Topic_5_rc Beam Design [Compatibility Mode]

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RC BEAM DESIGNTABLE OF CON TENTS5.1 Introduction5.2 Sizing5 .3 Design for flexure5.4 Design for shear5.5 Checking of Deflect ion5 .6 Checking of Cracking5.7 Design for torsion

CHAPTER 5

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5 .1 Int roduct ion The purposes o f th is chapter is to compi le the design pr inc ip les and

procedures that have been discussed previously in order to form acomp lete design procedures of reinforced concrete beam.

Basical ly, beam is the structural e lement which subjected transverseload in t he form bend ing mo ment , shear force and torsion. Thereforebeam is d esigned t o resist a l l t hat part icular factors.

Beside o f tha t , beams a lso been check to fu l f i l l the serviceab i l i tyrequirement in order to p roduce an adequate and safe beam d esign.

Therefore, the complete design procedures of reinforced concretebeam are stated as fo l lows;

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5 .1 Int roduct ion

The comple te beam design procedures can be sta te as fo l lows;

Durabil i t y and fire resistance requirement

Preliminary beam sizing

Loading est im ation

Structural analysis (bending mo ment , shear force and torsion)

Design of main reinforcem ent

Bar curta i lment and anchorage

Design of shear l ink

Checking of deflection

Checking of cracking

Detai l ing

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5 .2 Beam sizing Beam ar rangement and s izing normal ly control by an archi tectural

detai l ing and for mechanical device and equipm ent.

An eng ineers shou ld ensure tha t the proposed beam s ize is enoughto carry the loads.

As a guidel ine, the prel iminary sizing can be determine using tr ia land error approach or deflect ion l imit basis.

Overa ll dep t h, h = s pa n / 1 5

B re ad t h , b = 0 . 3 h 0 . 6 h or b as ed o n d ur ab i li t y

and fire resistance requirem ent.

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5 .2 Beam sizing

M i n im u m s ec ti on d i m en si on s

RC BEAM DESIGN

Fireresistance

Minimum Dimension

Beam Floor Fully exposed

hrs. Width Thickness column width

(b mm) (h mm) (b mm)

0.5 200 75 150

1.0 200 95 200

1.5 200 110 250

2 200 125 300

3 240 150 400

4 280 170 450

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5 .3 Desi gn for Fl exu re Design procedures of reinforced concrete rectangular beam based on

Clause 3.4.4.4, BS 8110.

1) Calculate K = M / fcubd2

2) Calculateb =

3) Calculate K

K = 0.156 Redistribution < 10%

K = 0.405(b - 0.4) 0.18(b - 0.4)2 Redistribution > 10%

4) If K < K ; Compression reinforcement not required

i) Calculatez = d { 0.5 + 0.25 K/0.9 } but z < 0.95d

ii) Calculatex = (d z) / 0.45 and check ratio of x/d

iii) CalculateAs = M / 0.95fyz

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5 .3 Desi gn for Fl exu re

5) If K > K ; Compression reinforcement is required

i) Calculatez = d { 0.5 + 0.25 K/0.9 }ii) Calculatex = (d z) / 0.45

iii) Check the ratio of x/d and d/x

iv) Calculate;

If d/x < 0.37

If d/x > 0.37

v) Calculate;

RC BEAM DESIGN

)'(95.0

)'('

2

ddf

bdfKKA

y

cu

s

)/'1(700

)'('

2

xd

bdfKKA cu

s

'95.0

2's

y

cu

s A

zf

bdfKA

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5 .3 Desi gn for Fl exu re Design procedures of reinforced concrete flange beam based on

Clause 3.4.4.4, BS 8110.

