DESIGN CHECK FOR SS-5 SUBSTATION FLOOR SLAB - (LV Switgear)- TWO LONG EDGE DISCONTINUOUS
Data:-
Shoreter Span = lx = 4 m
Longer Span = ly = 8.2 m
ly/lx = 8.2/4 = 2.05
Equipment load (Refer Clause 5.0 DESIGN LOADS)= wl = 9.2 kN/m2
Assumed width = b = 1000 mm
Overall depth = h = 175 mm
Thickness of screed = hs = 50 mm
Grade of concrete = fcu = 20 N/mm2
Grade of main steel = fy = 336 N/mm2
Clear cover = c = 50 mm
Elastic modulus of steel = Es = 200000 N/mm2
Elastic modulus of concrete = Ec = 28000 N/mm2
Design allowable crack width = w = 0.3 mm ( As per BS8110)
Equip.Load = 9.2 kN/m2
Analysis
DL = 5.58 kN/m2
The moments in continuous Two-way spanning slab is calculated using the coefficients given in
- Table 3.14 of BS 8110
Self weight of slab Tk.of slab x concrete Density= ws = 175x25 = 4.38 kN/m2
Weight of Screed Tk.of screed x screed Density= wsc = 50x24 = 1.2 kN/m2
Total dead load on slab = wd = 5.58 kN/m2
Service load on slab = w = 14.78 kN/m2
Ultimate load on slab = wu = 22.1625 kN/m2
Coefficients for Ultimate Moment given in Table 3.14 for Coefficient x Wl x l
Considered coefficient of TWO LONG EDGE DISCONTINUOUS
Shorter Span Moment Coefficients
- ve Moment Coefficients @ Cont. Edge = = 0
+ ve Moment Coefficients @ Mid Span = = 0.1
Longer Span Moment Coefficients
- ve Moment Coefficients @ Cont. Edge = = 0.045
+ ve Moment Coefficients @ Mid Span = = 0.034
Shorter Span Moment
Mu (-Ve) M1 = 0 X22.1625x4x4
= 0 kN m
Mu (+Ve) M2 = 0.1 X22.1625x4x4
35.46 kN m
Longer Span Moment
Mu (-Ve) M3 = 0.045 X22.1625x4x4
= 15.957 kN m
Mu (+Ve) M4 = 0.034 X22.1625x4x4
= 12.056 kN m
Reinforcement along shorter direction
@ Supports
Mu (-Ve) = M1 = 0 kN m
Effective depth (Refer DWG no : SS5-E20-0-3269 ) = d = h-C
= 175 - 50- 8
= 117 mm
The ratio = k = M1 / bd2fcu
= 0/(1000x117^2 x 20)
= 0
Lever arm = z = d[0.5+SQRT(0.25-k/0.9 ]
= 117[0.5+√(0.25-0-/0.9]
= 117 mm
Lever arm maximum = Zmax = 0.95 x d
= 0.95 x 117
= 111.15 mm
Therefore the design value of lever arm = 111.15 mm
Area of steel required = (As)req1 = M1/0.95 fy z
= 0/(0.95x336x111.15)
= 0 mm2
Minimum area of steel = (As)Min. = 0.13% b D
= (0.13/100)x1000x175
= 227.5 mm2
Maximum area of steel = (As)Max. = 0.04bD
= 0.04 x 1000x175
= 7000 mm2
T 16 @ 250
Area of steel provided = (As)PRO1 = 803.84 mm2 OK
C/C spacing of the bottom most main bar = 250 mm
@ Mid Span
Mu (+Ve) = M2 = 35.46 kN m
Effective depth (Refer DWG no : SS5-E20-0-3269 ) = d = h-C
= 175 - 50- 8
= 117 mm
The ratio = k = M2 / bd2fcu
= 35.46/(1000x117^2 x 20)
= 0.13
Lever arm = z = d[0.5+SQRT(0.25-k/0.9 ]
= 117[0.5+√(0.25-0.13)/0.9]
= 96.5125 mm
Lever arm maximum = Zmax = 0.95 x d
= 0.95 x 117
= 111.15 mm
Therefore the design value of lever arm = 96.51 mm
Area of steel required = (As)req2 = M2/0.95 fy z
= 35.46/(0.95x336x96.51)
= 1151.07 mm2
Minimum area of steel = (As)Min. = 0.13% b D
= (0.13/100)x1000x175
= 227.50 mm2
Maximum area of steel = (As)Max. = 0.04bD
= 0.04 x 1000x175
= 7000 mm2
T 16 @ 250
Area of steel provided = (As)PRO2 = 803.84 mm2
C/C spacing of the bottom most main bar = 250 mm Not Ok
Reinforcement along longer direction
@ Supports
Mu (-Ve) = M3 = 15.957 kN m
Effective depth (Refer DWG no : SS5-E20-0-3269 ) = d = h-C
