Project :
Woman & Child Hospital, Kuala Lumpur
SLAB DESIGN (FLAT SLAB)
PRC
Design by :
Date:
Location/slab mark: Typical panel
18/11/13
Key in the RED field
(Refer to Clause 3.7, BS 8007) ly = 8.00 m lx = 8.00 m Thickness of Slab, h
=
200 mm
Self Weight,
=
4.8 kN/m2
Superimposed dead loads
=
6.05 kN/m2 2
Dead Load, DL
= 10.85 kN/m
Live Load, LL
=
n = 1.4DL + 1.6LL UDL, w
Drop panel size
=
35 N/mm2
Reinforcement, fy
=
460 N/mm2
Effective depth, d (middle strip)
3 kN/m2
As min = bh/100
= 19.99 kN/m2 (from slab) = 19.99 kN/m
(x-dir)= 2700 (y-dir)
Concrete, fcu
2700 mm
Concrete cover Rebars size used
0.13% = =
= 159 mm 260 mm2
=
25 mm clear 16 mm
The average drop panel load per bay 0.68 kN/m2
Drop panel thickness, hdrop = Column head, ho = Column head, hc =
450 mm 800 mm 0 mm
Effective depth, d (support)
Effective span, L
=
8.00 mm
As min = bh/100
Width of column strip, b Width of middle strip, bm
= =
2700 mm 5300 mm
Propotion to span Propotion to span
=
=
0.13%
=
409 mm 585 mm2
= 0.675 = 1.325
A) Design loads and moment Total dead load, Gk
= 11.53 kN/m2
Total live load, Qk
=
Design ultimate laod, n
= 20.95 kN/m2
3.00 kN/m2 Total design load per bay, F
The moments are calculated using coefficient from Table 3.19, BS 8110 The moments distribution into panels of flat slab as per Table 3.20 Near center of 1st span, M1span = +0.083xFxL 1st Interior Support, M1supp = -0.063xFxL Center of Interior Span, Mspan = +0.071xFxL Interior Support, Msupp = -0.055xFxL
= = = =
890.2 675.7 761.5 589.9
= 1340.6 kN
Moment distribution Column Strip Middle Strip 55% 359.4 kNm 45% 530.8 kNm 75% 451.8 kNm 25% 223.8 kNm 55% 307.4 kNm 45% 454.0 kNm 75% 394.5 kNm 25% 195.4 kNm
kNm kNm kNm kNm
M1sup = - 675.7 kNm Msup = - 589.9 kNm
Mspan= + M1span= +
761.5 kNm
890.2 kNm Moment Envelop (kNm)
Mspan= +
761.5 kNm
B) Column Strip Design Negitive Moments:1st Interior Support M1supp = 451.8 kNm k z
As
Interior Support Msupp = 394.5 kNm
2 = M / bd fcu = 0.0286
k
= [ 0.5 + (0.25 - k/0.9)1/2] d = 0.967 d > 0.95 d = 0.95 x 409 = 388.6 mm
z
= m x / 0.87 fy z =
2
2906 mm
As =
2
1076 mm /m
2/3 of the bars are placed in the center half of column strip As = 1937 mm2 = 1435 mm2/m As req Provide As prov
= T
1435 mm2/m 16 @
125 mm c/c within 1/2 strip
=
1608 mm2/m
> As req ==> OK !
Remining bars required As = 969 mm2 As req Provide As prov
= T
717 mm2/m 16 @
= 717.5 mm2/m 250 mm c/c within 1/2 strip
2
> As req ==> OK !
= 804.2 mm /m
2 = M / bd fcu = 0.0250
= [ 0.5 + (0.25 - k/0.9)1/2] d = 0.971 d > 0.95 d = 0.95 x 409 = 388.6 mm = m x / 0.87 fy z =
2537 mm2
= 939.6 mm2/m
2/3 of the bars are placed in the center half of column strip As = 1691 mm2 = 1253 mm2/m As req Provide As prov
= T
1253 mm2/m 16 @
150 mm c/c within 1/2 strip
=
1340 mm2/m
> As req ==> OK !
Remining bars required As = 846 mm2 As req = 626 mm2/m T 16 @ Provide As prov = 804 mm2/m
= 626.4 mm2/m 250 mm c/c within 1/2 strip > As req ==> OK !
Positive Moments:Center of 1st span M1span = 359.4 kNm k z
As
2 = M / bd fcu = 0.1504
= = = =
k
[ 0.5 + (0.25 - k/0.9)1/2] d 0.788 d < 0.95 d 0.788 x 159 125.3 mm
z
= m x / 0.87 fy z =
As req Provide As prov
Center of interior span Mspan = 307.4 kNm
= T =
2
7169 mm
As =
2
2655 mm /m
2
2655 mm /m 20 @ 2
3142 mm /m
100 mm c/c > As req ==> OK !
