Serviceability of beams and one way slabs

Strength limit states are based on the safety or load-carrying capacity of structure and include buckling, fracture, fatigue, overturning, and so on. Serviceability limit states refer to the performance of structures under normal service loads and are concerned with the uses and/or occupancy of structures. Serviceability is measured by considering the magnitudes of deflections, cracks, and vibrations of structures as well as by considering the amounts of surface deterioration of the concrete and corrosion of the reinforcing. You will note that these items may disrupt the use of structures but do not usually involve collapse.

The magnitudes of deflections for concrete members can be quite important. Excessive deflections of beams and slabs may cause sagging floors, ponding on flat roofs, excessive vibrations, and even interference with the proper operation of supported machinery.Such deflections may damage partitions and cause poor fitting of doors and windows. Inaddition, they may damage a structure's appearance or frighten the occupants of the building, even though the building may be perfectly safe. Any structure used by people shouldbe quite rigid and relatively vibration-free so as to provide a sense of security.

Perhaps the most common type of deflection damage in reinforced concrete structuresis the damage to light masonry partitions. They are particularly subject to injury due toconcrete's long-term creep. When the floors above and below deflect, the relatively rigidmasonry partitions do not bend easily and are often severely damaged. On the other hand,the more flexible gypsum board partitions are much more adaptable to such distortions.

CONTROL OF DEFLECTIONSOne of the best ways to reduce deflections is by increasing member depths (but

designersare always under pressure to keep members as shallow as possible). Reinforced concrete specifications usually limit deflections by specifying certain minimum depths or maximum permissible computed deflections.

Minimum ThicknessesTable 9.5(a) of the ACI Code, provides a set of minimumthicknesses for beams

and one-way slabs to be used, unless actual deflection calculationsindicate that lesser thicknesses are permissible. These minimum thickness values, whichwere developed primarily on the basis of experience over many years, should be usedonly for beams and slabs that are not supporting or attached to partitions or other members likely to be damaged by deflections.

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Maximum Computed DeflectionsIf the designer chooses not to meet the minimum thicknesses given in Table 9.5-a. the deflection should be estimated and the values determined may not exceed the values specified in Table 9.5(b) of the ACI Code 318-2011.

CALCULATION OF DEFLECTIONS1- Instantaneous or immediate deflections

Deflections for reinforced concrete members can be calculated with the usual deflection expressions, several of which are shown in Fig.1. A few comments should bemade about the magnitudes of deflections in concrete members as determined by theexpressions given in this figure. It can be seen that the center line deflection of a uniformlyloaded simple beam [Fig.1(a)] is five times as large as the center line deflection of thesame beam if its ends are fixed [Fig.1(b)]. Nearly all concrete beams and slabsare continuous, and their deflections fall somewhere between the two extremes mentioned here.

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Because of the very large deflection variations that occur with different end restraints,it is essential that those restraints be considered if realistic deflection calculations are to bemade. For most practical purposes it is sufficiently accurate to calculate the central line deflectionof a member as though it is simply supported and to subtract from that value the deflection caused by the average of the negative moments at the member ends. (This can bedone by using a combination of expressions taken from Fig.1. For instance, the deflection equation of part (a) may be used together with the one of part (g) applied at one orboth ends as necessary.)Effective Moment of InertiaRegardless of the method used for calculating deflections, there is a problem in determining the moment of inertia to be used. The trouble lies in the amount of cracking that hasoccurred. If the bending moment is less than the cracking moment (that is, if the

flexuralstress is less than the modulus of rupture of about for normal-

weight concrete),the full un-cracked section provides rigidity, and the moment of inertia

for the gross section is available. When larger moments are present, different size

tension cracks occurand the position of the neutral axis varies.Fig.2illustrates the problem involved in selecting the moment of inertia to beused

for deflection calculations. Although a reinforced concrete beam may be of constantsize (or prismatic) throughout its length, for deflection calculations it will behave asthough it is composed of segments of different-size beams.

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For the portion of a beam where the moment is less than the cracking moment,

the beam can be assumed to be un-cracked and the moment of inertia can be

assumed toequal .When the moment is greater than the tensile cracks that

develop in the beamwill, in effect, cause the beam cross section to be reduced, and the

moment of inertia maybe assumed to equal the transformed value, . It is as though the

beam consists of the segments shown in Fig.2 (d).It is true that atcross sections where tension cracks are actually located, the moment of inertia is probablyclose to the

transformed but in between cracks it is perhaps closer to Furthermore,diagonal

tension cracks may exist in areas of high shear, causing other variations. As a result, it is

difficult to decide what value of should be used.

