Worked Examples for Eurocode 2 Draft Version
3.3 Continuous one-way solid slab
3.4.8 Indirect supports
As the ribs of the slab are not supported at the top of the supporting beam sections (A, B, C, D), additional vertical reinforcement should be provided in these supporting beams and designed to resist the reactions. This additional reinforcement should consist of links within the supporting beams (see Beams design, Section 4. 4.9).
<9.2.5, Fig. 9.7>
Support A (and D) at solid/rib interface: VEd = 26.5 kN / rib
As,req = 26.3 × 1000 / (500 / 1.15) = 60 mm2
This area is required in links within h / 6 = 300 / 6 = 50 mm of the ribbed/solid interface and within h / 2 = 300 / 2 = 150 mm of the centreline of the rib.
Support B (and C) at solid/rib interface, VEd = 37.9 kN / rib
As,req = 37.9 × 1000 / (500 / 1.15) = 87 mm2 placed similarly
<Fig. 9.7>
3.4.9 Summary of design
Figure 3.15 Summary of design
3.4.10 Other design/detailing checks
a) Minimum area of reinforcement in flange
As,min = 0.26 (fctm/ fyk) btd ≥ 0.0013 btd where
bt = width of tension zone
fctm = 0.30 × fck 0.666 <9.3.1.1 9.2.1.1, Exp. (9.1N)> <Table 3.1> As,min = 0.26 × 0.30 × 350.666 × 1000 × 100 / 500 = 166 mm2 / m (0.17%)
∴ Use A142 in flange (say OK) <BS 8666[28]> b) Secondary reinforcement
Not applicable
c) Maximum spacing of bars, < 3 h < 400 mm
By inspection OK <9.3.1.1.(3)>
d) Crack control
Loading is the main cause of cracking ∴ use Table 7.2N or Table 7.3N for wmax = 0.3 mm and max. σs = 200 MPa (see deflection check)
Max. bar size = 25 mm or max. spacing = 250 mm <7.3.3(2) 7.3.1.5> <Table 7.2N> <Table 7.3N> OK by inspection.
e) 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,req = 15% × 525 = 79 mm2 / rib
<9.3.1.2(2)> For the rib in tension:
As,min = 0.26 × 0.30 × 300.666 × 159 × 257 / 500 = 55 mm2 / rib
3.4.11 Curtailment
Wherever possible simplified methods of curtailing reinforcement would be followed. The following is intended to show how a rigorous assessment of curtailment of
reinforcement might be undertaken. End support A: bottom steel at support Check anchorage
As simply supported, 25% of As should be anchored in support.
25% × 595 = 148 mm2 <9.3.1.1(4), 9.3.1.2(1) & Note, 9.2.1.4(1) & NA>
Use 1 no.H20 / rib (314 mm2 / rib) Check anchorage length
Envelope of tensile force:
To resist envelope of tensile force, provide reinforcement to al or lbd beyond centreline of support.
<9.3.1.1(4), 9.2.1.3(1), 9.2.1.3(2), 9.2.1.3(3) Fig. 9.2>
For members without shear reinforcement, al = d = 232 By inspection, σsd = 0, lbd = lbd,min = max(10φ, 100 mm)
<9.2.1.3>
Indirect support:
As anchorage may be measured from face of indirect support, check, force to be resisted at solid/rib interface:
Fs = MEd / z + FE
<9.3.1.1(4), 9.2.1.4(2), 9.2.1.4(3), Fig. 9.3b>
where MEd = 18.3 kNm / rib z = 220 as before FE = VEd × al / z where VEd = 29.3 kN / rib al = z cot θ / 2 ∴ FE = VEd cot θ / 2 = 29.3 × 1.25 = 36.6 kN / rib <Exp. (9.3)> <9.2.1.3, Exp (9.2)> Fs = 18.6 × 106 / (220 × 103) + 36.6 = 121.1 kN Anchorage length lbd = αlb,rqd ≥ lb,min where: α = conservatively 1.0 lb,rqd = (ϕ / 4) (σsd / fbd) where ϕ = 20
σsd = design stress in the bar at the ULS
= 121.1 × 1000 / 314 = 385 MPa
fbd = ultimate bond stress = 2.25 η1 η2 fct,d where
η1 = 1.0 for good bond conditions η2 = 1.0 for bar diameter ≤ 32 mm
fct,d = αct fct,k / γc = 1.0 × 2.2 / 1.5 = 1.47 MPa fbd = 2.25 × 1.47 = 3.31 MPa ∴ lb,rqd = (20 / 4) (385 / 3.31) = 581 mm lb,min = max[10ϕ; 100 mm] = 200 mm <8.4.4, Exp. (8.4)> <Exp. (8.3)> <8.4.2 (2)>
<3.1.6 (2),Tables 3.1 and 2.1 and NA>
<Fig. 9.3> ∴ lbd= 581 mm measured from solid/rib intersection.
i.e. 31 mm beyond centreline of support‡‡‡‡. End support A: top steel
Assuming partial fixity exists at end supports, 15% of As is required
to extend at least 0.2 × the length of the adjacent span§§§§. <9.3.1.2(2)>
As,req = 15% × 525 = 79 mm2 / rib
As,min = 0.26 × 0.30 × 350.666 × 159 × 257 / 500 = 68 mm2 / rib <9.3.1.1
9.2.1.1(1), Exp. (9.1N)> Use 2 no. H12 T1 / rib in rib and 2 H10 T1 / rib between ribs
(383 mm2 / rib) Support B (and C): top steel
At the centreline of support (2 H16 T + 3 H12 T) / rib are required. The intention is to curtail in two stages, firstly to 2 H16 T / rib then to 2 H12 T / rib.
