Unit 6 (Design & Detailing of Beams)

7/28/2019 Unit 6 (Design & Detailing of Beams)

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UNIT 6

Structure6.1

6.1 INTRODUCTIONThis unit is meant to design different types of beams based upon the mechanics ofreinforced concrete and principles involved in design & detailing processes- such assafety and serviceability requirements discussed in Units 1 to 5. Through the design of asimply supported beam, a cantilever beam, a continuous beam and a beam with overhang,analysis, design and detailing have been explained in a systematic sequence.ObjectivesAfter going through this unit a-student will be able to design & detail all types of beams,One will learn the following :

a sequence to be followed in the design of beams,a analysis of beam, and

DESIGN AND DETAILING OFBEAMS

IntroductionObjectivesDesign of a Simply Supported BeamDesign of a Cantilever BeamDesign of a Continuous BeamDesign of a Beam with OverhangSummaryAnswers to SAQs

a structural design & detailing of beams.

6.2 DESIGN OF A SIMPLY SUPPO RTED BEAM

Design a simply supported beam of 6 m clear span. The beam is supported on375 thick wall and loaded with a super-imposed dead weight of 16 kN/m as wellas a live load of 12 kN/m. Use M 15 concrete and Fe 250 steel.

SolutionDepth CD)(i) Thumb Rule

lef lefD lying between --to-10 20Assuming l e , 6 + 0.375 = 6.375m to start with

lCf 6 . 3 7 5 ~ 1 0 "and taking D =- = 63810 10


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Figure 6.1 :A Simply Supported Beam(ii) From Control of Deflection Criteria

d l : 1, fK B K , K ? K ,

where KB= 20For M 15 concrete and Fe 250 steel, the amount of balanced section steel= 1.32% and correspondingly K ,= 1.3

K2= ,= 1Substituting these values in the above equation

Adopted D = 800 and.d = 800 - 40 = 7601 2b = 400 (i.e. lying between -rd to - d of D)3 3

(iii) From Moment of Resistance ConsiderationLoads

Total DLLLTotalDL +LL = 36.0 kN/mDesign Load, w u 1.5 x 36.0 = 54.0 kN/m

2Maximum B. M., Mu = !!!dd- - 54 6.3752 = 274 . 3 2mm8 8(vide Figure 6.2)


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Design and Detailing of Beamsr wU=54kNm

B.M.D.Figure 6.2 :S. F. D. and B. M. D. for the Beam

274.32 x l o 6or, d = = $55.8 c 7600.148 x 15 x 400Now ler = lesser of

(i) 6 + 0.375 = 6.375 m, and(ii) 6 + 0.76 = 6.76 m

Thus le ,= 6.375 mCheck for b from Lateral Stability Consideration

( 0 60h = 60 x 400 x lo-' = 24m >> le , (6.375m), andP

250b2 250 x 4002-ii) - = 52.632 m >> le, (6.375 m)d 760

A,, x 250or, 274.32 x lo6=0.87 x 250 x As[ x 760 400 x 760 x 15

or, 9.06 AS:- 165300AS1+ 74.32 x lo6= 0or, As, ='1846.38 mm2


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Limit State Metbod1'033.6mm2 < 1846.38mm2

~ U I Z . U ~ ~1Ulb.3Umm

mm2> 1846.38mm2) n onerovided 400 x 800 section with 6$22 (A , = 1888.4layer at d =760Curtailment and Detailing of Tension ReinforcementIf three out of six bars are curtailed, then moment of resistance of the section withcontinuing 3920,

Let x be the distance of theoretical cut-off point from the L.H. supports. then fromM u =R.x - u 2 where R = Reaction at L.H. support

or, 147.74 = 172.13 x - 5 4 x2/2or, x' = 1.02 m and 5.355 m from L.H. upport

i.e. 2.17 m from either side of mid point of the span.Actual cut-off point'from mid section

=2.17+(>of 129ord)= 21.7 + 0.76=2.93 m

Lengtm available beyond the actual cut-off point towards supports

Provision of Shear Reinforcement(a) Shear reinforcement at theface of the support


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Dcslgn and DetaUlng of Bunu

( 0 . 4 6 - 0 . 3 5 ) NAccordingly, 'c = 0 .35 + x ( 0 . 3 - 0 . 2 5 ) = 0 .3 8--i-(0 .5 - 0 . 2 5 ) m mtc< tn:hence shear reinforcement shall be provided for

or sV= 357.52 c/cCheck for minimum shear reinforcementsVshall be the least of(i) 0.75d = 0.75 x 760 = 570(ii) 450, and

o.87fy$v 0.87 x 250 x 100.53(iii) - = 136.650.4b 0.4 x 400Hence f 8-2 legged vertical stirrups 8135 dc may be provided.(b) d e a r reinforcement at the cut-off section

This shear reinforcement shall be provided over a distance of

from the cut-off section.


