DISEÑO DE VÍAS - Curva Circular Espiral CircularDescripción completa
This work modelled circular footing on lateritic soil. This was done by studying the relationship between bearing capacity of circular footing and fines content of lateritic soil. Lateritic soil samples were collected and subjected to laboratory test
ansys
The article discusses about the "Circular Economy. A circular economy is a contrasting option to a linear economy influence, use, to arrange in which we keep assets being used for whatever length of time that conceivable, remove the most extreme ince
Descripción: EJERCIOSIOS DE CANALES
partes de la carta circular uso y esquemasDescripción completa
IVRCL INFRASTRUCTURES AND PROJECTS LIMITED DESIGN OF PIER UNDER AQUEDUCT AT KM. 0.334
2.200 +
203.626
+ + + + +
201.426 201.386 200.986 200.236 199.136
0.040 0.400 0.750 1.100
MFL + GL
m m m m m
4.439 5.339 2.589
198.647 +
197.647 m
+
196.547 m
+
195.647 m
1.100 2.000
0.900 1.200 4.800 CROSS SECTION OF THE PIER
4.350 3.750
0.200
0.075
0.7 + 203.626 0.350
0.04
+
201.426 + 201.386 200.986 + 200.236 + 199.136
GL
+
197.647
1.200 1.100 0.900
+ + 4.800 LONGITUDINAL SECTION OF PIER
196.547 195.647
1.800
1.200
1.800
1.800
4.800
1.200
1.800
4.800
## ## 79 ##
Clear span of trough Effective span of trough Overall length of slab Depth of slab Thickness of the wearing coat Diameter of the pier at top Diameter of pier at bottom Width of abutment Width of pier cap Length of pier cap Width of trough slab Width of clear trough Width of foundation offset Depth of Water in the trough Free board Height of side walls Width of side wall Clear cover for pier and pier cap Main reinforcement provided for pier cap reinforcement for stirrups Distribution steel Hanger bars Effective depth for the pier cap (steel provided in two rows) = Half of the effective depth = =
0.849 m
= = = = = = = = = = = = = = = = =
=
8.800 m 9.4 m 10.000 m 0.350 m 0.04 m 1.200 m 1.200 m 1.200 m 1.200 m 4.550 m 4.350 m 3.650 m 1.800 m 2.200 m 0.000 m 2.200 m 0.350 m 40.000 mm 25.000 mm 12.000 mm 10.000 mm 12.000 mm 0.75 + 1.1 - 40/1000 - 25/1000 -25/ 2000 1.7725 m 1.7725 / 2 0.886 m
1.200 m
0.175 Side of square inscribed in a circle = SQRT(0.6² + 0.6²) = 0.849 m Half of the side of a square = 0.425 m Distance from the outremost point of circle to side of the square = 0.6 - 0.425 = 0.175 Clear span for LL = 1.675 + 0.175 = 1.850 m Effective span for LL = 1.85 + 0.886 = 2.736 m
DESIGN OF PIER CAP 4.550 0.750 1.100
1.675
1.200
Self wt of the pier cap Wt of major rectangular portion Wt of triangular portion Wt of minor rectangular Portion Total load
= = =
1.675
0.75 x 1.2 x 4.55 x 2.5 1/2 x 1.1 x 1.675 x 1.2 x 2.5 x 2
1.2 x 1.1 x 1.2 x 2.5
= = = =
10.238 5.528 3.96 19.726
t t t t
=
178.12 t
Weight of trough slab:Reaction from side walls from staad out put Load due to self weight of cantelever portion Wt. triangular portion Wt. major rectangular portion Wt. minor rectangular portion
Max. B.M at face of the support BM due to LL BM due to DL
