a) Minimum areas
Minimum area of longitudinal tension (flexural) reinforcement
As,min= 0.26(ffctm/ff ) byk bb dt ≥ 0.0013 bbb dt where
bt
bb = width of tension zone fctm f = 0.30 × ffck0.667 As,min= 0.26 × 0.30 × 300.6677 × 1000 × 144/500 = 216 mm2/m (r = 0.15%)r = H12 @ 225 B1 OK Cl. 9.3.1.1, 9.2.1.1 Table 3.1
Secondary (transverse reinforcement)
Minimum 20% As,req
20% As,reqq = 0.2 × 502 = 100 mm2/m
Consider As,min to apply as before. As,min= 216 mm2/m
Try H10 @ 350 B2 (224 mm2/m)
Cl. 9.3.1.1(2) SMDSC[9]
Check edge.
Assuming partial fixity exists at edges, 25% of As is required to extend 0.2 × the length of the adjacent span.
As,reqq = 25% × 639 = 160 mm2/m As,minas before = 216 mm2/m
= Use H10 @ 350 (224 mm2/m) U-bars at edges
Cl. 9.3.1.2(2)
46
Curtail 0.2 × 5975 = 1195 mm, say 1200 mm measured from face of support‡.
Cl. 9.3.1.2(2)
Maximum spacing of bars
Maximum spacing of bars < 3h < 400 mm OK Cl. 9.3.1.1.(3)
Crack control
As slab < 200 mm, measures to control cracking are unnecessary. Cl. 7.3.3(1) However, as a check on end span:
Loading is the main cause of cracking,
= use Table 7.2N or Table 7.3N for wwwmax = 0.4 mm and ss = 241 MPas (see deflection check).
Max. bar size = 20 mm or max. spacing = 250 mm = H12 @ 150 B1 OK. Cl. 7.3.3(2), 7.3.1.5 Table 7.2N & interpolation, Table 7.3N & interpolation End supports: effects of partial fixity
Assuming partial fixity exists at end supports, 15% of As is required to extend 0.2 × the length of the adjacent span.
As,reqq = 15% × 639 = 96 mm2/m But, As,min as before = 216 mm2/m
(r = 0.15%)r One option would be to use bob bars, but choose to use U-bars
Try H12 @ 450 (251 mm2/m) U-bars at supportspp
Cl. 9.3.1.2(2)
Cl. 9.3.1.1, 9.2.1.1
Curtail 0.2 × 5975 = say, 1200 mm measured from face of support.‡ Cl. 9.3.1.2(2)
b) Curtailment
i) End span, bottom reinforcement
Assuming end support to be simply supported, 50% of Asshould extend into the support.
50% × 639 = 320 mm2/m
Try H12 @ 300 (376 mm2/m) at supportspp
Cl. 9.3.1.2(1)
In theory, 50% curtailment of reinforcement may take place al from where the moment of resistance of the section with the remaining 50% would be adequate to resist the applied bending moment. In practice, it is usual to determine the curtailment distance as being alfrom where MEd= MEd,max/2.
Cl. 9.3.1.2(1) Note, 9.2.1.3 (2)
‡ Detail MS2 of SMDSC[9], suggests 50% of T1 legs of U-bars should extend 0.3l (= say 1800 mm) from face of support by placing U-bars alternately reversed.
a) Load arrangementrr Tensile for TT ce in reinforcement, FFs 633 50% 50% 633 (say 500) Tensile r TT esistance of reinforcement X MEd,max MEdx = RAX –X nXXX /22 MEdx/z lbd lbd n A B A B A B A B
b) Bending moment MEdx
c) Tensile TT force in bottom reinforcement
d) Curtailment of bottom reinforcement
987
987 (say 850)
Figure 3.5 Curtailment of bottom reinforcement: actions, bending moments, forces in reinforcement and curtailment
Thus, for a single simply supported span supporting a UDL of n, MEd,max= 0.086nl2; RA = 0.4nl
At distance, X, from end support, moment,XX MEd@X =X RAX –X nX2/2
= when M@X =X MEd,max/2: 0.086nl2/2 = 0.4nlX – X nX2/2
48 Assuming X =X xl 0.043nl2 = 0.4nlxl –l nxxx l22/2 0.043 = 0.4x –x xxx /22 0 = 0.043 – 0.4x + x xxx /22 x = 0.128 or 0.672, say 0.13 and 0.66 x
= at end support 50% moment occurs at 0.13 x span 0.13 × 5975 = 777 mm
Shift rule: for slabs, almay be taken as d (= 144 mm),d = curtail to 50% of required reinforcement at 777 – 144 = 633 mm from centreline of support.
Say 500 mm from face of support App
Cl. 9.2.1.3(2), 6.2.2(5)
= in end span at 1st internal support 50% moment occurs at 0.66 × span
0.66 × 5975 = 3944 mm
Shift rule: for slabs al may be taken as d (= 144 mm),d = curtail to 50% of required reinforcement at 3944 + 144 = 4088 mm from support A
or 5975 – 4088 = 987 mm from centreline of support B.
Say 850 mm from face of support Bpp
Cl. 9.2.1.3(2), 6.2.2(5)
ii) 1st interior support, top reinforcement
Presuming 50% curtailment of reinforcement is required this may take place alfrom where the moment of resistance of the section with the remaining 50% would be adequate. However, it is usual to determine the curtailment distance as being alfrom where MEd= MEd,max/2.
