4 Beam Design

4. Design of Reinforced Concrete Beam

• This chapter will discuss the following topics:

• Outline of the procedures for member sizing.

• Design procedures for bending, deflection and shear.

D i f i l t d b d ti• Design of simply supported beams and a continuous beam using moment and shear force coefficients.

• Special requirements for link when there is compressionSpecial requirements for link when there is compression steel.

• Definition of T and L beam and the determination of effective flanged width.

• Design of T and L beam for the case where the stress bl k li ithi th flblock lies within the flange.

• Curtailment of bars by calculation and simplified rules.

RC Design and Construction – HKC 2004 (2nd Edition)4-1

Preliminary Analysis and Member Sizing

– The preliminary analysis requires the values of max.moments and max. shears in order to estimatereasonable dimensions. Beam dimensions requiredare:-

i l C i f• Nominal Cover to reinforcement

• Breadth (b)

Eff ti d th (d)• Effective depth (d)

• Overall depth (h)

– The strength of a beam is affected more by its depththan its breadth A suitable breadth may be 1/3 1/2 ofthan its breadth. A suitable breadth may be 1/3 ∼ 1/2 ofthe depth.



• The dimensions for ‘b’ and ‘d’ can be obtained bya few trial calculations as follows:-a few trial calculations as follows:– Without compression reinforcement,

M/bd2f ≤ 0 156 (for f ≤ 45 MPa)M/bd2fcu ≤ 0.156 (for fcu ≤ 45 MPa)

– With compression reinforcement,

/bd2f 10/f if h f b di i fM/bd2fcu ≤ 10/fcu if the area of bending reinf

is not to be excessive.

– Shear stress v = V/bd and v > or 7 N/mm20 8. fcuwhichever is the lesser. To avoid congested shearreinf., v shall be in about 1/2 of the max. allowablevalue


value.

Preliminary Analysis and Member Sizing• Deflection (Span-Effective Depth Ratio Approach)

– Unlike structural steel members, deflection of R.C. beamsis normally checked by comparing the allowable span-effective depth ratio with the actual span-effective depthratioratio.

Th ll bl ff ti d th ti– The allowable span effective depth ratio

= (Basic span-effective depth ratio – see Table 7.3)

* ( difi ti f t f t i i f T bl 7 4)* (modification factor of tension reinf. - see Table 7.4)

* (modification factor of comp. reinf. - see Table 7.5)

– The actual span-effective depth ratio

( ff i l)/( ff i d h d)


= (Effective span l)/(Effective depth d)


– The basic span-effective depth ratio for span ≤ 10m should be as given below.

• Cantilever Beam 7 } Basic span-} p

• Simply Supported Beam 20 } effective

• Continuous Beam 26 } depth ratio.

• End Span 23 }

Refer to Table 7.3 of the Code

– For span greater than 10m, the basic ratios shall be multiplied by 10/spanmultiplied by 10/span.



If the allowable span-effective depth ratio

≥ the actual span-effective depth ratio,≥ p p ,

⇒ Deflection O.K.

If the allowable span-effective depth ratio< the actual span-effective depth ratio,⇒ Deflection NOT O.K.



• Overall depth– The overall depth of the beam is given byp g y

h = d + nominal cover + twhere t = estimated distance from the outside ofthe link to the centre of the tension barthe link to the centre of the tension bar.

b

d

h

Nominal Cover

t


Effective Span of Beam

The effective span, l of a member should be calculated as follows:follows:

l = ln + a1 + a2

where:• ln is the clear distance between the faces of the

supports,• a1 and a2may be determined from the appropriate ai

in figure 5.3• S is the width of the supporting element• Sw is the width of the supporting element

Continuous slabs and beams may generally be y g yanalysed on the assumption that the supports provide no rotational restraint.


Figure 5.3 - Effective span (l) for different support conditions

R d f HK C d


Reproduce from HK Code

Design for Bending Reinforcement

• An excessive amount of reinforcement indicatesthat a member is undersized (too small) and it maythat a member is undersized (too small) and it mayalso cause difficulty in fixing the bars and pouringof concrete.of concrete.

For rectangular beam– For rectangular beam

Code stipulates 100As/bh ≤ 4.0%

and 100A /bh ≥ 0 24% (for R)and 100As/bh ≥ 0.24% (for R)

≥ 0.13% (for T)

( f t T bl 9 1 d l 9 2 1 3 f th C d )– (refer to Table 9.1 and cl.9..2.1.3 of the Code)


Singly Reinforced Rectangular Section for fcu ≤ 45 MPa ( βb ≥ 0.9)

0 67 fcu

9x

εccb =0.0035

S/2

0.67 fcuγm = 0.45fcu

dNeutralAxis

S=0.

9

xFcc

S/2

As

Axis

ε Fst

z = la*d

εst

STRAINS STRESS BLOCKSECTION STRAINS STRESS BLOCK

Fig.4.1 Singly reinforced section


Fig.4.1 Singly reinforced section

Design Procedures for Bending Reinforcement

• The design procedures are summarised below:-– Calculate K = M/bd2fCalculate K M/bd fcu

– Determine the lever-arm, z, from ‘design table 1’ or byusing equation.g q

Design Table 1:

K M/bd2f ≤ 0 043 0 05 0 06 0 07 0 08 0 09 0 10 0 11 0 12 0 13 0 14 0 15 ≥ 0 156K= M/bd2fcu≤ 0.043 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 ≥ 0.156

la = z/d 0.950 0.941 0.928 0.915 0.901 0.887 0.873 0.857 0.842 0.825 0.807 0.789 0.775

Equation( )[ ]z d K= + −0 5 0 25 0 9. . / . Equation

– Pls note the upper and lower limit for z. (z ≤ 0.95d & z

( )[ ]


≥0.775d)

Design Procedures for Bending Reinforcement

– The area of tension steel is:-M

A =

– Select suitable bar sizes. (Available bar size Φ = 40,

zfA

ys ⋅=

87.0

Select suitable bar sizes. (Available bar size Φ 40,32, 25, 20 , 16, 12 & 10 mm) In practice, Φ = 12 & 10mm would NOT be used in beam as mainreinforcement.

– Check that the area of steel provided is within the limitsCheck that the area of steel provided is within the limitsrequired by the code, i.e.

• 100As/bh ≤ 4.0%

• and 100As/bh ≥ 0.24% (for R)

≥ 0 13% (f T)


• ≥ 0.13% (for T)

Example 4.1 - Design of tension reinforcement for a rectangular section.

