canal syphon programme
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Program by- Sunil K Goyal Hydraulic 1 Checked by - P.P.Pareek
DESIGN OF A CANAL SYPHON
NAME OF PROJECT : Case study, Design example 9.4/p394
DESIGN INPUT DATA :
(a) Canal
(i) Full supply discharge of canal 40.00 Cumec
(ii) Bed width of canal 18.00 m
(iii) Full supply depth of canal 2.10 m
(iv) Bed level of canal (C.B.L.) at D/S 250.00 m
(v) Side slope of canal (s) 1.50 :1
(vi) Free board of canal 0.75 m
(b) Drain
(i) Max. observed flood discharge 100.00 cumec
(ii) Bank level 254.00 m
(iii) Bed level 251.80 m
(iv) Highest Flood Level (H.F.L.) 253.25 m
(v) slope 1/600
HYDRAULIC DESIGN :
(1) Section of the drainage channel
According to Lacey's formula
P = 4.83 X= 4.83 X 10= 48.3
Provide bed width of the drain at the crossing = 44.50 m.
(2) Canal waterway
Bed width of canal = 18.00 m.
Normal X-area of the channel = BD +(A) = 44.42 Sq.m.
Velocity in the normal section = Q/A= 0.90 m/sec
Width = 3.00 m , Wall thickness = 0.30 m
Q 1/2
sD 2
Adopt size of the barrel as
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Program by- Sunil K Goyal Hydraulic 2 Checked by - P.P.Pareek
Height = 2.50 m , No. of barrels = 2 No.
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Program by- Sunil K Goyal Hydraulic 3 Checked by - P.P.Pareek
Reduce the canal waterway from 18.00 m to 6.30 m
Velocity through the barrels =40
= 2.67 m/sec15 < 6.00 m/sec
The velocity is within the recommended rangeThe size of the barrel is, therefore okayIt should be checked that the flow is subcritical in the barrel, .i.e., Froude number (F) shouldbe less than unity
Now F =Where,
V = 2.67 m/secg = 9.81 m/secd = 2.50 m
Therefore,F = 0.538
Since the value of F is less than 1 the flow will be subcritical in the barrel
(3) Head loss and bed levels at different sections :
Width of canal in the flumed portion = 6.3 mProvide 2 in 1splay in contraction and 3 in 1 splay in expansion transition
11.70Length of contraction transition = -------- X 2 = 11.7 m
211.70
Length of expansion transition = -------- X 3 = 17.55 m 2
Assume,Length of the barrels in the flumed portion= 70.0 m > 68.50 massumed length is O.K.
In the transitions, the side slopes of the section shall be warped from 1.50 : 1 to vertical.
1 2 3 4
0.3 m Thick wall
Canal
18.0 6.3 18.0
R.C.C. barrels
11.70 70.00 17.55
1 2 3 4
Canal waterway (All dim in meter)
V/(gd)1/2
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Program by- Sunil K Goyal Hydraulic 4 Checked by - P.P.Pareek
At section 4-4
Area of section = 44.42 Sq.m.(Normal channel section)Velocity = Q/A = 0.901 m/sec
Velocity head = 0.0414 mR.L. of bed = 250.00 m (given)R.L. of water surface = 250.00 + 2.1
= 252.10 mR.L. of T.E.L. = 252.10 + 0.0414
= 252.141 m
At section 3-3
Water depth at the entry and exit of the barrel should be kept slightly higher than the depth of the barrels so as to keep the ends of the barrel submerged for proper syphoningProvide water depth equal to 3.00 mArea of section = 3.00 X 6.3
= 18.9 Sq.m.Velocity = Q/A = 2.116 m/sec
0.228 mLoss of head in expansion from section 3-3 to section 4-4
== 0.056 m
Hence elevation of T.E.L. at section 3-3= 252.141 + 0.056= 252.197 m
R.L. of water surface = 252.197 - 0.228= 251.969 m
R.L. of bed = 251.969 - 3.00= 248.969 m
From section 3 - 3 to section 2 - 2, area and velocity are constant.
