02207421
.
. . 2555
1.
OUTLET TRANSITION
OUTLET TRANSITION
2.
BENCH FLUME
BENCH FLUME
3.
5.
4.
i
(
)
6. DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
CONTROL NOTCH
STILING BASIN
TRANSITION
WEIGHTED CREEP RATIO
UPLIFT
7. DESIGN OF VERTICAL DROP WITH WEIR CONTROL
DUCKBILL WEIR
STILING BASIN
WEIGHTED CREEP RATIO AND UPLIFT
TRANSITION
8. DESIGN OF INCLINED WITH WEIR CONTROL
DUCKBILL WEIR
TRANSITION
TRANSITION
WEIGHTED CREEP RATION
ii
(
)
9.
(Retaining Wall)
10.
(The Stem)
(The base slab)
(Counterfort)
11.
(TYPE A)
(Counterfort Retaining Wall)
(TYPE B)
iii
OUTLET TRANSITION
OUTLET TRANSITION
OUTLET
: 1.50
,γ , γ sat
γ
COVERING
/
.
2.2
/
.
10
/
.
2.4
/
=
2
R.b.d
vc
=
u
=
3
.
2
3
5
As.fs.j.d
3
0.15%
-
-
1.9
30
,
φ
,
Bearing Capacity
-
Mc
=
2
-
Ms
2.10
-
-
1
-
-
1.79
-
-
2.60
-
-
3.00
2.20
-
-
-
TRANSITION
Working Stress
fs
=
1,500
R
=
11.995
./
.
./
.
2
j =
0.885
V bd
V
Σo. j.d
OUTLET TRANSITION
, vc
3.8
2
1
Length of transition
(1.5 × 1.79 + 0.5 × 2.00 ) − 0.5 × 2.20
=
tan 22 .5
=
L
=
6.97
m.
7.00
m.
PLAN
LONGITUDINAL SECTION
Design reinforcement of
-
B -B
0.20
2
SURCHARGE 0.3 T/m.
P2
3.0 P1
O+
0.30
2.20
0.30
Internal friction angle, φ
∴
1,900 30
=
1 3
Su
=
300
P1
= =
P2
0.5 ×
1 3
0.3 × 1900 × 3 + 2
3142.13 ka.Su.H
=
1
Mmax Mmax
transition
= =
=
× 300 × (3 + 0.15)
4126.11
2
kg/m. =
3142.13 + 315 O
=
2
kg/m
2 1 Ka γ H SOIL 2
3
Vmax
3
o
ka
kg/m
=
315 =
kg/m. 3457.13 kg/m.
3.15 3 + 0.15 3142.13 × + 315 × 3 2
(kg-m)/m.
OUTLET TRANSITION
2
dm
Use
=
=
M / Rb
=
18.55
dv
=
V v c .b
t
=
. 3457.13 = 3.8 × 10
=
30
.
Covering =
5
.
d
=
25
.
As
=
M
Use
=
φ 16 @
0.20 +
As
=
1500 × 0.885 × 25
Use
10.05 + 5.65
=
φ 12 @
2
φ12 @ 0.20
Ast
.
12.43
0.15%
As
=
Asmin
0.20
.
9.10
4126.11 × 100
=
fs. j.d
As
4126 .11 × 100 11 .995 × 100
2
.>
15.70
0.0015 × 100 × 30
=
4.50
2
OK.
=
=
.
12.43
.
2 2
.>
5.65
.
4.50
2
OK.
C L
P2 3.15
P1
O
1.25 1800 kg
χ 4126.11 kg-m 1285.71 kg/m Weight of side wall
=
1 2
× (0.20 + 0.30) × 3.00 × 2400
=
1800
Upward soil reaction
=
2 × 1800 = 2 .80 × 1 .00
Moment at CL
=
− 4126.11 − 1285.71 ×
Moment at CL
kg/m. 1285.71 1.25 2 2
=
− 4126 .11 − 1004 .46 + 2250
=
-2880.57
OUTLET TRANSITION
kg/m.
2
+ 1800 × 1.25
kg-m/m.
3
As Use
φ 16 @ 0.20
2880 .57 × 100
=
1500 × 0.885 × 25
=
As
.
10.05
2
=
8.68
.
>
8.68
.
2
2
OK.
χ M for bars φ16 @ 0.20
− 4126 .11 − 1285 .71
χ2 2
=
3335.34 kg-m
=
-3335.34
χ
=
0.546
=
1
=
2 1
3
Use
Mmax
=
As
=
φ 12 @ 0.20
.
25 100
Use
0.80
=
1026
kg
180
kg
.
P2
10 .05 × 1500 × 0.885 ×
+ 1800 χ
P1
=
× × 1900 × 1.8
2
3
× 300 × 1.8
=
1. 8 1 .8 × 180 × 3 2 777 .60 × 100 = 1500 × 0 .885 × (26 − 5 )
1026 ×
As
1.80 m.
1
=
5.65
Design reinforcement of
=
777.6
kg-m
.
2.79
.2 (
2
<
Ast
Ast)
-
Use section - for designing Height of vertical wall
=
1.8
m.
Height of sloped wall
=
0.85
m.
Thickness of slab
=
0.25
m.
C L
P2 = 180
1.80
P1= 1026 1.275 SOIL
SURCHARGE
0.85 1.25
918 kg. 972 kg.
748.51 kg/m.
OUTLET TRANSITION
4
P1
=
1 1.8 2 × 1900 × = 3 2
P2
=
1 × 300 × 1.8 = 3
Weight of sloped wall
=
1 × (0.20 + 0.25 ) × 1.8 × 2400 2
=
972
=
0.25 ×1.53 × 2400
=
918
1 .8 3
+ 180 ×
kg/m.
kg/m.
972 + 918 (1.25 + 1.275 )
=
1026 ×
kg/m.
=
= MA
kg/m.
180
Weight of vertical wall
∴ soil reaction Upward
1026
748.51
1. 8 = 2
777.6
2
kg/m. /m. kg-m.
( As
=
777 .6 × 100
=
) .
2.93
2
Use
12 @ 0.20
1500 × 0.885 × (25 − 5 )
Ast
=
0.0015 × 25 × 100 =
MB
=
1026 ×
=
1487 .70 + 608.40 + 315 − 585 .23 − 1239 .3
=
586.57
As
=
586 .57 × 100 1500 × 0.885 × (25 − 5 )
MCL
=
1026 ×
MCL
=
1487.70 + 315 + 2386.11 − 1732.73 − 2454.30
=
+1.78
=
1.78 × 100 1500 × 0.885 × (25 − 5 )
As
.
3.75
2
As = 5.65
.
2
1.8 + 0.85 + 748 .51× (1.275 )2 +180 × 1.8 + 0.85 − 918 ×1.275 − 972 ×1.275 2 2 3 2
kg-m/m (
=
2.21
.
1 .8 1 .8 (2.525 ) 2 + 0.85 + 180 + 0.85 + 748 .51 3 2 2
kg-m/m (
OUTLET TRANSITION
=
) 2
12 @ 0.20
1 .275 − 918 + 1.25 − 972 (1.275 + 1.25 ) 2
0.0067
Use
) .
2
Use
12 @ 0.20
5
A 0 .0 1
L C
5 .1 0
0 .3 1
5 1 . 3
1 :1.5
0.20
1 :1.5
0 .0 7
N IO S N A R T F O N A L P F L A H
1
0 .1 0
0 2 . 0
5 .2 0
0 1 . 0
L C
A
OUTLET TRANSITION
6
3 .2 2 2 + 5 0 . 0
0 1 . 2
3 .1 0 2 +
0 .2 0
0 .0
0 .2 0
1 0 .2 0
F O T N E M E C R O F N I E R
1:1.5
C
C
A A N IO T C E S
0 .0 7
2
-
. m 6 8 . 2
F O T N E M E C R O F IN E R
R E L IL F IC T S A L E
0 3 . 0
0 1 . 0
B
7 .6 9 1 +
7 .6 2 2 + 3.0
OUTLET TRANSITION
0 .3 0
B
0 .2 0
0 5 .3 .2 0 0
7
0.20
12 @ 0.20
1.80
12 @ 0.20 12 @ 0.20
3.0
16 @ 0.20
12 @ 0.20 1.20
0.30
0.95
0.95
0.30
2.20
SECTION B-B 0.20 0.30
12 @ 0.20 12 @ 0.20
1.80 VARIED
12 @ 0.20
16 @ 0.20
VARIED 12 @ 0.20 12 @ 0.20 VARIED 12 @ 0.20 12 @ 0.20
VARIED
16 @ 0.20
REINFORCEMENT OF
OUTLET TRANSITION
-
8
0.20
12 @ 0.20 12 @ 0.20 VARIED 12 @ 0.20 VARIED
VARIED 12 @ 0.20 12 @ 0.20 12 @ 0.20
VARIED
12 @ 0.20
REINFORCEMENT OF
OUTLET TRANSITION
-
9
3 2 . 2 2 +
0 .1 2
5 0 . 0
0 2 . 0 @ 2 1
F O T N E M E C R O F IN E R
S E C A F H T O B
3 .1 0 2 +
S E C A F H T O B
0 .2 0 @ 2 1
0 .2 0 @ 2 1
C
-
. m 6 .8 2
0 .2 0
0 .0 1
0 2 . 0 @ 2 E 1 C A F R A F
F O T N E M E C R O F IN B E R
R E L IL F C I T S A L E
0 .2 0 @ 2 1
0 2 . 0 @ 2 1
0 2 . 0 @ 6 1
C
0 2 . 0 @ 2 1
0 0 . 7
-A A N O I T C E S 3
0 2 . 0 @ 6 1
0 2 . 0 @ 2 1
1 : 1
7 .6 9 1 +
0 .3 0
B
0 .2 0
6 7 . 2 2 + .0 3
OUTLET TRANSITION
0 3 . 0
5 2 . 0
10
BENCH FLUME
BENCH FLUME
FLUME
BENCH FLUME
b
=
5.00
m.
d
=
2.50
m.
A
=
21.875 m.
P
=
14.013 m.
R
=
1.561
Side Slop n
15.20 cms.
2
m.
1 : 1.5 =
0.025
bed slope
1 : 6000
V
=
0.695
m/s
Q
=
15.20
cms.
0.35
m.
Allowable losses
MS
= AS.fS.j.d
vC
=
V
u
=
b.d V
Working stress
fs = 1,500
.
.
2
j
= 0.885
Σo. j.d
MC = R.b.d
2
R
= 11.995
El. 102.50 El. 102.15
El. 100.00 El. 99.65 Flume Sta. 1+220
Sta. 1+000
and structure profile 1 Canal BENCH FLUME
1
flume = 3.0 m/s
Vmax
∴
A
Q
=
15.20
=
=
b
5.067
m.
2
3
V
1
3
d
b
=
2
d
=
2d
A
=
bd
d
∴d
= b
∴
1.592 =
b d× = V
2
=
2d =
m.
use
m.
2
m.
3.18 3.2 1.6×
3.20
=
5.12
2.97
m/s.
15.20=
=
5.067
m. 2
m.
5.12
Manning , n = 0.016 ( V
=
Flume) 1
R
……….
R 2 / 3 S1 / 2
n
5.12
=
=
0.8
m.
3.2 + 2 (1.6 )
1/2
S
= =
n V R 2/3 0.016 × 2.97 =
0.05514
(0.8) 2 / 3
S
=
Flume
0.003041
1:329
=
1 : 328.84
=
2d
A
=
2d
P
=
4d
R
=
A P
BENCH FLUME
Flume 1 : 1,000 b
1 : 500
2
=
2d 2 4d
=
d 2
2
Manning V
1
=
n 15.20
0.016
8/3
d =
2/3
(1 / 1,000)
1/ 2
6.104
d
=
1.97
b
=
1.97 2×
1/ 2
d 2
1
=
2d 2
d
R 2 / 3 SO
m. =
3.94
m.
4.0
.
Manning Q
1
=
n
15.20
AR 2 3 S O1 2 /
1
=
0.016
/
(4d )
4d 4 + 2d
2/3
(1 / 1,000)1 / 2
trial
error d
=
1.95
V
=
Q
m.
A
15.2
=
=
1.95
m/s.
4 × 1.95
Flume
Free board
Flume
Flume F
=
0.2 + 0.1d
F
=
0.2 + 0.1(1.95) =
0.395
d+F
=
0.395 + 1.95
2.345
(Spce.
=
RID) m.
2.4
.
Broken – back transition Inlet Transition
θ=
27.5
o
loss
=
0.3
∆hv
loss
=
0.5
∆hv
1
Onlet Transition θ 2 =
BENCH FLUME
22.5
o
3
3.75
3.75
5.0
2.5 1 1.5
5.0
1.95
4.0
3.75
3.75 4.25
4.25 1
6.25 2.0
2
6.25 2.5
2.5
2.0
C
INLET T.
L
FLUME OUTLET T.
FLUME
2 Inlet Transition , L
=
Transition
4.25
=
8.16
9.0
.
