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If your sheet is macro disable then do it enable.This Excel Sheet is The Solution For Different Type of Slab Design.In this Excel Sheet Seen Yellow Cell is the Data that we have to Input Man…Full description
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The aim of the present work was to develop a safe and reliable roll pass design for producing ribbed bars. For this purpose slit rolling method with use of dog-bone and slit-pass grooves was appl...
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Project Details:
Page #
Standardized Ribbed Slab Designs to BS 8110-97 8110-97::
Prepared:
Case-3) Continuous 5m Long Spans & Imposed Loading 2.5kPa
Checked:
CALCULATIONS
REF ► Slab
1 of 4 Yusuf
OUTPUT
Geometry:
5m
5m
5m
- Design a one-way ribbed slab with 3 equal continuous spans BS8110§3.5.2.3 & §3.5.2.4
- It satisfies the criteria of BS8110 for using Table 3.12 For +ve moment, design the most critical section: i.e. End Span, Span (upport c-r) L =
5
m
For -ve moment, design the most critical section: i.e. First Interior Support ► Material
Allowable L/d ratio = Therefore, dmin= Bottom cover to links =
= 20 . 8
31 . 3 159.9 9.9 mm 25 mm (includes 5mm for deviation)
bo bottom skin =
15 mm
Dia of main bars =
14 mm
Dia of links = hreq =
8 mm 214.9 4.9 mm Try an overall depth of slab h =
224 mm
Project Details:
Page #
Standardized Ribbed Slab Designs to BS 8110-97:
Prepared:
Case-3) Continuous 5m Long Spans & Imposed Loading 2.5kPa
Checked:
CALCULATIONS
REF ► Ulitmate
Yusuf
OUTPUT
Load & Moment Calculation:
Width of slab carried by one rib =
750
mm
Self-weight of slab =
2.50 kN/m
Design Dead Load = 1.4xDL =
5.18 kN/m
Design Implosed Load = 1.6xLL =
3.00 kN/m
Total Design Load, w =
8.18 kN/m
Total Design Load on a span, F = wL =
BS8110§3.5.2.3 & §3.5.2.4
2 of 4
40.91 kN
Ultimate bending moment and shear force as p er Table 3.12
0.086FL
0.063FL BMD
0.063FL
0.086FL 0.5F
0.4F
0.5F
► Design
SFD
0.5F
0.6F of End Span (Bottom Reinforcement):
Ultimate Moment M = 0.086 FL =
T section, b = Assume 0.9x = hf =
750
mm, d =
100
mm
Mf = 0.45 f cu b hf (d - 0.5hf ) = 120.49 ∴Neutral
mm
kNm
>M
750
mm, h =
224
mm
K = M/(f cu b d2) = 0.027 < 0.156 ∴ hence compression reinforcement is not required z/d = 0.5+√ (0.25 - K/0.9) =
0.97
≯ 0.95
∴ z
BS8110Table 3.25
kNm
axis lies in the flange; treat as a rectangular section, having width b =
BS8110§3.4.4.4
169
17.6
= 0.95d = 160.55 As req = M/(0.87 f y z) = 274 mm2 Minimum reinforcement check: web in tension, bw/b≥0.4 Asmin =
0.13% bw h =
67
∴ Provide
Bottom Reinforcement 2Y 14
mm2 Asprov(mm2)= 308
Project Details:
Page #
Standardized Ribbed Slab Designs to BS 8110-97:
Prepared:
Case-3) Continuous 5m Long Spans & Imposed Loading 2.5kPa
Checked:
CALCULATIONS
REF ► Design
3 of 4 Yusuf
OUTPUT
of First Interior Support (Top Reinforcement): Ultimate Moment M = 0.086 FL =
17.6
kNm
Hogging moment over support; slab bottom face in compression. Ribs are not terminated before the support, hence: Solid section: b = BS8110§3.5.2.3 & §3.5.2.4
BS8110§3.4.4.4
230
mm, h =
224
mm
Table coefficient include 20% redistribution of moment at support ∴
βb = 0.8
K' = 0.402(b b – 0.4) – 0.18(bb – 0.4)2 =
0.132
K = M/(f cu b d2) = 0.051 < K' ∴ hence compression reinforcement is not required z/d = 0.5+√ (0.25 - K/0.9) =
0.94
≯ 0.95
∴ z
= 0.95d = 160.55 As req = M/(0.87 f y z) = 274
BS8110Table 3.25 §3.12.11.2
∴ Provide
Top Reinforcement
2
mm Minimum reinforcement check: for rectangular sections Asmin = 0.13% b h = 67 mm2
2Y 10 +1Y 12
Asprov(mm2)= 270 ► Shear
Check: Max. shear force at first interior support = 0.6F =
24.5
kN
Ribs are not terminated before the support, hence critical section is at the face of the load-bearing : x=
62
Shear Force V = (0.6F) - (wx) =
mm from support centerline 24.0
BS8110Eq. 22
Calculate design applied shear stress: v = V/bvd = 0.62
'Where two or more bars are used in a rib, the use of link reinforcement is ∴ Provide Double recommended to ensure correct cover to reinforcement. The spacing of the Legged Links links can generally be of the order of 1 m to 1.5 m depending on the size of Y8@500mm c/c the main bars'
Project Details:
Page #
Standardized Ribbed Slab Designs to BS 8110-97:
Prepared:
Case-3) Continuous 5m Long Spans & Imposed Loading 2.5kPa
Checked:
CALCULATIONS
REF
4 of 4 Yusuf
OUTPUT
► Deflection
BS8110§3.4.6.3 Table 3.9
Check (Middle of End Span): Code says that Table 3.9 ratios " are based on limiting the total deflection to span/250 and this should normally ensure that the part of the deflection occurring after construction of finishes and partitions will be limited to span/500 or 20 mm, whichever is the lesser, for spans up to 10 m"
bw/b =
0.31
Interpolated Basic L/d =
20.8
M/bd2 =
0.82
βb ≈
0.95
f s =
287
N/mm²
tension m.f =
1.47
≯2
As'prov =
157
mm²
compression m.f =
1.04
≯1.5
∴
BS8110Eq. 8
> 0.3, N/mm²
BS8110Eq. 7
=
1.47
As'prov/bd = =
Allowable L/d = Basic L/d*m.f =
31.9
Actual L/d =
29.6
1.04 < Allowable L/d
No further checks are required. ► Reinforcement
0.12
Deflection Check Satisfactory
in Topping:
Single layer of welded steel fabric is needed for the topping: Use Mesh A252
BS8110§3.6.6.2
Taking 1m width of the topping, b = As = 0.12% b hf = Maximum spacing Smax = rib spacing/2 = ► Detailing
