the value specified by the designer are “+10

mm and –0 mm” (No reduction in clear cover is permitted; an increase in clear cover up to 10

mm above the specified nominal cover is allowed).

Table 3.1: Nominal cover requirements based on exposure conditions

The clause in the earlier version of the Code, limiting the maximum clear cover in any

construction to 75 mm, for some reason, has been eliminated in the revised code. The general

message underlying the revised recommendations in the code pertaining to clear cover seems to

be: “the more the cover, the more durable the concrete”. Unfortunately, the code does not also

convey the message that the provision of very large covers (100 mm or more) is undesirable, and

can be counter-productive, causing increased crack-widths, particularly in flexural members

(such as slabs and beams). Large crack-widths (greater than 0.3 mm) permit the ingress of

moisture and chemical attack to the concrete, resulting in possible corrosion to reinforcement and

deterioration of concrete. There is little use in providing increased cover to reinforcement, if that

cover is cracked, and the likelihood of cracking increases with increased cover. It is therefore

necessary to impose an upper limit to clear cover (usually 75 mm), and to enforce the checking

for the limit state of cracking when large covers are provided.

It may be noted that in the earlier version of the Code, the clear cover requirements were based

on the type of structural element (for example, 15 mm in slabs, 25 mm in beams, 40 mm in

columns, etc.). The clear cover specifications are now made applicable for all types of structural


elements. However, certain minimum clear cover requirements have been specified in Cl.

26.4.2.1 of the Code for columns (for longitudinal bars, 40 mm in general) and in Cl. 26.4.2.2 for

footings (50 mm in general).

In addition, the Code has introduced nominal cover requirements, based on fire resistance (in

terms of hours) required. They are described in Cl. 26.4.3 of the Code. In general, for a nominal

1 hour fire resistance, the nominal cover specified is 20 mm for beams and slabs, and 40 mm for

columns. Larger cover is required only if the structural element under consideration has to be

specially designed for fire resistance.

3.2.2 Spacing of Reinforcing Bars

The Code specifies minimum and maximum limits for the spacing between parallel reinforcing

bars in a layer. The minimum limits are necessary to ensure that the concrete can be placed easily

in between and around the bars during the placement of fresh concrete. The maximum limits are

specified for bars in tension for the purpose of controlling crack-widths and improving bond.

The minimum spacing limits can be met without difficulty in slabs in general, because of the

large widths available and the relatively low percentage of flexural reinforcement required.

However, in the case of beams, which have limited widths and are required to accommodate

relatively large areas of flexural reinforcement, the minimum spacing requirements can

sometimes govern the selection of the widths of the beams. If all the reinforcing bars cannot be

accommodated in a single layer with the necessary clearance between the bars and the clear

cover on the two sides [Fig. 3.1], the options are:

• to increase the beam width;

• to place the bars in two or more layers, properly separated [Fig. 3.1(a)]; and

• to bundle groups of parallel bars (two, three or four bars in each bundle) [Fig. 3.1(c)].

While fixing the overall size of the beam or the thickness of the slab, it is desirable to use

multiples of 5 mm for slabs and 50 mm (or 25 mm) for beams. This will be convenient in the

construction of the formwork. The requirements for placement of flexural reinforcement are

described in Cl. 26.3 of the Code. The salient features of these specifications are summarized in

Fig. 3.1. The requirements for singly reinforced beams, slabs and doubly reinforced beams are

depicted in parts (a), (b) and (c) respectively of Fig. 3.1.

Stirrups provided in beams serve as transverse shear reinforcement. In singly reinforced beams,

they may be provided as U-shaped stirrups, with two hanger bars at top [Fig. 3.1(a)]. However, it

is more common to provide fully closed rectangular stirrups [Fig. 3.1(c)], for both singly and

doubly reinforced sections; this is mandatory in the doubly reinforced sections for the effective

functioning of the compression steel. Stirrups required for resisting torsion must also be of the

closed form.


