1
Table of Contents 1. introduction 2. Basic Assumptions 3. DESIGN OF SOLID TOW WAY SLAB 3.1. SLAB 3.2. BEAM 3.3. FORM WORK 3.4. BAR SCHEDULE 4. DESIGN OF FLAT SSLAB 4.1. Slab 4.2. FORM WORK 4.3. BAR SCHEDULE 5. DESIGN OF RIBBED SLAB 5.1. DESIGN OF RIB 5.2. DESIGN OF TOPPING SLAB 5.3. Design of GIRDER 5.4. FORM WORK 5.5. BAR SCHEDULE 6. CONCLUSION and SUMMERY
1.
INTRODUCTION
2
This structural design document presents all the assumptions analysis and design Calculations done in the design Material In the design of the building material properties of concrete C25 and steel S300 are used for structural elements. Slab and beam design and method of analysis The slab and beam is designed according to EBCS-1 and EBCS-2 1995 using direct design method. The partition wall load on each slab panel is also considered properly for each panel area. After calculation of the design actions, shear fore and bending moments, the proper reinforcement is provided for moment and the slab panel against shear. Objective:-
comparision between solid tow way slab, flat slab and ribbed slab
-
To help and identify least cost and most economical
-
To know which is ease to construct and needed minimum construction material
-
Help the designer to identify and understand the baste preferable in terms of design, economy and way of construction
-
To provide the first estimation for the designer
2.
BASIC ASSUMPTION
Dead Load: - Reinforced concrete
=25KN/m 3
- Cement Screed
=23KN/m 3
3
- HCB- 200mm thick
=14KN/m3 =23N/m 3
- Terrazzo tile
Partial Safety Factor for Dead Load =1.30 Live Load: LL= 4KN/m2 Partial Safety Factor for Live Load =1.50 Material Properties Concrete: Grade C-25
Fck = 20 Mpa Fctk =1.5 Mpa Partial Safety Factor = 1.5
F cd = 0.85[20/1.5] = 11.33 Mpa
Fctd = 1.5/1.5 = 1 Mpa Ecm = 29Gpa Reinforcing Steel: Fyk = 300 Mpa Partial Safety Factor =1.15 Es = 200Gpa
Spacing of column: Lx= 5.0m
Ly= 6.0m
Size of Column: 0.5m X 0.5m
3. DESIGN OF SOLID TOW WAY SLAB 3.1
SLAB
4
Fyd = 300/1.15 = 260.87 Mpa
f yk l x d ≥ 0.4 + 0.6 400 βa 300 5000 d ≥ 0.4 + 0.6 = 112mm 400 38
,
Use d = 115mm Total depth D = 115 + 20 + 5 = 140 mm Design Load: Dead Loads Selfe weight = 0.14 * 25 = 3.5 KN/m2 Design constant
Ceiling plaster(15mm) =23 *0.015= 0.345KN/m2
For S-300, C-25, l x = 5m l y = 6m ,
Cement scread(30mm) = 23*0.03 = 0.69KN/m2
φ =10mm Clear cover = 20mm Live Load LL = 4 KN/m2
f cd =
0.68 f c y
γs
=
Terrazzo tile(20mm) = 23*0.02 = 0.46 KN/m2
0.68 * 25 = 11.33 N mm 2 1 .5
Partition wall = 2 KN/m2 Total Gk = 7 KN/m2
fyk 300 fyd = = = 260.87 N / mm 2 γs 1.15 0.21 fck fctd = γs
2 3
0.21( 20) = 1.15
2 3
= 1.03 N / mm
Pd = 1.3DL +1.6 LL = 1.3( 7 ) +1.6( 4 ) Pd = 15.49 KN / m 2
2
oment alculate
αxs = 0.063 αxf = 0.047
Depth determination by Serviceability Limit State method
f yk l x d ≥ 0.4 + 0.6 400 βa
αys = 0.047 αyf = 0.036
,
2 Mxs = α xs * Pd * l x =0.063*15.49 *
The value of βa can be determined: using
ly
52=24.40
6 = 1.2 ⇒ is end lx 5 span. And from EBCS-2 , ratio 2 βa = 30, and ratio 1 βa = 40 the ratio of length
M
=
Mxf = α xf * Pd * l x =0.047*5.49 * 2
52=18.20
Mys = α ys * Pd * l x =0.047*5.49* 52=18.20 2
For ratio 1.2
βa =
( 2 − 1.2)( 30 − 40) + 30 = 38...... by ( 2 − 1)
Myf = α yf * Pd * l x =0.036*5.49* 52=13.94 2
interpolation
5
αxs = 0.056 αxf = 0.042
Mmax.= 24.40KNm
αys = 0.039
d chec =
αyf = 0.030
Mxs = α xs * Pd * l x
M = 0.2952 * b * fcd
24.40 * 10 6 0.295 * 1000 * 11.33
d chec = 85mm < d servc = 115mm Use d servc = 115mm
2
=0.056*15.49* 52=21.69
Calculate Moment Redistribution for longer span
Mxf = α xf * Pd * l x =0.042*5.49 * 2
1. Section 1-1
52=16.27
Mys =18.20, Myf =13.94 ….. for S1
Mys = α ys * Pd * l x =0.039*5.49* 52=15.10 2
Mys =15.04, Myf =11.62 ….. for S1
Myf = α yf * Pd * l x =0.030*5.49* 5 =11.62 2
2
αxs = 0.048 αxf = 0.036 αys = 0.039
∆Mxs 18.20 − 15.10 *100% = *100% = 17% < 20% M max 18.20
αyf = 0.029 2 Mxs = α xs * Pd * l x =
0.048*15.49* 52=18.59
We take average moment method to adjusts support moment.
Mxf = α xf * Pd * l x =0.036*5.49* 52=13.94
Mys =
2
Mys = α ys * Pd * l x =0.039*5.49* 52=15.10 2
18.20 +15.10 = 16.65 KNm 2
Adjustment of moment for span (field) For Panel S1: Moment adjustment coefficients from the table by using .
Myf = α yf * Pd * l x =0.029*5.49* 52=11.23 2
αxs = 0.042 αxf = 0.032
ly lx
α ys = 0.032 α yf = 0.024
=
6 = 1 .2 ⇒ 5
Cx = 0.338, cy = 0.172 ∆M =18.20 −16.65 =1.55 KNm
2 Mxs = α xs * Pd * l x =0.042*15.49*
M adj x = c x ∆M = 0.338 * 1.55 = 0.523KNm
52=16.27
M adj y = c y ∆M = 0.172 * 1.55 = 0.266 KNm
Mxf = α xf * Pd * l x =0.032*5.49* 52=12.39 2
Adjusted span moments
Mys = α ys * Pd * l x =0.032*5.49* 52=12.39 2
Mxf adj = M adj x + Mxf Mxf adj = 0.523 +18.20 =18.723KNm
Myf = α yf * Pd * l x =0.024*5.49* 52=9.30 2
Myf adj = M adj y + Myf Myf adj = 0.266 +13.94 =14.206 KNm
Depth check
6
Myf adj = M adj y + Myf
For Panel S2:
Myf adj = 0.241 +11.231 =11.472 KNm
The support moment is increases, according to EBCS-2 no adjustment requires for span moments. Therefore
For Panel S2: The support moment is increases, according to EBCS-2 no adjustment requires for span moments.
Mxf adj = Mxf =16.27 KNm Myf adj = Myf =11.62 KNm
2. Section 2-2
Therefore
Mys =15.10, Myf =11.23 ….. for S3
Mxf adj = Mxf =12.39 KNm Myf adj = Myf = 9.29 KNm
Calculate Moment Redistribution for shorter span
Mys =12.39, Myf =9.29 ….. for S4
1. Section A-A
Mxs = 24.40, Mxf =18.20 ….. for S1 ∆Mxs 15.10 − 12.39 *100% = *100% = 18% < 20% M max 15.10
Mxs = 18.59, Mxf =13.94 ….. for S3
We take average moment method to adjust support moment.
Mys =
15.10 +12.39 = 13.75KNm 2
∆Mxs 24.40 − 18.59 *100% = *100% = 24% > 20% M max 24.40
Adjustment of moment for span (field)
Therefore we use Moment Distribution Method
For Panel S3: Moment adjustment coefficients from the
Relative stiffness: K1 = K 2 = K 3 =
table by using
ly lx
=
6 = 1 .2 ⇒ 5
I I = l 5
Distribution factors:
DF1 = DF 2 =
Cx = 0.338, cy = 0.172 ∆M =15.10 −13.75 =1.355 KNm M adj x = c x ∆M = 0.338 * 1.355 = 0.458 KNm M adj y = c y ∆M = 0.172 * 1.355 = 0.241KNm
I
5 = 0.5 I +I 5 5
Adjusted support Moment: DF
0.5
0.5
0.5
0.5
Adjusted span moments
Mxf adj = M adj x + Mxf
FEM
-24.40
18.59
-18.59
24.40
Mxf adj = 0.458 +13.942 =14.40 KNm
Adj.