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Limit State Method

- d -760

SArea of bars cut-off =0.5)V, max - 190 C/C where fi b =$P i 8 x 0 .5 Total area of bars

Hence provided 4 8 - 2 legged vertical reinforcement @ 110 c/ctowards both supports. As the distanc e 570 > ( 3000 - 930 = 70), thesame reinforcement shall continue upto the face of support. The shearreinforcements may be provided in an usual way (Unit 3) beyond cut-off point towards centre of beam.Detailing Near the Supports( 9 Ld = 1088

Length of bars inside the support for L, beyond the cut-off point= 1088- 3000- 2930) = 1018

The bars shall be bent up with a standard hook of 9 8 bend.L34(ii) Again L =- +L,v

where M , = 147.74 kNm

1.3x 147.74 x l o 6or, i080 = 172.13 x lo " + Loor, Lo= (-) 27.8This means that Lo is not required for the diam eter of bars provided.,

(iii) Minimum length of bar inside the support

Side Face Reinforcement

Provided 2910 bars (A, = 156mm2> 30412mm2) n each faceThe details have been shown in Figure 6.3.


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! rA Ir 0 10 (Hmgcrbars) ,

SECTIONA-AI Figure 6.3 :Details of the Beam DeslgnedSAQ 1

Design and detail a simply supportedR.C. beam of 5m clear span supported onwalls 380 thick to cany a total dead and live load (excluding self load) of25 kN/m. Use M 15 concrete and Fe 415 steel.

6.3 DESIGN OF A CANTILEVER BEAMExample 6.2

Design a cantilever beam for the superimposed dead and live loads shown inFigure 6.4 below. Use MI5 concreate and Fe 25 0 steel.-

I SolutionDepth (D)(i) From Control of Deflection Criteria

where K,, 7For M I 5 concrete and Fe 250 steel, the balanced section steel area= 1.32%for which


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Limit State Method K,=1.3K2= K,= 1

Substituting above values4000d k = 439.56 rnm7 ~ 1 . 3 ~ 1 ~ 1

Adopted 6 = 50 at fixed end and b = 400 throughout.

Figure 6.4 :A Cantilever Beam(ii) From Moment of Resistance Consideration

Loads

Superimposed LoadTotal loadDesign loads, w, = 1 . 5 ~ 4 6 = 69 kNlm

w 1 6 9 x 4 'Design Moment, MU = -KL- --- = 55.2 kNlm2 2Effective depth (6) or resisting applied mom ent is obtained byMu.~,m = 0 . 1 4 8 4 , bd2552x lo6 = 0 . 1 4 8 ~ 5 x 4 0 0 x d 2or , d = 7 8 8 . 4 3


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Assuming effective cover 50,d = 850- 50 = 800> 788.43

Check for b from lateral stability consideration

ASt 250or. 55.2 x lo6 = 0.87x 250 x Ast 800 (I - 4oo l 5or, 9.06 s: - 174000 Ast 552x lo6= 0or, All 4009.46mm2

Hence provided 400 x 850 section at fixed end with 5$32( A , = 4021.24 mml)

Curtailment and Detailing of Tension ReinforcementLet 2$32 be curtailed at x rom the free end , then mom ent of resistance of thesection of cqntinuing 3402 with 50 cover throughout the length

where de= effective depth at free end

= 1.0467x 10' + 65596.456 xApplied B.M. at x from free end

Equating M, with Mu34.3 x2 = ( 1.046x lo8 65596.456~)

or x= 1.74m

Design and Detailing of Beams


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Llmit State Method

Actval cut-off point fromfixed end

The tensile reinforcement will extend both sides of the face of the support, alength 2 L,, and the continuing 3g3 2 towards free end from cut-off section shallalso extend 2 L, (Figure 6.5).Provision of Shear Reinforcement0 ) At support

