Base on B.C.Punmia book RCCdesign exmpale 8.

5
DESIGN OF STAIR CASE WITH CENTRAL STRINGER BEAM

Project name :- pkn

Space available for staire case 7.30 x 2.90 m
Vetical distance of floor 2.53 m 2528 mm
width of stair case 1.40 mtr 1400 mm
Risers 0.16 mtr 158 mm
Treads 0.30 mtr 300 mm
Live load 4000 N/m2
Conrete M- 25 m 10.98
wt. 25000 N/mm3 scbc 8.5 N/mm2
Steel fy 500 sst 275 N/mm2
Nominal cover 25 mm Effective cover 37.5 mm
West slab thickness 80 mm
assume width of stringer beam 300 mm
Reinforcement
Main Bars 25 mm F Bars Required 2 Nos.
Anchor bars 20 mm F Bars Required 2 Nos.
Waist slab bars 10 mm F Sapcing c/c 300 mm
Strirrups 10 mm F Sapcing c/c 170 mm

1.35 4.50 1.45

450
2- 20 mm anchore bars
Bottom of waist
10 mm f @
2.20 300 mm c/c
stirrups 2 ldg.
10 mm F @ 300
170 mm c/c 158 10 mm f @
stirrups 2 ldg. 300 mm c/c
10 mm F @ Landing
340 mm c/c 200
mid span 8 mm F @
2.20 300 mm c/c
stirrups 2 ldg.
L - section 10 mm F @
170 mm c/c

10 mm f @ 300 mm c/c
1450

200 mm
 200 mm

2 - 20 mm f
10 mm f @ 170 mm c/c Anchor bars

10 mm F @ 340 mm c/c 2 -25 mm f

stirrups 2 ldg. main bars
300
pk_nandwana@yahoo.co.in Cross section at mid span
 Genral lay out of stair case
Note:- Red color data are
theoretical calculation data
where black data are as per
1.40 4.50 1.40 design data.
treads= 15
1.35 4.50 1.45 7.30

D C 1.45 1.40

2.90
Treads= 0 0.00 0.10
2.90
treads= 0

A B 1.45 1.40

0.00 1.45
1.45
7.30
No of riser required = 2528 / 158 = 16 Space Space
require requir
No of tread required = 16 - 1 = 15 d (mm) ed (m)

No of treads accomeded 4500 / 300 = 15 x 300 4500 4.50

in long direction=

No of tread accomeded in 100 / 300 = 0 x 300 0 0.00

short direction
No of treads Required in 15 - 15 = 0 x 250 0 0.00
other long direction=
15 115 20 45 33 44 18 40 20 41 20 40 18
DESIGN OF STAIR CASE WITH CENTRAL STRINGER BEAM

width of stair case 1.40 mtr 1400 mm

Risers 0.16 mtr 158 mm Treads
live load 4000 N/m2
Conrete M- 25 Steel fy-
Nominal cover 25 mm Effective cover

1 Design Constants:- For HYSD Bars Cocrete M = 25

sst = = 275 N/mm2 wt. of concrete = 25000 N/mm2

scbc = = 8.5 N/mm3
m = 10.98
m*c 10.98 x 8.5
k= = = 0.253
m*c+sst 10.98 x 8.5 + 275
j=1-k/3 = 1 - 0.253 / 3 = 0.916
R=1/2xc x j x k = 0.5 x 8.5 x 0.916 x 0.253 = 0.986

2 Loading on waist slab:-

Assume waist slab thickness = 200 mm
The weight of waist slab on the slope should be multi plied by the factor
Ö R2 +T2 where R= 158 mm and T =
T
= Ö 158 +
2
300 2
= 1.13 to get the equivalent weight of horizontal p
300
Considered 1 m width of slab. Load per metre horizontal run will be as follows.

Self weight = 0.2 x 1 x 1 x 25000 x

Weight of steps = 0.5 x 0.16 1 x 1 x
Laoding of finishing = L.S.
Live load =

The loading on landing will be lasser : however , for simplicity , we will take the same loading throught.

