Design of Minaret

7.2 Design of minaret: Geometry: The figure below shows the longitudinal section in the minaret and the cross section will be shown when start calculate the self-weight of each section.

Figure 7.2.1 Longitudinal section in minaret

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178

Analysis and Design: We should calculate the base shear from the wind and the earthquake load respectively And decide about the load combination which will be used in analysis and design procedure.

Calculation of the base shear from the wind load: 1- Calculate design wind pressure According to UBC97 Code. As following: P = Ce × Cq × qs × I w

A) Basic wind speed = 1.61x90 = 144 .9 mph From table A − 1( Appendix ) q s = 0.0479 x 20 .8 = 0.996 ≈ 1KN / m 2 From table A − 1( Appendix ) B) For chimneys and solid tower : C q = 0.8 for any direction. From table A − 5( Appendix ) D) For chimneys and solid tower I w = 1 From table A − 6( Appendix ) E) Choose Exposure C for Mutah University Zone from table A-2(Appendix). Height above average level of adjoining ground(m) 0.0 1.0 2.0 3.0 4.0 4.57 6.09 7.62 9.14 12.19 18.29 24.38 30.48 36.58

Ce for EXPOSURE C

q(KN/m2)

1.06 1.06 1.06 1.06 1.06 1.06 1.13 1.19 1.23 1.31 1.43 1.53 1.61 1.67

0.848 0.848 0.848 0.848 0.848 0.848 0.904 0.952 0.984 1.048 1.144 1.224 1.288 1.336

Table 7.2.1 Design wind pressure for exposure C

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Wind Pressure Distribution 40 35 Height(m)

30 25 20 15 10 5

0. 84 8 0. 84 8 0. 84 8 0. 84 8 0. 84 8 0. 84 8 0. 90 4 0. 95 2 0. 98 4 1. 04 8 1. 14 4 1. 22 4 1. 28 8 1. 33 6

0

q(KN/m2)

Figure 7.2.2 Design wind pressure for exposure C

2- Distribute the wind pressure for range of height as following: Height Outer Pressure Distribution force Range(m) Diameter(m) (KN/m2) (KN/m') 0 - 2.1 4.8 0.848 4.1 2.1 - 17.5 4.6 1.144* 5.26 17.5 - 19 4.0 1.224 4.5 19 – 23.5 4.6 1.224 5.63 23.5 - 25 4.0 1.288 5.15 25 – 28.75 3.0 1.288 3.86 28.75 - 31 3.0 1.336 4 31 – 34.5 3.0 1.336 4 34.5 - 35.5 3.0 1.336 4 *Also use this for the elevated water tank (from 12 - 16) m Table 7.2.2 Design wind pressure distribution over the range of height

3- Calculation of the Base shear: V= 4.1 x (2.1 - 0) + 5.26 x (17.5 - 2.1) + 4.5 x (19 - 17.5) + 5.63 x (23.5 - 19) + 5.15 x (25 - 23.5) + 3.86 x (28.75 - 25) + 4 x (35.5 - 28.75) =170.9KN

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After calculating the earthquake base we should decide about the load combination which will be used in analysis and design procedure, because according to UBC97 criteria, we should use the greatest of wind or earthquake load.

Calculation of the base shear from earthquake load: 1- Calculate the weight of the structure: The minaret will be will be divided to more than one section according to the varying in the section area at each elevation as follow: Assume the footing is at elevation -3 below the ground level. Section No.1 Elevation from -3m to 2.1m From the architectural drawing thickness of the stone facing = 10 cm. π  π  2 2 W1 = Wconcrete + W stone =   × Dout − Din2 concrete × γ concrete +   × Dout − Din2 stone × γ stone 4 4

[

]

[

]

π  π  =   × 4.8 2 − 3.4 2 × 25 +   × 5 2 − 4.8 2 × 23 = 266.88 KN / m' 4 4

[

I=

π 64

(

)

4 × Dout − Din4 =

π 64

(

]

[

]

)

× 4.8 4 − 3.4 4 = 19.5m 4

Figure 7.2.3 Section No.1

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Section No.2 Elevation from 2.1m to 17.5m and from 19m to 23.5m From the architectural drawing thickness of the stone facing = 10 cm. π  π  2 2 W1 = Wconcrete + Wstone =   × Dout − Din2 concrete × γ concrete +   × Dout − Din2 4 4    

