5.0 DESIGN CALCULATIONS

5.0 DESIGN CALCULATIONS Load Data Reference Drawing No. 2-87-010-80926 Foundation loading for steel chimney 1-00-281-53214 Boiler foundation plan sketch : Figure 1 Quantity Unit Dia of Stack, d 6.00 m Dia. of Raft, D 13.00 m Area of Raft, A = 3.14 xdxd/4 132.67 m^2 Section Modulus, z = 3.14xDxDxD/32 215.58 m^3 Height of soil above Raft, h 4.00 m Thickness of the Raft, t 2.00 m Height of pedestal, h1 4.50 m Thickness of pedestal t1 1.00 m Load from chimney Mimimun Dead Load, W1 170.00 ton Maximun Dead Load and Live Load, W2 320.00 ton Wind Load Shear due to wind on main stack, H1 125.00 ton Moment due to wind on main stack, M1 5450.00 ton-m Hence Addition Moment due to shear at founding level, M2 = H1 x (h1+t) 812.50 ton-m Load from Boiler Foundation Dead Load and live(hrsg Foundation) Dead Load + Live Load + Vertical wind Load, W3 = 79.9+26.8+34 140.70 ton Hence moment, M3 = W3 x a 717.57 ton-m where 'a' = 5.1, distance from center line of the stack to central line of HRSG pedestal Wind shear and Thermal Shear = 6.5+16 22.50 Additional Moment, M4 = (6.5+16)xb 78.75 ton-m where 'b' = 3.5, distance from the founding level of HRSG to top of HRSG Pedestal Net Safe Bearing Pressure = S.B.C. 50.00 ton/m^2 Gross Bearing Pressure (DL+LL) = S.B.C. + (1.8x(h+t)) 60.80 ton/m^2 Gross Bearing Pressure (in DL+LL+WL) = S.B.C. + (1.8x(h+t)) 60.80 ton/m^2 No increase in stresses as chimney is a critical structure Weight of pedestal, W5 = 3.14xdxt1xh1 211.95 ton Weight of soil above the raft, W6 = ((3.14x(D^2)/4)-(Area of pedestal)) Xhx unit wt of soil 819.54 ton =(3.14x13^2/4-(3.14x6x1))x4x1.8 Weight of raft, W7 =3.14xD^2/4 x H X uint wt of concrete 663.33 ton W self = W5+W6+W7 1694.82 ton 5

Case 1 : Check for Bearing Pressure in Pure Dead Load Live load Total Axial Weight (gross) Wg = W2+W3+W self 2155.52 ton Total Moment = Mg1 = M2 717.57 ton-m Maximum Base Pressure, q1 = Wg/A + Mg1/z 19.58 ton/m^2< 50 ok Minimum Base Pressure, q2 = Wg/A-Mg1/z 12.92 ton/m^2 > 0 Ok Case 2 : Check for Bearing Pressure in Dead Load+Live Load+Wind Load Total Gross Weight Wg = W2+W3+Wself Total Moment = Mg2 = M1+M2+M3+M4 2155.52 ton 7058.82 ton-m Maximum Base Pressure, q3 = Wg/A + Mg2/z Minimum Base Pressure, q4 = Wg/A-Mg2/z 48.99 ton/m^2 <60.8 OK -16.50 ton/m^2 Modified Base Pressure e = Mg2 / Wg 3.27 m e/ D = 3.28 / 13 = 0.25 Crossponding to e/d, C2 = 3.55 (refer Mark Fintel, Fig 5-14) q5 = Wg / A 16.25 ton/m^2 Modified Base Pressure, q p = C2xq5 57.68 <60.8 ton/m^2 ok Check for overturning : Vertical load V = W1 + W self 1864.82 ton lever arm = D/2 6.50 Moment of Rsistance, MR = 0.9x(V)xD/2 10909.17 Overturning Moment, MO = Mg2 7058.82 Factor of safety F.O.S. 1.55 <1.5 Check for sliding ( µ = 0.5) sliding =(V)xµ/ Η1 7.46 <1.5 Analysis and Design of Raft: Case A : Full Raft with axial laod condition For Bending Moment Calculation in raft, Selfweight and weight above soil not considered Axial load without soil and weight of raft 460.70 ton 4607.00 Kn p = P / A 34.73 Kn/m^2 a= 6.5 β =(c/c of chimney rad,r)/(radius of raft,a) =3/6.5 0.46 Refer table 6.9, Tall chimneys by S.N. Manohar Y2 =-2.52-(8Xln β)-(2.96xβxβ) 3.13 Y3 = Y4 0.00 Y6 =5.48-2.96xβxβ 4.74 Y8-8.00 Y7 =-8XβXβ -1.70 6