The beam section shown in Fig. 4.2 is subject to a designsagging moment (i.e. at the ultimate limit state) of 170 kNm.Design the main reinforcement for the beam. Given that thecharacteristic material strength fcu = 35 N/mm2 and fy = 460N/ 2 b 250N/mm2. b=250

= 5

50

d =

490

h

As = 2-T25


Fig. 4.2

Solution: Ex. 4.1

081.035490250

101702

6

2===

x

fbd

MK

This is less than 0.156, therefore compression steel is NOT

35490250 22 xxfbd cu

s s ess t a 0. 56, t e e o e co p ess o stee s NOrequired.

From ‘design table 1’, la = 0.90g a

∴ lever arm z = la*d = 0.90*490 = 441 mma

26

96410170

mmxM

A === 96444146087.087.0

mmxxzf

Ay

s


Solution: Ex. 4.1

Provide TWO nos. T25 bars, area = 982 mm2.

(see design table 2)( g )

100 100 982

250 5500 71

A

bh

x

xs = = .

∴ 0.13 < < 4.0

250 550bh x

100As∴ 0.13 < < 4.0

∴ steel percentage is within the limits specified by the code

bh

∴ steel percentage is within the limits specified by the code.


Example 4.2 - Span-Effective Depth Ratio Check

A rectangular continuous beam 11.5m span (interior span)with a mid-span ultimate moment of 380 kNm. The breadth(b) and overall depth (h) are 300 mm and 650 mmrespectively. Check deflection by using span-effective ratio.Gi th tGiven that:

(i) f 460 N/ 2(i) fy = 460 N/mm2

(ii) Nominal cover = 30 mm

(iii) Li k i 10(iii) Link size = 10 mm

(iv) The size of main bar assumed to be 20 mm

( ) b l d h h(v) Two T16 bars are located within the compression zone.


Solution: Ex. 4.2 (Modification Factor of Tension Steel)

Basic span-effective depth ratio (design table 7.3) = 26

As span > 10m,p ,

∴ adjusted basic l/d ratio = 26 x 10/11.5 = 22.6

d 650 30 10 20/2 600d = 650 - 30 - 10 - 20/2 = 600 mm

Tension reinforcement modification factor (design table 7.4):

M

bd

x

x2

6

2

380 10

300 6003 52= = .

From design table 7.4, for fy = 460 N/mm2 (assume As,req =As prov , fs = 307 N/mm2, ), modification factor = 0.87

bd x300 600

s,prov , s , ),


Solution: Ex. 4.2 (Modification Factor of Compression Steel)

Compression reinf. modification factor (design table 7.5):100 100 402

0 22A xs

'

From design table 7.5, modification factor = 1.07

300 6000 22

bd xs .= =

o des g tab e 7.5, od cat o acto .07

Hence, the allowable span-effective depth ratioHence, the allowable span effective depth ratio

= 22.6 x 0.87 x 1.07 = 21.0

Actual span-effective depth ratio = 11.5x103/600 = 19.2

< allowable span-effective depth ratio = 21 0< allowable span-effective depth ratio 21.0

∴ Deflection O.K.


Doubly Reinforced Concrete Beamfor f ≤ 45 MPa ( β ≥ 0 9)for fcu ≤ 45 MPa ( βb ≥ 0.9)

• Compression steel is required whenever the concretein compression is unable to develop the necessarymoment of resistance (i.e. K > 0.156).

d' 0.9x

b 0.0035 0.45fcu

FscAs'

d

d

NeutralAxis

S=0

x = d/2Fsc

Fcc

z

As εst

εsc

Fst

SECTION STRAINS STRESS BLOCK

RC Design and Construction – HKC 2004 (2nd Edition)4-20Fig.4.3 Section with compression reinforcemen

Doubly Reinforced Concrete Beamfor fcu ≤ 45 MPa ( βb ≥ 0.9)for fcu ≤ 45 MPa ( βb ≥ 0.9)

Moment Redistribution Factor βb ≥ 0.9 and d’/x ≤ 0.43

• Compression reinforcement is required if,

K = M/(fcubd2) > 0.156( cu )

• Area of compression steel

1560 2bdfM

( ) '87.0

156.0'

2

ddf

bdfMA

y

cus −

−=

• Area of tension steel,

1560 2bdfand z = 0.775d '

87.0

156.0s

y

cus A

zf

bdfA +=


Doubly Reinforced Concrete Beamfor f ≤ 45 MPa ( βb ≥ 0 9)

• If > 0 43 the stress in compression steel( )d x'/

for fcu ≤ 45 MPa ( βb ≥ 0.9)

• If > 0.43, the stress in compression steelshould be determined by the stress-strainrelationship e g f = E * ε

( )d x'/

relationship, e.g. fsc = Es * εsc.


Containment of Compression Reinforcement:Link requirements ( Cl 9 2 1 10 )Link requirements ( Cl. 9.2.1.10 )

Links should be provided according to the following rules.• The links should pass round the corner bars and each alternate• The links should pass round the corner bars and each alternate

bar.

• The link size ≥ 1/4 of the size of largest compression bar orThe link size ≥ 1/4 of the size of largest compression bar or6mm whichever is greater. (Compression steel).

• Transverse spacing should not exceed 12 times the diameter ofTransverse spacing should not exceed 12 times the diameter ofthe smallest longitudinal bar. (Compression steel).

• The spacing of links ≤ 0.75dp g

• No longitudinal bar is more than 150mm from a restrained bar.

• At right-angles to the span, the horizontal spacing should besuch that no longitudinal tension bar is more than 150 mmfrom a vertical leg; this spacing should in any case not exceed


g p g yd.

Link requirements ( 9.2.2 )

CornerBar

CornerBaralt. bar alt. bar

alt.bar

alt.bar

Compression Steel

CornerCorner alt. baralt. bar alt.b

alt.b

TensionSteel

BarBar bar bar

Arrangement of Links when there are Compression Steel Bars


Example 4.3 - Design of Tension and Compression Reinforcement, βb>0.9

The beam section shown in Fig. 4.4 is subject to an ultimatehogging moment of 190 kNm. Design the main reinforcementfor the beam. Given that the characteristic material strengthof fcu = 35 N/mm2 and fy = 460 N/mm2.

b = 250

As=

340

d' =

50

h =

400

As'

d = d h

Fig. 4.4 Beam doubly reinforced to resist a hogging moment


Solution: Ex. 4.3 (Area of Compression Reinforcement)

> 0.156M x2

6

2

190 100 188= = .

