DACS_UNIT-3- RCC Flat slab ARP.pptx

Sanjivani Rural Education Societies‘
Sanjivani College of Engineering, Kopargaon
(An Autonomous Institute, affiliated to SPPU,Pune)
A/P:-Sahajanandnagar, Tal:-Kopargaon, Dist:- Ahmadnagar-423603
Department of Civil Engineering
Design of Advanced Concrete Structures
B.E. Civil
by
A. R. Pabale
(Assistant Professor)

Text Books Refered:
1. Limit state theory and design of reinforced - Dr. V. L. Shah and Dr S. R. Karve –
Structures Publications, Pune.
2. Fundamentals of Reinforced Concrete- N.C. Sinha, S.K. Roy – S. Chand & Co. Ltd
3. Advanced design of structures- Krishnaraju - Mc Graw Hill.
4. Design of Prestressed concrete structures- T. Y. Lin.
5. Prestressed Concrete- N. Krishna Raju – Tata Mc Graw Hill Publication Co.
Reference Books:
1. Comprehensive RCC Design - Punmia, Jain & Jain - Laxmi Publications.
2. Design of design of reinforced Concrete structures- M. L. Gambhir –PHI.
3. Reinforced Concrete, Vol I- Dr.H J. Shah Charotar Publishing House
4. Prestressed Concrete – A Fundamental Approach- Edward Nawy – PHI..
5. Reinforced concrete design- Pillai and Menon TMH.
6. Elementary Structural Dynamics-Selvam, Dhanpatrai Publications.
I.S. Codes:
1. IS: 456: Indian Standard code of practice for plain and reinforced concrete, BIS, New
Delhi.
2. IS: 1343: Indian Standard code of practice for Prestressed concrete, BIS, New Delhi.

Unit 3 : Syllabus
Design of Flat slab:-
Introduction to flat slab, Design of
prestressed two way flat slab by direct
design method. (IS 456:2000)

Classification of PC Slab:
1. One-way slabs:
2. Two-way slabs: rectangular slab panels supported
on all four edges by either walls or beams.
3. Flat slabs:
A) Flat slab without drop and column head:
(continuous slab of constant thickness
supported by a rectangular grid of columns).
B) Flat slab without drop but with column head
C) Flat slab with drop and without column head
D) Flat slab with drop and column heads.

Flat slabs:
• The term flat slab means a reinforced concrete slab
with or without drops, supported generally without
beams, by columns with or without flared column
heads (cl.no.:31 & Fig. 12, pg. 54,IS:456-2000).
• A flat slab may be solid slab or may have recesses
formed on the soffit so that the soffit comprises a
series of ribs in two directions. The recesses may be
formed by removable or permanent filler blocks.

Fig. 12: Critical Sections for Shear in
Flat Slabs (IS : 456-2000,Pg.No. 54)
Note:-
1. Do = the diameter
of column or column
head to be
considered for design
and
2. d = is effective
depth of slab or drop
as appropriate,

Fig: Classification of PC Slab

Clause. No.:- 31.1.1 pg. 53 [IS:456-2000]
For the purpose of this clause, the following definitions shall apply:
a) Column strip:- Column strip means a design strip
having a width = 0.25 l2 ,but It should not be
greater than 0.25 l1 on each side of the column
centre line,
Where,
l1, - Span in the direction moments are being
determined (measured centre to centre of supports).
l2 - Span transverse to l1,
(measured centre to centre of supports).

Clause. No.:- 31.1.1 pg. 53 , IS:456-2000
b) Middle strip:- Middle strip means a design strip
bounded on each of its opposite sides by the
column strip.
c) Panel:- Panel means that part of a slab bounded
on each of its four sides by the centre-line of a
column or centre-lines of adjacent spans.

Direct Design Method (Cl.No.: 31.4, IS:456-2000)
31.4.1 Limitations:
Slab system designed by the direct design method
shall fulfill the following conditions:
a) There shall be minimum of three continuous spans
in each direction,
b) The panels shall be rectangular, and the ratio of
the longer span to the shorter span within a panel
shall not be greater than 2.0 (Two way slab)

31.4 Direct Design Method, 31.4.1 Limitations………………………….
c) It shall be permissible to offset columns to a
maximum of 10 percent of the span in the
direction of the offset not withstanding the
provision in (b),
d) The successive span lengths in each direction shall
not differ by more than one-third of the longer
span. The end spans may be shorter but not longer
than the interior spans.
e) The design live load shall not exceed three times
the design dead load.