[

]

[

]

stone

× γ concrete

π  π  =   × 4.6 2 − 3.4 2 × 25 +   × 4.8 2 − 4.6 2 × 23 = 222.46 KN / m' 4 4

[

I=

π 64

(

)

4 × Dout − Din4 =

π 64

]

(

[

]

)

× 4.6 4 − 3.4 4 = 15.42m 4

Figure 7.2.4 Section No.2

Section No.3 Elevation from 17.5m to 19m and from 23.5m to 25m π  2 W1 = Wconcrete =   × Dout − Din2 concrete × γ concrete 4

[

]

π  =   × 4.0 2 − 3.4 2 × 25 = 87.18 KN / m' 4

[

I=

π 64

(

)

4 × Dout − Din4 =

Minaret design

]

π 64

(

)

× 4 .0 4 − 3 .4 4 = 6 m 4

182

Figure 7.2.5 Section No.3

Section No.4 Elevation from 25m to 28.75m There is no stone facing but there are stone wall a round the concrete section with more than one diameter. Use the average diameter of the stone wall and calculate the weight of the wall π  2 W1 = Wconcrete + W stone =   × D out − Din2 4

[

]

concrete

π  2 × γ concrete +   × D out − Din2 4

[

]

stone

× γ concrete

π  π  =   × 3.0 2 − 2.4 2 × 25 +   × 4.6 2 − 4 2 × 23 = 156.83KN / m' 4 4

[

I=

π 64

(

)

4 × Dout − Din4 =

π 64

(

]

[

]

)

× 3.0 4 − 2.4 4 = 2.35m 4

Figure 7.2.6 Section No.4

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Section No.5 Elevation from 28.75m to 31m There is an empty space in this elevation because there are four windows with height equal to 2.25m. W1 = Wtotal − Wempty space

π  π  2 Wtotal =   × Dout − Din2 concrete × γ concrete =   × 3 2 − 2.4 2 × 25 = 63.62 KN / m' 4 4  1   1 Wempty space =   × rout × Lout −   × rin × Lin  × γ concrete 2  2    π  Lout = rout × θ = 1.5 × 15 ×   = 0 .4 m  180   π  Lin = rin × θ = 1.2 × 15 ×   = 0.3m  180   1   1 Wempty space =   × 1.5 × 0.4 −   × 1.2 × 0.3 × 25 = 3KN / m' 2  2   W1 = Wtotal − Wempty space = 63.62 − 4 × 3 = 51.62 KN / m'

[

I=

π 64

(

]

)

4 × Dout − Din4 =

π 64

[

(

]

)

× 3.0 4 − 2.4 4 = 2.35m 4

Figure 7.2.7 Section No.5

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Section No.6 Elevation from 31m to 34.5m π  2 W1 = Wconcrete =   × Dout − Din2 4  

[

]

concrete

× γ concrete

π  =   × 3 2 − 2.4 2 × 25 = 63.62 KN / m' 4

[

I=

π 64

(

)

4 × Dout − Din4 =

]

π 64

(

)

× 3.0 4 − 2.4 4 = 2.35m 4

Figure 7.2.8 Section No.6

Section No.7 Elevation from 34.5m to 35.5m π  2 W1 = Wconcrete =   × Dout − Din2 4

[

]

concrete

× γ concrete

π  =   × 3.0 2 − 1.8 2 × 25 = 113.1KN / m' 4

[

I=

π 64

(

)

4 × Dout − Din4 =

]

π 64

(

)

× 3.0 4 − 1.8 4 = 3.46m 4

Figure 7.2.9 Section No.7

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Total gravity loads: W gravity = 260.82 × (2.1 − (− 3)) + 222.46 × (17.5 − 2.1) + 222.46 × (23.5 − 19 ) + 87.18 × (19 − 17.5) + 87.18 × (25 − 23.5) + 156.83 × (28.75 − 25) + 51.62 × (31 − 28.75) + 63.62 × (34.5 − 31) + 113.1 × (35.5 − 34.5) = 7058.7 KN

