/;THE INSTITUTION OF ENGINEERS, SRI LANKA

PART II EXAMINATION - DECEMBER 2013

206 STRUCTURAL DESIGNS - REINFORCED CONCRETE

Time Allowed: Three (03) Hours

Answer Three (03) questions. Relevant code or extracts from code will be provided.Design Charts are provided at the end of the paper.

All designs are to be done according to the recommendations of the Standard BS 8110.

Ql.Plan view of the second floor slab of a three story office building is depicted below. Slab panels aresupported on a grid of beams and cast monolithically. Reinforced Concrete columns support thissuspended slab/beam system at the intersections of beams. Material specifications and loads are alsogiven below.

6300 6300 6300 6300- 30 MPa-460 MPa

Grade of concreteCh. strength of steel(deformed Type II)

sDead loads on slab:Weight of finishesWeight of partitionsUnit weight of concrete

1 Imposed loads on sJ~.b:I under normal service - 3.0 kN/m'

Open to Sky -l.skN/m'- 2.0 kN/m-- 24 kN/m3

4200

v,4200,

42001I

,i

_JL. ._ Thickness of the slabNominal cover for reinforcement

-rrz rnrn-zo mrn

Loads are uniformly distributed.

Design the panel PQUV using simplified method indicated in the code, along the following steps.

i.) Slab is to be designed for mild LXP()~·.I·e conditions and fire resistance for a period of onehour. Check whether the given I') niia! cover for reinforcement satisfies the coderequirements. (04 marks)

ii.) Using the uniformly distributed load d at a. evaluate the characteristic loads on the panel.(05 marks)

iii.) Categorize the slab panel type based on its spanning condition and calculate the relevantmid span and over the supports moments (use factors from the table in BS code).

(07 marks)

iv.) Design reinforcement to resist bendins, .it mid spans and over supports for the slab panel.(Assume 10 mm diameter HYS t~pe [I be 5) (08 marks)

v.) Check the slab panel for deflection Iimita. ons and only propose modifications if necessary(05 marks)

vi.) Curtail reinforcement using the sirn pl ied method indicated in the (ode and sketch thereinforcement on a plan and a cross -l .tion of the slab panel using the standard method ofdetailing. (05 marks)

The Institution of Engineers, Sri Lanka ?l • cccer-iber 2013 206 Structural Designs - Reinforced Concrete

Nominal cover for reinforcement -25mm'Design vehicle' weight (two axle)Distance between axles

-10 kN-1·5m

Q2_To ease access problems in rural areas where crossing over wide streams are involved a single girderpre-cast bridge system for pedestrian and small capacity vehicular traffic (hand tractor/trailer) wasproposed by an Engineering Consultant. The main pre-cast bridge girder spans the streamsupported by abutments on the two banks while the bridge deck is made up of pre-cast slabscantilevering to either side of the girder bolted on to it. Diagrams given below show the plan and across section of the proposed bridge.

I Cross slab bridge deck IIRubble Masonry Abutment:

.-~

~"--- I Cross slab bridge deck

~ ...

~\

!'1.5 m

1.5 mI Bridge Girder

>1I Bridge Girder I

12.0 m '-1--'~<~--------------~~---------------»Cross section 01 the Bridge Plan of the Bridge

Assuming that when a 10 kN, two axle vehicle with a wheel base of 1.5 m and equal loaddistribution to front and rear axles, comes to the exact center of the bridge is the most critical case ofimposed load, design relevant sections of the girder along the given steps using the data givenbelow.

Material PropertiesGrade of concrete (feu)Ch. strength of main steel (fy)Ch. strength of stirrups (MS)(fsy)

-40-460 MPa- 250 MPa

Loads\leglect the load by handrailUnit weight of Concrete

Size of the Bridge girder - 600 (deep) x 450 (wide)""fective span of the girder - 12.0 m

i.) Evaluate the characteristic loads transferred on to the girder, by self weight, weight of cross slabsas well as the 'design vehicle' positioned at the critical location and indicate them on a linediagram.

(07 marks)

ii.) Calculate the design bending moment at mid span of the girder and the design shear force at thesupport of the girder based on the loads calculated for step i.).