.
tan 27.5
Outlet Tansition, L
=
4.25
=
10.26
11.0
tan 22.5
Sta 1 + 000 L
Sta 1 + 220 =
Flume, Lf = Δhv =
1220 - 1000 220 - 11 - 9
(1.95
2
= =
− 0.695 2 )
.
200
.
1 = 2 x 9.81
Inlet Transition loss
=
0.3 ∆hv
=
Oultet Transition loss
=
0.5 ∆hv =
0.086
Loss in flume
=
SfLf
=
BENCH FLUME
220
0.169 0.3(0.169)
. =
0.051
.
. SoLf
4
1
=
=
x 200
0.20
.
1000
Total loss
=
0.34
0.051 + 0.20 + 0.086
=
.
0.35 .
1.3 ∆hv
=
1.3(0.169)
=
0.22
.
=
0.5(0.169)
=
0.085
.
Broken -back transition Inlet Transition ∆WS
=
Outlet Transition ∆WS
0.5 ∆hv
=
El. 102.50 El. 102.265 El. 102.065 El. 102.15
2.5 So = 1 : 1000 El. 100.115
1.95
El. 100.315
El. 99.65 2.5
El. 100
Sta. 1+ 000
Sta. 1+ 009
Sta. 1+ 209
INLET
BENCH FLUME
Sta. 1+220
OUTLET
TRANSITION
TRANSITION
3 Profile
Flume
Transition Inlet transition
=
Outlet transition =
w1
=
Vmax =
γh w
P
BENCH FLUME
=
100.315 − 100
=
1:28.57
1:6
9 100.115 − 99.65 = 11
OK
1:23.66
1:6
OK
Flume =
× 1,000
2.4
1 wh
1.00 m.
0.25
=
2,400
=
0.5 × 2400 × 2.4
=
2,880
kg/m
2
kg
5
Mmax
=
2.4 0.25 + 2 3
=
2,800
2,664
kg-m
2.4 W1 0.25
2,664 kg.-w
2,880 kg. FORCE
4 dm
=
dv
=
M
(
V
SHEAR
2,664 × 100
=
=
14.9
11.995 × 100
Rb
MOMENT
SHEAR
MOMENT
.
vc = 3.8 ksc)
vc b
=
2,880 3.8 × 100
7.60
= = d =
covering =
.
.
, w2 Vmax Mmax
=
flume
=
.
=
2,304
=
2,304=
523.64 ×
4.40 4.2
kg 523.64 kg/m =
1099.64 kg.
2 4.2 2 523.64 × 8 + 2,664
2
dv
.
2 (0.2 × 2.4 × 1.0 ) × 2,400
w 2 l + 2,664 = 8 1154.63 + 2,664
=
15
=
=
dm
.
flume
=
20
5
Mmax =
.
3818.63 kg-m 3818.63 × 100 = 11.995 × 100 1099.64 =
3.8 × 100
BENCH FLUME
17.84 2.89
. .
6
4.40
W2 = 523.64 kg/m 2,664
2,664
4.20
1099.64
1099.64
3818.63 2,664
2,664
5
SHEAR
d covering
As
=
25
=
20
.
=
5
.
.
M fs ⋅ j ⋅ d
=
2,664 × 100 1500 × 0.885 × 15
=
13.38
.
2
0.2% Ast
BENCH FLUME
BENCH FLUME
=
MOMENT
=
× 0.002 × 20100
=
4
.
2
7
As
1.41,000 ×
=
1 l wl 2 3
=
0.5 × 1,400 × 1.4 ×
As
Ast=
A=s
12 mm @ 0.28 20 mm @ 0.28
u
=
Σo
2
=
13.46
.
2
9.34
.
Σo
2
.
=
22.44
.
1.40
As =
4.04
Σo
=
13.46
As =
11.22
Σo
=
22.44
15.26
Σo=
15.26
.
2
20 mm (
35.90
13.38
(Bond stress)
Bond stress
2
As = As =
.
11.22
.
4.04
13.38 – 4.04
20 mm @ 0.28
As =
Ast
4.0
=
2
1.40
1.40
.
12 mm @ 0.28
As
=
.2
.
kg/m
2.30
.
4
3
457 33×100 = 1 500 × 0 885×15
4
1. 4
457.33 kg-m
-
1,400
=
Ast
=
=
,
=
w M
1.40 .
= =
BENCH FLUME
21.4
./
2
.
ksc
)
V
Σo. j.d 2,880 35.90 × 0.885 × 15
=
6.04 < 21.4 ksc
OK
8
As As
= =
3818.63 × 100
=
1500 × 0.885 × 20 2664 × 100 = 1500 × 0.885 × 20
12 mm @ 0.28
A=s
14.38
.
10.03
.
2
2
15.26
.
2
Σo
=
35.9
.
20 mm @ 0.28
Bond stress u
=
12 mm @ 0.22
OK
=
5
.
2
A=s
2
5.14
. > 5
.
2
0.002 × 25 100 ×
OK
. >4
=
1.73 < 15.1 ksc
2
4.04
12 mm @ 0.22
=
st
A=s
)
A
12 mm @ 0.28
(
35.9 × 0.885 × 20
Ast
1099.64
.
2
OK
12 mm @ 0.28
12 mm @ 0.22
Bearing Bearing load
=
flume . flume
BENCH FLUME
. Flume +
9
.
flume
.
flume
Bearing load
=
2(2.4 × 0.2 × 1.0) 2400 + (4.4 × 0.25 ×1.0) 2400
=
2304 + 2640
=
(4 × 2.41)1000 ×
=
Bearing load
Ton/ m
flume
foundation
2
3305.45 kg/m
flume
kg/1 m.
Bearing Capacity
kg/1 m.
2
Bearing capacity
9600
=
4.4 ×1.0
3.305
4944
=
4944 + 9600
=
Bearing load
=
0.20
12 mm. @ 0.22 12 mm. @ 0.28 1.40 12 mm. @ 0.28 2.40
20
mm. @ 0.28 12 mm. @ 0.28
1.00
0.25
4.40
6
BENCH FLUME
FLUME
4.00
10
INLET TRANSITION E.G.L hL
2
V1 /2g
E.G.L
2
V2 /2g
FLOW y1
FLOW y2
Z
∆ +
(
V1
2
=
2g
−
z y
1
)
y
2
y 2 − (∆z + y1 )
FLUME
∆z + y1 +
= =
y2 +
2g
+ hL
V22 V12 L 2g − 2g + h - ∆ hv – 0.3 ∆ hv =
BENCH FLUME
V2
2
-1.3
∆ hv
1.3 ∆ hv
11
OUTLET TRANSITION hL
E.G.L
2
V2 /2g
2
V1 /2g
FLOW
y2
FLOW y1
Z
FLUME
V2 2
∆z + y 2 +
−hL
=
∆ z + y 2 − y1
(∆z + y 2 )− y1
=
∆h v − 0.5∆h V
V12 2g
V1
2
=
y1 +
2g
−
V22 2g
BENCH FLUME
2g
+hL
=
+ 0.5
∆ hv
0.5 ∆ h
12
.
ACI–Code
1.
fs
fc′ =
fc
=
=
1,500.00 175.00
./
.
./
.
./
.
2 2
′
0.45 f
c
=
78.75
=
Es/ Ec
=
6 2.04×10 / (15,210 f ′ )
=
10.138
=
n/ (n + fs/ fc)
=
0.347
j
=
1- k
R
=
0.5j.k fc
vc
=
0.292
n
2
c
k
( =
= =
f ′ c
5 8
= 12.078
./
.
= 3.86
./
.
.
0.53 (
= 0.884
3
f
c
′
. .
= 7.01
(318–77)
Spillway
2
)
./
2
.
2
)
1
2.
A36
1,526
./
.
ASTM
fy = 2,450
1,040T
=
=
=
.-
Ast
Ast
=
=
500
./ .
./ . ./ .
2,400
./ .
./ . ./ .
1,900
./ . ./ .
7,900
./ .
15
2
/ .
3 3 3 3 3 3 3 3 3
2
24
2
0.10 (Anchorage) 48
0.002 bt
2,300
0.0025 bt
0.6 fy
./ .
1,700
=
2,600 =
fs
E37
2,100
=
)
T
.
(
1,800
=
=
.
1,000 =
4.
./
=
./
2
Surcharge Load
2
ASTM
3.
……….(1)
……….(2)
2
=
Ast
b
= t
Chute
1.00
(Uplift Pressure)
5.1
5.2
1
(Pi) P1 = S.H3.Ka 2
P2 = 0.5 γmH2 .Ka
3
P3 =
γ
m
.H2.Ka.H4 2
12
.
Shear
A
A (Vi)
H3/2
P1.a1
P1
H2/3 + H4
P2.a2
P2
H4/2
P3.a3
P3
P4.a4
P4
5
P5 = 0.5 γw. H 2 4
H4/3 P5.a5
(Mi)
H4/3
(ai)
P4 = 0.5 γsub.H4 .Ka
Moment
4
2
A
1
5.3
(i)
0.10
(U-Shape Retaining Wall) 1 Chute 0.90 0.20
.
5.
100
=
=
P5
3
P1
=
Surcharge load
P2, P3, P4
=
/ .
/ .
/ .
=
Surcharge load = 0.5
/ .
m
=
Unit Weight of Moist Soil = 2.0
/ .
γs
=
Unit Weight of Saturated Soil
/ .
γsub
= =
2.10 Unit Weight of Submerged Soil
/ .
=
(γs–γw) = 1.10
γw
=
Unit Weight of Water = 1.00
Ka
=
Active Earth Pressure Coefficient
=
tan (45–θ/2)
P5
=
S
γ
θ
=
H
=
H1
=
H2
=
H3
=
H4
=
ai
=
1
A;
A ;
3 3
3
3
/ .
2
30
(
Granular Soil)
Pi
MA
=
VA
=
A
5
∑ Pi.a
– ./ .
……….(3)
5 ∑ Pi i =1
/ .
……….(4)
i =i
2
Angle of Friction of Soil =
A
.-
4
1
5
dm
=
dv
=
As
=
.
……….(5)
VA×10/vc
.
……….(6)
MA×1000×100/(fs.jd)
.
MA × 1000 / R
=
dm
dv
=
R
=
12.078
vc
=
./
=
3.86
fs
=
1,500
j
=
0.884
d
=
dm
1
(
2
3)
A
B
(Pi)
(bi)
(Mi)
P1 = S.H.Ka
H/2
P1.b1
2
P2 = 0.5γm H2 Ka
H/3
P2.b2
Shear
1
(i)
2
Moment B
2
2
2
0.10
.
.
.) dv
(
5.4
./
.
./
……….(7)
2
B (VB)
W1
6
2
7
CL
(
)
W1
C
w1
A
MB B w2
=
3
+
8
B; B;
MB
=
VB
=
2
……….(8)
∑ Pi.bi i =1
W1 + w1.T3 – w2. T3 2
C; C;
W1
MC
=
MB+(w2–w1)(0.5L1) / 2–W1.L1/2
……….(9)
Vc
=
w1 + w1.L1/2 – w2.L1/2
……….(10)
= w1
=
w2
=
L
=
bi
=
γm
.
/ .
Pi
=
Unit Weight of Moist Soil
=
2.0
2
1
2
2
(
2 2
/ .
<
15
/ .
2
B
3
/ .
4)
(i)
(Pi)
1
(ai)
P1 = S.H.Ka
H/2 2
2
P2 = 0.5 γm.H2 .Ka
3
P3 =
γ
m
.H2.Ka.H1 2
P4 = 0.5 γsub.H1 .Ka
4
2
P5 = 0.5 γw.H1
5
B;
MB
5 ∑ Pi.ai i =1
=
H2/3+H1 H1/2 H1/3 H1/3
……….(11)
9
4
C;
MC
2
=
2
[MB+w2.(0.5L1) / 2–w1.(0.5L1) / 2–W1.L1/2]
……….(12)
10
VB
=
W1 + w1.T3 – w2. T3 2
VC
=
W1 + w1.L1/2 – w2.L1/2
=
Pi
=
L
=
w2
=
1
1
1
2
5 ( 7 .
1.
2100 vc
=
u
=
)
(γ ) 1900 ./
3
10
/ .
2
2.29
5.1
3.2 6. Surcharge
φ
f c′
φ
3.23
u
f′ c
0.5
/ .
./ 25 35
.
2
./ ./
. .
2
2
2
11
3
) .
5
m
B
V bd V ∑ o.j.d
5.
(
./ .
3. Bearing capacity 4.
2.
B
/ .
Pi
C
……….(14)
ai
……….(13)
2
=
tan (45- θ )
P1
=
1 3
×0.5×4.5
P2
=
1 2
× 13 ×1.9×(2.5)
=
1 3
×1.9×2.5×2.0
P4
=
P5
=
1 2 1 3
0.73
2
2.0
, VA
=
8.63
M2
=
1.98 ( 2.5 +2) 3
M3
=
3.17× 2
=
3.17
-
M4
=
0.73× 2
=
0.49
-
M5
=
2.0× 2
=
1.33
-
1.69
2
dm
=
dv
=
2 3
3
=
12.29×1000×100 12.078×100
V vc b
=
=
12.29
-
31.9
.