850 . - 350where tan = = 0.1254 0 0 0

Taking 0 8 - 2 legged vetical stirrups as sh ear reinforcement

(ii) At cut-off sectionActual cut-off section from free end= 4 - 2.78 = 1.22 m: Shear force and B.M. t cut-off section,


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I

t @ 135sa -2 leg. I

SECTION A-AFigure 6.5 :Details of the Designed Beam

Provided @S- 2 legged stirrups@ 135 throughout


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Limit State Metbod


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(ii) From Control of Deflection Criteria

support and the other is continuous. C .For M15 and Fe 250, pB% = 1.32% andcorrespondingly, K , =1.3K ,= K3=

6000: d 4 = 200.672 3 x 1 . 3 x I x IIAssuming D = 600 ;d = 600- 55 = 545 and taking b (between - d to3

2- d D) = 3003Check b from Lateal Stability Consideration

Let the beam be laterally supported at 6 m interval i.e. at supports,ze, < 60b; or lef(6m) < 60 x 300(= 18m)


LoadsSelf = 0.6 x 0.3 x 1x 25 = 4.5 kN/mDLTotal DLLLTotal (DL+LL)Design DL = 1.5x 14.5 = 21.75 kN/mDesign LL = 1.5x 12.0 = 18.0 kN/mDesign (DL+LL) = 1.5 x 26.5 = 39.75 kN/m

* Evaluation of Design B.Ms & S. Fs. at Critical SectionsMs Ks u 1 and M n Kn u 2

where M S nd M,,re span and support B.Ms and Ks nd Kn re their corresponding momentcoefficients respectively. Similarly VU S.F. at the centre of support ( Figure 6.7 (a) to (f)).

* Refer Appendix A.


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Limit State MethodDue to DL only

(a) Span and SupportB.Ms.

VU= . 3 8 ~ 2 1 . 7 5 ~ 61 = 49.59,(b) Maximum S.Fs. at the Supports for DL

Due to LL only

(c) Span and Support B.Ms. for LL

(d) Maximum S.F. at the Supports for LLDue toDL+LL only

(e) Span and SupportB.Ms. forDL+LL

(f) Maximum S.F. at the Supports for DL+LLFigure 6.7 : Evaluation of B. Ms and S. Fs at Critical Sections


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(iii) Ffom Moment of Resistance Considerationxu, max 1 - 0.42 -)u rnax bd2fckM u = Mu,iirn = 0.36- d

(where adoted d = 600 - 25 - 20 - 12.5 = 542.5 for mainreinforcement of $20 in two layers and a clear spacing of 25between two layers of bars)Moment of resistance for balanced secion

Mu,lim= 0.36 %d?E Xu rnaxd I - 0.42 -1 d bPfrk= 0.36x053(1-0.42x0,53)x 3a)x 542.5' x 15 x 10-6= 196.01 kN m > 178.87 kNm

Hence the section is under-reinforced.

Hence provided 6 Q 16 (A,= 1206.37 mm2) n tw o layersAt Intermediate support

and Detailing of Beam


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Llmlt State Metbod Provided 10 Q, 16 (A,,=2010.62mm3 in two layersDetaling ofMain ReinforcementAt Simple SupportLe t 4 4 16 be curtailed and the rem aining 2616 be continued in the simplesupport. Moment of resistance of the section with 2416 ,

Let xmbe the distance of curtailment of 4@16 rom the outer support, then

or- 19.875 xZ- 97.47 x + 45.49 = 0or x = 0.52 m and 4.38 mFrom cen tre line of end supp ort, the actual cutloff point

= 0.52 - > o f 1 2 0 or d )E 0.52 - 0.5425 = - 0.0225 m

The continuink bars must extend into the support by a distance

V at the centre line of support= 97,47 kN

or Lo 263.28Maxim um straight length available from centre line of suppo rt to the outer face ofthe bent bars- 375- - - 25 = 162 .5 (Figure 6.8)


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Now 263.28 = 162.5 + 8 0 + xor x =- 27.22

Hence only standard hook of 90" bend shall be sufficient (Figure 6.8).

