4 Beam Design
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Transcript of 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.
RC Design and Construction – HKC 2004 (2nd Edition)4-2
Preliminary Analysis and Member Sizing
• 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
RC Design and Construction – HKC 2004 (2nd Edition)4-3
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)
RC Design and Construction – HKC 2004 (2nd Edition)4-4
= (Effective span l)/(Effective depth d)
Preliminary Analysis and Member Sizing
– 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.
RC Design and Construction – HKC 2004 (2nd Edition)4-5
Preliminary Analysis and Member Sizing
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-6
Preliminary Analysis and Member Sizing
• 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
RC Design and Construction – HKC 2004 (2nd Edition)4-7
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-8
Figure 5.3 - Effective span (l) for different support conditions
R d f HK C d
RC Design and Construction – HKC 2004 (2nd Edition)4-9
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)
RC Design and Construction – HKC 2004 (2nd Edition)4-10
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
RC Design and Construction – HKC 2004 (2nd Edition)4-11
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
( )[ ]
RC Design and Construction – HKC 2004 (2nd Edition)4-12
≥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)
RC Design and Construction – HKC 2004 (2nd Edition)4-13
• ≥ 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
RC Design and Construction – HKC 2004 (2nd Edition)4-14
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
RC Design and Construction – HKC 2004 (2nd Edition)4-15
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-16
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-17
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 , ),
RC Design and Construction – HKC 2004 (2nd Edition)4-18
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-19
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 +=
RC Design and Construction – HKC 2004 (2nd Edition)4-21
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-22
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
RC Design and Construction – HKC 2004 (2nd Edition)4-23
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
RC Design and Construction – HKC 2004 (2nd Edition)4-24
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
RC Design and Construction – HKC 2004 (2nd Edition)4-25
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
RC Design and Construction – HKC 2004 (2nd Edition)4-26
= 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)
RC Design and Construction – HKC 2004 (2nd Edition)4-27
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-28
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-29
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
RC Design and Construction – HKC 2004 (2nd Edition)4-30
Fig. 4.6 - T-Beam and L-Beam
bf bfbf bf
hf
d h
bw bw
Fig. 4.6 T-beam and L-beam
RC Design and Construction – HKC 2004 (2nd Edition)4-31
Effective flange width parameter
RC Design and Construction – HKC 2004 (2nd Edition)4-32
Reproduce from HK Code
lpi : Distance between points of zero moment
RC Design and Construction – HKC 2004 (2nd Edition)4-33
Reproduce from HK Code
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-34
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
RC Design and Construction – HKC 2004 (2nd Edition)4-35
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-36
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-37
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
RC Design and Construction – HKC 2004 (2nd Edition)4-38
• 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.
RC Design and Construction – HKC 2004 (2nd Edition)4-39
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
RC Design and Construction – HKC 2004 (2nd Edition)4-40
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-41
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
RC Design and Construction – HKC 2004 (2nd Edition)4-42
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-43
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
RC Design and Construction – HKC 2004 (2nd Edition)4-44
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-45
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-46
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
RC Design and Construction – HKC 2004 (2nd Edition)4-47
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
RC Design and Construction – HKC 2004 (2nd Edition)4-48
= 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.
RC Design and Construction – HKC 2004 (2nd Edition)4-49
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)
RC Design and Construction – HKC 2004 (2nd Edition)4-50
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)
RC Design and Construction – HKC 2004 (2nd Edition)4-51
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
RC Design and Construction – HKC 2004 (2nd Edition)4-52
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-53
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-54
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 −
RC Design and Construction – HKC 2004 (2nd Edition)4-55
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-56
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*
RC Design and Construction – HKC 2004 (2nd Edition)4-57
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)
RC Design and Construction – HKC 2004 (2nd Edition)4-58
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
RC Design and Construction – HKC 2004 (2nd Edition)4-59
(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.