Head loss through barrels
Head loss through barrels is given by
=
= 0.080 ,for bell mouthed syphon
= a(1+b/R)
Where, a = 0.00316 , b = 0.10000
L = 70.0 mR = A/P
= 0.682 mHence,
= 0.00362Therefore loss of head in barrels = 0.526 m
V2/2g =
Velocity head = V2/2g =
0.3 (V22 - V1
2) /2g
( 1 + f1 + f2 xL/R )V 2/2g
Where, f1
f2
f2
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Program by- Sunil K Goyal Hydraulic 5 Checked by - P.P.Pareek
At section 2-2
R.L. of T.E.L. = T.E.L. at section 3-3 + head loss through barrels= 252.197 + 0.526= 252.724 m
R.L.of water surface = R.L. of T.E.L. - Velocity head at section 3-3= 252.724 - 0.228= 252.495 m
R.L. of bed = 252.495 - 3.00= 249.495 m
At section 1-1
Loss of head in contraction transition from section 1-1 to section 2-2
== 0.037 m
R.L. of T.E.L. = T.E.L. at section 2-2 + head loss in contraction transition= 252.724 + 0.037= 252.76 m
R.L. of water surface = 252.76 - 0.041 = 252.72 m
R.L. of bed = 252.72 - 2.10= 250.62 m
(4) Transitions
The general method of Hinds shall be applied for designing the transitions, as the water depths inthe transitions vary from 2.10 m to 3.00 m
(a) Contraction transitionw.s.profile
1
R.L. 252.72 2
0.112
5.85 R.L. 252.50
1
11.70 2
(All dim in m)
Contraction transition
=Water level at section 1-1 - Water level at section 2-2
2
=252.72 - 252.50
2= 0.1122
= Length of contraction transition2
=11.70
2= 5.85 m
0.2 (V22 - V1
2) /2g
y1 =
x1=
y1
x1
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Program by- Sunil K Goyal Hydraulic 6 Checked by - P.P.Pareek
C = =0.112
= 0.0032834.223
y = 0.00328 (equation of water surface profile in contraction transition)
The details of the contraction transition have been worked out in Table 1
(b) Expansion transitionw.s.profile 4
3 R.L. 252.10
0.065
R.L. 251.97
8.775
17.55 4
3
Expansion transition
=Water level at section 4-4 - Water level at section 3-3
2
=252.10 - 251.97
2= 0.0654 m
=Length of expansion transition
2
=17.55
2= 8.775 m
Hence,
C = =0.06541
= 0.0008577.0006
y = 0.00085 (equation of water surface profile in expansion transition )
The details of expansion transition have been worked out in Table 1
y1
x12
x2
y1 =
x1=
y1
x1
y1
x12
x2
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Program by- Sunil K Goyal Hydraulic 7 Checked by - P.P.Pareek
Table 1
water Elevation Velocity Velocity side Area A= Bed Depth(D) Bed
Distance surface of slope 's' Q/V Level Col.(3) width
elevation T.E.L. (4) - (3) -Col(9) B = A/D
m m m m -sD
1 2 3 4 5 6 7 8 9 10 11
From section 1-1 to 2-2 CONTRACTION TRANSITION
0 0.0000 252.720 252.761 0.041 0.901 1.500 :1 44.392 250.62 2.10 18.00
3.0 0.0295 252.690 252.752 0.062 1.099 1.115 :1 36.406 250.34 2.35 12.86
5.85 0.1122 252.608 252.742 0.135 1.627 0.750 :1 24.593 250.06 2.55 7.73
8.7 0.0295 252.525 252.733 0.208 2.021 0.385 :1 19.794 249.78 2.75 6.30
11.7 0.0000 252.495 252.724 0.228 2.116 0.00 :1 18.900 249.50 3.00 6.30
From section 3-3 to 4-4 EXPANSION TRANSITION
0 0.0000 251.969 252.197 0.228 2.116 0 :1 18.900 248.969 3.00 6.30
3 0.0076 251.977 252.188 0.211 2.035 0.256 :1 19.656 249.145 2.83 6.30
6 0.0306 252.000 252.178 0.179 1.872 0.513 :1 21.372 249.322 2.68 6.61
8.775 0.0654 252.035 252.169 0.135 1.627 0.750 :1 24.593 249.485 2.55 7.73
11.55 0.0306 252.069 252.161 0.091 1.337 0.987 :1 29.913 249.648 2.42 9.96
14.55 0.0076 252.092 252.151 0.059 1.072 1.244 :1 37.300 249.824 2.27 13.62
17.55 0.0000 252.100 252.141 0.041 0.901 1.500 :1 44.392 250.00 2.10 18.00
(5) Invert level
Bed level of drain = 251.80 m
Provide 0.30 m thick concrete slab and 0.60 m thick earth fill over the slab
Invert level of the concrete = 251.80 - ( 0.6 + 0.3 + 2.5 )= 248.40 m
Invert level at the entrance and exit of the barrel shall be the same as the bed levels alreadyworked out at sections 2-2 and 3-3 respectively.