2.905
2.905
-2.905
-2.905
21.495 -21.495
21.495
Madj -21.495
7
Adjusted support Moment Mxs = 21.495
Distribution factors:
Adjustment of moment for span (field) For Panel S1:
DF1 = DF 2 =
Moment adjustment coefficients from the table by using
ly lx
=
I
5 = 0.5 I +I 5 5
Adjusted support Moment:
6 = 1 .2 ⇒ 5
DF
0.5
0.5
0.5
0.5
Cx = 0.344, cy = 0.364 FEM
-21.69
16.27
-16.27
24.40
M adj x = c x ∆M = 0.344 * 2.905 = 0.999 KNm
Adj.
2.712
2.712
-2.712
-2.712
M adj y = c y ∆M = 0.364 * 2.905 =1.057 KNm
Madj -18.978
18.978 -18.978
18.978
Adjusted span moments
Adjusted support Moment Mxs = 18.978
Mxf adj = M adj x + Mxf
Adjustment of moment for span (field)
Mxf adj = 0.999 +18.20 =19.20 KNm
For Panel S2:
∆M = 24.4 − 21.495 = 2.905 KNm
Myf adj = M adj y + Myf
Moment adjustment coefficients from the table by using
Myf adj =1.057 +13.94 =15.00 KNm
ly
For Panel S3:
lx
The support moment is increases, according to EBCS-2 no adjustment requires for span moments. Therefore
=
6 = 1 .2 ⇒ 5
Cx = 0.344, cy = 0.364 ∆M = 21.69 −18.978 = 2.712 KNm M adj x = c x ∆M = 0.344 * 2.712 = 0.933KNm
Mxf adj = Mxf =13.94 KNm
M adj y = c y ∆M = 0.364 * 2.712 = 0.987 KNm
Myf adj = Myf =11.23KNm
1. Section B-B
Adjusted span moments
Mxs = 21.69, Mxf =16.27 ….. for S2
Mxf adj = M adj x + Mxf Mxf adj = 0.933 +16.27 =17.203KNm
Mxs = 16.27, Mxf =12.39 ….. for S4
Myf adj = M adj y + Myf Myf adj = 0.987 +11.62 =12.607 KNm For Panel S4: The support moment is increases, according to EBCS-2 no adjustment requires for span moments.
∆Mxs 21.69 − 16.27 * 100% = * 100% = 25% > 20% M max 21.69 Therefore we use Moment Distribution Method
Therefore
I I Relative stiffness: K1 = K 2 = K 3 = = l 5
Mxf adj = Mxf =12.39 KNm Myf adj = Myf = 9.29 KNm
Maximum Moments on the panel
8
Panel -S1
Mxs = 21.495KNm
Mxf =19.20KNm
Mys =16.652KNM 14.997KNm
Myf =
2M ρcalc = 1 − 1 − 2 bd fcd
Panel- S2
Mxs = 18.978KNm
Mxf =
17.203KNm Mys =16.652KNM 12.607KNm
Myf =
Panel- S3
Mxs = 21.495KNm
Mxf =
14.409KNm Mys =13.75KNM
Myf =11.472KNm
Panel- S4
Mxs = 18.978KNm
Mxf =
13.749KNm Mys =13.75KNM
Myf =9.295KNm
Reinforcement b= 1000mm,
d = 115mm
φ =10mm
Minimum Reinforcement:
ρ min =
0.5 0.5 = = 0.002 fyk 300
As min = ρbd = 0.002 * 1000 * 115 = 230mm 2 Calculated reinforcement:
S=
b * ab As
9
fcd fyd
PANEL
MOMENT TYPE
ρ
VALUE
Mxs Mxf PANEL1 Mys Myf Mxs Mxf PANEL2 Mys Myf Mxs Mxf PANEL3 Mys Myf Mxs Mxf PANEL4 Mys Myf Load transfer to beam:
21.50 19.39 16.65 15.15 18.98 17.36 16.65 12.73 21.50 14.54 13.75 11.58 18.98 12.51 13.75 9.38
Ascal (mm2)
0.0068 0.0060 0.0051 0.0046 0.0059 0.0054 0.0051 0.0039 0.0068 0.0044 0.0042 0.0035 0.0059 0.0038 0.0042 0.0028
Vx = βvx.Pd .Lx Vy = βvy.Pd .Lx
776.92 694.64 589.83 533.49 678.79 616.74 589.83 444.07 776.92 510.78 481.55 402.19 678.79 436.03 481.55 323.12
SPACING (mm)
SPACING PROVIDED (mm)
145.63 162.88 191.82 212.08 166.68 183.45 191.82 254.78 145.63 221.51 234.96 281.32 166.68 259.48 234.96 350.16
145 160 190 210 165 180 190 250 145 220 230 280 165 255 230 350
Panel S-3: Shear force coefficients for uniformly loaded rectangular panel is:
Panel S-1:Shear force coefficients for uniformly loaded rectangular panel is: Lx = 5m Pd=, =15.491KN
Lx = 5m , Pd = 15.491KN
βvcx =0.42 βvcy =0.36 6 = = 1.2 ⇒ βvdx =0 lx 5 βvdy =0.24
βvcx =0.47 ly 6 βvcy =0.40 = = 1.2 ⇒ βvdx =0.31 lx 5 βvdy =0.26
ly
Vxd = βvdx.Pd .Lx = 0 Vxd = βvdx.Pd .Lx = 0.31 * 15.491 * 5 = 24.01 Vxc = βvdx.Pd .Lx = 0.42 * 15.491 * 5 =32.53 Vxc = βvdx.Pd .Lx = 0.47 * 15.491 * 5 =36.40 Vyd = βvcy.Pd .Lx = 0.24 * 15.491 * 5 =18.59 Vyd = βvcy.Pd .Lx = 0.26 * 15.491 * 5 = 20.14 Vyc = βvcy.Pd .Lx = 0.36 * 15.491 * 5 = 27.88 Vyc = βvcy.Pd .Lx = 0.40 * 15.491 * 5 = 30.98
Panel S-2:
Panel S-2:
Shear force coefficients for uniformly loaded rectangular panel is:
Shear force coefficients for uniformly loaded rectangular panel is:
Lx = 5m , Pd = 15.491KN
Lx = 5m , Pd = 15.491KN
βvcx =0.44 βvcy =0.36 6 = = 1.2 ⇒ βvdx =0.29 lx 5 βvdy =0
ly
Vxd = βvdx.Pd .Lx = 0.29 * 15.491 * 5 = 22.46 Vxc = βvdx.Pd .Lx = 0.44 * 15.491 * 5 = 34.08 Vyd = βvcy.Pd .Lx = 0 Vyc = βvcy.Pd .Lx = 0.36 * 15.491 * 5 = 27.88
10
βvcx =0.39 ly 6 βvcy =0.33 = = 1.2 ⇒ βvdx =0 lx 5 βvdy =0
Vxd = βvdx.Pd .Lx = 0 Vxc = βvdx.Pd .Lx = 0.39 * 15.491 * 5 =30.21 Vyd = βvcy.Pd .Lx = 0 Vyc = βvcy.Pd .Lx = 0.33 * 15.491 * 5 = 25.56
Ly- longer direction axis Vcx 1-discontineous 2-contineous 36.4 1-discontineous 2-contineous 34.08 2-contineous 32.53 2-contineous 32.53 2-contineous 30.21 2-contineous 30.21
Panel S-1 S-2 S-3 S-4
3.2
0.68 f c y
fyd = fctd =
γs
Mu = µ sd * f cd * bw
d≥
=
0.68 * 20 = 9.07 N mm 2 1 .5
2 3
=
Vdy 20.14 18.59 -
95.84 *10 6 0.295 * 11.33 * 2500
d ≥ 338.67 mm Use d = 357mm
fyk 300 = = 260.87 N / mm 2 γs 1.15 0.21 fck γs
Vdx 24.01 22.46 -
check depth:
Beam design
design constant
f cd =
Lx- Shorter direction Vcy 1-discontineous 2-contineous 30.98 2-contineous 27.88 2-contineous 27.88 1-discontineous 2-contineous 27.88 2-contineous 25.56 2-contineous 25.56
Beam span AB & CD , Msd= 79.79 KNm
2 3
0.21( 20) = 1.03 N / mm 2 1.15
µ sd
M sd 79.79 *10 6 = = = 0.22 < 0.295.....ok! bd 2 f cd 250 * 357 2 *11.33
From GDC for μsd = 0.22, { Kz = 0.87
Longer Direction
Reinforcment
As =
M sd 79.79 *10 6 = = 984.77 mm 2 k z df yd 0.87 * 357 * 260.87