A concrete section that is fully cracked on its tension side will have a rigidity of anywhere from one-third to three-fourths of its full section rigidity if it is un-cracked. At different sections along the beam, the rigidity varies depending on the moment present. If it is desired to obtain the immediate deflection of an un-cracked prismatic

member,the moment of inertia may be assumed to equal along the length of the

member.

Section 9.5.2.3 of the ACI Code gives a moment of inertia expression that is to be used fordeflection calculations. This moment of inertia is an average value and is to be

used at anypoint in a simple beam where the deflection is desired. It is referred to as

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the effectivemoment of inertia, and is based on an estimation of the probable amount of crackingcaused by the varying moment throughout the span:

(ACI Equation 9-8)

Where:Ig is the gross amount of inertia (without considering the steel) of thesection, Mcr

is the cracking moment , with for normal-weight concrete

(different for lightweight concrete, as per Section 9.5.2.3 of the Code),yt is the distance from neutral axis to extreme fiber in tension. Ma; is themaximum serviceload moment

occurring for the condition under consideration, and , isthe transformed moment of

inertia of the cracked section.The values of the effective moment of inertia vary with differentloading conditions.

This is because the service load moment, Ma used in the equation forIe is different for each loading condition.

2- LongTerm Deflections(sustained) Long-term or sustained loads, however, cause significant increases in thesedeflections due to shrinkage and creep. The factors affecting deflection increases includehumidity, temperature, curing conditions, compression steel content, ratio of stress tostrength, and the age of the concrete at the time of loading.

If concrete is loaded at an early age, its long-term deflections will be greatly increased. Excessive deflections in reinforced concrete structures can very often betraced to the early application of loads. The creep strain after about five years (afterwhich creep is negligible) may be as high as four or five times the initial strain whenloads were first applied seven to ten days after the concrete was placed, while the ratiomay only be two or three when the loads were first applied three or four months afterconcrete placement.The Code (9.5.2.5) states that to estimatethe increase in deflection due to these causes, the part of the instantaneous deflection thatis due to sustained loads may be

multiplied by the empirically derived factor and the result added to the instantaneous

deflection.

(ACI equation 9-11)

Where: is atime-dependent factor that can be determined from Table and figure below.

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The effect of compression steel on long-term deflections is taken into account in

the expression with the term and is to be computed at midspan for simpleand

continuous spans and at the supports for cantilevers.The full dead load of a structure can be classified as a sustained load, but the

type ofoccupancy will determine the percentage of live load that can be called sustained. For anapartment house or for an office building, perhaps only 20 to 25% of the service live loadshould be considered as being sustained, whereas perhaps 70 to 80% of the service liveload of a warehouse might fall into this category.

It is logical to assume that the live load cannot act on a structure when the dead load is not present. As a resultcompute an effectiveIe, anda deflection of dead

loads for the case where the dead load alone is acting. Then compute an Ie,and a

deflection of dead and live loads for the case where both dead and live loads are

acting. Then:

The long-term deflection will equal the initial live load deflection will be:

The steps involved in calculating instantaneous and long-term deflections can besummarized as follows:

(a) Compute the instantaneous or short-term deflection for dead load only.

(b) Compute instantaneous deflection for dead plus full live load.

(c) Determine instantaneous deflection for full live load only.

(d) Compute instantaneous deflection due to dead load plus the sustained part of the

live load

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(e) Determine instantaneous deflection for the part of the live load that is sustained.

(1) Determine the long-term deflection for dead load plus the sustained part of thelive

load

Ex-1: Calculate the total (dead+ live load) deflection of the cantilever beam shown, the service dead load including its own weight (30 kN/m) and service live load (60 kN). use f'c =21 MPa.

Ex-2: The beam of figure below has a simple span of6m and supports a dead load including its own weight of 14kN/m and a live load of 10kN/m , use f'c = 21 MPa.

a) Estimate the instantaneous deflection for D+L.b) Estimate the sustained deflection assuming that 30% of the live load is contiuously

applied for three years.

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Solution:

a) instantaneous deflection for dead load :

Instantaneous or short-term deflection for dead + full live load

Initial deflection for full live load :

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Initial deflection due to dead load + 30% live load ( )

Initial deflection due to 30% live load ( )

b) Long-term deflection for dead load plus three years of 30% sustained live load (

)

Ex-3:- Calculate the total deflection of beam shown, the service dead load including its own weight (25 kN/m) and service point live load (50 kN). use f'c =25MPa, n=9, 50% if live load was continued.