22.4
.
d = 43
.
=
21.55
Ast
0.002×100×40
)
12.29×1000×100 1500×0.884×43
40
-
50
=
8.63×1000= 3.86×100
25
=
A
=
5.61
A
, MA
(
-
×
A
3.17
0.75× 4.5 =
Ast
=
× ×(2.1-1)×(2.0)= ×1.0×(2.0) =
1.98
=
M1
0.75
2
1 3
=
1 3
=
2
= 2
A
2
tan (45- 30 )
=
2
1.0
2
Ka
P3
As
5
=
.
40
8.0
.
.
.
.
12
5
13
50 .
=
1 3
× 0.5 × 5.0 × 5.0 2
M2
=
1 2
× 13 ×1.9 × (2.5)2( 2.5 +2.5) 3
=
6.6
-
M3
=
1 × 1.9 3
× 2.5 × 2.5 × 2.5 2
=
4.95
-
=
0.95
-
M4
1 2
=
M5
1
=
,
1
× 13 × (2.1-1.0) × (2.5) × 2.5 3 2
× 1.0 × (2.5) × 2.5
W1
=
2.08
-
2.60
-
3
, MB
= 0.5(0.25+0.50)×4.5×2.4
=
17.18
=
4.05
-
= (4.05 ×2)+(5.0+0.5+0.5)×0.5×2.4
.
=
2
2
.
B
M1
= 15.3
14
≈ (2.0+0.5) × (5.0+0.5+0.5)×1.0 (
(Bouyance)
(
=
15.0
=
15.0×1.0
=
15.0
=
<
15.3
(15.3-15.0 = 0.3
8.10 + 7.20
15.30 6.0×1
=
=
=
2.55
, w1
B
B
=
2
+
2
=
.
=
0.52.4 ×
=
4.05+(1.2×0.25)-(2.55×0.25)
=
3.71
=
17.18+2.55×
2
/ .
2.75
=
1.2
(
(0.25) 2 2
/ .
-1.2×
2
)
(0.25) 2 2
0.5
5.0
.
1.20
/ .
L1
15.30
2.55
)
×2) + (5+0.5+0.5) × 0.5 × 2.4
(4.05
, w2
)
1.20 × 15.0 = 18.0 ) (Uplift)
=
.
.
.
− ( 4.05 × 0.75 )
15
C
C
16.21
=
4.05 + (1.2×2.75) – (2.55×2.75)
=
0.34
=
17.18+2.55
=
11.15
dm
=
dv
=
.
C
=
As
B
=
B
(
)
(2.75) 2
-1.2
2
2
− 4.05 × 0.75
-
16.21×1000×100 12.078×100
3.71×1000 3.86×100
d = 43
As
-
(2.75) 2
50
=
=
=
36.63
9.62
. < 43
. < 43
.
11.15×1000×100 1500×0.884×43
16.21×1000×100 1500×0.884×43
C
=
19.56
.
.
=
28.43
.
.
16
Uplift
B
M1
=
1 3
×0.5×0.5× 5.0 2
M2
=
1 2
× 13 ×1.9×5.0× 5.0× 5.0 3
=
2.08
MB
C (
.
. .
1
, , ,
C
Uplift C
C
w2 w1
=
1.2
W1
=
4.05
=
15.3/(6.0×1) / .
-
=
15.27
-
9.24
=
4.05 + (1.2
=
14.23
B
=
3.71
(
(
B
0.34
(
C
)
×2.75) – (2.55×2.75) =
2
×
-
/ .
(2.75) 2 - 4.05 2.75 2
-
2.55
=
2
15.27+(2.55-1.2)
B
13.19
2
/ .
=
Uplift)
=
=
-
(
)
)
)
B
17
.
=
=
15.3+(5.0
37.8/(6.0 =
×4.5×1.0×1.0) ×1.0) 2
6.30
/ .<
1
38.89
Ast
0.002×100×38.89
=
As
φ16 φ22
/ .
2
2.5
.
37.8
10
1.
=
=
7.73
.
=
7.78
2.50
. (As
.
φ
As
=
7.73
.
.
mm. @ 0.26
As
=
14.62
.
.
.
.
=
Ast
As
φ 16 mm.
.
As
@ 0.25
.
mm. @ 0.26
φ16 mm.
5.13
)
4.5
>
16 mm. @ 0.26
Ast
2.5
.
22.35
=
0.002
=
8.04
.
=
8
=
19.33
.
=
26.58
.
=
45.91
.
.
.
.
@ 0.25
×100×40
∑o ∑o ∑o
18
(
1
P1
)
(
P2 )
(
P3 )
(
P4 )
(
P5
)
(
VA
)
(
)
M1 (
- .)
M2 (
- .)
M3 (
- .)
M4 (
- .)
M5 (
- .)
MA (
- .)
1.50
0.25
0.71
-
-
-
0.96
0.19
0.36
-
-
-
0.54
2.00
0.33
1.27
-
-
-
1.60
0.33
0.84
-
-
-
2.50
0.42
1.98
-
-
-
2.40
0.52
1.65
-
-
-
3.00
0.50
1.98
0.79
0.05
0.13
3.44
0.75
2.64
0.20
0.01
3.50
0.58
1.98
1.58
0.18
0.50
4.83
1.02
3.63
0.79
4.00
0.67
1.98
2.37
0.41
1.13
6.56
1.33
4.62
4.50
0.75
1.98
3.17
0.73
2.00
8.63
1.69
5.61
dm (
.)
dv (
.)
Eff.Depth (
.)
As (
.
.)
6.71
2.49
26.33
1.56
1.18
9.87
4.15
29.11
3.05
2.17
13.40
6.21
31.89
5.13
0.02
3.62
17.30
8.92
34.67
7.86
0.06
0.17
5.67
21.66
12.51
37.44
11.42
1.78
0.21
0.56
8.50
26.53
16.99
40.22
15.94
3.17
0.49
1.33
12.29
31.89
22.36
43.00
21.55
19
8.63
A (
u
3.23
=
u
V
=
=
3.23 2.2
=
19.42
o
2
φ16 mm. @ 0.26 φ16 φ22
4.95
.
./
.
2
2
< 19.42
.
.
Σ
o
=
19.33
.
mm. @ 0.26
As
=
7.73
.
.
Σ
o
=
19.33
.
mm. @ 0.26
As
=
14.62
.
.
Σ
o
=
26.58
.
.
Σ
=
30.08
.
o
=
.
3.71
=
u
=
o Σ
3.23 φ
>
28.43
φ 16 mm. @ 0.25
B
u
65.24
(1.50-0.25 = 1.25)
7.73
As
./
=
175
As
.
Σ
2.
22
45.91×0.884×43
=
.
8.63×1000
=
∑ o.j.d
16
)
f c′
φ
f c′
V ∑ o.j.d
16
=
3.23 2.2
=
19.42
=
3.71×1000 65.24×0.884×43
=
1.50
. 16
.
22
.
175
./
.
./
.
2
2
< 19.42
20
2
D (
)
(
dm
- .)
(
.)
dv (
Eff. depth .)
(
As
.)
(
.
.)
0.00
4.05
17.18
37.71
10.49
43
30.10
0.25
3.71
16.21
36.63
9.62
43
28.40
0.50
3.38
15.32
35.62
8.74
43
26.84
1.00
2.70
13.81
33.81
6.99
43
24.18
1.25
2.36
13.17
33.02
6.12
43
23.08
1.50
2.03
12.62
32.33
5.25
43
22.11
1.75
1.69
12.16
31.73
4.37
43
21.30
2.00
1.35
11.78
31.23
3.50
43
20.64
2.25
1.01
11.48
30.84
2.62
43
20.12
2.50
0.68
11.27
30.55
1.75
43
19.75
2.75
0.34
11.15
30.38
0.87
43
19.53
L
=
fs.φ 4u
6
=
1500×1.6 4×19.42
=
30.90
.
50
.
21
Footing
' fc
= 175
ksc.
fc
= 78.8
ksc.
fs =
1,500 =
10
k
=
0.344
j
=
0.885
R
=
11.995
d
=
4.51
kd
=
1.55
(N.A)
(Covering = 4 cm.)
/2 m.
=
0.325
m.
from top < 4.15 m.
elevated flume = 2 b'
m.
(0.25+0.40)
ksc.
n
b' =
1
flume shape beam 1.0 u-
flume
=
4.15(6.5)(1,000)
flume
w
M =
kg/m.
=
26,975
kg/m.
=
14,130
=
1,500 + 26,975 + 14,130
2
kg/m.
WL2 8
R(2
b' ) d
2
<
Mc
As
=
M = f s .j.d
1,500
[0.4(7.3)+2(1.0)(0.2) + 1 (0.25+0.40)(3.95)] (2,400)
M
=
=
=
Mc
750 2×
=
=
42,605
=
42,605(10)2 = 8(1,000)
=
11.995(2×0.325)(451) 2 (1,000)
=
1585
kg/m.
532.56
t.m.
t.m
532.56(1,000) (1,500)(0.885)(4.51)
=
88.95
cm
=
13.69
cm
=
42,605(10) 2
=
=
(2×32.5)(451) =
2
/ 6.5 m width
2
/1 m width
2
DB 20 @ 0.20 ( A s = 15.71 cm ) V
WL 2
=
V
v
=
v c= Vc = V'
=
3.8 (2 V-
kg.
213,025
(2b' )d
3.8
213,025
7.27
ksc.
ksc.
×32.5)(451) Vc =
=
111,397
kg.
213,025 – 111,397 =
101,628
kg.
2
S
A v .f v .d V'
=
S
(4.02)(1,500)(451) 101628
= =
26.76
=
=
flume
w
>
20
cm.
=
OK.
span. beam 10.0 Simple 750 kg/m (6.5)
4.15 =
2
(1,000)
14,130/2
750 + 13,487.5 + 7065
=
13,487.5
kg/m
=
7,065
kg/m
=
21,302.5
kg/m
21,302.5 (10)2
M
=
WL2 8
Mc
=
2 R B . R b'. d ;
=
0.981 (11.995)(0.325)(415+20)2/(1,000) =
M
<
Mc
As
=
M fs. j. d
=
8 (1,000) R
B
=
266.28 t.m
723.65 t.m
=
266.28 (1,000) (1,500)(0.885)(4.35)
=
=
21,302.5 (10) 2
106,512.5
46.11
cm
2
2
10 – DB 25 ( A s = 49.09 cm ) V vc V'
cm
DB 16 @ 0.20
2
(DB 16, A s = 4.02 cm )
wL 2
= = =
3.8 (32.5) (435) V-
=
=
Vc
(
53,722.5
= kg
106,512.5 – 53,722.5 =
U-shape beam)
kg
52,790 kg
3
LL
750
=
DL w
1 1.0×
=
0.2 × 1.002,400 ×
=
750 + 480
M
=
(wl)2 2
As
=
fsM . j. d
=
= =
750 =
kg/m.
480 =
kg/m.
1,230
1,230 (0.75)2
kg/m.
=
2
345.94 kg/m.
345.94 (1,500)(0. 885)(0.16) =
1.63
cm 2
2
DB 12 @ 0.20 ( A s = 5.65 cm ) V
=
dv
=
wL V b vc
=
1,230(0.75)
=
922.5 (100) (3.8)
=
(H)
=
P
=
(4.15) (4.15) ½
M
=
8.611 (4.15/3)
=
922.5
=
Cantilever beam
p
γw
=
kg
2.43 cm < 16 cm
O.K.
1.0 (4.15)
=
4.15
/ .
2
8.611 =
11.912
-
4
=
Mc
Rb
d
11.995 (1.0) (36)2 (1,000)
2
=
15.54
> M
M Rb
=
dm
11.912 (1,000) = 11.995
=
31.5
. < 36 O.K.
V b vc
=
dv
8.611 (1,000) (100) (3.8)
=
=
22.7
. < 36 O.K.
M fs. j. d
=
As
11.912 (1,000) = (1,500) (0.865) (0.36)
=
24.93
.
2
2
DB 16 @ 0.20 + DB 20 @ 0.20 ( A s = 25 76 cm )
Σ o = 25.13 + 31.42 = 56.55
w=1
.
flume
3.95(0.25)(2.4) =
w2
=
2.370
1 2 (0.15)(3.95)(2.4)
=
0.711
w= 3
0.4(0.4)(2.4)
=
0.384
w=4
0.2(1.0)(2.4)
=
0.48
w=5
750(1.0)/1,000
=
0.75
Σw dv
,
u
=
V b vc
=
12.36
u= =
=
<
=
V o.j. d
4.8 ksc.
<
4.695
4.695(1,000) 100(3.8)
= .
36
.
O.K.
8.611(1,000) (25.13+31.42)(0.885)(36)
21.4
ksc.
O.K.
5
(1) = (2) ;
M1
=
A s . f s .j . d ;
M
=
½
2
2
DB 16 @ 0.20 ( A s = 10.05 cm )
γ
d
=
0.25-0.04+
3
(2)
- 2.882H – 16.81 H
2.50 m.