3 Design of waist slab:-

the waist slab is supported on central stringer beam . Hence the worst condition
when we considred concentrated live load of 4000 N to act to one side only.
Dead weight = 5650 + 1975 + 100
Assume width of stringer beam = 300 mm
1.40 - 0.3
Projection of slab beyond the rib of beam =
2
wL2 7725 x( 0.55 )2
B.M. due to dead load =
2 2
wL2 4000 x( 0.55 )2
B.M. due to U.D. live load =
2 2
B.M. due to concentreted live load = 4000 x 0.55
Max. B.M. M = 1168 + 2200
 BM 3368000
Effective depth required = = = 59
Rxb 0.986 x 1000
However , keep minimum total depth = 200 mm . Efective depth =
BM x 1000 3368000
Ast = = = 76
sst x j x D 275 x 0.916 x 175
3.14xdia2 3.14 x
using 10 mm F bars A = =
4 x100 4
spacing of Bars = A*1000/Ast = 79 x 1000 / 76
However , keep spacing = 300 mm , one bar per step
Distribution reinforcement = 1.2 x 200 = 240 mm2
3.14xdia 2
3.14 x
using 8 mm F bars A = =
4 x100 4
spacing of Bars = A*1000/Ast = 50 x 1000 / 240
Maimum permissible spacing = 45 x 8 = 360 mm say
However , keep spacing = 300 mm, Maximum

4 Design of stringer beam :-

The stringer beam will act as T- beam. Flight CD is longest,
Hence we will design the stringer beam CD
0.3 1.45
Effective span = 1.35 - + 4.50 + =
2 2
The loading on stringer beam will be as follows,
asssuming the web to be 300 mm wide and 300 mm deep
(a) Weight of rib /m run = 0.30 x 0.30 x 25000 x 1.13 =
(b) Load from waist slab = 11725 x 1 x 1.35 =
Total =
Assuming partial fixidity at ends,
wL2 18400 x 6.43 2
x 1000
M = = = 75956350
10 10
Taking lever arm = 0.9 x d , balance depth is given by Eq.
2kcd - D where bf = flange width of isolated T-Beam given b
M = 0.45 bf. scbc. Df
kc
l0 where lo = L= 6.43 m ;b = actual width
bf = + bw,
l0 and bw = 0.30 m
+ 4
b
6.43
bf = + 0.30 = 1.034 m =
6.43
+ 4
1.35
2x 0.253 d-
Hence, = 0.45 x 1034 x 8.5 x 200
0.253
2 x 0.253 d- 200
\ 791010 = 75956350
0.253
200
\ 2d - = 96.02
0.253
96.02 + 789.31
d = = 443.00 mm
2
Ve where tc max =1.8 N/mm2 for m-20 concrete
Also, d =
bw . Tc max
 T wL 18400 x
Ve = V+ 1.6 Where V = =
bw 2 2
T = torsional moment, which will be induced due to live load acting only to one side of step
4000 x 0.55 2
x 6.43
T = x 1000 = 1943562.5
2 2
6.4
or T =( 4000 x 0.55 )x x 1000 = 7067500
2
\ T = 7067500 which ever is more
7067500
Ve = 59110 + 1.6 x = 96803.333333
300
96803.3333333333
Hence, d = = 202 mm However, keep total depth
300 x 1.6
using 25 mm main bars, 10 mm F ring and cover
Net available d = 450 - 12.5 - 10 - 25 = 403 mm
BM x 1000 75956350
Ast = = = 750
sst x j x D 275 x 0.916 x 403
3.14xdia 2
3.14 x
using 25 mm F bars A = =
4 x100 4
spacing of Bars = Ast/A = 750 / 491 = 2 Nos. Say
Actual Ast provided = 2 x 491 = 981.25 mm2
Note:- the above reinforcement is for bending requirements only. there will be additional longitudinal reinforce
later.
Location of N.A. Assuming the N.A. falls within the flange, we have
1034 x n2
= = 10.98 x 981.25 x( 403 -
2
\ 1034 x n2 = 8673171 - 21548.25 n
\ n 2
= 8388 - 20.84 n
\ n 2
+ 20.84 n - -8388 = 0
20.84 + ( 434 + 33552 )0.5
\ n =
2*1
\ n = 81.8 mm Hence the resultant falls inside the flanges y
\ L.A. a = d-y = 403 - 27.252 = 375.2
M 75956350
Stress in steel = = = 206
Ast . A 981.25 x 375
Corresponding stress in concrete is given by
txn 206 81.8
c = = x = 4.79
m d-n 10.98 403 - 81.8
5 Design for torsion :-
As computed earlier, T= 7067500 N-mm
v = 59110 N and ve = 96803.333333 N
Ve 96803.3333333333
\ tve = = = 0.80 N/mm2
bw.d 300 x 403
100 x Ast 100 x 981.25
= = 0.8126 % hence from table 3.2 tc
bxd 300 x 403
Since tve > tc shear reinforcemnt required
(a) Longitudinal reinforcement:-
M e1 = M + MT Where M= 75956350
(1+ D/bw) 1 + 450 /
MT = T = 7067500
 MT = T = 7067500
1.7 1.7
\ Me1 = 75956350 + 10393382 = 86349732 N-mm
Me1 86349732
Ast = = = 837 mm2
sst x j x D 275 x
375.25
Hence the provision of 2 bars of 25 mm f, giving 981.25 mm2 is
Near the column D, take the bars straight up. Provide 2- 25 mm f bars at the lower face under the

