Design Examples 1 2 of Circular Silo 1 PDF [PDF]

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Design Example 1



Design the wall and hopper of a wheat silo with an internal diameter of 10 meter and with the height of cylindrical portion of 40 m. The central hopper is supported by eight columns monolithic with the lower walls. The Roof load ( DL = 150 kg/m2 and LL= 100 kg/m2) Use the following parameter f c'  350 kg / cm 2 f y  4200 kg / cm 2



  800 kg / m 3   25o  '  0.444



1.5m



20m



10m



60m



40m



D= 10m



1.5m



ENGC6353



Dr. Mohammed Arafa



Page 1



Solution Assume angle of response  = =25 hs  5 tan 25  2.33







2 hs 3



1.5m



k  1  sin 25  0.577



 R



4D 2 



 D 



 D 4  10 / 4  2.5m



H / D  40 /10  4 Overpressure Factor Cd



H / D  40 /10  4 From Table 1 upper H1



cd  1.5



lower 2/3 H cd  1.85 Hooper use cd  1.5 ACI313-4.4.3.2 allows to use cd =1.35 for the Hooper



At the bottom of the silos At the bottom of the silos Y=40-1.5=38.5m q



R    ' kY 1e    'k



R



  7.65 t/m 2 



P  kq  4.42 t/m 2



Ring Tension C d Pu D 1.85 1.7  4.42   10   69.5 ton 2 2 69.5 T A st    18.4 cm 2 /m ie. 9.2 cm 2 /m for each side  f y  0.9  4200



T 



use [email protected] If slip forming will be used:



A st 



T 69.5   19.4 cm 2 /m ie. 9.7 cm 2 /m for each side 0.95  f y  0.95  0.9  4200



Minimum Thickness 0.0003  200 104  1680  8  35    4.42 10   ε sh E s  f s  nfct t= T    =7.5 cm 100f s fct 100  1680  35 2   ENGC6353



Dr. Mohammed Arafa



Page 2



The thickness of silo walls shall be not less than 150 mm for cast-in-place concrete. Use Wall thickness t=20cm Vertical Loads



Weight of the wall Wt  2.5  0.2  60  30 ton / m Friction



V   Y  q  R



atY  38.5



V   0.8  38.5  7.65   2.5  57.9 ton/m



Roff



0.15   D 2 / 4 =0.15 10 4   0.375 ton / m D LL  0.10 10 / 4   0.25 ton / m DL=



Pver  1.7  57.9  0.375   1.4  30  0.25   141.4 ton Check for Buckling



141.4  101 kg/cm 2 0.7  20  100  0.55 f c'  0.55  0.7  350  134.75  f c ,vert



f c ,vert  Pnw



The buckling does not control



A st  0.002  20 100  4 cm 2 /m



5.0m



Design for the Hopper q y  q0   hy at h y  1.0 m 4.1



q y  7.65  0.8 1  8.45 t/m 2 W L = weight of the material in hopper 0.8  2 2 WL= 4.1   0.75    5.8   84.4 ton   3  2.5 Wg=  2  4.1 0.2  2  0.75  0.2  5.8  29.5 ton 3 Merdional forces and required reinforcing



5.8m



0.75



 qy D  Wg  WL  Fmu  1.7     1.4    4sin   D sin     D sin   1.5  8.45  2  4.1    84.4 29.5 Fmu  1.7     1.4    59.2 ton/m 4sin 60   2  4.1 sin 60      2  4.1 sin 60  59.2 A st   16.5cm 2 /m 0.9  4200



ENGC6353



Dr. Mohammed Arafa



Page 3



Hoop Reinforcement 1.5  q  D  Ftu  1.7    2sin   q  P sin 2   q cos 2  where P  kq  0.577  8.45  4.87 t/m 2 q  4.87 sin 2 60  8.45cos 2 60  5.765t/m 2 assume  '  25 or q   p n 



q y tan  tan   tan  '







8.45 tan 30  4.67t/m 2 tan 30  tan 25



use q  p n  4.67t/m 2 1.5  5.765   2  4.1  Ftu  1.7    59.6 ton/m 2sin 60   69.6  19.4 cm 2 /m A st, hopper  0.9  4200



ENGC6353



Dr. Mohammed Arafa



Page 4



Design of the Circular Beam a1  100



33cm



b1  90 a2  100



33



b 2  57



28.5 100cm



28.5



r=467cm



100



A r  6150



R=4.67m 32.9



32.9 90



x  32.9cm , y  42.3cm a  87.2cm b  74.5cm M t  0.285  684  19.5 t .m



90cm



R  5  32.9 /100  4.67m q y  7.65  0.8 100  42.3 /100  8.1 t / m 2 W L  0.8



3



 4.67



2



 0.752   6.24  116.5ton







 2  4.1 0.2  2  0.75  0.2   5.8  29.5ton 3  q D  Wg  WL   1.7  y   1.4   D sin    4sin   D sin   



W g  2.5 Fmu







1.5  8.1 10   116.5  29.5 Fmu  1.7     1.4    68.4 ton  10  sin 60   4sin 60   10  sin 60  Fx  Fmu cos   68.4 cos 60  34.2ton Fy  0.615  2.5 1.4  68.4sin 60  61.5ton



Location



Shear



Comp. Force Bending Moment due to Fx due to Mt Due to Fy



Mt due to Fy



Support



112.5



159.4



91



69.4



0



Midspan



0



159.4



91



34.86



0



64.7



159.4



91



0



5.34



9 33 form support



ENGC6353



Dr. Mohammed Arafa



Page 5



Example 2 If the silo’s bottom in Example 1 is a circular slab with central opening on the lower walls and carrying hopper forming concrete fill.



Load on the slab a) Load from wheat in Hopper (assume uniform)  3  5   5  0.8  2



WL 



  5



2



 1.3 t/m 2



at y=38.5 m ie. h=40m q=7.65 t/m2



10m



p=kq=4.42 t/m2 Total LL=7.65+1.3=9 t/m2



40m



b) Dead Load



5m



Weight of Hopper forming fill 2 3  5   5  2.5 



50cm



2



Wg 



  5



2



 8.33 t/m 2



7m



Slab weight assume 40 cm slab thickness W slab  0.4  2.5  1.0 t/m 2 DLtotal  8.33  1.0  9.33 t/m 2 W u  1.7  9  1.4  9.33  28.4 t/m 2 Design of the slab Holes



Slabs with holes may be designed in two ways 



By computing bending moments for slabs with no holes and reinforcing with a steel member with adequate strength and of stiffness equal to that of removed slab.







By considering the hole and reinforcing for bending moments obtained using tables or Timoshenko equations.



ENGC6353



Dr. Mohammed Arafa



Page 6



Check for shear on slab



Vu 



28.4  5  0.35  2  5  0.35 



2



 66 ton



V c  0.53  0.85  300  35   2  5  0.35    798 ton V u



Total reaction at the bottom wall must includes



From Roof, Material above the Hopper, Material in the Hopper, Hopper filling form, Bottom Slab, Upper Wall, and Lower Wall



ENGC6353



Dr. Mohammed Arafa



Page 7