# exampl_conv_slabs by xiaoyounan

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```									                                   Example Conv slab - WJL

Design of r.c. slabs
Example (with solution)

A floor slab rests on circular columns 400 mm
diameter. The columns are equally spaced at 6.5m in
each direction. The slab has three equal spans in both
x and y directions. Figure below shows the interior
span.
h=250

Ly=6.5 m               Ly=Lx

The slab supports a variable load of 5 kN/m2. Char.
strengths of materials are: fck = 25 N/mm2 ;
fyk = 500 N/mm2. The effective depth d=205 mm.
a)    Design bending reinforcement for the slab
b)    Check and design the slab for punching shear
Ans: (a) Table 5.4 (Flat slab)
Col. Strip: support BM=43.4 kNm (75% of
M=57.9kNm); interior BM=23.3 kNm (25% of M=42.25
kNm). Use H10 at 150 c/c for hogging bending and
H10 at 200 c/c for sagg. bending (min. reinft)
Middle strip: Require H10 at 200 c/c (min.reinft)
(b) V=404 kN; reaction from column=2 kN Net V=
402 kN; Manual, p.34 Veff=462.3 kN;
vEd=0.59 N/mm2> vRd,c=0.54 N/mm2shear reinfort
required, radial spacing sr=150 mm; perimeter spacing
st= 300 mm; Asw=235 mm2. Place H8 links in the zone
0.3d=61.5mm and 1.5d=307 mm from the face of the
column.
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Example Conv slab - WJL

Reinforcement for r.c. slabs

Solution
per panel.
Total load per 1 m width of panel:
673.3/6.5=103.6 kN/m.
Let’s consider Ly direction (Fig.4 below).

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Example Conv slab - WJL

Conditions in M, 5.2.3.4 are satisfied, hence use
Table 5.4.
Centre of interior span
M=+0.063FL=0.063103.66.5=42.42 kNm/m
(sagging moment)
Interior support:
M=-0.086FL= -57.9 kNm/m (hogging moment)
Hence, with reference to Figure 4 below (or Fig.
4.2, Manual):
- division of moments into strips

Column strip along Ly     (Fig. 4.2, Manual)

Support moment (75% of M=-57.9 kNm)
Msupp=0.7557.9=43.4 kNm/m (hogging)

Interior moment (55% of M=+42.42kNm)
Mint= 0.5542.42=23.3 kNm/m (sagging)

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Example Conv slab - WJL

5.4 are

43.4 kNm

23.3 kNm

Resultant BM’s in the slab. Column strip- Ly direction

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Example Conv slab - WJL

As the bending moments are small, will need to
compare with the requirement of minimum
reinforcement.

Column strip- Reinforcement for hogging bending

M        43.4  106
K                          0.04
fck bd2 25  1000  2052

From Fig. 5.5, M  z/d=0.96
M             43.4  106
As                                        507mm2 / m
0.87fyk z 0.87  500  (0.96  205)

Select H10 bars @ 150mm c/c
As,prov=523mm2/m.

Middle strip along Ly
Support moment (25% of M)
Msupp=0.2557.9=14.5 kNm/m (hogging)

Interior moment (45% of M)
Mint= 0.4542.42=19.1 kNm/m (sagging)

Reinforcement for sagging bending:

M        19.1  106
K                           0.018  z/d=0.98
fck bd2 25  1000  2052

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Example Conv slab - WJL

M             19.1  106
As                                      219mm2 / m
0.87fyk z 0.87  500  (0.98  205)

Check min. reinforcement:

0.00014fck2/3bh=
.00014252/31000250=299mm2

or 0.0015bh=0.00151000250=375 mm2

Select H10 at 200 c/c , As,prov=393 mm2.
The value of Msupp is even lower (14.5 kNm/m),
hence use the min. reinforcement requirement
again, i.e., Select H10 at 200 c/c .
The sagging BM for the column strip, equal to
23.3 kNm/m will also require min.
reinforcement, i.e. H10 @ 200 C/c.

Column strip along Ly

H10-150                       H10-150

each way                      each way

H10-200
each way          Reinforcement to follow
detailing rules,
Ly
Manual 5.12.

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Example Conv slab - WJL

Repeat the same reinforcement along Lx.
For Middle strip along Ly use minimum
reinforcement for both the column and the
middle strips.
b) Punching shear
From Table 5.4, Manual,
V=0.6F=0.6673.3= 404 kN
Reaction from the column:
0.42/4[673.3/(6.5)2]= 2 kN
Net shear force: 404-2=402 kN
To get the effective shear force, increase the
above by 15% (Manual, p.34).
Hence Veff= 4021.15=462.3 kN.
Check column perimeter
u0=2200=1256.6 mm
1000Veff 1000  462.3
vE d                            1.71 N/mm2
u0d     1256 .6  205
Check the limit:

0.2[1-(fck/250)]fck=

=0.2[1-25/250]25=4.5        N/mm2

vEd=1.71 N/mm2<4.5 N/mm2                OK.
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Example Conv slab - WJL

Check shear at the basic control perimeter,
u1

2d=2205=410 mm

u1=2(200+410)=3832.7 mm

2d     u1

400

1000Veff   1000  462.3
vE d                            0.59 N/mm2
u1d      3832.7  205

Longitudinal bars used for bending
reinforcement, H10 at 200 mm spacing, give
As=393 mm2/m,

so 100As/bd=100393/(1000205)=0.19%,
which gives vRd,c=0.54 N/mm2 (Table 5.7,
Manual).

vEd=0.59 N/mm2> vRd,c=0.54 N/mm2. Hence
shear reinforcement is required.

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Example Conv slab - WJL

Shear reinforcement, Asw

The required area of shear reinforcement along
one perimeter, in the zone between the

column face and 1.5d inside the outer
perimeter (Manual p. 41):

v
vRd,cs  0.7 5 Rd,c
Asw 
fywd, ef
1.5
sru1

fywd,ef =250+0.25d=250+0.25205=301 N/mm2

sr=0.75d=0.75205=153.75 mm 150 mm

and putting: vRd,cs=vEd

0.59  0.75  0.54
Asw                         235mm2
301
(1.5              )
150  3832.7

Minimum shear reinforcement (M, p.44)

With the radial spacing sr =150mm and assumed
perimeter spacing st of 1.5d=307300mm, the
minimum area of one leg of the shear link is:

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Example Conv slab - WJL

0.08 fck sr st        0.08 25  150  300
As w,min                                                24mm2
1.5fyk                   1.5  500

The smallest bar that can be used is 8 mm dia.,
giving Asw=50.3 mm2 .

Outer perimeter, uout

vEd=vRd,c=Veff/(uoutd)

Hence

uout= Veff/(vRd,cd)

=462.3103/(0.53205)=4255mm;

a circle of radius 4255/2 = 677.2mm.

With the radius of the column=200mm,

uout< 3d from the face of the column. Thus, the
reinforcement must be placed in the zone not
less than 0.3d=61.5 mm (up to 0.5d) and
1.5d=307 mm from the face of the column.

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