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Slab 6, 7, 8

This document provides the design details for slabs 6 & 7 of Case 9. It determines that the slab is a one-way slab based on span ratios. Minimum thickness is calculated to be 150mm. Service loads, factored loads, and steel ratios are calculated. The main reinforcement is determined to be 6 bars of 12mm diameter spaced at 200mm. Shrinkage and temperature reinforcement is 4 bars of 10mm diameter spaced at 300mm.

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Ayreesh Mey Spnt
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122 views3 pages

Slab 6, 7, 8

This document provides the design details for slabs 6 & 7 of Case 9. It determines that the slab is a one-way slab based on span ratios. Minimum thickness is calculated to be 150mm. Service loads, factored loads, and steel ratios are calculated. The main reinforcement is determined to be 6 bars of 12mm diameter spaced at 200mm. Shrinkage and temperature reinforcement is 4 bars of 10mm diameter spaced at 300mm.

Uploaded by

Ayreesh Mey Spnt
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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DESIGN OF SLAB 6 & 7

CASE 9
fy = 276 Mpa
f’C = 20.7 Mpa
La = 4m
Lb = 1.20 m

CHECK IF THE SLAB IS ONE - WAY SLAB:


La 4m
= = 3.30 ¿ 2
Lb 1.20 m
Therefore, the design is One way slab:
MINIMUM THICKNESS:
l fy 1.20 276
t= (0.40+ ¿= ( 0.4+ ) = 0.048mm
20 700 20 700

Assume t = 150mm.
SERVICE LOADS: (CONSIDER 1m STRIP)
A.)DEAD LOAD(WDL):
Wslab = 23.54 KN/m3(0.15m)(1m) = 3.531 KN/m
Wfloor finish = 0.60 KN/m2(1m) = 0.60 KN/m
Welect.& mech. = 0.25 KN/m2(1m) = 0.25 KN/m
Wplumbing. = 0.25 kN /m2(1m) = 0.25 kN /m
Wceiling = 0.20 KN/m2(1m) = 0.20 KN/m
TOTAL = 4.831 KN/m
B.)LIVE LOAD (WLL) :
WLL = 2.9 Kn/m2(1m) = 2.90 kN/m
SOLVE FOR THE EFFECTIVE DEPTH (d):
d = h – CC – 1/2φ bar (Use 12 mm φ bar)
d = 150 mm – 20mm – ½(12mm)
d = 124 mm
FACTORED LOADS:
WU = 1.2WDL + 1.6WLL (EQ. 409-2, NSCP 2010)
WU = 1.2(4.831 KN/m) + 1.6(2.90 kN/m)
WU = 10.437 Kn/m

Wu L2 (10.437 Kn )¿ ¿
MU = = = 1.879 kn-m
8 m

SOLVING FOR STEEL RATIO:

Mu 1.879 ( 106 ) N −mm


Rn = =
ϕb d 2 (0.90)(1000 mm)(124 mm)2

Rn = 0.136

0.85 f ' c 2 Rn
ρ=
fy √
(1− 1−
0.85 f ' c
)

0.85 (20.7 Mpa) 2(0.136 Mpa)


ρ=
(276 Mpa) √
(1− 1−
0.85(20.7 Mpa)
)

ρ=0.00049

CHECK IF ρmax > ρ> ρmin:


1.4 1.4
ρmin= = = 0.00507 ;ρmin> ρ
fy 276

Therefore, use ρ=¿ 0.00507


SOLVING FOR THE AREA OF STEEL IN 1m STRIP (ASt):
AS = ρbd=¿0.00507(1000mm)(124 mm) = 628.68 mm2

 No. OF STEEL BARS REQUIRED OF 1m STRIP:

AS 628.68 mm 2
N= = =5.559 pcs . say 6 pcs
Abar 113.097 mm2

SPACING OF BARS:
b 1000 mm
S= = =200 mm say 200 mm
n−1 6−1
Therefore, use 12 mm bar diameter spaced @ 200 mm O.C.

 FOR SHRINKAGE AND TEMPERATURE REINFORCEMENT:

SOLVING FOR THE AREA OF STEEL IN 1m STRIP:


As = 0.002bh (section 407.13.2.1 – NSCP 2010)
As = 0.002(1000mm)(150mm)
As = 300 mm2
NO. OF STEEL BARS REQUIRED FOR 1m STRIP:

As 300mm 2
N= = =3.8197 pcs . say 4 pcs
Abar 78.53980mm 2

SPACING OF BARS:
b 1000 mm
S= = =333.333 mm
n−1 4−1

S = 333.333 mm say 300 mm


Therefore, use 10 mm bar diameter spaced @ 300mm O.C.

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