Moment calcluation due to axial laod. 1 0.15-171.81-168.12-17.36 0.00 f<b 2 0.31-192.32-177.56-34.73 0.00 f<b 3 0.46-215.79-193.29-52.09 0.00 f<b 4 0.62-74.09-151.46 113.95 0.00 f>b 5 0.77-9.93-117.23 59.90 0.00 f>b 6.5 1.00 5.72-83.76 0.00 0.00 f>b Case B : Full Raft with Moment laod condition Y2 =3/β^2-5.46-0.81β^2 8.45 Y3 = Y4 0.00 Y6 =-5.46-0.81xβxβ -5.63 Y8 12.00 Y7 =3XβXβ 0.64 Refer : Tall chimney by S.N.Manohar TABLE 6.10 Moment calcluation due to Moment laod. When θ is 0 cos 0 = 1 Sin 0 = 0.00 r f Mr Mt Mtor Qr Qt Qtor=1.6 xmtor/b m - Kn-m Kn-m Kn-m Kn Kn Kn 1 0.15-595.57-273.50 0.00-768.30 0.00 0.00 f<b 2 0.31-1223.56-558.02 0.00-824.98 0.00 0.00 f<b 3 0.46-1916.42-864.57 0.00-919.43 0.00 0.00 f<b 4 0.62-724.16-673.16 0.00 899.76 0.00 0.00 f>b 5 0.77-173.43-444.88 0.00 476.84 0.00 0.00 f>b 6.5 1.00-0.10-243.48 0.00-32.59 0.00 0.00 f>b q = M / Z 327.43 Kn/m^2 Total (max at f=β ) Radial Moment Mr Total (max at f=β ) Tengantial Moment Mt -2132.20 Kn-m -1057.87 Kn-m When θ is 45 cos 45 = 0.707 sin 45 = 0.71 Qtor=1.6 r f Mr Mt Mtor Qr Qt xmtor/b m - Kn-m Kn-m Kn-m Kn Kn Kn 1 0.15-421.07-193.36-113.22-543.19 525.38 181.15 f<b 2 0.31-865.06-394.52-230.23-583.26 512.03 368.36 f<b 3 0.46-1354.91-611.25-354.80-650.04 489.77 567.68 f<b 4 0.62-511.98-475.92-347.07 636.13 72.30 555.31 f>b 5 0.77-122.62-314.53-257.51 337.13-146.57 412.02 f>b 6.5 1.00-0.07-172.14-149.68-23.04-353.14 239.49 f>b 7

When θ is 90 cos 90 = 0 sin 90 = 1.00 r f Mr Mt Mtor Qr Qt Qtor=1.6 xmtor/b m - Kn-m Kn-m Kn-m Kn Kn Kn 1 0.15 0.00 0.00-160.14 0.00 743.12 256.23 f<b 2 0.31 0.00 0.00-325.64 0.00 724.23 521.02 f<b 3 0.46 0.00 0.00-501.84 0.00 692.74 802.94 f<b 4 0.62 0.00 0.00-490.90 0.00 102.27 785.44 f>b 5 0.77 0.00 0.00-364.23 0.00-207.31 582.77 f>b 6.5 1.00 0.00 0.00-211.71 0.00-499.49 338.74 f>b Summery at different angle When θ is 0 1 0.15 767.37 441.62 785.67 0.00 2 0.31 1415.88 735.57 859.70 0.00 3 0.46 2132.20 1057.87 971.52 0.00 4 0.62 798.25 824.62 1013.70 0.00 5 0.77 183.36 562.11 536.74 0.00 6.5 1.00 5.82 327.24 32.59 0.00 When θ is 45 1 0.15 706.09 474.70 741.71 706.54 2 0.31 1287.60 802.30 986.35 880.39 3 0.46 1925.49 1008.00 1269.80 1057.45 4 0.62 933.14 974.45 1305.38 627.61 5 0.77 390.06 689.27 809.05 558.59 6.5 1.00 155.47 405.58 262.52 592.63 When θ is 90 1 0.15 331.95 328.26 273.59 999.35 2 0.31 517.95 503.19 555.75 1245.25 3 0.46 717.62 695.13 855.03 1495.68 4 0.62 564.99 642.36 899.39 887.71 5 0.77 374.16 481.46 642.68 790.08 6.5 1.00 217.43 295.47 338.74 838.23 Sample calculation Grade of concrete = M25 Grade Of steel = Fe415 Design Moment and Reinforcment Calculation Dia of bar to be used = 25 r Mr Mu/bd^2 pt min pt ast spacing Provided m Kn-m SP-16 (TABLE 1) mm^2 required 1 767.374 0.318853 0.09 0.12 2280 215.19 200 2 1415.881 0.588316 0.17 0.12 3230 151.90 200 3 2132.205 0.885958 0.25 0.12 4750 103.29 100 4 798.2542 0.331685 0.09 0.12 2280 215.19 200 5 183.3625 0.076189 0.09 0.12 2280 215.19 200 6.5 5.817362 0.002417 0.09 0.12 2280 215.19 200 8