= = =
1/2 x 1.1 x 1.675 x 1.2 x 2.5 = 0.75 x (1.675 + 0.6) x 1.2 x 2.5= 0.6 x 1.1 x 1.2 x 2.5 =
= =
2.764 5.119 1.98 9.863
= = =
(178.12) x (2.736 - 0.1 - 0.175 )
9.863 x 2.736/2 Total Load
DESIGN CONSTANTS For M25 grade concrete and Fe415 Fck sst scbc m k j Q
25 N/mm2 2 190 N/mm 2 8.5 N/mm
= = = = = = =
10.980 0.329 0.89 1.244
Max required depth of pier cap
= =
(4518.46 X 10^6/(1.244 X 1200))^0.5 1740 mm
Max depth provided
=
1772.5 mm HENCE SAFE
Area of stel required
= =
4518.46 x 10^6/(190 x 0.89 x 1772.5) 2 15075.113 mm
No of 25 mm dia bars required
=
Area of steel provided
=
31.000
No's
15223.214 HENCE SAFE
mm2
t t/m t/m t/m
438.353 t-m 13.493 t-m 451.846 t-m
Check for shear Critical plane occurs at distance of d/2 from the face of the support = 1.675-1.7725/2 = 0.789 m
Total depth of section at critical section
Shear force at d/2 from face Due to dead load Due to live load Total
= =
= =
Nominal shear stress tv
=
(1.1 x 0.789 / 1.675) + 0.75 1.268 m
19.726 x 1.268/2 178.12
= = =
1906.26 x 10^3 1772.5 x 1200
=
= Area of steel provided % of steel provided
=
Permissible shear stress =
15223.2 x 100 1850 x 1200
0.31 + ((0.36-0.31) / (0.75 - 0.5) x 0.216)
12.506 t 178.12 t 190.626 t 1906260 2127000 0.896 N/mm
=
15223.2 mm
=
0.716 %
=
= = Vs = V-(tv x bd) = =
1772.5 x 1200 x 0.3531 751043.7 N 1906260 - 751043.7 1155216.3 N
Provide 12 mm dia 4 legged stirrups Spacing
Sv
=
(175 x 113.143 x 4 x 1772.5) / 1155216.3 120 mm Adopt Spacing of 120 mm As per article 26.5.1.5 of IS : 456 - 2000, maximum spacing of shear reinforcement is (I) 0.75d = 0.75 x 1850 = 1387.5 mm ( ii ) 300 mm whichever is less. Hence provide a spacing of = Hence provide
300 mm
12 mm dia @ 120 mm c/c 4-legged stirrups
Dead load due to super structure:-
Dead Load due to superstructure
=
2
0.3531 N/mm2
Hence shear reinforcement is to be provided
Shear force carried by concrete
2
356.24 t/m
MOMENT DUE TO WIND FORCE:-
As per clause 212.2 of IRC:6 - 2000 CASE (I) DRAIN FULL CONDITION
As per clause 212.3 of IRC :6 - 2000 Height of side wall
=
2.2 m
Exposed height over trough bottom
= =
0.35 + 0.04 + 2.2 + 0.05+0.35+0.7 3.69 m
Center of gravity of exposed area
=
3.69 x 3.69 / 2 3.69
= Center of gravity of exposed area acts at a elevation of = = Height above FSL
1.845 m
200.986 + 1.845 202.831
200.986 - 198.647 + 1.845 =
4.184 m
Height above base of pier 202.831 - 196.547
=
6.284 m
Wind pressure for a height 4.184m
=
2 63.92 Kg/m
Wind force at a height of 4.184m
= =
10 x 3.69 x 63.92 / 1000 2.359 t
Moment due to wind force at top of foundation of pier = =
2.359 x 6.284 14.824 t-m
Moment due to wind force at bottom of foundation of pier = =
2.359 x 7.184 16.947 t-m
CASE (II) DRAIN EMPTY CONDITION
Center of gravity of exposed area acts at a elevation of Height above CBL 200.986 - 197.647 + 1.845= Wind pressure for a height 5.184m = - + 1.845 Wind force at a height of 5.184m = = Moment due to wind force at top of foundation of =pier = Moment due to wind force at bottom of foundation of pier = = Condition:- I canal full condition
