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.
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
con
tin
uo
us
4 @ 3 m
8 m
con
tin
uo
us
dis
con
tin
uo
us
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 kN/m2
Design load, n = 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 =11.57 kN/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
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
ii) Shear force and bending moment diagram
a) Bay area = ( 8 x 3 ) x 4 = 96 m2 > 30 m2 OK!
b) qk / gk = 4 / 4.125 = 0.97 < 1.25 OK!
c) qk = 4 kN/m2 < 5 kN/m2 OK!
4 @ 3 m
8 m
1 m
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
F = wL
= 11.57 x 3 m = 34.71 kN
0.46F
0.6F
0.5F 0.5F
0.5F 0.5F
0.075FL
0.086FL
0.063FL
0.063FL
0.063FL
+ + +
- - -
+ + +
- -
3 m 3 m 3m
w = 11.57 kN/m
0.04FL
3 m
+
-
0.6F
0.4F
0.086FL
0.086FL
+
-
(1.35 Gk + 1.5 Qk)
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
iii) Effective depth, d
Assume bar = 10 mm
d = h β c - bar/2 = 125 β 25 β 10/2 = 95 mm
a) At outer support
M = 0.04FL = 0.04 (34.71)(3) = 4.17 kNm/m
πΎ =π
ππ2πππ=
4.17 π₯ 106
1000 π₯ 952 π₯ 30= 0.015 < 0.167
Compression reinforcement is not required.
π§ = π 0.5 + 0.25 βπΎ
1.134 = 0.99π > 0.95π
π΄π ,πππ = π
0.87ππ¦π π§=
4.17 π₯ 106
0.87 π₯ 500 π₯ 0.95 95 = 106 ππ2/π
Provide: H10 β 300 (As,prov = 262 mm2/m) > π΄π ,πππ 143
b) At middle of end span
M = 0.075FL = 0.075 (34.71)(3) = 7.81 kNm/m
πΎ =π
ππ2πππ=
7.81 π₯ 106
1000 π₯ 952 π₯ 30= 0.03 < 0.167
Compression reinforcement is not required.
π§ = π 0.5 + 0.25 βπΎ
1.134 = 0.97π > 0.95π
π΄π ,πππ = π
0.87ππ¦π π§=
7.81 π₯ 106
0.87 π₯ 500 π₯ 0.95 95 = 199 ππ2/π
Provide: H10 β 300 (As,prov = 262 mm2/m) > π΄π ,πππ 143 OK! β bottom
Slab : 8 mm β 12 mm
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
πΎ =π
ππ2πππ=
9.75 π₯ 106
1000 π₯ 952 π₯ 30= 0.04 < 0.167
Compression reinforcement is not required.
π§ = π 0.5 + 0.25 βπΎ
1.134 = 0.97π > 0.95π
π΄π ,πππ = π
0.87ππ¦π π§=
9.75 π₯ 106
0.87 π₯ 500 π₯ 0.95 95 = 248 ππ2/π
Provide: H10 β 300 (As,prov = 262 mm2/m) > π΄π ,πππ 143 OK!
d) At middle interior spans and interior supports
M = 0.063FL = 0.063 (37.81)(3) = 7.14 kNm/m
πΎ =π
ππ2πππ=
7.14 π₯ 106
1000 π₯ 952 π₯ 30= 0.03 < 0.167
Compression reinforcement is not required.
π§ = π 0.5 + 0.25 βπΎ
1.134 = 0.97π > 0.95π
π΄π ,πππ = π
0.87ππ¦π π§=
7.14 π₯ 106
0.87 π₯ 500 π₯ 0.95 95 = 182 ππ2/π
Provide: H10 β 300 (As,prov = 262 mm2/m) > π΄π ,πππ 143 OK!
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
Calculate As min and As max
π΄π ,πππ = 0.26 (2.9) 1000 (95)
(500)= 143 ππ2/π β₯ 0.0013 1000 95 = 123.5ππ2/π
π΄π ,πππ₯ = 0.04 π΄π = 0.04 1000 125 = 5000 ππ2/π
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
π = 1 + 200
95= 1.45 < 2.0 π ππ ππ β΄ ππΎ!
ππ =π΄π π
ππ€π=
262
1000 π₯ 95= 0.0028
ππ π ,π = 0.12π 100πππππ 13ππ€π β₯ ππππ
= 0.12 1.45 100 0.0028 30 13 1000 95 = 33.6 ππ
ππππ = 0.035 π3/2πππ1/2 ππ€ π
= 0.035 1.45 32 30
12 1000 π₯ 95 = 31.8 ππ
VRd,c > Vmin
Use VRd,c = 33.6 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
π =π΄π ,πππ
ππ€π=
248
1000 π₯ 95= 2.61 π₯ 10β3
π0 = 30π₯ 10β3 = 5.48 π₯ 10β3
< o
π
π= πΎ 11 + 1.5 πππ
πππ
+ 3.2 πππ πππβ 1
3/2
= 1.3 11 + 1.5 30 5.48
2.61 + 3.2 30
5.48
2.61β 1
32
= 63
From table 7.4N, K = 1.3 (one way continuous slab)
(i) Calculate the modification factor
310
ππ =
500
ππ¦π π΄π ,πππ
π΄π ,ππππ£
= 500
500 248262
= 1.06
(ii) Calculate (L/d)allowable
πΏ
π πππππ€ππππ
= πΏ
π πππ ππ
π₯ πππππππππ‘πππ ππππ‘ππ
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
πΏ
π πππππ€ππππ
= 63 π₯ 1.06 = 66.78
a) Calculate (L/d)actual
πΏ
π πππ‘π’ππ
=ππππππ‘ππ£π π πππ πππππ‘π
ππππππ‘ππ£π ππππ‘π =
3000
95 = 31.6
b) Compare with (L/d)actual with (L/d)allowable
(L/d)actual < (L/d)allowable
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 !
i) (L/d)actual β€ (L/d)allowable β Beam is safe against deflection (OK!)
ii) (L/d)actual > (L/d)allowable - Beam is not safe against deflection (Fail!)
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
CONTINUOUS ONE WAY SLAB EXAMPLE 3.4
3. Draw the detailing.
Plan view
Cross-section
H10-300 (T)
H10-300 (B)
H10-300 (T)
H10-300 (B)
H10-300 (T)
T10-300 (B)
H8
-30
0 (
B)
H8
-30
0 (
T)
H8
-30
0 (
T) H
8-3
00
(B
)
H8
-30
0 (
T)
H10-300 (T)
H10-300 (B)
H10-300 (T) H10-300 (T)
H8-300 (B)
CONTINUOUS ONE WAY SLAB APPENDIX A