Provide 4 Φ20 bar
Beam span BC, Msd= 20.98 KNm
µ sd =
M sd 20.98 *10 6 = = 0.058 < 0.295.....ok bd 2 f cd 250 * 357 2 *11.33
From GDC for μsd = 0.22, { Kz = 0.96 Reinforcment
b = 250mm, D = 400mm, cc=25mm, Φmain=20mm, stirrup Φ=8mm ,d= 357mm
As =
11
M sd 20.98 *10 6 = = 234.96mm 2 k z df yd 0.96 * 357 * 260.87
Provide 2 Φ20 bar
For beam section 2,3,4,5
vc = 0.25 f ctd k1k 2 bd = 48.42 KN
For support B & C, Msd = 95.84
Design shear for span AB & CD M sd 95.84 *10 6 = = 0.265 < 0.295.....ok ! = 88.07 Vc,max = 48.42 Vsd,max 2 2 bd f cd 250 * 357 *11.33 Vs = Vsd − Vc = 88.07 − 48.42 = 39.65 KN From GDC for μsd = 0.265, { Kz = 0.83
µ sd =
Reinforcment
S=
M sd 95.84 *10 6 As = = =1239.87 mm 2 k z df yd 0.83 * 357 * 260.87 Provide 4 Φ20 bar
adf yd Vs
=
100 * 357 * 260.87 = 234.88mm 39.65 * 103
af yk 100 * 300 = 300mm = 0.5b 0.5 * 250 0.5d = 178.5mm
S max = <,
Shear design S max = 175mm Use
φ8c / c...175mm Design shear for span BC Vsd,max = 68.59
Vs = Vsd − Vc = 20.17 KN
By similarity of triangle: design shear
66.12( 2.5 − 0.357 ) Vsd1 = Vsd 6 = = 56.68KN 2.5 Vsd 2 = Vsd 5 =
98.07( 3.5 − 0.357 ) = 88.07 KN 3.5
Vsd 3 = Vsd 4 =
77.85( 3 − 0.357 ) = 68.59 KN 3
Resistance shear
Vrd = 252.80KN >> Vsd (allo) Shear Capacity
v c = 0.25 f ctd k1 k 2 bd
k 2 = 1.6 − d = 1.243 ≥ 1.....ok! φ = 20
k1 = 1 + 50
S=
As = 1239.87 mm 2 As 1239.87 ρ= = bd 250 * 357
1239.87 = 1.695 ≤ 2.....ok! 250 * 357 12
adf yd Vs
=
max
100 * 357 * 260.87 = 461.73mm 20.17 * 10 3
af yk 100 * 300 = 300mm = 0.5b 0.5 * 250 0.5d = 178.5mm
S max = <,
S
Vrd = 0.25 f ctd bd = 0.25 * 11.33 * 250 * 357
At support
Vc,max = 48.42
=
175mm
φ8c / c...175mm
……….Use
As =
M sd 235.66 *10 6 = = 2325.55mm 2 k z df yd 0.85 * 457 * 260.87
Provide 8 Φ20 bar Shear design b = 400mm, D = 500mm, cc=25mm, Φmain=20mm, stirrup Φ=8mm ,d= 457mm
check depth:
Mu = µ sd * f cd * bw
d≥
257.67 * 10 6 0.295 * 11.33 * 400
d ≥ 439.04mm Use d = 457mm
By similarity of triangle: design shear
Vsd1 = Vsd 6 =
Beam span AB & CD , Msd= 215.68 KNm
µ sd =
6
M sd 215.68 *10 = 2 bd f cd 400 * 457 2 *11.33
Vsd 2 = Vsd 5 = = 0.228 < 0.295.....ok! Vsd 3 = Vsd 4 =
From GDC for μsd = 0.22, { Kz = 0.86
Shear Capacity
Beam span BC, Msd= 55.52 KNm
v c = 0.25 f ctd k1 k 2 bd
k 2 = 1.6 − d = 1.143 ≥ 1.....ok! M sd 55.52 *10 = = 0.057 < 0.295.....ok! 2 2 bd f cd 400 * 457 *11.33 φ = 20 6
From GDC for μsd = 0.22, { Kz = 0.96
At support
Reinforcment
As =
208.8( 3 − 0.457 ) = 176.99 KN 3
Vrd = 0.25 f ctd bd = 0.25 * 11.33 * 400 * 457 M sd 215.68 *10 6 = = 2103.64mm 2 Vrd = 517.80KN >> Vsd (allo) k z df yd 0.86 * 457 * 260.87
Provide 7 Φ20 bar
µ sd =
264.46( 3.5 − 0.457 ) = 229.93KN 3.5
Resistance shear
Reinforcment
As =
178.57( 2.5 − 0.457 ) = 145.93KN 2.5
k1 = 1 + 50
M sd 55.52 * 10 6 = = 485.10mm 2 k z df yd 0.96 * 457 * 260.87
As = 2325.55mm 2
2325.55 = 1.636 ≤ 2.....ok! 400 * 457
For beam section
vc = 0.25 f ctd k1k 2 bd = 88.02 KN
Provide 2 Φ20 bar
Design shear for span AB & CD Vsd,max = 299.93 Vc,max = 88.02
For support B & C, Msd = 235.66
M sd 235.66 *10 6 Vs = Vsd − Vc = 141.91KN = = 0.249 < 0.295.....ok! 2 2 bd f cd 400 * 457 *11.33 adf yd 100 * 457 * 260.87 From GDC for μsd = 0.249, { Kz = 0.85 S= = = 84mm Vs 141.91 * 103
µ sd =
Reinforcment
13
af yk 100 * 300 = 300mm = 0 . 5 b 0 . 5 * 250 0.5d = 228.5mm
S max = <,
S max = 225mm
.Use
Beam span AB & CD , Msd= 48.41 KNm
V sd,max = 176.99
µ sd =
adf yd Vs
=
Reinforcment
100 * 457 * 260.87 = 129mm 91.97 * 10 3
S max = <,
As =
af yk 100 * 300 = = 300mm 0.5b 0.5 * 250 0.5d = 228.5mm
M sd 48.41*10 6 = = 565.86mm 2 k z df yd 0.92 * 357 * 260.87
Provide 2 Φ20 bar
Beam span BC, Msd= 20.98 KNm
µ sd =
S max = 225mm Use
M sd 48.41 *10 6 = = 0.133 < 0.295.....ok bd 2 f cd 250 * 357 2 *11.33
From GDC for μsd = 0.133, { Kz = 0.92
Vs = Vsd − Vc = 91.97 KN
S=
67.69 *10 6 0.295 * 11.33 * 250
d ≥ 284.62mm Use d = 357mm
φ8c / c...80mm
Design shear for span BC: Vc,max = 88.02
Mu = µ sd * f cd * bw
d≥
φ8c / c...125mm
M sd 12.34 *10 6 = = 0.058 < 0.295.....ok bd 2 f cd 250 * 357 2 *11.33
From GDC for μsd = 0.22, { Kz = 0.96
Shorter Direction
Reinforcment
As =
M sd 20.98 *10 6 = = 234.96mm 2 k z df yd 0.96 * 357 * 260.87
Provide 2 Φ20 bar For support B & C, Msd = 67.69
µ sd =
M sd 67.69 *10 6 = = 0.185 < 0.295.....ok bd 2 f cd 250 * 357 2 *11.33
From GDC for μsd = 0.185, { Kz = 0.91 Reinforcment
As =
M sd 67.69 *10 6 = = 7948.71mm 2 k z df yd 0.91* 357 * 260.87
Provide 3 Φ20 bar b = 250mm, D = 400mm, cc=25mm, Φmain=20mm, stirrup Φ=8mm ,d= 357mm
Shear design
check depth:
14
af yk 100 * 300 = 300mm = 0 . 5 b 0 . 5 * 250 0.5d = 178.5mm
S max = <,
S max = 175mm Use
By similarity of triangle: design shear
φ8c / c...175mm Design shear for span BC
Vsd1 = Vsd 6 =
48.04( 2 − 0.357 ) = 39.465 KN 2
Vsd 2 = Vsd 5 =
71.11( 3 − 0.357 ) = 62.65 KN 3
Vsd 3 = Vsd 4 =
56.02( 2.5 − 0.357 ) = 48.02 KN 2.5
Vsd,max = 48.02
Vs = Vsd − Vc = 6.68 KN
S=
adf yd Vs
=
100 * 357 * 260.87 = 1394.17mm 6.68 * 10 3
Resistance shear
Vrd = 0.25 f ctd bd = 0.25 * 11.33 * 250 * 357
Vc,max = 41.34
S max = <,
Vrd = 252.80KN >> Vsd (allo) Shear Capacity
S max = 175mm
v c = 0.25 f ctd k1 k 2 bd
Use
af yk 100 * 300 = = 300mm 0.5b 0.5 * 250 0.5d = 178.5mm
φ8c / c...175mm
k 2 = 1.6 − d = 1.243 ≥ 1.....ok!