(07 marks)

iii.) Design the reinforcement to resist bending at the mid span of the girder. (You may assume 25mrn dia. tor steel tensile reinforcement & 10 111m dia. tor steel shear stirrups.)

(05 marks)

iv.) Check for shear at supports of the girder and pr~\ ide shear reinforcement if necessary. (You mayassume two or four legged 10 mm tor steel shear <tirrups.)

(05 marks)

v.) Check for deflection of the girder based on conditions at the mid span_ Propose modifications ifthis check fai Is.

(05 marks)

vi.) Use an elevation and necessary cross sections of the beam, to sketch reinforcement using thestandard method of detailing. You should md i, lte curtailment lengths.

(05 marks)

"" YOoS .vld bvcsfkare wdh.kh The Institution of Engineers, Sri Lanka Part II Exami'l3")r Dcccr-ber 2013 206 Structural Deslgns . Reinforced Concrete

____________________________________________________________________________________________________r

Q3.The plan of a two storey Car park cum Office building is shown below. The ground level car parkhas to have a clear height of 3.0 m to allow for parking of larger vehicles and the office floor height is2.5 m to the eaves level of the hipped roof from the floor level. Assuming that the columnssupporting the structure are founded 1.2 m below finished ground level on individual pad footings(300 mm thick) not capable of resisting moments, design the column segment from footing to upperfloor level at D, along following steps. Use the load and material data as given below.

-- ---- ----------------~

Grade of concreteCh. strength of steel

· 35· 460 MPa

Dead loads;Weight of the roofWeight of finishes on slabWeight of partitions on slabUnit weight of concrete

Upper floor beam sizes are indicated on planAll columns extend up to roof eaves level

· 0.6 kN/m'·1.okN/m'·1.0 kN/m'·24 kN/m3

Imposed loads:On slab under normal serviceNeglect imposed load on roof

· 2.5 kN/m'DB .\L Thickness of slab

Nominal cover for reinforcement·138mm-zo mrn

4400 I____________ O- ~

6600 ~)~(----'-'66'-'-OO"----~ Size of the column at DAssume that the Ratio of d/h

225 x 225· 0.85

All dimensions are r ''111'1'

')'."tprmine the type of column (braced/unbraced), its eft 'ctl'

ratio and determine the slenderness condition (slender/short)msv, ers.

b~lhht, and the slendernessClearly state reasons for your

(10 marks)

ii ) Lvaluate the design axial load and design bending mornonr-, m am') actine on the column.'} OLI should pay due consideration to the moments created eccentricity of loading andpossibility of slenderness buckling in evaluating these values.

(10 marks)

iii.) Design the column main reinforcement and tie requireme-nt. il<;<;uming a symmetricalarrangement of reinforcement. (Assume 16 mm HTS type II bars tor main reinforcement and10 rum MS plain bars for ties.)

(07 marks)

"r iuce a detailed reinforcement sketch with column in... .ons. You should adopt the standard method of detailing.

iti m and required cross

(07 marks)

The Irstitution of Engineers, Sri Lanka Part II Examination - December 2013 206 s« :> Reldorced Conc.rete 3/4

Q4.A square pad footing is to be designed for a corner column in a school building to be constructed inMannarama. Structural Engineer in-charge has evaluated the loads on the footing and provided youwith the data, requesting you to conduct the necessary design checks. The column transfers asignificant bending moments to the footing apart from normal axial load, about the two orthogonalaxes of the footing as shown. Check the adequacy of the given section along the steps given belowand design the footing reinforcement.

i.) Check the footing against overall failure. (Failure modes are the soil beannz failure and possibilityof lifting of the tooting, under service loads.) If failure occurs in any mode, prope<;e modifications.

(10 marks)

Axial Loaddue to;Deadload . 60 kNLive load - 40 kN

Bending Moment Bending Momentabout x-x by; about y-y by;Deadload - 05 kNm Deadload - 05 kNm

Live load - 03:~~-r~v_e I;ad - 03 kNm

2000~ \ x Y --J 200:>«/300 I

t,

Grade of concreteCh. strength of steel

-30-460 MPa

Unit weight of concrete - 24 kN/m3

Nominal cover for reinforcement -40mm

Soil type - Silty sandAllowable Soil bearing capacity - 210 kN/m'

Assume: 10 mm dia. for main rlfAll dimensions are in "rnrn"

ii.) Since the footing as well as loading are symmetrical, design reinforcement to resist bending at thecritical section of the footing under design loads about one orthogonal axis onlv .

