You can find reinforcement requirement and crack width using this spread sheet

1. ROAD DEVELOPMENT AUTHORITY
BRIDGE DESIGN DIVISION Sheet No.
Reference Calculations Out put
Bending moment Mx = 202 kNm
b = 1000 mm
Thickness t = 400 mm
Effective depth d = 342 mm
d' = 50 mm
cover = 50 mm
Main R/F =T 16 mm
Distribution R/F =T 10 mm
=
fcu 25 N/mm2
fy
= 460 N/mm2
Size of aggregate hagg = 20 mm
BS 5400 K = M K' = 0.156
Cl 5.3.2.3 bd2fcu
K = 0.06908 K<K' Compression reinforcement not required
BS 5400
Cl 5.3.2.3 Leaver arm
Eqn:5 Z = 0.5d 1+ 1- 5Mu
Modified bd2fcu
Concrete tec
clark Z = 309.35 mm
0.95d = 324.9
Z = 309.35 Z < 0.95d Use Z from formula
BS 5400 Calculation of main reinforcement
equation 1 As = M
0.87fyZ
As = 1631.63 mm2
Minimum areas of main reinforcement
BS 5400 As ≥ 0.15*b*d/100 = 513 mm2 Tore steel used
Part IV
CL 5.8.4.1
Calculated minimum reinforcement area is sufficient then use As From formula
Required reinforcement area = 1631.63 mm2/m

2. ROAD DEVELOPMENT AUTHORITY
BRIDGE DESIGN DIVISION Sheet No.
Reference Calculations Out put
Main R/F
use T 16 123.28 mm Distance 4.9 inch T 16
100
Actual reinforcement area A = 2011.43 mm2/m
Calculation of secondary reinforcement
B.S 5400 As ≥ 0.12bd/100
Part IV As = 410.4 mm2/m
CL 5.8.4.2
use T 10 191.45 mm Distance ### inch T 10
200
Actual reinforcement area A = 392.86 mm2/m
B.S 5400 Maximum reinforcement area in member
Part IV As ≤ bh*4/100
CL 5.8.5 = 16000 mm2
Therefore
Condition satisfies
Calculation of reinforcement for cmpression zone
B.S 5400
Part IV Provide minimum of main reinforcement for compression zone
CL 5.8.4.2
As ≥ 0.12bd/100
Therefore provide
= 410.4 mm2/m
use T 12 275.69 mm Distance ### inch T 12
200
B.S 5400 Minimum distance between bars ≥ hagg + 5 mm
Part IV
CL:5.8.8.1 ≥ 25 mm Condition satisfies
B.S 5400 Check for shear resistance
Part IV
cl:5.3.3.1 Ultimate shear force V = 83.5 kN/m
Shear stress ν = V
bd
= 0.2442
0.75 fcu = 3.75
0.75fcu>v condition satisfied for maximum shear

3. ROAD DEVELOPMENT AUTHORITY
BRIDGE DESIGN DIVISION Sheet No.
Reference Calculations Out put
0.27  100 As  1
B.S 5400 Vc =  ( f cu ) 3
Part IV γ m  bw



Table 8&9
Vc = 0.529 N/mm2
Depth factor
ξ s = (500 / d )1 / 4 or 0.70, whichever is the grater
ξs = 1.100
ξs vc = 0.582 N/mm2
ν < ξ s v c Condition satisfied
Check for crack width control
Moment at servicibility limite state SLS = 122.42 kNm
Copressive
strength fcu Static Modulus ( Ec)
N/mm2 kN/mm2 kN/mm2
20 25 21 to 29
25 26 22 to 30
30 28 23 to 33
40 31 26 to 36
50 34 28 to 40
60 36 30 to 42
Actual R/F
use T 16 100 mm Distance As = 2011.43 mm2
No compression reinforcement
Єc fcb
x
h d
Єs
fs
Є1
Actual Steel area As = 2011.43 mm2

4. ROAD DEVELOPMENT AUTHORITY
BRIDGE DESIGN DIVISION Sheet No.
Reference Calculations Out put
Grade of concrete Fcu = 25 N/mm 2
Static moduulus of concrete Ec = 26
Consider half modulus = 13 KN/mm2
Es = 200 KN/mm2
α = Es/Ec
= 15.38
propotion of reinforcement φ = As/bd
= 0.00588
Depth to neutral axis
Step 1
x = αφ 1+ 2 - 1
d αφ
X = 117.80 mm
Step 2
z = d - x/3
= 302.73 mm
Step 3 Steel stress
fs = M
AsZ
= 201.05 N/mm2
fs < 0.87fy pass
Step 4 Concrete stress
fcb = 2M
xbz
= 6.87 N/mm2
fcb < 0.45fcu pass
Step 5
Є1 = fs h-x

5. ROAD DEVELOPMENT AUTHORITY
BRIDGE DESIGN DIVISION Sheet No.
Reference Calculations Out put
Es d-x
= 0.001265
Step 6
Єs = fs
Es
= 0.001005
B.S 5400
Part IV  3.8bt h( a ' − d c )   Mq  −9 
equation 25 ε m = ε1 −   1 −

10 

 ε s As ( h − d c )   Mg  

but not greater than ε1
Where
 3.8bt h( a ' − d c )  Mq  −9 
ε2 =  1 −

10 

 ε s As (h − d c )   Mg  

ε1 - is the calculated strain at the level where
cracking is being considered, ignoring the
stiffening effect of the concrete in the
tension zone;
bt - is the width of the section at the level of
the centroid of the tension steel;
a' - is the distance from the compression face
to the point at which the crack width is
being calculated;
Mg - is the moment at the section considered
due to permanent loads;
Mq - is the moment at the section considered
due to live loads;
εs - is the calculated strain in the tension
reinforcement, ignoring the stiffening
effect of the concrete in the tension zone;
As - is the area of tension reinforcement.
Live Mq = 120.00 KNm
Dead Mg = 2.42 KNm
Є2 = -0.04 mm
Єm = Є1 - Є2
= 0.0377247 mm
Є2<0 so tht Єm=Є1

6. ROAD DEVELOPMENT AUTHORITY
BRIDGE DESIGN DIVISION Sheet No.
Reference Calculations Out put
Єm = 0.00127 mm
= d-x
r
acr acr = 68.58 mm
Step 8
Design crack width = 3acrЄm
1 + 2(acr-cnom)/(h-dc)
BS 5400-4
Table-1 where
Severe acr is the distance from the point (crack)
condition considered to the surface of the nearest
bar which controls the crack width;
Cnom is the required nominal cover to the
outermost reinforcement given in
Table 13; where the cover shown on the
drawing is greater than the value given
in Table 13, the latter value may be
used;
dc is the depth of the concrete in
compression (if dc = 0 the crack widths
should be calculated using equation 26);
h is the overall depth of the section;
Єm is the calculated strain at the level
where cracking is being considered,
allowing for the stiffening effect of the
concrete in the tension zone; a negative
value of Єm indicates that the section is
uncracked. The value of Єm should be
obtained from the equation:
Therefore design crack width
= 3acrЄm
1 + 2(acr-cnom)/(h-dc)
0.2300 < 0.25mm Crack width ok

7. ROAD DEVELOPMENT AUTHORITY
BRIDGE DESIGN DIVISION Sheet No.
Reference Calculations Out put