CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
Figure below shows a first floor plan of an office building. It is estimated that the 125 mm thick slab will carry 4.0 kN/m2 variable action and 1.0 kN/m2 load from finishes & suspended ceiling. This building is exposed to XC1 exposure class. Using concrete class C30/37 and high yield steel, prepare a complete design and detailing for this slab.
discontinuous
continuous
8m
continuous
4@3m
Design data: For all slab panels, Ly/Lx = 8/3
= 2.67
>
2.0
one / two way continuous slab. Variable action, qk = 4.0 kN/m2 Loads from finishes & suspended ceiling = 1.0 kN/m2 fck = 30 N/mm2 fyk = 500 N/mm2 h = 125 mm
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
SOLUTION: 1. Calculate the design load acting on the slab. Self-weight of slab = 25 x slab thickness
= 3.125 kN/m2
Finishes & ceiling
= 1.0 kN/m2
=
Total charac. permanent action, gk
= 4.125 kN/m2
Total charac. variable action, qk
= 4.0
Design load, n
kN/m2
= 1.35 gk + 1.5 qk = 1.35 ( 4.125 ) + 1.5 (4) = 11.57 kN/m2
Consider 1 m width of slab, w
= 11.57 x 1 m
2. Design the main reinforcement. i)
Nominal cover Minimum cover (bond), Cmin,b = bar = 10 mm Minimum cover (durability), Cmin,dur = 15 mm Minimum value = 10 mm Cmin = maximum value = 15 mm Cnom = Cmin + Cdev = 15 + 10 = 25 mm
=11.57
kN/m
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
ii)
Shear force and bending moment diagram > 30 m2 OK!
a) Bay area
= ( 8 x 3 ) x 4 = 96 m2
b) qk / gk c) qk
= 4 / 4.125 = 0.97 < 1.25 = 4 kN/m2 < 5 kN/m2
OK! OK!
4@3m
1m 8m
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4 w = 11.57 kN/m
3m
3m
0.46F
3m
0.5F
+
+
-
0.086FL -
+
0.063FL
F = wL = 11.57 x 3 m = 34.71 kN
(1.35 Gk + 1.5 Qk)
0.4F
0.5F
0.063FL
-
-
-
0.5F
0.086FL
0.075FL
+
-
0.6F
+
0.6F
0.5F
+
0.04FL
3m
+
0.063FL
+
0.086FL
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
iii)
Effective depth, d
Assume bar = 10 mm d = h – c - bar/2
Slab : 8 mm – 12 mm
= 125 – 25 – 10/2
= 95
mm
a) At outer support M = 0.04FL = 0.04 (34.71)(3) = 4.17
kNm/m
𝑀 4.17 𝑥 106 𝐾= 2 = = 0.015 < 0.167 𝑏𝑑 𝑓𝑐𝑘 1000 𝑥 952 𝑥 30 Compression reinforcement is not required. 𝑧 = 𝑑 0.5 + 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾 = 0.99𝑑 > 0.95𝑑 1.134
𝑀 4.17 𝑥 106 = = 106 𝑚𝑚2 /𝑚 0.87𝑓𝑦𝑘 𝑧 0.87 𝑥 500 𝑥 0.95 95
Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143 b) At middle of end span M = 0.075FL = 0.075 (34.71)(3) = 7.81 kNm/m 𝑀 7.81 𝑥 106 𝐾= 2 = = 0.03 < 0.167 𝑏𝑑 𝑓𝑐𝑘 1000 𝑥 952 𝑥 30 Compression reinforcement is not required. 𝑧 = 𝑑 0.5 + 0.25 −
𝐴𝑠,𝑟𝑒𝑞
𝐾 = 0.97𝑑 > 0.95𝑑 1.134
𝑀 7.81 𝑥 106 = = = 199 𝑚𝑚2 /𝑚 0.87𝑓𝑦𝑘 𝑧 0.87 𝑥 500 𝑥 0.95 95
Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143 OK! – bottom
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
c) At first interior support and near middle of end span M = 0.086FL = 0.086 (37.81)(3) = 9.75 kNm/m 𝑀 9.75 𝑥 106 𝐾= 2 = = 0.04 < 0.167 𝑏𝑑 𝑓𝑐𝑘 1000 𝑥 952 𝑥 30 Compression reinforcement is not required. 𝑧 = 𝑑 0.5 + 0.25 −
𝐴𝑠,𝑟𝑒𝑞
𝐾 = 0.97𝑑 > 0.95𝑑 1.134
𝑀 9.75 𝑥 106 = = = 248 𝑚𝑚2 /𝑚 0.87𝑓𝑦𝑘 𝑧 0.87 𝑥 500 𝑥 0.95 95
Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143 OK! d) At middle interior spans and interior supports M = 0.063FL = 0.063 (37.81)(3) = 7.14 kNm/m 𝑀 7.14 𝑥 106 𝐾= 2 = = 0.03 < 0.167 𝑏𝑑 𝑓𝑐𝑘 1000 𝑥 952 𝑥 30 Compression reinforcement is not required. 𝑧 = 𝑑 0.5 + 0.25 −
𝐴𝑠,𝑟𝑒𝑞 =
𝐾 = 0.97𝑑 > 0.95𝑑 1.134
𝑀 7.14 𝑥 106 = = 182 𝑚𝑚2 /𝑚 0.87𝑓𝑦𝑘 𝑧 0.87 𝑥 500 𝑥 0.95 95
Provide: H10 – 300 (As,prov = 262 mm2/m) > 𝐴𝑠,𝑚𝑖𝑛 143 OK!