=
202.831 m 5.184 m 2 68.92 Kg/m = 6.284 m 10 x 3.69 x 68.92 / 1000 2.543 t 2.543 x 6.284 15.98 t-m
Moment due to wind force at top of foundation
=
14.824 t-m
Moment due to wind force at bottom of foundation
=
16.947 t-m
2.543 x 7.882 18.269 t-m
t-m
Condition:- II canal empty condition Moment due to wind force at top of foundation
=
15.98 t-m
Moment due to wind force at bottom of foundation
=
18.269 t-m
CASE(2)
As per clause212.6 of IRC: 6 - 2000 The wind force as calculated should not be les than that obtained by 450 Kg/m on the trough 10 x 0.450 = 4.5 t This force acts at center of the slab Therefore lever arm from top of foundation of pier 4.439 + 1.845
=
6.284 m
Moment due to wind force at top of foundation 4.5 x 6.284
=
28.278 t-m
There fore lever arm at bottom of foundation 5.339 + 1.845
=
7.184 m
Moment due to wind force at bottom of foundation 4.5 x 7.184
=
32.328 t-m
CASE(3)
As per clause 212.7 of IRC-6-2000, Wind load on pier considered is = Span on which wind load acts = x Lateral wind force = =
240 10 10 x 3.69 x 240 / 1000 8.856
Moment at base of the pier
8.856 x 6.284
=
55.651
Moment at bottom of foundation
8.856 x 7.184
=
63.622
THE MAXIMUM OF THE ABOVE THREE CASES IS CONSIDERED FOR DESIGN
Moment due to wind force at top of foundation
=
55.651
Moment due to wind force at bottom of foundation
=
63.622
MOMENT DUE TO WATER CURRENTS:CONDITION I: (IN DRAIN FLOW DIRECTION) CASE: I
As per clause 213.2 of IRC:6 - 2000 Force due to water currents in direction of drain flow is given by p 52 k V 2
V
=
1.414 x v
v
= =
V
=
Velocity of flow in drain 1.5 m/sec 2.121 m/sec
Max velocity at the base of the pier = 2 x 2.121² / 3 = 2.999 m/sec K
=
0.66
Pressure due to water currents at FSD P
= =
52 x 0.66 x 2.121² 2 154.393 Kg/m
Pressure due to water currents at drain bed level = 52 x 0.66 x2.999² 2 = 308.674 Kg/m
1.2 0.489
4.439 1.2
198.647
2.589
1 1.2
197.647 1.2 =
2 154.393 Kg/m
Pressure due to water currents at bottom of drain =
2 308.674 Kg/m
Pressure due to water currents at FSD
Area of pier profile immersed in water 1.2 x 1
=
2 1.2 m
Total pressure due to water currents. (154.393 + 308.674) / 2 x 1/1000 x 1.2 =
0.278 t
lever arm
0.444 m
(154.393 x 2 + 308.674)
(154.393 + 308.674)
x
1 3
=
Depth of pier from bottom level of drain to top level of foundation = Moment due to water currents at top of foundation 0.278 x (0.444 + 1.1) = Depth of pier from canal bed level to bottom level of foundation = Moment due to water currents at bottom of foundation 0.278 x (0.444 + 2) =
1.1 m
0.4292 t-m
2m
0.6794 t-m
CASE ( II ) (A)
Force parrllel to the length of the pier :Due to the cross currents as per para 213.5 of IRC - 6 - 1966 52 x K x V2 P = K = 0.66 V = 20.5 x V x cos(A) Maximum variation in the direction of currents Pressure at FSL
= = =
=
20
52 x K x V2 x cos20 52 x 0.66 x 2.121² x cos (20) 2 145.082 Kg/m
Max velocity at the base of the pier = 2 x 2.121² / 3 = 2.999 m/sec Pressure at canal bed level