Standard 90" bend0 Design and Detailing of BeamsFigure 6.8 : Detailing of Positive Reinforcement at End Support

PositiveBend ing Moment Reinforcement at Intermediate supp ortActual cut-off section from centre line of end support

= 4.38 + ( >12 0 or d)= 4.38 + 0.5425- 4.923 mi.e. 1.077 m from intermediate support. The continuing bars shall extend into the

Ld 8 7 0 - 2 9 0upport by at least - -3 3Let point of inflexion be at x from end support, then

x20 = 9 7 . 4 7 ~ 39.75 -- ;or x =4.9 m2S.F. at point of inflexion,V = 97.4- 39.75 x 4.9

= 97.305 kNM IL', >- Lov

45.49 x l o 6or 8 7 0 > + Lo97.305 x 1 0 "or Lo> 0.402 m< 1.18 m (Figure 6.9)

Negative B.M. Reinforcement at Intermediate SupportLet half of the (-) tive B.M. reinforcement be curtailed, then moment of resistanceof the section with remaining 5 016

M u 0.87.fyAstd[1- *)d .


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Limit State Method

71x 250 x 5 x - 16' x 542.54 3 0 0 ~ 5 ~ 5 4 2 . 5 ~5Let this B.M. occur at x from centre line of end support then

39.75x2-106.41 = 97.47~- or, x = 5.824 m2i.e. theoretical cut-off point from centre line of intermediate support= 6-5.824=0.176mTherefore, actual cut-off point from centre line of intermediate support

=0.176+(>of 124)ord)= 0.176 + 0.5425 = 7 19 m < 0.87 m (L,)

Hence, all (-)tive B.M. reinforcement shall extend upto 0.87 m from intermediatesupport. The remaining 5 $16 must extend beybnd the inflexion point for adistance greater of d = 542.5or l a @ = 2 x 16 = 192

L 1or =- 6000-375) = 351.5616 16Therefore remaining 5 Q16 shall extend upto (1.077 + 0.5425) m = 1.62 m(Figure 6.9)

Figure 6.9 : Showing Details of Reinforcement Near Intermediate Support

Detailbng of Transverse Reinforcement at Intermediate SupportS.F. at theface of support,


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TC100 x 10 x - x 16'bd 300 x 542.5

Accordingly,

(0.64 - 0.6)2 , = 0.6 + x 0.235 = 0.C38(1.25 - 1.0)V,,, = V , - ~ , b d 140.45- 0.638 x 300 x 542.5

= 36.616 kNTaking 0 -2 leg. stp

S"

0.87 x 250 x 100.53x 542.5or;sv = 36.616x lo-"= 323.95< 450 < 0.75d(406)

From minimum shear reinforcement considerationA,, 0.4- -bsV 0.87 y

100.53x 0.87 x 2505 1 182.21< 323.950.4 x 300Shear reinrbrcement of cut-off point of (+) tiveB.Mnear intermediate supportS.F.= V, ,= 147.9- 39.75 x 1.077= 105.0kN

105.09x I000 N7 , = = 0.65-7300 x 542.5 ni m -

100A, 100 x 2 x 201- -Accordingly T, = 0.35 NlmmZIfuS= Vu- tcbd = 105.09- 0.35 x 30Q x 542.5 x

Desiw:and Detailing of Beams

. einforcement


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Limit State Method 0.87fyA, ,d""\ =

v

- "us s 1Or A'' 0 . 8 7 J y d

0 . 4 0 ~ ~A,, = ---. f y

: Total area of transverse reinforcemetiA, = As, + A, , = 10b.53 =

100.53or; S , = [ 48.12 x l o 30.87 x 250 x 542.5P h = 0.5

d 542.5. - - - -- - 135.625 > 1 13.5S,, 8 x 0 . 53 3- ( I = - X 542.5 = 406.874 4

Hence provided48 -2 legged step @ 110for adistance 406 from thecutoff point.(Figure6.9)Shear reinf ~rc em ent s or other portion of the beam may be provided in usual way(Unit 3)3Design and detail a continuous tieam of two spans each of 6 m between theirsuppo rts fob a total load (excluding self w eight) of 30 kN/m. The beam isrnonolithicallly su pported on co lumn s. Use M 20 concrete and 415 steel


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6.5 DESIGN OF A BEAM WITH OVERHANG Design and Detailing of BeamsExample 6.4

Design an overhang beam loaded with a dead load of 16 kN/m and a live load of12 kN/m shown in Figure 6.10. Use M 15 concrete and Fe 415 steel.