Shear in Flat Slab:-
(Cl. No. 31.6., Pg. No.57 & 58, IS 456-2000)
31.6.1
a) The critical section for shear shall be at a distance
d/2 from the periphery of the column/capital/drop
panel and perpendicular to the plane of the slab
where, d is the effective depth of the section / slab
(see.Fig.13).

b) The shape in plan is geometrically similar to the
support immediately below the slab
(see Fig.13A and 13B).,
Fig:13A Fig:13B
Note :- d is the effective depth of the flat slab/drop.
Fig 13A & 13B:-Critical Sections in Plan For Shear
in Flat Slabs (Pg. No.58, IS 456-2000)

c) For column sections with re-entrant angles. the
critical section shall be taken as indicated in
Fig.13C and 13D.
Note - d is the effective depth, of the flat slab/drop.
Fig 13C & 13D:-Critical Sections in Plan For Shear in Flat Slabs
(Pg. No.58, IS 456-2000)

d) 31.6.1.1 In the case of columns near the free edge
of a slab, the critical section shall be taken as
shown in Fig. 14.
Fig:14 Effect of Free Edges on Critical Section For Shear
(Pg. No.58, IS 456-2000)

e) 31.6.1.2 When openings in flat slabs are located at
a distance less than 10 times thickness of the
slab from a concentrated reaction or when the
openings are located within the column strips, the
critical sections specified in 31.6.1 shall be
modified so that the part of the periphery of the
critical section which is enclosed by radial
projections of the openings to the centroid of the
reaction area shall be considered ineffective.
(see Fig. 15), and openings shall not encroach
upon column head.

Fig:15A Fig:15B
Fig:15C Fig:15D
Fig. 15 Effect of Openings on Critical Section for Shear
(Pg. No.59, IS 456-2000)

Data Required:
Size of panel, Position of column (L1 & L2) , WL , fp , fck
Requirements:
1. Thickness of slab (D), Moments,
2. size of drop / Capital / Column head.
3. Reinforcement (in middle and column strip).
Design of Flat Slab: (Direct design Method)
(Ref. Code: IS:456-2000 & IS:1343-2012)

1] Width of column strip (Along longer span):
(Cl. No. 31.1.1, P.No.53, IS 456-2000)
Width of column strip (2 bc)
bc = Width on either side of Column centre line
= 0.25L1 or 0.25 L2 which ever is less
⚫Width of column strip = 2bc
⚫Width of middle strip along longer span = L1-2bc
⚫Width of middle strip along shorter span =L2-2bc

2] Thickness of slab (D) : Cl. No. 31.2.1 P.No. 53
a) If Drop is not provided to slab
D = ≥ 125 mm
b) If Drop is provided
D = ≥ 125 mm

3] Size of Drop (Cl. No.31.2.1 Pg.No.53)
⚫Size of Drop = (For interior panel)
⚫Size of Drop = (For exterior panel)
⚫Thickness of drop = (D = thickness of slab)

4] Size of column heador Capital
(cl. No.31.2.1 Pg. No.53, 54, fig. no. 12)
⚫Diameter of column head =
⚫Where ,
L - Average span length in mm.

5] Load on Slab:-
⚫D.L. = Wd = 25 x D
F.F.L = Wf = Assume 1 to 2 kN/m2 if not given
L.L. = WL = Given ORAs per the IS: 875 Part-II
Total Load = WT
Design Load = Wu = 1.5 WT

6] Tofind BendingMoment:
A) Along longer span (take L1 = Ly, L2 =Lx)
⚫Clear span = (Ln) = L1 – Dc > 0.64 L1
⚫Load on Panel =(Wo) = Wu.Ln.L2
⚫Total moment = Mo = kN.m
⚫Total - ve moment = (Mo)- =0.65 (Mo)
⚫Total + ve moment = (Mo)+ = 0.35(Mo)

Distribution of moments along longer span
Moments Column Strip Middle Strip
-ve
Moments
0.75((Mo)-
0.25 (Mo)-
+ve
Moments
0.60(Mo)+
0.40(Mo)+