2 290.059 Kg/m
=
Length of the pier = Height of water column = Area on which water pressure acts = lever arm
(145.082 x 2 + 290.059) x (145.082 + 290.059)
Total force
1.2 m 1m 2 1.2 m 1= 3
=
0.444 m
0.261 t
Moment due to this force at the top foundation of the pier = 0.261 x ( 0.444 + 1.1) = 0.403 t-m Moment due to this force at the bottom of foundation concrete = 0.261 x ( 0.444 + 2) = 0.638 t-m CASE ( II ) (B)
Force perpendicular to the length of the pier :Due to the cross currents as per para 213.5 of IRC - 6 - 1966 52 x K x V2 P = K = 1.5 V = 20.5 x V x sin(A) Maximum variation in the direction of currents Pressure at FSL
Pressure at canal bed level Length of the pier
= = = = =
=
20
52 x K x V2 52 x 1.5 x 2.121² x sin 20 2 120.013 Kg/m 2 80.006 Kg/m 1.2 m
Height of water column = Area on which water pressure acts = lever arm
(120.013 x 2 + 80.006) x (120.013 + 80.006)
Total force
1m 2 1.2 m 1 3
=
0.533 m
=
0.12 t
Moment due to this force at the top foundation of the pier = 0.12 x ( 0.533 + 1.1) = 0.196 t-m Moment due to this force at the bottom of foundation concrete = 0.12 x ( 0.533 + 2) = 0.304 t-m Maximum moments of both the cases are considered for the design Moment at the base of the pier = Moment at the base of the foundation concrete
0.403 t-m 0.638 t-m
CONDITION II: (IN CANAL FLOW DIRECTION)
Water force in road way direction due to 250mm difference in water levels between the opposite forces of the pier. Depth of flow in drain = 1m Length of pier immersed in water = 1.2 m Depth of pier from CBL to the top of foundation 1.1 m Moment at top of foundation M1 1/2 x (1²) x (1 /3 + 1.1) x 1.2
=
0.86 t-m
M2
=
1.422 t-m
=
0.562 t-m
Moment at bottom of foundation M1 1/2 x ( 1²) x (1 /3 + 2) x 1.2
=
1.4 t-m
M2
=
2.266 t-m
=
0.866 t-m
1/2 x (1.25²) x (1.25 / 3 + 1.1) x 1.2 M2-M1
1/2 x (1.25²) x (1.25 /3 + 2) x 1.2 M1-M2
DRAIN FULL CONDITION:(I) TOTAL MOMENTS IN CANAL FLOW DIRECTION
Water currents Total moments
TOP OF FOUNDATION CASE(I) 0.562 0.562
(II) TOTAL MOMENTS IN DRAIN FLOW DIRECTION
BOTTOM OF FOUNDATION CASE(I) 0.866 0.866
TOP OF FOUNDATION CASE(I) 55.651
Wind force
BOTTOM OF FOUNDATION CASE(I) 63.622
Water currents
0.429
0.679
Total moments
56.080
64.301
DRAIN EMPTY CONDITION:(II) TOTAL MOMENTS IN DRAIN FLOW DIRECTION TOP OF FOUNDATION CASE(I)
BOTTOM OF FOUNDATION CASE(I)
Wind force
55.651
63.622
Total moments
55.65
63.622
SELF WEIGHT OF PIER:
Weight of bracket = Weight of circular pier
=
19.726 t 7.323 t 27.049
( 22/7) / 4 x (1.2)² x 2.589 x 2.5
Weight of displaced water due to pier (with 100% buoyancy) (22/7)/4 x 1.2² x 1 x 1
=
1.131 t
Self weight of pier with (100% buoyancy) = 19.726 + 7.323 - 1.131
25.918 t
Direct loads at top of foundation (when drain is full) Case(i)
=
25.918 + 356.24
=
382.158 t
=
383.289 t
Direct loads at top of foundation (when drain is empty) Case(ii)
=
27.049 + 356.24
SELF WEIGHT OF FOUNDATION
Weight of foundation including top soil
=
79.758 t
=
43.912 t
=
426.07 t
=
463.047 t