φ = 20
At support
As = 794.26mm 2
k1 = 1 + 50
794.26 = 1.445 ≤ 2.....ok! 250 * 357
For beam section
vc = 0.25 f ctd k1k 2 bd = 41.34 KN Design shear for span AB & CD Vsd,max = 62.65 Vc,max = 41.34
Vs = Vsd − Vc = 21.31KN
S=
adf yd Vs
=
100 * 357 * 260.87 = 437 mm 21.31 * 10 3 b = 350mm, D = 400mm, cc=25mm, Φmain=20mm, stirrup Φ=8mm ,d= 357mm
check depth:
15
Mu = µ sd * f cd * bw
d≥
147 * 10 6 0.295 * 11.33 * 350
d ≥ 354.48mm Use d = 357mm Beam span AB & CD , Msd= 125.36 KNm
µ sd =
M sd 125.36 *10 6 = = 0.248 < 0.295.....ok! By similarity of triangle: design shear bd 2 f cd 350 * 357 2 *11.33
From GDC for μsd = 0.248, { Kz = 0.85
124.07( 2 − 0.357 ) = 101.92 KN 2
Vsd1 = Vsd 6 =
Reinforcment
182.87( 3 − 0.357 ) = 161.12 KN 3
Vsd 2 = Vsd 5 = M sd 125.36 *10 6 As = = =1583.61mm 2 k z df yd 0.85 * 357 * 260.87 Provide 5 Φ20 bar
Vsd 3 = Vsd 4 =
Beam span BC, Msd= 29.34 KNm
Resistance shear
µ sd
141.07( 2.5 − 0.357 ) = 120.92 KN 2.5
Vrd = 0.25 f bd = 0.25 * 11.33 * 350 * 357
ctd M sd 29.34 *10 6 = = = 0 . 058 < 0 . 295 ..... ok ! Vrd = 353 . 92 KN >> Vsd (allo) bd 2 f cd 350 * 357 2 *11.33
From GDC for μsd = 0.058, { Kz = 0.96
Shear Capacity
Reinforcment
v c = 0.25 f ctd k1 k 2 bd
M sd 29.34 *10 6 As = = = 328.17 mm 2 k z df yd 0.96 * 357 * 260.87
k 2 = 1.6 − d = 1.243 ≥ 1.....ok!
φ = 20
At support
As = 1924.91mm 2
Provide 2 Φ20 bar
k1 = 1 + 50
For support B & C, Msd = 147KNm
µ sd =
For beam section M sd 147 *10 6 = = 0.291 < 0.295.....ok! 2 2 bd f cd 350 * 357 *11.33 v = 0.25 f k k bd = 70.79 KN c
From GDC for μsd = 0.291, { Kz = 0.82
ctd
1
2
Design shear for span AB & CD Vsd,max = 161.12 Vc,max = 70.79
Reinforcment
As =
1924.91 = 1.77 ≤ 2.....ok! 350 * 357
Vs = Vsd − Vc = 90.33KN M sd 147 *10 6 = =1924.91mm 2 k z df yd 0.82 * 357 * 260.87
S=
Provide 7 Φ20 bar Shear design
16
adf yd Vs
=
100 * 357 * 260.87 = 103.10mm 90.33 * 10 3
Ab1=(0.3*(15.05+18.05)2)=19.83 ….for (250x400) Ab2=(0.4*15.05)2)=12.04 …..for (350x450) Ab3=(0.45*18.05)2)=16.245 …..for (400x500)
af yk 100 * 300 = 300mm = 0 . 5 b 0 . 5 * 250 0.5d = 175mm
S max = <,
As - Ab=271.6525-19.83-12.0416.245=223.54m2
S max = 175mm Use
φ8c / c...100mm
For beams:
Design shear for span BC Vsd,max = 120.92
Vc,max = 70.79
=(0.3*(15.05+18.05)2)+2(0.4*(15.05+18.05)*2=72.82
Ab2=(0.4*15.05)*2)+(2*(0.4*15.05)*2)=36.12
Vs = Vsd − Vc = 50.13KN
S=
adf yd Vs
=
100 * 357 * 260.87 = 185.78mm 50.13 * 10 3
S max = <,
af yk 100 * 300 = 300mm = 0.5b 0.5 * 250 0.5d = 175mm
Total area of form work = 377.603m2
3.4
S max = 175mm Use
Ab3=(0.45*18.05*2)+(2*0.4*18.05*2)=45.125
φ8c / c...175mm
3.3 FORM WORK FOR SOLID TOW WAY SLAB For Slab: As - Ab As = (15.05*18.05)=271.6525m2
17
BAR SCHEDULE
Ref Location
No
Panel1
Panel2 slab Panel3
Panel4
No of Bars
Dia
Dia
Dia
Dia
Dia
Dia
Dia
6
8
10
12
14
16
20
810.00
0.00
0.00
810.00
0.00
0.00
0.00
0.00
1056.12 448.00
0.00 0.00
0.00 0.00
1056.12 448.00
0.00 0.00
0.00 0.00
0.00 0.00
0.00 0.00
2
572.76
0.00
0.00
572.76
0.00
0.00
0.00
0.00
1 1 1
252.00 392.66 0.00
0.00 0.00 0.00
0.00 0.00 0.00
252.00 392.66 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
x
1
253.08
0.00
0.00
253.08
0.00
0.00
0.00
0.00
x
1
307.80
0.00
0.00
307.80
0.00
0.00
0.00
0.00
2 2 0 1 1 0
x x x x x x
1 1 1 1 1 1
290.00 184.00 0.00 96.00 125.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00
290.00 184.00 0.00 96.00 125.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00
0
x
1 0.00 m Kg/m
0.00 0.00 0.222
0.00 0.00 0.395
0.00 4787.42 0.617
0.00 0.00 0.888
0.00 0.00 1.208
0.00 0.00 1.578
0.00 0.00 2.468
0
0
2954
0
0
0
0
Di a
Length
T. Length
10
8.10
25
x
2
x
2
10 10
6.77 4.00
39 28
x x
2 2
x x
2 2
10
3.33
43
x
2
x
10 10 10
6.00 6.77 0.00
21 29 0
x x x
2 2 0
x x x
10
3.33
38
x
2
10
8.10
19
x
2
10 10 10 10 10 10
5.00 4.00 0.00 6.00 5.00 0.00
29 23 0 16 25 0
x x x x x x
10
0.00
0
x
No of Bars
Kg
18
BAR SCHEDULE FOR BEAMS Location
Ref No
Di a
Length
No of Bars No of Bars
total Length
19
Dia 6
Dia 8
Dia 10
Dia 12
Dia 14
Dia 16
Dia 20
axis1,4
Span support
shear rfmt
axis2,3
Span support
shear rfmt
axis A&B
Span support
shear rfmt
axis C&D
Span support
shear rfmt
AB,CD BC B,C AB,CD BC AB,CD BC AB,CD BC B,C AB,CD BC AB,CD BC AB,CD BC B,C AB,CD BC AB,CD BC AB,CD BC B,C AB,CD BC AB,CD BC
20 20 20 20 20 8 8 20 20 20 20 20 8 8 20 20 20 20 20 8 8 20 20 20 20 20 8 8
6.00 6.00 4.00 6.35 6.00 1.11 1.11 6.00 6.00 4.00 6.45 6.00 1.70 1.70 5.00 5.00 3.33 5.35 5.00 1.11 1.11 5.00 5.00 3.33 5.35 5.00 1.30 1.30
4 2 2 2 2 36 36 7 2 6 2 2 76 49 2 2 1 2 2 30 30 5 2 5 2 2 51 30
x x x x x x x x x x x x x x x x x x x x x x x x x x x x
2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 1 2 2 1 2 1
x x x x x x x x x x x x x x x x x x x x x x x x x x x
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
96.00 24.00 32.00 50.80 24.00 159.84 79.92 168.00 24.00 96.00 51.60 24.00 516.80 166.60 40.00 20.00 13.32 42.80 20.00 133.20 66.60 100.00 20.00 66.60 42.80 20.00 265.20 78.00
m Kg/m Kg
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 159.84 79.92 0.00 0.00 0.00 0.00 0.00 516.80 166.60 0.00 0.00 0.00 0.00 0.00 133.20 66.60 0.00 0.00 0.00 0.00 0.00 265.20 78.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
96.00 24.00 32.00 50.80 24.00 0.00 0.00 168.00 24.00 96.00 51.60 24.00 0.00 0.00 40.00 20.00 13.32 42.80 20.00 0.00 0.00 100.00 20.00 66.60 42.80 20.00 0.00 0.00
0.00 0.222 0
1466.16 0.395 579
0.00 0.617 0
0.00 0.888 0
0.00 1.208 0
0.00 1.578 0
975.92 2.468 2409
Live Load LL = 4 KN/m2
f cd =
4. 4.1
DESIGN OFFLAT SLAB
0.68 f c y
fyd =
SLAB
fctd =
γs
=
0.68 * 25 = 11.33 N mm 2 1 .5
fyk 300 = = 260.87 N / mm 2 γs 1.15 0.21 fck γs
2 3
=
2 3
0.21( 20) = 1.03 N / mm 2 1.15
PROPORTIONING AND DIMENSHINING VARIOUS PARTS Select Slab thikness: Depth determination by Serviceability Limit State method
f l d ≥ 0.4 + 0.6 yk x 400 βa ly Design constant
=
6 = 1.2 ⇒ From EBCS-2 for End span 5
lx βa = 24, Le = Ly
For S-300, C-25, l x = 5m l y = 6m ,
φ =14mm Clear cover = 20mm, Column size: 500 X 500mm
20
,
f yk l y d ≥ 0.4 + 0.6 400 βa
Depth of drop panel in shorter direction dd2 = 380 – 20 – 1.5(14) = 339mm
300 6000 d ≥ 0.4 + 0.6 = 212.5mm 400 24
dd effective = ds eef =
353 + 339 = 346mm 2
Design Load:
Total depth D =212.5 + 20 + 7= 239.5 mm
Dead Loads :
Use Ds= 250mm
Average effective depth
Dav =
Slab Strips: According to EBCS-2
Ds + Dd 250 + 380 = = 315mm 2 2
Shorter direction:
Self weight = 0.315 * 25 = 7.875 KN/m 2
Half column strip =
Ceiling plaster (15mm) =23 *0.015= 0.345KN/m2
0.25l x = 0.25 * 5000 = 1250mm Middle strip =
Cement screed (30mm) = 23*0.03 = 0.69KN/m2
Longer direction:
Terrazzo tile(20mm) = 23*0.02 = 0.46 KN/m2
Half column strip =
Partition wall = 2 KN/m2 outer wall (66m length) = 3*0.2*14*66/(18*15)= 2KN/m2
l x − 2( 0.25l x ) = 5000 − 2 * 0.25 * 5000 = 2500mm
0.25l x = 0.25 * 5000 = 1250mm Middle strip =
Total Gk = 13.37 KN/m2, Qk = 4KN/m2
l x − 2( 0.25l x ) = 6000 − 2 * 0.25 * 5000 = 3500mm
Drop panel: Dd, from center of column Minimum length =
Pd = 1.3DL +1.6 LL = 1.3(13.37 ) +1.6( 4 ) Pd = 23.781 KN / m 2
l x 500 = = 1667mm 3 3
Check of shear
Use 1700mm
The governing factor for flat slab is punching shear.