RC Design and Construction – HKC 2004 (2nd Edition)4-60
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
RC Design and Construction – HKC 2004 (2nd Edition)4-61
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
RC Design and Construction – HKC 2004 (2nd Edition)4-62
∴ 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.( )
RC Design and Construction – HKC 2004 (2nd Edition)4-63
Example 4.6 - Fig. 4.9
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
RC Design and Construction – HKC 2004 (2nd Edition)4-64
Example 4.6 - Fig. 4.10
3.25
m
Continuous Beam
3.25
m
5.5m 5.5m 5.5m
Fig. 4.10
RC Design and Construction – HKC 2004 (2nd Edition)4-65
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
RC Design and Construction – HKC 2004 (2nd Edition)4-66
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
RC Design and Construction – HKC 2004 (2nd Edition)4-67
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
RC Design and Construction – HKC 2004 (2nd Edition)4-68
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
RC Design and Construction – HKC 2004 (2nd Edition)4-69
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
RC Design and Construction – HKC 2004 (2nd Edition)4-70
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
RC Design and Construction – HKC 2004 (2nd Edition)4-71
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
RC Design and Construction – HKC 2004 (2nd Edition)4-72
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 ===
RC Design and Construction – HKC 2004 (2nd Edition)4-73
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)
RC Design and Construction – HKC 2004 (2nd Edition)4-74
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
RC Design and Construction – HKC 2004 (2nd Edition)4-75
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
RC Design and Construction – HKC 2004 (2nd Edition)4-76
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
RC Design and Construction – HKC 2004 (2nd Edition)4-77
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=+=+=
RC Design and Construction – HKC 2004 (2nd Edition)4-78
∴ 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
RC Design and Construction – HKC 2004 (2nd Edition)4-79
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=+
−=+
−=
RC Design and Construction – HKC 2004 (2nd Edition)4-80
∴ 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
RC Design and Construction – HKC 2004 (2nd Edition)4-81
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-82
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
RC Design and Construction – HKC 2004 (2nd Edition)4-83
Summary of Design Procedures for Bending (βb ≥ 0.9, Rectangular Section)ec gu Sec o )
– 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.
RC Design and Construction – HKC 2004 (2nd Edition)4-84
Summary of Design Procedures for Bending (βb ≥ 0.9, Rectangular Section)ec gu Sec o )
– 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
RC Design and Construction – HKC 2004 (2nd Edition)4-85
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-86
Summary of Design Procedures for Deflection
• 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)
RC Design and Construction – HKC 2004 (2nd Edition)4-87
steel, if applicable)
Summary of Design Procedures for Deflection
• 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
RC Design and Construction – HKC 2004 (2nd Edition)4-88
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
RC Design and Construction – HKC 2004 (2nd Edition)4-89
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-90
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
RC Design and Construction – HKC 2004 (2nd Edition)4-91
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
RC Design and Construction – HKC 2004 (2nd Edition)4-92
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
RC Design and Construction – HKC 2004 (2nd Edition)4-93
Table 6.1 - Design Ultimate Bending Moments and Shear Forces for Beams
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Reproduce from HK Code
Table 6.2 – Form and area of Shear Reinforcement in Beams
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Reproduce from HK Code
Table 6.3 - Values of vc Design Concrete Shear Stress
Reproduce from HK Code
RC Design and Construction – HKC 2004 (2nd Edition)4-96
Reproduce from HK Code
Table 7.3 - Basic Span/Effective Depth Ratio
Reproduce from HK Code
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Reproduce from HK Code
Table 7.4 - Modification Factor for Tension Reinforcement
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Reproduce from HK Code
Table 7.5 - Modification Factor for Compression Reinforcement
R d f HK C d
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Reproduce from HK Code
Table 9.1 - Minimum Percentage of Reinforcement
R d f HK C d
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Reproduce from HK Code
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
RC Design and Construction – HKC 2004 (2nd Edition)4-101
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
RC Design and Construction – HKC 2004 (2nd Edition)4-102
(c) 20
Self-Assessment Questions
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
RC Design and Construction – HKC 2004 (2nd Edition)4-103
Self-Assessment Questions
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
RC Design and Construction – HKC 2004 (2nd Edition)4-104
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-105
Assignment No. 4
325 325
580
d =
514
580
d =
514
d
2T32 1T20
d
3T322T32 + 1T20 3T32
Figure AQ1 Figure AQ2
RC Design and Construction – HKC 2004 (2nd Edition)4-106
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-107
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-108
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.
RC Design and Construction – HKC 2004 (2nd Edition)4-109
Assignment No. 4
b = 1000
30
f
3T16
8
h =
13
f
d =
488
420
3T32
4
b = 300w
Figure AQ4
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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)
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(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
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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.
RC Design and Construction – HKC 2004 (2nd Edition)4-113
(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
RC Design and Construction – HKC 2004 (2nd Edition)4-114
(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
RC Design and Construction – HKC 2004 (2nd Edition)4-115