Reinforcment Due to Radial Moment (max at f= 0.5) Mr 2132.20 Kn-m Factored Moment 3198.31 Kn-m Depth of Raft D 2000.00 mm Effective Depth d 1900.00 mm Mu/bd2 0.89 For the Mu/bd2 and fck pt 0.23 Using SP-16 (table-1) Pt Min 0.12 Ast = ptxbxd 4370.00 mm^2 Dia of bar to be used 25.00 Spacing required 112.27 mm Provide 25 mm dia bars @ 100 mm C/C Reinforcment Due to Radial Moment (at f = 0.67) Mr 798.25 Kn-m Factored Moment 1197.38 Kn-m Depth of Raft D 2000.00 mm Effective Depth d 1900.00 mm Mu/bd2 0.33 For the Mu/bd2 and fck pt 0.09 Using SP-16 Pt Min 0.12 Ast = ptxbxd 2280.00 mm^2 Dia of bar to be used 25.00 Spacing required 215.19 mm Provide 25 mm dia bars @ 200 mm C/C Design Moment and Reinforcment Calculation Dia of bar to be used = 25 r Mt Mu/bd^2 pt min pt ast spacing Provided m Kn-m SP-16 (TABLE 1) mm^2 required 1 441.6152 0.183497 0.1 0.12 2280 215.19 200 2 735.575 0.305641 0.1 0.12 2280 215.19 200 3 1057.868 0.439557 0.12 0.12 2280 215.19 200 4 824.6215 0.34264 0.1 0.12 2280 215.19 200 5 562.1095 0.233564 0.1 0.12 2280 215.19 200 6.5 327.2372 0.135971 0.1 0.12 2280 215.19 200 Reinforcment Due to (Max) Tangential Moment Mt 1057.87 Kn-m Factored Moment 1586.80 Kn-m Depth of Raft D 2000.00 mm Effective Depth d 1900.00 mm Mu/bd2 0.44 For the Mu/bd2 and fck pt 0.09 Using SP-16 Pt Min 0.12 Ast = ptxbxd 2280.00 mm^2 Dia of bar to be used 25.00 Spacing required 215.19 mm Provide 25 mm dia bars @ 200 mm C/C 9

Check for One Way Shear :: Shear Force (max at d at any angle ), 1000.00 KN per m length Factored shear force, Vu 1500 KN per m length Calculated Shear Stress 0.789 N/mm2 Min % Of Steel provide 0.258 Shear strength of conc. 0.42 N/mm2 Enhancement of shear strength,2xdxtc/av 0.840 As per clause 40.5.1 Of IS-456 Hence Safe Check for Two-Way Shear :: Max. vertical load on Raft = W2 + W3 Shear Sress Permissible stress 460.70 KN 0.363711 N/mm2.t c 0.25 X fck N/mm 2 1.25 N/mm 2 ok Design of Pedestal Mg2 70588.2 Kn-m Wg 21555.15 kn e = M / W 3.2747719 m r 7 m e / r 0.467825 β 0 µ 0 value of φ 150 ο from chart 7.1 value of C Assume % of reinforcment 3 % Fc= (CW)/( r t (1-p)) 0.84 N/mm^2 <0.38xfck 9.5 ok Fs = (SxmxW)/(r t (1-p) S = C ((cos F) + cos m))/((cos b) - cos f)) 0.0244 Maximum steel stress, Fs Provide 20 mm dia bars @ 150 mm C/C 0.64 N/mm^2 <0.57fy 236.55 ok 10