Design and Drawing of Reinforced concrete structures

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 L
eff = 0.65 L
Slenderness ratio =
L
eff
b
< 12 for square column
{Note L
eff and b are in mm}
Slenderness ratio =
L
eff
D
< 12 for circular column
{Note L
eff and D are in mm}
Hence, it may be designed as short column
2. Minimum eccentricity
Clause 25.4 in IS: 456
e
min =
L
500
+
b
30
for square column
{Note L
and b are in mm}
e
min =
L
500
+
D
30
for circular column
{Note L
eff and D are in mm}
i.e., e
min < 20 mm
Hence, it may be treated as axially loaded column
3. Main Reinforcement
Factored load P
u=1.5 P
Gross area A
g = b x b for square column
Gross area A
g =
π
/4 D
2
for circular column
A
g = A
sc + A
c
A
c = A
g - A
sc
Clause 39.3 in IS: 456
P
u= [0.4 f
ck A
c + 0.67 f
y A
sc]
P
u= [0.4 f
ck (A
g - A
sc) + 0.67 f
y A
sc]
{Note P
u in N/mm
2
}
Then find Provide A
sc value it should be with limits
From IS: 456 clause 26.5.3 columns
Minimum A
sc= 0.8% of A
g = 0.008 x A
g
Maximum A
sc= 6% of A
g = 0.06 x A
g
Minimum A
sc < Provide A
sc < Maximum A
sc
{Note If Minimum A
sc > Provide A
sc, then take Provide
A
sc value
If Provide A
sc > Maximum A
sc, then take Maximum A
sc
value}
[n
π
/4 ϕ
2
] = A
sc
ϕ is not less than 12 mm so ϕ = 20 mm for A
sc < 1000
mm
2
ϕ = 25 mm for A
sc < 2000 mm
2

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 P
u=1.5 P
Gross area A
g =
π
/4 D
2

A
g = A
sc + A
c
A
c = A
g - A
sc
P
u= 1.05 [0.4 f
ck A
c + 0.67 f
y A
sc]
P
u= 1.05 [0.4 f
ck (A
g - A
sc) + 0.67 f
y A
sc]
{Note P
u in N/mm
2
}
Then find Provide A
sc value it should be with limits

2. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
2

Minimum A
sc= 0.8% of A
g = 0.008 x A
g
Maximum A
sc= 6% of A
g = 0.06 x A
g
Minimum A
sc < Provide A
sc < Maximum A
sc
[n
π
/4 ϕ
2
] = A
sc
ϕ is not less than 12 mm so ϕ = 20 mm for A
sc < 1000
mm
2
ϕ = 25 mm for A
sc < 2000 mm
2

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 A
k=
π
/4 ψ
2
Volume of core V
k= A
k x S
Area of spiral A
H =
π
/4 8
2

Length of one spiral S
HL= π x [D – (2 x 40 + 8)]
Volume of helical reinforcement per pitch of the column,
V
H = A
H x S
HL
From IS: 456 clause 39.4.1
V
H
V
k
≥ 0.36 [
A
g
A
k
-1] x
f
ck
f
y

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 = b
1 x b
2
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 = B
1 x B
2
{Note B
2> B
1 hint B
1=0.93B
2}
2. Upward Soil pressure
Factored load P
u= 1.5 x P
Soil pressure at ultimate load
q
u= [P
u/Area of footing]
3. Depth of footing from Bending Moment
For square footing M
u= q
u
B(B−b)
2
8

For rectangle footing M
u= q
u
B
2(B−b)
2
8

For square footing M
ulim = 0.36 f
ck Bd
2

x
u
d
[1- (0.42
x
u
d
)]
For rectangle footing M
ulim = 0.36 f
ck B
2d
2

x
u
d
[1- (0.42
x
u
d
)]
x
u
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 M
u = 0.87 f
y A
st d’ [1-
f
yA
st
f
ckBd′
]
For rectangle footing M
u = 0.87 f
y A
st d’ [1-
f
yA
st
f
ckB
2d′
]
Find area of steel A
st
Use ϕ =12 mm diameter bars
a
st =
π
/4 ϕ
2

Spacing S= [a
st/ A
st] xB
5. Check for one way shear
For square footing V
u1= q
u B [
B(B−b)
8
- d' ]
Clause 40.1 in IS: 456
τ
v1= V
u1/ (Bd')
For rectangle footing V
u1= q
u B
2 [
B(B−b)
8
- d' ]
τ
v1= V
u1/ (B
2 d')
P
t= (
a
st
S d′
) x 100
τ
c1 = from table 9 in IS: 456

τ
c2 =0.5 x 2.8 for M20 from table 10 in IS: 456
τ
c3=
K τ
c1
K = from clause 40.2.1.1 in IS: 456
τ
v1 > τ
c1
τ
v1 > τ
c2
τ
v1 > τ
c3
Hence, O.k.,
6. Check for two way shear
Two way shear stress τ
v2 = V
u2/A
2
For square footing V
u2= q
u [B
2
-(b+d)
2
]
A
2= 4 (b+d) d
For rectangle footing V
u2= q
u [B
1xB
2 - (b
1+d) x (b
2+d)]
A
2= [(b
1+d) + (b
2+d)] x d
τ
p= 0.25 √f
ck
τ
p > τ
v2
7. Check for development length
Length available beyond the column face =0.5(B-b)
L
d = [0.87 fy ϕ]/[4 τ
bd ]
τ
bd= 1.6 x 1.2
0.5 (B-b) > L
d
Hence, O.k.




