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MANUAL CALCULATION UNIVERSITY OF ILORIN DEPARTMENT OF CIVIL ENGINEERING PROJECT Input Calculation Output Tank capacity = 300,000litres = 300m3 Height of tower from ground level = 10.0m Live load on the dome = 1.5kN/m2 Nimet 2014 Works dept unilorin (2014) Wind intensity = 1.504kN/m2 Weight of water = 10kN/m2 Soil bearing capacity = 150kN/m2 Concrete grade = M30 Concrete unit weight = 24kN/m Resistance to crack = 1.2kN/mm2 Tensile stress 150N/mm2 Nominal cover = 25mm Effective cover = 35mm Dimension of Tank D = inside diameter of tank Assuming the average depth of 0.75D π×π· 2 We have ( 4 × 0.75π·) = 300 3.142 × π·2 × 0.75π· = 300 4 π· = 7.98 π ππ¦ ≅ 8π D= 8m CALCULATION INPUT OUTPUT Height of the cylindrical portion of the tank 0.78 X 7.99 = 5.9 Depth of conical dome = D/5 or D/6 = 8/5 = 1.6m Diameter of the supporting tower = 4.5m DESIGN OF TOP DOME Thickness of the dome slab = 100mm or 0.10m Slab self-weight = 0.10 x 1 x 1 x 24 = 3kN/m2 Live load = 1.5kN/m 2 Finishes = 0.10kN/m 2 Total = 4.60kN/m 2 If R is the radius of the dome D = diameter of tank = 8m D= 8m r = central rise which is the depth of the conical dome D/5 r = 1.6m or D/6 = 8/5 = 1.6m 2 2 The radius r is given by π = cos π = π = cos (π·⁄2) +π 2 2π = (8⁄2) +1.62 2×1.6 = 5.5π 5 = 0.9091 5.5 −1 (0.9091) = 24.62° ≅ 25° Meridional thrust at the edge T = Cos = 250 π‘ππ‘ππ ππππ ππ ππππ ×π 1+cos π 4.60 × 5.5 = 13.25π/π 1 + 0.9091 1.00 Circumferential force = ππ ( cos π − 1+cos π) (4.6 × 5.5 × 0.9091 − R = 5.5m 1.00 ) = 22.48ππ/π 1 + 0.9091 Thrust = 13.25N/m INPUT CALCULATION 13.25×1000 Meridional stress = ( 100×1000 ) = Hoop stress= 22.48×1000 100×1000 = 0.23π ππ2 0.132π ππ2 OUTPUT < 5π/ππ2 < 5π/ππ2 Safe Safe So far the stress is still within the safe limit, provide minimum reinforcement at 0.2% of area in each direction Ast = 0.2 100 × 1000 × 100 = 200ππ2 Provide Y8mm diameter bar @175mm c/c (287mm2) in both Provide direction Y8m@175mm Design Of Top Ring Beam π cos π π· Hoop tension = ( 2 ) 13.25 × 0.9091 × 8 = 48.18ππ 2 hoop tension = 48.18kN Permissible stress in high yield strength deformation bars = 150N/mm2 48.18×1000 = 321ππ2 provide Provide 4Y12mm(452mm2) diameter ring bar 4Y12mm Ast = 150 If Ac = cross section of ring beam Equivalent area of composite section of beam 48.18 × 1000 π΄π + π × 452 π€βπππ π = 13 π΄π = 34.274ππ2 Provide ring beam 300 X 300 = 9000mm Provide 2leg 8mm diameter stirrups @300mm c/c Design Of Cylindrical Tank Wall The tank is assumed to be free on top and bottom and maximum hoop tension occur at the base of wall Y8@300mm stirrups INPUT CALCULATION OUTPUT ππ»π· Maximum hoop tension at the base = ( = 4 ) 10 × 5 × 8 = 200ππ/π 2 200×1000 Area of ring beam = 150 = 1333ππ2 πππ πππ‘ππ Resisting the hoop tension at 1meter below Ast= 1.25×667 5 × 2 = 333.5ππ2 Provide Y10 diameter @ 175mm (449mm) c/c