Thus the invert level at the entry = 249.495 mThe invert level at the exit = 248.969 m
The invert level of the barrel would be kept at 248.40 m in a length of 44.5 m(under base of drain) after which it would meet the respective bed levels at the entrance andexit, so as to obtain a slope of about 1in 15 in the barrel at either side.Thus, Length of barrel upstream = 16.00 m
Length of barrel downstream = 8.00 m
The length of the pucca floor on either end should be adequate to provide safe hydraulic gradientand its thickness sufficient to counterbalance the total uplift pressure by gravity. The barrel shallbe made of reinforced concrete box construction and its structural design is given subsequently.
y=cx2 head (hV) V= Ö2ghV
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Program by- Sunil K Goyal Hydraulic 8 Checked by - P.P.Pareek
(6) Pucca Floor
Provide pucca floor in half the transition length in the upstream and 3/4 th the length of expansiontransition in the downstream.Length of pucca floor upstream = 1/2 X 11.7
= 5.85 msay 6.00 m
Length of pucca floor downstream = 3/4 X 17.55= 13.16 m= say 13.00 m
The floor shall be subjected to static uplift and seepage head; it is maximum when high flood ispassing through the drain and there is no flow in the barrel. The seepage head would becalculated by Bligh's theory.
(7) Uplift pressures on the barrel floor and pucca floor
(a) Static pressure
At bottom of barrel floor
Deepest invert level of the barrel = 248.40 mThe thickness of the barrel is = 0.30 mThe bottom level of the barrel floor = 248.10 mAssuming the sub-soil water level upto the bed level i.e.at R.L. 250.00 m, the maximumstatic head = 250.00 - 248.10
= 1.90 m
At the downstream end of barrel
Floor level at d/s end of barrel = 248.969 mAssuming floor thickness at this point 2.00 m,The bottom level of pucca floor = 248.969 - 2.00
= 246.969 m Hence,
Static head = 250.00 - 246.969= 3.031 m
At the upstream end of barrel
Floor level at u/s end of barrel = 249.495 mAssuming floor thickness at this point 1.50 m,The bottom level of pucca floor = 249.495 - 1.50
= 247.995 m Hence,
Static head = 250.62 - 247.995= 2.624 m
(b) Seepage head
The seepage head will be maximum when the drain is running full and there isno flow in the canal. Thus total seepage head =
= H.F.L. in the drain - Bed level of canal
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Program by- Sunil K Goyal Hydraulic 9 Checked by - P.P.Pareek
= 253.25 - 250.00= 3.25 m
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Program by- Sunil K Goyal Hydraulic 10 Checked by - P.P.Pareek
At bottom of barrel floor
The residual seepage head at point 'a' in the centre of the first barrel has been calculated byBligh's theory. The seepage line would follow the path indicated by the line x a b y. Its total length(neglecting floor depression or thickness) is the sum of the following
