Example 3 Rectangular Silo

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Design Example 3

Rectangular Silo

Design a single rectangular concrete silo for storing peas. The bottom is a symmetrical Design a single rectangular concrete silo for storing peas. The bottom is a symmetrical pyramidal Hopper. The silo walls rest on the Hopper base which is supported by four pyramidal Hopper. The silo walls rest on the Hopper base which is supported by four columns. The Roof load ( DL = 150 kg/m

columns. The Roof load ( DL = 150 kg/m22and LL= 100 kg/mand LL= 100 kg/m22 Use Use ). ). ' ' 2 2 22 3 3550 / 0 / , , 4422000 / 0 / c c yy f f

=

kg kg cm cm f f

=

kg kg cmcm Solution Solution For Peas For Peas 3 3 '' 8 8000 0 / / 25 25 0.296 0.296 o o kg kg mm γ γ φ φ µ µ

=

b=6m b=6m a a = = 4 4 m m An Above Hopper An Above Hopper b=6m b=6m a=6m a=6m Openning0.5x0.5m Openning0.5x0.5m 30m 30m 5m 5m 7m 7m φφ50cm50cm 3m 3m

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Assume angle of response = =25 2 3 tan 25 1.4 1.0 3 1 sin 25 0.577 4 1.0 4 4 ' 1.0 4 2 4 6 ' ' 4.8 1.2 4 6 4 s s a b b h m h m k a R m a R m a a R m ρ φ

=

⇒

≅

= −

=

= = =

=

× ×

=

+

 Overpressure Factor C 1 d d / 40 /10 4 upper H c 1.5 lower 2/3 H c 1.85 Hooper _d 1.5 H D c

=

d

At the bottom of the silos

( ' )

2

2 2

2

At the bottom of the silos 30 -1.0 29.0 1 ' ( 1.0) 4.65 t/m 0.577 4.65 2.7 t/m ( 1.2) 5.53t/m 0.577 5.53 3.2 t/m kY R Y m R q e k p kq

For short wall R q

p kq For long wall R q

p kq µ γ µ −

=





=

_

−

_

=

= =

× =

=

= =

× =

Vertical Loads Due to Friction

(

)

(

)

(

)

Short Wall 0.8 30 4.65 1.0 19.35 ton Long Wall 0.8 30 5.53 1.2 22.16ton Friction V Y q R V V γ

=

−

= × −

× =

= × −

× =

Wall tension and bending moment

(

)

(

)

, ,

Short Wall 1.7 1.85 3.2 6 2 30.2ton/m Long Wall 1.7 1.85 2.7 4 2 17.0ton/m

a u b u F F

=

×

=

×

=

Frame action analysis using moment distribution Analysis Assume wall thickness h=30cm

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The moment distribution is computed for an idealized rectangular frame 6.3 by 4.4 m Using symmetry

a

2 2

Short Wall =0.465I

4.3 2 2

Long Wall =0.317I

6.3 0.465 DF 0.6 0.465 0.317 0.4 a a b b b I I K L I I K L DF

=

≅

+

≅

Short Wall Long Wall

DF 0.6 0.4

FEM 4.16 -10.6

Balancing 3.86 2.58

FINAL 8.0 -8.0

Assume fillit (hunch) at the corner =25cm

Negative moment will be calculated at the face of the hunch

2 b,-ve 2 a,-ve M 8.0 3.2 0.4 / 2 10.1 0.4 4.2 . M 8.0 2.7 0.4 / 2 5.8 0.4 5.9 . t m t m

= + ×

− × =

= + ×

− × =

Check for thickness

( )( )

(

)

( ) ( )

( )( )

(

)

( ) ( )

' 2 , 2 5 3 2 , 2 5 3 2 , 2 6 2 2 350 37.4 / For long Wall

6 4.2 10 1 17 10

11 / 1.7 1.85 30 100 _{30 100}

For short Wall

6 5.9 10 1 30.2 10

15.7 / 1.7 1.85 30 100 _{30 100}

The wall thicknes

t b r c t b r t b r T M f f kg cm bt bt kg cm f kg cm f

= + ≤ =

=



_×







=

+

=

<

×

_

_

_

_



_×







=

+

=

<

×

_

_





sisoK Design for Reinforcement Long Wall

negative moment M _-ve

3.2 t/m2 2 . 7 t / m 2 6.3m 4.3m 7.9 t.m -8.0 -8.0 - 1 . 5 25cm

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(

)

4.2 100 24.7 '' 15 5.7 9.3 17 2 u u M h e d F

= =

= > − = − =

Small eccentricity approach can not be used

3

2 st

'' 15 5.7 9.3 2

Direct tension reinforcement 17 10 A 4.5 / 0.9 4200 y h e d T cm m f φ

= − = − =

×

=

×

Bending Moment Reinforcement

( )

(

)

(

)( )( )

' , 5 2 2 ( ) 2 , 4.2 17 9.3/100 2.6 . d=30-5.7=24.3 2.61 10 2.6 0.85 350 1 1 0.00117 4200 _{100 24.3 350} 0.00117 100 24.3 2.85 4.5 2.85 7.35 / 14@20 u ve ve s ve s total M t m A cm A cm m use cm ρ φ − − −

