Example 3 Rectangular Silo

ENGC6353 Dr. Mohammed Arafa

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

columns. The Roof load ( DL = 150 kg/m2 and LL= 100 kg/m2

Use

). ' 2 2350 / , 4200 /c yf kg cm f kg cm= =

Solution

For Peas 3

'

800 /250.296

o

kg mγ

φ

µ

=

=

=

b=6m

a=4m

An Above Hopper

b=6m

a=6m

Openning0.5x0.5m

30m

5m

7m

φ50cm

3m


Assume angle of response = =2523tan 25 1.4 1.03

1 sin 25 0.5774 1.0

4 4' 1.0

42 4 6 '' 4.8 1.2

4 6 4

s s

a

b

b

h m h m

kaR m

aR m

aa R m

ρ φ

= = ⇒ ≅

= − =

= = =

= =

× ×= = = =

+

Overpressure Factor C

1 d

d

/ 40 /10 4upper H c 1.5lower 2/3 H c 1.85Hooper 1.5d

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/m0.577 4.65 2.7 t/m

( 1.2) 5.53t/m0.577 5.53 3.2 t/m

kY R

Y mRq ek

p kqFor short wall R q

p kqFor long wall R q

p kq

µγµ

−

= =

= −

=

= =

= = × =

= =

= = × =

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 30.2 ton/m

Long Wall 1.7 1.85 2.7 4 2 17.0ton/ma u

b u

F

F

= × × =

= × × =

Frame action analysis using moment distribution Analysis

Assume wall thickness h=30cm

25cm


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

Using symmetry

a

2 2Short Wall =0.465I4.3

2 2Long Wall =0.317I6.3

0.465DF 0.60.465 0.3170.4

aa

bb

b

I IKLI IK

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

2a,-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

532

, 2

532

, 2

6 2 2 350 37.4 /

For long Wall

6 4.2 101 17 10 11 /1.7 1.85 30 100 30 100

For short Wall

6 5.9 101 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 Mf f f kg cmbt bt

f kg cm f

f kg cm f

= + ≤ = = =

×× = + = <

×

×× = + = <

× s is oK

Design for Reinforcement

Long Wall

negative moment M

Check for small eccentricity

-ve

3.2 t/m2

2.7

t/m2

6.3m

4.3m

7.9 t.m

-8.0

-8.0

-1.5

25cm


( )4.2 10024.7 '' 15 5.7 9.3

17 2u

u

M he dF

= = = > − = − =

Small eccentricity approach can not be used

32

st

'' 15 5.7 9.32

Direct tension reinforcement17 10A 4.5 /

0.9 4200y

he d

T cm mfφ

= − = − =

×= = =

×

Bending Moment Reinforcement

( )( )

( )( )( )

',

5

2

2( )

2,

4.2 17 9.3 /100 2.6 .d=30-5.7=24.3

2.61 10 2.60.85 350 1 1 0.001174200 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 muse 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 6.320.85 350 1 1 0.002894200 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 muse cm

ρ

φ

+

+

+

= − × =

⋅⋅ = − − =⋅ ⋅

= =

= + =

Design for Short Wall

negative moment M

Check for small eccentricity

-ve

( )5.9 10019.5 '' 15 5.7 9.3

30.2 2u

u

M he dF

= = = > − = − =

Small eccentricity approach can not be used


32

st

'' 15 5.7 9.32

Direct tension reinforcement30.2 10A 8.0 /0.9 4200y

he d

T cm mfφ

= − = − =

×= = =

×

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 3.00.85 350 1 1 0.001374200 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 muse 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 3.6 t/m

1.35 5.53 7.47 t/m

1.35 0.577 5.53 4.3 t/m

a des

a des

b des

b des

q

p

q

q

= × =

= × × =

= × =

= × × =

1

1

Angleof slopes3tan 48

3 0.33tan 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/ma des

b des

q

qα

α

= + =

= + =


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/mtau

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 38L

L

W ton

W ton

π

π

= × =

= × × =

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

can be calculated as:

( )

( )

2

2

2

1 4 3 621 6 2 62

61/ 4

a

b

a b

a b

A m

A m

A A mc c

= × =

= × =

= == =

( ) ( ) ( )( )

( ) ( ) ( )( )

,

,

1.7 1.4 1.7 0.25 60 6 6.28 1.4 0.25 3834.6

sin 4sin 48

1.7 1.4 1.7 0.25 60 6 7.47 1.4 0.25 3822

sin 6sin 60.5

a L a a des b gmau

a

b L b b des b gmbu

b

c W A q c WF ton

a

c W A q c WF 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 nx = -0.209 and ny

( )( )( )( )

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

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


Design of the edge beam

Dowels are provided to transfer the vertical loads from hopper edge beam into the vertical walls

32

,

sin 34.6sin 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 simplify 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

Example 3 Rectangular Silo

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