Thickness of drop panel:= (1.25 to 1.50)Ds Consider Dd = 1.5Ds = 1.5 * 250 = 375mm. Use Dd = 380mm
According to ACI and EURO cod the critical point considers to be 0.5d distance from face of column.
Effective depth:
Case -1: middle column and drop panel
Depth of slab in Longer direction ds1 = 250 – 20 – 7 = 223mm Depth of slab in shorter direction ds2 = 250 – 20 – 1.5(14) = 209mm ds effective = dseef =
223 + 209 = 216mm 2
Depth of drop panel in Longer direction dd1 =380 – 20 – 7 = 353mm
21
Ud =4(1700+216)=7664 Punching shear area: Around column: Acolumn = Uc*ddeff=3384*346=1170864mm2 Around drop: Adrop = Ud * dseff= 7664*216=1655424mm2 Total area for load to punching shear consider is A = 6 * 5 = 30m2 Punching force area: Around column: ac = (0.846)2 Around drop: For column:
ad = (1.916)2
Capacity of concrete for shear
Punching force: Around column:
(
ρ min =
)
Vdv = ( A − a c ) Pd = 30 − ( 0.846 ) 23.78
v c = 0.5 f ctd k1 k 2
2
Vdv = 696.38KN
0.5 0.5 = = 0.002 fyk 300
Punching stress: Around column:
k1 =1 + 50( 0.002 ) ≤ 2 Vd =
k1 =1.083 ≤ 2.....ok!
Vdv 696.38 * 10 3 = puncingsheararea 1170864
For column:
Vd = 0.595 N / mm 2
k 2 = 1.6 − d deff = 1.6 − 0.346 = 1.254 ≥ 1
Vd < Vc ……. Is ok!
ok!
Punching force: Around drop:
For drop
(
)
Vdv = ( A − a c ) Pd = 30 − (1.916 ) 23.78
k 2 =1.6 − d deff =1.6 − 0.216 =1.384 ≥1
Vdv = 696.38 KN
ok!
2
Punching stress: Around drop:
For column:
vc = 0.5 * 1.03 *1.083 * 1.253 = 0.699 N / mm For column:
2
Vd =
Vdv 626.102 * 10 3 = puncingsheararea 1655424
Vd = 0.378 N / mm 2
vc = 0.5 * 1.03 * 1.083 * 1.384 = 0.772 N / mm 2 Vd < Vc ……. Is ok! Punching perimeter:
Hence the depth of the drop and the slab is adequate against punching.
Around column:
Case – 2: At the edge column and drop panel
Uc = 4(Cl + dd-eff) =4(500+346) = 3384mm Around drop: Ud=4(droplength+ds)
22
Vd =
Vdv 343.16 * 10 3 = puncingsheararea 758432
Vd = 0.452 N / mm 2 Vd < Vc ……. Is ok! Punching force: Around drop:
Vdv = ( A − a c ) Pd = (15 − 2.314) 23.78 Vdv = 301.66 KN Punching stress: Around drop: Punching perimeter:
Vd =
Around column:
Vdv 301.66 * 10 3 = puncingsheararea 935712
Vd = 0.322 N / mm 2
d d .eff Uc = ( Cl + d d .eff ) + 2 Cl + 2 346 Uc = ( 500 + 346 ) + 2 500 + = 2192mm 2
Vd < Vc ……. Is ok! Hence the depth of the drop and the slab is adequate against punching. Case – 3: At the coner of column and drop panel
Around drop:
l + d d .eff Ud = ( l + d s .eff ) + 2 2 (1700 + 216) = 4332mm Ud = (1700 + 216) + 2 2 Punching shear area: column:
Around
Acolumn = Uc*ddeff=2192*346=758432mm2 Around drop: Adrop = Ud * dseff= 4332*216=935712mm2 Total area for punching shear consider is A = 3 * 5 = 15m2
Punching perimeter:
Punching force area: Around column:
d d .eff Uc = 2 Cl + 2
ac = (0.846*0.673)=0.5694m2
346 = 2 500 + = 1346mm 2
Around drop:
Around drop: ad = (1.916*1.208)=2.314m
Around column:
l + d d .eff Ud = 2 2
2
(1700 + 500 + 216) = 2416mm = 2 2
Punching shear area:
Punching force: Around column:
Around column:
Acolumn = Uc*ddeff=1346*346=465716mm2
Vdv = ( A − a c ) Pd = (15 − 0.5694) 23.78
Around drop:
Vdv = 343.16 KN
Adrop = Ud * dseff= 2416*216=521856mm2
Punching stress: Around column:
23
Total area for punching shear is A = 3 *2.5 = 7.5m2
Moment along the longer direction:
L = ly −
Punching force area: Around column: ac = (0.673)2 Around drop: ad = (1.208)2 Punching force: Around column:
(
For interior Msy = 0.055 FL = 0.55 PdAL Msy = 0.055 * 23.78 * 30 * 5.624 = 220.67 KNm
)
Vdv = ( A − a c ) Pd = 7.5 − ( 0.673) 23.78 2
Vdv = 167.58 KN
Mfy =0.071FL =0.071PdAL Mfy =0.071* 23.78 * 30 * 5.624 = 284.86 KNm
Punching stress: Around column:
For end span
Vdv 167.58 *103 = V Vd = puncingsheararea 465716
Msy A = 0.040 FL = 0.040 PdAL
d
Msy A = 0.040 * 23.78 * 30 * 5.624 = 160.49 KNm
2 < Vd = 0.360 N / mm Vc . ok!
Msy B = 0.063FL = 0.063PdAL Msy B = 0.063 * 23.78 * 30 * 5.624 = 252.77 KNm
Punching force: Around drop:
(
)
Vdv = ( A − a c ) Pd = 7.5 − (1.208) 23.78 Vdv = 143.65KN
2hc 2 * 0.564 =6− = 5.624m 3 3
2
Mfy = 0.083FL = 0.083PdAL Mfy = 0.083 * 23.78 * 30 * 5.624 =333KNm
Moment along the shrter direction:
Punching stress: Around drop:
2hc 2 * 0.564 =5− = 4.624m 3 3
Vdv 143.65 * 10 3 Vd = = puncingsheararea 521856
L = lx −
Vd = 0.275 N / mm 2
For interior span
Vd < Vc ……. Is ok!
Msx = 0.055 FL = 0.55 PdAL Msx = 0.055 * 23.78 * 30 * 4.624 =181.43KNm
Hence the depth of the drop and the slab is adequate against punching.
Mfx = 0.071FL = 0.071PdAL Mfx = 0.071 * 23.78 * 30 * 4.624 = 234.21KNm
DESIGN FOR FLEXURE
For end span
Msx A = 0.040 FL = 0.040 PdAL Msx A = 0.040 * 23.78 * 30 * 4.624 = 131.95 KNm Msx B = 0.063FL = 0.063PdAL Msx B = 0.063 * 23.78 * 30 * 4.624 = 207.82 KNm
F (KN)= (1.3 Gk + 1.6 Qk)A = Pd*A
Mfx = 0.083FL = 0.083PdAL Mfx = 0.083 * 23.78 * 30 * 4.624 = 273.80 KNm
A = 5 x 6 = 30m2
L = li − hc =
2hc 3
LATERAL DISTRIBUTION OF MOMENTS
, where
2
4dc = Π
From EBCS-2
4 * 500 Π
2
= 564mm
Negative moment:-for column strip=75% :- for middle strip=25%
dc= column width
Positive moment:- for column strp= 55%
24
:- for middle strip=45%
Span
Distri. span
Location End span LONG Middle span End span SHORT Middle span
Total moment
Moment in
Adjusted moment
Support1 Center
-160.49 333
column strip 120.37 183.15
Support2
-252.77
189.58
63.19
0.23
175.04
77.73
Support1
-220.67 284.86 -131.95 273.82 -207.82 -181.43 234.21
165.50
55.17
0.23
152.81
67.86
156.67 98.96 150.60 155.87 136.07 128.82
128.19 32.99 123.22 51.96 45.36 105.39
0.23 0.32 0.32 0.32 0.32 0.32
127.19 88.41 111.17 139.24 121.56 95.09
157.67 43.54 162.65 68.58 59.87 139.12
Point
Center Support1 Center Support2 Support1 Center
Middle strip 40.12 149.85
Adjustment factor of moment 0.23 0.23
column strip 111.14 148.68
Middle strip 49.35 184.32
Note:
Column strip is 1.7m and the rest lengthe divied by middle length of each side to get adj. factor
- adjustment factor is multiplied by middle strip moment. And add at middle strip moment and subtracts from column strip moment.