= 175 - 50- 8
= 117 mm
The ratio = k = M3 / bd2fcu
= 15.957/(1000x117^2 x 20)
= 0.058
Lever arm = z = d[0.5+SQRT(0.25-k/0.9 ]
= 117[0.5+√(0.25-0.058/0.9]
= 108.899 mm
Lever arm maximum = Zmax = 0.95 x d
= 0.95 x 117
= 111.15 mm
Therefore the design value of lever arm = 108.9 mm
Area of steel required = (As)req3 = M3/0.95 fy z
= 15.957/(0.95x336x108.9)
= 459.05 mm2
Minimum area of steel = (As)Min. = 0.13% b D
= (0.13/100)x1000x175
= 227.5 mm2
Maximum area of steel = (As)Max. = 0.04bD
= 0.04 x 1000x175
= 7000 mm2
T 10 @ 300
Area of steel provided = (As)PRO3 = 261.67 mm2 Not Ok
C/C spacing of the bottom most main bar = 300 mm
@ Mid Span
Mu (-Ve) = M4 = 12.056 kN m
Effective depth (Refer DWG no : SS5-E20-0-3269 ) = d = h-C
= 175 - 50- 8
= 117 mm
The ratio = k = M3 / bd2fcu
= 12.056/(1000x117^2 x 20)
= 0.044
Lever arm = z = d[0.5+SQRT(0.25-k/0.9 ]
= 117[0.5+√(0.25-0.044/0.9]
= 110.969 mm
Lever arm maximum = Zmax = 0.95 x d
= 0.95 x 117
= 111.15 mm
Therefore the design value of lever arm = 110.97 mm
Area of steel required = (As)req4 = M4/0.95 fy z
= 12.056/(0.95x336x110.97)
= 340.36 mm2
Minimum area of steel = (As)Min. = 0.13% b D
= (0.13/100)x1000x175
= 227.5 mm2
Maximum area of steel = (As)Max. = 0.04bD
= 0.04 x 1000x175
= 7000 mm2
T 10 @ 300
Area of steel provided = (As)PRO4 = 261.67 mm2 Not Ok
C/C spacing of the bottom most main bar = 300 mm
Check for Crack width (Refer Section 3.12.11.2.7 - BS8110)
No further crack check is required when the following is used,
1) grade 250 steel is used and the slab depth does not exceed 250 mm; or
2) grade 460 steel is used and the slab depth does not exceed 200 mm; or
3) The reinforcement percentage (10As /bd) is less than 0.3 %.
Max of (As)PRO1,(As)PRO2,(As)PRO3,(As)PRO4, = (803.84,803.84,261.67,261.67)
= 803.84 mm2
Checking for reinforcement %:
10As 10 x 803.84
= = 0.068704 < 0.3 % Ok
bd 1000x117
Check for shear
Shear stress at the face of the support(Actual) = V/bvd
= 0.38 N/mm2
< 0.8 x fcu)1/2 or 5.0
< 3.58 or 5
= 0.38 < 5
100As / bd = 0.69 %
(400/d) = 3.42
Therefore design value of (400/d) = 1
Ultimate shear stress (Allowable) Vc = 0.53 N/mm2
> v
NO SHEAR REINFORCEMENT REQUIRED
Span / Effective depth ratio
Basic span / depth ratio = 34.19
Modification factor for tension reinforcement = 477 - fs + 0.55
120(0.9+(Mu/bd2))
= 0.62
Modified span / depth ratio = 21.2 mm
Minimum effective depth required 188.7 mm
Provided effective depth = 117 mm Not Ok
Result
@Shorter Span
Ast Req. @Support = 0 mm2 < Ast Provided = 803.84 mm2
Ast Req. @Mis Span = 1151.07 mm2 > Ast Provided = 803.84 mm2
@Longer Span
Ast Req. @Support = 459.05 mm2 > Ast Provided = 227.5 mm2
Ast Req. @Mis Span = 340.36 mm2 > Ast Provided = 261.67 mm2
Ultimate shear stress (Allowable)= 0.53 N/mm2 > Actual Shear stress = 0.38 N/mm2
Min. effective depth required = 188.7 mm > Pro. effective depth = 117 mm
Depth of Slab = 175 mm
Bottom Reinforcement = 16 @ 250 = 804 mm2
Top Reinforcement = 16 @ 250 = 804 mm2
Distribution Reinforcement = 10 @ 300 = 262 mm2
Based on preliminary loads the structural design verification calculation and results observed that the existing
slab of Substations SS-5 is safe to accommodate the new switchgear panel loads.