Figure 6.10 : An Overhanging BeamSolution

Depth (D)(i) From Thum b Rule

For simply supported position

1 . 6000Taking D = A- =-- 600 at the start .10 10(ii) From Deflection Criter ia

For simply supported portion

= 23 since one of the supports is simply supportedand the other is continuous.


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pmax 0.72% for M 15 concrete and F e 415 stee l accordingly K, 1.05K2 K, = 1Substituting these values in the above equation

For cantilever portion

where K, =7

As maximum (+)tive bending moment in simply supported portion is, tosome extent, balanced by the overhang, the depth required will not be thesame as that for simply supported beam (Example 6.1). Thus adoptedD = 600 and b = 30 0

Check for b from Lateral Stability ConsiderationLet the beam be laterally supported at supportsFor simply supported portion60 b = 6 0 x 3 0 0 x l o - '= 1 8> 6m

For cantilever portion25b = 25 x 30 0 = 7.5 m > 2m

Hence O.K.(iii) D from Moment of Resistance Consideration

LoadsSelf = 0.6 x 0.3 x 1 x 25 = 4.5 kN/mDLTotal DLLLTotal (DL +LL)Design Load, w U =1.5 x 32.5 = 48.75 kN/m


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AnalysisofBeamE M , = O (Figure 6.1 1)or, 48:75 x 8 x 4- R , x 6 =0

.or, R, =48.75 x 8 x 416 =26 0 kN:. R, = 48.75 x 8 - 260= 130kN

(a) Beam Loaded With @L+LL) onBoth Spans

(b)S.F.D.

(c) B.M.D.Figure 6.11 :S. F. D and B. M.D. or the Beam

Let Mu,_x occur at x from A in the span, then S.F, t that section will be zeroi.e.,

130- 48.75 x =0.or. x =2.67 m

4 8.75 ~ ' 2 . 6 7 "Thus MU,MI 1 30 x 2.67 - = 173.33kNm2B.M. at support section

Let the point of inflex ion be at x from simple support i.e.


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Llmrt ,State M ethod or x r 5.33 m i.e. 0.67 from supportB.The resulting S.F.D.and B.M.D.re as shown in (Figure 6.11)

or, d= 528.38Assuming two layers of # 16 steel,d = 600- 25- 16-2512 = 546.5 > 528.38

Hence, provided b x D = 30 0 x 600Provision of M ain Reinforcement

(i) For span

or, 173.33x 1o6 = 0.87 x 415x 4,x 546.5 300x 546.5x 15or, 33.296833 A,2 - 197313.83Ant+ 173.33 x 106= 0

Cor,^:*- 5925.9 As, + 5205600.1 = 05925.9 t J 5 9 2 5 . 9 ~ 4 x 1x 5205600.1or, A,, = 2

or,.As,= 1072.6mm2

Hence provided 6 # 16 ( A , = 1206.37 mm2> 1072.6 mm2)( i i) At support B

Mu = 0.87. Astd(l - "/-Id . Ck A,, x 415or, 97.5 x l o b = 0.87 x 415 x A,, x 546.5 300 x 546.5 x 15or,As> 5925.9 An t+ 2928206.8 = 0

5925.9 + d5925.9? - 4 x 1 x 2928206.8or, A,, =-2

5925.9 f 837.7 2or; Ast = = 544.lmrn > 368.67(A,,,,, )2


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Hence provided 3#16 ( A , = 603.2 mma)Curtailment and Detailing of Tension Reinforcement(i) At simple support

If three out of six (+)tive reinforceing bars are curtailed, then moment ofresistance of the section with remaining three bars,

= 106.904 kNmLet this B.M. e at x from L.H. support, then106.904= 130 x - 48.75 x2/2or x2- 5.33 x + 4.3858 = 0

or; x = 1.017 m and 4.317 m(a) These curtailed bars will be extended on both ends bygreater of 120 (192) or d(546.5)(b) The curtailed bars have length greater than L, ( 903)towards the maximum bending moment i.e. towards the

critical section(c) The continuing bars, from cut-off point should havetotal length encluding Lo greater than L, and must

Ld 903extend into the support by - - 30 13 3(d) The length of straight portion of into the support

= 380-25- (4 + 1) x 16= 275 < 301Hence, the bars shall be bent with 90 bend.