25
COMPUTE REINFORCMENT & C/C SPACING
Span
Edge
Interior
Moment (KNm) col L(-ve) col L(+ve) col L(-ve) col S(-ve) col S(+ve) col S(-ve) Mid L(-ve) Mid L(+ve) Mid L(-ve) Mid S(-ve) Mid S(+ve) Mid S(-ve) col L(-ve) col L(+ve) col S(-ve) col S(+ve) Mid L(-ve) Mid L(+ve) Mid S(-ve) Mid S(+ve)
1.
d (mm)
b (mm)
ρ
339 209 339 353 223 353 209 209 209 223 223 223 339 209 353 223 209 209 223 223
1700 1700 1700 1700 1700 1700 3300 3300 3300 4300 4300 4300 1700 1700 1700 1700 3300 3300 4300 4300
0.0022 0.0085 0.0036 0.0016 0.0054 0.0026 0.0013 0.0052 0.0021 0.0008 0.003 0.0012 0.0031 0.0072 0.0023 0.0046 0.0018 0.0044 0.0011 0.0026
111.14 148.68 175.04 88.41 111.17 139.24 49.35 184.32 77.73 43.54 162.65 68.58 152.81 127.19 121.56 95.09 67.86 157.67 59.87 139.12
EDGE SPAN REINFORCMENTE
COLUMN SRIP LONGER DIRECTION 1. Φ14 C/C 540,L=(0.33*5.5)+0.69 = 2.5m 1’.Φ14 C/C 540,L=(0.33*5.5)+0.69 = 2.5m 2, Φ14 C/C 105,L=6-(0.125*5.5)+0.10=5.21m 3. Φ14 C/C 540,L=2(0.33*5.5)+0.5 = 4.13m 3’.Φ14 C/C 540,L=2(0.33*5.5)+0.5 = 2.7m SHORTER DIRECTION
Ascalc. (mm)2 1289.98 3023.21 2064.49 978.40 2037.01 1558.65 919.26 3596.49 1461.25 755.35 2896.64 1196.06 1792.14 2542.24 1355.27 1724.92 1271.55 3046.82 1042.23 2464.37
Asmin (mm)2 960.5 592.17 960.5 1000.2 631.83 1000.2 1149.5 1149.5 1149.5 1598.2 1598.2 1598.2 960.5 592.17 1000.2 631.83 1149.5 1149.5 1598.2 1598.2
Sreq (mm)
S-Max (mm)
202.95 86.597 126.81 267.58 128.52 167.97 552.84 141.3 347.78 876.67 228.61 553.65 146.08 102.98 193.17 151.78 399.67 166.8 635.37 268.71
272.57 442.11 272.57 261.76 414.35 261.76 442.11 442.11 442.11 414.35 414.35 414.35 272.57 442.11 261.76 414.35 442.11 442.11 414.35 414.35
SProv. (mm) 200 85 125 260 125 165 440 140 345 410 225 410 145 100 190 150 395 165 410 265
REMARK OK OK OK Smax OK OK Smax OK OK Smax OK Smax OK OK OK OK OK OK Smax OK
8,.Φ14 C/C 240,L=6-0.075+0.15-0.25 = 5.83m 8’.Φ14 C/C 240,L=6-0.15*5+0.15-0.25= 5m 9. Φ14 C/C 210,L=2(0.22*5.5)+0.5= 2.92m SHORTER DIRECTION 10. Φ14 C/C 410,L=(0.22*4.5)+0.69 = 1.68m 11,.Φ14 C/C 420,L=5-0.075+0.15-0.25=4.83m 11’.Φ14 C/C 420,L=5-0.15*5+0.15-0.25= 4m 12. Φ14 C/C 370,L=2(0.22*4.5)+0.5= 2.48m 2.
4. Φ14 C/C 520,L=(0.33*4.5)+0.69 = 2.18m 4’.Φ14 C/C 520,L=(0.33*4.5)+0.69 = 1.60m
INTERIOR SPAN REINFORCMENTE COLUMN SRIP LONGER DIRECTION
5, Φ14 C/C 140,L=6-(0.125*4.5)+0.10=4.34m
13. Φ14 C/C 400,L=2(0.33*5.5)+0.5=4.13m 13’.Φ14 C/C 400,L=2(0.33*5.5)+0.5 =2.7m
6. Φ14 C/C 440,L=2(0.33*4.5)+0.5 = 3.47m 6’.Φ14 C/C 440,L=2(0.33*4.5)+0.5 = 2.30m MIDDEL STRIP
14.Φ14 C/C 120,L=6-2(0.125*5.5)= 4.625m L=4.3+2(24*0.014)=4.97m
LONGER DIRECTION
SHORTER DIRECTION
7. Φ14 C/C 340,L=(0.22*5.5)+0.69 = 1.9m 26
15. Φ14 C/C 510,L=2(0.33*4.5)+0.5=3.47m 15’.Φ14 C/C 510,L=2(0.33*4.5)+0.5 =2.30m
20. Φ14 C/C 500,L=5-2(0.075)=4.85m C/C 500,L=5-2(0.15*5)=3.5m
16.Φ14 C/C 165,L=5-2(0.125*4.5)=3.88m L=3.3+2(24*0.014)=3.97m MIDDEL STRIP
4.2 SLAB
20’.Φ14
FORM WORK FOR FLAT
FOR DROP PANEL
LONGER DIRECTION
=9*(1.25*1.25)+ 9*(1.7*0.13)*4
17. Φ14 C/C 245,L=2(0.22*5.5)+0.5=2.92m
=14.06+7.956=22.0185m2
18. Φ14 C/C 290,L=6-2(0.075)=5.85 18’.Φ14 FOR SLAB C/C 290,L=6-2(0.15*6)=4.20m SHORTER = (15.05*18.05)-(1.7*1.7)+0.25(18.05+15.05)2 DIRECTION =262.1925m2 19. Φ14 C/C 410,L=2(0.22*4.5)+0.5=2.48m TOTAL = 284.211m2
27
4.3
BAR SCHEDULE
Flat slab Reinforcement Ref Location
longer span
Cl.strip
Mdl.strip
No of Bars
Dia
Dia
Dia
Dia
Dia
Dia
Dia
6
8
10
12
14
16
20
0.00
0.00
0.00
78.30
0.00
0.00
0.00
0.00
0.00
54.00
0.00
0.00
0.00
0.00
0.00
0.00
656.46
0.00
0.00
0.00
0.00
0.00
0.00
173.46
0.00
0.00
113.40
0.00
0.00
0.00
0.00
113.40
0.00
0.00
3
148.68
0.00
0.00
0.00
0.00
148.68
0.00
0.00
3
97.20
0.00
0.00
0.00
0.00
97.20
0.00
0.00
x
3
270.00
0.00
0.00
0.00
0.00
270.00
0.00
0.00
2
x
3
91.20
0.00
0.00
0.00
0.00
91.20
0.00
0.00
x
2
x
3
417.60
0.00
0.00
0.00
0.00
417.60
0.00
0.00
x
2
x
3
360.00
0.00
0.00
0.00
0.00
360.00
0.00
0.00
x
2
x
3
175.20
0.00
0.00
0.00
0.00
175.20
0.00
0.00
x
2
x
3
156.60
0.00
0.00
0.00
0.00
156.60
0.00
0.00
x
1
x
3
175.50
0.00
0.00
0.00
0.00
175.50
0.00
0.00
x
1
x
3
126.00
0.00
0.00
0.00
0.00
126.00
0.00
0.00
x
2
x
3
26.16
0.00
0.00
0.00
0.00
26.16
0.00
0.00
Di a
Length
1 1'
14
2.61
5
x
2
x
3
78.30
0.00
14
1.80
x
2
x
3
54.00
0.00
2 3 3' 13 13'
14
5.21
5 2 1
x
2
x
3
656.46
14
4.13
7
x
2
x
3
173.46
14
2.70
7
x
2
x
3
14
4.13
6
x
2
x
14
2.70
x
2
x
14 7
14
5.00
6 1 8
x
1
14
1.90
x
8
14
5.80
8'
14
5.00
9 17
14
2.92
8 1 2 1 2 1 0
14
2.90
18
14
5.85
18' 4
14
4.20
9 1 0 1 0
14
2.18
4
No
total Length
No of Bars
28
Shorter span
Cl.strip
Mdl.strip
4'
14
1.60
5 6 6' 15 15'
14 14
x
2
x
3
38.40
0.00
0.00
0.00
0.00
38.40
0.00
0.00
4.34
4 1 5
x
2
x
3
390.60
0.00
0.00
0.00
0.00
390.60
0.00
0.00
3.47
6
x
2
x
3
124.92
0.00
0.00
0.00
0.00
124.92
0.00
0.00
14
2.30
6
x
2
x
3
82.80
0.00
0.00
0.00
0.00
82.80
0.00
0.00
14
3.47
5
x
2
x
3
104.10
0.00
0.00
0.00
0.00
104.10
0.00
0.00
14
2.30
x
2
x
3
69.00
0.00
0.00
0.00
0.00
69.00
0.00
0.00
16
14
3.97
x
1
x
3
154.83
0.00
0.00
0.00
0.00
154.83
0.00
0.00
10
14
1.68
x
2
x
3
110.88
0.00
0.00
0.00
0.00
110.88
0.00
0.00
11
14
4.83
x
2
x
3
289.80
0.00
0.00
0.00
0.00
289.80
0.00
0.00
11'
14
4.00
x
2
x
3
240.00
0.00
0.00
0.00
0.00
240.00
0.00
0.00
12
14
2.48
x
2
x
3
163.68
0.00
0.00
0.00
0.00
163.68
0.00
0.00
19 20 20'
14
2.48
5 1 3 1 1 1 0 1 0 1 1 1 2
x
2
x
3
178.56
0.00
0.00
0.00
0.00
178.56
0.00
0.00
14
4.85
9
x
1
x
3
130.95
0.00
0.00
0.00
0.00
130.95
0.00
0.00
14
3.20
9
x
1
x
3
86.40
0.00
0.00
0.00
0.00
86.40
0.00
0.00
0.00
0.00
0.00
0.00
5284.68
0.00
0.00
0.222
0.395
0.617
0.888
1.208
1.578
2.468
0
0
0
0
6384
0
0
m Kg/m Kg
5.
