Example 3 Rectangular Silo

Design Example 3 Rectangular Silo 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 2

2

columns. The Roof load ( DL = 150 kg/m and LL= 100 kg/m ). Use  f c' = 350 kg / cm 2 , f y = 4200 kg / cm   2

b=6m 30m

  m    4   =   a

An Above Hopper

5m

b=6m

3m 7m

a=6m

Openning0.5x0.5m

φ50cm

Solution

For Peas γ  = 800 kg / m   3 φ  = 25o µ ' = 0.296

ENGC6353

Dr. Mohammed Arafa

Page 1

Assume angle of response  ρ =φ =25 hs = 3 tan 25 = 1.4 m

⇒

2 3



hs ≅ 1.0 m

k  = 1 − sin 25 = 0.577  R a =

a

=

4

= 1.0m 4 4 a'  R b = = 1.0m 4 2× 4×6 a'= = 4.8 4+6

Rb =

a'

4

= 1.2 m

Overpressure Factor C  d  25cm

 H  / D = 40 /10 = 4

upper H1 c d = 1.5 lower 2/3 H

c d = 1.85

Hooper c d  = 1.5

At the bottom of the silos At the bottom of the silos Y = 30 -1.0 = 29.0m q=

γ  R

1 − e −( µ ' kY µ ' k  

R)

 

 p = kq For short wall ( R = 1.0)

2 q = 4.65 t/m

 p = kq = 0.577 × 4.65 = 2.7 t/m For long wall ( R = 1.2)

q = 5.53t/m

2

2

 p = kq = 0.577 × 5.53 = 3.2 t/m

2

Vertical Loads Due to Friction Friction

V = ( γ Y − q ) R

Short Wall

V  = ( 0.8 × 30 − 4.65 ) ×1.0 = 19.35 ton

Long Wall

V  = ( 0.8 × 30 − 5.53 ) ×1.2 = 22.16ton

Wall tension and bending moment

Short Wall

F a ,u = 1.7 (1.85 × 3.2 ) × 6 2 = 30.2ton/m

Long Wall

F b ,u = 1.7 (1.85 × 2.7 ) × 4 2 = 17.0ton/m

Frame action analysis using moment distribution Analysis

Assume wall thickness h=30cm ENGC6353

Dr. Mohammed Arafa

Page 2

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

Short Wall

K b =

Long Wall DFa =

0.465

0.465 + 0.317  DF b ≅ 0.4

2 I  La

2 I  Lb

2I 

=

4.3

=0.465I    2

=

2I  6.3

  m    /    t    7  .    2

=0.317I

4.3m 2

3.2 t/m

≅ 0.6 6.3m 7.9 t.m

Short Wall DF

Long Wall

0.6

0.4

FEM

4.16

-10.6

Balancing

3.86

2.58

FINAL

8.0

-8.0

-8.0 -8.0    5  .    1   -

Assume fillit (hunch) at the corner =25cm Negative moment will be calculated at the face of the hunch M b,-ve = 8.0 + 3.2 × 0.4 2 / 2 −10.1 × 0.4 = 4.2 t . m M a,-ve = 8.0 + 2.7 × 0.4 / 2 − 5.8 × 0.4 = 5.9 t . m 2

Check for thickness T 6M    2 + 2 ≤  f r = 2 f c ' = 2 350 = 37.4 kg / cm t ,b = bt bt   For long Wall

25cm

5  17 ×103 6 ( 4.2 ×10 )    = 11kg / cm 2 < f r   = + 2 1.7 ×1.85  (30 )(100 ) ( 30 ) (100 )   

1

t ,b

For short Wall

 30.2 ×10 3 6 ( 5.9 ×10 5 )  2   = 15.7 kg / cm   = + < f r   2 1.7 ×1.85  (30 )(100 ) ( 30) (100 )    1

t ,b

The wall thicknessisoK Design for Reinforcement Long Wall

negative moment M -ve Check for small eccentricity ENGC6353

Dr. Mohammed Arafa

Page 3

e

=

 M u F u

=

4.2 (100 ) 17

h

= 24.7 >

2

− d  '' = 15 − 5.7 = 9.3

Small eccentricity approach can not be used h e = − d  '' = 15 − 5.7 = 9.3 2 Direct tension reinforcement Ast =

T 

φ  f  y

=

17 ×103 0.9 × 4200

= 4.5 cm 2 / m  

Bending Moment Reinforcement '  M u , −ve = 4.2 − 17 × 9.3/100 = 2.6 t . m

d=30-5.7=24.3  ρ −ve

2.61⋅105 ( 2.6 )  0.85 ⋅ 350  1 − 1 −  = 0.00117 = 2 4200  100 ⋅ ( 24.3 ) ⋅ 350  

 As ( −ve ) = ( 0.00117 )(100 )( 24.3 ) = 2.85 cm

2

 As ,total = 4.5 + 2.85 = 7.35 cm /  m 2

use φ 14@20cm

Design for Positive Moment at Midspan  M u' , +ve = 7.9 − 17 × 9.3/100 = 6.32 t. m

d=30-5.7=24.3  ρ +ve

2.61⋅105 ( 6.32 )  0.85 ⋅ 350  1 − 1 −  = 0.00289 = 2 4200  ⋅ ⋅ 100 24.3 350  ( ) 