Weight of foundation considering 100% Buoyancy including top soil
Direct loads at bottom of foundation Direct loads at bottom of foundation (when drain is full)
Case(i)
=
382.158 + 43.912
Direct loads at bottom of foundation (when drain is empty)
Case(ii)
=
383.289 + 79.758
SECTIONAL PROPERTIES:-
At the base of the pier Area (a)
=
(22/7)/4 x(1.2²)
=
1.131 m
2
Section modulus: Zxx = Zyy
=
PI x 1.2³ / 32
=
0.17 m
3
At the base of the foundation Area =
3.147 x 4.8² / 4
=
2 18.127 m
Section modulus Zxx
=
3.147 x 4.8³ / 32
=
10.857 m
=
3.147 x 4.8³ / 32
=
3 10.857 m
Zyy
CONDITION - I STRESSES AT TOP OF FOUNDATION (CONCRETE) (when drain is full) Case(I) = 382.158 1.131 =
337.894
Max
=
671.084
t/m
Min
=
4.704
t/m2
±
0.562 0.17
±
56.080 0.17
±
3.306
±
329.884
±
0.866 10.857
±
64.301 10.857
±
0.08
±
5.923
±
±
55.651 0.17
±
±
327.359
2
STRESSES AT BOTTOM OF FOUNDATION (SOIL) (when drain is full) Case(I) = 426.07 18.127 =
23.505
Max
=
29.508
t/m2
Min
=
17.502
t/m2
CONDITION - II STRESSES AT TOP OF FOUNDATION (CONCRETE) (when drain is empty) Case(I) = 383.289 1.131 =
338.894
Max
=
666.253
t/m2
Min
=
11.535
t/m2
3
STRESSES AT BOTTOM OF FOUNDATION (SOIL) (when drain is empty) Case(I) = 463.047 18.127 =
25.545
Max
=
31.405
t/m2
Min
=
19.685
t/m2
±
±
63.622 10.857
±
±
5.86
STRESS TABLE CONDITION(I) CASE(I)
TOP OF FOUNDATION MAX 671.084
CONDITION(II) CASE(I)
666.253
MIN 4.704
BOTTOM OF FOUNDATION MAX MIN 29.508 17.502
11.535
31.405
2 337.894 t/m 2 800 t/m
19.685
Calculated direct compressive stress = 382.158 / 1.131
=
Permissible direct stress
=
Permissible direct stress x 1.33 Calculated bending compressive stress
The section is cracked and therefore it is designed as per IS : 456 - 2000 = =
1.200 m 120.000 cm
Axial load coming on to the pier = 382.158 t (value corresponding to max. bending compressive stress) Pu Factored axial load = 382.158 x 1.5 = 573.237 t Moment acting about X - axis = 0.562 t-m (value corresponding to max. bending compressive stress) Factored moment about X - axis = 0.562 x 1.5 = 0.843 t-m Moment acting about Y - axis = 56.080 t-m (value corresponding to max. bending compressive stress) Factored moment about Y - axis = 56.0802 x 1.5 = 84.12 t-m Unsupported length of the pier
l
Effective length of the pier
1.2 x l
Cross - sectional area of the pier
sco
2
333.189 = 1330
Diameter of the pier
sco(cal)
= = = = = =
199.136 - 196.547 2.589 m 258.900 cm 2.589 x 1.2 3.107 m (22/7 x 1.2²/4)
sc <1
Hence safe
Moment of inertia about X - axis Moment of inertia about Y - axis Radius of gyration about X - axis Radius of gyration about Y - axis Effective length / Least radius of gyration
= = = = = = = = = = =
1.131 m2 PI x 1.2^4 /64 4 0.102 m PI x 1.2^4 /64 4 0.102 m SQRT (0.102 / 1.131) 0.3 m SQRT (0.102 / 1.131) 0.3 m 3.107 / 0.3 10.357