6] Tofind bending Moment
B) Along shorter span (L1 = Lx, L2 =Ly)
⚫Clear span = (Ln) = L1 – Dc > 0.64 L1
⚫ Load on Panel =(Wo) = Wu.Ln.L2
⚫Total moment = Mo = kNm/m width
-
⚫Total -ve moment = (Mo) = 0.65 (Mo)
+
⚫Total + ve moment = (Mo) = 0.35(Mo)

Distribution of moments along shorter span
Moments ColumnStrip Middle Strip
-ve Moments 0.75((Mo)-
0.25 (Mo)-
+ve Moments 0.60(Mo)+
0.40(Mo)+

7] Design of Reinforcement
A) Along Longer Span:-
a) In column strip Along Longer Span
⚫ For max. –ve B.M.
mm
⚫ Effective depth; deff,y =
⚫ Assuming dcy’ = 35 mm
st,reqd
fy
4.6 X 𝑀𝑢
𝑐𝑘
𝑓 .𝑏.𝑑eff,y
2
⚫ A = 0.5 X fck X {1- 1 − }.b .deff,y
⚫ Spacing of Bars in Column strip
⚫Assuming,ϕ -----mmTMT bar (ast =------ mm2)

Ast,min= 0.2 % x Gross C/S Area
= 0.2
x b x D
100
=……….mm2
Check Ast,reqd > Ast,min , if < Then Provide Ast,min
Provide …..mm 𝛟bars @ ………..mm C/C.
b) In middle Strip Along Longer Span
⚫For max. –ve B.M.
⚫Effective depth; deff,y= mm
⚫Ast,reqd
fy
= 0.5 X fck X {1- 1 −
4.6 X 𝑀𝑢
𝑓𝑐𝑘.𝑏.𝑑
2
eff,y
}.b .deff,y

Spacing of Bars in Middle strip
Assuming,ϕ -----mmTMT bar (ast =------ mm2)
Assuming, ϕ -----mm TMT bar (ast =------ mm2)
= 0.2
x b x D
100
=……….mm2

c) In column strip Along Longer Span
⚫ For max. +ve B.M.
mm
⚫ Effective depth; deff,y =
⚫ Assuming dcy’ = 35 mm
st,reqd
fy
4.6 X 𝑀𝑢
𝑐𝑘
𝑓 .𝑏.𝑑eff,y
2
⚫ A = 0.5 X fck X {1- 1 − }.b .deff,y
⚫ Spacing of Bars in Column strip
⚫Assuming,ϕ -----mmTMT bar (ast =------ mm2)

B) Along Shorter Span:-
a) In column strip AlongShorter Span
⚫ For max. –ve B.M.
mm
⚫ Effective depth; deffx =
⚫ Assuming dcx’= 25 mm
st,reqd
fy
⚫ A = 0.5 X fck X {1- 1 −
4.6 X 𝑀𝑢
𝑓𝑐𝑘.𝑏.deffx 2
}.b .d
⚫Spacing of Bars in Column strip
⚫Assuming, ϕ -----mm TMT bar (ast =------ mm2)

= 0.2
x b x D
100
=……….mm2
b) In middle Strip Along Shorter Span
⚫For max. –ve B.M.
⚫Effective depth; deffx = mm
⚫Ast,reqd
fy
= 0.5 X fck X {1- 1 −
4.6 X 𝑀𝑢
𝑐𝑘
𝑓 .𝑏.deffx
2
}.b .d

8] Reinforcement details:
Fig : Slab without Drops

Width of column strip
…….
s
d /2 s
d /2
Column (B x D)
Critical
section for
shear
ds
D= ……… mm
s
d = …….. mm
D
bc/2
bc/2
B
D
LY/2 LY/2
L
X
/2
L
X
/2

12] Details of Main & SecondaryReinforcement in Columnstrip
Reinf. details along
short span (direction) c
Span= Lx = ……. m
Span= Ly = ………… m
Long span (direction)
Provide …….mm dia. bar @......mm C/C at top in both
Directions as non-tension reinforcement. (straight)
Ds=…….mm
Ds=…….mm
dey=…….mm
dex=…….mm
c
Provide …….mm dia. Alt. Bent up bar @......mm C/C in column strip.

12] Details of Main & Secondary Reinforcement in Middlestrip
Ds=…….mm
Ds=…….mm
dey=…….mm
dex=…….mm
c
Provide …….mm dia. Alt. Bent up bar @......mm C/C in Middle strip.