∴ compression steel required, and = 50/170 = 0.30

bd f x xcu2 2250 340 35

( )d x'/∴ compression steel required, and 50/170 0.30

< 0.43

∴ f = 0.87f

( )d x/

∴ fsc 0.87fy

Compression steel( )

( ) 156.0

'2

bdfMA cu

s−

=Compression steel

=

( ) '87.0 ddf ys −

( )34025035156.010190 26 − xxxx

= 278 mm2

( ) 5034046087.0 −xx


= 278 mm

Solution: Ex. 4.3 (Area of Tension Reinforcement)

Tension steel '870

156.0 2

scu

s Af

bdfA +=

=

87.0 sy

s zf

27834025035156.0 2

+xxx

=1775 mm2

340775.046087.0 xxx

1775 mm

Provide TWO T16 bars as compression reinf (A ’= 402 mm2)Provide TWO T16 bars as compression reinf. (As 402 mm )and 2T32 + 1T16 bars as tension reinf. (As = 1809 mm2)


Solution: Ex. 4.3 (Check Steel % and Link Requirements)

> 0.2100 100 4020 40

A xs '= =

< 4.0 ∴ O.K.250 4000 40

bh x.= =

> 0.13

< 4.0 ∴ O.K.

100 100 1809

250 4001 81

A

bh

x

xs = = .

4.0 ∴ O.K.


4.4 T-Beam and L-Beam

• Fig. 4.5 and Fig. 4.6 show the arrangement of T-beams and L-beams. When the beams resistsagging moment, part of the slab will be incompression and it acts as a compression flange ofh b Th b b d i d Tthe beam. The members may be designed as a T-

beam or L-beam.

• When the beams resist hogging moment, the slabwill be in tension and assumed to be cracked (i.e.no contribution of the flexural strength of thebeam), ∴ the beam must be designed as a

l i i h id h b d llrectangular section with a width bw and an overalldepth h.


Fig. 4.5 - Arrangement of T-Beams and L-Beams

amSp

an o

f B

ea

Bea

m

Bea

m

Bea

m

Bea

m

Bea

mS1 S1 S1 S1 PLAN

S

L-beamL-beam T-beam T-beam T-beam

hA B C D EF G H I f

span of slab span of slab span of slab span of slab

SECTION


Fig. 4.6 - T-Beam and L-Beam

bf bfbf bf

hf

d h

bw bw

Fig. 4.6 T-beam and L-beam


Effective flange width parameter



lpi : Distance between points of zero moment



4.4 T-Beam and L-Beam

• When the stress block falls within the flange, designthe T-beam or L-beam as an equivalent rectangularq gsection of breath bf *h.

bf bf

hf

d h

bw bw

• Transverse reinforcement should be placed across the

Equivalent Rectangular Section of Breath bf

ptop flange with an area of not less than 0.15% of thelongitudinal cross-section of the flange.


Design Procedures for T or L Beam

• Determine the effective flange width bf

• Calculate M/bfd2fcu and determine the lever-arm z from designt bl 1table 1.

• If d - z ≤ hf/2, the stress block lies within the flange depth, andh d i i d d f l i i hthe design is proceeded as for a rectangular section with a

section of bf * h.

If d h /2 h bl k li b l h l b h l• If d - z > hf/2, the stress block lies below the slab, then consultthe clause 6.1.2.4(d) of HKC2004 for the design procedures. Itwill not be taught here.

• In either case, provide transverse steel (0.15% of thelongitudinal cross-section area) in the top of the flange.g ) p g

Area = 0.15hf x 1000/100= 1 5hf mm2 per meter length of the


1.5hf mm per meter length of theslab.

4.5 Curtailment (cutting) of Bars

• The Code states that in every flexural member every barshould be extended beyond the theoretical cut-off point(TCP) (theoretically the bar is no longer required at thispoint where the design resistance moment of the section

l t th d i t) f di t l tequals to the design moment) for a distance equal togreater of:-

– (i) The effective depth of the member or

– (ii) 12φ

• This applies to compression and tension reinforcement, butreinf in the tension zone should satisfy the followingreinf. in the tension zone should satisfy the followingadditional requirements.


4.5 Curtailment of Bars

– (iii) The bars extend a full anchorage bond lengthKA*Φ (as discuss in chapter 3) beyond the theoreticalcut-off point.

– (iv) At the physical cut-off point (PCP - the bar isactually cut at this point), the shear capacity of theb t PCP i t l t t i th t l h fbeam at PCP is at least twice the actual shear force.

( ) A PCP h l b di PCP i– (v) At PCP, the actual bending moment at PCP is notmore than half the moment at TCP.


4.5 Curtailment of Bars (Example)

MR of continuing bars, MAMR of continuing bars, MA B

Bending Moment Diagram

• For condition (i) and (ii), AB is the greater of d or 12φ.

g g

• For condition (iii), AB equals the full anchorage bond

l hlength.

F diti (i ) At B he it ≥ 2*(the he f e)• For condition (iv), At B, shear capacity ≥ 2*(the shear force)

• For condition (v) At B moment < half moment at A


• For condition (v), At B, moment < half moment at A.

4.5 Curtailment of Bars (Simplified Rules)

• In design office, the simplified rules for curtailmentsare normally used for conventional (typical) R.C.y ( yp )structures. But for special structural members,detailed checking as discussed before will be used.

• Where the loads on a beam are substantially uniformlyy ydistributed, simplified rules for curtailment may beused. These rules only apply to continuous beam ifthe characteristic imposed load does not exceed thecharacteristic dead load and the spans are equalFig.4.7 shows the rules in a diagrammatic form.


Fig. 4.7 Simplified rules for curtailment of bars in beams

100 %50 % 50 %

0 08L 0.08L0.08L 0.08LL

Simply Supported Beam

c =0.15L

c = 0.25L

c < 45 Φ

100 %30 % 30 %20 % 60% 100%

L

0.1L 0.15L

Continuous Beam


Continuous Beam

4.6 Design for Shear ( for fcu ≤ 40 MPa )

• If V is the shear force at a section, then the shearstress v is given by v = V/bd.stress v is given by v V/bd.

t d th l f• v must never exceed the lesser of

(i) orcuf8.0

(ii) 7 N/mm2.

• For fcu < 40 N/mm2, will govern thedesign

cuf8.0design.


4.6 Design for Shear (Shear Links)

• Rules:– ( Min. spacing sv of links should not be less than 80 mm.)v

– Max. spacing sv of links should not exceed 0.75dlongitudinally along the span.longitudinally along the span.

– At right angles to the span, the horizontal spacing should besuch that no longitudinal tension bars is more than 150 mmsuch that no longitudinal tension bars is more than 150 mmfrom a vertical leg, this spacing should in any case notexceed d.

– Nowadays high-tensile steel (T) is often used for shear linksbecause of its high strength per unit cost But for narrowbecause of its high strength per unit cost. But for narrowmembers, mild steel (R) is normally used because it may bebent to a smaller radius than high-yield steel (T). Thisallo s correct positioning of the tension reinforcement


allows correct positioning of the tension reinforcement.

4.6 Design for Shear (Shear Links)

• The size (Φ) and spacing sv of the shear links isgiven by the following equation.given by the following equation.

( )cvsv vvbA −≥

yvv fs 87.0≥

where Asv= Area of the all legs of a link

sv = Spacing of the link

bv = Breadth of the beam

v = V/bd = shear stress

vc = Ultimate shear resistance of conc.

fyv= Characteristic strength of the link reinf.