(b) Transverse reinforcement:-

Transverse reinforcement will be provided in the form of vertical stirrups.
Let provide 25 mm clear cover all round
b1 = center to center distence between corner bars in the direction of width
= 300 - 2 x 25 - 10 = 240 mm
d1 = center to center distence between corner bars in the direction of depth
= 450 - 2 x 25 - 10 = 390 mm
mm F stirrups 3.14xdia2 3.14 x
using 10 A= 2x = 2x
bars 4 x100 4
T.sv + Vsv
now, Asv = = or
b1d1ssv2.5d1 ssv
7067500 59110
157 = +
240 x 390 x 275 2.5 x 390 x
157 =( 0.2746 + 0.2205 )Sv or Sv = 157 /
However, the spacing should not exceed the least of x 1, (x1+y1)/4 and 300 mm
where x1= short diamension of stirrups = 240 + 20 + 10 =
y1 = 390 + 20 + 10 =
(x1+y1)/4 =( 270 + 420 )/ 4 = 172.5 mm
Hence Sv = 317.15 mm is not permissible. Keep Sv = 170 mm c/c . Incase the
200 mm c/c in the mid span where both transverse shear as well as torsional shear are minimum.
provide 2 - 10 mm f holding bars. Keep the same setion for other flight.
 40 16 34 16 34 26
RINGER BEAM

0.30 mtr 300 mm

500 N/mm2
37.5 mm

N/mm2

300 mm

ent weight of horizontal plane .

l be as follows.

1.13 = 5650 N
25000 = 1975 N
= 100 N
= 4000 N
Total = 11725 N
ng throught.

ence the worst condition may be

= 7725 N

= 0.55 mtr

= 1168 N-m

= 605 N-m

= 2200 N-m
= 3368 N-m
 mm But available= 200 mm

200 - 25 = 175 mm

mm2

10 x 10
= 79 mm2
x 100
= 1027 mm

8 x 8
= 50 mm2
x 100
= 209 mm
mm say 300 mm

6.43 m

mm deep
2542.5 N
15828.75 N
18371.25 N say 18400 N/m

75956350 75.956 x 10 6 N-mm

solated T-Beam given byEq.

m ;b = actual width = 1.35 m

1034 mm

200
= 75956350
0.253

75956350
 6.43
= 59110 N
2
only to one side of step.

N-mm

N-mm

owever, keep total depth = 450 mm

25 mm

mm2

25 x 25
= 491 mm2
x 100
Nos. Say 2

nal longitudinal reinforcement for torsion, as computed

-b + (b2-4*a*c)0.5
b=
2*a
n )

20.84 - ( 434.3 - 33552 )0.5

N=
2*1
N= 81.8 mm

lls inside the flanges y = 27.252 mm

mm

N/mm2 < 275 Hence safe

N/mm2 < 8.5 Hence safe

able 3.2 tc = 0.35 N/mm2

300
= 10393382.4 N-mm
 = 10393382.4 N-mm
.7

mm2

mm2 is O.K.
t the lower face under the landing.

vertical stirrups.

10 x 10
= 157 mm2
x 100

110
Sv
275
0.495 or Sv = 317.15 mm

270 mm
420 mm

mm c/c . Incase the spacing to

are minimum.
 Genral lay out of stair case
Note:- Red color data are theoretical calculation da
where black data are as per design data.