Fig. 3.1: Code requirements for flexural reinforcement placement

In addition to the requirements indicated in Fig. 3.1, the Code specifies limits to the maximum

spacing of tension reinforcing bars for crack control [refer Table 15 of the Code]. It may be

noted that, for a given area of tension reinforcement, providing several small-diameter bars (that

are well distributed in one or more layers in the extreme tension zone) is more effective in

controlling cracks and improving bond than providing fewer bars of larger diameter. For this


reason, the Code (Cl. 26.5.2.2 & 26.3.3b) limits the maximum diameter of reinforcing bars in

slabs to one-eighth of the total thickness of the slab, and the maximum spacing of such main bars

to 3d or 300 mm (whichever is less) [Fig. 3.1(b)]. However, it may be noted that when large

cover is provided, more stringent bar spacing may be required to achieve the desired crack

control.

The horizontal distance between two parallel main bars shall usually be not less than the greatest

of the following

i. Diameter of the bar if the diameters are equal

ii. The diameter of the larger bar if the diameters are unequal

iii. 5mm more than the nominal maximum size of coarse aggregate

Greater horizontal spacing than the minimum specified above should be provided wherever

possible. However when needle vibrators are used, the horizontal distance between bars of a

group may be reduced to two thirds the nominal maximum size of the coarse aggregate, provided

that sufficient space is left between groups of bars to enable the vibrator to be immersed.

Where there are 2 or more rows of bars, the bars shall be vertically in line and the minimum

vertical distance between the bars shall be of the greatest of the following

i. 15 mm

ii. Maximum size of aggregate

iii. Maximum size of bars

3.2.3 Minimum and Maximum Areas of Flexural Reinforcement

A minimum area of tension reinforcing steel is required in flexural members not only to resist

possible load effects, but also to control cracking in concrete due to shrinkage and temperature

variations.

Minimum Flexural Reinforcement in Beams

In the case of beams, the Code (Cl. 26.5.1.1) prescribes the following:

So,

In the case of flanged beams, the width of the web bw should be considered instead of b.

It can be shown that the (Ast)min given by the above equation results in an ultimate moment of

resistance that is approximately equal to the „cracking moment‟ of an identical plain concrete

section. Thus, the minimum reinforcement requirement ensures that a sudden failure is avoided

at M = Mcr.


Minimum Flexural Reinforcement in Slabs

As specified in Cl. 26.5.2, the minimum reinforcement (Ast)min in either direction in slabs is

given by

where Ag denotes the gross area of the section (b×D).

In the design of one-way slabs, this minimum reinforcement is also to be provided for the

secondary (or distributor) reinforcement along the direction perpendicular to the main

reinforcement, with the spacing of such bars not exceeding 5d or 450 mm (whichever is less)

[Fig. 3.1(b)]. It may be noted that in the case of slabs, sudden failure due to an overload is less

likely owing to better lateral distribution of the load effects. Hence, the minimum steel

requirements of slabs are based on considerations of shrinkage and temperature effects alone,

and not on strength. Accordingly, the specified value of (pt)min is somewhat smaller in the case of

slabs, compared to beams. However, for exposure conditions where crack control is of special

importance, reinforcement in excess of that given by the above equation should be provided.

Maximum Flexural Reinforcement in Beams

Providing excessive reinforcement in beams can result in congestion (particularly at beam-

column junctions), thereby adversely affecting the proper placement and compaction of concrete.

For this reason, the Code (Cl. 26.5.1) restricts the area of tension reinforcement (Ast) as well as

compression reinforcement (Asc) in beams to a maximum value of 0.04bD. If both Asc and Ast are

provided at their maximum limits, the total area (Asc + Ast) of steel would be equal to 8 percent

of the gross area of the beam section; this is rather excessive. It is recommended that such high

reinforcement areas should be generally avoided by suitable design measures. These include:

• increasing the beam size (especially depth);

• improving the grades of concrete and steel.

3.3 GUIDELINES FOR SELECTION OF MEMBER SIZES

As explained in Section 3.2, the selection of flexural member sizes (from a structural viewpoint)

is often dictated by serviceability criteria (need to control deflections and crack-widths) as well

as requirements related to the placement of reinforcement. However, there are other structural,

economic and architectural considerations that come into play in the design of reinforced

concrete beams.