2- Quake force parameters: a) From Jordan seismic hazard map, Karak is on the 2B zone and for this zone get Seismic zone factor Z = 0.2 from table (A-3) (Appendix) b) There is no available information about the soil, so use SD soil profile type from table (A-4) (Appendix). c) The important factor I = 1 from table (A-6) (Appendix). e) Response modification factor R: for Cast-in-place concrete silos and chimneys having walls continuous to the foundations R = 3.6 from table (A- 7) (Appendix). f) Seismic coefficient Ca and Cv: based on the soil profile type and the seismic zone factor Ca = 0.28 and Cv = 0.4 from table and table (A- 8) and (A-9) (Appendix). g) Fundamental period T: for the non-building structure (self supporting structure) like chimney, silo, minaret, we should use method 1 as an initial fundamental period 3/ 4 3/ 4 The initial fundamental period T = C t (hn ) = 0.0488 × (35.5) = 0.71sec ond

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3- Quake force initial base shear: Based on the initial fundamental period T = 0.75second C I   0 .4 × 1  V =  v  ×W =   × 7058.7 = 1104.65 KN  3.6 × 0.71   RT 

V min = (0.11C a I ) × W = (0.11 × 0.28 × 1) × 7058.7 = 217.4 KN

 2.5C a I   2.5 × 0.28 × 1  V max =   ×W =   × 7058.7 = 1372.53KN 3 .6    R  Extra load Ft = 0.07 T V = 0.07 × 0.71 × 1104.65 = 54.9 KN It should be less than 0.25 V = 0.25 x 1045.73 = 261.4KN This base initial base shear (V) will be initially distributed over the height as shown in figure below: according to the following equation:

Figure 7.2.10 Lateral force distributed

4- Then calculate the exact fundamental period using the structural properties and deformational characteristics of the resisting elements in a properly substantiated analysis (method 2).

We prepare a computer program to calculate the exact fundamental period T that needs trail and error. From the computer program we get T = 0.5 second. And the table as in the next page.

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Mass No. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Weight (KN/m') 0.000 266.88 266.88 266.88 266.88 266.88 266.88 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 87.180 87.180 222.640 222.460 222.460 222.460 222.460 87.180 87.180 87.180 156.830 156.830 156.830 156.830 51.620 51.620 51.620 63.620 63.620 63.620 63.620 113.100 113.100

Wi(KN) 130.44 266.88 266.88 266.88 266.88 146.784 113.451 211.337 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 222.460 166.845 77.410 65.385 154.82 222.46 222.460 222.460 166.845 77.410 65.385 65.385 61.003 117.623 156.830 137.226 65.264 32.263 51.620 57.620 63.620 63.620 47.715 44.180 56.550 28.275

Accumulative Wi(KN) 7051.786 6921.346 6654.466 6387.586 6120.706 5853.826 5707.042 5593.591 5382.254 5159.794 4937.334 4714.874 4492.414 4269.954 4047.494 3825.034 3602.574 3380.114 3157.654 2935.194 2712.734 2490.274 2323.429 2246.019 2180.634 2025.814 1803.354 1580.894 1358.434 1191.589 1114.179 1048.794 983.409 922.406 804.783 647.953 510.727 445.463 413.200 361.580 303.960 240.340 176.720 129.005 84.825 28.275

4

I(m )

Elevation(m)

Hi(m)

Wi*Hi

Fi(KN)

19.500 19.500 19.500 19.500 19.500 19.500 19.500 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420 15.420

-3 -2 -1 0 1 2 2.1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 17.5 18 19 20 21 22 23 23.5 24 25 25.5 26 27 28 28.75 29 30 31 32 33 34.000 34.500 35.000 35.500

0.000 1.000 2.000 3.000 4.000 5.000 5.100 6.000 7.000 8.000 9.000 10.000 11.000 12.000 13.000 14.000 15.000 16.000 17.000 18.000 19.000 20.000 20.500 21.000 22.000 23.000 24.000 25.000 26.000 26.500 27.000 28.000 28.500 29.000 30.000 31.000 31.750 32.000 33.000 34.000 35.000 36.000 37.000 37.500 38.000 38.500