=
H3 3
flume
=
960
kg/m
4,150 + 960
=
-11,912 + 5,110
(6.9)2 8
wL 2
Rb =
5,110 =
2 d =
11,995(1.0)
kg/m
18,499
5,110 (6.9) = 2
=
kg.m
17,630
(36)
2
kg
= 15,545
kg.m
+−
(M Mc ) Mc + f s (d - d') fs. j. d 15,545
+
(1,500)(0.885)(0.36)
(18,499−15,545)
=
38.7 cm
2
(1,500)(0.36)−0.04)
2
DB 20 @ 0.20 + DB 25 @ 0.20 ( A s = 40.25 cm ) −
As
2.934 m.
=0.40 (2,400)
=
=
kg/m
Mc =
0
4,150
=
+
=
=
=
V
As
½ (1,000)
4.15 (1,000)
= +
=
Fixed end beam
w M
(1)
2 H .H 3
10.05(1,500)(0.885)(0.21+0.036H) H
0.15 (H) 4.15
=
M− fs. j. d
11,912 = (1,500)(0.885)(0.36)
=
24.93
cm
2
2
DB 16 @ 0.20 + DB 20 @ 0.20 ( A s = 25.76 cm )
6
=
Vc V
'
=
vc . b . d
V-
Vc
= =
3.8(100)(36) 17,630-13,680
=
13,680
3,950
kg
kg
2
: DB 16 @ 0.20 + DB 20 @ 0.20
( A = 25.76 cm ) s
=
2
: DB 20 @ 0.20 + DB 25 @ 0.20
=
( A s = 40.25 cm )
25.76 + 40.25
=
2 cm /m
66.01
V' As
= =
3,950 66.01
=
60 ksc < = f v1,500
ksc.
O.K.
Cap Beam
=
750 (2)(10)
=
/1,000
4.15(6.50)(10)(1.0)
Flume
=
=
=
269.75
[0.2(1.0)(10)(2) + 1 (0.25 + 0.40)(3.95)(10)(2) + 2
0.4 (7.3)(10)] 2.4
beam Cap
=
Cap beam Cap beam
0.5 (0.7)(2.4) =
= =
=
141.30 0.84
15 + 269.75 + 141.30 =
15
=
/
426.05
426.05 (1,000) (50) (730)
11.70 ksc
'
< 0.25 f c=
43.8 ksc
O.K.
7
Beam
∴ =
M
+
M (cantilever)
= =
1 WL'2 2
(
2
Rb d
As
=
M f s .j . d
V
=
Σo
=
/
m.
=
0.40
×0.40
=
=
59.20
1.20 – 0.40
kg-m.
=
3,444
kg-m.
=
1 (59.20×1,000) (0.35) 2 = 2
3,626
kg-m.
V u . j. d
kg-m. (66)
4,210 1,500×0.885×0.66
=
1.15
2
m. 4,210
11.995(0.50)
' 1.15 w L =
0.80
1 (59.20×1,000) (0.80) 2= 11
4,210
=
=
m.
2 1 9 (59.20×1,000) (0.80) =
=
∴M(max) Mc
= 1.20
, L′ =
1 WL'2 11
(15+ 269.75+141.30) + 0.84 7.3
=
'2 1 9 WL
−
M
=
=
=
26,125 kg-m.
4.81
×59.20×1,000×0.80/2
27,232 = 12.1×0.885×66
=
2
cm
=
38.53
2
27,232
kg.
cm
DB 25 mm
u = 12.10 ksc)
2
5–DB25, A s = 24.54 cm , Σo = 39.27 cm. v Vc= V
V bd
=
'
3.8 =
V-
×50×66
=
Vc =
=
Av . fv . d V′
=
=
=
2.54
8.25
12,540
27,232-12,540
Stirrup 2-RB9, A s = 2×1.27 S
27,232 50×66
=
cm
=
3.8
ksc.
kg. 14,692
kg.
13.70
cm.
2
2.54×1,200×66 = 14,692
>
8
Stirrup 2-RB9 @ 0.125
= 2 V = 2 ×27,232 = 54,464 kg.
′
P = 0.85 A g (0.25 f c + A s f s / A g ) Ag
4-DB 25, A s = 19.63 cm P
40×40
=
=
1,600
cm
2
2
=
0.85×1,600(0.25×175+19.63×1,500/1,600)
=
84,528 kg. > 54,464 kg. (
7,232 ×2) 2
O.K.
4 –DB25
RB 9 @ 0.20
9
h
=
Height of Column
r
=
Radius of gyration
=
0.3×0.40
=
P(long) P(short)
R
P(long)
3.00
=
0.12
m. h r
1.07-0.008
1.07-0.008× 3.00 0.12
=
R × P(short)
=
0.87×84,528
m.
= 0.3 t =
=
=
=
0.87
=
73,539 kg. > 54,464
O.K.
5 1 . 4
0 4 . 0
Σ M=A As
Cap beam
req.
(4.15)2 = 2
50 =
M fs . j . d
0.40
2
50 kg/m
431
kg-m. 431 (1,500)(0.885)(0.36)
= 2 cm /m.
=
0.9
<
DB 16 @ 0.20 + DB 20 @ 0.20 ( A s = 25.76 cm )
2
Shear = =
O.K.
10.00 m.
50 (4.55)(10.00) 2,275
kg / 1 cap beam
10
2,275 = 7
= v
=
V bd
=
0.23
325
kg / 1
=
325 (40)(36)
ksc
<
3.8
ksc.
O.K.
=
0.75(2)(10)
=
=
=
1.0(4.15)(6.50)(10)
15 =
[0.2(1.0)(10)(2)+0.4(7.3)(10) + (3.95)(10)(2)](2.4)
Cap beam
=
270
½ (0.25+0.4)×
=
[0.7(8.3)(0.5)+2(0.4)(0.5)(0.5)](2.4)
142 =
8
11
Column
=
7(0.4)(0.4)(11)(2.4)
=
30
Bracing
=
12(0.8)(0.4)(0.4)(2.4)
=
4
=
Pile cap
[8.3(2.2)(2.0) – 7(0.4)(0.4)(2.0)](2.15) =
8.3(0.6)(2.2)(2.4)
=
= M
= =
2.275(14.575)
74
27
570
=
2.275(1,000) = 2(7)
=
33.16
≈
-
. /pile
162.5
< 1 kip/pile (≈ 453
34
. /pile)
O.K.
Σ d2
=
P
=
V N
P 1
=
P 2
=
1 ( 570 - 34(3.6) ) 7 40.32 2
=
39.20
P3
=
P4
=
1 ( 570 - 34(2.4) ) 7 40.32 2
=
39.70
P5
=
P6
=
1 ( 570 - 34(1.2) ) 7 40.32 2
=
40.21
P7
=
P8
=
1 ( 570 - 34(0) ) 7 40.32 2
=
40.71
P 9
=
P10
=
1 ( 570 - 34(1.2) ) 7 40.32 2
=
41.22
P11
=
P12
=
1 ( 570 - 34(2.4) ) 7 40.32 2
=
41.73
=
1 ( 570 - 34(3.6) ) 7 40.32 2
=
42.23
2
2
2( (1.2) + (2.4) + (3.6) 2 ) ±
=
40.32
Σ M.d Σ d2
570 7
P13
=
P14
12
P (t)
H (t)
N (t)
N1 = N2
30.20
3.775
30.44
N3 = N4
39.70
4.9625
40.01
N5 = N6
40.21
5.026
40.52
N7 = N8
40.71
5.089
41.03
N9 = N10
41.22
5.153
41.54
N11 = N12
41.73
5.216
42.05
N13 = N14
42.23
5.279
42.56
40.71
( V
=
570 – 270 – 74
=
=
P1
pile cap)
P1
226
P2
1 ( 226 - 34(3.6) ) 7 40.32 2
P2
=
P1
P 2
compression
=
14.625 O.K.
13
(footing)
Footing
=
M Rb
=
16,286×100 11.995×120
60
.
V
=
40,715
.
As
u
' fc
0.53 =
V bd
=
6.28
=
M fs. j. d
=
0.40
1×40,715×0.40
=
v
.
0.60 -
Footing
v c=
=
0.40
2
= dm
2
40,715 =
=
0.40×0.40
∴
ksc.
3.23 f c' φ
=
16,286
.- .
=
33.64
.
d
=
54
=
0.53
=
40,715 120×54
175 =
<
7.01
=
16,286×100 1,500×0.885×54
=
22.72
>
35
.
7.01
ksc.
O.K
.
.
ksc.
14
25
.
u
=
3.23 175 = 2.5
Σo
=
V u . j. d
25
ksc.
<
35
=
40,715 17.1×0.885×54
=
49.82
. /1.20
=
41.52
./
O.K.
. @ 0.18, As
=
27.27
Σo
Check deflection
17.1
=
. 43.63
.
.
flume
∆
=
5 WL4 384 EI
W
=
DL of flume
E
=
2 × 10
5
=
141.3 t. /10 m =
141.3
kg.
ksc.
15
N.A. (
x [0.4(7.3)+2(1.0)(0.2)+ =
flume)
( 1 ) (0.25+0.4)(3.95)(2)] 2
0.4(7.3)(0.2)+2(0.2)(1.0)(4.45)+2(0.25)(3.95)(2.175)+ 2 ( 1 ) (0.15)(3.95)(1.717) 2
x
=
1.304
m.
INA
=
3 2 3 1 1 12 (7.3) (0.4) +(7.3)(0.4) (1.104) +2( 12 )(1.0) (0.2) + 2 2 2(1.0)(0.2) (3.146) +2( 1 )(0.25) (3.95) 3 +2(0.25)(3.95) (1.071) + 12 2 2 ( 1 ) (0.15) (3.95) 3 +2 ( 1 ) (0.15)(3.95) (0.413) 2 36
∆
∆
4
=
13.006
=
5(141.3)(10)4 384(2×105 )(13.006)
=
0.0071
cm.
=
2.8
cm.
= =
m
<
L/360 O.K.
5(269.75+141.3)(104 ) 384(2×105 )(13.006)
0.0206
cm.
<
2.8
cm.
O.K.
16
17
18
1.
-
,d
-
(
:
)
1 : 1.50
0.014
.(
.)
. . .2
+279.721
.(
.)
.1
+278.211
.(
.)
.2
+278.061
.(
.)
.5
+280.842
.(
.)
TRANSITION
-
-
+279.871
2.00
. . .1
-
-
1.66
2.
-
18.0
1
-
2.5
./
,b
-
-
3
4.134
, γ
-
Bearing Capacity
-
Friction factor
,φ
.
4.0
1.9
/
2.2
/
3
3
/
2
0.6
0.90
10
./
30
,γ
1.50
1
,γ
-
2.4
0.2
/ %
3
1
5 1 . 0
1
5 1 . 0
2
2
-
COVERING
-
Ms
=
vc
=
u
=
Mc
=
5
Working stress
A s .f s .j.d
f s = 1,500
2
./
.
j = 0.885
V b⋅d V ∑ o ⋅ j⋅ d R.b.d 2
R = 11.995
-
3.
, vC
A
=
(TS1)
Q/V
3.8
0.25 .
=
./
.
4.134/1.5
=
2
(TS2)
2.756
0.25
.
2
2
H) (L ×
=
A P=
4.00
VP=
4.134/4.00
× 2.00
2.00 .
2
=
1.034
./
hVP
.
1.50 ./
OK
2
=
VP /2g
=
1.034 /(2 ×9.81)
2
=
0.0545
.
INLET TRANSITION
5
1.
LTU
θ
θ
1.66
= 27.5
o
2.50
2.30
3
LTU
2.5 2.3 o − 1.5 × 1.66 + / tan (27.5 )= 2 2
=
4.98
.
LTU
=
3H
= LTU=
(2.5 + 1.5 × 1.66) 1.66
V
=
Q A
=
4.134 8.283
=
.
8.283
.
=
0.499
2
./
Velocity head
hVC
=
V2 = 2g
∆hV
=
hVP - hVC
=
0.0545 – 0.0127
=
0.062
.3
.3
=
0.08 8
.
. (H)
10
–
=
279.871 – 0.10 -2.00
=
277.771
.(
.)
0.90 .
.3
.5 – 0.90 – 0.25 – 2.0
=
280.842 – 0.90 – 0.25 – 2.0
=
277.692
.(
.)
277.661
.(
.)
=
0.041
.
.
. . .1 –
=
.3
0.0127
<
=
6.00
6.00
=
1.5∆hV
=
A
hVC =
×
3 2.0
INLET TRANSITION (SIN) SIN
=
(278.211 – 277.661) / 6 =
1
10.91
1
4.00
OK
4
HHU =
H+ =
2.0 + 0.25 +0.30 +0.10
HHU=
+
2.65
=
2.65
+ 0.10
.
.
Wetted perimeter R
Sf
P
=
A/P
=
(22)/8.0 ×
n
=
0.014
Sf
=
=
= 8.0
.
=
0.50
VP2 n 2 / R 4 / 3
.
………..