4.6 Design for Shear (Minimum / Nominal Links)

• If v is less than vc, minimum (nominal) links must still beprovided unless (i) the beam is a very minor one and (ii)v < vc/2 .

• The minimum links should be provided for fcu ≤ 40 MPaaccording to:according to:

sv

f

bA

870

4.0≥

Th h i t f th t l th i i

yvv fs 87.0

• The shear resistance of the concrete plus the minimumlinks for fcu ≤ 40 MPa

V ≈ (0 4 + v )bd


Vn ≈ (0.4 + vc)bd.

4.6 Design for Shear (Minimum / Nominal Links)

• If the design shear force V > Vn, then shear links should be provided.

• If the design shear force V ≤ Vn, then minimum links n

would be adequate. At this section, the shear links necessary to resist shear can be stopped and replaced by th i i li kthe minimum links.


Example 4.4

A simply supported beam as shown in Fig. 4.8 is subject to adesign loading w = 65 kN/m. Design the shear reinforcementfor the beam. Given that the characteristic strength of themild steel links is fyv = 250 N/mm2 and concrete is fcu = 35N/ 2N/mm2.


Example 4.4 - Fig. 4.8

4T252T25 2T252R10

5-R10-links @ 225 c/c 5-R10-links @ 225 c/cR10 links @ 300 c/c300

6-R10-links @ 200c/c R10 links @ 275c/c 6-R10-links @ 200c/c

4T252T25 2T25

300

550

560 5607000

227.5 kNVn = 163 kN

Vs = 218 kN

Vs = 218 kN

Vd = 182 kN

Shear Force Diagram 227.5 kN

Vn = 163 kNVd = 182 kN


Fig. 4.8 Beam - Shear Reinforcement

Solution: Ex. 4.4 (Check max. shear stress)

• Check max. shear stress

Total load on span, F = w * span = 65 *7

= 455 kN455 kN

At face of support

Shear V = F/2 w * support width/2Shear Vs = F/2 - w support width/2

= 455/2 - 65 * 0.15 = 218 kN

Shear stress, vs = Vs/bd

= 218*103 / (300*550)= 218*103 / (300*550)

1 32 N/ 2 < 0 8 f


= 1.32 N/mm2 < 0.8 fcu

Solution: Ex. 4.4 (Design shear stress)

• Shear links

Distance 1d from face of support

Shear Vd = Vs - w*d

218 65*0 55 182 kN= 218 - 65*0.55 = 182 kN

Sh St 182*103 / (300*550)Shear Stress vd = 182*103 / (300*550)

= 1.10 N/mm2.


Solution: Ex. 4.4 (Design shear links)

Shear LinksOnly two 25 mm bars extend a distance d past the criticalOnly two 25 mm bars extend a distance d past the criticalsection. Therefore for determining vc

982100100 xA

F d i t bl 6 3 0 53 (35/25)1/3 0 59 N/ 2

60.0550300

982100100==

x

x

bd

As

From design table 6.3, vc = 0.53 (35/25)1/3 = 0.59 N/mm2

( ) ( )7030

59.010.1300 −− vvbA csv ( ) ( )703.0

25087.087.0===

xfs yvv

From design table 4, provide R10-links.-200 c/c.

(Asv /sv = 0.785)


Solution: Ex. 4.4 (Design Minimum Links)

• Minimum linksFor mild steel linksFor mild steel links

551.0250870

3004.0

870

4.0===

x

f

bAsv

From design table 4 provide R10 links 275 c/c

25087.087.0 xfs yvv

From design table 4, provide R10-links-275 c/c.

(Asv /sv = 0.571)


Solution: Ex. 4.4 (Extent of Shear Links)

Shear resistance of nominal links + conc. is( )V v bdn c≈ +0 4.

= (0.4 + 0.59) 300*550 = 163 kN

( )n c

Shear reinf. is required over a distance s given byV V− −218 163

sV V

wms n= = =

218 163

650 85.

No. of R10 links at 200 c/c required at each end of the beamis:is:

1 + (s/200) = 1 + (850/200) = 6


Example 4.5 - Design of a simply supported beam

A simply supported R.C. beam has an effective span of 8mand size of 300 * 500 mm deep (overall depth h). The beamis subject to characteristic dead load including an allowancefor self-weight of 20 kN/m, and characteristic imposed load of15 kN/ D i th i d h i f t d15 kN/m. Design the main and shear reinforcement, andcheck deflection of the beam by using span-effective depthratio approach. Given that:ratio approach. Given that:

(i) fcu = 35 N/mm2, fy = 460 N/mm2, fyv = 250 N/mm2;

(ii) Effective depth = 450 mm;(ii) Effective depth 450 mm;

(iii) Inset to compression reinf. d’= 55mm;

(iv) Link size assumed to be 10 mm(iv) Link size assumed to be 10 mm.


Solution: Ex. 4.5 (Design load, shear force and bending moment)

Design load = (1.4*20) + (1.6*15) = 52 kN/mg ( ) ( )

Design moment = 52 * 82/8 = 416 kNm.es g o e t 5 8 /8 6 N .

Shear force at support = 52*8/2 = 208 kN.Shear force at support 52 8/2 208 kN.


Solution: Ex. 4.5 (Design Bending Reinforcement)

KM

bd f

x

x xcu

= = = >2

6

2

416 10

300 450 350 195 0 156. .

∴ compression steel req’d.p q

= 55/225 = 0.24 < 0.43, ∴ comp. steel yielded.

)1560( 2bdfK

( )d x'/

) '(87.0

)156.0( '

2

ddf

bdfKA

y

cus −

−=

( )( )

22

52555450460870

45030035156.0195.0mm

x

xx=

−−

=

Provide 2T25 Top (As’ = 982 mm2)

( )5545046087.0 x −


Solution: Ex. 4.5 (Design Bending Reinforcement)

( ) '775.087.0

156.0 2

sy

cus A

df

bdfA += ( )yf

22

290252545030035156.0

mmxxx

=+=

Provide 2T40 + 1T25 Bottom (As = 3005 mm2)

2902525450775.046087.0

mmxxx

=+=

Provide 2T40 1T25 Bottom (As 3005 mm )

3005*100100A13.000.2

500*300

3005*100100>==

bh

As

100 100 982

300 5000 65 0 2

A

bhs ' *

*. .= = > ∴ and < 4.0 O.K.


Solution: Ex. 4.5 (Design Shear Reinforcement)

Assume the width of support = 0 (Support Width Not Given)

The max. shear stress at the support is:pp

O.K.v N mm= =208 10

300 4501 54

32*

*. / < 0.8 fcu O. .