3.3.1 General Guidelines for Beam Sizes

The design problem does not have a unique solution. Many choices of beam sizes are feasible in

any given design situation. In general, for the purpose of designing for flexure, it is economical


to opt for singly reinforced sections with moderate percentage tension reinforcement (pt ≈ 0.5 to

0.8 times pt,lim).

Given a choice between increasing either the width of a beam or its depth, it is always

advantageous to resort to increase in depth. This results not only in improved moment resisting

capacity, but also in improved flexural stiffness, and hence, less deflections, curvatures and

crack-widths. However, very deep beams are generally not desirable, as they result in a loss of

headroom or an overall increase in the building height. In general, the recommended ratio of

overall depth (D) to width (b) in rectangular beam sections is in the range of 1.5 to 2. It may be

higher (up to 3 or even more) for beams carrying very heavy loads. The width and depth of

beams are also governed by the shear force on the section. Often, architectural considerations

dictate the sizes of beams. If these are too restrictive, then the desired strength of the beam in

flexure can be provided by making it „doubly reinforced‟ and/or by providing high strength

concrete and steel. In the case of beam-supported slab systems which are cast integrally, the

beams can be advantageously modelled as „flanged beams‟.

In the case of building frames, the width of beams should, in general, be less than or equal to the

lateral dimension of the columns into which they frame. Beam widths of 200 mm, 250 mm and

300 mm are common in practice world-wide. Where the beam is required to support a masonry

wall, the width of the beam is often made such that its sides are flush with the finished surfaces

of the wall; thus, beam widths of 230 mm are also encountered in practice in India. In design

practice, the overall depths of beams are often fixed in relation to their spans. Span to overall

depth ratios of 10 to 16 are generally found to be economical in the case of simply supported and

continuous beams. However, in the case of cantilevers, lower ratios are adopted, and the beams

are generally tapered in depth along their lengths, for economy. Such traditional methods of

fixing the depth of beams are generally satisfactory from the viewpoint of deflection control for

the normal range of loads.

Note: From practical considerations, it is desirable to limit the number of different beam sizes in the same

structure to a few standard modular sizes, as this will greatly convenience the construction of formwork, and

permit reusability of forms.

3.3.2 General Guidelines for Slab Thicknesses

In the case of slabs, whose thicknesses are very small in comparison with the depths of beams,

the limiting span/depth (l/d) ratios will generally govern the proportioning. In practice, Fe 415

grade steel is most commonly used, and for such steel, a pt value of about 0.4 – 0.5 percent may

be assumed for preliminary proportioning. This gives a kt (Fig. 4 of IS:456-2000) value of about

1.25; accordingly, the required effective depth (for preliminary design) works out to about

span/25 for simply supported slabs and about span/32 for continuous slabs.


In order to determine the thickness of the slab, the clear cover (based on exposure, refer Table

3.1) plus half the bar diameter of the main reinforcement (usually along the shorter span) have to

be added to the effective depth, as indicated in Fig. 3.1(b). The calculated value of the thickness

should be rounded off to the nearest multiple of 5 mm or 10 mm.

3.4 REQUIREMENTS FOR DEFLECTION CONTROL

Excessive deflections in slabs and beams are generally undesirable as they cause psychological

discomfort to the occupants of the building, and also lead to excessive crack-widths and

subsequent loss of durability and ponding in roof slabs.

The selection of cross-sectional sizes of flexural members (thicknesses of slabs, in particular) is

often governed by the need to control deflections under service loads. For a given loading and

span, the deflection in a reinforced concrete beam or slab is inversely proportional to its flexural

rigidity. It is also dependent on factors related to long-term effects of creep and shrinkage. From

the point of view of design, it is the ratio of the maximum deflection to the span that is of

concern, and that needs to be limited. The Code (Cl. 23.2a) specifies a limit of span/250 to the

final deflection due to all loads (including long-term effects of temperature, creep and

shrinkage). Additional limits are also specified in Cl. 23.2(b) of the Code ⎯ to prevent damage to

partitions and finishes.