If ratio of effective length to least radius of gyration is less than 12 PEDESTAL If ratio of effective length to least radius of gyration is greater than 12 and less than 50 SHORT column If ratio of effective length to least radius of gyration is greater than 50 and less than 150 long column column (a) Check for minimum eccentricity:Minimum eccentricity about X - axis Minimum eccentricity as per IS: 456 - 2000 Minimum moment about X - axis Factored moment about X - axis Minimum eccentricity about Y - axis Minimum moment about Y - axis Factored moment about Y - axis
(b) Check for combined Bi - axial bending and thrust:Percentage of steel Gross - cross sectional area Puz =
= = = = = = = = = = = = =
258.9 / 500 + 120/30 4.518 cm 2 cm 5732.37 x 0.04518 258.988 KN - M 258.988 x 1.5 388.482 KN - M 258.9 /500+ 120/30 4.518 cm 5732.37 x 0.04518 258.988 KN - M 560.802 x 1.5 841.203 KN - M
= =
0.40 % 1131000 mm2
[0.45 x 25 x (1131000 - 1131000 x 0.4/100) + 0.67 x 415 x 1131000x 0.4/100)] /1000 = 13930.753 KN Pu/Puz
= =
an
=
Clear cover provided Reinforcement provided
= =
5732.37 / 13930.753 0.411 1.352 40 mm 20 mm
Uniaxial moment capacity about X - axis d'
= =
40 + 10 50 mm
D
=
d' / D
= =
50 / 1200
=
0.4 / 25
P / fck
1200 mm
0.042
= Pu
=
0.016 5732.37x 10³ x /( 25 x 1.2² x 10^6)
fck d
2
=
0.159
Mu fckd3
=
0.04
MuX1
=
Refer chart 55
Muxfckd3 6
10 = =
0.04 x 25 x 1200³/10^6 1728 KN - m
Uniaxial moment capacity about Y - axis
Check
M ux M ux1
n
d'
=
d
=
d' b
= =
p fck
=
50 mm 1.200 m 50 / 1200 0.042 m
0.016
Pu
=
Fckd2
=
0.159
Mu Fckbd2
=
0.04
Muy1
= =
M uy M uy1
5732.37 x 10³ / ( 25 x 1.2² x 10^6)
0.04 x 25 x 1200³ / 10^6 1728 KN - m
n
=
388.482 1728
= Area of steel required
Diameter of bar considered Number of bars required Area of steel provided
= =
1.
0.511 < 1
+
841.203 1728 Hence safe
1.131 x 0.4/100 x 10^6 4524 mm2
= 20 mm = 15 Nos = 4712.389 mm2 2 4712.389 > 4524 mm HENCE SAFE
Transverse reinforcement:(a) Diameter of transverse reinforcing bar is the largest of the following
1
(i) 1/4 th of the dia of longitudinal bar
= =
(ii) or 8 mm Hence provide 8 mm dia bars (b) Spacing of transverse reinforcement is the least of (i) Minimum spacing (ii) Least lateral dimension of the member (iii) 12 times the dia of longitudinal bar
= = = = =
20/4 5 mm
300 mm 1.2 x 1000 1200 mm 12 x 20 240 mm
Hence provide a pitch of 240 mm 0
CHECK FOR PERMISSIBLE STRESSES IN STEEL AND CONCRETE
X N
R
1
2
r
X A
Ast Area of steel provided = Radius of the pier R = Clear cover provided = Dia of steel provided = Radius excluding cover r = Maximum axial load = Moment acting on the column about X- axis = Hence eccentricity about X-axis ex = 5.62 / 3821.58 Moment acting on the column about Y- axis = Hence eccentricity about Y-axis ey = 560.802 / 3821.58 Resultant Eccentricity, e = √ex2 + ey2 = Resutant Moment, M = Pxe = = Modular ratio
Assume the thickness of equivalent steel shell placed at a radius of r be Z Z
=
4712.389 2pr
=
1.364 mm
e
=
561.772 3821.580
=
147 mm
n = kd The value of k may be assumed between 0.4 - 0.7, smaller value for larger eccentricity d
=
1200 - 40 - 40 - 10 - 10
=
1100 mm
Assume K
=
0.700
n
=
770 mm
a1
n cos 1 1 R
=
cosa1 a2
R cos 1 cos 1 r
=
=
73.541
=
0.283
=
72.017
sina1
=
0.959