2  As ( +ve ) = ( 0.00289 )(100 )( 24.3 ) = 7.0 cm 2  As ,total = 4.5 + 7 = 11.5 cm /  m

use φ 16@15cm

Design for Short Wall

negative moment M -ve Check for small eccentricity e=

 M u F u

=

5.9 (100 ) 30.2

= 19.5 >

h

2

− d  '' = 15 − 5.7 = 9.3

Small eccentricity approach can not be used

ENGC6353

Dr. Mohammed Arafa

Page 4

e=

h

− d  '' = 15 − 5.7 = 9.3 2 Direct tension reinforcement T 

Ast =

φ  f  y

=

30.2 ×10

3

0.9 × 4200

= 8.0 cm 2 / m  

Bending Moment Reinforcement '  M u , −ve = 5.9 − 30.2 × 9.3/100 = 3.0 t. m

d=30-5.7=24.3  ρ −ve

2.61⋅105 ( 3.0 )  0.85 ⋅ 350  1 − 1 −  = 0.00137 = 2 4200  ⋅ ⋅ 100 24.3 350  ( ) 

2  As ( −ve ) = ( 0.00137 )(100 )( 24.3 ) = 303 cm 2  As ,total = 8.0 + 3.3 = 11.3 cm /  m

use φ 16@15cm

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 1.35 × 4.65 = 6.28 t/m

2

q a ,des

=

 p a ,des

=

1.35 × 0.577 × 4.65 = 3.6t/m

q b ,des

=

1.35 × 5.53 = 7.47 t/m

q b ,des

=

1.35 × 0.577 × 5.53 = 4.3t/m

2

2

2

Angleof slopes α a α a

3  = tan −1   = 48  3 − 0.3  3  = tan −1   = 60.5  2 − 0.3  

2



2

2

q α a ,des

=

3.6sin 48 + 6.28cos 48 = 4.8t/m

q α b ,des

=

4.3sin 60.5 + 7.47cos 60.5 = 5.0 t/m

ENGC6353

2

2

2

Dr. Mohammed Arafa

Page 5

Horizontal Ultimate tensile forces F tau

=

1.7 ( 5.0 )( 6/2 ) sin ( 48) =19.0t/m

F tbu

=

1.7 ( 4.8 )( 4/2) sin ( 60.5) =14.2t/m

2

2

The own weight of the Hopper and its contents W  L = π  3 ( 4 × 6 )( 3 )( 0.8 ) = 60 ton W  L = π  3 ( 4 × 6 )( 3 )( 0.2 × 2.5 ) = 38 ton

For simplicity neglect the opening area at the bottom of the hopper. Hopper side  A a and  A b can be calculated as:  A a =  Ab =

1 2 1 2

( 4 × 3) = 6 m 2 (6 × 2) = 6 m 2

 A a = Ab = 6 m

2

c a = cb = 1/ 4 F

mau =

Fmbu

=

(

1.7 c aW L

+

A aq a , des ) + 1.4c bW g   a sin α a

(

1.7 cbW L

+

Ab q b ,des ) + 1.4c bW g  b sin α b

=

=

1.7 ( 0.25 × 60 + 6 × 6.28 ) + 1.4 ( 0.25 )( 38 ) 4 sin 48 1.7 ( 0.25 × 60 + 6 × 7.47 ) + 1.4 ( 0.25 )(38 ) 6 sin 60.5

=

=

34.6 ton

22 ton

Hopper wall bending can be computed using Tables for triangular slabs: For Hopper wall A a = 4.3m c=

2

( 6.3 / 2 ) + 32 = 4.35m

a / c ≅ 1.0

From table 16.4 in Appendix At the centre of the top edge n x = -0.209 and n y =-1.255

( = 1.7 (1.255 ( 4.8 ) 4.3

) / 64 ) = 2.89 t . m

2  M  xau = 1.7 0.209 ( 4.8 ) 4.3 / 64 = 0.493 t . m

 M  yau

2

This slab is to be designed for bending and tensile force similarly as shown above.

ENGC6353

Dr. Mohammed Arafa

Page 6

Design of the edge beam

Dowels are provided to transfer the vertical loads from hopper edge beam into the vertical walls T = Fmau sin α a = 34.6 sin 48 = 25.7 ton /  m  A st ,dowels =

25.7 ×103 0.9 × 4200

  = 6.8 cm 2 / m

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)= Total =

ENGC6353

31.5 ton/m 90 ton/m

Dr. Mohammed Arafa

Page 7