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Ex-4:-A simply supported beam with 6m length, support 20kN/m service dead load including its own weight and two concentrated sevice live load each one 50kN.Detrmine the immediate live load deflection.Using f'c= 25MPa, n=9.

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A= 800*100*50 = 4*106 and B = 9* 2121*(530-100)=8.208*106mm3 then T-section

X= 140.4mm

Ex.5: Check the requirments for deflection for the beam shown, that supports a service distributed deadload including its own weight of 6 kN/m and service live load of 9 kN/m and concetrated dead load at free end of the beam 40 kN , the beam is part of floors not supporting or attached to non-structural elementslikely to be damaged by large deflections. Use f'c = 30MPa and n=8.

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and X= 181.95 mm

Ex.6: Check the deflection for the beam shown , it supports a service distributed dead load of 14 kN/m and service live load of 12 kN/m, the beam is part of roof supporting or attached to nonstructural elementslikely to be damaged by large deflections . Use f'c = 25MPa and n=9 and 50% of live load in continued for 24 months.

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Not acceptable

CONTINUOUS-BEAM DEFLECTIONSA continuous T beam subjected to both positive and negative moments. As in

Fig.2 , the effective moment of inertia used for calculatingdeflections varies a great deal

throughout the member. For instance, at the center of the spanat section 1-1 where the

positive moment is largest, the web is cracked and the effective section consists of the

hatched section plus the tensile reinforcing in the bottom of the web. Atsection 2-2 in the

figure, where the largest negative moment occurs, the flange is cracked andthe effective

section consists of the hatched part of the web (including any compression steelin the

bottom of the web) plus the tensile bars in the top. Finally, near the points of

inflection,the moment will be so low that the beam will probably be uncracked, and thus

the wholecross section is effective, as shown for section 3-3 in the figure. (For this case I

is usually calculated only for the web, and the effect of the flanges is neglected).

From the preceding discussion it is obvious that to calculate the deflection in a

continuous beam, theoretically it is necessary to use a deflection procedure that takes into

account the varying moment of inertia along the span. Such a procedure would be very

Fig.2: Deflections for a continuous T beam.

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lengthy, and it is doubtful that the results so obtained would be within + 20% of the

actualvalues. For this reason the ACI Code (9.5.2.4)permits the use of a constant

moment of inertia throughout the member equal to the average of the I, values

computed at the criticalpositive and negative moment sections. The I, values at the

critical negative-momentsections are averaged with each other, and then that average is

averaged with L, at thecritical positive-momentsection. It should also be noted that the

multipliers for long-term deflection at these sections should be averaged, as were the I,

values.

For distributed loads :

For constreated loads:

Where:Mm : is positive moment at mid span. Ma and Mb are negative moments at supports.

If the beam is continuous from two sides or one side and connected with structural

member in the other so there will be a nagative moment then:

When the beam is continuous from one side and simply on the other then:

Where:Im : at mid span. Ia and Ib are moment of inertia at supports.

The ACI Code permits for prismatic members, Ie shall be permitted to be taken as the value obtained at midspan for simple and continuous spans, and at support for cantilevers. ACI Committee 435 has shown that better results for the deflections in continuousmembers can be obtained if an I, is used that gives greater weight to the midspan values.

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The committee suggests the use of the following expressions in which Im , Ia and Ib arethe computed effective moments of inertia at the midspan and the two ends of the span,respectively. Beams with two ends continuous :

Beams with one end continuous

Ex-7: Check the deflection for the continuouse beam shown, supports a service distributed dead load including its own weight of 25 kN/m and service live load of 35 kN/m , the beam is part of floors supporting and attached to nonstructural elements likely to be damaged by large deflections . Use f'c = 21MPa and n=8 and 20% of live load to be continued .

N.A. position for T-section a-a:

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Then:

For rectangular

Determine Icr for T-section positive moment:

A= moment area above neutral axis=

B= moment area below neutral axis=

Then T- section:

For section b-b: tension steel (As= 3885mm2 the transformed area= nAs=34965mm2.)compression steel (As'= 1473mm2 the transformed area=( n-1)As=11784mm2.)

For positive moment:

Continuous loads: Wsus = 25 + 0.2*35= 32 kN/m

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For nagative moment:

Continuous loads:

The result will be summeraized in table:

Positive momentNegative moment

Immediate deflectionfor total loads:

Immediate deflectionfor continuous loads:

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then the deflection is acceptable.