cosa1
=
0.283
sin2a1
=
0.543
sina2
=
0.951
sin4a1
=
-0.912
cosa2
=
0.309
sin2a2
=
0.587
C1
C1
=
sin 3 1 1 cos 1 2 R 2 c1 cos 1 sin 21 1 cos 1 3 2 4
2 x 600²C'/(1+0.283) x [0.959³/3+3.14-1.284/2x0.283-0.283/4x0.543] =
290842 C'
M1 M1
=
2R 3 1 sin 4 1 cos 1 sin 3 1 1 cos 1 8 32 3
2x600³C'/(1+0.283) x [(3.14 - 1.284)/8 -0.912/32+0.283/3 x 0.959³] = 96534846 C'
C2 C2
=
2r 2 Z mc 1c1 sin 2 ( 2 ) cos 2 R r cos 2
2 x 550²x1.364(16.47-1)x C'/(600+550x0.309) x (0.951 + (3.14 -1.257) x 0.309) = 25415.37 C'
M2 M2
=
2r 3 Z mc 1c' 2 sin 2 2 R r cos 2 4 2
2 x 550³ x 1.364(16.47 - 1) c' / (600 + 550x0.309) x ((3.14 - 1.257)/2+0.587/4)) =
9924049 C'
T
=
2 r 2 Z m c' sin 2 2 cos 2 R r cos 2
T
=
2 x 550² x 1.364 x 10.98 x c' / (600 + 550x0.309) x (0.951-1.257 x 0.309) = 6620.629 C'
M3
2r 3 Zmc' R r cos 2
2 sin 2 2 4 2
M3
=
2 x 550³x1.364x10.98/(600+550x0.309) x (1.257 / 2 -0.587/4) =
3118128 C'
Equating the sum of internal forces to the external forces C1 + C2 - T = P 290841.979C' + 25415.365C' - 6620.629C' = 3821580 309636.7 C' = 3821580 C' = 12.342 N/MM2 Equating the sum of internal moments to the external moment M1 + M2 + M3 = M 1.1E+08 C' C' Adopt C'
t t
7.215
= =
561772000 N-MM 2 5.127 N/MM
= =
1/2 x (12.342 + 5.127) 2 8.735 N/MM HENCE OK
mc' d n n = =
10.98 x 8.735/770 x (1100 - 770) 41.104 N/MM2 HENCE OK
DESIGN OF PIER FOOTING:Maximum load coming on to the foting Max. Moment
= =
Diameter of the pier Thickness of the footing at the face of support Thickness of footing at edge Footing offset Diamter of footing Reinforcement provided Distribution reinforcement Clear cover provided height of soil above footing (at face of support) height of soil above footing (at edge of support) Average height of soil (above sloped portion)
= = = = = = = = = = =
Area of footing Area of major circular portion
=
3832.89 KN 636.22 Kn - m
1.200 0.9 0.3 1.8 4.8 25 16 75 1.2 1.8 1.5
Average diameter of the sloped portion of footing
2
Area of footing Thickness of footing in sloped portion
= = = = = = =
Weight of footing in sloped portion Total weight of footing
=
135.72 + 144.315
Area of soil above footing
=
m2 2.67 = 1.2 2.67 x 1.2 x 21 67.284 KN
Area of soil above sloped portion Average height of soil (above sloped portion) Weight of soil above footing =
(2.2 + 4.8) / 2 3.5 m 3.147 x 3.5² / 4 2 9.621 m 0.6 m
18.096 x 0.3 x 25 135.72 KN 9.621 x 0.6 x 25 144.315 KN 280.035 KN
3.147 (2.2² - 1.2²) / 4
= height of soil above footing (at face of support) Weight of soil = =
Total weight of soil
491.719 201.408
18.096 m 0.3 = = =
Weight of major rectangular portion of footing
m m mm mm mm m m m
3.147 x 4.8² / 4
= =
Thickness of footing for major portion
m m
= = = =
67.284 + 450.261 = 517.545
Total weight footing + soil
m
2 14.294 m 1.5 m 14.294 x 1.5 x 21 450.261 KN
KN
weight of footing + soil + pier
= = =
280.035 + 517.545 797.58 KN 797.58 + 3832.89 =
Exposed area of footing in plan
=
3.142 x 4.8² / 4
=
2 18.096 m
Section modulus of footing
=
3.142 x 4.8³ /321
=
3 10.857 m
P max.