Example 1: Design an interior panel of flat slab of
size 5m x 7m without providing drop and capital
head. Column size 500 x 500 mm and live load on
Panel is 4 kN/m2. take F.F.=1 kN/m2, use Fck=35 MPa,
Fy= 500MPa, Use IS:456 and direct method for
design. Assume density of RCC= 25kN/m3
Given Data :
Ly=7 m, Lx= 5 m, F.F.=1 kN/m2, Fck =35 MPa,

1] Thickness of slab (D) : (Cl. No.31.2.1 Pg. No.53)
a) If Drop is not provided then
D = ≥ 125 mm
Say, D= 245 mm
Assuming effective cover = 35 mm
Therefore, effective depth of slab(d);
d = 245 – 35=210 mm

Say, bc = Width on either side of Column centre line
bc = 0.25L1 or 0.25 L2 which ever is less
bc = 0.25 x L1 = 0.25 x 7 = 1.75 m
OR
bc =0.25 x L2 = 0.25 x 5 = 1.25 m
Provide bc = 1.25 m
Total Width of column strip = 2bc = 2 x 1.25 =2.5 m
Width of middle strip along longer span = L1- 2bc = 7-2x1.25 = 4.5 m
Width of middle strip along shorter span =L2- 2bc = 5-2x1.25= 2.5 m
2] Width of column strip (2bc) (Along longer span):
(Cl. No. 31.1.1, P.No.53, IS 456-2000)

3] Load calculations:-
i) D.L. = 0.245 x 24 = 5.88 kN/m2
ii) L.L. =
iii) F.F. =
= 4.00 kN/m2
= 1.00 kN/m2
WT = 10.88 kN/m2
Therefore, Ultimate load = 1.5 x WT = 16.32 kN/m2
4] To Find B.M.(maxi)
A) Along long span (L1= 7 m, L2 = 5 m)
Ln = L1 – (column side along L1) = 7- 0.5 = 6.5 m
Total design moment Mo =
W = 16.32 x 7 x 5 = 571.2 kN

5] Distribution of Moments:- (Along long span )
-ve Moments
= 0.75 x (Mo) -
= 0.75 x 301.67
= 226.25 kN.m
= 0.25 x (Mo) -
= 0.25 x 301.67
= 75.42 kN.m
+ve Moments
=0.60 x (Mo)+
=0.60 x 162.43
= 97.46 kN.m
=0.40 x (Mo)+
=0.40 x 162.43
= 64.97 kN.m

B) Along Short span (L1= 5 m, L2 = 7 m)
Total –ve Moment = 0.65 x 321.3 = 208.845 kN.m
Total +ve Moment = 0.35 x 321.3 = 112.455 kN.m
Ln = L1 - (column side along L1) = 5 - 0.5 = 4.5 m
W = 16.32 x 7 x 5 = 571.2 kN

-ve Moments
= 0.75 x (Mo) -
= 0.75 x 208.845
= 156.63 kN.m
= 0.25 x (Mo) -
= 0.25 x 208.845
= 52.211 kN.m
+ve Moments
=0.60 x (Mo)+
=0.60 x 112
= 67.20 kN.m
=0.40 x (Mo)+
=0.40 x 112
= 44.80 kN.m
6] Distribution of Moments:- Along short span

7] Design of Reinforcement
a) In column strip Along Longer Span
⚫ For max. –ve B.M. = 226.25 kN.m
⚫ Assuming dcy’= 35 mm
⚫ Effective depth; dyeff = 210 mm.
st,reqd
fy
⚫ A = 0.5 X fck X {1- 1 −
4.6 X 𝑀𝑢
𝑐𝑘
𝑓 .𝑏.dyeff
2
}.b .dyeff
= 0.5 X
35
500
x {1- 1 −
4.6 X 226.25 𝑋 106
35 𝑋2500𝑋2102
}X 2500 X 210
=2672.30mm2

