Aqueduct Design
Aqueduct Design
Aqueduct Design
Design of aqueduct length of aqueduct(L) = Height(h)= width(B) = water depth(y) = M20 concrete HYSD steel bars 1. Design Constants For M20 conctrete, cbc = For HYSD steel on waterside, st = hence, K= j= R= 15 3.5 3.5 clear 2.98
3. Design of horizontal slab let the thickness of slab be Effective span of slab = Load of water of slab = wH self weight of slab = Therefore, Total load(p) = Total water pressure on the vertical wall = Fixing Moment at the end of slab = Simply supported B.M. at the centre of span of slab = pL /8 Net B.M. at the centre = the slab is to be designed for this B.M. since tension face is outside, st =
2 190 N/mm 2
R= Effective depth = Cover = Total thickness = Ast = take, diameter of bars as then, spacing required = Spacing provided =
Area of steel required at the end (near supports) This is more than half
2480.814 mm
4. Design of side walls as beam Length of Beam = Assumed Breadth of beam = Load from slab = Self load = Total load = Max B.M. = Effective depth = Total depth = Actual total depth = cover Actual Effective depth = Ast = 15 0.5 64232 43750 107982 3036994 2451.102 3800 3800 75 3725 m m N/m N/m N/m N/m mm mm mm mm
2 4810.599 mm
try, diameter of bars as area of each bar = Number of bars required = Number of bars provided = Ast Provided = Shear force = v = Permissible shear stress: Ast/bd*100 = c = Hence shear reinforcement is necessary. Vc = shear resistance of concrete = Vs = V-Vc = Spacing of stirrups is given by Sv = (sv*d*Asv)/Vs Minimum spacing is given by, Sv <= 2.5Asvfy/b Using 8 mm dia. 2 legged stirrups, Asv = therefore, Sv = hence, provide stirrups at the spacing of
28 mm
2 615.7522 mm 7.812557 8
428375 N 381490 N
Design of aqueduct length of aqueduct(L) = Height(h)= width(B) = water depth(y) = M20 concrete HYSD steel bars 1. Design Constants For M20 conctrete, cbc = For HYSD steel on waterside, st = hence, K= j= R= 15 c/c 3.5 clear 3.5 clear 2.98
3. Design of horizontal slab Depth required for deflection control: (l/d)max=(l/d)basic *(kt*kc) assuming kt*kc =1.3, and (l/d)basic =23 (l/d)max = dmin = calculation of B.M. mid span moment(kN/m) short long
29.9 0.501672
Support moment(kNm)
Here, Mx =
let the thickness of slab be Effective span of slab = Load of water of slab = wH self weight of slab = Therefore, Total load(p) = Total water pressure on the vertical wall = Fixing Moment at the end of slab = Simply supported B.M. at the centre of span of slab = pL2/8 Net B.M. at the centre = the slab is to be designed for this B.M. since tension face is outside, st = R= Effective depth = Cover = Total thickness = Ast = take, diameter of bars as then, spacing required = Spacing provided =
Area of steel required at the end (near supports) This is more than half
2480.814 mm
4. Design of side walls as beam Length of Beam = Assumed Breadth of beam = Load from slab = Self load = Total load = Max B.M. = Effective depth = Total depth = Actual total depth = cover Actual Effective depth = Ast = try, diameter of bars as area of each bar = Number of bars required = Number of bars provided = Ast Provided = Shear force = v = Permissible shear stress: Ast/bd*100 = c = Hence shear reinforcement is necessary. Vc = shear resistance of concrete = Vs = V-Vc = Spacing of stirrups is given by Sv = (sv*d*Asv)/Vs Minimum spacing is given by, Sv <= 2.5Asvfy/b 428375 N 381490 N 15 0.5 64232 43750 107982 3036994 2451.102 3800 3800 75 3725 m m N/m N/m N/m N/m mm mm mm mm
2 4810.599 mm
28 mm
2 615.7522 mm 7.812557 8
Using 8 mm dia. 2 legged stirrups, Asv = therefore, Sv = hence, provide stirrups at the spacing of
Es = b=
200000 300
D 700 700
cover 50 50
d 650 700
fck 20 25
fy