300 450* cu

The shear at d = 450 mm from the support is:

Vd = 208 - 0 45 * 52 = 184 6 kNVd 208 0.45 52 184.6 kN

v N mmd = =184 6 10

300 4501 37

32. *

*. /d 300 450*


Solution: Ex. 4.5 (Design Shear Reinforcement)

The area of steel at the section is 3005 mm2.

3005*100100A26.2

450*300

3005100100==

bd

As

From design table 6.3, vc = 0.92 N/mm2.vd > vc + 0.4 ∴ shear links required.

Shear Links:

( ) ( )( ) ( )620.0

250*87.0

92.037.1300

87.0=

−=

−=

yv

c

v

sv

f

vvb

s

A

From design table 4, provide R10-links-250 c/c(Asv/sv=0.628)


Solution: Ex. 4.5 (Extent of Shear Links)

Vn = (0.4 + vC) bd = (0.4 + 0.92)300*450= 178.2 kN.

s = (208 – 178.2)/52 = 0.57 m

Therefore the no. of links required is:1 + s/0.25 = 1 + 0.57/0.25 = 4

id 4 10 li k 2 0 / h dProvide 4 nos. R10-links-250 c/c at each end.

Minimum linksMinimum links

55.0250870

3004.0

870

4.0===

x

f

bAsv

From design table 4, provide R10-links.-275 c/c

25087.087.0 xfs yvv


(Asv/sv = 0.571)

Solution: Ex. 4.5 (Shear Links)

• As 2T25 compression steel bars are used, special requirements of links, in addition to shear, shouldrequirements of links, in addition to shear, should be observed.

– Link size ≥ 1/4*Φmax. = 1/4*25 = 6.25 mm.• Use R10, ∴ O KUse R10, ∴ O.K.

– Link spacing ≤ 12*Φmin = 12*25 = 300 mm.Link spacing ≤ 12 Φmin. 12 25 300 mm.• Use spacing of 250mm & 275mm ≤ 300 mm

∴ O.K.


Solution: Ex. 4.5 (Deflection Check)

Basic l/d ratio = 20

Modification factor for tension steel:

M 6416 10*M

bd 2 2

416 10

300 4506 85= =

*.

2

,

, /2963005*3

2902*460*21

3

2mmN

A

Aff

bprovs

reqsys ==⋅=

β

• Moment redistribution is not considered, therefore

, p

11=

bβ

From design table 7.4, M.F. for tension steel = 0.74


Solution: Ex. 4.5 (Deflection Check)

Modification factor for compression steel:

100 100 982A' *100 100 982

300 4500 73

A

bds prov *

*., = =

From design table 7.5, M.F. for compression steel = 1.20From design table 7.5, M.F. for compression steel 1.20

∴ the allowable l/d ratio = 20*0 74*1 20 = 17 8∴ the allowable l/d ratio 20 0.74 1.20 17.8

the actual l/d ratio = 8000 / 450 = 17 8the actual l/d ratio 8000 / 450 17.8

= allowable l/d ratio

∴ Deflection O K


∴ Deflection O.K.

Example 4.6 - Design of a Continuous Beam

A continuous beam has three equal spans of 5.5 m and issubject to characteristic imposed load qk of 45 kN/m andcharacteristic dead load gk, including self-weight of 70 kN/m.The continuous beam is 350*650 mm deep. In the traversedi ti th b d t 3 25 t ith 150direction, the beams are spaced at 3.25 m centres with a 150mm thick slab as shown in the Fig. 4.9 & Fig. 4.10. Designthe main and shear reinforcement for the continuous beam bythe main and shear reinforcement for the continuous beam byusing force coefficients of HKC2004.

Given that:-

(i) fcu = 35 N/mm2, fy = 460 N/mm2, fyv = 250 N/mm2

(ii) Nominal concrete cover = 40 mm.( )



Moment M =0 -0.11FL -0.11FL 0

Shear V = 0 45F 0 6F 0 55F 0 55F 0 6F 0 45F

0.09FL 0.07FL 0.09FL

L=5.5 m L=5 5 m L=5 5 m

0.45F 0.6F 0.55F 0.55F 0.6F 0.45F

L 5.5 m L=5.5 m L=5.5 m

F = 1.4 Gk + 1.6 Qk

Fig. 4.9 Continuous Beam with Ultimate B.M and Shear Force Co



3.25

m

Continuous Beam

3.25

m

5.5m 5.5m 5.5m

Fig. 4.10


Solution: Ex. 4.6 (Design load, bending moment & shear)

For each span

Ultimate load w = (1.4gk + 1.6qk) = (1.4*70 + 1.6*45)( gk qk) ( )

= 170 kN/m

Total ultimate load on a span is

F = 170 * 5.5 = 935 kNF 170 5.5 935 kN

As the loading is uniformly distributed qk < gk and the spansAs the loading is uniformly distributed, qk < gk, and the spansare equal, the coefficients shown in Fig. 4.9, which areextracted from Table 6.1 of the HKC2004, are used tocalculate the design moments and shears.

– Observe the requirements in the Code before using thecoefficients


coefficients.

Solution: Ex. 4.6 (Bending - 1st & 3rd spans)Design as a T-section for the span

Moment M = 0.09FL = 0.09*935*5.5 = 463 kNm

Actual flange width, b = 3.25 mb1 = b2 = ( b - bw ) / 2 = (3.25 – 0.35) / 2 = 1.45 m1 2 w

Distance between points of zero moment, lpi

= 0 85L = 0 85*5 5 = 4 675m= 0.85L = 0.85 5.5 = 4.675m

Consider beff,i = 0.2*1.45 + 0.1*4.675 = 0.7575 m < 0.2*4.6750 935= 0.935m

< 1.45 mHence beff = Σ beff i + bweff eff,i w

= 2*0.757 + 0.35 = 1.864 m

Effective depth d 650 40 12 32/2 582 mm


Effective depth, d = 650 - 40 - 12 - 32/2 = 582 mm

Solution: Ex. 4.6 (Bending - 1st & 3rd spans)

043.0021.035*582*1864

10*4632

6

2<==

cuf fdb

M

From design table 1, la = z/d = 0.95,

cuf f

∴ z = 0.95 * 582 = 553 mm

d - z = 582 - 553 = 29 mm < hf/2 = 150/2 = 75 mm

∴ stress block lies within the flange.

26

2092553*460*870

10*463

870mm

f

MAs ===

Provide 2T32 + 1T25 bottom bars (A = 2099 mm2)

553*460*87.087.0 zf ys

Provide 2T32 + 1T25 bottom bars (As 2099 mm )

0.18 > 92.0650*350

2099*100100==

bh

As


OK 4.0 <

Solution: Ex. 4.6 (Bending - Interior support)

Design as a Rectangular Section for the interior supportM = 0.11FL = 0.11*935*5.5 = 566 kNm (Hogging)

Effective depth d = 650 -40 -12-32-16 = 550 mm

1560153010*566 6

<==M

156.0153.035*550*350 22

<==cufbd

∴ No compression steel reqd.

