CAPITULO II PANORAMA GENERAL DE LAS TECNOLOGÍAS DE INFORMACIÓN (TI),
2.3 A UDITORÍA EN I NFORMÁTICA
2.3.4 Funciones de la Auditoría en Informática
2.3.4.2 Funciones del área de Auditoría en Informática
influence zone for the counterfort walls= 1.775 m
(this is the width from which each counterfort receives the eatrh pressure)
unit weight of soil = 1.80 t/m3 Fst = 24000 t/m2
earth pressure coeffiecnt = 0.279 j = 0.891
surcharge height = 1.20 m
maximum soil height = 16.786 m
pressure due to soil at specified depth = 0.502 X H (t/m2) (varying) pressure due to surcharge at specified depth = 0.603 t/m2 (constant)
pressure diagram will be somewhat like this
0.603 t/m2
X 1.775 =
16.786 m
9.033 t/m2 16.04 t/m2
D = 9.662 m 32 φ
d = 9.596 m 50 mm
b= 0.850 m
φ nos
1 16.786 16.04 854 4160.876 32 5.00
2 13.000 12.66 695 3388.114 32 5.00
3 10.000 9.99 570 2775.782 32 3.00
4 6.500 6.87 423 2061.395 32 3.00
5 3.000 3.75 276 1347.008 32 3.00
1.07 t/m2
6.786 m and lap with 28φ therafter extending to top
16.786 16.04 t/m2 71.80 0.09 0.2
13.000 12.66 t/m2 44.63 0.05 0.2
10.000 9.99 t/m2 27.65 0.03 0.2
6.500 6.87 t/m2 12.90 0.02 0.2
3.000 3.75 t/m2 3.61 0.00 0.2
τv < τc at all positions, so provide nominal shear reinforcement :
8 φ = 100.531 mm2
Spacing = 200 mm (horizontal ties)
Average downward pressure = 16.91 t/m2
Total area of reinforcement required to resist this tension = 704.59 mm2/m
Provide 2 legged 8φ = 100.53 mm2
spacing = 100.53 x 1000 / 704.59 = 142.679
(vertical ties)
Two layers of 32φ from the base to height of
Depth (m)
Pressure at this depth
(t/m2)
Vu (t) τv (N/mm2) τc (N/mm2)
To secure connection between counterfort and heel slab, counterfort needs to be designed for tension arising out of the net outward pressure of heel slab.
Provide 2 legged stirrup
Provide 8φ 2 legged stirrup at a spacing of 120 c/c
3413
9728
300
modular ratio m = 280/(3σcbc) for M35, σcbc = 11.67
= 8
assuming effective cover = 100 mm d = 10478 mm
finding the neautral axis (X)
1) assuming the Neutral axis lies in the flange
Using Ast= 9 nos 32 Φ
b*X^2/2= m*Ast(d-X) ====> X^2 + (2mAst/b)X - (2mAst/b)d
X^2 + (34)X - 356252 a = 1
Solving we get X = 580 mm b = 34
c = -356252
for moment M = 1641
Stress in concrete (Fc) = M/(0.5*b*X*(d-X))
∴ Fc = 1.68 Mpa < 11.67 Mpa Stress in steel (Fs) = mFc*(d-X)/X
∴ Fs = 228.64 Mpa < 240 MPa
850
7238.23 mm2
9817
850
modular ratio m = 280/(3σcbc) for M35, σcbc = 11.67
= 8
assuming effective cover = 100 mm d = 10567 mm
finding the neautral axis (X)
1) assuming the Neutral axis lies in the flange
Using Ast= 3 nos 32 Φ
b*X^2/2= m*Ast(d-X) ====> X^2 + (2mAst/b)X - (2mAst/b)d
X^2 + (22)X - 232474 a = 1
Solving we get X = 471 mm b = 22
c = -232474
for moment M = 423
Stress in concrete (Fc) = M/(0.5*b*X*(d-X))
∴ Fc = 1.00 Mpa < 11.67 Mpa Stress in steel (Fs) = mFc*(d-X)/X
∴ Fs = 171.66 Mpa < 240 MPa
2412.74 mm2 1775
850
Thickness of slab = 0.30 m wearing course = 56.00 mm
B= span in transverse direction = 3.550 m (short) L= span in longitudinal direction = 5.275 m (long)
weight of deck slab = 0.720 t/m2 weight of wearing course = 0.123 t/m2 total weight = 0.843 t/m2
along short span
K = short span/long span = 0.67 ===>> m1 = 0.047
from Pigeud's curve
1/K = 1.49 ===>> m2 = 0.018
along long span
0.15 for reinforced concrete bridges
15.79 t
impact factor= 25 %
u= 0.962 m
v= 3.712 m limited to 5.275 m
K= 0.67 using pigeud's curve=
u/B= 0.271 m1= 0.12
v/L= 0.704 m2= 0.048
SIZE OF PANEL of deck slab=
width of load spread along short span=
width of load spread along long span=
Design of ROOF SLAB by Pigeud's curve
Live load bending moment due to IRC Class AA tracked vehicle = Maximum bending moment due to dead load =
poisson's ratio µ =
Total dead weight W =
ONE TRACK
35.00 t
total load per track including impact = 1.25*35 = 43.75 t
effective load on span= 43.75*3.712/3.712 = 43.75 t
moment along short span= (m1+µ*m2)*43.75 = 5.57 t-m
moment along long span= (m2+µ*m1)*43.75 = 2.89 t-m
Y
2.6375
4 1 2 3
X 3.75 t 6.25 t 3.75 t X
2.6375 8 5 6 7
3.75 t 6.25 t 3.75 t
1.775 Y 1.775
B = 3.550 m
L = 5.275 m
load per track of AA=
Live load bending moment due to IRC Class AA wheeled vehicle =
6.25 t
6.25 t
The Class AA wheeled vehicle is placed as shown to produce the severest moments. The front axle is placed along the centre line with 6.25t wheel at centre of panel.