(i) Half the barrel span = 1.5 m(ii) Length of barrel in indicated portion = 8.00 m(iii) Length of pucca floor = 13.00 m
Thus total creep length = 22.50 m and creep length upto point 'a' i.e. centre of firstbarreel= 1.5 m
8 13.00
1.5
Barrel Plan
Residual seepage head point 'a' = 3.033 m
Thus total uplift in the barrel = 1.90 + 3.033= 4.93 m
say 4.93
At the downstream end of barrel floor
total creep length upto the end of barrel floor i.e. at point b= 1.5 + 8.0= 9.5 m
Hence , Residual seepage head at this point = 1.88 m
Thus total uplift = Static uplift + residual seepage head= 3.031 + 1.88= 4.909 m
4.909The floor thickness required (sp.gr.=2.22) = ----------- = 2.211 m
2.22Say 2.20 m
Provide 2.20 m thick c.c.floor d/s and reduce it to thickness 0.90 m at the end of floor
t/m2
c.c.cutoff
c.c.floor
R.C.C.Barrel
x
a by
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Program by- Sunil K Goyal Hydraulic 11 Checked by - P.P.Pareek
At the upstream end of barrel floor
total creep length upto the end of barrel floor = 1.5 + 16.00= 17.5 m
Hence , Residual seepage head at this point = 0.83 m
Thus total uplift = Static uplift + residual seepage head= 2.624 + 0.83= 3.454 m
3.454The floor thickness required (sp.gr.=2.22) = ----------- = 1.556 m
2.22Say 1.60 m
Provide 1.60 m thick c.c.floor u/s and reduce it to thickness 0.70 m at the end of floor
**********************
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Program by - Sunil K Goyal Drawing 12 Checked by- P P Pareek
DR
AIN
4.28TRANSITION WING TRANSITION WING
4.28B
R.C.C. BARRELS 0.30 M THICK
A A
18.0
0 3.00
18.0
0
CANAL 3.00
4.28 16.00 44.50B
8.00 4.28
CONTRACTION TRANSITION EXPANSION TRANSITION
11.7 68.50 17.55
PLAN
0.60 TH. EARTH FILL
TOP OF WING WALL TOP OF WING WALL
R.L. 253.47R.L. 254.00 H.F.L. 253.25
254.00 R.L. 252.85
U/S F.S.L. 252.72 D/S F.S.L. 252.10
CANAL
U/S BED R.L. 250.62249.50 2.50 248.97
D/S R.L. 250.00
248.40
U/S TOE WALL U/S CUT OFF C.C. BLOCK 0.30 TH. R.C.C. BARREL LEAN CONCRETE C.C. BLOCK D/S CUT OFF D/S TOE WALL
DRY BRICK PITCHING 5.70 6.00 16.00 44.50 8.00 13.00 4.55 D.B. PITCHING
11.7 68.50 17.55
TOP OF BANK 254.00 SECTION AT A-AH.F.L. OF DRAIN 253.25 DETAILS OF PROTECTION WORKS
BED LEVEL OF DRAIN 251.80 TOP OF BARREL ROOF 1 U/S TOE WALL 0.40 X 0.80
R.L. 251.20 2 U/S CUT OFF 0.50 X 1.00
R.L. 248.403.10 3 C.C. BLOCK 0.50 X 0.50
4 D/S CUT OFF 0.50 X 1.50
6.90 5 D/S TOE WALL 0.40 X 1.00
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Program by - Sunil K Goyal Drawing 13 Checked by- P P Pareek
SECTION AT B-B 6 LEAN CONCRETE 0.15 M THICK
7 D.B.PITCHING 0.40 M THICK
DETAILS OF TRANSITION WINGS
4.275
TRANSITION WING TRANSITION WING
4.275
18.0
0
18.0
0
###
7.73
6.30
6.30
6.30
6.61
7.73
9.96
13.6
2
6.00
3.00 8.77
4.275
5.85 11.55
4.2758.70 14.55
11.70 17.55
CONTRACTION TRANSITION EXPANSION TRANSITION
DETAILS OF PUCCA FLOOR
0.70
1.15
1.60
2.20 1.