1.40 4.50 1.40

treads= 15
1.35 4.50 1.45 7.30

D C 1.45 1.40

2.90
Treads= 0 0.00 0.10
2.90
treads= 0

A B 1.45 1.40

0.00 1.45
1.45
7.30
No of riser required = 2528 / 158 = 16 Space required Space required
(mm) (m)
No of tread required = 16 - 1 = 15
No of treads accomeded in 4500 / 300 = 15 x 300 4500 4.50
long direction=

No of tread accomeded in short 100 / 300 = 0 x 300 0 0.00

direction
No of treads Required in other 15 - 15 = 0 x 250 0 0.00
long direction=

1.35 4.50 1.45

450
2- 20 mm anchore bars
Bottom of waist
10 mm f @
2.20 300 mm c/c
stirrups 2 ldg.
10 mm F @ 300
170 mm c/c 158
stirrups 2 ldg.
10 mm F @ Landing
340 mm c/c
mid span
2.20
stirrups 2 ldg.
10 mm F @
L - section 170 mm c/c
 10 mm f @ 300 mm c/c
1450

200

2 - 20 mm f
10 mm f @ 170 mm c/c Anchor bars

10 mm F @ 340 mm c/c 2 -25 mm f

stirrups 2 ldg. main bars
300
Cross section at mid span
pk_nandwana@yahoo.co.in
e theoretical calculation data
er design data.

Space required
(m)

4.50

0.00

0.00

1.45

10 mm f @
300 mm c/c
Landing
0
8 mm F @
300 mm c/c
mm
 VALUES OF DESIGN CONSTANTS
Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40 Grade of concrete
Modular Ratio 18.67 13.33 10.98 9.33 8.11 7.18 tbd (N / mm2
scbc N/mm 2 5 7 8.5 10 11.5 13
m scbc 93.33 93.33 93.33 93.33 93.33 93.33
kc 0.4 0.4 0.4 0.4 0.4 0.4
(a) sst = De
140 jc 0.867 0.867 0.867 0.867 0.867 0.867
N/mm2 Rc 0.867 1.214 1.474 1.734 1.994 2.254 Grade of
(Fe 250) P (%) 0.714 1 1.214 1.429 1.643 1.857 concrete
c

kc 0.329 0.329 0.329 0.329 0.329 0.329 M 15

(b) sst = jc 0.89 0.89 0.89 0.89 0.89 0.89 M 20
190
N/mm2 Rc 0.732 1.025 1.244 1.464 1.684 1.903 M 25
Pc (%) 0.433 0.606 0.736 0.866 0.997 1.127 M 30
kc 0.289 0.289 0.289 0.289 0.289 0.289 M 35
(c ) sst =
230 jc 0.904 0.904 0.904 0.904 0.904 0.904 M 40
N/mm2 Rc 0.653 0.914 1.11 1.306 1.502 1.698 M 45
(Fe 415) P (%) 0.314 0.44 0.534 0.628 0.722 0.816 M 50
c

kc 0.253 0.253 0.253 0.253 0.253 0.253

(d) sst =
275 jc 0.916 0.916 0.916 0.914 0.916 0.916
N/mm2 Rc 0.579 0.811 0.985 1.159 1.332 1.506
(Fe 500) P (%) 0.23 0.322 0.391 0.46 0.53 0.599
c

Permissible shear stress Table t v in concrete (IS : 456-2000)

100As Permissible shear stress in concrete tv N/mm2 Permissi
bd M-15 M-20 M-25 M-30 M-35 M-40
0.18 0.18 0.19 0.2 0.2 0.2 Grade of
< 0.15
concrete
0.25 0.22 0.22 0.23 0.23 0.23 0.23
0.50 0.29 0.30 0.31 0.31 0.31 0.32 M 10
0.75 0.34 0.35 0.36 0.37 0.37 0.38 M 15
1.00 0.37 0.39 0.40 0.41 0.42 0.42 M 20
1.25 0.40 0.42 0.44 0.45 0.45 0.46 M 25
1.50 0.42 0.45 0.46 0.48 0.49 0.49 M 30
1.75 0.44 0.47 0.49 0.50 0.52 0.52 M 35
2.00 0.44 0.49 0.51 0.53 0.54 0.55 M 40
2.25 0.44 0.51 0.53 0.55 0.56 0.57 M 45
2.50 0.44 0.51 0.55 0.57 0.58 0.60 M 50
2.75 0.44 0.51 0.56 0.58 0.60 0.62
3.00 and above 0.44 0.51 0.57 0.6 0.62 0.63