FRICTION SLOPE
2
4/3
=
(1.034 × 0.14) / (0.5)
=
0.000528
18.0 .4
277.571
.4
= =
+277.561
HT=
.)
1
180
1
OK
∆hV + ( L × Sf ) + 0.7 ∆hV
=
0.4 ( 0.041 ) + ( 18.0 × 0.000528 ) + 0.7 ( 0.041 )
=
0.0546
ALLOWABLE LOSS
200
(HT) 0.4
= =
.(
(277.661 – 277.561) /18.0 =
TOTAL LOSS
0.005
.3 – (18.0 × 0.005)
= =
. .1 –
.2
279.871 – 279.721
5
=
0.15
0.0528
OK
OUTLET TRANSITION = 22.5
o
θ
θ
5
1.
2.30
2.50
1.66
5
1.
1.66
LTD
LTD
SOU
=
2.5 2.3 1.5 × 1.66 + 2 − 2 / tan 22.5
=
6.25
6.50
OUTLET TRANSITION (SOU) =
(
.2 –
=
(278.061 – 277.561) / 6.5
=
1
13
.4) / L TD
1 : 4.0
OK
. .
0.4 W .
=
.4 –
=
280.842 – 277.561 -2.0 -0.25
=
1.031
.5 –
1900
./ .
–
3.382
3
1.031900 × = 19.57 600 0.25 ×2400 = 2 3 = 16,000 = 7273 27273 × = 2672
./ . ./ .
2
2
. ./ .
2
3.02 (1.03 × 1.75)
6
Load Impact
3
0.4 W 0.4W 0.4W 0.4W 6’ 4’ 6’
1.03 3.02
.
FLOW
3
*
38
3.
WEEL LOAD
AD
BC
4
7
w1
=
1957 + 600 +2672
5229
INTERNAL FRICTION ANGLE w2
Ka γSOIL h
=
1 × 1900 × 1.03 + 0.25 3 2
=
731.5
( w3
w4
1900
. .)
1 0.25 × 1900 × 1.03 + 0.25 + 2.0 + 3 2
=
2156.5
=
./ .
2
5229 + 960
=
2 × 2 × 0.25 × 2400 (2.50 × 1.00)
=
960
./ .
=
6189
./ .
2 × 2 × 1 × 1000 = (2.50 × 1.00)
=
./ .
600
./ .
6189 + 1600 + 600
=
2
BEARING LOAD =
1600
0.25 × 1.00 × 2400 =
=
1/3
3
=
1 − sin 30 = 1 + sin 30
=
./ .
Ka
2
30
=
./ .
8389
./ .
9 / MOMENT DISTRIBUTION
K
2
K/EI
4 EI / 2.25
1.778
BC
4 EI / 2.25
1.778
CD
4 EI / 2.25
1.778
AB
8
DA
4 EI / 2.25
AB, BC, CD
3
1 W1L21 12
=
M FBA
=
=
1 2 12 × 5229 × 2.25 =
2205
.- .
M FCD
=
M FDC
=
1 × 6189 × 2.25 2 12
M FBC
=
M FAD
=
1 1 2 2 × 731.5 × 2.25 + × (2156.5 − 731.5) 2.25 12 30
=
M FDA
.- .
.–
2206
.
-
w1 = 5229
.
A
.- .
1 1 × 731.5 × 2.25 2 + × (2156.5 − 731.5) 2.25 2 12 20
=
669
+2206
2611
.- .
= =
.–
=
549
M FCB
549
0).25
FIXED END MOMENT
M FAB
-
I = 121 (1.00)(
DA ,
1.778
.–
.
./ .
+549
A
B
B
.–
.
w3 = 731.5
./ .
w4 = 2156.5 C
D
D +669.3
C
.- .
669.3
-
2611
.–
.
5
w4 = 6189
+2611
.–
.
.–
.
./ .
9
./ .
1
Moment Distribution
A
JOINT
B
C
D
MEMBER
AD
AB
BA
BC
CB
CD
DC
DA
K
1.778
1.778
1.778
1.778
1.778
1.778
1.778
1.778
CYCLE
DF
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
MOMENT
-549.07
2206.00
-2206.00
549.07
-669.30
2611.00
-2611.00
669.30
DISTRIBUTION
-828.47
-828.47
828.47
828.47
-970.85
-970.85
970.85
970.85
CARRY OVER
485.43
414.23
-414.23
-485.43
414.23
485.43
-485.43
-414.23
DISTRIBUTION
-449.83
-449.83
449.83
449..83
-449.83
-449.83
449.83
449.83
CARRY OVER
224.91
224.91
-224.91
-224.91
224.91
224.91
-224.91
-224.91
DISTRIBUTION
-224.91
-224.91
224.91
224.91
-224.91
-224.91
224.91
224.91
CARRY OVER
112.46
112.46
-112.46
-112.46
112.46
112.46
-112.46
-112.46
DISTRIBUTION
-112.46
-112.46
112.46
112.46
-112.46
-112.46
112.46
112.46
1
2
3
4
CARRY OVER
56.23
56.23
-56.23
-56.23
56.23
56.23
-56.23
-56.23
DISTRIBUTION
-56.23
-56.23
56.23
56.23
-56.23
-56.23
56.23
56.23
5
6
CARRY OVER
28.11
28.11
-28.11
-28.11
28.11
28.11
-28.11
-28.11
DISTRIBUTION
-28.11
-28.11
28.11
28.11
-28.11
-28.11
28.11
28.11
-1341.94
1341.94
-1341.94
1341.94
SUM
1342 1342
.–
.–
1675.75 1675.75
w1 = 5229 1342 ./ .
.
.–
1675.75 1675.75
-
.
1342
.
A
A
1209
B
5883
2040
6963 D
D
1676
-
.
.–
.
6
5883
6963
.
.
W3 = 731.5
./ .
.
1676 1676 w4 = 6189 ./
.- .
1676
.
B
1209 .
.–
W4 = 2156.5
. C
C
.– .–
.
.
.
2040
SHEAR
10
./ .
DIAGRAM 7 SHEAR
DIAGRAM 8 MOMENT
11
MOMENT BENDING
SHEAR
AB M Rb
dm
=
dv
=
V bv
d
=
1967 × 100 11.995 × 100
=
5883 100 × 3.8
=
20
AS
.
+
M fs ⋅ j ⋅ d
=
1967 × 100 1500 × 0.885 × 20
-
15.8
=
.
5.00
.
=
7.41
.
=
5.05
.
2
)
1342 × 100 1500 × 0.885 × 20
=
2
=
0.002 × 100 × 25
2241× 100 = 11.995 × 100
13.67
ASt
.
( AS
COVERING
=
AS
=
12.8
)
( AS
=
=
5.00
.
2
CD
dm
=
dv
=
6963 = 100 × 3.80
d
18.32
20
=
AS +
-
ASt
=
=
5.00
.
) =
8.44
.
=
6.31
.
2
)
( AS
COVERING
2241 × 100 1500 × 0.885 × 20
=
AS
.
. (
AS
.
1676 × 100 1500 × 0.885 × 20
2
=
0.002 × 100 × 25
=
5.00
.
2
12
AD
BC
dm
=
dv
=
1676 × 100 = 11.995 × 100
2040 = 100 × 3.80
5.37
20
.
d
=
AS
( AS
11.82
COVERING
1676 × 100 1500 × 0.885 × 20
=
.
=
5.00
.
)
-
.
=
.
5.00
=
6.31
.
2
2
AB
-
16 @ 0.26
AS=
7.73
.
AS=
6.70
.
AS=
8.74
.
AS=
6.70
.
AS=
6.70
.
AS=
5.14
.
2
ΣO=
19.33
.
2
ΣO=
16.76
.
2
ΣO=
21.85
.
2
ΣO=
16.76
.
2
ΣO=
16.76
.
2
ΣO=
17.14
.
16 @ 0.30
CD
16 @ 0.23
-
16 @ 0.30
BC
AD
16 @ 0.30
12 @ 0.22
13
BONDING STRESS (u)
-
AB u
= =
V
19.83
ΣO AB
5883 16.76 × 0.885 × 20
=
∑o ⋅ j⋅ d ./
.2
16 @ 0.30
2
1) 16 @ 0.30
18.90
./
.2
2) 16 @ 0.30
CD
u
u
AB
=
BC
=
16.76 × 0.885 × 20
∴
6963
=
23.47 > 18.9
16 @ 0.30 2
6963 33.52 × 0.885 × 20
=
Σ O = 33.52
11.47 < 18.9
.
2
OK
AD
u
=
2040 = 16.76 × 0.885 × 20
6.87
18.9 >/
OK
14
9
15
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
1.
,Q
0.907
0.907
0.80 0.80
0.80 0.80
3
./
.
,d
,b
, H
, Hc
b
, 1: m ,L , n
s
2.
, INV
,
.
.
1.00
1.00
1.30
1.30
1: 1.5
1: 1.5
1: 4000
1: 4000
0.016
0.016
+91.592
+90.592
.(
.)
+92.392
+91.392
.(
.)
.
CONTROL NOTCH
,B
≥
CONTROL NOTCH
bu
B
NOTCH, T
=
du
T
T + (Hcu - du)
=
≥
.
0.8 =
.
0.85
Z
=
WING WALL
0.85 + (1-0.8)
L
WING WALL
.
1.05
0.25 Qd Qw=
WING WALL
=
3
./
0.227 =
B + 3T
=
0.8 + 3(0.85)
=
Hcu - T
=
1 - 0.85
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
=
3.35
.
=
0.15
.
1
WING WALL Q
∴
Q
=
Q
2
=
3
Cd L H3 2 /
=
1.82 × 3.35 × 0.15 3 2
=
0.354 > 0.227
/
0.2 Qd
d
NOTCH (S)
NOTCH
Q
Q
0.189
u
=
OK.
MANNING .
0.37
P
3
./
Q
1.0 Qd
0.2 Qd
4 C d 2g T 3 2 P + ⋅ S ⋅ T 5 /
Cd
0.189
0.25
0.65
0.37 + 0.05
0.33
0.907
0.25
0.65
0.80 + 0.05
0.52
0.55
T
du
P
3.
NORMAL DEPTH
=
S
=
P
N
=
P + 2⋅S⋅T
=
0.25 + 2(0.55)(0.85)
S
+0.05
=
0.25
=
1.185
.
STILLING BASIN
CRITICAL DEPTH Q2
=
2
=
9.81
0.0839
A 3c N
g ( 0.907 )
CONTROL NOTCH
=
+ Sy c ) y c ] 3 [ P + 2Sy c ]
[( P
[( 0.25 + 0.55 y c ) y c ]
3
0.25 + 2(0.55) y c
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
2
yc = 0.7 .
Trial + Error Ac=
0.444
Vc
=
hvc
=
∆h
=
Q
=
.1 -
.2
92.392 - 91.392
=
.
1.0
.
R
=
F
=
∆h + R
=
1.0+0.20
0.20
1.20 .
=
0.212 .
=
2g
./
2.04
0.444
Vc2
=
0.907
=
Ac
(1)
NOTCH
STILLING
BASIN
V2
2
F + yc + Vc
=
d1 +
1.2 + 0.7 +0.212
=
d1 +
=
d1 +
2g
2.112
1
2g 1 2 × 9.81
[08 ( .
0.907 ( 0.8 + 1.5d 1 ) d 1
0.0419 + 1.5d 1 )
d1
2
]
2
d1 = 0.143 .
TRIAL AND ERROR
FROUDE NUMBER V1
A
=
0.907 / [(0.8+1.5 x 0.143)0.143]
=
0.907/0.145
=
Bw =
0.145
=
6.25
/
2
0.8+ 2(1.5)(0.143) =
1.229
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
3
Fr1
V
=
=
gA / B w
MOMENTUM M1 Q
=
+ d1A1
=
Q2
d1
+ 3B)
=
gA1
6 Q
+ B)
=
( 2 md1
2
gA1
+
d12 2
(d1
A1
Q2
+
gA 2 Q2 gA 2
=
+
d 22 6
( 0.907 )
9.81x 0.145
+
d2
d 22 2
( 2 md 2 + 3B)
…..… (1)
( d 2 + B)
(0.8 + 1.5 x 0.143)(0.143)
(1) 2
+ d2A2
gA 2
2
+
HYDRAULIC JUMP
M2
2
gA 1
Q2
5.81 >1 SUPERCRITICAL FLOW
=
.
0.145
2
( 0.143)
2
2
TRIAL AND ERROR
+ 0. 8)
=
0.598
=
( 0.143
d2
=
( 0.907 )
2
9.81(0.8 + 1.5d 2 )d 2 0.0839 ( 0.8
+ 1.5d 2 )d 2
+
+
d22 2
d22 2
( 0.8
( 0. 8
+d2 )
+d2 )
.
0.82
FROUDE NUMBER A2 Bw2
= =
0.8 + 2 x 1.5 x 0.82
V2 = Fr
2
(0.8 + 1.5 x 0.82) 0.82
0.907 / 1.665
9.81×1.665 / 3.26
L3
L3 =
= =
0.545
=
=
=
.