= 0.2
x 2500 x 245
100
= 1225 mm2 <Ast,reqd
Therefore, provide Ast,reqd = 2672.30 mm2
∴ Spacing of Bars in Column strip
Assuming 12 mm dia. Bar, (ast =113 mm2)
Spacing = 113 X 2500
= 105.714 mm
2672.30
Provide 12 mm ϕ TMT bar @ 100 mm c/c (Straight Bar)

b) In Middle strip Along Longer Span
⚫ For max. –ve B.M. = 75.42 kN.m
⚫ Assuming dc’= 35 mm
⚫ Effective depth; deff = 210 mm.
st,reqd
fy
⚫ A = 0.5 X fck X {1-
𝑐𝑘
𝑓 .𝑏.𝑑2
1 − 4.6 X 𝑀𝑢
}.b .d
35
= 0.5 X X {1-
500 35 𝑋2500𝑋2102
1 − 4.6 X 75.42 𝑋 106
} X 2500 X 210
=845 .66 mm2
100
= 0.2
x 2500 x 245
= 1225 mm2 > Ast,reqd

Therefore, provide Ast,Min = 1225 mm2
∴ Spacing (S) = 𝑠𝑡
𝑎 𝑋 2500
Ast mi𝚗
,
= 230.61mm
1225
∴ Provide 12 mm ϕ TMT bar @ 225 mm c/c
(Straight Bar)

c) In Column strip Along Longer Span
⚫ For max. +ve B.M. = 97.46 kN.m
st,reqd
fy
⚫ A = 0.5 X fck X {1-
𝑐𝑘
𝑓 .𝑏.𝑑2
1 − 4.6 X 𝑀𝑢
}.b .d
35
= 0.5 X X {1-
500 35 X2500X2102
6
1 − 4.6 X 97.46 X 10
} X 2500 X 210
=1100.36mm2
100
= 0.2
x 2500 x 245

𝑎 𝑋 2500
Ast mi𝚗
,
= 230.61mm
1225
∴ Provide 12 mm ϕ TMT bar @ 225 mm c/c ( Alternate
bar Bent up)

d) In middle strip Along Longer Span
⚫ For max. +ve B.M. = 64.97 kN.m
st,reqd
fy
⚫ A = 0.5 X fck X {1-
𝑐𝑘
𝑓 .𝑏.𝑑2
1 − 4.6 X 𝑀𝑢
}.b .d
35
= 0.5 X X {1-
500 35 X2500X2102
6
1 − 4.6 X 64.97 X 10
} X 2500 X 210
= 725.915mm2
100
= 0.2
x 2500 x 245

B) Reinforcement along Shorter Span:-
a) In column strip along Shorter Span
⚫ For max. –ve B.M. = 156.63 kN.m
⚫ Effective depth; deffx = 225 mm.
⚫Ast,reqd
fy
= 0.5 X fck X {1- 1 −
4.6 X 𝑀𝑢
𝑐𝑘
𝑓 .𝑏.deffx
2
}.b . deffx
= 0.5 X
35
500
x {1- 1 −
4.6 X 156.63 𝑋 106
35 𝑋2500𝑋2252
}X 2500 X 225
=1670.11mm2

= 0.2
x 2500 x 245
100
= 1225 mm2 < Ast, Reqd
Therefore, Provide Ast,Reqd = 1225 mm2
𝑎 𝑋 2500
Ast
,
mi𝚗
1670.11
= 169.15mm
∴ Provide 12 mm ϕ TMT bar @ 165 mm c/c (Straight bars)

b) In Middle strip along Shorter Span
⚫ For max. - ve B.M. = 52.226 kN.m
⚫ Effective depth; deffx = 225 mm.
⚫Ast,reqd
= 0.5 X fck X {1- 1 −
fy
4.6 X 𝑀𝑢
𝑐𝑘
𝑓 .𝑏.deffx
2
}.b . deffx
= 0.5 X
35
500
x {1- 1 −
4.6 X 52.226𝑋 106
35 𝑋4500𝑋2252
}X 4500 X 225
=537 mm2

= 0.2
x 4500 x 245
100
∴ Spacing (S) = 𝑎𝑠𝑡
𝑋 4500
Ast mi𝚗
,
= 230.61mm
2205
∴ Provide 12 mm ϕ TMT bar @ 225 mm c/c ( straight
bars)

09) Non-tension Reinforcement :-( Cl.19.6.3.3 pg. No.27 IS: 1343-2012)
Use 10 mm dia.TMT bar as non tension Reinforcement (secondary
Reinforcement column and middle strip)
Provide 10mm dia.bar @160mm C/C in both Directions as non-tension
reinforcement in column and middle strip at Top & Bottom
(Secondary Reinforcement)
= 0.2
x 1000 x 245
100
= 490 mm2
𝑎 𝑋 1000
Ast
,
mi𝚗
490
∴ Spacing (S) = 78.54 𝑋 1000
=160.28 𝑚𝑚