Maximum shear stress tc.max in concrete (IS : 456-2000)

Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40

tc.max 1.6 1.8 1.9 2.2 2.3 2.5
 table-3.2
Shear stress tc Reiforcement %
100As 100As
M-20 M-20
bd bd
0.15 0.18 0.18 0.15
0.16 0.18 0.19 0.18
0.17 0.18 0.2 0.21
0.18 0.19 0.21 0.24
0.19 0.19 0.22 0.27
0.2 0.19 0.23 0.3
0.21 0.2 0.24 0.32
0.22 0.2 0.25 0.35
0.23 0.2 0.26 0.38
0.24 0.21 0.27 0.41
0.25 0.21 0.28 0.44
0.26 0.21 0.29 0.47
0.27 0.22 0.30 0.5
0.28 0.22 0.31 0.55
0.29 0.22 0.32 0.6
0.3 0.23 0.33 0.65
0.31 0.23 0.34 0.7
0.32 0.24 0.35 0.75
0.33 0.24 0.36 0.82
0.34 0.24 0.37 0.88
0.35 0.25 0.38 0.94
0.36 0.25 0.39 1.00
0.37 0.25 0.4 1.08
0.38 0.26 0.41 1.16
0.39 0.26 0.42 1.25
0.4 0.26 0.43 1.33
0.41 0.27 0.44 1.41
0.42 0.27 0.45 1.50
0.43 0.27 0.46 1.63
0.44 0.28 0.46 1.64
0.45 0.28 0.47 1.75
0.46 0.28 0.48 1.88
0.47 0.29 0.49 2.00
0.48 0.29 0.50 2.13
0.49 0.29 0.51 2.25
0.5 0.30
0.51 0.30
0.52 0.30
0.53 0.30
0.54 0.30
0.55 0.31
0.56 0.31
0.57 0.31
0.58 0.31
0.59 0.31
0.6 0.32
0.61 0.32
0.62 0.32
0.63 0.32
0.64 0.32
0.65 0.33
0.66 0.33
0.67 0.33
0.68 0.33
0.69 0.33
0.7 0.34
0.71 0.34
0.72 0.34
0.73 0.34
0.74 0.34
0.75 0.35
0.76 0.35
0.77 0.35
0.78 0.35
0.79 0.35
0.8 0.35
0.81 0.35
0.82 0.36
0.83 0.36
0.84 0.36
0.85 0.36
0.86 0.36
0.87 0.36
0.88 0.37
0.89 0.37
0.9 0.37
0.91 0.37
0.92 0.37
0.93 0.37
0.94 0.38
0.95 0.38
0.96 0.38
0.97 0.38
0.98 0.38
0.99 0.38
1.00 0.39
1.01 0.39
1.02 0.39
1.03 0.39
1.04 0.39
1.05 0.39
1.06 0.39
1.07 0.39
1.08 0.4
1.09 0.4
1.10 0.4
1.11 0.4
1.12 0.4
1.13 0.4
1.14 0.4
1.15 0.4
1.16 0.41
1.17 0.41
1.18 0.41
1.19 0.41
1.20 0.41
1.21 0.41
1.22 0.41
1.23 0.41
1.24 0.41
1.25 0.42
1.26 0.42
1.27 0.42
1.28 0.42
1.29 0.42
1.30 0.42
1.31 0.42
1.32 0.42
1.33 0.43
1.34 0.43
1.35 0.43
1.36 0.43
1.37 0.43
1.38 0.43
1.39 0.43
1.40 0.43
1.41 0.44
1.42 0.44
1.43 0.44
1.44 0.44
1.45 0.44
1.46 0.44
1.47 0.44
1.48 0.44
1.49 0.44
1.50 0.45
1.51 0.45
1.52 0.45
1.53 0.45
1.54 0.45
1.55 0.45
1.56 0.45
1.57 0.45
1.58 0.45
1.59 0.45
1.60 0.45
1.61 0.45
1.62 0.45
1.63 0.46
1.64 0.46
1.65 0.46
1.66 0.46
1.67 0.46
1.68 0.46
1.69 0.46
1.70 0.46
1.71 0.46
1.72 0.46
1.73 0.46
1.74 0.46
1.75 0.47
1.76 0.47
1.77 0.47
1.78 0.47
1.79 0.47
1.80 0.47
1.81 0.47
1.82 0.47
1.83 0.47
1.84 0.47
1.85 0.47
1.86 0.47
1.87 0.47
1.88 0.48
1.89 0.48
1.90 0.48
1.91 0.48
1.92 0.48
1.93 0.48
1.94 0.48
1.95 0.48
1.96 0.48
1.97 0.48
1.98 0.48
1.99 0.48
2.00 0.49
2.01 0.49
2.02 0.49
2.03 0.49
2.04 0.49
2.05 0.49
2.06 0.49
2.07 0.49
2.08 0.49
2.09 0.49
2.10 0.49
2.11 0.49
2.12 0.49
2.13 0.50
2.14 0.50
2.15 0.50
2.16 0.50
2.17 0.50
2.18 0.50
2.19 0.50
2.20 0.50
2.21 0.50
2.22 0.50
2.23 0.50
2.24 0.50
2.25 0.51
2.26 0.51
2.27 0.51
2.28 0.51
2.29 0.51
2.30 0.51
2.31 0.51
2.32 0.51
2.33 0.51
2.34 0.51
2.35 0.51
2.36 0.51
2.37 0.51
2.38 0.51
2.39 0.51
2.40 0.51
2.41 0.51
2.42 0.51
2.43 0.51
2.44 0.51
2.45 0.51
2.46 0.51
2.47 0.51
2.48 0.51
2.49 0.51
2.50 0.51
2.51 0.51
2.52 0.51
2.53 0.51
2.54 0.51
2.55 0.51
2.56 0.51
2.57 0.51
2.58 0.51
2.59 0.51
2.60 0.51
2.61 0.51
2.62 0.51
2.63 0.51
2.64 0.51
2.65 0.51
2.66 0.51
2.67 0.51
2.68 0.51
2.69 0.51
2.70 0.51
2.71 0.51
2.72 0.51
2.73 0.51
2.74 0.51
2.75 0.51
2.76 0.51
2.77 0.51
2.78 0.51
2.79 0.51
2.80 0.51
2.81 0.51
2.82 0.51
2.83 0.51
2.84 0.51
2.85 0.51
2.86 0.51
2.87 0.51
2.88 0.51
2.89 0.51
2.90 0.51
2.91 0.51
2.92 0.51
2.93 0.51
2.94 0.51
2.95 0.51
2.96 0.51
2.97 0.51
2.98 0.51
2.99 0.51
3.00 0.51
3.01 0.51
3.02 0.51
3.03 0.51
3.04 0.51
3.05 0.51
3.06 0.51
3.07 0.51
3.08 0.51
3.09 0.51
3.10 0.51
3.11 0.51
3.12 0.51
3.13 0.51
3.14 0.51
3.15 0.51
 Permissible Bond stress Table t bd in concrete (IS : 456-2000)
M-10 M-15 M-20 M-25 M-30 M-35 M-40 M-45 M-50
-- 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4