1.3MI(e) Ld > + Lo at L.H. supportv

Design, and Detailing of Beerm

where M, = 0 .87&~, ,d


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-Limit State Method

V = 130 at centre line of su pport1.3x 106.904or; Lo < - 0.903 = 0.16 6 m130

Let Lo= Equiva lent length of 90 bend + xor ; 1 6 6 = 8 x 16+ xor; x = 38The detailing on L.H. support has been done as shown in(Figure 6.12).

Figure 6.12 :Detailing of the Simple SupportCu rtailm ent of (+)t ive Reinforcement towards R.H. S u p p o r t(a ) The continuing three bars shall extend upto6 - .317 - .5465 = 1.1365m> 0.903 m (L,) (vide Figure 6.13)(b ) These three bars must extend into the support at length not less than

upto [ ? d l - F) 111 beyond the centre line of the support.


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(c) As the point of inflexion is away from the support, factor 1.3multiplication for M , V will not applyDesign

whereM, = 106.904 kNm as before, andV = S.F. at point of inflexion i.e. at 0.67m from centre line ofsupport = 97.5- 0.67 x 48.75 = 64.838Substituting M, & V in the above equation.

The detailing of positive reinfo~cement f R.H. upport is shown in(Figure 6.13).

Figure 6.13 :Detailing of Positive Reinforcements at R. H. Supports

Curtailment of (Stive Reinforcement at R.H. Support1#16 is curtailed out of 3#16,(a) The continuing 2#16 shall extend beyond the point of inflexion for alength greater of

(i) d = 546.5(ii) 1 2 @ = 1 2 ~ 1 6 = 1 9 2 , a n d


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1 1(iii) - h of clear span = - 6 - 0.38)x lo3= 35 116 16i.e. the continuing bars shall extend ( 0.67 + 0.5465 ) - 1.22 m fromcentre line of R.H. support (vide Figure 6.14).

Figure 6.14 : Detailing of Fegative Reinforcement over R. H. Supportsout of 3#16 let 1#16 bar be curtailed then M.R. of the section with 2#16

300 x 546.5 x 15

If x be the distance fromfree end to the point where My -76.652 kNm,then

or; 24.375 x2- 260 x + 443.348 = 0

= 2.131 from free end i.e. (2.131m - m) = 0.131 from centre line of R.H.support.The curtailed bars 2#16 shall extend greater of 12$(196) or d (546.5)beyond the theoretical cut-off point i.e. the bars will extend ( 0.131+0.5465 ) m = 0.6775m from centre line of support into the simple span orLd(0.903m)fromface of support whichever is greater.


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Curtailment of Bars in Cantilever Portion(a) Let 1#16 be curtailed at x from free end then from eqn.

or x = 1:254 m i.e. 2-1.254 = 0.746 m from support.The 2#16 bars must extend greater of

(ii) 1241= 196, and(iii) d = 546.5froin centre line of R.H.upport.

(b) The continuing 1#16 shall extend beyond cut-off point a distance equalto LdThe detailing has been shown in Figure 6.15.

Figure 6.15 :Details of Cantilever PortionThe shear reinforcement for the beam shall be designed as done in theprevious examples.

SAQ 4Design and detail an overhang beam of 5.5 m clear span between the supports and1.75 m of cantilever portion beyond the centre line of support. I t is loaded with20 kN/m of superimposed load. The supports consists of walls 380 thick. UseM 15 concrete and Fe 4 15 steel.

Design and Detailing of Beam s


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6.6 SUMMARYDesign and detailing of a R.C. rectangular beam is a process of fixing size of concretesection, determination of area of tensile and sheer reinforcements of its different crosssections to resist different internal forces due to applied loadings keeping in view itsserviceability requirements. The various compon ents of design such as design of beamsfor flexure, shear, bond & anchorage as well as detailing to m eet the serviceabilityrequirements discussed from Unit 1 to 5 have been knitted together to ge t a completedesign and details for beam elements. Design of various types of beams of rectangularcross section have been illustrated with detailed drawings in this unit so that a studentmay be able tlo design a beam in a systema tic mann er and put forward a comprehensivedrawing for execution.6.7 ANSWERS TO SAQsSAQ 1