As req Provide As prov
2 = M / bd fcu = 0.1287
= [ 0.5 + (0.25 - k/0.9)1/2] d = 0.827 d < 0.95 d = 0.83 x 159 = 131.5 mm = m x / 0.87 fy z =
5841 mm2
= T
2
=
2163 mm /m 20 @ 2
2513 mm /m
=
2163 mm2/m
125 mm c/c > As req ==> OK !
C) Middle Strip Design Negitive Moments:1st Interior Support M1sup = 223.8 kNm k z
As
2 = M / bd fcu = 0.0477
= = = =
k
[ 0.5 + (0.25 - k/0.9)1/2] d 0.944 d < 0.95 d 0.944 x 159 150.1 mm
z
= m x / 0.87 fy z =
As req Provide As prov
Interior Support Msup = 195.4 kNm
= T =
2
3727 mm
As 2
= 703.1 mm /m
2
703 mm /m 12 @
150 mm c/c
2
> As req ==> OK !
754 mm /m
As req Provide As prov
2 = M / bd fcu = 0.0417
= [ 0.5 + (0.25 - k/0.9)1/2] d = 0.951 d > 0.95 d = 0.95 x 159 = 151.1 mm = m x / 0.87 fy z =
3232 mm2
= T
2
=
610 mm /m 12 @
= 609.9 mm2/m 175 mm c/c
2
> As req ==> OK !
646 mm /m
Positive Moments:1st Span M1sup k z
As
2 = M / bd fcu = 0.1132
= = = =
k
[ 0.5 + (0.25 - k/0.9)1/2] d 0.852 d < 0.95 d 0.852 x 159 135.5 mm
z
= m x / 0.87 fy z =
As req Provide As prov
Interior Span Msup = 454.0 kNm
= 530.8 kNm
= T =
2
9784 mm
As =
2
1846 mm /m
2
1846 mm /m 20 @ 2
2094 mm /m
150 mm c/c > As req ==> OK !
As req Provide As prov
2 = M / bd fcu = 0.0968
= = = =
[ 0.5 + (0.25 - k/0.9)1/2] d 0.877 d < 0.95 d 0.877 x 159 139.5 mm
= m x / 0.87 fy z =
8132 mm2
= T
2
=
1534 mm /m 20 @ 2
1571 mm /m
=
1534 mm2/m
200 mm c/c > As req ==> OK !
(D) Deflection Check (base on basic ratio in Table 3.10, BS8110) (calculations are made using total moment at mid-span and the average of the column and middle strip tension steels) Near center of 1st span Near center of 1st span, M1span = +0.083xFxL M / bd
2
=
= 890.2 kNm
4.40
Total rebars provided, Asreq aveg
=(
2655
+
1846 )/2
=
2 2251 mm /m
Total rebars provided, Asprov aveg
=(
3142
+
2094 )/2
=
2 2618 mm /m
fs = (5/8) x fy x (As / As prov) 247 N/mm2 2 Modification = 0.55 + (477 - fs) / [120 (0.9 + mx/bd )] = 0.91 < 2.00 = 0.91 =
Allowable l/d = 0.91 x 26 = 23.69 Actual l/d = 8000 / 159 = 50.31
==> Deflection check, FAIL !!!
Near center of interior span Center of Interior Span, Mspan = +0.071xFxL M / bd
2
=
= 761.5 kNm
3.76
Total rebars required, Asreq
=(
2163
+
1534 )/2
=
2 1849 mm /m
Total rebars provided, Asprov
=(
2513
+
1571 )/2
=
2 2042 mm /m
fs = (5/8) x fy x (As / As prov) 260 N/mm2 2 Modification = 0.55 + (477 - fs) / [120 (0.9 + mx/bd )] = 0.94 < 2.00 = 0.94 =
Allowable l/d = 0.94 x 26 = 24.36 Actual l/d = 8000 / 159 = 50.31
(E) Shear Check
==> Deflection check, FAIL !!!
(base on basic ratio in Clause 3.7.6, BS8110)
Internal Columns Punching shear check Shear, V
= 1.15xnx(Lx x Ly - ho2) =
2 = 1.166 N/mm
Vmax
2 < 0.8xfcu1/2 = 4.733 N/mm ==> Punching shear check, OK !
1526 kN
Shear check @ 1.5d from column face = 0.435 N/mm2
perimeter, u =(2 x 1.5 x d) + ho)x4
Shear stress, v
= 8108 mm shear plane, ls =(2 x 1.5 x d) + ho) =
Top reinforcement provided = 100As/bd = 0.328
Shear, V
2
2.027 m
1340 mm2/m
= 1.15xnx(Lx x Ly - ls )
vc = 0.79x(100As/bd))1/3(400/d)1/4(fcu/25)1/3/1.25
=
vc
1443 kN
= 0.485 N/mm2 ==> 1.5d shear check, OK !