2
. ./
0.243 < 1
SUBCRITICAL
Ld + Lj
=
Vc 2F / g + 3d2
=
2.04 2 ×1.2 / 9.81 + 3 × 0.82
=
3.469
3.26 0.545
=
3.6
1.665
*
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
4
4.
TRANSITION
L1
=
3 du
=
0.8 3×
=
L2
5.
=
4 dd
T
WEIGHTED CREEP RATIO
=
0.8 4×
3.5 .
*
=
3.2
.
=
0.85
.
1.0 .
*
UPLIFT
SAND COURSE CUTOFF
2.4
2.5 .
L4
=
Cw = 5.0
H1
=
0.6
H2
=
H3
=
0.7
H4=
0.6 0.7
VERTICAL PATH, Lv
=
(2H1 - 0.15) + (2H 2 - 0.15 - 0.20) + (F-0.2+0.20) + (H3 - 0.20) + (H3 + R -0.15) + (2 H4 - 0.15)
Lv =
2(0.6) - 0.15 + [2(0.6) - 0.15 -0.20] + [1.2 - 0.2 + 0.20] + [0.7 - 0.20] + [0.7 + 0.20 - 0.15] + [2(0.7-0.15)]
= HORIZONTAL PATH, LH =
5.60 (L1 + L2 + L3 + L4) / 3 =
(2.5 + 1.0 + 3.6 +3.5) / 3
=
3.53
Max. Head
=
F+T
Cw
= =
1.2 + 0.85 5.60 + 3.53
.
.
CREEPING PATH, L
=
2.05
2.05
=
4.45 < 5.0
=
2.05(5) - (5.60+3.53)
=
1.12
.
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
5
Lv
CUTOFF H1
=
0.8
.
H2
H3
=
0.8
.
H4 =
Lv Cw
6.80
=
=
.
0.8
*
.
0.8
.
6.80 + 3.53
=
2.05
=
5.04 > 5.0
OK.
SHORT PATH
Min L=1
1.2 (H - 0.15) 1
=
0.78 < 2.5
OK.
Min L=2
1.2 (H - 0.20) 2
=
0.72 < 1.0
OK.
Min L=4
1.2 (H +3 R - 0.15)
INLET TRANSITON ( OUTLET TRANSITION (
L4)
1: 4
=
1.02 < 3.5
) L)
OK.
(
1
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
1: 4 L1
L4
6
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
7
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
8
NOTCH
N
T
P
B
COMPLETE FLOW Q
=
Cd
= =
P
=
S
=
d
=
FREE FLOW .2
3
C d 2g d 3 2 ( P + /
4 5
⋅ S ⋅ d)
0.70
0.65
NOTCH
DESIGN OF VERTICAL DROP WITH NOTCH CONTROL
NOTCH
9
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
1.
,Q
0.907
0.907
0.80 0.80
0.80 0.80
3
./
.
,b
, H
,d
1.00, Hc
1.00
b
1: m , LS , n
2.
, INV
,
.
(CHECK)
DUCKBILL WEIR (
C
.
1.30
1.30
.
1:1.5
1:1.5
.
1:4000
1:4000
0.016
0.016
+91.592
+90.592
.(
.)
+92.392
+91.392
.(
.)
)
DUCKBILL WEIR
(ROUNDED CREST)
0.36
Q
y1
CL 2 g H crt ( 3 / 2 )
=
=
+ du0.1
B
.
b u
≥
=
0.8 + 0.1
, B, bu
0.80 .
,
B
≥
0.40
B = 1.0
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
α
45
=
0.9
.
75
. .
1
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
2
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
3
Type A
B
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
4
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
5
H
S
H-S
=
y1 + (Hcu-du) + 0.05
=
0.9 + (1-0.80) + 0.05
=
0.85 y1
=
0.85 0.9 ×
=
0.77
=
1.15 – 0.77
W1=
1.5 =
Q2
.
=
0.38
.
=
0.57
.
=
0.13
.
(H-S)
=
0.5(1.0) + 1.5(0.77) – 0.20
=
1.455
=
y1 – s
=
0.9 -0.77
2
(UNROUNDED)
C
, L
CL 2g H crt ( 3 / 2 )
=
0.32 (0.90)
=
0.06
=
Qd – Q1
=
0.907 – 0.06
3
(q)
=
C 2g H crt
=
0.36
=
0.0747
C = 0.32
0.90
,L
= 2A + 0.40 -0.90
.
2 × 9.81 (0.13)1.5 ./
= 0.36
=
q
0.765
Q1
.
.
(ROUNDED)
2.
1.15
0.5B + 1.5S - 0.20
1.
=
1.5 (0.38)
W2 =
Hcrt
=
=
0.847
3
./
(3 / 2 )
2 × 9.81 (0.13)1.5 3
./
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
6
(Bt)
Bt
=
0.847/0.0747
Bt
=
11.35
A
=
0.5 Bt – 0.20 + 0.45
=
0.5 (11.35) – 0.20 + 0.45
=
5.925
α
.
=
cos (1.455/5.925)
=
75.78
-1
=
cos α =
3.
W2 / A
B = 1.20
W2
=
.
75
*
*
0.5 (1.20) + 1.5(0.77) - 0.20 = 1.555
1.555/5.925 =
α
.
α
B
11.338
.**
α
cos
=
=
74.81
0.262
.
75
L1
≥
y1
L1
=
1.00
L2
=
A sin
=
5.925 sin (74.81)
α
=
5.717
.
STILLING BASIN DROP
(P) = 1.20
d1 +
2 1
V
=
y1
y1 + P
.*
d 1
STILLING BASIN ….. (1)
2g A1
=
(B + 1.5 d 1) d1 =
V1
=
(1.20 + 1.5 d1) d1
0.907 (1.2 + 1.5d 1 )d 1
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
7
(1)
[0.907 /(1.2 + 1.5d 1 ) d 1 ]
d1 +
2
=
2 × 9.81
Trial and Errors
d1 = 0.106
A1
=
V=1
0.144
. .
0.907/0.144
d1
Bw =
0.9 +1.20
=
6.299
./
1.2 + 2(1.5)(0.106)
=
1.518
.
FROUDE NUMBER Fr1
V
=
gA / B
6.299
=
9.81×
=
1.518
SUPERCRITICAL
d d STILLING BASIN HYDRAULIC JUMP 1
Q
2
+
d
2 1
6
gA 1
=
(2m d1 + 3B)
Q2
+
( 0.106)
Q2
=
d 22 6
+
d 22 2
gA 2
2
2
+
gA 2
2
9.81× 0.144
(2m d2 + 3B)
(d2 + B)
( 0.907 )
(0.106 + 1.20) =
2
9.81(1.20 + 1.5d 2 )d 2 ( 0.084)
=
0.589
M2
=
Q 2 + d 1 (d + B) 1 2 gA 1 2
MOMENTUM
2
M1
(0.907 )
> 1
6.53
0.144
(1.20 + 1.5d 2 ) d 2
+
+
d 22 2
d 22 2
(d2 + 1.20)
(d2 + 1.20)
Trial and Error d2 y+3 R ∴
=
0.745
=
0.80 + 0.20
d2 <
( y3 + R )
.
=
1.00
FROUDE NUMBER
.
O.K. d2
1.0
SUBCRITICAL FLOW
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
8
STILLING BASIN Ld
=
Vc
Lg
=
3d2
L3
=
Ld + Lj
Z
=
S+P
=
0.77 + 1.20
2Z / g
=
Vc 2Z / g + 3 d2
=
1.97
.
(CRITICAL DEPTH)
Hc
=
3
Hc
=
11.35 + 0.45 + 0.45
=
3
( 0.907 /12.25)
0.907/ (12.25 0.0735
Lj
= 0.745 3×
L3
=
2
=
12.25
/ 9.81 =
0.082) ×
= =
.
0.0735 =
./
0.047
.
2.235
0.047 + 2.235
)
.
0.082
=
2 ×1.97 / 9.81
Ld + Lj
(
UNROUNDED CREST
Bt + 0.45 + 0.45
Ld =
3
=
Vc =
4.
L L
L
2
(Q / L) / g
.
=
2.282
.
2.5 L1
2.5 L1 =
2.51.0 ×
=
L3
=
2.5
Qd
=
0.907
P
=
1.20
.
B
=
1.20
.
y1
=
0.90
.
H
=
1.15
.
S
=
0.77
.
2.50
.
**
3
/
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
9
5.
W1=
0.57
.
W2=
1.555
.
L1 =
1.00
.
L2 =
5.72
.
L3 =
2.50
.
WEIGHTED CREEP RATIO AND UPLIFT
Max
∆H
N
=
P+S
=
1.20 + 0.77
=
N
SAND COURSE
CUTOFF
L1 + L2
.
=
0.15
.
t2
=
0.20
.
t4
=
0.20
.
t5
=
0.20
.
hl
=
0.60
.
h2
=
0.80
.
LH
Cw
N(2h
) – N(2t4) + t4 + (P - t4 + t5) + (h2 – t5) + (h2 + R)
1
=
2(2 × 0.6) – 2(2 × 0.20) + 0.20 + (1.2-0.20 + 0.20) + (0.8-0.20) + (0.8 + 0.20)
=
2.4 + 0.8 + 0.2 + 1.20 + 0.6 + 1.0
=
(L1 + L2 + L3 + t2) / 3
=
(1 + 5.72 + 2.5 + 0.20) / 3
=
(Lv + LH) /
=
(4.6 + 3.14)/1.97
=
=
=
9.85 – (4.6 + 3.14) =
=
4.6
.
=
3.14
.
∆H
= 1.97 5×
h1 = 0.8 .
3
ti
2
2
Lv =
1.97
=
=
Cw = 5.0
3.93
9.85
5.0
.
2.11
CUTOFF
.
CUTOFF
3
t1 = 0.20 .
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
10
Lv
=
3(2 × 0.8) – 3(2 × 0.20) + 0.20 + (1.20 - 0.2 + 0.20) + (0.8 - 0.20) + (0.8 + 0.20)
=
6.60
- 4.60 = 2) (6.60
L1
=
1.50
L
=
(1.5 + 5.72 + 2.5 + 0.20) / 3
=
3.31
.
H
Cw
= (6.60 + 3.31)/1.97
SHORT PATH
X
≥
X
=
1 (1.5 + 5.72 - 4 × 0.20) 3
=
1.2(0.80 - 0.20)
>
1.2 (h1 - t4)
1.2 (h1 - t4)
X
1.2 (h
1
=
5.028
5.0
O.K.
- t4) ** =
2.14 =
0.72
. .
O.K.
A
UPLIFT Lv =
3(2 =
LH=
×
0.80) - 3(2 × 0.20) + 0.20 + (1.20 - 0.20 + 0.20)
5.0
.
(1.5 + 5.72)/3
=
2.41
.
A
UPLIFT
UA =
6.
.
[1.97 -
] ×1,000
.
5.028
=
496.25
t5 =
496.25
t5
( 5 + 2.41)
./
2
=
0.21
.
2400
=
0.22
=
3du
=
3(0.80)
=
2.5
.
TRANSITION Lu
Lu
=
2.4
.
.*
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
11
Ld
7.
=
4du
=
4(0.80)
=
3.2
.1
=
INV.1
=
+91.592
.(
.)
=
.1
=
+92.392
.(
.)
.2
=
.1 + Hcu
=
.5
=
.
.*
.
=
91.592 + 1 - y1
=
+92.592
.(
.)
=
+92.392 - 0.90
+91.492
.(
.)
UPSTREAM TRANSITION (Lu)
.3 . (2) .4
=
( .1 - .5) / Lu
=
(91.592 - 91.492) / 2.5
=
1:25
=
INV.2
=
+91.392
=
= .(
91.592
=
.5 - P
=
+90.292
+90.592
.(
.)
.(
.)
.)
.3 + Hcd
=
OK.
1:4
= .(
90.592 + 1
.)
BASIN .6
= .(
91.492 - 1.20
.)
DOWNSTREAM TRANSITION =
[ .3- ( .6 + R)] / L d
=
(90.592 - 90.292 - 0.20) / 3.2
=
1: 32
1: 40
DESIGN OF VERTICAL DROP WITH WEIR CONTROL
OK.
12
DESIGN OF INCLINED DROP WITH WEIR CONTROL
1. (u)
,Q ,b
,d 0.65 , H , H0.95 , 1: m , LS , n
,
(d)
. /
0.60
0.60
.
0.482
.
0.65
c
.
0.95
.
1:1.5
1:1.5
1/1000
1/1000
0.018
0.018
+104.45
+103.05
+104.932
3
0.47
0.482
, INV
0.47
b
+103.532
.(
.)
.(
.)
DUCKBILL WEIR
2.
WITH WEIR CONTROL
Y1 = B
0.75 =
1.00
DUCKBILL WEIR
.
S
=
0.64
.
.
H
=
1.00
.
W1=
0.615
.
W2=
1.26
.
L1 =
0.80
.
L2 =
3.485
.
L =
2.00
VERTICAL DROP
.
3
L1 + L 2 + L 3
= 6.29
.