=
4630.47 18.096
+
636.22 10.857
4630.47
2 314.484 Kn/m
= P min.
=
4630.47 18.096
_
636.22 10.857
2
197.284 Kn/m
=
1.200
1.300
0.500
0.500 1.300
0.6 0.9 0.3 1.800
1.800 2.200 4.800 197.284
314.484
Stresses at foundation level Maximum stress intensity
= = There fore BM at the face of support Effective depth required
Effective depth provided
Required area of steel
p0
= =
( R r )(3R 2 2 Rr 2r 2 )
2.4 1.1
270.534 x 3.142 / 20 (2.4 - 1.1) x (3 x 2.4² - 2 x 2.4 x 1.1 - 2 x 1.1²) 529.237 Kn - m = 732.79 =
732.79 x 10^6 1.244 X 10^3
=
767.502 mm
= =
0.9 - 75 / 1000 - 25 /2000 0.813 m
=
732.79 x 10^6 / (190 x 0.89 x 0.813x1000)
= Required spacing of 25 mm dia bars
20
= =
Hence safe
2 5330.223 mm
491.719 x 1000 / 5330.223 90 mm c/c
(REQUIRED)
Provided spacing of 25 mm dia bars
=
Area of steel provided
=
Min area of steel required
=
90 mm c/c 2
5464 mm
(PROVIDED) Hence safe
0.12 x 1000 x 813 100
=
975.6
<
2
5464 mm Hence safe
Distribution steel: 0.2% of the cross - sectional area is provided as distribution steel = 0.2 x1000 x 813 100 2 1626 mm
= Required spacing of 16 mm dia bars
= =
201.408 x 1000 / 1626 120 mm C/C
(REQUIRED)
Provide spacing of 16 mm dia bars
=
120 mm C/C
(PROVIDED)
Area of steel provided
=
2 1678 mm
Hence safe
TWO WAY SHEAR:Critical section is at d/2 from the face of the pier
As this is circular footing around the circular column, the punching shear is at a distance d/2 from the face of the column all around the column. ro distance of punching shear from = 0.813 / 2 + 1.2/ 2 centre of the column = 1.007 m
F P0 (R 2 r02 )
Punching shear F
Hence from punching shear ponit of view
= =
270.534 x 3.142 (2.4² - 1.007²) 4033.62 KN
F 0.16 f ck 2tr0 d 0
4033.62 x 10^3 / (2 x 3.142 x 1007 x 813) 0.784
≤ ≤
0.16 x SQRT(25) 0.8 Hence SAFE
8 mm dia at 240 mm c/c 1.800
15 Nos of 20 mm dia bars 1.200
1.800
1.800
4.800
1.200
1.800
4.800 25 mm dia bars at 90 mm c/c spacing
16 mm dia bars at 120 mm c/c spacing 4.800 m
4.800
m 25 mm dia bars at 90 mm c/c spacing
16 mm dia bars at 120 mm c/c spacing REINFORCEMENT DETAILS OF FOOTING