From design table 1, Z = 0.783*550 = 431 mm


Solution: Ex. 4.6 (Bending - Interior support)

26

3281431*460*870

10*566

870mm

zf

MAs ===

Provide 3T32 - top 1 and 2T25 - top 2 (As = 3394 mm2)

431*460*87.087.0 zf y

ov de 3 3 top a d 5 top ( s 339 )

0.13> 49.13394100100

==xAs

< 4.0 ∴ O.K.650350xbh


Solution: Ex. 4.6 (Bending - 2nd span)

Design as a T-section for the span

M = 0.07FL = 0.07*935*5.5 = 360 kNm

Actual flange width b = 3 25 mActual flange width, b = 3.25 mb1 = b2 = ( b - bw ) / 2 = (3.25 – 0.35) / 2 = 1.45 m

Distance between points of zero moment, lpi

= 0.7L = 0.7*5.5 = 3.85m

Consider beff,i = 0.2*1.45 + 0.1*3.85 = 0.675 m < 0.2*3.85 = 0.77m< 1 45 m< 1.45 m

Hence beff = Σ beff,i + bw

= 2*0.675 + 0.35 = 1.7 m


Solution: Ex. 4.6 (Bending - 2nd span)

043.0018.035*582*1700

10*3602

6

2<==

cuf fdb

M

z = 0.95*582 = 553 mm

d 583 553 29 h / 2 150/2 75d - z = 583 - 553 = 29 < hf / 2 = 150/2 = 75 mm

∴ Stress block lies within the flange.6

26

1627553*460*87.0

10*360

87.0mm

zf

MA

ys ===

Provide 2T32 +1T16 Bottom (As = 1809 mm2) - Bottom

1809*100100A

OK4 0

0.18 > 80.0650*350

1809*100100

<

==bh

As


OK 4.0 <

Solution: Ex. 4.6 (Shear - max. shear stress)

Check the max. shear stress

Max. shear force at face of support(width = 300 mm) is:

Vs = 0.6F - w*support width/2Vs 0.6 w suppo t w dt /

= 0.6*935 - 170*0.15 = 534 kN

Shear stress,310*534V

cu2

3

f0.8 < /77.2550*350

10*534mmN

bd

Vv s

s ===


Solution: Ex. 4.6 (Shear - minimum links)

Minimum Links

350*4040 bA64.0

250*87.0

350*4.0

87.0

4.0===

yvv

sv

f

b

s

A

From design table 4, provide R10-links-225 c/c

(A /s = 0.698)(Asv/sv 0.698)


Solution: Ex. 4.6 (Shear - End support)

Shear force at a distance 1d from face of support is

Vd = 0.45F - w*(d + support width / 2)d ( pp )

= 0.45*935 - 170*(0.582 + 0.15) = 296.3 kN

Shear stress, 23

/45.1582*350

10*3.296mmN

bd

Vv d

d ===

03.1582*350

2099*100100==

bd

As

From design table 6 3 v = 0 71 N/mm2

582350bd

From design table 6.3, vc 0.71 N/mm ,

∴ vd > vc + 0.4


Solution: Ex. 4.6 (Shear - End support)

Provide shear links

( ) ( )191

71.045.1350 −− cdsv vvbA

From design table 4, provide R10-links-125 c/c.

( ) ( )19.1

250*87.0

71.045.1350

87.0===

yv

cd

v

sv

f

vvb

s

A

From design table 4, provide R10 links 125 c/c.(Asv / sv = 1.26)

Shear resistance of nominal links + conc. is

Vn ≈ (0.4 + vc)bd = (0.4 + 0.71)*350*582 = 226 kN.

Sh i f t th th th i i i i dShear reinforcement other than the minimum is required over

a distance, mdVV

s nd 00.1582.0170

2263.296=+

−=+

−=

∴ No. of shear links required = 1 + (1.00/0.125) = 9

w 170


q ( )

Solution: Ex. 4.6 (Shear - 1st & 3rd spans interior supports)

Distance d from face of support,

Vd = 0.6*935 - 170(0.55 + 0.15) = 442 kNd ( )

Shear stress, 23

/30210*442

mmNV

v d ===S ea st ess, /30.2550*350

mmNbd

vd ===

3394*100100A76.1

550*350

3394100100==

bd

As

From design table 6.3, vc = 0.85 N/mm2,

∴ vd > v + 0 4∴ vd > vc + 0.4


Solution: Ex. 4.6 (Shear - 1st & 3rd spans interior supports)

Provide shear links.( ) ( )

33285.03.2350 −− cdsv vvbA

From design table 4 provide R12 D S 175 c/c (4 Legs)

( ) ( )33.2

250*87.087.0===

yv

cd

v

sv

fs

From design table 4, provide R12-D.S.-175 c/c. (4 Legs)(Asv / sv = 2.586)

Shear resistance of nominal links + conc. isVn ≈ (0.4 + vc)bd = (0.4 + 0.85)*350*550 = 240.6 kN.

Shear reinforcement other than the nominal is required over a

distance dVVd 6.240442−−distance,

N f h li k i d 1 (1 73/0 175) 11

mdw

VVs nd 73.155.0

170

6.240442=+=+=


∴ No. of shear links required = 1 + (1.73/0.175) = 11

Solution: Ex. 4.6 (Shear - 2nd span)

Distance d from face of support,

Vd = 0.55*935 - 170(0.55 + 0.15) = 395 kNd ( )

Shear stress, 2

3

/05.2550*350

10*395mmN

bd

Vv d

d ===S ea st ess,550*350bdd

7613394*100100As 76.1550*350

3390000==

bds

From design table 6.3, vc = 0.85 N/mm2,

∴ vd > v + 0 4∴ vd > vc + 0.4


Solution: Ex. 4.6 (Shear - 2nd span)

Provide shear links.

( ) ( )931

85.005.2350 −− cdsv vvbA ( ) ( )93.1

250*87.0

85.005.350

87.0===

yv

cd

v

sv

f

vvb

s

From design table 4, provide R12-links-100 c/c.(Asv / sv = 2.262)

Shear resistance of nominal links + conc. is

Vn ≈ (0.4 + vc)bd = (0.4 + 0.85)*350*550 = 240.6 kN.