DESIGN OF INCLINED DROP WITH WEIR CONROL
1
3.
DESIGN OF INCLINED DROP WITH WEIR CONROL
2
.5
.6
INCLINED DROP
=
.1 - Y1
=
104.932 - 0.75
B
=
+104.182
.( . . .)
+102.78
.( . . .)
STILLING POOL 18.4656 Q
=
1
Q + 9.912
18.4656
=
0.47
0.47 + 9.912 =
yA
=
Q/ B1
=
0.47/1.30
=
=
YC
=
[ ( 0.362 ) /9.81]
=
0.24
0.362 3
=
V
2
A
3
2.80
Fb =
q /g
V
=
2 C
YC
=
/s
=
0.24 + 0.30
FREE BOARD
0.30
=
0.119
.
2
2g
2g
.
2
1/ 3
2
Q
.
A
CRITICAL FLOW Q
1.30
1.219
.
Hs
=
y A + Fb
=
0.54
0.60
.
B ZA + yA +
V
2 A
2g
ZA
VB=
yB + V
=
.
A
B
2 B
……………(1)
2g
=
.5 - .6
=
104.182 - 102.78
=
1.402
.
Q/A
DESIGN OF INCLINED DROP WITH WEIR CONROL
3
0.47/(1.3 y B )
=
0.47/( B1 ⋅ y B )
=
=
y B + y + 0.47 × 1 B 1.3y B 2 × 9.81
=
y B + 0.0067 / (y B )2
2
1.402 + 0.24 + 0.12 1.762
TRIALAND ERROR
y B=
0.063
DROP = =
STILLING POOL
.
POOL STILLING
103.05 - 102.78 0.27
.
.2
=
DROP + d d
=
0.27 + 0.482
=
0.752
CONJUGATE DEPTH
y BB
=
VB
=
FrB
=
.
yB yB
(
1 + 8 FrB
2
0.47
2
-1)
=
………………..(2)
5.739
./
1.3( 0.063) V gy B
5.739
=
9.81 × 0.063 =
7.3
>
1
SUPERCRITICAL FLOW
(2) y BB
=
0.063
(
1 + 8 ( 7. 3 )
2
-1)
2 =
0.62
=
4 y BB=
.
0.755
OK.
STILLING POOL
LH
DESIGN OF INCLINED DROP WITH WEIR CONROL
2.48
.
3.00
.
4
SMALL CANAL STRUCTURE Q
102 (
VB y B
( 0.47 × 3.28 )( 5.739 × 3.28 )( 0.063 × 3.28 )
=
2
A
(1.3 × 0.063 × 3.28 )
FREE BOARD , F STILLING BASIN
=
73.22
=
1.56
=
Hp
Hp
0.48
.
dd
=
0.48 + 0.27 + 0.482
=
1.232
1.30
.
DROP + =
4.
=
F + DROP +
Hp =
)
3
H bd
0.27 + 0.95
H p=
=
3du
=
3(0.482)
=
1.446
=
( B + 1.5y − B1 )/ tan 27.5o 1
1.12
=
1.12
.
.
TRANSITION
Lu
L TU
2
=
[
1. 0
1.6
.
L TD
1.87
=
[
=
[
bd 2 0. 6 2
=
.
2
+ 1.5( 0.75) -
1. 3
2 =
.
] / tan 27.5°
2
2.0 + 1.5d d -
B1
.
]/tan 22.5°
2
+ 1.5( 0.482 ) -
1.3
]/tan 22.5°
2
0.9
DESIGN OF INCLINED DROP WITH WEIR CONROL
1.50
5
5.
.1
=
. .2
.5
= =
INV.1
.1
=
+104.45
.(
.)
=
+104.932
.(
.)
=
+105.100
.(
.)
.1 + Hcu
=
104.45 + 0.65
=
+104.182
.(
.)
UPSTREAM TRANSITION (Lu) SLOPE = =
(104.45-104.182) / 1.6
=
1: 5.97
1:4.0
OK.
LI
( .1- .5) / Lu
(SLOPE 1:2) =
2 ( .5 – .6)
=
2.804
.*
LL
=
Lu + L1 + L2 + L3 + LTU + 2.5 + L I + LH + LTD
LL
=
1.6 + (6.29) + 2.0 + 2.5 + 2.804 + 3.0 + 1.50
=
19.694
.3
=
+103.05
.4
=
.6
.(
.)
.3 + HCD
=
103.05 + 0.65
=
+102.78
=
+103.70 .(
.(
.)
.)
DOWNSTREAM TRANSITION (LTD) SLOPE =
.7
=
(103.05 - 102.78) / 1.50
=
1: 5.56
= =
.8
( .3 - .6) / LTD
= =
1:4.0
OK.
.6 + Hp 102.78 + 1.12
=
+103.90
.(
.)
=
+105.40
.(
.)
.1 + Hbd 104.45 + 0.95
DESIGN OF INCLINED DROP WITH WEIR CONROL
6
HF
6.
=
.8 - .5
=
105.40 - 104.182
=
1.218
.
WEIGHTED CREEP RATIO
COURSE SAND
C = 5.0 w
Max. ∆ H
LH
CUTOFF h1 Lv
Cw
=
.5 + S - .3
=
104.182 + 0.64 - 103.05
=
1.772
=
(L1 +L2 + L3 + LTU +2.50 + LI + LH + LTD)/3
=
(6.29 + 2.0 + 2.50 + 2.804 + 3.0 + 1.50)/3
=
6.03
=
0.60 .
=
h1 + 7(h1 - 0.15) + 3(h2 - 0.15) + h 2
=
5.70
=
6.03 + 5.70
.
h2
=
0.60
.
. =
6.62 > 5.0
OK.
1.772
DESIGN OF INCLINED DROP WITH WEIR CONROL
7
DESIGN OF INCLINED DROP WITH WEIR CONROL
8
(Retaining Wall)
1.
2.
Surcharge
0.60
1
1
4
3
Surcharge
Rankine
P
2)
0.6
0.65
Toe
1)
0.5
0.70 101
3. 4.
.
20
=
P
γ
1 − Sinφ 1 + Sinφ
=
θ
2
2
θ =
φ
h
γ
cosθ − cos 2θ − cos 2 φ h2 cosθ 2 cosθ + cos 2θ − cos 2 φ
Internal friction angle
=
1.
2.
4.
5. 6.
3.
Stress
(Sliding)
Key wall
8.
Toe
9.
Heel
Middle third
7.
(Overturning)
1
(Surcharge Load)
0.4
SURCHARGE 0.4 T/m.
2
0.25
5.50
A
C
D
E 0.50
B
O 2.10
2. 3.
γ = 1.9
/
γ = 2.2
/
3
3
4. Bearing Capacity
/
γ = 2.4
3
0.6
6. 7.
φ = 30
20
5. Friction factor
0.2 %
/
Covering 7
3
9.
Ms
0.60
1.
8.
0.50
=
A s .f s .j.d
vc
=
V b⋅d
u
=
Mc
=
Working stress
f s = 1,500
./
2
.
j = 0.885
V ∑ o ⋅ j⋅ d
R.b.d
2
R=
11.995
2
1.0
W3
0.4
T/m.
2
0.25
5.50
P2 P1
W1
W4
A
C
W5
D
E 0.50
W2 B
O 2.10
1
0.50
0.60
O
(
W2
W3
W4
W5
1.90 × 2.10 × 5.50
Surcharge Load
)
=
P1
=
P2
3.2/2
6.144
(2.1 × 1.0) × 0.4
0.84
2.15
1.806
3.30
0.975
3.218
1.65
0.767
1.266
× (0.25 × 5.5 × 1.0) × 2.4
=
=
11.40
= =
59.616
1 3
=
1 × 1.9 × 6.0 2 = 3
1 × 0 .4 × 6 .0 3
-
3.84
1 2 2 Ka γ soil H
Ka ⋅ Su ⋅ H
(
(2.1 + 0.5 + 0.6) × 2.4 × 0.5
2
1 - Sin 30
=
)
47.182
1
1 + Sin 30
=
(
1 2
2.15
31.575
Ka
21.945
(0.25 × 5.5 × 1) × 2.4
W1
Surcharge Load
0.80
3
)
O ΣM
P
P2
1
=
6 6.0 11.40 × + 0.8 × 3 2
=
22.80 + 2.40
=
25.20
=
11.40 + 0.80
=
12.20
=
31.575
=
ΣM
-
ΣP ΣP
1 ΣW
=
59.616
-
F.S × ΣP ≤ C.ΣW
F.S
=
C.∑ W ∑P
=
0.6 × (31.575) = 12.20
F.S × ΣM
F.S
(Overturning)
=
∑M ∑M
=
59.616 = 25.20
2.366
2.0
Middle Third
Tensile
Stress
= =
1 Middle Third 2
(
Stress
(∑ M
- ∑M ∑M
59.616 − 25.20
=
31.575
)
1.09
(Eccentricity : e) e
e
O X
)
=
3.2 2
=
3.2 6
0.533
B S
1.5
≤ ΣM
1.553
O =
− 1.09 =
+0.51
Tensile Stress
∑ W ± 6(∑ W ) ⋅ e l
± 0.533
=
l2
1
4
l
SB
SO
=
)( ) 0.51 31.575 6(31.575 = − 3.2 (3.2)2
0.431
/ .
=
)( ) 0.51 31.575 6(31.575 = + 3.2 (3.2)2
19.303
/ .
C
Stress
=
D
Heel
Toe
2
2
2.10 A
C
E
D
B
O
2
0.431 kg./m.
2
19.303 kg./m.
2.60
SC
=
SD
=
2.1 (19.303 − 0.431) 3.2 2.6 0.431 + (19.303 − 0.431) 3.2 0.431 +
0.60
=
12.816
/ .
=
15.765
/ .
2
2
2
0.4 T/m
P2 5.50 P1
C
P1
=
1 1 × 1.9 × 5.5 (5.5) × 1.0 23
P2
=
5.50(0.4 ) ×
Mmax
=
9.579 ×
1 × 1.0 3
5.5 5.5 + 0.733 × 3 2
=
=
D
9.579 Vmax =
0.733 =
19.577
ΣP
= 10.312
-
5
dm
=
dV
=
M Rb
19.577 × 1000 × 100 = 11.995 × 100
=
Vmax
10.312 × 1000 3.8 × 100
=
vc b
d
= 43 t
=
AS
=
Amax S
.
50
.
.
27.14
.
Covering =
7
.
M fs ⋅ j ⋅ d
19.577 × 1000 × 100
=
1500 × 0.885 × 43
3.5
(
=
34.30
t
.
11 1 × 1.9 × 3.5 (3.5) + 3.5 × 0.4 × 23 3
=
=
3.879 + 0.467
4.346
M
=
3.5 3.5 + 0.467 = 3.879 × 3 2
AS
=
5.343 × 1000 × 100 = 1500 × 0.885 × (40.91 − 7 )
3.50
=
, A 0.002 × 40 × 100
=
11.869
2
.
8
Toe
.
40.91
=
TOE
-
40
.
St
ASt
V
5.343
2
.)
= 40.94
=
ΣP
=
40.40
Uniform load
.
2
Cantilever Beam .
=
Stress
2400 × 0.5 = 1200 Kg./m. Toe
2
D
15765 -1200 = 14565 Kg./m.
2
19303 -1200 = 18103 Kg./m.
0.6 m.
2
Toe
6
Mmax
14.565 × 0.6 ×
=
2.622 + 0.425
=
3.047 × 1000 × 100 1500 × 0.885 × (50 − 7 )
AS
=
Heel
.
=
.
0.5
St
.
2
=
8.0
.
.)
Toe
Cantilever Beam
=
10.45
T/m
=
1.2
T/m
. Surcharge Load
Toe
( A
Heel
5.5
-
5.337
ASt
3.047
2 0.6 × 0.6 3
0.6 1 ) + (18.103 − 14.(565 2 2
=
=
0.4
2
2
T/m
W = 12.05 T/m
= 12.05 T/m
2
2
2
C C
B 0.431
12.816 2.10 m.
C, Mc
12.05 × 2.1 ×
=
2.1 2.1 1 () − 0.431× 2.1× − (12.816 − 0.431 2 2 2
= 25.62 – 9.10 dm d
=
16.52 × 1000 × 100 = 11.995 × 100
=
= 50 -7
-
=
43
)
(50 - 7) OK
.
=
1500 × 0.885 × (43)
(
37.11
16.52 × 1000 × 100
AS =
16.52
2.1 3
2.1
28.94
.
2
2
ASt = 8
. :
Heel
-
20 @ 0.26
3.50
AS=
12.08
.
.
Σo
= 24.17
.
7
-
20 @ 0.26
S
12.08
.
.
Σo
= 24.17
.
AS=
23.68
.
.
Σo
= 33.86
.
AS =
35.76
.
.
o = 58.00
.
28 @ 0.26
AS
3.50 A =
34.30
.
.
Bonding Stress u
V
=
∑ o ⋅j ⋅ d 10.312 × 1000 58.00 × 0.885 × 43
=
15.3
./
.
Σo
2
-
./
.
2
.
.