Ex-8: Check the immediate deflection for the beam shown , the beam is part of floors not supporting or attached to nonstructural elements likely to be damaged by large deflections. Use f'c = 24MPa.

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Acceptable

Cracks Controls:Types of cracks:

1- Flexural crack: are vertical cracks that extend from the tension sides of beams up to the region of their neutral axes as in figure (a). Should beams have very deep webs; the cracks will be very closely spaced, with some of them coming together above the reinforcing and some disappearing there. These cracks may be wider up in the middle of the beam than at the bottom.

2- Inclined cracks due to shear can develop in the webs of reinforced concrete beams either as independent cracks or as extensions of flexural cracks. Occasionally, inclined cracks will develop independently in a beam, even though no flexural cracks are in that locality. These cracks, which are called web-shear cracks and which are illustrated in Figure (b), sometimes occur in the webs of pre-stressed sections, particularly those with large flanges and thin webs.

3- The usual type of inclined shear cracks is the flexure-shear cracks, which are illustrated in Figure(c). They commonly develop in both pre-stressed and non pre-stressed beams.

4- Torsion cracks, which are illustrated in Figure (d), are quite similar to shear cracks except they spiral around the beam. Should a plain concrete member be subjected to pure torsion, it will crack and fail along 45° spiral lines due to the diagonal tension corresponding to the torsional stresses. For a very effective demonstration of this type of failure. Although torsion stresses are very similar to shear stresses, they will occur on all faces of a member. As a result, they add to the shear stresses on one side and subtract from them on the other.

5- Sometimes bond stresses between the concrete and the reinforcing lead to a splitting along the bars, as shown in Figure (e).

Of course, there are other types of cracks not illustrated here. Members that are loaded in axial tension will have transverse cracks through their entire cross sections. Cracks can also occur in concrete members due to shrinkage, temperature change, settlements, and soon.

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CONTROL OF FLEXURAL CRACKSCracks are going to occur in reinforced concrete structures because of concrete's low tensile strength. Many factors affect the crack width and cause to steel corrosion some of them are:

a) Steel stress: for members with low steel stresses at service loads, the cracks may be very small and in fact may not be visible except upon careful examination. Such cracks, called microcracks, are generally initiated by bending stresses. When steel stresses are high at service load, particularly where high-strength steels are used, visible cracks will occur. These cracks should be limited to certain maximum sizes so that the appearance of the structure is not spoiled and so that corrosion of the reinforcing does not result. The use of high-strength bars and the strength method of design have made crack control a very important item indeed.

b) Size of reinforced bars: rather natural for designers to specify approximately the same size bars as they are accustomed to using, but fewer of them. The result has been more severe cracking of members. Although cracks cannot be eliminated, they can be limited to acceptable sizes by spreading out or distributing the reinforcement. In other words, smaller cracks will result if several small bars are used with moderate spacing rather than a few large ones with large spacing.

c) Location of members and type of structure: the maximum crack widths that are considered to be acceptable vary depending on the location of the member in

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question, the type of structure, the surface texture of the concrete, illumination, and other factors. Somewhat smaller values may be required for members exposed to very aggressive environments such as deicing chemicals and saltwater spray.

ACI Committee 224, in a report on cracking presented a set of approximately permissible maximum crack widths for reinforced concrete members subject to different exposure situations. These values are summarized in Table:

Results of laboratory tests of reinforced concrete beams to determine crack sizes vary. The sizes are greatly affected by shrinkage and other time-dependent factors. Thepurpose of crack control calculations is not really to limit cracks to certain rigid maximum values but rather to use reasonable bar details, as determined by field and laboratory experience, that will in effect keep cracks within a reasonable range. In 1968 the following equation was developed for the purpose of estimating the maximum widths of cracks that will occur in the tension faces of flexural members. It is merely a simplification of the many variables affecting crack sizes.

Where: w = the estimated cracking width in (mm).

= ratio of the distance to the neutral axis from the extreme tension concrete fiber to the

distance from the neutral axis to the centroid of the tensile steel (values to be determined by the working-stress method).

= steel stress, in N/mm2 or MPa at service loads (designer is permitted to use 0.6fy, for

normal structures).= the cover of the outermost bar measured from the extreme tension fiber tothe center

of the closest bar or wire. (For bundled bars is measured to the c.g. of the bundles)

mm.