Shear reinforcement other than the minimum is required over

VV 6240395a distance,

∴ No of shear links required = 1 + (1 46/0 125) = 13

mdw

VVs nd 46.155.0

170

6.240395=+

−=+

−=


∴ No. of shear links required 1 + (1.46/0.125) 13

Solution: Ex. 4.6 (R.C. details of the beam)

9R10-S.S.-125 10R12-D.S.-200R10-S.S.-250R10-S.S.-225 12R12-D.S.-1759R10-S.S.-125 11R12-D.S-175

2T32 + 1T252T32 3T323T322T32

2T25 2T25

3002T32 + 1T202T32 + 1T202T32 + 1T25

L=5.5 m

R.C. Details for 1st Span


Summary of Design Procedures for R.C. Beamsfor fcu ≤ 40 MPa

• The design of R.C. Beam consists of three main gsteps:-

– Design for Bending Reinforcement at ULS.

– Design for Shear Reinforcement at ULS.

– Check deflection by using span-effective ratio.

• The design procedures of each step will be summarised here.


Summary of Design Procedures for Bending (βb ≥ 0.9, Rectangular Section)ec gu Sec o )

• Calculate K = M/(bd2fcu)– Case 1Case 1

• If K ≤ 0.156, ⇒ no compression steel required.

• Find z from the design table 1 or by formulag y

• Find As,zf

MA

ys 87.0=

• Select suitable bar size from design table 2.

• Check the steel percentage

y

Check the steel percentage.

0 13100

4 0 460 2. . /≤ ≤ =A

bhf N mms

y for bh



– Case 2• If K > 0.156, ⇒ compression steel required.• z = 0.775d

• Find As’, AK f bd

scu'

( . )=

− 0 156 2

– If d’/x ≤ 0.43, fsc = 0.87fy

If d’/ 0 43 f E *

f d dssc ( ' )−

– If d’/x > 0.43, fsc = Es*εsc

2

• Find As, '775.0*87.0

156.0 2

sy

cus A

df

bdfA +=

• Select suitable bar size from design table 2.



– Case 2 (cont’d)

• Check the steel percentage

– Tension Steel:

0 13100

4 0 460 2. . /≤ ≤ =A

bhf N mms

y for

– Compression Steel:

bh y

p

0 20100

4 0 460 2.'

. /≤ ≤ =A

bhf N mms

y for


Summary of Design Procedures for Deflection

• Select the suitable basic span-effective depth ratio from Table 7.3.– Note that these ratios apply to beams with spans ≤ 10m.

For span > 10m, modification to the basic ratio has to b dbe made.

• Determine the modification factor of tension steel.C l l t M/bd2– Calculate M/bd2

– Calculate fs, f fA

As ys req

s prov b= ⋅ ⋅ ⋅

2

3

1,

β• For βb ≥ 0.9, (1/ βb) is taken as 1.• For conservative design, one may assume fs =307 N/mm2

As prov b3 , β

- Determine the modification factor of tension steel from table 7.4.



• Determine the modification factor of compression steel.– Note that this factor may be ignored if the modification

factor of tension steel is large enough.

– Calculate 100As’/(bd)

– Determine the modification factor of compression steel from table 7.5.

D i h ll bl ff i d h i• Determine the allowable span-effective depth ratio.(allowable span-effective depth ratio)

= (basic span-effective depth ratio) * (modification factor of tension steel) * (modification factor of compression steel if applicable)


steel, if applicable)


• Determine the actual span-effective depth ratio(actual span-effective depth ratio)(actual span effective depth ratio)

= (effective span of beam)/(effective depth of beam)

= l/d l/d

• Compare the allowable and the actual span• Compare the allowable and the actual span-effective depth ratio.

If th t l ff ti d th ti ≤ th ll bl– If the actual span-effective depth ratio ≤ the allowable span-effective depth ratio, ⇒ deflection O.K.

– If the actual span-effective depth ratio > the allowable span-effective depth ratio ⇒ deflection NOT O K


span-effective depth ratio, ⇒ deflection NOT O.K.

Summary of Design Procedures for Shear

• Determine the design shear force V(kN).

• Determine the design shear stress v = V/bd• Determine the design shear stress, v = V/bd– Design shear stress must NOT exceeds the smaller of

Oth i h t i th b i

2N/mm 7or 8.0 cuf

– Otherwise one has to increase the member size.

i h h i f• Determine the shear resistance of concrete, vc

– Calculate 100As/bd

– Read vc from table 6.3.

– Multiply the value of table 6.3 by (fcu/25)1/3 if the 2


concrete grade is NOT 25 N/mm2.

Summary of Design Procedures for Shear

• If v > vc + 0.4, ⇒ use shear link.

bA )(

yv

c

v

sv

f

vvb

s

A

87.0

)( −≥

• Normally, if v ≤ vc + 0.4, ⇒ use minimum (nominal) link(nominal) link.

sv bA 4.0≥

yvv

sv

fs 87.0≥

• Select suitable link size and spacing from design table 4.


DESIGN FORMULAE (Based on HK Code 2004)

z dK

= + −⎡

⎣⎢

⎤

⎦⎥0 5 0 25

0 9. .

⎣ ⎦0 9.

Table 1:

K= M/bd2fcu≤ 0.043 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 ≥0.156

la = z/d 0.950 0.941 0.928 0.915 0.901 0.887 0.873 0.857 0.842 0.825 0.807 0.789 0.775


Table 2 Area of Steel Reinforcement

Steel Reinforcement

Bar Size Area of one Bar (mm2)

8 50.310 78 510 78.512 11316 20120 31420 31425 49132 80440 125740 1257


Table 4 - Asv/sv Ratio for Shear Link

A /s Ratio for Shear Link (2 legs)Asv/sv Ratio for Shear Link (2 legs)

Spacing of Links in (mm)Link Size (mm) 80 100 125 150 175 200 225 250 275 300 325

8 1.257 1.005 0.804 0.670 0.574 0.503 0.447 0.402 0.366 0.335 0.309

10 1.964 1.571 1.257 1.047 0.898 0.785 0.698 0.628 0.571 0.524 0.483

12 2 827 2 262 1 810 1 508 1 293 1 131 1 005 0 905 0 823 0 754 0 69612 2.827 2.262 1.810 1.508 1.293 1.131 1.005 0.905 0.823 0.754 0.696

16 5.027 4.021 3.217 2.681 2.298 2.011 1.787 1.608 1.462 1.340 1.237


Table 6.1 - Design Ultimate Bending Moments and Shear Forces for Beams



Table 6.2 – Form and area of Shear Reinforcement in Beams



Table 6.3 - Values of vc Design Concrete Shear Stress




Table 7.3 - Basic Span/Effective Depth Ratio




Table 7.4 - Modification Factor for Tension Reinforcement



Table 7.5 - Modification Factor for Compression Reinforcement

R d f HK C d



Table 9.1 - Minimum Percentage of Reinforcement

R d f HK C d



Minimum and Maximum Percentages of Reinforcement in Beams, Slabs and ColumnsReinforcement in Beams, Slabs and Columns

• Minimum % : Table 9.1 of the code• Maximum %

(a) Beam (9.2.1.3)Neither the area of tension reinforcement nor the area of

compression reinforcement should exceed 4% of the gross cross-sectional area of concrete.