The explicit computation of maximum deflection can be rather laborious and made difficult by

the need to specify a number of parameters (such as creep coefficient and shrinkage strain as

well as actual service loads), which are not known with precision at the design stage. For

convenience in design, and as an alternative to the actual calculation of deflection, the Code

recommends certain span/effective depth (l/d) ratios which are expected to satisfy the

requirements of deflection control (Δ/l < 1/250). Nevertheless, explicit calculations of deflections

become necessary under the following situations:

• when the specified l/d limits cannot be satisfied;

• when the loading on the structure is abnormal; and

• when stringent deflection control is required.

3.4.1 Deflection Control by Limiting Span/Depth Ratios

For a rectangular beam, made of a linearly elastic material, the ratio of the maximum elastic

deflection to the span (Δ/l) will be a constant if the span /overall depth ratio (l/D) is kept

constant. This can be proved as follows for the case of a simply supported rectangular beam,

subjected to a uniformly distributed load „w‟ per unit length:


where σ is the bending stress at service loads, Z=bD

2/6 is the section modulus and I =bD

3/12 is

the moment of inertia.

Substituting the value of „w‟ and „I‟ in „‟, we get,

where, in the present case of a simply supported beam with uniformly distributed loading, the

„constant‟ works out to 5/24E.

The above equation is generalized, and holds good for all types of loading and boundary

conditions (with appropriately different constants). It is thus seen that, by limiting the l/D ratio,

deflection (in terms of Δl) can be controlled.

The above equation is not directly applicable in the case of reinforced concrete, because it is not

a linearly elastic material and the parameters σ, Z and E are not constants, being dependent on

such factors as the state of cracking, the percentage of reinforcement, as well as the long-term

effects of creep and shrinkage. The Code however adopts this concept, with suitable

approximations, and prescribes limiting l/d ratios for the purpose of deflection control.

3.4.2 Code Recommendations for Span/Effective Depth Ratios

The check for deflection is done through the following two 456:2000 (Refer clause 42.1)

1. Empirical Method

In this method, the deflection criteria of the member is said to be satisfied when the actual value

of span to depth ratio of the member is less than the permissible values. The IS code procedure

for calculating the permissible values are as given below:

a. Choosing the basic values of span to effective depth ratios (l/d) from the following,

depending on the type of beam

1. Cantilever = 8

2. Simply supported = 20

3. Continuous = 26

b. Modify the value of basic span to depth ratio to get the allowable span to depth ratio.

(l/d)Allowable = (l/d)Basic x kt x kc x kf


Where, kt = Modification factor obtained from Fig. 4 of IS 456:2000. It depends on the area of

tension reinforcement provided and the type of steel.

kc = Modification factor obtained from Fig. 5 of IS 456:2000. This depends on the area of

compression steel used.

kf = Reduction factor got from Fig. 6 of IS 456:2000.

Note: The basic values of l/d mentioned above are valid upto spans of 10m. The basic values are

multiplied by 10/span in meters except for cantilever. For cantilevers whose span exceeds 10 m the

theoretical method shall be used.

2. Theoretical method of checking deflection

The actual deflections of the members are calculated as per procedure given in annexure „C‟ of

IS 456:2000. This deflection value shall be limited to the following:

i. The final deflection due to all loads including the effects of temperature, creep and

shrinkage shall not exceed span/250.

ii. The deflection including the effects of temperature, creep and shrinkage occurring after

erection of partitions and the application of finishes shall not exceed span/350 or 20 mm

whichever is less.

3.5 CRACKING IN STRUCTURAL MEMBERS

Cracking of concrete occurs whenever the tensile stress developed is greater than the tensile

strength of concrete. This happens due to large values of the following:

1. Flexural tensile stress because of excessive bending under the applied load

2. Diagonal tension due to shear and torsion

3. Direct tensile stress under applied loads (for example hoop tension in a circular tank)

4. Lateral tensile strains accompanying high axis compressive strains due to Poisson‟s effect

(as in a compression test)

5. Settlement of supports

In addition to the above reasons, cracking also occurs because of

1. Restraint against volume changes due to shrinkage, temperature creep and chemical

effects.