One way simply supported RCC slab design
l
y
l
x
> 2
1. Depth of slab
l
x
d
= 25
{Note l
x in mm}
d =
l
x
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. l
x+d
ii. l
x+w
t where w
t is wall thickness
Smaller value = l
eff
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 w
u= 1.5 w
4. Moment and Shear
For simply support udl load,
M
u =
w
u
8
l
eff
2

V
u=
w
u
2
l
eff
5. Check for moment
M
ulim = 0.36 f
ck bd
2

x
u
d
[1-(0.42
x
u
d
)]
x
u
d
= 0.48 for Fe 415
b=1000 mm
M
ulim > M
u
Hence safe
6. Main reinforcement
M
u = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd
]
Find area of steel A
st
Use ϕ = 10 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τ
v=
V
u
bd

P
t= (
a
st
S d
) x 100
τ
c1 = from table 9 in IS: 456
τ
c2 =0.5 x 2.8 for M20 from table 10 in IS: 456
τ
c3=
K τ
c1
K = from clause 40.2.1.1 in IS: 456
τ
v > τ
c1
τ
v > τ
c2
τ
v > τ
c3
Hence, O.k.,
8. Check for deflection
Provide
l
d
=25
P
t= (
a
st
S d
) x 100
f
s= 0.58 fy
Modification factor F
1=1.5 from figure 4 in IS: 456
Max
l
d
=20 F
1=30
Hence safe
9. Distribution Steel
A
g= 0.15%bd =0.0015bd for Fe250
A
g= 0.12%bd =0.0012bd for Fe415 and Fe 500
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] 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
l
y
l
x
> 2
1. Depth of slab
l
xh =
l
x
2
ie., intervals

l
xh
d
= 30
{Note l
x in mm}
d =
l
xh
30

D = d + C
{Note C=Cover =25 mm or 30 mm hint:better use round
figure 0 or ends 5}
2. Effective span
l
eff = l
x+d-w
t where w
t 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 = w
d = Dead load or Self weight +Floor
Finish
Total live load = w
l= Live Load or Super imposed load
Factored dead load w
ud = 1.5 w
d
Factored live load w
ul=1.5 w
l
4. Moment and shear
Table 12 of IS: 456
Support Moments (Negative)
At support to next to end support
M
u max-ve = [
w
ud
10
l
eff
2
]+[
w
ul
9
l
eff
2
]
Span moments (Positive)
M
u max+ve = [
w
ud
12
l
eff
2
] + [
w
ul
10
l
eff
2
]
M
u= M
u max-ve
Shear
Table 13 of IS: 456
V
u= 0.6[(w
ud l
eff)+( w
ul l
eff)]
5. Check for moment
M
ulim = 0.36 f
ck bd
2

x
u
d
[1- (0.42
x
u
d
)]
x
u
d
= 0.48 for Fe 415
b=1000 mm
M
ulim > M
u
Hence safe
6. Main reinforcement
M
u = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd
]
Find area of steel A
st
Use ϕ = 10 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τ
v=
V
u
bd

P
t= (
a
st
S d
) x 100
τ
c1 = from table 9 in IS: 456

6. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
6

τ
c2 =0.5 x 2.8 for M20
τ
c3=
K τ
c1
K = from clause 40.2.1.1 in IS: 456
τ
v > τ
c1
τ
v > τ
c2
τ
v > τ
c3
Hence, O.k.,
8. Check for deflection
Provide
l
d
=30
P
t= (
a
st
S d
) x 100
f
s= 0.58 f
y
Modification factor F
1=1.6 from figure 4 in IS: 456
Max
l
d
=26 F
1=40
Hence safe
9. Distribution Steel
A
g= 0.15%bd =0.0015bd for Fe250
A
g= 0.12%bd =0.0012bd for Fe415 and Fe 500
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 5d=
iii. Spacing 450 mm
Use spacing which is minimum


Two way RCC slab design
l
y
l
x
< 2 slab is supported all
around on wall
1. Depth of slab
l
x
d
= 25
{Note l
x in mm}
d =
l
x
25

D = d + C
{Note C=Cover =25 mm or 30 mm hint:better use round
figure 0 or ends 5}
2. Effective span
l
xeff = l
x+d
l
yeff = l
y+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 w
u = 1.5 w
4. Moment and shear
ANNEX D of IS: 456 table 26 case 9
α
x= and α
y=
M
ux = α
x w
u (l
xeff)
2