0.000 266.880 533.760 800.640 1067.520 733.920 578.600 1268.022 1557.220 1779.680 2002.140 2224.600 2447.060 2669.520 2891.980 3114.440 3336.900 3559.360 3781.820 4004.280 4226.740 3336.900 1586.905 1373.085 3406.040 5116.580 5339.040 5561.500 4337.970 2051.365 1765.395 1830.780 1738.586 3411.067 4704.900 4254.006 2072.132 1032.416 1703.460 1959.080 2226.700 2290.320 1765.455 1656.750 2148.900 1088.588

0.000 3.702 7.403 11.105 14.806 10.179 8.189 17.996 22.100 25.257 28.414 31.571 34.728 37.885 41.043 44.200 47.357 50.514 53.671 56.828 59.985 47.357 22.521 19.487 48.366 72.643 75.771 78.928 61.564 29.113 25.054 25.982 24.674 48.409 66.771 60.372 29.407 14.652 24.175 27.803 31.601 32.504 25.055 23.512 30.497 15.449

Sum

110603.0

Table 7.2.3 (lateral force distribution)

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Hi (m) 0.00 1.00 2.00 3.00 4.00 5.00 5.10 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 15.00 16.00 17.00 18.00 19.00 20.00 20.50 21.00 22.00 23.00 24.00 25.00 26.00 25.50 27.00 28.00 28.50 29.00 30.00 31.00 31.75 32.00 33.00 34.00 35.00 36.00 37.00 37.50 38.00 38.50

Fx (KN) 0.00 3.31 6.62 9.93 13.24 9.10 7.18 15.73 19.31 22.07 24.83 27.59 30.35 33.11 35.87 38.63 41.39 44.15 46.90 49.66 52.42 41.39 19.68 17.03 42.24 63.46 66.22 68.98 53.80 25.44 21.90 22.71 21.56 42.31 58.35 52.76 25.70 12.80 21.13 24.30 27.62 28.41 21.90 20.55 26.65 13.50

Accumulative Fx (KN) 1371.76 1371.76 1368.45 1361.83 1351.90 1338.66 1329.56 1322.38 1306.66 1287.34 1265.27 1240.44 1212.85 1182.50 1149.39 1113.52 1074.90 1033.51 989.36 942.46 892.80 840.37 798.99 779.31 762.28 720.03 656.57 590.36 521.38 467.58 442.13 420.24 397.53 375.97 333.66 275.31 222.55 196.85 184.05 162.92 138.62 111.00 82.60 60.70 40.15 13.50

Moment (KN.m) 30510.16 29138.40 27769.94 26408.11 25056.20 23717.54 23584.58 22394.44 21087.78 19800.44 18535.16 17294.72 16081.87 14899.37 13749.98 12636.46 11561.57 10528.06 9538.69 5896.23 7703.44 6863.06 6463.57 6073.92 5311.64 4591.61 3935.04 3344.68 2823.30 2589.52 2368.45 1948.21 1749.44 1561.46 1227.80 952.49 785.57 736.36 552.32 389.40 250.78 139.78 57.18 26.83 6.75 0.00

Deflection (mm) 0.00 0.11 0.28 0.53 0.83 1.19 1.22 1.61 2.11 2.67 3.29 3.96 4.68 5.45 6.26 7.11 8.00 8.91 9.86 10.83 11.83 12.85 13.36 13.89 15.00 16.11 17.23 18.36 19.51 20.08 20.66 21.83 22.42 23.02 24.26 25.51 26.46 26.77 28.05 29.34 30.63 31.19 33.22 33.87 34.51 35.16

Table 7.2.3 (cont.)

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From the table in previous page the accumulative Fi = 1371.76KN Compare to the wind load base shear = 170.9KN the quake force will be used in analysis and design procedure.

Compute the safety factor against overturning: Over turning moment = 30510.16KN.m Restoring moment = restoring force x half the base width Self weight of the minaret (total gravity load) = 7058.7KN Assume footing dimension = 10 x 10 x 1 Weight of the soil surround by the minaret π  π  2 W =   × Dout − Din2 × 18 × 3 =   × 4.8 2 − 3.4 2 × 18 × 3 = 486.88 KN 4 4 Self Weight of column = (0.62 x π) / 4 x 25 x 28.75 = 203.22KN Total restoring weight =7058.7 + 10 x 10 x 1 x 25 + 468.88 + 203.22 = 10230.8KN Restoring moment = 10230.8 x 5 = 51154KN.m Re storing moment 51154 Factor of safety against overtruning = = = 1.7 > 1.5 ok . overturning moment 30510.16

[

]

[

]

Check the allowable deflection: From the above table the maximum deflection equal to 35.16mm The allowable deflection equal to 0.0025 × h = 0.0025 × 35.5 × 1000 = 88.75mm So it is ok.