Lifting Design using Rebars
Details of panel
Panel thickness (Refer : P5693-A020-DWG-16-60-220) = tp = 550 mm
Length of Panel (Refer : P5693-A020-DWG-16-60-220) hp = 5000 mm
Width of Panel (Refer : P5693-A020-DWG-16-60-220) = bp = 1275 mm
Lifting dimensions
Lifting Hook Distance along Longitudinal = h1 = 300 mm (< h/4)
Lifting Hook Distance along Transverse = b1 = 260 mm (< b/4)
Angle of lifting line inclination to the vertical = β = 60 °
Grade of concrete (Refer : P5693-A020-DWG-16-60-220) = fcu = 40 N/mm2
Grade of main steel (Refer : P5693-A020-DWG-16-60-220) = fy = 460 N/mm2
Clear cover (Refer : P5693-A020-DWG-16-60-220) = c = 75 mm
Elastic modulus of steel = Es = 200000 N/mm2
Elastic modulus of concrete = Ec = 28000 N/mm2
Design allowable crack width = w = 0.3 mm ( As per BS8110)
Unit Weight of Concrete = ρ = 25 kN/m3
Lifting Hook Details
Dia of Lifting Hook Rebar(Refer : P5693-A020-DWG-16-60-220) φrebar = 25 mm
No. of Rebar (Refer : P5693-A020-DWG-16-60-220) = 1 nos
Grade of lifitng Hook Rebar (Refer : P5693-A020-DWG-16-60-220) = fyr = 460 N/mm2
Bond coefficient = β = 0.28 ( Table:3.26 - BS8110)
Lifting Factor = 1.5
No. of lifting hook = 4 Nos
Width of Hook (Refer : P5693-A020-DWG-16-60-220) b1 = 100 mm
Projection of Rebar above Concrete (Refer : P5693-A020-DWG-16-60-220) h1v = 50 mm
Provided rebar leg length La = 160 mm
Analysis & Design
Self weight of Panel Tk.of Panel x concrete Density= w= 550x25 = 13.75 kN/m2
Section modulus of panel Zxx = 1000x tp2 / 6
= 1000 x 550^2 / 6
= 5E+07 mm3
Lh = (5000-(2*300))/2
2200.00 mm
Lb = (1275-(2*300))
755 mm
Pv_hdir =((2200/2+300)x((755/2)+260)x550)x25
12.27 kN
Ph_hdir = (12.27x1000xTAN(60*PI()/180)/(755/2+260))
= 33.34 kN
Ph_bdir = (12.27x1000xTAN(60xPI()/180)/(2200/2+300)))
15.18 kN
Moment @ Cantilever Portion (Longi. Direc.) Muh_cant = w x h12/2
= 13.75x300^2 / 2
= 0.62 kN m
Moment @ Fixed Portion (Longi. Direc.) Muh_end span = w x Lh2/12
(Lifting Hook location considered as a support) = 13.75x2200^2 / 12
= 5.55 kN m
Moment @ Mid Span (Longi. Direc.) Muh_mid span = w x Lh2/8 -Muh_cant +Muh_end span/2
(Lifting Hook location considered as a support) = (13.75x2200^2 / 8 ) - 0.62+5.55/2
= 6.62 kN m
Moment @ Cantilever Portion (Transv. Direc.) Mub_cant = w x b12/2
= 13.75x260^2 / 2
= 0.46 kN m
Moment @ Fixed Portion (Transv. Direc.) Mub_end span = w x Lb2/12
(Lifting Hook location considered as a support) = 13.75x755^2 / 12
= 0.65 kN m
Moment @ Mid Span (Transv. Direc.) Mub_mid span = w x Lb2/8 -Mub_cant +Mub_end span/2
(Lifting Hook location considered as a support) = (13.75x755^2 / 8 ) - 0.46+0.65/2
= 0.58 kN m
Shear Force @ Cantilever Portion (Longi. Direc.) Vuh_cant = w x h1
= 13.75x300
= 4.13 kN
Shear Force @ Fixed Portion (Longi. Direc.) Vuh_end span = w x Lh/2
(Lifting Hook location considered as a support) = 13.75x2200/2
= 15.13 kN
Shear Force @ Cantilever Portion (Transv. Direc.) Vub_cant = w x b1
= 13.75x260
= 3.58 kN
Shear Force @ Fixed Portion (Transv. Direc.) Vub_end span = w x Lb/2
(Lifting Hook location considered as a support) = 13.75x755/2