1) Calculate Mf= 0.45fcubhf(d hf/2)

2) If M < Mf ; Neutral axis lies in flange and compression reinforcement

not required.

i) CalculateK = M / fcubd2

ii) Calculatez = d { 0.5 + 0.25 K/0.9 } but z < 0.95diii) Calculatex = (d z) / 0.45 and check ratio of x/d

iv) CalculateAs = M / 0.95fyz

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5 .3 Desi gn for Fl exu re

3) If M > Mf ; Neutral axis lies in web

i) Calculatefusing equation 2 or Table 3.7 BS 8110

ii) CalculateMuf =ffcubd2

iii) CompareMwithMuf

iv) Check hf < 0.45d

4) If M < Muf; Compression reinforcement not required

i) Calculate

RC BEAM DESIGN

)5.0(95.0

)45.0(1.0

fy

fwcu

s

hdf

hddbfMA

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5 .3 Desi gn for Fl exu re 5) If M > Muf; Compression reinforcement is required

i) Calculate;

ii) Calculate;

RC BEAM DESIGN

)'(95.0

)('

ddf

MMA

y

uf

s

'95.0

))(45.02.0(s

y

wfcuwcu

s Af

bbhfdbfA

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5 .4 Design f or Shear

As we have a lready seen from

the examp les of fa i lure mod esfor RC beams due to b ending,now we must consider thecapaci ty of the b eam withrespect to shear. Shear failu rema y be one of t wo types asshown in the figures:

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5 .4 Design f or Shear The f irst of these, d iagonal tension, can be prevented by the provision

of shear l inks; while the second, diagonal com pression, can beavoided by l imit ing the maximum shear stress to 5 N/ mm 2 orwhichever is the lesser.

The design shear stress v, at any cross section is g iven by:

Equation 3 (BS 8110 )

where V, is the design shear force on the section, bv is the width of thesection and d is the effect ive depth of the section.

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From the research done by Ta i lor 19 74 show that the shear

resistance provided b y the reinforced concrete section without anyshear reinforcement given by 3 internal com ponent o f forces whichare;

Concret e in the com press ion zone, Vcz Aggregate inter loc k a cross t he cra ck ed zon e, Va Do we l a ct ion of t he t en si on rei nf orcem en t , Vd

The complexi ty o f these factors has led to the adopt ion o f aexpression for design con crete shear resistance vc as shown:

RC BEAM DESIGN

Refer Table 3.8 BS 8110

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5 .4 Design f or ShearWhere;

As = Area of tension reinforcementbv = Width of beam = bd = Effective dept h

m = Part ia l safety factor = 1.25

Should not be taken as greater than 3

Should not be taken as less than 1

If may be mult ip l ied by the valueoffcu should not be taken as greater than 40 N/ mm

2

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Generally, when the design shear stress is greater than the design

concrete shear stress, enhancement of the shear resistance is gainedby the provision of shear reinforcement: either l inks or a comb inatio nof l inks and bent up b ars.

As the use o f l inks is fa r simpler and more usua l th is method wi ll beadopted for our purposes.

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5 .4 Design f or Shear Shear l ink is p rovided in te rm o f the i r spacing. But to eva luate the

spacing a diameter of l inks must f i rst be chosen.

Al though we have arrived at only two possibi l i t ies for the calculat ionof shear reinforcement provision is should b e noted t hat the BS givesus three al ternatives to consider as shown in Table 3 .7 BS8 11 0:

RC BEAM DESIGN

vy

v

vs

fbSA

95.04.0

vy

v

vs

f

bSA

95.0

4.0

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A fu r ther l imi ta t ion p laced on the spacing of l inks by the BS is tha t themaximum spacing should be less than 0 .75 d, which is obviouslynecessary to avoid a failure pla ne forming which m isses the l inksaltogether.

RC BEAM DESIGN

vy

cv

vs

f

vvbSA

95.0

)(

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5 .4 Design f or Shear Area o f shea r rein fo rcemen t , Asv

RC BEAM DESIGN

Single Leg Link Double Leg Link

Area of steel Area of steel

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Exam ple 5 .1 : Des ign of M a in and S hear Re in forcement .

Figure below show the beam t hat has been designed t o resist bendingmom ent. I f the character ist ic strength of shear l ink,fyv i s 2 5 0 N / m m

2

and characteristic strength of concrete,fcu i s 3 0 N / m m2 . Design the

main and shear re inforcement of t he beam.

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5.5 Checking of Def lect ion As a l ready ment ioned many t imes, we must remember to check

serviceabil i t y as well as ultim ate l im it states.

The SLS for deflect ion considers the performance of the structureunder Working Loads such that the structure, or part of i t underconsideratio n, do not deflect excessively causing unsightly crackingand loss of durabi l i ty.

BS 81 10 deta i ls how def lect ions and the accompanying crack wid thsmay be calculated.

