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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
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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;
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)'(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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5 .4 Design f or Shear
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:
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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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5 .4 Design f or Shear
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:
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vy
v
vs
fbSA
95.04.0
vy
v
vs
f
bSA
95.0
4.0
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5 .4 Design f or Shear
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.
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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
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Single Leg Link Double Leg Link
Area of steel Area of steel
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5 .4 Design f or Shear
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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5.5 Checking of Def lect ion
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
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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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5.5 Checking of Def lect ion
M FTR
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5.5 Checking of Def lect ion M FCR
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5.5 Checking of Def lect ion
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 ).
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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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5.6 Checking of Cracking
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
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Control
crack bar
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5.6 Checking of Cracking
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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5.7 Design for Torsion
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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5.7 Design for Torsion
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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