Σo
= 20.94
.
AS = 30.79
.
.
Σo
= 43.98
.
10.47
.
.
Σo
= 20.94
.
20 @ 0.30
AS = 10.47
Heel
-
4.67
Toe
-
=
A =
28 @ 0.20
20 @ 0.30
*
S
20 @ 0.26 20 @ 0.30
28 @ 0.26
28 @ 0.20 HEEL
TOE 20 @ 0.30
8
(Counterfort Retaining Wall)
(Cantilever) 6 (Slab)
(The Stem)
1.
wl 2 ± 12
(The base slab)
2.
2 5H
Toe
(Counterfort)
(Cantilever)
wl 2 ± 10
2 3H
3.
Heel
1 3H
(d)
2.1
2 3H
1
1) 2)
1760
./ .
4) fc = 79
/
. , fs = 1400
5) v = 5.2
/
.,
.
/ .
.
7.30×
=
4.9
.
45
.
45
=
.
3.0
.
(Overturning) :
C.G.
f = 0.5
=
/ .
=
.
18.0
2
2
3
0.67
:
n = 12
/
2400
2
u = 13.0
1.20 .
/
6) 7)
2
2
3
φ = 30
3)
1
=
155712 57215
=
1 2 1 - sinφ 2 γ H 1 + sinφ
=
1 2
=
× 17 60 ×7.3
M
=
15630
M
=
155712 =
CD
7.33 .
1
2.72
2
×
1 3=
.
15630
.
×
7.3 3
=
38033
.– . .– .
155712 = 38033
4.09
>
1.5
2
0.15
A
1 6.85
C
B
F
1.20
0.45 D
E
0.30
0.60
4.90
C
B
G
F
0.45 D
E 0.60
0.30
4.00
0 6 0 6 2
. / . 4 2 9 5 1
0 0 3 7 1
6 3 2 5 1
2
3
1
D
D
( 1 2
4
.)
( .)
1233 2466
0.825
2034
5292
2.45
12965
48224
2.90
139850
57215
155712
(Sliding)
ΣP =
=
15630
=
.
57215 ×
0.5
28608 15630
e
S
D E
Smin
1.83
=
2.45 – (155712-38033) 57215
=
0.393 < 6
1.5
4.9
ΣW
A ±
6(ΣW)e bl 2
=
57215 1×4.9
=
11676.5 ± 5619
=
17295.5 ≈ 17300
±
6×57215 1×(4.9) 2
×
0.393
./ .
6057.5 ≈ 6060
>
=
=
.
./ .
2
< 18.0
2
wl 2 ± 12
w
28608
4.9 ΣM 2 – ΣW
=
Smax
=
=
.
.– .) 863
(
0.70
3
.
=
γh
1-sin 30o 1+sin 30o
4
G
w
1 max.
d = 25
=
3015
.– .
=
14.83
.
9.90
.
M Rb
=
3015×100 13.7×100
φ 12 mm. @ 0.11
As
0.15 %
Ast
2
4020
=
= 5
./ .
4020×(3) 2 12
d
Covering
4020
=
=
.
w
M
1
1760×6.85× 3 =
=
./ .
. 3015×100 0.87×1400 ×25
=
As = 10.28
.
2
=
∑0 = 34.27
2
.
0.0015×30×100
=
φ 12 mm. @ 0.24
As = 4.71
.
=
4.5
.
2
2
max.
3−0.45 2
V
=
4020×
v
=
5125 0.87×100 ×25
=
5125
.
=
2.36
./
.
./
.
5.2
2
2
./
.
u
=
5125 0.87×34 .27×25
=
6.88
2)
0 2.00 . 2.00 4.00 .
3)
4.00
1)
.
6.85 .
2
./
3
6.85 .
–4.00 .
max.
13.0
2
w
=
1760×4× 1 3
=
M
=
1 12
2
× 2347 ×(3)
2347 =
1760
./ . .– .
5
=
15+ 6.85 × 4 − 5 =
18.8
.
As
=
1760×100 0.87×1400×18.8
7.68
.
φ 12 mm. @ 0.14 –2.00 . max.
15
d
As = 8.08
.
2
Σ0
= = 26.93
2
.
w
=
1760×2× 1 3
M
=
1 12
2 =
882
.– .
d
=
15 15+ 6.85 × 2 − 5 =
14.4
.
As
=
882×100 0.87×1400×14.4
5.1
.
φ 10 mm. @ 0.15
=
× 1175 ×(3)
As = 5.24
.
2
1175
=
./ .
0 = 20.94
.
Toe
=
D
B d
d = 38
.
(
0 = 20.94
2
=
16220
./ .
=
15924–1080
=
14844
./ .
=
14844×0.6× 2 + 2 (16220–14844)×0.6×0.6× 3
=
2671.9+165.2
0.6
=
2837
2837 × 100
=
=
.– .
14.39
100 × 13.7
2
2
1
.
. =
1 2 (16220+14844)
v
=
9319 0.87×100 ×38=
2.82
As
=
2837×100 1400×0.87×38=
6.13
Ast
=
0.0015×40×100
×0.6
40
.)
=
9319
.
< 5.2
./ .
=
6.00
.
.
2
2 2
φ 12 mm. @ 0.18
As = 6.28
.
2
. u
Toe
17300–1080
V
=
Covering = 7 max.
×2400
17300–0.45
B
=
9319 0.87×20.94×38
= 13.46
≈
13.0
./
.
2
6
Heel
E
Heel
max.
=
1×6.85×1760+1×0.45×2400–6060
=
12056+1080–6060
M
=
7076
./ .
wl 2 12
=
±
=
7076× 12
(3)2
d
d
=
38
As
5307×100 13.7×100
=
.
5307
.– .
19.68
.
Covering =
5307×100 0.87×1400 ×38
=
φ 16 mm. @ 0.16 max.
= =
.
7076×
v
=
9022 0.87×38×100
u
=
9022 0.87×31.42×38 =
φ 12 mm. @ 0.18
=
./
.
< 13
./
.
8.69
P
2 1 −sin φ 1 + sinφ
=
1 2
×
=
1 2
× 1760 ×
=
41292×
γ H
×
2 685 .× ×
1
1 3
3
= 41292
.
94238
.– .
.
M
=
d
=
tanθ
=
θ =
2
6.85 3
max.
d
2
P
2.73 < 5.2
T
.
2
.
9022
.
0 = 31.42
(3− 0.45) = 2
=
.
11.5
2
As = 12.56
V
=
7
=
6.85 3
T–Covering
400
=
.
400 cosθ –Covering 4.0 6.85
=
0.583
30.282
7
d
=
400 cosθ –10
d
=
345.42–10
As
=
94283×100 = 0.87×1400×335
8–φ 20 mm.
=
As = 25.13
.
2
335.42
.
23.11
.
2
0 = 50.27
θ
=
P– M d sinθ
=
41292–
=
41292–14192
94283×100 335
0.504
×
=
27100
B
θ T
d1 =
B–covering
d1 =
335 cos
θ 400
.
v
=
289
=
27100 0.87×289 ×45
θ
d cos
θ
=
.
. =
2.40
5.2
./
.
./
.
2 2
n1 n
n
6 6 8
=
h1 =
h
h 1
5.93
4 8
h12 (6.85)2
h1 =
h12 h2
=
n1
5.6 4
h12 (6.85)2
=
.
.
4.84
4.5
8
0
.
4.5 .
4.5
6.85
P1
.
(P1)
P1 =
As
(1−sin φ ) ×(3–0.45) γ h (1 +sin φ )
=
1.0
(3–0.45)
h
4–φ 20 mm. 8–φ 20 mm.
×6.85× 13 ×(3–0.45)
1760
=
φ 12 mm.
=
10248
10248 1400
2
. =
7.32
As = 2×1.13
100×2 ×1.13 = 7.32
=
. .
2
2
30
. 2
(
φ 12 mm. @ 0.18
As = 6.28
(
.
)
)
(3–0.45)
1.0
EF P1 =
As
φ 12 mm.
=
=
18044
18044 1400
×(3–0.45)
7076
=
2
. =
12.89
As = 2×1.13
100×2 ×1.13 12.89 =
17
. .
2
2
.
φ 12 mm. @ 0.18
9
Heel
Toe
3
10
4
(
- )
11
5
(
- )
12
1.
(TYPE A)
•
•
•
•
Q
. . .1
. . . “1”
•
“3”
•
“6”
•
•
•
•
•
dU
HU
•
2. 3. 4.
,D
, VP
,
2 P
V
2g
BL
BR
TR
TL
= Q/A
HCU
bU
•
1.
, Q
1.50
/
HVP
1
)
(
1 .
2
) -
(
2
3
5.
Friction Slope
, Sf
Sf
n 2 VP2 R 4/3
=
2
6.
Q(
HA
2
106
188
294
424
0.30
0.40
0.50
0.60
B1 ( .)
0.75
0.75
0.80
0.90
B2 ( .)
0.50
0.60
0.70
0.80
t1 ( .)
0.15
0.15
0.20
0.20
t2 ( .)
0.10
0.10
0.15
0.15
/ D ( .)
7.
0.1575 D 4/3
t
B1 , B2 , t1
2
n VP
=
)
Inlet)
“2” (
Inlet
Cutoff
(Suppressed Rectangular Weir) 3/2
Q Q 8.
9.
11.
12. 13.
=
Inlet)
.
=
“4” “2” = “2” “2” “2”
. . .(
“4” “4”
10.
“4” (
“4”
CLH 1.71 B1 H 3/2 A
= =
. .1) – (1.78HVP + 0.08) – D – 0.10
0.90
.
“6” – 0.90 –
8
9
“4”
“3”
– D - 0.10
“4”
. . . – HA
11
“1”
12
“2”
“4”
4
14.
LU, LTU
LC
LU
=
0.5 b u + 1.5 (
LTU
=
1.5 (
)
=
Bm + 3.0 (
LC (
“2” –
“3” –
“2”) “6”) – 1.5 (
TR - 1.5 ( Bm 15.
Hu =
16.
“3” –
Entrance Loss + Friction Loss + Exit Loss
17.
. . .2 = . . . - HL
18.
. . .2 =
20.
=
=
“4” + 0.10
HHD =
. . .2 + 0.20 – 0.65 - D
“5”
20,
2 1
22
7 = . . .2 + Freeboard
HHD =
“7”–
HHD
“5”
“5”
LTD = 1.5 HHD
26.
1:1
28.
“9” = . . .2 – dd
29.
“9” =
30.
“4”
“5”
24.
27.
18
. . .2 - D - 0.05
“5”
“5”
25.
17
“5”
23.
+ 0.30
“5”
22.
0.78 HVP + SfLC + 1.0 HVP
“5” 21.
, HL
= =
. . .2
“4” + 0.10
“3”) + TL
“7”) + 0.5 (Min.)
=
HL
19.
“1”)
L1=6 (
(dd)
(bd)
“8” “8” –
“5”)
5
31.
L2
1.50
LC
32.
Weighted Creep Ratio
33.
( ) Weighted Creep 5.0 Collar 2
Full
Supply
2.
Collar Collar
Piping
1.50
(TYPE B)
•
•
•
•
Q
. . .1
. . . “1”
•
“3”
•
“6”
•
•
•
•
•
HU
•
BL
BR
T
TR
L
1
6 6
(dC)
7.
dU HCU
•
bU
TYPE A
(V )
C
6
)
(
3
7
) -
(
4
8
8.
“4”
. . .(
∆HV
=
VC2 V12 − 2g 2g
∆HV
=
9.
+ d C + 0.10
2 1
2g
V / 2g
“6” – 0.90 –
“4”
11.
2g
2 C
=
0.90
“4”
10.
VC2
V
“4”
. . .1) – 0.5 ΔH V +
=
“4”
8
7
13
“4”
11
-0.10
9
“3”
“2”
“4”
TYPE A
12.
LU
13.
HHU =
14.
HVD
15.
LTU LU
=
0.5 bU + 1.5 (
LTU
=
1.5 (
“3” –
Vd2 2g
17.
=
18.
0.30
0.30
HHD =
1 :1
(bd)
. . .2 - D - 0.05
“5” “5”
(N.G.L )
N.G.L +
“5”
“2”)
(dd)
=
“5”
“1”)
“4” + 0.10
. . .2 16.
“3” –
. . .2
“2” –
. . .2 + 0.20 – 0.65 - D
“5”
19.
“7” = . . .2 + Freeboard
20.
“8” = . . .2 + dd
21.
“9” =
16
17
“8”
22.
HHD =
23.
LTD = 1.50 HHD
“7” -
“5”
9
L1 = 6 (
24. 25.
“5”)
“8” -
L2
1.50
26.
27.
F
14
=
. . .2 + HVCD – (d2 + HVP) – 0.1d2 (LS)
Outlet Transition (LH +LS)
30.
29.
Pipe Drop
“10” “10”
28.
4 d2
1:2
(LH)
1:3
(LC)
5D
Weighted Creep Ratio
33
TYPE A
1.8
31.
10
3.
4.
•
.-
-04
•
.-
-04
38/41 39/41
11