(08 marks)

iii.) Check for Direct Shear at the critical section about one orthogonal axis only.

iv.) Check for Punching Shear at the critical perimeter and propose modifications if necessary,(08 marks)

(08 marks)

iv.) Produce a sketch of the reinforcement, in the form of a plan and a cross section of the footing, usingthe standard method of detailing with curtailment of main reinforcement.

(08 marks)

w6:;l!udul" D~~,~,,"·".fr crcec t.cncre te.lULu' lUll uf fl1ghll,:cl~, SI! LJIlIw P...•rt H EJh.lll1h, ...•tiull DI.!LI.:IIIUL'I olUIJ

To be used as reference for 206 Structural Designs - Reinforced ConcreteNotto be taken out from the exam hall. Return this to the supervisor after the exam

Table 3.2 Classification of exposure conditionsEnvironment Exposure conditionsMild Concrete surface protected against weather or aggressive conditions

Moderate Exposed concrete surface but sheltered from severe rain or freezing whilst wet Concretesurface continuously under non-aggressive water Concrete in contact with non-aggressive soil (see sulfate class 1 of table 7a in BS 5328: Part 1: 1997) Concrete subject tocondensation

Severe Concrete surfaces exposed to severe rain, alternate wetting and drying or occasionalfreezing or severe condensation

Very severe Concrete surfaces occasionally exposed to sea water spray or de-icing salts (directly orindirectly) Concrete surfaces exposed to corrosive fumes or severe freezing conditionswhilst wet

Most severe Concrete surfaces frequently exposed to sea water spray or de-icing salts (directly orindirectly) Concrete in sea water tidal zone down to 1 m below lowest low water

Abrasive'! Concrete surfaces exposed to abrasive action, e.g. machinery, metal tyred vehicles orwater carrying solids

Table 3.3 Nominal cover to all reinforcement (including links to meet durability requirements (Note 1)Conditions of exposure (see 3.3.4) Nominal cover (Dimensions in millimeters)t-.1ild 2:> 20 201) 201) 201'Moderate - 35 30 25 20Severe - - 40 30 25Very severe - - 502) 402) 30Most severe - - - - 50Abrasive - - - See Note 3 See Note 3Maximum free water/cement ratio 0.65 0.60 0.55 0.50 0.4:>Minimum cement concrete (kg/ m3) 275 300 325 350 400Lowest grade of concrete C30 C35 C40 C45 C501) These covers may be reduced to 15 mm provided that the nominal maximum size of aggregate does not exceed 15

mm2) ""here cannel' is subject to freezing whilst wel, air-entrainment should be used (see 5.3.3 of BS 5328. Part 1: 1997)

and the strength grade may be reduced by 5.Note 1: This table relates to normal-weight aggregate of 20 mm nominal size. Adjustments to minimum cement

rontont-, for aggregates other than 20 mm nominal maximum size are detailed in table 8 of BS5328:Part 1: 1997.Note 2: Use of sulfate resisting cement conforming to BS 4027. These cement have lower resistance to chloride ion

migration. If they are used in reinforced concrete in very severe or most severe exposure conditions. The coversin table 3.3 should be increased by.

Note 3: Cover should be not less than the nominal value corresponding to the relevant environmental category plusany allowance for loss of cover due to abrasion.

Table 3.5 Design ultimate bending moments and shear forces for Continuous BeamsAt outer support Near middle of At first interior At middle of interior At interior

end span suppuIl spans suppurb .-Moment 0 0.09Fl -O.l1Fl O.D7Fl -G.08FfShear Force 0.45f - 0.6F - 0.55FNOTE. I is the effective span;

F is the total design ultimate load (1.4Gk + 1.6QJ.No redistribution of the moments calculated from this table should be made.