Refer Section 6. 2SAQ 2

Refer Section 6.3SAQ3

Refer Section 6.4. Since the supports are monolithic with the beam, the endsupports must be provided with the (-)tive reinforcement and detailed as perrules. The maximum bending mom ent and shear forces may be calculated as perC 21.4.1( as of BIS :45 6 - 1978)

SAQ4Refer Section 6.5

FURTHER READINGS P 1 6 : 1980 "Des ign Aids for Reinforced Concrete to IS : 56-1978BIS :456-1978, "Code of Practice for Plain and Reinforced Concrete". Bureau of IndianStandard$, Manak Bhaw an, 9 Bahadur S hah Zafar Marg, New Delhi - 110 002.Ashok K. Jain, "Reinforced Concrete Limit State Design", New Chand & Bros, Roorkee.S. K. Mallick & A. P. Gupta, "Reinforced Concrete", Oxford & IBH Publishing Co. Pvt.Ltd.Sinha S. N. "Reinforced Concrete Design", Tata McGraw-Hill Publishing CompanyLimited.14112Asaf A li Road, New Delhi - 110 002.S P : 24-1983, "Explanatory Handbook on Indian Standard Code of Practice of Plain andReinforced Concrete (IS : 56-1978)", Burau of Indian Standards, Manak Bhawan,9 Bahadur Shah Zafar Marg, New Delhi - 110 002.S P : 3 4 4 s & T)-1978, "Handbook on Concrete Reinforcement and Detailing", Bureau ofIndian Standards, Manak Bhaw an, 9 Bahadur Shah Zafar Marg, New D elhi - 110 002.Ram Chandra, "Limit State Design", Standard Book House, 1705-A, Nai Sarak,Delhi - 110 006.Arthur H. ilson & George W inter, "Des ign of Concrete S tfuctures", McGraw-Hill, Tnc.


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Design and Detailing of Beams6.9 APPENDIX ABENDING M OMENT AND SHEAR FORCE

Table A.1 Maximum Bending Moment and Shear in ContinuousBeams of Equal Span

A11 beams freely sup por ted at ends.- For all spans equally loaded For incidental load causingsimultaneously the worst effect--ao ,125 ,125t",2 -a." & D.070 D.070 D 11.096 D.096 Da -

c .i! 3 . lo0 . lo0 . 1 17 ,1172 k .c D.080 D. 025 D.0 80 D U.101 D.075 D . I O ID -+a$ 8 . ,107 ,071 .I07 ,121 ,107 . I21"3 D.077 D.036 D ,036 D .077 d.- D.099 D.081 D.081 D.099 L)

' 5 g ,105 ,080 ,080 ,105 .I 20 ,111 .I1 1 .120T~ D.078 D.033 D.046D .033 D.078 D ~ . i o o . 08 0 D. 08 6 D .0 80 D .1 00 Dc- 5Note : Bending moment = coefficient x w x 1

W = Total load on one spanl = effective span

The coefficients written above the span are for negative mom ent at supports and thosewritten below the span are for positive moment at midspan.

Table A.2 : Maximum Shear Force in Continuous Beams ofEqual Span- - Al l beams freely s uppo rted at ends

For all spans equally loaded For incidental load causingsimultaneously the worst effectm

Note : Shear force = coefficient x total load on one span

Note 1. S.F. coefficients above line apply to SF. at right hand side of sup port.S.F. coefficients below line apply to S.F. at left hand side of su pport.


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l i m i t State Method Table Bending Moment in Continuous Beams of EqualSpans Freely Supported at Ends due to Unit MomentA p p l i e d a t E n d sBending moment = coefficient x applied moment

Bending Moment CoeficientsUni t moment app l i ed a tone end o n l y

'b y DI .@ .267 - .067D J D D D1.m ,268 - , 0 1 8DJ D D D1 . ,2 68 - .0 7 2 , 0 1 9 -.OD5' D D D D D

Unit moment applied ateach of the two ends' b W O1 . k . 2 0 .20 -1 .0DJ D D D1.w . 2 6 8 - ,286 -1.0D D ID D D

1 . , 2 6 3 - . 0 5 3 - .0 5 3 - . 2 6 3 $ . OD ' D D D D D

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Unit 6 (Design & Detailing of Beams)