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A = the effective tension area of concrete around the main reinforcing (having the same centroid as the reinforcing) divided by the number of bars (mm2).

This expression is referred to as the Gergely-Lutz equation after its developers. In

applying it to beams, reasonable results are usually obtained if is set to 1.20. For thin

one-way slabs, however, more realistic values are obtained if is set to 1.35. The

number of reinforcing bars present in a particular member decidedly affects the value of A to be used in the equation and thus the calculated crack width. If more and smaller bars are used to provide the necessary area, the value of A will be smaller, as will the estimated crack widths. Should all the bars in a particular group not be the same size, their number (for use in the equation) should be considered to equal the total reinforcing steel area actually provided in the group divided by the area of the largest bar size used.

According to no. of steel layers

For different bar size no. of bars =

ACI CODE PROVISIONS CONCERNING CRACKS: In the ACI Code, Sections 10.6.3 and 10.6.4 require that flexural tensile

reinforcement be well distributed within the zones of maximum tension so that the

center-to-center spacing of the reinforcing closest to a tension surface is not greater than

the value computed with the following expression:

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: the computed tensile stress at working load. It may be calculated by dividing the un-

factored bending moment by the beam's internal moment arm, or it may simply be taken

equal to 0.6*fy. The term represents the clear cover from the nearestsurface in tension

to the surface of the tensile reinforcement in mm.

Ex. 9: Is the spacing of the bars in figure below within requirements of ACI Code from the standpoint of cracking if fy =420MPa?

Ex. 10: The T-beam shown supports service load 661kN.m, determine the crack width expected. Then check the reinforcement details for the crack width that the beam exposed to wet weather. Use n=8, fy =400MPa the cover for 10mm diameter web reinforcement 40mm from all sides.

As 4826mm2 nAs= 38608mm2

Check section: (If the N-A within the flange)

A=750*150*75 = 8437500

B=38608*(650-150) = 19304000

A is less than B then T- section

y = 217mm

h2= 750-217 = 533mm h1= 533-100 = 433mm

A= 200*300 /6 =10000 mm2

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Note: for simplicity if MPa and

Then

For ACI- Code requirements:

cover Cc =70 - 32/2 = 54mm

Ex. 11:- Check the requirements for control crack width for the simply supported beam (exposed to dry air). Use , [cover = 40mm all around

and for stirrups bars with 10 mm diameter were used].

As= 2412.74 mm2 =8 nAs = 19302 mm2

then X2 +154.416 X – 81840.48= 0 and X = 219.1 mm

h1 = d –x = 530 -219.1 =310.9 mm and h 2 = h – x = 600 – 219.1 = 380.9 mm

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for dry air.

Spacing of bars:

Sheet No. 1Note: Use f ‘c= 21MPa and fy= 350 MPa for solving the questions.Q.1: Calculate the long term deflection due to continuous loads, for the simply supported

beam having the cross section as shown. It supports service distributed (dead load including its own weight of 6 kN/m and live load of 8 kN/m) witha concentrated service live load of 20 kN.

,

Q.2: Check the immediate deflection due to live load if the maximum allowable limit for immediate live load deflection is (i)L.L= L/360 mm . The beam supports distributed service dead as well as to its own weight of 10kN/m and service live load of

25kN/m.

Q.3: A simply supported beam with 6m length, support 20kN/m service dead load including its own weight and two concentrated sevice live load each one 40kN. Check:

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1- The deflection if the beam is part of floor attached to nonstructural element not likely to be damaged by large deflection , and if the 30% of live load is to

sustained for 24 months. ( )

2- The crack width if the beam is interior one .

Q.4: A simply supported beam with 6m length, support 30kN/m service dead load including its own weight and one concentrated sevice live load each one 50kN. Check :

1- The deflection if the beam is part of floor attached to nonstructural element not likely to be damaged by large deflection , and if the 40% of live load is to

sustained for 3 years.( )

2- The crack width if the beam is interior one .

Q.5:Find maximum immediate deflection of the free end of the beam shown due to total load, the distributed dead load in addition to its weight is 10 kN/m and a concentrated

live load of 50 kN, and , d = 630mm?

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Q.6:Calculate the immediate deflection due to live loads for the simply supported beam ABC having the cross section as shown. It is part of one way slab of 150mm thickness, the slab supports service distributed (service dead load including its own weight of 5 kN/m2and service live load of 6 kN/m2 ) witha concentrated service live load of 30 kN at point B.Use d =530m, area of tension steel used 4 bars with 30mm diameter.

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