Limi ts for SLS of de f lect ion are set ou t in BS811 0: Par t 2 , c lause3 .2.1 . The deflection is not iceable if i t exceeds where L= span of abeam or length of a canti lever. The code also stat es that dam age topartit ions, cladd ing and finishes wil l generally occur if the deflectionexceeds

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5.5 Checking of Def lect ion

L/ 5 0 0 or 2 0 m m wh ich eve r i s l esse r f or b ri t t le f in ish es

L/ 3 5 0 or 2 0 m m wh ich eve r i s l esse r for non -b ri t t le f in is hes .

One o f the me thod fo r check ing that def l ec t ion is not excessi ve isby l im iting t he span-to-effective depth ratio as set out in Clause3 .4 .6 .

The a l lowab le va lue fo r the span -to -e ffec t ive dep th rat i o dependson:-

The b asi c span -t o- ef fec t ive d ep th r at io f or rect an gu la r o rf langed beams and the support condit ions.

The am ount of t en si on st ee l a nd i t s s t ress es

The am ount of c om press ion st ee l

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5.5 Checking of Def lect ion The code sta tes that t he basic span-to-ef fect ive depth ra t ios fo r

rectangular and f langed beam s are so determined as to l imit the totaldeflect ion to L/ 25 0 as shown in Table 3.9.

Note: i f bw/b >0.3 l inear interpolat ion between 0.3 and 1 can be used

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Compare the span -to -e ffec t ive depth rat i o .

I f the above cond i t ion app l ied , that mean the def lect ion is underal lowable l imitat ion and t he structures are safe to use.

is defined as;

If L < 1 0 m

RC BEAM DESIGN

permissible

permissibleactual

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5.5 Checking of Def lect ion If L > 1 0 m

Where;can be obtained from Table 3.9 BS 81 10

M FTR : M odif icat ion Factor for Tension ReinforcementCan be obtained from Table 3.10 BS 81 10

MFCR : Modi f ica t ion Factor fo r Compression Rein forcementCan be obtained from Table 3.1 1 BS 81 10

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M FTR

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5.5 Checking of Def lect ion M FCR

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If the beam is fa iled in term of

deflect ion and the fo l lowing steps should be taken. In crea se t he a rea o f t en si on re in forcem en t (As ).

In crea se t he e ffec t ive d ep th o f t he b ea m (d ).

RC BEAM DESIGN

actual permissible

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5.5 Checking of Def lect ionExample 5.2 : Checking of Deflection .

A simply supported beam with span of 6 m have a dimension of 25 0 x42 5 m m as shown in f igure. The beam subjected to bending m omentof 17 9.1 kNm. Check t he deflect ion of th is beam , assume there areno moment redistr ibut ion.

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5.6 Checking of Cracking

Cracking Check (cl . 3.12 .11 .2 BS 81 10 )

The excessive crack of the structures may inf luence the esthetic valueand durabi l i ty of reinforced concrete structures.

There are two methods are app l icab le to ensure the crack wid th notexceed the l im it, which are;

1) Calculat ion of actual crack width

The wid ths of f lexu ra l c ra ck s a t par t ic ul ar poi nt on the surf aceof a m ember can be est imat ed using the method given in cl .3 .8 .3 BS 8110 : Pa rt 2 : 19 85 .

No rm al s t ru ct ures crac k wi d th s ho uld be < 0 .3 m m

Wa ter ret ain in g st ru ct ur e < 0 .2 m m

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5.6 Checking of Cracking2) Limitat ion of maximum spacing of bars

The req ui rem en t t o l im it the m axim um sp aci ng o re in forcem en tis to min imi ze surface cracking.

The l im itat ion is gi ven in c l. 3 .1 2 .1 1 .2 BS 8 1 1 0 :Par t 1 : 1 9 9 7and required as follows;

The clear hor izontal spacing between bar, S1 should be notgreater than t he value given in Table 3 .2 8 dep ending onthe amount of redistr ibut ion and the characterist ic strength

of reinforcement . The distance between edge face of the beam and the

nearest longitud inal bar in tension, S2 should not begreater than half clear d istance given in Table 3 .28 .

