calculation of water current forces :-
DESCRIPTION
CALCULATION OF WATER CURRENT FORCES :-TRANSCRIPT
CALCULATION OF WATER CURRENT FORCES :-
Mean velocity of water current at HFL ,V = 7.0 m/s
= 9.90 m/s
= 98.0 m/s
Case 1:
Value of "K" = 0.66 Water current force diagram
3.363t/m2Pier shaft width at top (m) = 2.68
HFLPier shaft width at base (m) = 2.0 94.875
FIntensity of pressure due to water
= 3.363
Intensity of pressure due to water
= 1.853 1.853t/m2 80.200
Water current force on pier uptoend of varying portion, F (t) = 47.5
Level at which force F acts = 89.876
62.200
Case 2:
Value of "K" along transverse direction = 0.66
= 41.95 t
Value of "K" along longitudinal direction = 1.50
Water current force on pier uptoend of varying portion, F (t) = 56.2
Level at which force F acts = 94.162
Max. velocity at surface , V = (2)0.5*V
V2 at HFL
Water current force for normal case, water flowing at 0o
current at HFL, 52KV2 (t/m2)
at well cap top level (t/m2)
Top of well cap
Normal scour level
Water current force for normal case, water flowing at 20o
Force at base of pier along transverse direction ( taking cosine component of the value as calculated above & multiplying with the appropriate obstruction width)
Force at base of pier along transverse direction ( taking sine component of the value as calculated above & multiplying with the appropriate obstruction width)
3.90mPier cap top Level
RL 95.950
3.46m HFLRL 94.875
1.00m
pier centre line
H= 14.67m
2.00m RL 80.200Top of well cap
OBSTRUCTION DIAGRAM IN TRANSVERSE DIRECTION
11.00mPier cap top Level
RL 95.95HFL
RL 94.875 10.10m
5.00m
pier centre line
14.67m
RL 80.200 6.00mTop of foundation / well cap
OBSTRUCTION DIAGRAM IN LONGITUDINAL DIRECTION
= 98.0 m/s
Case 3:
Value of "K" = 0.66 Water current force diagram
3.363t/m2Pier shaft width at top (m) = 2.14
HFLPier shaft width at base (m) = 2.00 94.875
FIntensity of pressure due to water
= 3.363
Intensity of pressure due to water
= 1.685 1.685t/m2 80.200
Water current force on pier uptoend of varying portion, F (t) = 45.7
Level at which force F acts = 88.886
65.460
Case 4:
Value of "K" along transverse direction = 1.50
= 40.34 t
Value of "K" along longitudinal direction = 1.50
Water current force on pier uptoend of varying portion, F1 (t) = 54.7
Level at which force F1 acts = 94.303
V2 at HFL
Water current force for seismic case, water flowing at 0o
current at HFL, 52KV2 (t/m2)
at well cap top level (t/m2)
Top of well cap
Seismic scour level
Water current force for seismic case, water flowing at 20o
Force at base of pier along transverse direction ( taking cosine component of the value as calculated above & multiplying with the appropriate obstruction width)
Force at base of pier along transverse direction ( taking sine component of the value as calculated above & multiplying with the appropriate obstruction width)