Maximum moments in the short span and long span directions are computed for individual wheel loads taken in the order shown
6.25 t
tyre contact dimensions= 300 X 150 mm
u= sqrt((0.3+2*0.056)^2+0.3^2)= 0.5097
v= sqrt((0.15+2*0.056)^2+0.3^2)= 0.3984
u/B = 0.144 v/L = 0.076 K = 0.70
m1 = 0.221 Total load allowing for 25 % impact = 7.813 t
m2 = 0.190
Moment along short span = 1.950 t-m Moment along long span = 1.744 t-m
6.25 t
Y
X 6.25 t 6.25 t X
0.3 0.3
Y
Intensity of loading = (6.25*1.25)/(0.5097*0.3984) = 38.47 t/m2
Consider the loaded area of 0.150 X 2.300
u = sqrt((2.3+2*0.056)^2+0.3^2) = 2.4306 v = sqrt((0.15+2*0.056)^2+0.3^2) = 0.3983
u/B = 0.685 v/L = 0.076 K = 0.70
m1 = 0.112
m2 = 0.119
Moment along short span = 4.578 t-m Moment along long span = 4.788 t-m Bending Moment due to wheel 1 =
Bending Moment due to wheel 2 =
Here wheel load is placed unsymmetrical to YY axis. But Pigeuds Curves are derived for symmetrical loading. Hence we place an equal dummy load symmetrical about the YY axis and consider the whole loading area. Then we deduct the area beyond the actual loaded area. Half of the resulting value is taken as the moment due to actual loading.
0.85 0.85
2.3
Now consider the area beyond the actual loading = 1.7 X 0.15
u = sqrt((1.7+2*0.056)^2+0.3^2) = 1.8367 v = sqrt((0.15+2*0.056)^2+0.3^2) = 0.3983
u/B = 0.517 v/L = 0.076 K = 0.70
m1 = 0.132
m2 = 0.14
Moment along short span = 3.517 t-m Moment along long span = 3.674 t-m
0.531 t-m 0.557 t-m
3.75 t
By similar procedure as for previous case, we get B.M along short span = 0.052 t-m B.M along long span = 0.121 t-m
3.75 t
By similar procedure as for previous case, we get B.M along short span = 0.520 t-m B.M along long span = 0.607 t-m
6.25 t
B.M along short span = 0.823 t-m B.M along long span = 0.195 t-m
6.25 t
B.M along short span = 0.823 t-m B.M along long span = 0.195 t-m Bending Moment due to wheel 4 =
Bending Moment due to wheel 5 =
In this case loading is eccentric w.r.t XX axis. By similar procedure as for previous case but with load area extended w.r.t. XX axis, we get
Bending Moment due to wheel 6 = Bending Moment due to wheel 3 = Net moment along short span =
Net moment along long span =
In this case loading is eccentric with respect to both XX and YY axes. A strict simulation would be very complicated and laborious. For A reasonable approximation, eccentricity w.r.t. only XX axis is considered and calculations made as for previous case.
3.75 t
By similar procedure as for previous case, we get B.M along short span = 0.486 t-m B.M along long span = 0.115 t-m
3.75 t
By similar procedure as for previous case, we get B.M along short span = 0.486 t-m B.M along long span = 0.115 t-m
total bending moment along short span = 5.671 t-m
total bending moment along long span = 3.649 t-m
To allow for continuity, the computed momnts are multiplied by a factor of 0.8
Design Bending Moment=
along short span= 5.16 t-m along long span= 3.24 t-m
Grade of conc. 35
Grade of steel 500
Dia of bar used 16
Q for concrete grade used 170 t/m2
k value for concrete 0.327
j value for concrete 0.891
Permissible stress in steel 24000 t/m2 Permissible stress in concrete 1167 t/m2
cover 50 mm
effective depth required= 0.174 m
provided deff = 0.242 m OK
998 mm2 ϕ 16 @ 150 c/c 1340 mm2 OK
625 mm2 ϕ 12 @ 150 c/c 754 mm2 OK longitudinal reinforcement required =
main reinforcement required=
So provide
Bending Moment due to wheel 7 =
Bending Moment due to wheel 8 =
So provide
Y
X
0.225 m
2.781 m
= 6257.25 mm2
3128.625 mm2
Provide ϕ 25 @ 150 c/c = 3272.492 mm2
0.225 m
1.000 m
2250 mm2
steel on both top and bottom face = 1/2 of steel red= 1125 mm2
Provide ϕ 12 @ 150 c/c = 1507.964 mm2
==> 2 legged stirrup Along Y direction
Steel reqd = 1% of cross section