Development Length in tension

Plain M.S. Bars H.Y.S.D. Bars

tbd (N / mm2) kd = L d F tbd (N / mm2) kd = L d F
0.6 58 0.96 60
0.8 44 1.28 45
0.9 39 1.44 40
1 35 1.6 36
1.1 32 1.76 33
1.2 29 1.92 30
2.0
1.3 27 2.08 28
1.4 25 2.24 26
1.4

Modification factore
1.2

0.8

0.4
Permissible stress in concrete (IS : 456-2000)
Permission stress in compression (N/mm 2) Permissible stress in bond (Average) for
Bending acbc Direct (acc) plain bars in tention (N/mm2) 0.0
(N/mm2) Kg/m2 (N/mm2) Kg/m2 (N/mm2) in kg/m2
3.0 300 2.5 250 -- --
5.0 500 4.0 400 0.6 60
7.0 700 5.0 500 0.8 80
8.5 850 6.0 600 0.9 90
10.0 1000 8.0 800 1.0 100
11.5 1150 9.0 900 1.1 110
13.0 1300 10.0 1000 1.2 120
14.5 1450 11.0 1100 1.3 130
16.0 1600 12.0 1200 1.4 140
 0.4 0.8 1.2 1.6 2
Percentage of tension reinforcement
2 2.4 2.8