= 5.19 kN
Max. Bending Moment @Long. Direc. Muh = 6.62 kN m
Max. Bending Moment @Transverse Direc. Mub = 0.65 kN m
Max. Shear Force @Long. Direc. Vuh = 15.13 kN
Max. Shear Force @Transverse Direc. Vub = 5.19 kN
Max allowable compressive stress , σmax = 0.33 x fcu
= 0.33x40
= 13.2 N/mm2
Compression in Longitudinal direction
including bending stress σh_dir =(33.34/550)+((6.62x1000000)/50416666.67)
0.19 < 13.2N/mm^2)
Compression in Transverse direction
including bending stress σh_dir =(15.18/550)+((0.65x1000000)/50416666.67)
0.04 < 13.2N/mm^2)
Hence OK
Design tension rebars for Lifting,
Muh = 6.62 kN m
Effective depth d = h-C
= 550 - 75- 12
= 463 mm
The ratio = k = Muh / bpd2fcu
= 6.62/(1275x463^2 x 40)
= 0.000
Lever arm = z = d[0.5+SQRT(0.25-k/0.9 ]
= 463[0.5+√(0.25-0/0.9]
= 463 mm
Lever arm maximum = Zmax = 0.95 x d
= 0.95 x 463
= 439.85 mm
Therefore the design value of lever arm = 439.85 mm
Area of steel required = (As)req = Muh/0.95 fy z
= 6.62/(0.95x460x439.85)
= 34.44 mm2
Minimum area of steel = (As)Min. = 0.13% b tp
= (0.13/100)x1000x550
= 715 mm2
Maximum area of steel = (As)Max. = 0.04 b tp
= 0.04 x 1000x550
= 22000 mm2
T 12 @ 125
Area of steel provided = (As)PRO. = 904.32 mm2 OK
C/C spacing of the bottom most main bar = 125 mm
Check for shear
Shear stress at the face of the support(Actual) = V/bvd
= 0.03 N/mm2
< 0.8 x fcu)1/2 or 5.0
< 5.06 or 5
= 0.03 < 5.06
100As / bd SR1 = 0.2 < 3
(400/d)1/4 SR2 = 0.22 < 1
0.79 x (SR11/3) x SR2 x 1/γm x (fcu/25)1/3
Ultimate shear stress (Allowable) Vc = 0.005 N/mm2
> v
NO SHEAR REINFORCEMENT REQUIRED
Lifting Design using reinforcement bars
Dia of Lifting Hook Rebar φrebar = 25 mm
No. of Rebar Ars = 1 nos
Area of Rebar = 490.87 mm2
Total Area of rebar Art = Area of Rebar x No. of Rebar x 2 Legs
= 490.87x1x 2
= 981.74 mm2
Moment of Inertia of Rebar Ixx,rebars = π x d4/ 64
= 19174.8 mm4
Section Modulus of Rebar zxx,rebars = Ixx,rebars /(d/2)
= 19174.76/ (25/2)
= 1533.98 mm3
Tensile strength of Rebar = 0.87 x fyr x Ars
= 0.87x460x490.87
= 196.45 kN / leg
Shear strength of Rebar = 0.57 x fyr x Ars
= 0.87x460x490.87
= 128.71 kN / leg
Moment Capacity of Rebar = zxx,rebars x fyr
= 1533.98x460
= 0.71 kN m / leg
Total Weight of Panel W = 550 x5000 x1275 x 25
= 87.66 kN
No of lifting hook = 4.00 Nos
Load per lifting hook = 87.66/4
= 21.92 kN
Factored Load Wv = 1.5x21.92
= 32.88 kN
Lifting Angle θ1 = 60 º to θ2 = 60 º
Maximum Ver. force component Pv = Wv x Sin(θ2)
= 32.88x sin(60)
= 28.47 kN
Maximum Hor. force component Ph = Wv x Cos(θ2)
= 32.88x Cos (60)
= 16.44 kN
Tensile Force in the Rebar = Pv / No.of legs + ( Ph x h1 )/b1
= (28.47/2)+((16.44x50)/100)
= 14.24 kN
< 196.45 kN Safe
Shear Force in the Rebar = Ph / No .of legs
= 16.44/ 2
= 8.22 kN
< 128.71 kN Safe
Moment in the Rebar = Ph x h1 / No. of legs
= 16.44x50/2
= 0.41 kN m
< 0.71 kN m Safe
Rebar combined capacity (14.24/196.45 ) + (8.22/128.71) + (0.41 / 0.71) = 0.71
< 1.2 Safe
=
Anchorage Check =
fbu = β √fcu
= 0.28√40
= 11.2 N/mm2
Lanchorage = 98.13 mm per rebar leg
Provided rebar leg length per side = 160 mm Safe