Cl. 9.3.1.2(1) Note, 9.2.1.3(2)
Thus, for the 1st interior support supporting a UDL of n, MEd,maxTT= 0.086nl2; R
B = 0.6nl
At distance Y from end support, moment,Y MEd@Y =Y MEd,maxTT–RAY + Y nY2/2
= when M@Y =Y MEd,maxT/2
0.086nl2/2 = 0.086nl2– 0.6nlY +Y nY2/2 Assuming Y =Y yl 0.043nl2= 0.086nl2– 0.6nlyl +l ny2l2/2 0 = 0.043 − 0.6y + y y2/2 y = 0.077 y (or 1.122), say 0.08
= at end support 50% moment occurs at 0.08 × span 0.08 × 5975 = 478 mm
Shift rule: for slabs, almay be taken as d 144 mmd = curtail to 50% of required reinforcement at 478 + 144 = 622 mm from centreline of support.
50% of reinforcement may be curtailed at, say, 600 mm from either face of support Bpp
Cl. 9.2.1.3(2), 6.2.2(5)
100% curtailment may take place al from where there is no hogging moment. Thus,
when M@Y = MEd,maxT/2 0 = 0.086nl2 – 0.6nlY + nY2/2
Assuming Y = yl
0 = 0.086 – 0.6y + y2/2 y = 0.166 (or 1.034), say 0.17
= at end support 50% moment occurs at 0.17 × span 0.17 × 5975 = 1016 mm
Shift rule: for slabs, al may be taken as d
= curtail to 100% of required reinforcement at 1016 + 144 = 1160 mm from centreline of support.
100% of reinforcement may be curtailed at, say, 1100 mm from either face of support B.
iii) Support B bottom steel at support
At the support 25% of span steel required Cl. 9.3.1.1(4), 9.2.1.5(1), 9.2.1.4(1)
0.25 × 639 = 160 mm2 As,min as before = 216 mm2/m
For convenience use H12 @ 300 B1 (376 mm2/m)
Cl. 9.3.1.1, 9.2.1.1
c) Anchorage at end support
As simply supported, 50% of As should extend into the support. This 50% of As should be anchored to resist a force of
Cl. 9.2.1.2(1) & Note, 9.2.1.4(2)
FE = VEd × al/z where
Exp. (9.3)
VEd = the absolute value of the shear force
al = d, where the slab is not reinforced for shear
z = lever arm of internal forces FE = 29.4 × d/0.95‡ d = 30.9 kN/m Cl. 9.2.1.3(2) Anchorage length, lbd: Cl. 8.4.4 lbd = alb,rqd ≥ lb,min Exp. (8.4) where a = conservatively 1.0
lb,rqd = basic anchorage length required
= (f/4) (ssd/fbd) Exp. (8.3)
where
f = diameter of the bar = 12 mm
ssd = design stress in the bar at the ultimate limit state = FE/As,prov
= 30.9 × 1000/376 = 81.5 MPa
50
fbd = ultimate bond stress
= 2.25 n1 n2 fct,d Cl. 8.4.2(2)
where
n1 = 1.0 for ‘good’ bond conditions and 0.7 for all other conditions = 1.0
n2 = 1.0 for bar diameter ≤ 32 mm fct,d = design tensile strength
= act fct,k/ gC. For fck = 30 MPa = 1.0 × 2.0/1.5 = 1.33 MPa Cl. 3.1.6(2) & NA, Tables 3.1 & 2.1N =f[]= 2.25 × 1.33 = 3.0 MPa lb,rqd = (12/4) (81.5/1.33) = 183 mm lb,min = max(10d, 100 mm) = 120 mm lbd = 183 mm measured from face of support
By inspection, using U-bars, OK
Exp. (8.6) Fig. 9.3 d) Laps
Lap H12 @ 300 U-bars with H12 @ 150 straights.
Tension lap, l0 = a1 a2 a3 a5 a6 lb,rqd a l0min Exp. (8.10)
where a 1 =1.0 (straight bars) a2=1 − 0.15 (cd − f)/f Table 8.2 where
cd = min(pitch, side cover or cover) = 25 mm
Fig. 8.4
f = bar diameter = 12 mm a2=0.84
a3=1.0 (no confinement by reinforcement) a5=1.0 (no confinement by pressure)
Table 8.2
a6=1.5 Table 8.3
lb,rqd=(f/4) ssd/fbd Exp. (8.3)
where
ssd = the design stress at ULS at the position from where the anchorage is measured.
Assuming lap starts 500 mm from face of support (587.5 mm from centreline of support): MEd= 29.5 × 0.59 − 12.3 × 0.592/2
= 15.2 kNm ssd = MEd/(Asz)
= 15.2 × 106/(376 × 144/0.95) = 267 MPa fbd = ultimate bond stress
= 2.25 n1 n2 fct,d
where
n1 =1.0 for ‘good’ conditions n2 =1.0 for f < 32 mm fct,d=act fct,k/gC where act =1.0 fct,k=2.0 gC =1.5 =f[]= 2.25 × 2.0/1.5 = 3.0 MPa Cl. 3.1.6 (2) & NA Table 3.1 Table 2.1N & NA lb,rqd =(f/4) ssd/fbd =(12/4) × (267/3) = 267 mm l0min b =max[0.3 a6 lb,rqd; 15f/ 200 mm] =max[0.3 × 1.5 × 229; 15 × 12; 200] =max[124; 180; 200] = 200 mm = l0=a1 a2 a3 a5 a6 lb,rqd ≥ l0min =1.0 × 0.84 × 1.0 × 1.0 × 1.5 × 329 ≥ 200 = 414 mm Exp. 8.6
But good practice suggests minimum lap of max[tension lap; 500] = lap with bottom reinforcement = 500 mm starting 500 from face of support.