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S1 = b 2(clear cover) 2(link) n(bar)

n = number of bars

S2 = y2 + y2 -bar/ 2

y = clear cover + link + (bar)/ 2

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5.6 Checking of Cracking I f the overa l l depth o f beam exceed ing 750 mm , the control crack ing

longi tudinal bars should be provided and distr ibuted at spacing notexceeding 25 0 m m over a distance of two-thirds (2/ 3h) of the beamoveral l depth (cl . 3.1 2.5 .4).

bar = Sb.b / fy

Where;

Sb = bar spacing

b = breadth of the sectionNot exceeding 500 mm

RC BEAM DESIGN

Control

crack bar

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Examp le 5 .3 : Check ing o f Cracking

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5.7 Design for Torsion Torsion re fers to the twist ing o f s t ructura l member when i t i s loaded by

couples that produce rotat ion about i t s longi tudinal axis.

Torsional moments produce shear stresses which resul t in pr incipaltensile stresses inclined at a pproxima tely 4 5 o to t he longi tudinal axisof the mem ber. The cracks wil l form a spiral around t he mem ber asshown in figure below.

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5.7 Design for Torsion

Torsion design for rein forced concre te beams is based on BS811 0-

2:1 98 5 clause 2.4. In most of the normal slab-and-beam or framedconstructio n, torsion calculations are not usually necessary. Normall y,the torsional cracking is cont rolled by shear reinforcement. However, i fthe main effect is due t o torsion, then i t must be considered.

The torsiona l shear st ress equat ion , vtof any reinforced concrete

rectangular beam is derived from sand heap analogy and stated inequa t ion 2 , BS 811 0 : 2 :198 5 ;

Where;

T= Torsional mom ent due to ul t imate load

hmin = smaller dimension of rectangular sectionhmax = larger dimension of rectangular section

RC BEAM DESIGN

3/(

2

minmaxmin

2 hhh

Tv

t

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5.7 Design for Torsion For T, L or I sect ions, i t should be divided to their component rectangles

which can p rovide the highest t orsional stiffn ess. This wil l generally beachieved if the widest rectangle is mad e as long as possible.

The to rs ional shea r s t ress , vtcarried by each of these component

rectangles may be calculated b y treating them as rectangular sectionssubjected to a torsional mom ent of :

For examp le a T section shown in f igure below is subjected to atorsional mo ment of T. Determine the to rsional shear stresses.

RC BEAM DESIGN

)( max

min3

min3

hh

hhT xma

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5.7 Design for TorsionRC BEAM DESIGN

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5.7 Design for TorsionTorsion reinforcemen t

Acco rd ing to BS8110 -2 :1985 c lause 2 .4 .6 , i f the to rs ional shea rstress, vtgreater than vt min in Table 2 .3, torsion reinforcement should

be provided. Recomm endations for re inforcement for combinat ions ofshear and torsion are given in Table 2 .4.

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The torsion reinforcements (vert ical or hor izontal re inforcement) are

addit ional to any requirements for shear and bending and should b esuch that:

x1 = sm al ler d imension of the l ink

y1 = larger dimension of the l ink

Asv = area of two legs of the l ink

fyv = character ist ic of the l inkSv = longi tudinal spacing of t he l inks.

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5.7 Design for Torsion The spacing o f l ink, sv should not exceed the least o f x1 , y1/ 2 or 2 0 0

mm . The l inks should be a closed shaped.

Longi tud inal to rs ion rein forcement shou ld be d is t r ibu ted even ly roundthe inside perim eter of the l inks. The clear distance between these barsshould not exceed 30 0 m m and at least four bars, one in each cornerof the l inks, should be used.

Add i t iona l long itud ina l re inforcement requ i red a t the leve l o f thetension or compression reinforcement m ay be provided by using larger

bars than those required for bending alon e. The torsion rein forcement shou ld extend a d is tance a t least equa l tothe largest dimension of t he section beyond where it theoreticallyceases to be required.

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BS81 10 : Pa rt 2 , c lause 2 .4 .10 , s tates tha t the l i nk cages should

inter lock in T- and L-sections and t ie t he component rectanglestogether as shown in figure below. If t he torsional shear stress in ami nor component recta ngle does not exceed vt min then no t orsional

shear reinforcement need b e provided in that rectangle.

RC BEAM DESIGN

(a) Closed link; (b) torque reinforcement for a T-beam.

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