M
uy= α
y w
u (l
xeff)
2

Shear
V
u= w
uxλ x(
l
xeff
2
)
Where λ =
r
4
1+r
4
and r =
l
yeff
l
xeff

5. Main reinforcement in xdirection
M
ulim = 0.36 f
ck bd
2

x
u
d
[1- (0.42
x
u
d
)]
x
u
d
= 0.48 for Fe 415
b=1000 mm
M
ulim > M
ux
M
ux = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd
]
Find area of steel A
st
Use ϕ = 10 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

7. DESIGN AND DRAWING OF REINFORCED CONCRETE STRUCTURES
Syed Jeelani Basha (Assistant Professor)
7

iv. Spacing S= [
a
st
A
st
] xb
v. Spacing 3d=
vi. Spacing 300 mm
Use spacing which is minimum
Main reinforcement in y direction
** d’ = d - 8
M
uy = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd′
]
Find area of steel A
st
Use ϕ = 8 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τ
v=
V
u
bd

P
t= (
a
st
S d
) x 100
τ
c1 = from table 9 in IS: 456

τ
c2 =0.5 x 2.8 for M20
τ
c3=
K τ
c1
K = from clause 40.2.1.1 in IS: 456
τ
v > τ
c1
τ
v > τ
c2
τ
v > τ
c3
Hence, O.k.,
8. Check for deflection
Provide
l
d
= 25
P
t= (
a
st
S d
) x 100
f
s= 0.58 f
y
Modification factor F
1=1.6 from figure 4 in IS: 456
Max
l
d
=20 F
1=32
Hence safe
9. Distribution Steel
A
g= 0.15%bd =0.0015bd for Fe250
A
g= 0.12%bd =0.0012bd for Fe415 and Fe 500
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 5d=
iii. Spacing 450 mm
Use spacing which is minimum

Two way RCC slab design
l
y
l
x
< 2 slab, two adjacent
edges are continuous and other two discontinuous
1. Depth of slab
l
x
d
= 32
{Note l
x in mm}
d =
l
x
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

l
xeff = l
x+d
l
yeff = l
y+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 w
u = 1.5 w
4. Moment and shear
ANNEX D of IS: 456 table 26 case 4
α
x-ve= and α
y-ve=
M
u x-ve = α
x-ve w
u (l
xeff)
2

M
uy-ve= α
y-ve w
u (l
xeff)
2

α
x+ve= and α
y+ve=
M
ux+ve = α
x+ve w
u (l
xeff)
2

M
uy+ve= α
y+ve w
u (l
xeff)
2

Shear
V
u= w
uxλ x(
l
xeff
2
)
Where λ =
r
4
1+r
4
and r =
l
yeff
l
xeff

5. Main reinforcement in x–ve reinforement
M
ux-ve = 0.36 f
ck bd
2

x
u
d
[1- (0.42
x
u
d
)]
x
u
d
= 0.48 for Fe 415
b=1000 mm
M
ulim > M
ux
M
ux-ve = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd
]
Find area of steel A
st
Use ϕ = 10 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

vii. Spacing S= [
a
st
A
st
] xb
viii. Spacing 3d=
ix. Spacing 300 mm
Use spacing which is minimum
Main reinforcement in y-ve direction
** d’ = d - 8
M
uy-ve = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd′
]
Find area of steel A
st
Use ϕ = 8 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
6. Main reinforcement in x+ve reinforement
M
ux+ve = 0.36 f
ck bd
2

x
u
d
[1- (0.42
x
u
d
)]
x
u
d
= 0.48 for Fe 415
b=1000 mm
M
ulim > M
ux+ve
M
ux+ve = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd
]
Find area of steel A
st
Use ϕ = 10 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] 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

M
uy+ve = 0.87 f
y A
st d [1-
f
yA
st
f
ckbd′
]
Find area of steel A
st
Use ϕ = 8 mm diameter bars
{Note for A
st> 200 use ϕ =8 mm
A
st< 200 use ϕ =10,12 mm }
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 3d=
iii. Spacing 300 mm
Use spacing which is minimum
7. Check for Shear
τ
v=
V
u
bd

P
t= (
a
st
S d
) x 100
τ
c1 = from table 9 in IS: 456

τ
c2 =0.5 x 2.8 for M20
τ
c3=
K τ
c1
K = from clause 40.2.1.1 in IS: 456
τ
v > τ
c1
τ
v > τ
c2
τ
v > τ
c3
Hence, O.k.,
8. Check for deflection
Provide
l
d
= 32
P
t= (
a
st
S d
) x 100
f
s= 0.58 f
y
Modification factor F
1=1.65 from figure 4 in IS: 456
Max
l
d
=20 F
1=33
Hence safe
9. Distribution Steel
A
g= 0.15%bd =0.0015bd for Fe250
A
g= 0.12%bd =0.0012bd for Fe415 and Fe 500
a
st =
π
/4 ϕ
2

i. Spacing S= [
a
st
A
st
] xb
ii. Spacing 5d=
iii. Spacing 450 mm
Use spacing which is minimum