10) Check for deflection under service load:
Deflection (long term along ly) is given:
le𝑦 = l𝑦 + side of column along the ly
= 7+ 0.5
= 7.5 m

= 29580 MPa
b = width of slab = 1000mm
h = overall thickness of slab = 245 mm

δ(permissible) = 30 mm
…………..ok.
δ𝑚𝑎𝑥
= 0.00772 × (10.88×10 ×7.5×7500
3 3
3.71×1013
)
δ𝑚𝑎𝑥 = 7.1633 𝑚𝑚
δ
(𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒) =
7500
250

longterm Deflection alonglx is given:
q = 10.88
b= width of slab= 1000mm
h = overall thickness of slab= 245 mm

le𝑦 = l𝑥 + side of column along the lx
= 5+ 0.5
= 5.5 m
10.88×103
×5.5×55003
3.71×1013 )
δ𝑚𝑎𝑥 = 0.00772 × (
δ𝑚𝑎𝑥 = 2.072 𝑚𝑚
δ(permissible) = 22 mm
δ
(𝑝𝑒𝑟𝑚𝑖𝑠𝑠𝑖𝑏𝑙𝑒) =
5500
250
…………..ok.

500 500
1250 1250
3500
3500
2500
2500
D= 245 mm
s
d = 210 mm 500
s
d /2 s
d /2
Column (500 x 500)
Critical
section for
shear
ds D
Width of column strip
11) Check for Shear: (Limit state of shear)

The critical section for shear shall be at a distance
ds/2 from periphery of column perpendicular to plane of
slab.
Where, d = effective depth of section
ds = effective depth of slab
Now here,
Upward force = downward force
shear resistance=load from drop OR load from slab to column
(Shear stress x shear area) = (load x shaded area)
Permissible shear stress (clause 31.6.3 pg.58 I.S.456-2000)

When shear reinforcement is not provided the
calculated shear stress at critical section shall not
exceed
ks = (0.5 + )

ks = (0.5 +1)
=1.5 < 1
ks = 1
shear resistance at critical section =
= kN/m2

Shear resistance = load from slab to column
(Shear stress x shear area) = (load x shaded area)
1480x [4 x (0.5 + ds) x ds] =16.32 x [(7x5)-(0.5+ds)2]
1480x[4 x (0.5ds + ds2)]=16.32 x [35-(0.25+1ds+ds2)]
5936.32 ds2+2976.32ds-567.12=0
ds=0.147 m = 147 mm < dprovided =210 mm
Therefore, section is safe in shear.

12] Reinforcement Detailing :
layout of column and middle strips

Column Strip Ly
lx
ly
Column Strip Ly
Column
Strip
Lx
Column
Strip
Lx
Middle Strip Ly
Middle
Strip
Lx
Flat slab

12 A] Details of Main & SecondaryReinforcement in Columnstrip
Ds=…….mm
Ds=…….mm
dey=…….mm
dex=…….mm
c

12 B] Details of Main & SecondaryReinforcement in Middlestrip
Ds=…….mm
Ds=…….mm
dey=…….mm
dex=…….mm
c

Assignment Example 2: The floor of a building
constructed of flat slabs is 30.0 m x 24.0 m. The
column centers are 6.0 m in both directions, and the
building is braced with shear walls. The panels are
to have drops of 3.0 m x 3.0 m. The depth of the
drop panel is 250 mm and the slab depth is 200
mm. The internal columns are 450 mm square and
the column heads are 900 mm with depth of 600
mm. The loads are as follows:

Dead load = self weight + 2.50 kN/m² for screed, floor
finishes, partitions and finishes
Imposed load = 3.50 kN/m²
The materials are grade M30 concrete and grade Fy250
reinforcement.
Design an internal panel next to an edge panel on two
sides and show the reinforcement details. Using
IS:456 and direct design method.
Use Magnel Cable is with 7-7 mm dia. Wire stressed
to 1000 MPa. Use IS:456 and direct method for design

DACS_UNIT-3- RCC Flat slab ARP.pptx

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DACS_UNIT-3- RCC Flat slab ARP.pptx