2. Bond and anchorage failures.

Cracking spoils the aesthetics of the structure and also adversely affect the durability of the

structure. Presence of wide cracks exposes the reinforcement to the atmosphere due to which the

reinforcements get corroded causing the deterioration of concrete. In some cases, such as liquid

retaining structures and pressure vessels cracks affects the basic functional requirement itself

(such as water tightness in water tank).


3.5.1 Permissible crack width

The permissible crack width in structural concrete members depends on the type of structure and

the exposure conditions. The permissible values are prescribed in clause 35.3.2 of IS 456:2000

and are shown in table below

Table: Permissible values of crack width as per IS 456:2000

3.5.2 Control of cracking

The check for cracking in beams are done through the following 2 methods specified in IS

456:2000 clause 43.1

1. By empirical method:

In this method, the cracking is said to be in control if proper detailing (i.e. spacing) of

reinforcements as specified in clause 26.3.2 of IS 456:2000 is followed. These specifications

regarding the spacing have been already discussed under heading general specifications. In

addition, the following specifications shall also be considered

i. In the beams where the depth of the web exceeds 750 mm, side face reinforcement shall be

provided along the two faces. The total area of such reinforcement shall not be less than 0.1%

of the web area and shall be distributed equally on two faces at a spacing not exceeding 300

mm or web thickness whichever is less. (Refer clause 25.5.1.3 IS456:2000)

ii. The minimum tension reinforcement in beams to prevent failure in the tension zone by

cracking of concrete is given by the following

As = 0.85bd/fy (Refer clause 26.5.1.1 of IS 456:2000)

iii. Provide large number of smaller diameter bars rather than large diameter bars of the same

area. This will make the bars well distributed in the tension zone and will reduce the width of

the cracks.

2. By crack width computations:

In the case of special structures and in aggressive environmental conditions, it is preferred to

compute the width of cracks and compare them with the permissible crack width to ensure the

safety of the structure at the limit state of serviceability. The IS 456-2000 has specified an

analytical method for the estimation of surface crack width in Annexure-F which is based on the

British Code (BS: 8110) specifications where the surface crack width is less than the permissible

width, the crack control is said to be satisfied.

Problems:

1. Given the following data of a simply supported T beam, check the deflection criteria by

empirical method

Width of the beam (b) = 230 mm

Effective depth (d) = 425 mm

Effective span = 8.0 m

Area of tension steel required = 977.5 mm2

Area of tension steel provided = 1256 mm2

Area of compression steel provided = 628 mm2

Type of steel = Fe 415

Width of flange (bf) = 0.9 m

Width of web (bw) = 0.3 m

Solution:

Basic �� = 20 for simply supported beam from clause 23.2.1

Allowable �� = Basic

�� x Mt x Mc x Mf …………. (1)

�� = 1265�100230�425 = 1.30%

�� = 0.58�� × �� !��"��# !�!

�� = 0.58 × 415 × 977.51256 = 187.3

From fig 4, for Pt = 1.3%, fs = 187.5 N/mm2

Mt = 1.1 ………. (a)

�& = 628 × 100230 × 425 = 0.65%

From fig 5, for Pc = 0.65%, Mc = 1.15 ………………(b)

From fig 6, for '(') =

*.+**.,* = 0.33, Mf = 0.80 ………(c)

Substituting a, b and c in equation (1)

We get allowable �� = 20 x 1.1 x 1.15 x 0.80 = 20.2

Actual �� =

-*../0 = 18.82 < allowable

��

Hence OK

2. A rectangular beam continuous over several supports has a width of 300 mm and overall

depth of 600 mm. The effective length of each of the spans of the beam is 12.0 m. The

effective cover is 25 mm. Area of compression steel provided is 942 mm2 and area of

tension steel provided is 1560 mm2. Adopting Fe 500 steel estimate the safety of the

beam for deflection control using the empirical method

Solution:


�� x Mt x Mc x Mf …………. (1)

Basic �� = 26 as the beam is continuous

�� = 0.58�� × �� !��"��# !�!