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Design of minaret sections: From the table 7.2.3 we got the moment at each elevation and we will check the stresses and calculate the reinforcement required for each section as follow:

Note:

Allowable tensile strength of concrete (f t ) = 0.5 f c' = 0.5 25 = 2.5MPa Allowable compressive strength of concrete (f c ) = 0.45 f c' = 0.45 × 25 = 11.25MPa

Allowable shear strength of concrete (V) ≈ 0.09 f c' = 0.09 25 = 0.45MPa

Elevation from -3m to 2.1m Vertical reinforcement: Maximum Moment (M) = 30510.16KN.m. at elevation -3 below the ground. Weight above elevation -3 (W) = 7058.7KN. Eccentricity (e) = M / W = 3540.48 / 7058.7 = 4.3m W M × c 7058.7 30510.16 × 2.4 ± = ± = 891.6 ± 3755.1 A I 8 19.5 σ max = 4646.7 KN / m 2 = 4.65MPa < 0.45 × f c' → it is ok .

σ=

σ min = −2863.5KN / m 2 = −2.86 MPa " Tension stress" The tensile stress is slightly greater than the allowable tensile strength of concrete And the minimum reinforcement will be ok. Area of steel /m = 0.0015 Ag = 0.0015 x (1000 x 700) = 1050 mm2 The steel will be arranged into two layers Steel for one layer = 1050 / 2 = 525mm2 Choose Ø14 154 spacing = × 1000 = 293.3mm 525 Provide Ø14@ 250mm

Horizontal reinforcement: V 1371.76 Applied shear stress = = = 171.47 KN / m 2 = 0.171MPa A 8 This is smaller than the shear strength of concrete. So minimum horizontal reinforcement will be provided Area of steel /m = 0.002 Ag = 0.002 x (1000 x 700) = 1400 mm2 The steel will be arranged into two layers Steel for one layer = 1400 / 2 = 700mm2 Choose Ø14 154 spacing = × 1000 = 220mm 700 Provide Ø14@ 220mm

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Elevation from 2.1m to 17.5 m Maximum Moment (M) = 23584.58KN.m. at elevation 2.1m Weight above elevation 2.1 (W) = 5728.52KN. Eccentricity (e) = M / W = 23584.58 / 5728.52 = 4.1 m

σ=

W M × c 5728.52 23584.58 × 2.3 ± = ± = 759.75 ± 3517.8 A I 7.54 15.42

σ max = 4277.55KN / m 2 = 4.28MPa < 0.45 × f c' → it is ok . σ min = −2758.1KN / m 2 = −2.76 MPa " Tension stress" The tensile stress is slightly greater than the allowable tensile strength of concrete And the minimum reinforcement will be ok. Area of steel /m = 0.0015 Ag = 0.0015 x (1000 x 600) = 900 mm2 Choose Ø14 154 spacing = × 1000 = 171.11mm 900 Provide Ø14@ 150mm at the center of the section

Horizontal reinforcement: V 1329.65 Applied shear stress = = = 176.3KN / m 2 = 0.176 MPa A 7.54 This is smaller than the shear stress of concrete. So minimum horizontal reinforcement will be provided Area of steel /m = 0.002 Ag = 0.002 x (1000 x 600) = 1200 mm2 Choose Ø16 200 spacing = × 1000 = 166.6mm 1200 Provide Ø16@ 160mm

Note: Use these horizontal Reinforcements for all other sections.