Curtailment of 2 H16 T / rib at support (capacity of 2 H12 T / rib + shift rule)
‡‡‡‡ Whilst this would comply with the requirements of Eurocode 2, it is common practice to take bottom bars 0.5 × a tension lap beyond the centreline of support (= 250 mm beyond the centreline of support; see MS1 in SMDSC[21]).
§§§§ It is usual to curtail 50% of the required reinforcement at 0.2l and to curtail the remaining 50% at 0.3l or line of zero moment (see MS2 in SMDSC[21]).
Assume use of 2 H12 T throughout in midspan: Assuming z = 211 mm as before,
MR2H12T = 2 × 113 × 434.8 × 211 = 20.7 kNm / rib (23.0 kNm / m) (Note: section remains under-reinforced)
From analysis MEd = 23.0 kNm / m occurs at 2250 mm
(towards A) and 2575 mm (towards B). Shift rule: αl = z cot θ / 2
Assuming z = 211 as before αl = 1.25 × 211 = 264 mm
∴ 2 no.H12 T are adequate from 2250 + 264 = 2513 mm from B towards A and 2575 + 263 = 2838 mm from B towards C.
∴ Curtail 2 no.H16 T @ say 2600 from BA and 2850 from BC Curtailment of 3 no.H12 T / rib at support (capacity of 2 H16 T / rib + shift rule)
MR2H16T = 2 × 201 × 434.8 × 211 = 36.9 kNm / rib (41.0 kNm / m) (Note: section remains under-reinforced)
From analysis MEd = 41.0 kNm / m occurs at 1310 mm (towards A) and 1180 mm (towards C).
Shift rule: αl = 263 mm as before
∴ 2 no.H16 T are adequate from 1310 + 263 = 1573 mm from B towards A and 1180 + 263 = 1443 mm from B towards C.
∴Curtail 3 no. H12 at say 1600 from B (or C).
Figure 3.16
Support B (and C) bottom steel at support
At the support 25% of span steel required <9.3.1.1(4), 9.2.1.5(1),
9.2.1.4(1)> 0.25 × 628 = 157 mm2
Try 1 no.H16 B / rib (201) This reinforcement may be anchored into indirect support or
carried through. <Fig. 9.4>
Support B (and C): bottom steel curtailment BA and BC To suit prefabrication 2 no.H20 / rib will be curtailed at solid/rib interface, 1000 mm from BA (B towards A) and BC. From analysis, at solid/rib interface sagging moment = 0.
From analysis, at solid/rib interface + al, i.e. at 1000 + 1.25 × 244 = 1303 mm
at 1305 mm from BA sagging moment = say 5 kNm / rib at 1305 mm from BC sagging moment = 0
Use 1 no.H16 B / rib (201)
3.4.12 Laps
At AB, check lap 1 no.H20 B to 2 no.H20 B in rib full tension lap
l0 = α1 α6 lb,rqd > l0,min where
<Exp. (8.10)> α1 = 1.0 (cd = 45 mm, i.e. < 3ϕ)
α6 = 1.5 (as > 50% being lapped)
lb,rqd = (ϕ / 4) (σsd / fbd) where <Table 8.2> ϕ = 20 σsd = 434.8 fbd = 3.0 MPa as before
l0,min = max. 10ϕ or 100 = 200 Exp. (8.6)
l0 = 1.0 × 1.5 × (20 / 4) × 434.8 / 3.0
= 1087 mm, say = 1200 mm <SMDSC[21]>
At BA and BC, check lap 2 no. H12 T to 2 no. H16 T in rib – full tension lap
l0 = α1 α6 lb,rqd > l0,min where
<Exp. (8.10)> α1 = 0.7 (cd = 45 mm, i.e. > 3ϕ)
α6 = 1.5 (as > 50% being lapped)
lb,rqd = (ϕ / 4) (σsd / fbd) where
<Table 8.2>
ϕ = 12
σsd = 434.8
fbd = 2.1 (3.0 MPa as before but η1 = 0.7 for not good bond conditions)
l0,min = max. 10ϕ or 100 = 120
<8.4.2 (2)> Exp. (8.6)
l0 = 0.7 × 1.5 × (12 / 4) × 434.8 / 2.1
= 651 mm, say = 700 mm <SMDSC[21]>
But to aid prefabrication take to solid/rib intersection 1000 mm from centre of support.
At BA and BC, check lap 1 H16 B to 2 H20 B in rib
By inspection, nominal say, 500 mm <SMDSC[21]>