DESIGN OF RIBBED SLAB
Assuming one way ribbed slab. Design strength:
f cd =
0.68 f c y
fyd = fctd =
γs
=
0.68 * 25 = 11.33 N mm 2 1 .5
fyk 300 = = 260.87 N / mm 2 γs 1.15 0.21 fck γs
2 3
=
2 3
0.21( 20) = 1.03 N / mm 2 1.15
assume that : width of joists, bw = 150mm and width of grider , b = 300mm
40mm Dt ≥ 1 10 clear..distnce..b / n..Ribs 40mm Dt ≥ 1 10 ( 550 − 150) = 40mm Dt ≥ 40mm ⇒ ...Use..Dt = 60mm
Depth of Joists and Toppings
Depth of Joists (rib):
Depth of Toppings:
Depth of ribs excluding any topping not more than 4 times of minimum width of the ribs Let, Dj (including toppings) = 300mm 29
Dj -Dt
≤ 4bw , 300 -60 ≤ 4*150…..Ok!
Transverse rib shall be provided if the span of the ribbed slab exceeds 6.0m Center to center spacing = 5.00 < 6.0m
Use Ø= 14mm, clear cover=15mm,
Therefore, Transverse strips is not necessary
d = 300 – 15 – 7 = 278mm check depth:
Design load on joists -Toppig self wt= 0.06*0.55*25=0.825KN/m -Joist self wt. = 0.240*0.15*25 = 0.9KN/m -cement screed=0.69*0.55 = 0.3795KN/m -terrazzo tile = 0.46*0.55 = 0.253 KN/m -plastering = 0.345*0.55 = 0.18975 KN/m -partition wall = 4*0.55 = 2.2 KN/m -HCB for floor=0.4*0.2*14=1.12KN/m total dead load = 4.747 KN/m
Mu = µsd * f cd * bw
d≥
20 * 10 6 = 199.73mm 0.295 * 11.33 * 150
Use d = 278mm Reinforcement for Joists: At the support ( -ve M): ,Msd= 20KNm
µ sd =
M sd 20 *10 6 = = 0.152 < 0.295.....ok bd 2 f cd 150 * 278 2 *11.33
From GDC for μsd = 0.152, { Kz = 0.0.91
Live load , LL = 4*0.55 = 2.2 KN/m As =
Pd = 1.3*4.747 + 1.6*2.2 = 9.69 KN/m
M sd 20 *10 6 = = 303.05mm 2 k z df yd 0.91* 278 * 260.87
Provide 2 Ø14mm bars
Assume that rib is continuous run on girder Then, Le = L – 2(0.5*bg) = 5 – 2(0.5*0.3) = 4.7m 2
At the span ( +ve M): ,Msd= 13.75KNm Assume N.A lies in the flange;
2
PdLe 9.96 * 4.7 = = 20 KNm 11 11 2 PdLe 9.96 * 4.7 2 +ve M max = = = 13.75 KNm 16 16 PdLe 9.96 * 4.7 Vmax = = = 23.41KN 2 2 −ve M max =
5.1
Le = 0.7L = 0.7*5000 = 3500mm
Design of Joists(RIB):
30
Shear reinforcement for rib
Le 3500 bw = = = 700mm beff ≤ 5 5 bactual = 550mm beff = 550mm µsd =
M sd beff d 2 f cd
Vs = Vsd − Vc = 20.81 − 19.353 = 1.46 KN
S=
adf yd Vs
=
56.55 * 278 * 260.87 = 2809mm 1.46 * 10 3
af yk 56.55 * 300 = = 262mm S max = <, 0.5b 0.5 *150 13.75 *10 0!.5d = 139mm = = 0.0286 < 0.295.....ok 550 * 278 *11.33 6
2
S max = 139mm
Kz = 0.97 From GDC for μsd = 0.152, Kx = 0.090
Use
5.2
Design of Topping:
According to EBCS-2 The topping shall be
Checking the Assumtion;
provided with reinforcement mesh providing in each direction a crosssectional area not less than 0.001 of the section of the slab
X =Kxd = 0.09*278 X =25.02mm < Df= 60mm……..Ok! As =
φ6c / c...135mm
M sd 13.75 *10 6 = =195.46mm 2As ≥ 0.001Le.Dt k z df yd 0.97 * 278 * 260.87 use Ø = 8mm
Provide 2 Ø14mm bars
and take 1m strip and
As = 0.001*1000*60 = 60mm2
Design for shear of Joists (rib):
S=
By similarity of triangle:
ab * b 50.265 * 1000 = = 837.76mm As 60
23.41( 2.5 − 0.278) = 20.81KN Smax = 400 mm…. provide Ø8 c/c 400mm. 2 .5 5.3 Design of girder: Resistance shear Vsd1 = Vsd 2 =
Vrd = 0.25 f cd bd = 0.25 * 11.33 * 150 * 278 Vrd = 117.80KN >> Vsd (allo) Shear,capacity v c = 0.25 f ctd k1 k 2 bd
Pd int =
k 2 = 1.6 − d = 1.322 ≥ 1.....ok! At support
k1 = 1 + 50
For interior girder the reaction transferred to the supporting girder may be taken twice end shear.
2 * Vmax 2 * 23.41 = = 85.127 KN / m c / c..rib 0.55
For edge girder taken end shear.
φ = 14
Pd int =
As = 303.05mm 2
303.05 = 1.3634 ≤ 2.....ok! 150 * 278
Vmax 23.41 = = 42.564 KN / m c / c..rib 0.55
Depth of girder determination
vc = 0.25 f ctd k1k 2 bd =19.353 KN
Edge (end) girder: 31
µ sd =
M sd 183.91 *10 6 = = 0.259 < 0.295.....ok bd 2 f cd 300 * 457 2 *11.33
From GDC for μsd = 0.2591, { Kz = 0.84
Let assume that , b = 300mm , D = 500mm, cc=25mm, ф = 20mm, stirrup ф= 8mm, d =457mm
Reinforcment
As =
M sd 183.91*10 6 = =1836.48mm 2 k z df yd 0.84 * 457 * 260.87
Provide 6 Φ20 bar Shear design
check depth:
Mu = µsd * f cd * bw
d≥
183.91 * 10 6 0.295 * 11.33 * 300
d ≥ 428.29mm Use d = 460mm Beam span AB & CD , Msd= 147.32 KNm
By similarity of triangle: design shear
122.91( 2.4 − 0.457 ) Vsd1 = Vsd 6 = = 99.50 KN M sd 147.32 *10 6 µ sd = 2 = = 0.207 < 0.295.....ok! 2.4 2 bd f cd 300 * 457 *11.33 171.84( 3.6 − 0.457 ) Vsd 2 = Vsd 5 = = 150.02 KN From GDC for μsd = 0.207, { Kz = 0.88 3.6 Reinforcment
Vsd 3 = Vsd 4 =
153.57( 3 − 0.457 ) = 130.18 KN 3
M sd 147.32 *10 6 2 shear = =1404.23mmResistance k z df yd 0.88 * 457 * 260.87 Vrd = 0.25 f ctd bd = 0.25 * 11.33 * 300 * 457 Provide 5 Φ20 bar As =
Vrd = 388.34KN >> Vsd (allo)
Beam span BC, Msd= 46.44 KNm
Shear Capacity
µ sd
M sd 46.44 *10 6 = = = 0.065 < 0.295 ..... ! f ctd k1 k 2 bd vc = 0ok .25 bd 2 f cd 300 * 457 2 *11.33 k 2 = 1.6 − d = 1.143 ≥ 1.....ok!