~Tactfrom <BS8110 -1997 1/7

l£X/ractfrom es 8110 -1997 2/7

Table 3.4 Nominal cover to all reinforcement (including links) to meet specified periods of fireresistance (see NOTES 1 and 2)Fire Nominal cover

resistance Beams-l Floors Ribs Columnshours Simply Continuous Simply Continuous Simply Continuous Imm

supported Mm supported nun supported rommm rom rom

0.5 2()2) 2()2) 2()2) 2()2) 202) 202) 2()2)1 2()2) 2()2) 20 20 20 202) 2()2)1.5 20 2()2) 25 20 35 20 202 40 30 35 25 45 35 253 60 40 45 35 55 45 254 70 50 55 45 65 55 251) For the purposes of assessing a nominal cover for beams and columns, the cover to main bars which would havebeen obtained from table 4.2 and 4.3 of BS 8110: Part 2: 1985 has been reduced by a notional allowance for stirrupsof 10 rom to cover the range 8 nun to 12 mm (see also 3.3.6)2)These covers may be reduced to 15 rom provided that the nominal maximum size of aggregate does not exceed15 rom (see 3.3.1.3).NOTE 1: The nominal covers given relate specifically to the minimum member dimensions given in figure 3.2Guidance on increased covers necessary if smaller members are used is given in section 4 of BS 8110: Part 2:P1085.NOTE 2: Cases that lie below the bold line require attention to the additional measures necessary to reduce therisks of spalling (see section 4 of BS 8110: Part 2: 1985).

IIBeam

I.• b ~I M I.• b ~It

~Floors .!:;

Plane soffit fColumns

I ~:EM MFully exposed 50% exposed One face exposed

Fire Min. Rib Min. Column width (b) Minimum wallresistance beam width thickness thicknessh width (b) of floors Fully 50% One face P< 04% < p>

(b) rom (Ii) mm exposed exposed exposed 0.4% P <1% 10'rom !.

nun rom mm nUll mm mm0.5 200 125 75 ISO 125 100 150 100 7~1 200 125 95 200 160 120 150 120 751.5 200 125 no 2::>0 200 140 175 140 1002 200 125 125 300 200 160 - 160 1003 240 150 150 400 300 200 - 200 1504 280 175 170 450 350 240 - 240 150NOTE1: These minimum dimensions relate specifically to the covers given in tables 3.4 and 4.9.NOTE2: P is the area of steel relative to that of concrete.---

Table 3.7 Form and area of shear reinforcement in beamsValue ofv Form of shear reinforcement to be provided Area of shear reinforcement to be providedN/mm2Less than o.s« See NOTE 1 -throughout the beam0.5ve < v < (v, + 0.4) Minimum links for whole length of beam Asv ~0.4bvsv/0.95fyv (NOTE 2)(v, + 0.4) < v <0.8..Jjcu Links or links combined with bentOup bars. Where links only provided:or5.ON/mm2 Not more than 50% of the shear resistance the Asv ~ b"s,,(v-vc)/O.95f1fV Where links and bent-

provided by the steel may be in the form of up bars provided: See 3.4.5.6bent-up bars (NOTE 3)

NOTE 1. While minimum links should be provided in all beams of structural importance, it will be satisfactory to omit them inmember of minor structural importance such as lintels or where the maximum design shear stress is less than half v,.NOTE 2. Minimum links provide a design shear resistance of 0.4 N/mm2.NOTE 3. See 3.4.5.5 for guidance on spacing of links and bent-up bars.

Table 3.8 Values of Vc design concrete shear stress100As Effective de oth mmbvd 125 150 175 200 225 250 300 ~4oo

N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2 N/mm2:$;0.15 0.45 0.43 0.41 0.40 0.39 0.38 0.36 0.340.25 0.53 0.51 0.49 0.47 0.46 0.45 0.43 0.400.50 0.67 0.64 0.62 0.60 0.58 0.56 0.54 0.500.75 0.77 0.73 0.71 0.68 0.66 0.65 0.62 0.571.00 0.84 0.81 0.78 0.75 0.73 0.71 0.68 0.631.50 0.97 0.92 0.89 0.86 0.83 0.81 0.78 0.722.00 1.06 1.02 0.98 0.95 0.92 0.89 0.86 0.80~3.00 1.22 1.16 1.12 1.08 1.05 1.02 0.98 0.91NOTE 1. Allowance has been made in these figures for a Ym of 1.25.NOTE 2. The values in the table are derived from the expression: 0.79 {lOOAs/ (bvd)}1/3 (400/ d)1/4JYmwhere; 100As / b.d should not be taken as erf'atE'r than ~,

(400/d/ -I should not be taken as less than 0.67 for members without shear reinforcement.(40Ojd/'" should not be taken as less than I for members with shear reinforcement providing design shearresistance ~ 0.4 . N/mm2

For characteristic concrete strengths than 25 N/mm2, the values in this table may be multiplied by {fclI/25)1/3.The value of fal should not be taken as greater than 40.