70 1.30
0.90
2.00
4.00 3.25 3.25 3.25 3.25
6.00
13.00
U/S PUCCA FLOOR D/S PUCCA FLOOR
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Program by - Sunil K Goyal Drawing 14 Checked by- P P Pareek
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Program by- Sunil K Goyal Structural 15 Checked by - P P Pareek
STRUCURAL DESIGN OF A CANAL SYPHON
NAME OF PROJECT : Case study, Design example 9.4/p394
DESIGN DATA :
1 UNIT WEIGHT OF DRY EARTH 1.60
2 UNIT WEIGHT OF SATURATED EARTH 2.00
3 UNIT WEIGHT OF SUBMERGED EARTH 1.00
4 UNIT WEIGHT OF CONCRETE 2.40
5 30 Degree
6 GRADE OF STEEL Fe 415
7 GRADE OF CONCRETE M 20
6 DIAMETER OF REINF. BARS Main 16 FDist 12 F
7 NUMBER OF BARRELS 2 Nos.
8 WIDTH OF EACH BARREL 3.00 m
9 HEIGHT OF EACH BARREL 2.50 m
10 THICKNESS OF BARREL 0.30 m
11 BANK LEVEL 254.00 m
11 DRAIN H.F.L. 253.25 m
12 LEVEL AT TOP OF BARREL 251.20 M
13 UPLIFT AT BASE OF BARREL 4.93
t/m3
t/m3
t/m3
t/m3
ANGLE OF INTERNAL FRICTION ( f )
t/m2
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Program by- Sunil K Goyal Structural 16 Checked by - P P Pareek
(1) Critical section of the barrel
The barrel shall be subjected to maximum loading under the bank at the lowest possiblelevel of the barrel as indicated below in the following section:-
Bank level 254.00 m
Saturation line 253.25 m
R.L. 251.20 m
2.50
R.L. 248.40 m
R.L. 248.10 m
6.90
(2) Design features
Following cosiderations have been made in the design of barrels :(i) Bottom slab : This design is for uplift pressure and reaction from soil resulting from
the loading on the foundations. Theoretically, the soil reaction is not uniform but for simplification it is assumed uniform.
(ii) side walls : Side walls would be tested in the following two critical conditions(a) High flood in the drain while barrels are empty(b) No water in the drain while barrels are full
(iii) Partition walls : The partition walls are subjected to equal pressures on either side, andtherefore,no reinforcement is required. Nominal reinforcement is, however,provided to take care of contingency arising due to unequal pressuresresulting from chocking up of any of the barrels.
(iv) Top slab : The loads considered for design of top slab are :(a) Earth load,(b) Weight of water below saturation line
As there is no roadway along the drain, no live load due to traffic shall be considered.
(3) Design
As the barrels are rigidly joined, they should be designed as a continuous structure. Hardy Crossmethod of moment distribution shall be used for design.The effective length of horizontal member = 3.30 mThe effective length of vertical member = 2.80 m
Distribution factors
At joint A
For member AB =2.8
= 0.462.8 + 3.3
For member AD =3.3
= 0.542.8 + 3.3
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Program by- Sunil K Goyal Structural 17 Checked by - P P Pareek
At joint D
For member DA =3.3
= 0.542.8 + 3.3
For member DC =2.8
= 0.462.8 + 3.3
(i) Dead loads
Consider one meter length of barrelDepth of dry earth over barrel = 254.00 - 253.25 = 0.75 mDepth of saturated earth = 253.25 - 251.20 = 2.05 m
Weight of dry and saturard earth = 0.75 X 1.60 + 2.05 X 2.00= 5.3 t/m2
Weight of top slab = 0.30 X 2.40 = 0.72