both direction from 0 - 2 meter from top of the tank wall To resist hoop tension at 3meter from top Ast= 2.5×667 5 × 2 = 667ππ2 Provide Y10@175mm Provide Y12mm diameter bar @ 150mm2 (754mm2) c/c for 2 to 3 meters from top To resist hoop tension at 5meters below top Ast = 5×667 5 × 2 = 1333ππ2 Provide Y16mm diameter bar @ 125mm2 (1610mm2) c/c for both direction from 3 to 5meters The thickness T is given by = 200×1000 1000+(π×ππππ ππ ππππ ππππ ππππ πππ) T= 200×1000 = 1.2 1000+(13×1336) Provide 200mm thickness at the bottom and 200mm thickness at the top Av thickness = 200mm 0.3 Distribution = 100 × 20 × 100 = 600ππ2 Provide half of the reinforcement near each face Asd = 300mm2 Provide Y10mm diameter bar @ 275mm c/c (285mm2) Provide Y16@125mm INPUT CALCULATION OUTPUT Design Of Bottom Ring Beam Assuming beam is 800 X 400mm Loading on the ring beam A). Load due to top dome = (meridional thrust X sinπ) = 13.25 x sin 25 = 5.6kN/m B). Load due to top ring beam = 0.3 x 0.3 x 24 = 2.16kN/m c). Load due to cylindrical wall = 0.2 x 5.0 x 24 = 24kN/m D). Self-load of ring beam = 0.8 x 0.4 x 24 = 7.68kN/m Total = 39.43kN/m ππππ π ×π· ) 2 =( Hoop tension = 39.4 × 0.9091 × 8 = 143.387ππ/π 2 This to be rested entirely by steel hoops, the area of which is π΄π π‘ = 143.38 × 1000 10 × = 1051.44ππ2 150 100 Provide 7Y20mm2 (2200mm2) diameter ring bar Steel in equivalent section = = 232.7 ×1000 800 ×400+13 ×2200 = 0.41π/ππ2 < 1.2 The 10mm diameter bar provided in the wall @ 150mm c/c should be taken round on the ring to act as stirrups Design Of Conical Dome Wall Loading on the conical dome Average diameter of conical dome = = Average depth of water = 5 + 1.33 2 8+5 2 = 13 2 = 6.50 = 5.67 Weight of water above conical dome = 3.14 x 6.5 x 5.67 x 1.33 x 10 =1542 Okay Y10@150mm stirrups INPUT CALCULATION OUTPUT Self –weight of slab 400mm thick = 3 x 6.5 x 2.12 x 0.4 x 24 = 416 Load from dome , top ring beam, wall and bottom beam are 3.14 x 8 x 39.43 = 991.1 Total load on conical slab = 2,949kN ππππ 2,949 π= = 187.8ππ/π π’πππ‘ ππππβπ‘ 3.14 × 5 Meridional thrust T = V 2 cosecπ = 187.8 x cosec45 = 187.8 x 1.41 = 265.6kN Meridional stress = 265.5 ×1000 400 ×1000 = 0.67π/ππ2 < 5π/ππ2 π· Hoop tension H (ππππ πππ + π cot 45) × 2 Weight of the conical slab per meter run is computed as Okay q = 0.4 x 24 = 9.6 f = 45° D = 8m π = 6.5 × 5 = 32.50ππ/π H = (32.50 × πππ ππ 45 + 9.6 × H = 222.25kN Ast = 223 ×1000 150 1 )× π‘ππ45 8 2 says 223kN = 1,487ππ2 Provide Y25mm @ 275mm2 (1784mm2) c/c 0.20 At the bottom Ast = (( 100 ) π₯ 400 π₯ 1000) =800mm2 Hoop tension Provide half of the reinforcement near each face = 400mm2 = 223kN Y10@ 200 (393mm2) both face along the meridional Provide Maximum tensile stress Y25@275mm 223 ×1000 = 400 ×1000+13×1487 = 0.532π/ππ2 < 1.2 ππππ¦ CALCULATION INPUT OUTPUT Analysis and Design of Column Member size for an unbraced column Column size = 600mm