Table 3.9 Basic span/effective depth ratio for rectangular or flanged beamsSupport conditions Rect. sections Flanged beams with bw/b<= 0.3Cantilever 7 5.6Simply sup. 20 16.0Continuous 26 20.8

Table 3.10 Modification factor for tension reinforcementService M/bd2

stress 0.50 0.75 1.00 1.50 2.00 s.on 4.00 .11.00 n.OO100 2.00 2.00 2.00 1.86 1.63 1.36 1.19 1.08 1.01150 2.00 2.00 1.98 1.69 1.49 1.25 1.11 0.01 0.94({y=250)167 2.00 2.00 1.91 1.63 1.44 1.21 1.08 0.99 0.92200 2.00 1.95 1.76 1.51 1.35 1.14 1.02 0.94 0.88250 1.90 1.70 1.55 1.34 1.20 1.04 0.94 0.87 0.82300 1.60 1.44 1.33 1.16 1.06 0.93 0.85 0.80 0.76

(fy=460)307 1.56 1.41 1.30 1.14 1.04 0.91 0.84 0.79 0.76

NOTE 1: The values in the table drive from the equation:Modification Factor = 0.55 + (477-fs)1120(0.9+MJbd2)::; 2.0..... eqn. 7Where; M is the design ultimate moment at the centre of the span or, for a cantilever, at the support.NOTE 2: The design service stress m the tension reinforcement in a member may be estimated from theequation: fs = 2fyAsreq/3AsprovPb ...... eqn. 8NOTE 3: For a continuous beam, if the percentage of redistribution is not known but the design ultimatemoment at mid-span is obviously the same as or greater than the elastic ultimate moment, the stress Is in thistable may be taken as 2/3fy .

'EJ(!ract from CBS8110 -1997 3/7

Table 3.12 Ultimate bending moment and shear forces in one-way spanning slabsEnd support/slab connection . , At first Middle InteriorSimple Continuous } • interior interior supportsAt outer Near mid. At outer Near mid. support spanssupport end span support In''dspan

Moment 0 0.086FI -O.04Fl - ,(j.0'fi5Fl 0.086Fl 0.063Fl -0.063FlShear Force O.4F - 0.46F - 0.6F - 0.5FNOTE. F is the total design ultimate load (l.4Gk + 1.6Q0; ~t

I is the effectivespan.

Table 3.14 Bending moment coefficient for rectangular panels supported on four sides withprovision for torsion corners

Type of panel and Short span coefficients, f3sx Long span

moments considered Values of lv/lx coeff., /3sy1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0 for all ly/lx

Interior panels. (-)moment at cont. edge 0.031 0.037 0.042 0.046 0.050 0.053 0.059 0.063 0.032(+)moment at mid span 0.024 0.028 0.032 0.035 0.037 0.040 0.044 0.048 0.024One short edge discont.(-)moment at cont. edge 0.039 0.044 0.048 0.052 0.055 0.058 0.063 0.067 0.037(+)moment at mid span 0.029 0.033 0.036 0.039 0.041 0.043 0.047 0.050 0.028One long edge discont.(-)moment at cant. edge 0.039 0.049 0.056 0.062 0.068 0.073 0.082 0.089 0.037(+)moment at mid span 0.030 0.036 0.042 0.047 0.051 0.055 0.062 0.067 0.028Two adj. edges discont.(-)moment at cont. edge 0.047 0.056 0.063 0.069 0.074 0.078 0.087 0.093 0.045(+)moment atmid span 0.036 0.042 0.047 0.051 0.055 0.059 0.065 0.070 0.034Two short edges discont.(-)moment at cant. edge 0.046 0.050 0.058 0.057 0.060 0.062 0.067 0.070 -(+)moment at mid span 0.034 0.038 0.040 0.043 0.045 0.047 0.050 0.053 0.034Two long edges discont.(-)moment at cont. edge - - - - - - - - 0.045(+)moment at mid span 0.034 0.046 0.056 0.065 0.072 0.078 0.091 0.100 0.034Three edges discont. (onelong edge continuous)(-)moment at cant. edge 0.057 0.065 0.071 0.076 0.081 0.084 0.092 0.098 -(+)moment at mid span 0.043 0.048 0.053 0.057 0.060 0.063 0.069 0.074 0.044Three edges discont. (oneshort edge continuous)(-)moment at cant. edge - - - - - - - - 0.058(+)moment at mid span 0.042 0.054 0.063 0.071 0.078 .0.084 0.096 0.105 0.044Four edges discontinuous(+)momentatmid span 0.055 0.065 0.074 0.081 0.087 0.092 0.103 0.111 0.056

