design of one way slab two way slab continuous slab square and circular column isolated square footing

1. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
1
Design of square column and circular column
1. Size of column b x b for square column
Diameter of column D for circular column
Unsupported length L
For column is effectively held in position and direction at
the both the ends,
From table 28 in IS: 456 Effective length Leff = 0.65 L
Slenderness ratio =
Leff
b
< 12 for square column
{Note Leff and b are in mm}
Slenderness ratio =
Leff
D
< 12 for circular column
{Note Leff and D are in mm}
Hence, it may be designed as short column
2. Minimum eccentricity
Clause 25.4 in IS: 456
emin =
L
500
+
b
30
for square column
{Note L and b are in mm}
emin =
L
500
+
D
30
for circular column
{Note Leff and D are in mm}
i.e., emin < 20 mm
Hence, it may be treated as axially loaded column
3. Main Reinforcement
Factored load Pu=1.5 P
Gross area Ag = b x b for square column
Gross area Ag = π
/4 D2
for circular column
Ag = Asc + Ac
Ac = Ag - Asc
Clause 39.3 in IS: 456
Pu= [0.4 fck Ac + 0.67 fy Asc]
Pu= [0.4 fck (Ag - Asc) + 0.67 fy Asc]
{Note Pu in N/mm2
}
Then find Provide Asc value it should be with limits
From IS: 456 clause 26.5.3 columns
Minimum Asc= 0.8% of Ag = 0.008 x Ag
Maximum Asc= 6% of Ag = 0.06 x Ag
Minimum Asc < Provide Asc < Maximum Asc
{Note If Minimum Asc > Provide Asc, then take Provide
Asc value
If Provide Asc > Maximum Asc, then take Maximum Asc
value}
[n π
/4 ϕ2
] = Asc
ϕ is not less than 12 mm so ϕ = 20 mm for Asc < 1000
mm2
ϕ = 25 mm for Asc < 2000 mm2
Find no. of bars n required
4. Lateral ties
From IS: 456 clause 26.5.3.2 c Transverse reinforcement
Diameter of lateral ties should not less than
i. ϕ/4
ii. 6 mm
Adopt 6 mm diameter bars
Pitch of the ties shall be minimum of
i. Least lateral dimension of column = b for
square column
Least lateral dimension of column = D for
circular column
ii. 16 x ϕ
iii. 300 mm
Circular column with helical reinforcement
Step 1 and step 2 same as above mentioned
3. Factored load Pu=1.5 P
Gross area Ag = π
/4 D2
Ag = Asc + Ac
Ac = Ag - Asc
Pu= 1.05 [0.4 fck Ac + 0.67 fy Asc]
Pu= 1.05 [0.4 fck (Ag - Asc) + 0.67 fy Asc]
{Note Pu in N/mm2
}
Then find Provide Asc value it should be with limits

2. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
2
Minimum Asc= 0.8% of Ag = 0.008 x Ag
Maximum Asc= 6% of Ag = 0.06 x Ag
Minimum Asc < Provide Asc < Maximum Asc
[n π
/4 ϕ2
] = Asc
ϕ is not less than 12 mm so ϕ = 20 mm for Asc < 1000
mm2
ϕ = 25 mm for Asc < 2000 mm2
Find no. of bars n required
4. {Note: If helical reinforcement not asks do same as
above step 4 Pitch and Diameter}
For Helical reinforcement Clause 39.4 in IS: 456
Take 8 mm spirals at pitch S with clear cover of 40 mm
Core diameter ψ = D – (2 x 40) + (2 x 8)
Area of core Ak= π
/4 ψ 2
Volume of core Vk= Ak x S
Area of spiral AH = π
/4 8 2
Length of one spiral SHL= π x [D – (2 x 40 + 8)]
Volume of helical reinforcement per pitch of the column,
VH = AH x SHL
From IS: 456 clause 39.4.1
VH
Vk
≥ 0.36 [
Ag
Ak
-1] x
fck
fy
Find pitch S
From IS: 456 clause 26.5.3.2 c Transverse reinforcement
The maximum pitch shall be minimum of
i. S
ii. (Core diameter)/6 = ψ/6
iii. 75 mm
Design of square footing and rectangle footing
1. Size of the footing
Column of size for square column = b x b
Column of size for rectangle column = b1 x b2
Load from column = P
Self weight of footing = 10% of P = 0.1 x P
Total load = Load from column + self weight of footing
Area of footing = [Total load/ Safe bearing capacity of
soil]
Area of square footing = B x B
B =√Area of footing
Area of rectangle footing = B1 x B2
{Note B2> B1 hint B1=0.93B2}
2. Upward Soil pressure
Factored load Pu= 1.5 x P
Soil pressure at ultimate load
qu= [Pu/Area of footing]
3. Depth of footing from Bending Moment
For square footing Mu= qu
B(B−b)2
8
For rectangle footing Mu= qu
B2(B−b)2
8
For square footing Mulim = 0.36 fck Bd2 xu
d
[1- (0.42
xu
d
)]
For rectangle footing Mulim = 0.36 fck B2d2 xu
d
[1- (0.42
xu
d
)]
xu
d
= 0.48 for Fe 415
Find depth d value

3. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
3
d' = 2 x d
D = d' + Cover
Cover = 50 mm
4. Reinforcement
For square footing Mu = 0.87 fy Ast d’ [1-
fy Ast
fck Bd′
]
For rectangle footing Mu = 0.87 fy Ast d’ [1-
fy Ast
fck B2d′
]
Find area of steel Ast
Use ϕ =12 mm diameter bars
ast =π
/4 ϕ2
Spacing S= [ast/ Ast] xB
5. Check for one way shear
For square footing Vu1= qu B [
B(B−b)
8
- d' ]
Clause 40.1 in IS: 456
τv1= Vu1/ (Bd')
For rectangle footing Vu1= qu B2 [
B(B−b)
8
- d' ]
τv1= Vu1/ (B2 d')
Pt= (
ast
S d′
) x 100
τc1 = from table 9 in IS: 456
τc2 =0.5 x 2.8 for M20 from table 10 in IS: 456
τc3= K τc1
K = from clause 40.2.1.1 in IS: 456
τv1 > τc1
τv1 > τc2
τv1 > τc3
Hence, O.k.,
6. Check for two way shear
Two way shear stress τv2 = Vu2/A2
For square footing Vu2= qu [B2
-(b+d)2
]
A2= 4 (b+d) d
For rectangle footing Vu2= qu [B1xB2 - (b1+d) x (b2+d)]
A2= [(b1+d) + (b2+d)] x d
τp= 0.25 √fck
τp > τv2
7. Check for development length
Length available beyond the column face =0.5(B-b)
Ld = [0.87 fy ϕ]/[4 τbd ]
τbd= 1.6 x 1.2
0.5 (B-b) > Ld
Hence, O.k.
One way simply supported RCC slab design
ly
lx
> 2
1. Depth of slab
lx
d
= 25
{Note lx in mm}
d =
lx
25
D = d + C
{Note C=Cover =25 mm or 30 mm hint:better use round
figure 0 or ends 5}
2. Effective span
Whichever is lesser use as effective span.

4. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
4
i. lx+d
ii. lx+wt where wt is wall thickness
Smaller value = leff
3. Loads
Dead load = Volume of Concrete x Density of concrete
= D x 25 KN/m
{Note D in meters}
Total load = w = Dead load or Self weight + Live Load
or Super imposed load + Floor Finish
Factored load wu= 1.5 w
4. Moment and Shear
For simply support udl load,
Mu =
wu
8
leff
2
Vu=
wu
2
leff
5. Check for moment
Mulim = 0.36 fck bd2 xu
d
[1-(0.42
xu
d
)]
xu
d
= 0.48 for Fe 415
b=1000 mm
Mulim > Mu
Hence safe
6. Main reinforcement
Mu = 0.87 fy Ast d [1-
fy Ast
fck bd
]
Find area of steel Ast
Use ϕ = 10 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τv=
Vu
bd
Pt= (
ast
S d
) x 100
τc1 = from table 9 in IS: 456
τc2 =0.5 x 2.8 for M20 from table 10 in IS: 456
τc3= K τc1
K = from clause 40.2.1.1 in IS: 456
τv > τc1
τv > τc2
τv > τc3
Hence, O.k.,
8. Check for deflection
Provide
l
d
=25
Pt= (
ast
S d
) x 100
fs= 0.58 fy
Modification factor F1=1.5 from figure 4 in IS: 456
Max
l
d
=20 F1=30
Hence safe
9. Distribution Steel
Ag= 0.15%bd =0.0015bd for Fe250
Ag= 0.12%bd =0.0012bd for Fe415 and Fe 500
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 5d=
iii. Spacing 450 mm
Use spacing which is minimum

5. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
5
One way continuous RCC slab design
ly
lx
> 2
1. Depth of slab
lxh =
lx
2
ie., intervals
lxh
d
= 30
{Note lx in mm}
d =
lxh
30
D = d + C
{Note C=Cover =25 mm or 30 mm hint:better use round
figure 0 or ends 5}
2. Effective span
leff = lx+d-wt where wt is wall thickness
3. Loads
Dead load = Volume of Concrete x Density of concrete
= D x 25 KN/m
{Note D in meters}
Total dead load = wd = Dead load or Self weight +Floor
Finish
Total live load = wl= Live Load or Super imposed load
Factored dead load wud = 1.5 wd
Factored live load wul=1.5 wl
4. Moment and shear
Table 12 of IS: 456
Support Moments (Negative)
At support to next to end support
Mu max-ve = [
wud
10
leff
2
]+[
wul
9
leff
2
]
Span moments (Positive)
Mu max+ve = [
wud
12
leff
2
] + [
wul
10
leff
2
]
Mu= Mu max-ve
Shear
Table 13 of IS: 456
Vu= 0.6[(wud leff)+( wul leff)]
5. Check for moment
Mulim = 0.36 fck bd2 xu
d
[1- (0.42
xu
d
)]
xu
d
= 0.48 for Fe 415
b=1000 mm
Mulim > Mu
Hence safe
6. Main reinforcement
Mu = 0.87 fy Ast d [1-
fy Ast
fck bd
]
Find area of steel Ast
Use ϕ = 10 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τv=
Vu
bd
Pt= (
ast
S d
) x 100
τc1 = from table 9 in IS: 456

6. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
6
τc2 =0.5 x 2.8 for M20
τc3= K τc1
K = from clause 40.2.1.1 in IS: 456
τv > τc1
τv > τc2
τv > τc3
Hence, O.k.,
8. Check for deflection
Provide
l
d
=30
Pt= (
ast
S d
) x 100
fs= 0.58 fy
Modification factor F1=1.6 from figure 4 in IS: 456
Max
l
d
=26 F1=40
Hence safe
9. Distribution Steel
Ag= 0.15%bd =0.0015bd for Fe250
Ag= 0.12%bd =0.0012bd for Fe415 and Fe 500
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 5d=
iii. Spacing 450 mm
Use spacing which is minimum
Two way RCC slab design
ly
lx
< 2 slab is supported all
around on wall
1. Depth of slab
lx
d
= 25
{Note lx in mm}
d =
lx
25
D = d + C
{Note C=Cover =25 mm or 30 mm hint:better use round
figure 0 or ends 5}
2. Effective span
lxeff = lx+d
lyeff = ly+d
3. Loads
Dead load = Volume of Concrete x Density of concrete
= D x 25 KN/m
{Note D in meters}
Total load = w = Dead load or Self weight +Floor Finis
h+ Live Load or Super imposed load
Factored load wu = 1.5 w
4. Moment and shear
ANNEX D of IS: 456 table 26 case 9
αx= and αy=
Mux = αx wu (lxeff)2
Muy= αy wu (lxeff)2
Shear
Vu= wuxλ x(
lxeff
2
)
Where λ =
r4
1+r4 and r =
lyeff
lxeff
5. Main reinforcement in xdirection
Mulim = 0.36 fck bd2 xu
d
[1- (0.42
xu
d
)]
xu
d
= 0.48 for Fe 415
b=1000 mm
Mulim > Mux
Mux = 0.87 fy Ast d [1-
fy Ast
fck bd
]
Find area of steel Ast
Use ϕ = 10 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2

7. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
7
iv. Spacing S= [
ast
Ast
] xb
v. Spacing 3d=
vi. Spacing 300 mm
Use spacing which is minimum
Main reinforcement in y direction
** d’ = d - 8
Muy = 0.87 fy Ast d [1-
fy Ast
fck bd′
]
Find area of steel Ast
Use ϕ = 8 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τv=
Vu
bd
Pt= (
ast
S d
) x 100
τc1 = from table 9 in IS: 456
τc2 =0.5 x 2.8 for M20
τc3= K τc1
K = from clause 40.2.1.1 in IS: 456
τv > τc1
τv > τc2
τv > τc3
Hence, O.k.,
8. Check for deflection
Provide
l
d
= 25
Pt= (
ast
S d
) x 100
fs= 0.58 fy
Modification factor F1=1.6 from figure 4 in IS: 456
Max
l
d
=20 F1=32
Hence safe
9. Distribution Steel
Ag= 0.15%bd =0.0015bd for Fe250
Ag= 0.12%bd =0.0012bd for Fe415 and Fe 500
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 5d=
iii. Spacing 450 mm
Use spacing which is minimum
Two way RCC slab design
ly
lx
< 2 slab, two adjacent
edges are continuous and other two discontinuous
1. Depth of slab
lx
d
= 32
{Note lx in mm}
d =
lx
32
D = d + C
{Note C=Cover =25 mm or 30 mm hint: better use round
figure 0 or ends 5}
2. Effective span

8. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
8
lxeff = lx+d
lyeff = ly+d
3. Loads
Dead load = Volume of Concrete x Density of concrete
= D x 25 KN/m
{Note D in meters}
Total load = w = Dead load or Self weight +Floor Finis
h+ Live Load or Super imposed load
Factored load wu = 1.5 w
4. Moment and shear
ANNEX D of IS: 456 table 26 case 4
αx-ve= and αy-ve=
Mu x-ve = αx-ve wu (lxeff)2
Muy-ve= αy-ve wu (lxeff)2
αx+ve= and αy+ve=
Mux+ve = αx+ve wu (lxeff)2
Muy+ve= αy+ve wu (lxeff)2
Shear
Vu= wuxλ x(
lxeff
2
)
Where λ =
r4
1+r4 and r =
lyeff
lxeff
5. Main reinforcement in x–ve reinforement
Mux-ve = 0.36 fck bd2 xu
d
[1- (0.42
xu
d
)]
xu
d
= 0.48 for Fe 415
b=1000 mm
Mulim > Mux
Mux-ve = 0.87 fy Ast d [1-
fy Ast
fck bd
]
Find area of steel Ast
Use ϕ = 10 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2
vii. Spacing S= [
ast
Ast
] xb
viii. Spacing 3d=
ix. Spacing 300 mm
Use spacing which is minimum
Main reinforcement in y-ve direction
** d’ = d - 8
Muy-ve = 0.87 fy Ast d [1-
fy Ast
fck bd′
]
Find area of steel Ast
Use ϕ = 8 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
6. Main reinforcement in x+ve reinforement
Mux+ve = 0.36 fck bd2 xu
d
[1- (0.42
xu
d
)]
xu
d
= 0.48 for Fe 415
b=1000 mm
Mulim > Mux+ve
Mux+ve = 0.87 fy Ast d [1-
fy Ast
fck bd
]
Find area of steel Ast
Use ϕ = 10 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
Main reinforcement in y+ve direction
** d’ = d - 8

9. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
9
Muy+ve = 0.87 fy Ast d [1-
fy Ast
fck bd′
]
Find area of steel Ast
Use ϕ = 8 mm diameter bars
{Note for Ast> 200 use ϕ =8 mm
Ast< 200 use ϕ =10,12 mm }
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τv=
Vu
bd
Pt= (
ast
S d
) x 100
τc1 = from table 9 in IS: 456
τc2 =0.5 x 2.8 for M20
τc3= K τc1
K = from clause 40.2.1.1 in IS: 456
τv > τc1
τv > τc2
τv > τc3
Hence, O.k.,
8. Check for deflection
Provide
l
d
= 32
Pt= (
ast
S d
) x 100
fs= 0.58 fy
Modification factor F1=1.65 from figure 4 in IS: 456
Max
l
d
=20 F1=33
Hence safe
9. Distribution Steel
Ag= 0.15%bd =0.0015bd for Fe250
Ag= 0.12%bd =0.0012bd for Fe415 and Fe 500
ast =π
/4 ϕ2
i. Spacing S= [
ast
Ast
] xb
ii. Spacing 5d=
iii. Spacing 450 mm
Use spacing which is minimum