x600mm Diameter = 4.5m Height of the column = 10m Axial load on the column = 121.1kN Area supported = 4.5 x 4.5 = 20.25m2 Total axial load = 121.1 x 20.25 = 2.45x103kN Axial load = 2.45 x 103kN Slenderness Le = Lo x B Where Lo is the clear height between resistance Le effective height The column satisfies the end condition 2:3 for an unbraced column whereas π½ = 1.80 Condition 2 state that; end condition at the top of the column is connected monolithically to the beam or slab on either side, which are shallower than the overall dimension of the column in the plane considered Condition 3 state that; end condition at the bottom of the column is connected to member which not specifically designed to provide restrain to rotation of the column will nevertheless, provided some nominal restrain. Le = 1.80 x 10000 = 18000mm Therefore Le/h = 18000 600 = 30 Since 30> 10 Axial loaded slender and unbraced column 600 x 600 Madd = N∅ = ∅ = π½π’πΎβ 1 πΏπ π½ = 2000 ( π ) = K = 1.0, h = 0.6 1 18000 2 ( 2000 600 ) = 0.45 π½ = 0.45 K = 1.0 h = 0.6 Q = 0.23 INPUT CALCULATION OUTPUT ∅=0.45 ×1.0×0.6=0.276 π½ = 0.74 Madd = (2.34 x 103) x 0.276 = 676.2kNm ∅= N/Fcubh= (2.45×γ10γ^6)/(30×600×600)=0.23 Since 0.23 < 0.3 use 0.3 π½ = 1 − 1.1644∅ = 1 − 1.1644(0.3) = 0.74 My = 1176.59kNm β′ My’ = Myy + π½ ( ) ππ₯π₯ π My = 676.2 + 0.74 (1) 676.2 My = 1176.59kNm d = 600 −40 = 560 560 d/h = 600 = 0.93 ∝= 0.2 π΄π = 0.0026 ∝Fcubh As = 5241.6mm2 π΄π = 0.0026 × 0.2 × 30 × 600 × 560 π΄π = 5241.6ππ 2 Provide 8Y32 π΄π + 10% ππ π΄π = 5241.6 + 524.16 = 5765.76ππ2 Provide 8Y32mm ∅ πππ @ area (6430mm) Links Y16 Maximum links = 32 x x 32 = 1024mm2 Use Y16 @ 1210c/c Foundation design Column load = 2,458.64kN Assumed depth of footing (h) = 500mm = 0.5m Loading = 2458.64kN INPUT CALCULATION Effective depth (d) = h – concrete cover - ∅/2 d = 500 – 50 -10 = 440mm column size = 600mm x 600mm concrete stress (fcu) = 30N/mm2 steel stress (fy) = 410N/mm2 soil bearing capacity = 150kN/m2 2458.64×1.1 The base area required = 150×1.46 = 12.35π2 OUTPUT d= 440mm h = 500mm Where ;1.46 carter for the serviceability limit state ; 1.1 carter for the load of the self-weight of pad footing using 10% of load Adedeji, A.A Table B9 Area required = 12.35m2 Square pad footing size = 4mx4m Provide 4000mmx4000mmx500mm square base Net pressure 1.1π€ Fnet = π΄πππ ππππ£ππππ − ππβπ Fnet = 1.1×2458.64 4 ×4 Area required = 12.35m2 − (1.4 × 0.4 × 24) = 155.59ππ/π Checking for punching shear Critical perimeter (punching shear) taken as 1.5h from column face Pcrt = Column perimeter + 3πβ 4(400) + 3(π)500 = 7112.29ππ Area within the critical perimeter (a1+3h)(a2+3h) − (4−π)(1.5β) BS8110 (400 + (3 × 500))(400 + (3 × 500)) − (4 − 3.142)(1.5 × 500)2 = 3.13m2 Load causing punching which equal total load outside perimeter (V) V = Fnet (Base area – Area within critical perimeter) Fnet =155.59kN/m2 INPUT CALCULATION π = 155.59((6 × 6) − 3.13) = 5053.56ππ π From table 3.8 BS8110 part 1:1997 OUTPUT V = 5053.56kN 5053.56 Vpunching = πππππππ‘ππ×π = 7112.29×44 = 0.02π/ππ2 Comparing this value with permissible shear of 0.36N/mm2 (assuming the minimum of 0.15% of steel) show that the base is safe against punching shear with depth of 400 x-x axis span = 1700mm moment (m) = WL2/2 = 155.59×1.7002 2 = 224.82πππ πππ ππ’π Vpunching = 0.02N/mm2 M= 224,82kNm Design of reinforcement 224.82 × 106 πΎ= = 0.039 30 × 1000 × 4402 0.039 La = 0.5 + √0.25 − ( 0.9 ) = 0.95 π = 0.95 × 440 = K =0.039 La = 0.95 Z = 420.03mm 420.03ππ 224.82×106 As = 0.87×410×420.03 = 1500.55ππ Minimum steel required = 0.15%bh 0.15 × 1000 × 500 = = 750ππ 100 Provide Y20mm @ 175 c/c btm (1800mm2) Checking for local bond stress Load for Local bond = Fnet x 1.0 x 1.7 =155.59 x 1.0 x 1.7 = 264.503kN No of rods per meter run = VLb = 1000 175 = 5.7 264.503×106 12(π)(5.7)440 = 2.79π/ππ2 Since 0.8√πππ’ = 4.38π/ππ2 > πππ = 2.79π/ππ2 Hence the local bond stress is still within the ultimate shear stress value Provide Y20@175mm VLb = 2.79N/mm2 INPUT CALCULATION Final checking on punching shear 100π΄π ππππ£ 100 × 1800 = = 0.41% ππ 100 × 440 By interpolation, Vc = 0.1724 The initial punching shear was not excessive the assumed 500mm depth is adequate Final checking for the shear stress Load causing shear is the load outside 1.5d from the column face Vmax = (1.7 − (1.5 × 0.44)) × 155.59 × 1.0 = 161.6 × 103 π V= 161.8×10^3 1000×440 = 0.37π/ππ2 Summary Provide 4000mmx4000mmx500mm base Provide Y20mm @ 175 c/c btm (1800mm2) OUTPUT Detailing for the Tank Detailing for Tank Column Foundation Detailing DESIGN OF TANK USING MIDAS Fig 4.1 Fig 4.3 Story Axail force sum results Concrete design force Axial forces sum of element 4.3 COMPARATION OF THE RESULTS Section Manual computation Software application Remark Top dome Y8@175mm(287mm2) Y10@175mm(449mm2) Software result is higher Top ring beam 4Y12mm(452mm2) 5Y12mm(565mm2) Software result is higher Cylindrical tank Y16@125mm(1610mm2) Y20@300mm(1050mm2) Manual result is wall Bottom ring higher 8Y20mm(2510mm2) 7Y20(2200mm2) beam Conical dome higher Y25@275mm(1784mm2) Y20@125mm(2510mm2) wall Column Manual result is Software result is higher 8Y32 @ 6460mm2 6Y40 @ 7540mm2 Software result is higher 4.4 DISCUSSION From the results above, it is shown that there are slight diffrences in some of the sections of the tank, which occured during manual computation reinforcement was provided considering the economic factor of which software will not do such because it is garbage in garbage out. As the result from manual is higher than the software application at the cylindrical section because the section was desiged as the water retaining structure using BS 8007 code of practice, and also the diffrences may comes from analysis employed and many approximation in manual calculation.
