KITUI SAND DAMS

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d) Reinforcement Bars.Types of reinforcement bars used.i) Round bar 12.5mm diameter and length of 12m.ii) Round bars 6.25mm diameter and length of 12mNature of placing.Φ=12.5Φ=6.250..3 m0.4m 0.4mSpacing columns. Columns are spaced 2m side to side.No of columns=Ld/2-1Note that it is unwise to place columns at the extreme like it done in fencing posts.That is why we use.Total lengthExample. When D e =2.0m, L d =20mNo of R12.5mm= No of columns x 4 x D e .Ld( − 1)4De24DeLdNumber of pieces of R12. = ( −112 2Note: One piece of reinforcementbar=12m.Ld∴= 0.33De( −1)26220Number of R12.5 bars = 0.33 x 20( − 1 )2=6.60-0.66=5.94.Length of one ring of R 6.250.40.4 0.40.4=0.4 x 4=1.6.
Number of rings per evaragecolumns.De De= = = 2DeSpacing 0.5Total length of R6.25=Length per Ring x No of Rings peraverage column x No of columns.Ld=1.6(2De+1)( 12 − )Ld= 1 .6( DeLd − 2De+ + 1)2No of R6.251. 6Ld= ( DeLd − 2De+ + 1)12 2Example.L d =20, D e =20.20=0.13 ( 20x2) − (2x2)− + 1)2=(5.2 –0.52+1.3+0.13)=6.11NB: 0.13 can be Negligible.HenceNo of R 6.25=Ld0.13 ( DeLd − 2De− −1)2Ld=0.13 ( DeLd − 2De+ + 1)2e) Barbed Wire.Gauge 16, Length=610m, Spacing= 0.45Average strands=10Total length of barbed wire= 10 x LdDeNo of layers of Barbed wire = 0.4510Ld.DeTotal length of Barbed wire =0.45Example.L d =20, D e =2No of rolls of barbed wire= 0.4 x 20 x 2= 1.6No of rolls of Barbed wire =10Ld.De0.45x610No of rolls of Barbed wire = 0.04L d D e63
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SASOL FOUNDATIONFIELD OFFICELIASON
Page 3 and 4:
SECTION 1INSTITUTIONAL FACTORS3
Page 5 and 6:
Metamorphic and igneous rocks of th
Page 7 and 8:
comes in have been the method of ex
Page 9 and 10:
Usually, many people only think abo
Page 11 and 12: anticipated sand dam project in the
Page 13 and 14: Cost Of Sand DamsContrary to the po
Page 15 and 16: TrainingTraining is an integral par
Page 17 and 18: Each site gives 5 participants for
Page 19 and 20: SECTION 11TECHNICAL CONSTRUCTION AN
Page 21 and 22: Sand filling the riverA12 B34CBedro
Page 23 and 24: infiltration, surface flow begins a
Page 25 and 26: decreases leading to a rise in the
Page 27 and 28: ∆i = ϖ =∆xwfxf− wi− xi‘K
Page 29 and 30: Discharge.This is the water that fl
Page 31 and 32: Soft Porous MaterialParent rockNatu
Page 33 and 34: It is assumed that the channel is a
Page 35 and 36: A, B, C Represent different energy
Page 37 and 38: Cross section area of dam is = A dA
Page 39 and 40: ) Pre Hydraulic Design.At this stag
Page 41 and 42: L j = Hydraulic Length.This length
Page 43 and 44: CONSTRUCTION.Setting the trench.Thi
Page 45 and 46: avoid filling the reservoir upstrea
Page 47 and 48: 5) Masonry comprising Hard core and
Page 49 and 50: 0.5mSection X-X49
Page 51 and 52: 0.73m56 26 30.73mH dH d1.0m 1.0m1 4
Page 53 and 54: Plan and section view of the finish
Page 55 and 56: Key.1 W dc 75cm when off the ground
Page 57 and 58: VOLUME DERIVED FORMULAS FOR MAJOR D
Page 59 and 60: v) Blinding slab V b0.10.6mLdV b =
Page 61: Cement is also in V e , V us , V b
Page 65: Summary of derived formulas.• V w
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