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
Calculate As min and As max 0.26 (2.9) 1000 (95) = 143 𝑚𝑚2 /𝑚 ≥ 0.0013 1000 95 = 123.5𝑚𝑚2 /𝑚 (500)
𝐴𝑠,𝑚𝑖𝑛 =
𝐴𝑠,𝑚𝑎𝑥 = 0.04 𝐴𝑐 = 0.04 1000 125 = 5000 𝑚𝑚2 /𝑚 Asmin < Asprov < As max => Ok!!
i)
Transverse reinforcement
Provide minimum = 143 mm2/m Provide: H8-300 (As,prov = 168 mm2/m)
1. Check the slab for shear VEd = Vmax = 0.6F = 0.6 (37.81) = 22.69 kN i)
Calculate VRd,c
𝑘 =1+
𝜌𝑙 =
200 = 1.45 < 2.0 𝑑 𝑖𝑛 𝑚𝑚 95
∴ 𝑂𝐾!
𝐴𝑠𝑙 262 = = 0.0028 𝑏𝑤 𝑑 1000 𝑥 95
𝑉𝑅𝑑,𝑐 = 0.12𝑘 100𝜌𝑙 𝑓𝑐𝑘
1 3 𝑏𝑤 𝑑
≥ 𝑉𝑚𝑖𝑛
= 0.12 1.45 100 0.0028 30
1 3
1000 95 = 33.6 𝑘𝑁
𝑉𝑚𝑖𝑛 = 0.035 𝑘 3/2 𝑓𝑐𝑘 1/2 𝑏𝑤 𝑑 = 0.035 1.45 VRd,c > Vmin Use VRd,c = 33.6 kN
3 2
30
1 2
1000 𝑥 95 = 31.8 𝑘𝑁
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
ii)
Compare VEd with VRd,c
VEd (22.69 kN) < VRd,c (33.6 kN) No shear reinforcement is required.
2. Deflection check Check only at mid span with maximum moment 𝜌=
𝐴𝑠,𝑟𝑒𝑞 248 = = 2.61 𝑥 10−3 𝑏𝑤 𝑑 1000 𝑥 95
𝜌0 = 30𝑥 10−3 = 5.48 𝑥 10−3 < o 𝑙 𝜌𝑜 𝜌𝑜 = 𝐾 11 + 1.5 𝑓𝑐𝑘 + 3.2 𝑓𝑐𝑘 −1 𝑑 𝜌 𝜌
3/2
5.48 5.48 = 1.3 11 + 1.5 30 + 3.2 30 −1 2.61 2.61 From table 7.4N, K = 1.3 (one way continuous slab) (i)
Calculate the modification factor
310 = 𝜎𝑠 (ii) 𝐿 𝑑
𝑓𝑦𝑘
500 500 = = 1.06 𝐴𝑠,𝑟𝑒𝑞 248 500 262 𝐴𝑠,𝑝𝑟𝑜𝑣
Calculate (L/d)allowable
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
=
𝐿 𝑑
𝑏𝑎𝑠𝑖𝑐
𝑥 𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
3 2
= 63
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
𝐿 𝑑
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 63 𝑥 1.06 = 66.78
a) Calculate (L/d)actual 𝐿 𝑑
𝑎𝑐𝑡𝑢𝑎𝑙
=
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑠𝑝𝑎𝑛 𝑙𝑒𝑛𝑔𝑡 3000 = = 31.6 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑒𝑝𝑡 95
b) Compare with (L/d)actual with (L/d)allowable (L/d)actual < (L/d)allowable
i) ii)
(L/d)actual ≤ (L/d)allowable – Beam is safe against deflection (OK!) (L/d)actual > (L/d)allowable - Beam is not safe against deflection (Fail!)
Therefore, slab is safe against deflection.
3. Crack check i)
h = 125 mm < 200 mm OK ! (Section 7.3.3 EC2) specific measures to control cracking is not necessary.
ii)
Maximum bar spacing, smax,slabs (Section 9.3 EC2) a) For main reinforcement: Smax, slabs = 3h 400 mm = 375 mm Actual bar spacing = 300 mm < Smax, slabs
OK !
b) For transverse reinforcement: Smax, slabs = 3.5h 450 mm = 437.5 mm Actual bar spacing = 300 mm < Smax, slabs
OK !
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
3. Draw the detailing.
H10-300 (T)
H10-300 (T)
H8-300 (B)
H8-300 (B)
H10-300 (T)
H10-300 (B)
H8-300 (T)
T10-300 (B)
H8-300 (T)
H8-300 (T)
H10-300 (B)
Plan view H10-300 (T)
H10-300 (T)
H10-300 (B)
H10-300 (T)
H8-300 (B)
Cross-section
CONTINUOUS ONE WAY SLAB APPENDIX A