�� = 0.58 × 500 × 15601560 = 290

From fig 4, for fs = 290, Pt = 0.90, Mt = 0.9……….(a)

From fig 5, for Pc = 0.54%, Mc = 1.15 ……………….(b)

From fig 6, for '(') = 1.0, Mf = 1 ……………………(c)

The equation (1) shall be multiplied by 2*

�345 . � 2*2/ as the span of the beam is greater

than 10.0 m

Allowable �� =

2*2/ x 26 x 0.9 x 1.15 x 1 = 22.4

Actual �� =

2/*.060 = 20.86 < allowable

��

Hence deflection control is satisfied.

3. Find the effective depth based on the deflection criteria of a cantilever beam of 6m span.

Take fy = 415 N/mm2, Pt = 1%, Pc = 1%.

Solution:


�� x Mt x Mc x Mf

Basic �� = 7 for cantilever beam

Assume 789:;<=>:;�7893:?@>�;� = 1.0

fs = 0.58 x 415 x 1= 240.7

From fig 4, for fs = 240, Pt = 1%, Mt = 1.0

From fig 5, for Pc = 1%, Mc = 1.25

From fig 6, for '(') = 1.0, Mf = 1

Allowable �� = 7 x 1.0 x 1.25 x 1.0 = 8.75

! = �-.60 =

A***-.60 = 685 mm

4. A simply supported beam of rectangular cross section 250mm wide and 450mm overall

depth is used over an effective span of 4.0m. The beam is reinforced with 3 bars of

20mm diameter Fe 415 HYSD bars at an effective depth of 400mm. Two anchor bars of

10mm diameter are provided. The self weight of the beam together with the dead load on

the beam is 4 kN/m. Service load acting on the beam is 10 kN/m. Using M20 grade

concrete, compute

a. Short term deflection

b. Long term deflection

Solution:

Data b = 250 mm, D = 450 mm, d = 400 mm, fy = 415 N/mm2

Ast = 3 x B.x 20

2 = 942 mm

2, l = 4.0 m, D.L = 4 kN/m, Service load = 10 kN/m,

Total load = 14 kN/m, fck = 20, Asc = 2 x B.x 10

2 = 158 mm

2

Es = 2.1 x 105 , Ec = 5000 C�&D = 22360 N/mm

2

m = /-*

+EFGF = /-*

+H6= 13.3

fcr = 0.7 C�&D = 0.7 √20 = 3.13 N/mm2

a. Short term deflection

To determine the depth of N.A

Equating the moment of compression area to that of the tension area, we get

b * x * H/ = m * Ast * (d-x)

‘m’ is used to convert the steel into equivalent concrete area

250 * HJ/ = 13 * 942 * (400-x)

Solving, x = 155 mm from the top

Cracked MOI Ir = /0*×200K

2/ + L250 × 155M x (155/2)2 + 13 x 942 (400 – 155)

= 10.45 x 108 mm

4

(2) Igr = Gross MOI = /0*H.0*K

2/ = 18.98 x 108 mm

4

(3) M = Maximum BM under service load

M= N�J

- = 2.×.J

- = 28 kN = 28 x 106 N-mm

(4) Cracked moment of inertia

Mr = OFPQRP�9 =

+.2+×2-.,-×2*S*.0×.0* = 26 x 10

6 N-mm

Lever arm = z = T! − H+V

= T400 − 200+ V = 348.34 mm

(5) Ieff = W XP2./YTZPZ VT[\VT2Y]\VTG(G V^

_ 2*..0×2*S2./Y`Ja×bca

JS×bcadTKeS.Keecc VT2YbffeccVL2Mg

Ieff = 14.93 x 108 mm

4

Further Ir < Ieff < Igr

(6) Maximum short term deflection

ai(perm) = h(N�eiFXj)) =

0+-. 2.×L.***MJ

//+A*×2..,+×2*S = 1.39 mm

Kw = 0

+-. for SSB with UDL

b. Long term deflection

(1) Shrinkage deflection (acs):

acs = K3 ψcs L2

K3 = 0.125 for simply supported beam from Annexure C-3.1

ψcs = Shrinkage curvature = k. TlF8m V

n&� = Ultimate shrinkage strain of concrete (refer 6.2.4) = 0.0003

�� = 2**×,.//0*×.** = 0.942

�& = 2**×20-/0*×.** = 0.158

Pt - Pc = (0.942 – 0.158) = 0.784

This is greater than 0.25 and less than 1.0 Hence ok.