3.8.1.6 Effective height of a column3.8.1.6.1 GeneralThe effective height, l", of a column in a given plane may be obtained from the following equation: Ie = PloValues of P are given in tables 3.19 and 3.20 for braced and unbraced columns respectively as a function ofthe end conditions of the column. It should be noted that the 'effective height of a column in the two plandirections may be different.

3.8.1.6.2 End conditionsThe four end conditions are as follows.

a) Condition 1. The end of the column is connected mono lithic ally to beams on either side which are atleast as deep as the overall dimension of the column in the plane considered. Where the column isconnected to a foundation structure, this should be of a formspecifically designed to carry moment.

b) Condition 2. The end of the column is connected monolithically to beams or slabs on either sidewhich are shallower than the overall dimension of the column ill the plane considered.

c) Condition 3. The end of the column is connected to members which, while not specifically designedto provide restraint to rotation of the column will, nevertheless, provide some nominal restraint.

f£:{Jractfrom<BS 8110-1997 4/7

d) Condition 4. The end of the column is unrestrained against both lateral movement and rotation (e.g.the free end of a cantilever column in an unbraced structure).

Table 3:19 Values of 13for braced columnsEnd condition at End condition at bottomtop 1 2 31 0.75 0.80 0.902 0.80 0.85 0.953 , 0,90 0.95 1.00

Table 3.20 Values of 13for unbraced columnsEnd condition at End condition at bottomtop 1 2 31 1.2 1.3 1.62 1.3 1.5 1.83 1.6 1.8 -4 2.2 - -

3.8.1.7 Slenderness limits for columnsGenerally, the clear distance, la, between end restraints should not exceed 60 times the minimum thickness ofa column.

3.8.3 Deflection induced moments in solid slender columns3.8.3.1 DesignIn general, a cross-section may be designed by the methods given for a short column (see 3.8.4) but in thedesign, account has to be taken of the additional moment induced in the column by its deflection.The deflection of a rectangular or circular column under ultimate conditions may be taken to be: au = PaKhIn this expression fl. has the value obtained from table 3.23 or, alternatively, from equation 34 from whichthe table is derived, where K is a reduction factor that corrects the deflection to allow for the influence of

N -Naxial load. K is derived from the following equation: K = 11= ~ ]

Nil: -»:Where: NU7 = 0.45fcuAc + 0.95fyAsc (including allowances, as appropriate for YIII)'The appropriate values of K may be found iteratively, taking an initial value of 1. Alternatively, it will

1 (1)1alwavs be conservative to assume that K = 1.Then a = -- .s:. fJa 2000 b'

NOTE. b' is generally the smaller dimension of the column (but see 3.8.3.6 for biaxial bending)The deflection induces an additional moment given by: Madd = Nail

3.12.10 Curtailment of reinforcement3.12.10.1 GeneralWhere a cantilever forms an extension beyond the end support of a continuous beam or slab, care should betaken to ensme that the top steel in the adjacent span extends beyond the point of contraflexure.3.12.10.2 Simplified rules for beamsThe simplified curtailment rules illustrated in figure 3.24 may be used for beams in the followingcircumstances.

a) The beams are designed for predominantly uniformly distributed loads.b) In the case of continuous beams, the spans are approximately equal.