Weight on the top slab including its own weight = 5.3 + 0.72
= 6.02
Weight of the barrels per metre of length=( 4 X 3.3 + 3 X 2.8 ) X 0.30 X 2.40
= 15.55 t
Total dead load/m length of barrels = 15.55 + 5.3 X 6.90= 52.12 t
Uplift/m length = 4.93 X 6.90 = 34.04 t
Net vertical load acting on foundation = 52.122 - 34.04= 18.08 t
Pressure on foundation soil =18.082
= 2.626.90
Pressure acting on the base slab = Soil reaction + uplift= 2.62 + 4.93
= 7.55
Net upward pressure on the base slab = 7.55 - 0.72
= 6.83
(ii) Earth pressure
The earth pressure shall comprise of the following :(a) dry earth pressure above saturation line from R.L. 253.25 to 254.00(b) saturated earth pressure from R.L. 253.25 to 248.25
t/m2
t/m2
t/m2
t/m2
t/m2
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Program by- Sunil K Goyal Structural 18 Checked by - P P Pareek
= 30 degree , = 1/3The pressure at A= Surcharge due to dry earth + Surcharge due to saturated earth
= 0.75 + X X 2.20+ w X 2.20
= 1/3 X 1.60 X 0.75 + 1/3 X 1.0 X 2.20+ 1 X 2.20
= 3.33
Pressure at D = 3.33 + X X 2.8 + w X 2.8= 3.33 + 1/3 X 1.00 X 2.8 + 1 X 2.8
= 7.07
Loads acting on different members are shown below :
6.02
3.33 3.33
3.30 3.30
2.80
7.07 7.07
6.83
(iii) Fixing moments
6.02 X 3.3(i) On span AB = 12 = 12
= 5.46 t-m
6.83 X 3.3(ii) On span CD = 12 = 12
= 6.20 t-m
(iii) On span AD fixed end moments in the wall at each end due to rectangular portion
3.33 X 2.8= 12 = 12
= 2.178 t-m
Fixed end moments due to triangular portion
3.73 X 2.8
= 30= 0.9756 t-m
= 3.73 X 2.820
Corresponding to f Cp
Cp X wd X Cp ws
t/m2
Cp ws
t/m2
t/m2
t/m2 t/m2
t/m2 t/m2
t/m2
wl2 2
wl2 2
wl2 2
2
MAD
MDA 2
B
D
A
CD
E
F
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Program by- Sunil K Goyal Structural 19 Checked by - P P Pareek
= 1.4635 t-m
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Program by- Sunil K Goyal Structural 20 Checked by - P P Pareek
Total fixed end moments at A = 2.178 + 0.976= 3.15 t-m
Total fixed end moments at D = 2.178 + 1.463= 3.64 t-m
(iv) Distribution of moments
Joints C D A BDistribution 0.46 0.54 0.54 0.46factorsFixed end -6.20 6.20 -3.64 3.15 -5.46 5.46momentsBalance -1.18 -1.39 1.25 1.06Carry over -0.59 0.62 -0.69 0.53Balance -0.29 -0.34 0.37 0.32Carry over -0.14 0.19 -0.17 0.16Balance -0.09 -0.10 0.09 0.08Carry over -0.04 0.05 -0.05 0.04Balance -0.02 -0.02 0.03 0.02
Total -6.98 4.63 -4.63 3.98 -3.98 6.19
(v) Net moments at centre and face
Span AB
6.02 X 3.3 6.02 X 0.15Sagging moments at face = ------------------- X 0.15 - -----------------------------
2 2
= 1.4222 t-m
Fixing moments at face= 3.98 +3.15
( 6.19 - 3.98 )3.3
= 6.0907 t-m
Net fixing moments at face = 6.0907 - 1.422= 4.6685 t-m
Sagging moments at centre = 6.02 X 3.38
= 8.1947 t-m
Fixing moments at centre = 3.98 + 6.192
= 5.0876 t-m
Net sagging moments at centre = 8.195 - 5.088= 3.107 t-m
2
2
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Program by- Sunil K Goyal Structural 21 Checked by - P P Pareek
Span CD
6.83 X 3.3 6.83 X 0.15Sagging moments at face = -------------------- X 0.15 - ----------------------------
2 2
= 1.6145 t-m
Fixing moments at face=4.63 +
3.15( 6.98 - 4.63 )3.3
= 6.8693 t-m
Net fixing moments at face = 6.8693 - 1.61451= 5.2548 t-m