(b) Column (9.5.1)( ) ( )The longitudinal reinforcement should not exceed the following

amounts, calculated as percentages of the gross cross-sectional area of the concrete :

~ vertically cast columns : 6%~ horizontally cast columns : 8%~ laps in the vertically or horizontally cast column : 10%~ laps in the vertically or horizontally cast column : 10%

(c) Wall (9.6.2)h f i l i f h ld d f h


The area of vertical reinforcement should not exceed 4% of the concrete cross-sectional area of the wall.

Self-Assessment Questions

Q1. What is the basic span-effective depth ratio of a continuous beam (end span) with a span of 12m?

Choices:

(a) 19.2

(b) 23

(c) 26( )

Q2. What is the basic span-effective depth ratio of a simply t d fl d b ith b /b 0 5?supported flanged beam with a bw/b = 0.5?

Choices:

( ) 16(a) 16

(b) 17.1

( ) 20


(c) 20


Q3. Determine the effective flange width of the end span of a continuous T-beam with effective spans of 8m. The width of the web is 400 mm and the spacing between the adjacent beams is 3.8m.

Choices:

(a) 2.20m

(b) 2.44m

(c) 3.8m



Q4. State the spacing requirements for links in beam.

AAnswer

Q5. In the design of shear reinforcement for beam, it is

normally to use Vd as the design shear force and not

Vs. Can you give an explanation for this?

Q6. What are the requirements for links when there is Q q

compression reinforcement?

Answer


Answer

Assignment No. 4

AQ1. A simply supported beam has an effective span of 8.8mand an effective section as shown in Fig. AQ1 Checkfor deflection of the beam by using span-effectivedepth ratio approach for the following cases.

(a) The beam is subjected to an ultimate saggingt f 320 kNmoment of 320 kNm.

(b) The beam is subjected to an ultimate saggingmoment of 320 kNm and the tension reinforcementmoment of 320 kNm and the tension reinforcementrequired is 1740 mm2.

(c) In addition to (b) two nos T16 bars are located in(c) In addition to (b), two nos. T16 bars are located inthe compression zone of the beam section.


Assignment No. 4

325 325

580

d =

514

580

d =

514

d

2T32 1T20

d

3T322T32 + 1T20 3T32

Figure AQ1 Figure AQ2


Assignment No. 4

AQ2 A continuous beam has effective spans of 10 m(interior span) and an effective section as shown inFig. AQ2. Check for deflection of the beam by usingspan-effective depth ratio approach for the followingcases.

( ) Th b i bj t d t lti t i(a) The beam is subjected to an ultimate saggingmoment of 370 kNm.

(b) The beam is subjected to an ultimate sagging(b) The beam is subjected to an ultimate saggingmoment of 370 kNm and the tension reinforcementrequired is 2175 mm2.q

(c) In addition to (b), two nos. T25 bars are located inthe compression zone of the beam section.


p

Assignment No. 4

AQ3 If a continuous beam has effective spans of 11m and issubjected to the conditions as stipulated in AQ2.Check for deflection of the beam by using span-effective depth ratio approach.


Assignment No. 4

AQ4 A simply supported flanged beam has an effectivespan of 8.6m and an effective section as shown in Fig.AQ4 Ch k f d fl ti f th b b iAQ4. Check for deflection of the beam by usingspan-effective depth ratio approach.

(a) Given that:(a) Given that:-(1) The design sagging moment is 390 kNm.(2) The area tension reinforcement required is 2250(2) The area tension reinforcement required is 2250mm2.(3) There are 3 nos. T16 bars located in compressionzone.

(b) Treat the flanged beam in (a) as a rectangular beam of( ) g ( ) g300 x 550 mm. Check for deflection of the beam byusing span-effective depth ratio approach.


Assignment No. 4

b = 1000

30

f

3T16

8

h =

13

f

d =

488

420

3T32

4

b = 300w

Figure AQ4


Assignment No. 4

AQ5. The simply supported beam as shown in Fig. AQ5 issubjected to uniformly distributed characteristic deadl d f 15 kN/ d h t i ti i d l d f 20load of 15 kN/m and characteristic imposed load of 20kN/m. Given that fcu = 30 N/mm2, fy = 460 N/mm2 andfyv = 250 N/mm2; nominal cover to reinforcement = 40yv ;mm; the width of the beam is 350 mm. Find:-(a) Reaction at the support.(b) Shear force at the face of the support (Vs)(c) Shear force at a distance of 1d from the face of

(V )support (Vd).(d) Shear resistance of concrete plus minimum links

(V )(Vn).(e) The spacing and number of links required.

(Use R10)


(Use R10)

Assignment No. 4

Centrelineof pier

Centrelineof pierCharacteristic Imposed Load = 20 kN/mp

Characteristic Dead Load = 15 kN/m

600

3T32

Supporting Pier Supporting Pier

8100200200 200 2008100200200 200 200

Fig AQ5


Assignment No. 4

AQ6 A simply supported beam has an effective span of6.5m and is subjected to a characteristic dead load of13.5 kN/m and a characteristic imposed load of 16.5kN/m. The clear span of the beam is 6m. Design the

i d h i f t f th b Ch k fmain and shear reinforcement for the beam. Check fordeflection of the beam by using span-effective depthratio approach.ratio approach.

Given that:-Given that:

(a) The overall size of the beam is 250 x 400 mm(deep).( p)

(b) The nominal cover = 30 mm.

(c) fcu = 35 N/mm2, fy = 460 N/mm2, fyv = 250 N/mm2.


(c) fcu 35 N/mm , fy 460 N/mm , fyv 250 N/mm .

Assignment No. 4

AQ7 A simply supported beam B1 has an effective span of6m as shown in Fig. AQ7. The overall size of theb i 250 400 Th b i i d t tbeam is 250 x 400. The beam is required to support a130 mm thick one-way spanning slab as shown.There is a 25mm thick finish on top of the slab.pDesign the beam B1 as a flanged beam. Determinethe main and shear reinforcement required. Check fordeflection of the beam by using span effective depthdeflection of the beam by using span-effective depthratio approach.

(a) The unit weight of the finishes is assumed to be24 kN/m3.(b) The imposed load on the slab is 5 kN/m2(b) The imposed load on the slab is 5 kN/m2.(c) The nominal cover is 35mm.(d) f = 35 N/mm2 f = 460 N/mm2 f = 250 N/mm2


(d) fcu 35 N/mm , fy 460 N/mm , fyv 250 N/mm

Assignment No. 4Beam (300 x 400)

00)

6000

1 (2

50 x

40

(130) (130)

B1

2900 2900

Beam (300 x 400)

Figure AQ7


Documents

4 Beam Design