3.12.10.3 Simplified rules for slabs3.12.10.3.1 GeneralThe simplified curtailment rules illustrated in figure 3.25 may be used for slabs in the followingcircumstances (but see 3.5.3.5 for details of torsion reinforcement at the comers of two-way slabs, 3.7.4.4 fordetailing rules at the edges of flat slabs and 3.1210.3.2 for end supports of continuous slabs):

a) the slabs are designed for predominantly uniformly distributed loads;b) in the case of conlinuous slabs, the design has been carried out for the single load case of maximum

design load on all spans and the spans are approximately equal.

3.12.10.3.2 Curtailment of bars at end support of slabs (where simple support has been assumed in assessment ofmoment.Despite this assumption, negative moments may arise which could lead to cracking. To control this, anamount of reinforcement equal to half the area of bottom steel at mid-span but not less than the minimumgiven in 3.12.S.3 should be provided in the top of the slab at the support. It should have a full effectivetensile anchorage into the su rt d extend not less than 0.1S1 or 45 times the bar size into the span.Bottom reinforcement may be detailed:

~ractfrom (}3S8110-1997 5/7

a) as indicated in figure 3.25 for a simply-supported end, in which case the shear strength at the!lupport may be based on the area of bottom steel continuing into the-support, or

b) as irtdicated in figure 3.25 for a simply-supported end except that th¢pottom steel is stopped at the,Jirl~'of effective support; in this case the shear strength at the supporf~hould be based on the area oftop' steel.

Normalized Design Charts for Beams & Columns

Graph No.1 Singly reinforced beams0.25.-----....-........:..........:..---,-----.-----r----.-----.,..------,----.-----,

fy:::~ _020r---~---~---4_---+_----~---~----~~~-=~~~-----4

x/d ~~~::::.:--

----~~-- -------- ------- -------- -~-~(>J•••••e 015F---.9.:.~-- I /V

~ V~~ .9.:.~ / .

0'0 /vO.05r------+~----+_----~------_+---+_--_4---~---r_--~

~~/o 030 0.450.05 0.10 0.15 0.20 025 035 0040

Graph No 10' R.ctangu~r columns

1.5 :""-'/"1-... <,~31.2 I

~ , ""-, ~I "<,1 I 0.4-4_-----j----+---+----+----4~ r-: / r-, / "<,

b

'_I

[

• A,e· ]• 2 d

Ale."2.

d/h·0.8~

I, .460a 'AacS4l1l/ •••

Design a, a beam in thi, region

-'-+---+-~-1

0.4. 0.5 0.6o 0.1 0.2 0.3. 2

1.11bh Ie.

6/7'E.:{Jractfrom (]3S8110 -1997

For Beams Curtailment of Reinforcement For Slabs

Effective span IClear span

0.251 Face of support

R/F for max. hoggingmoment

60%

20% through span

Mid span

.....................~?~.~~~~~ .1.III

Bottom steel

100%

R/F for max. sagging moment, (0.1 I for end support)

Effective span I.E

Clear span

Face of support

for max. hoggingmoment

,--.,

IIIIII

.....................~?~.~~~~~ !.III

Mid span50%

Bottom steel

100%0.21 R/F for max. sagging moment

a) Continuous member (approximately equal spans)

For Beams .r".I- I

<!J 12 or ~<d12 iequivaIent~t---~L..::::.. Bottom Steel I

anchorage 100% I

IR/F for max. moment

0.081

For Beams100%

1/2 ~\l)45

50%

R/F for max. momentTop Steel

<.dj2-----

NOTE: dL<P

is the effective depth;is the effective span;is the bar size.

Figure 3.24 Simplified detailing rules for BeamsFigure 3.25 Simplified detailing rules for Slabs

For Slabs.......................... +........... ; ~

<!J 12 orequivalentanchorage

Bottom Steel100%

'40%

'fu:jract from CBS8110 -1997 7/7

'<d/2

0.11R/F for max. moment

b) Simple support

For Slabs100%r.

112~45,~,

50%,,

R/F for max. moment,, Top Steel,,

. ---------------------------------,I~d/2 ;--

,c) Cantilever

(Combined)