= − ×

=



_⋅



⋅





=

−

=

⋅









=

+

=

Design for Positive Moment at Midspan

(

)

(

)

(

)( )( )

' , 5 2 2 ( ) 2 , 7.9 17 9.3/100 6.32 . d=30-5.7=24.3 2.61 10 6.32 0.85 350 1 1 0.00289 4200 _{100 24.3 350} 0.00289 100 24.3 7.0 4.5 7 11.5 / 16@15 u ve ve s ve s total M t m A cm A cm m use cm ρ φ + + +

= − ×

=



_⋅



⋅





=

−

=

⋅









=

+ =

Design for Short Wall negative moment M

Check for small eccentricity

-ve

(

)

5.9 100 19.5 '' 15 5.7 9.3 30.2 2 u u M h e d F

= =

= > − = − =

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3

2 st

'' 15 5.7 9.3 2

Direct tension reinforcement 30.2 10 A 8.0 / 0.9 4200 y h e d T cm m f φ

= − = − =

×

=

×

Bending Moment Reinforcement

( )

(

)

(

)( )( )

' , 5 2 2 ( ) 2 , 5.9 30.2 9.3/100 3.0 . d=30-5.7=24.3 2.61 10 3.0 0.85 350 1 1 0.00137 4200 _{100 24.3 350} 0.00137 100 24.3 303 8.0 3.3 11.3 / 16@15 u ve ve s ve s total M t m A cm A cm m use cm ρ φ − − −

=

−

×

=



_⋅



⋅





=

−

=

⋅









=

+

=

Design at Mid-span

Design of the Hopper Walls

The pressure changes very little with depth of the hopper, so use the pressure at the top of the hopper with Cd=1.35

2 , 2 , 2 , 2 , 1.35 4.65 6.28 t/m 1.35 0.577 4.65 3.6t/m 1.35 5.53 7.47 t/m 1.35 0.577 5.53 4.3t/m a des a des b des b des q p q q = × = = × × = = × = = × × = 1 1 Angleof slopes 3 tan 48 3 0.3 3 tan 60.5 2 0.3 a a α α − −





=

_

_

=

−









=

_

_

=

−





  2 2 2 , 2 2 2 , 3.6sin 48 6.28cos 48 4.8t/m 4.3sin 60.5 7.47cos 60.5 5.0 t/m a des b des q q α α = + = = + =

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Horizontal Ultimate tensile forces

( )( ) ( )

2 2 1.7 5.0 6/2 sin 48 =19.0t/m 1.7 4.8 4/2 sin 60.5 =14.2t/m tau tbu F F = =

The own weight of the Hopper and its contents

(

)( )( )

( )( )(

)

3 4 6 3 0.8 60 3 4 6 3 0.2 2.5 38 L L W ton W ton π π

=

×

=

×

=

For simplicity neglect the opening area at the bottom of the hopper. Hopper side Aaand Ab

can be calculated as:

(

)

(

)

2 2 2 1 4 3 6 2 1 6 2 6 2 6 1/ 4 a b a b a b A m A m A A m c c

=

× =

=

× =

=

(

)

(

) ( )( )

(

)

(

) ( )( )

, , 1.7 1.4 1.7 0.25 60 6 6.28 1.4 0.25 38 34.6 sin 4 sin 48 1.7 1.4 1.7 0.25 60 6 7.47 1.4 0.25 38 22 sin 6 sin 60.5 a L a a des b g mau a b L b b des b g mbu b c W A q c W F ton a c W A q c W F ton b α α + + × + × + = = = + + × + × + = = =

Hopper wall bending can be computed using Tables for triangular slabs: For Hopper wall A

(

)

2 2 4.3 6.3 / 2 3 4.35 / 1.0 a m c m a c

=

+

=

≅

From table 16.4 in Appendix At the centre of the top edge n x = -0.209 and n y

( )

(

)

( )

(

)

2 2 1.7 0.209 4.8 4.3 / 64 0.493 . 1.7 1.255 4.8 4.3 / 64 2.89 . xau yau M t m M t m

=

=-1.255

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Design of the edge beam

Dowels are provided to transfer the vertical loads from hopper edge beam into the vertical walls 3 2 , sin 34.6 sin 48 25.7 / 25.7 10 6.8 / 0.9 4200 mau a st dowels T F ton m A cm m α

=

×

=

×

Since the edge beam is to be joining the vertical wall using dowels. The upper wall shear and horizontal components of the hopper are assumed to be in equilibrium. Thus no horizontal load is carried by the edge beam. Its only purpose is to simplif y construction. Minimum longitudinal steel and shear stirrups are provided

Vertical Wall

The vertical walls are analyzed as deep girder (strut and tie analysis can be used) to carry vertical the following vertical loads:

From dowel 25.7 ton/m Friction 1.7(19.35) = 32.9 ton/m Wall weight,

1.4(2.5)(0.3)(30)= 31.5 ton/m Total = 90 ton/m