Sagging moments at centre = 6.83 X 3.38
= 9.3027 t-m
Fixing moments at centre = 6.98 + 4.632
= 5.8043 t-m
Net sagging moments at centre = 9.303 - 5.804= 3.498 t-m
Span AD
(a) Due to rectangular portion
=3.33 X 2.8
X 0.15 -3.33 X 0.15
2 2
= 0.6625 t-m
(a) Due to triangular portion
=3.73 X 2.8
X1
X 0.15 -0.2 X 0.15
X 0.052 3 2
= 0.2606 t-mTotal sagging moments at face = 0.662 + 0.261
= 0.92 t-m
Fixing moments at face = 3.98 + 2.65 ( 4.63 - 3.98 )2.8
= 4.598 t-m
2
2
2
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Program by- Sunil K Goyal Structural 22 Checked by - P P Pareek
Net fixing moments at face = 4.598 - 0.92= 3.67 t-m
Sagging moments at centre
(a) Due to rectangular portion = 3.33 X 2.808
= 3.27 t-m
(b) Due to triangular portion = 3.73 X 2.80X 2.8 X
12 7.81
= 1.87 t-m
Total sagging moments at centre = 3.27 + 1.87= 5.14 t-m
Fixing moments at centre =4.63 + 3.98
2
= 4.31 t-m
Net sagging moments at centre = 5.14 - 4.31= 0.83 t-m
The net moments at face and centre of the different spans are tabulated below :
SpanMOMENTS
At face At centreAB 4.67 t-m (hogging) 3.11 t-m (sagging)CD 5.25 t-m (hogging) 3.50 t-m (sagging)AD 3.67 t-m (hogging) 0.83 t-m (sagging)
(vi) Thickness of members
The maximum moment in the barrels is 5.25 t-m
For M 20 grade concrete and steel of Fe 415
70 1500.00m = 13k = 0.378j = 0.874
Q = 11.552The minimum effective thickness (d) required for bending moment
d = ÖM= Ö 5.25 X
bQ 100 X 11.55
= 21.33 cmAdopt overall thickness of all the members = 30.0 cmThen effective thickness = 25.20 cm
2
scbc = Kg/cm2 sst = Kg/cm2
105
Office of Director DamI.D.R.,Jaipur
document.xls
Program by- Sunil K Goyal Structural 23 Checked by - P P Pareek
(clear cover = 4.0 cm )
Office of Director DamI.D.R.,Jaipur
document.xls
Program by- Sunil K Goyal Structural 24 Checked by - P P Pareek
The reinforcement required at various points is given below :
(vii) Reinforcement
Span AB & BEMoments in t-m Reinforcement required
At face 4.67 t-m (hogging) = 14.13Provide 16 mm bars @ 14.0 cm c/c
At centre 3.11 t-m (sagging) = 9.40Provide 16 mm bars @ 21.0 cm c/c
Span CD & CFMoments in t-m Reinforcement required
At face 5.25 t-m (hogging) = 15.90Provide 16 mm bars @ 12.0 cm c/c
At centre 3.50 t-m (sagging) = 10.59Provide 16 mm bars @ 18.0 cm c/c
Span AD & EFMoments in t-m Reinforcement required
At face 3.67 t-m (hogging) = 11.12Provide 16 mm bars @ 18.0 cm c/c
At centre 0.83 t-m (sagging) = 2.52Provide 16 mm bars @ 25.0 cm c/c
Nominal steel shall also be provided on the outer face of the walls for the condition when there isno water in the drain and barrels are running full. The details of reinforcement are shown below
16 14.0 cm c/c
16 14.0 cm c/c
12 25.0 cm c/c
16 21.0 cm c/c
C.C. M - 20 16 25.0 cm c/c 12 25.0 cm c/c
16 18.0 cm c/c
16 12.0 cm c/c clear cover = 4.00 cm
16 18.0 cm c/c
0.30 3.00 m 0.30 3.00 0.30
DETAILS OF REINFORCEMENT IN SYPHON BARRELS
A t Cm2
A t Cm2
A t Cm2
A t Cm2
A t Cm2
A t Cm2
mm f @
mm f @
mm f @
mm f @
mm f @ mm f @
mm f @
mm f @
mm f @
Office of Director DamI.D.R.,Jaipur
document.xls
Program by- Sunil K Goyal Structural 25 Checked by - P P Pareek
*************
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