From GDC for μsd = 0.065, { Kz = 0.95 Reinforcment
As =
At support 6
M sd 46.44 *10 = = 410mm 2 k z df yd 0.95 * 457 * 260.87
k1 = 1 + 50
φ = 20 As = 1836.48mm 2
1836.48 = 1.67 ≤ 2.....ok! 300 * 457
For beam section
Provide 2 Φ20 bar
v c = 0.25 f ctd k1 k 2 bd = 67.38KN
For support B & C, Msd = 183.91KNm
Design shear for span AB & CD Vsd,max = 150.02 Vc,max = 67.38
32
Vs = Vsd − Vc = 82.64 KN
S=
adf yd Vs
=
100 * 457 * 260.87 = 144.26mm 82.64 *10 3
af yk 100 * 300 = = 200mm 0.5b 0.5 * 300 0.5d = 228.5mm
S max = <,
d ≥ 515.38mm Use d = 557mm
µ sd =
Reinforcment
Design shear for span BC Vsd,max = 130.18
As =
Vc,max = 67.38
Vs
100 * 457 * 260.87 = = 189.84mm 62.80 * 10 3
S max = <,
af yk 100 * 300 = = 200mm 0.5b 0.5 * 300 0.5d = 228.5mm
Beam span BC, Msd= 90.06 KNm
µ sd =
M sd 90.06 *10 6 = = 0.064 < 0.295.....ok bd 2 f cd 400 * 557 2 *11.33
From GDC for μsd = 0.064, { Kz = 0.96 Reinforcment
As =
S max = 175mm Use
M sd 284.84 *10 6 = = 2227.61mm 2 k z df yd 0.88 * 557 * 260.87
Provide 7Φ20 bar
Vs = Vsd − Vc = 62.80 KN
S=
M sd 284.84 *10 6 = = 0.202 < 0.295.....ok bd 2 f cd 400 * 557 2 *11.33
From GDC for μsd = 0.202, { Kz = 0.88
φ8c / c...140mm
adf yd
355.11 * 10 6 0.295 * 11.33 * 400
Beam span AB & CD , Msd= 284.84 KNm
S max = 200mm Use
Mu = µsd * f cd * bw
d≥
φ8c / c...185mm
M sd 90.06 *10 6 = = 645.63mm 2 k z df yd 0.96 * 557 * 260.87
Provide 2 Φ20 bar
interior girder:
For support B & C, Msd = 355.11KNm
µ sd =
M sd 355.11 *10 6 = = 0.252 < 0.295.....ok bd 2 f cd 400 * 557 2 *11.33
From GDC for μsd = 0.252, { Kz = 0.85
Let assume that , b = 400mm , D = 600mm, cc=25mm, ф = 20mm, stirrup ф= 8mm, d =557mm
Reinforcment
As =
M sd 355.11*10 6 = = 2875.18mm 2 k z df yd 0.85 * 557 * 260.87
Provide 9 Φ20 bar Shear design
check depth: 33
Design shear for span AB & CD Vsd,max = 300.89 Vc,max = 98.45
Vs = Vsd − Vc = 202.44 KN
S=
adf yd Vs
=
100 * 557 * 260.87 = 71.78mm 202.44 * 10 3
By similarity of triangle: design shear
Vsd1 = Vsd 6 = Vsd 2 = Vsd 5 =
237.60( 2.4 − 0.557 ) = 182.46 KNS 2.4
355.97( 3.6 − 0.557 ) = 300.89 KN 3.6 S max = 150mm
296.78( 3 − 0.557 ) = 241.68 KN 3
Vsd 3 = Vsd 4 =
max = <,
Use
Resistance shear
φ8c / c...70mm Design shear for span BC
Vrd = 0.25 f ctd bd = 0.25 * 11.33 * 400 * 557 Vrd = 631.08KN >> Vsd (allo)
Vsd,max = 241.68
S=
v c = 0.25 f ctd k1 k 2 bd
adf yd Vs
=
100 * 557 * 260.87 = 101.45mm 143.23 *10 3
k 2 = 1.6 − d = 1.043 ≥ 1.....ok!
k1 = 1 + 50
Vc,max = 98.45
Vs = Vsd − Vc = 143.23KN
Shear Capacity
At support
af yk 100 * 300 = = 150mm 0.5b 0.5 * 400 0.5d = 278.5mm
φ = 20 As = 2875.18mm 2
S max = <,
2875.18 = 1.645 ≤ 2.....ok! 400 * 557
af yk 100 * 300 = = 150mm 0 . 5 b 0 . 5 * 400 0.5d = 278.5mm
S max = 150mm
For beam section
Use
v c = 0.25 f ctd k1 k 2 bd = 98.45 KN
34
φ8c / c...100mm
5.5
FORM WORK
Topping=262.71 m2 ,
5.6
Ref
Di a
Length
top
14
5.30
2
x
34
x
3
bottm
14
5.00
2
x
34
x
shear bar
6
0.75
38
x
34
x
Lx
8
15.72
46
x
1
topping Ribe
No
Ly
Exterior
top bottom support shear bar top
Rib =137.36m2
BAR SCHEDULE
Location
Int
girder =111.2 m2,
No of Bars
total Length
Dia
Dia
Dia
Dia
Dia
Dia
Dia
6
8
10
12
14
16
20
1080.18
0.00
0.00
0.00
0.00
1080.18
0.00
0.00
3
1020.00
0.00
0.00
0.00
0.00
1020.00
0.00
0.00
3
2907.00
2907.00
0.00
0.00
0.00
0.00
0.00
0.00
x
1
722.89
0.00
722.89
0.00
0.00
0.00
0.00
0.00
No of Bars
8
18.72
39
x
1
x
1
729.89
0.00
729.89
0.00
0.00
0.00
0.00
0.00
AB,CD
20
6.45
2
x
2
x
2
51.60
0.00
0.00
0.00
0.00
0.00
0.00
51.60
BC
20
6.00
2
x
1
x
2
24.00
0.00
0.00
0.00
0.00
0.00
0.00
24.00
AB,CD
20
6.00
5
x
2
x
2
120.00
0.00
0.00
0.00
0.00
0.00
0.00
120.00
BC
20
6.00
2
x
1
x
2
24.00
0.00
0.00
0.00
0.00
0.00
0.00
24.00
B,C
20
4.00
4
x
2
x
2
64.00
0.00
0.00
0.00
0.00
0.00
0.00
64.00
AB,CD
8
1.50
44
x
2
x
2
264.00
0.00
264.00
0.00
0.00
0.00
0.00
0.00
BC AB,CD
8 20
1.50 6.55
34 2
x x
2 2
x x
2 2
204.00 52.40
0.00 0.00
204.00 0.00
0.00 0.00
0.00 0.00
0.00 0.00
0.00 0.00
0.00 52.40
35
erior
bottom support shear bars
BC
20
6.00
2
x
1
x
2
24.00
0.00
0.00
0.00
0.00
0.00
0.00
24.00
AB,CD
20
6.00
7
x
2
x
2
168.00
0.00
0.00
0.00
0.00
0.00
0.00
168.00
BC
20
6.00
2
x
1
x
2
24.00
0.00
0.00
0.00
0.00
0.00
0.00
24.00
B,C
20
4.00
7
x
2
x
2
112.00
0.00
0.00
0.00
0.00
0.00
0.00
112.00
AB,CD
8
1.90
87
x
2
x
2
661.20
0.00
661.20
0.00
0.00
0.00
0.00
0.00
BC
8
1.90
61
x
1
x
2
231.80
0.00
231.80
0.00
0.00
0.00
0.00
0.00
2907.00
2813.78
0.00
0.00
2100.18
0.00
664.00
0.222
0.395
0.617
0.888
1.208
1.578
2.468
645
1111
0
0
2537
0
1639
m Kg/m Kg
6. CONCLUSION SUMMERY CONPARISION OF DESIGN DETAILS
SLAB TYPE
ITEM
CONCRET volum(m3)
Solid Slab
Flat Slab
Ribbed Slab
slab beam total slab drop total topping rib girder total HCB
32.228 18 50.228 57.616 9.884 67.5 16.2 6.12 14.04 36.36 44.64
FORM WORK
STEEL (Kg) Dia 6
Dia 8
Dia 10
Dia 14
Dia 20
2953.8 579.13
2408.6 5941.54 6383.9 6383.9
573.85 645.35
2537 537.6
1638.8 5932.56
223.54 154.063 377.603 262.19 22.02 284.21 0 137.36 111.20 248.56
NOTE: According to the comparison of table, the most preferable and economical is ribbed slab. And secondly Solid tow way slab is preferable
36
RANK
Area(m2) 2
3
1