Therefore k. = 0.72 ×o9YoFCo9 = 0.72 ×*.,./Y*.20-√*.,./

K4 = 0.58

ψcs = *.0-×*.***+

.0* = 3.866 x 10-7

acs = K3 ψcs L2

= 0.125 x 3.866 x 10-7

x (4000)2

= 0.773 mm

(2) Creep deflection [acc(perm)]

Creep deflection acc(perm) = aicc(perm) – ai(perm)

Where, acc(perm) = creep deflection due to permanent loads

aicc(perm) = short term deflection + creep deflection

ai(perm) = short term deflection

aicc (perm) = kN ` N�KiFjXj))d

p&; = iFL2qrM = iFL2q2.AM

s = Creep coefficient = 1.6 for 28 days loading

aicc(perm) = 2.6 x short term deflection

= 2.6 x ai(perm)

= 2.6 x 1.39 = 3.614 mm

Creep deflection acc(perm) = 3.614 – 1.39 = 2.224 mm

Total long term deflection = shrinkage deflection + Creep deflection

= 0.773 + 2.224 = 3.013 mm

Total deflection = Short term deflection + Long term deflection

= 1.39 + 3.013 = 4.402 mm

5. A simply supported beam of rectangular section spanning over 6 m has a width of

300mm and overall depth of 600 mm. The beam is reinforced with 4 bars of 25 mm

diameter on the tension side at an effective depth of 550 mm spaced 50 mm centres. The

beam is subjected to a working load moment of 160 kN.m at the centre of the span

section. Using M-25 grade concrete and Fe-415 HYSD bars, check the beam for

serviceability limit state of cracking according to IS:456-2000 method. The beam is

protected and not exposed to aggressive environmental conditions.

a. Data:

b = 300 mm fck = 25 N/mm2

h = 600 mm fy = 415 N/mm2

d = 550 mm Es = 2 x 105 N/mm

2

M = 160 kN.m spacing between bars s = 5 mm

Ast = 1963 mm2

cover = 50mm

For fck = 25 N/mm2, from table 21 IS 456

t&'& = 8.5u/ww/

w = /-*+EFGF

= 11

b. Neutral axis depth

Let x = depth of neutral axis

Then we have 0.5 b�/ = m Ast (d-x)

0.5 x 300�/ = 11 x 1963 (550-x)

Solving, x = 220 mm

c. Cracked moment of area (Ir):

Ir = (bx3/3) + m Ast r

2

Where r = (d-x) = (550-220) = 330 mm

Ir = T+**×//*K+ V +L11 × 1963 ×330/M

= 34.1 x 108 mm

4

d. Maximum width of cracks

Cover = Cmin = (50 - 12.5) = 37.5

acr = [(0.5 S)2 + C

2min]

1/2

= [(0.5 x 50)2 + 37.5

2]1/2

= 45

Crack width will be maximum at the soffit of the beam

Distance of the centroid of steel from neutral axis

= r = (d-x) = (50-220) = 330 mm

Therefore n2 = TO8i8V yzYH�YH{

Where, �� = w y|�XP { = 11 y2A*×2*a×++*

+..2×2*S { = 170 N/mm2

Therefore, n2 = T 26*/×2*fV yA**Y//*

00*Y//*{ = 9.78 x 10-4

n} = n2 − y'9LzYHMLzYHM+i878L�YHM {

n} = L9.78 × 10Y.M − y +**LA**Y//*MLA**Y//*M+×/×2*f×2,A+L00*Y//*M{

= 8.67 x 10-4

Maximum width of crack is expressed as:

~&: = W +��єZ2q/��FP�FZ��] �^

~&: = W+×.0×-.A6×2*�e2q/� ef�K�.facc�JJc� ^

= 0.113 mm< Permissible crack width of 0.3mm from clause 35.3.2

page 67. Hence ok.

limit state of serviceability in rcc

Documents

design problem

design depth of concrete

concrete cover clear

analysis problem

increased cover requirements

design of rcc structures

durability requirements

complete design