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Elevation from 19m to 23.5 m Maximum Moment (M) = 5311.64KN.m. at elevation 19m Weight above elevation (W) = 2176KN. Eccentricity (e) = M / W = 5311.64 / 2176 = 2.44 m

σ=

W M × c 2176 5311.64 × 2.3 ± = ± = 288.6 ± 792.27 A I 7.54 15.42

σ max = 1080.9 KN / m 2 = 1.1MPa < 0.45 × f c' → it is ok . σ min = −503.67 KN / m 2 = −0.504MPa " Tension stress" There is a Tension stress and it is less than the allowable tensile strength of concrete So minimum reinforcement will be provided. Area of steel /m = 0.0015 Ag = 0.0015 x (1000 x 600) = 900 mm2 Choose Ø14 154 spacing = × 1000 = 171.11mm 900 Provide Ø14@ 150mm at the center of the section

Elevation from 17.5m to 19 m Maximum Moment (M) = 6463.57KN.m. at elevation 17.5m Weight above elevation 17.5 (W) = 2328.72KN. Eccentricity (e) = M / W = 6463.57 / 2328.72 = 2.8 m

σ=

W M × c 2328.72 6463.57 × 2 ± = ± = 665.35 ± 2154.52 A I 3 .5 6

σ max = 2819.87 KN / m 2 = 2.82MPa < 0.45 × f c' → it is ok . σ min = −1489.2 KN / m 2 = −1.49MPa " Tension stress" There is a Tension stress and it is less than the allowable tensile strength of concrete So minimum reinforcement will be provided. Area of steel /m = 0.0015 Ag = 0.0015 x (1000 x 300) = 450 mm2 Choose Ø12 113 spacing = × 1000 = 251.3mm 450 Provide Ø12@ 250mm at the center of the section

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193

Elevation from 23.5m to 25.5 m Maximum Moment (M) = 2589.52KN.m. at elevation 23.5m Weight above elevation 23.5 (W) = 1174.98KN. Eccentricity (e) = M / W = 2589.52 / 1174.98 = 2.2 m

σ=

W M × c 1174.98 2589.52 × 2 = 335.7 ± 863.2 ± = ± A I 3 .5 6

σ max = 1198.87 KN / m 2 = 1.2MPa < 0.45 × f c' → it is ok . σ min = −527.47 KN / m 2 = −0.53MPa " Tension stress" Provide Ø12@ 250mm at the center of the section

Note: For the vertical reinforcement: It is obvious that tension stress is small and the concrete can take it, so minimum reinforcement of 0.0015Ag will be provided for all sections.

Reinforcement Details: Section No. 1 2 3 4 5 6 7

Dimension Outer inner Diameter (m) Diameter (m) 4.8 3.4 4.6 3.4 4.0 3.4 3.0 2.4 3.0 2.4 3.0 2.4 3.0 1.8

Vertical Reinforcement

Horizontal Reinforcement

2 layerØ14@250mm 2layerØ14@ 220mm 1layer Ø14@150mm Ø16@ 160mm 1layer Ø12@250mm Ø16@ 160mm 1layer Ø12@250mm Ø16@ 160mm 1layer Ø12@250mm Ø16@ 160mm 1layer Ø12@250mm Ø16@ 160mm 1layer Ø14@150mm Ø16@ 160mm

Table 7.2.4 Minaret sections Reinforcement details

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This is a sample of the Reinforcement of the section (section No.1)

Figure 7.2.11 Reinforcement of section No.1

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195

7.3 Design of footing for minaret The footing maybe subjected to moment from all direction, so the footing will be designed to resist this moment by choosing square footing We assumed the footing dimension to be 10m x 10m x 1m

Loads: The normal force = 10230.8KN Bending moment = 30510.16 KN.m Check stress: M 30510.16 e= = = 2.98m P 10230.8 L 10 = = 1.67 m 6 6 Because e > L/ 6, so tension stress will be occurred P Mc 10230.8 30510.16 × 5 σ= ± = ± A I 100 833.33 Figure 7.3.1 Pressure distribution

σ 1 = 285.7 KN / m 2 σ 2 = −80.75 KN / m 2 When e > L/6, the length of the triangular distribution decreases to 1.5L _ 3e and the maximum pressure rises to: 2P 2 × 10230.8 q max = = = 337.65 KN / m 2 < 400 KN / m 2 → Ok . 1.5 B ( L − 2e) 1.5 × 10 × (10 − 2 × 2.98) As in figure 7.2.10 Put because we still have negative stress under the footing of the minaret and we should find any way to reduce it to zero or +ve value and we have two ways as follow: 1- Increase the size (Dimensions) of the footing: and this way isn’t economic 2- Use the pile foundation instead of the single footing and this is economic way and the design of the pile foundation for the Minaret as shown in Appendix B

Minaret design

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