Folien_SFPS_5

8/10/2019 Folien_SFPS_5

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Fig. 5.1

Prof. Dr. J. Tomas, chair of Mechanical Process Engineering

Fig_SFPS_5Storage and Flow of Particulate Solids Silo pressure calculations Prof. Dr. J. Tomas 05.05.2014 Figure 5.1

5. Silo and bunker pressure calculations5.1 shaft pressures

5.2 hopper pressures

5.3 wall thickness of concrete and metal sheet silos


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Fig. 5.2



Vertikal and Horizontal Pressures at Mass Flow Silo

height level of flattened

free bulk surface

intersection between

shaft and hopper

height level of

discharge opening

filling

discharging

siloheightH

silo pressures pv and ph

pv

pv

ph

ph

pv

ph

radial stress field r


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Fig. 5.3




4/13


5/13

Fig. 5.5



positi

ono

factive

sh

ear

pla

ne

Active and Passive Rankine's Stress State Limits

Yield locus: = tan i+c= tan i( + Z)

or R= sin i(M+ Z)

ph,a

activepassive

shea

rstress

tensile

strength Z

cohesion c

i

2,a

a=

i

4 2+

iR,a

2a

pv

1,a

yieldlocus

ph,p

1,p

p = i

4-

2

M,p =1,p+ 2,p

2

positionofpassiveshearplane

R,p=1,p 2,p

2

-

active passive

principal stresses

lateral or

horizontal

stress ratio

2,a= 1+ c1 - sin i

1 + sin i

2 cos i

1 + sin i

1,p= 2,p+ c1 + sin i

1 - sin i

2 cos i

1 - sini

lower limit for c= 0 upper limit for c= 0

ph,p

pv= = = tan2

1 + sin i

1 - sin i(

4+

i

2

p

ph,a

pv= = = tan2

1 - sin i

1 + sin i(

4-

i

2a

active

i4 2

+

pv

ph,a pv

ph,p

passive

i4 2

-

pv

ph,a

pvph,p

normal stress

given: pv= 1= b.g y

c, i

searched: 2 stress states which meet the yield condition

y

x

2,p


6/13

Fig. 5.6




7/13

Fig. 5.7




8/13

Fig. 5.8




9/13

Fig. 5.9



storage time tL = 22 h

mFill = 11,5 t

b = 0,8 t/m3

e = 33

w = 23 steel sheet (CF)

w

= 18 steel sheet (MF)

mass flow, switch load pn = (c4- c1) pn cos ,

TGL 32274/09

mass flow, switch load pn = b g (H or D),DIN 1055 part 6

0 5 10 15 20 25 30 35 40

wall normal pressures ph, pn in kPa

0

1

2

3

4

5

heigh

tH,

H'inm

mass flow (MF)

core flowflow

bondary

silo

G 807

pn

Wall Normal Pressures ph, pnof Wheat

measured, filling

measured, discharging

calculated, filling

calculated, discharging,

load factors c1= 1,4, c4= 3,0

calculated, DIN 1055 part 6

c3= 1.8(CF)

23

30

phD = 2,4 m

measurements according to Scholz


10/13

Fig. 5.10



Maximum Normal Pressures (Discharging) versus

Effective Angle of Internal Friction

1

2

3

4

5

6

7

8

9

10

11

w= 10

boundariesrough wall

w=10

2030 4050

0 10 20 30 40 50 60

p*h shaft

p*n hopper

20

30

405060

p*h

,max,

p*n,max

12

e in deg

Maximum Vertical Pressures (Filling) versus Effektive

Angle of Internal Friction

p*v,max

0 10 20 30 40 50 60

1

2

3

4

5

6

7

8

9

10

11

w= 10

20

30

40

50

boundary

rough wall

= 10 mass flow

=0,725

= 2,414

AUHD

shafthopper

w

=10

2030 40

50

12

13

14

e in deg

Maximum Vertical Pressures (Filling) versus

Wall Friction Angle

0 10 20 30 40 50 60

1

2

3

4

5

6

7

8

9

10

11

e=30 4050

12

13

isostatic pressure

boundaryrough wall

p*v,max

14

shaft

hopper

60

w in deg

Maximum Normal Pressures (Discharging)

versus Wall Friction Angle

1

2

3

4

5

6

7

8

9

10

11

30

4050

0 10 20 30 40 50 60


p*h,max

,

p*n,max

12

13

14

60e

=30

w in deg

605040

p*h shaft

p*n hopper

Maximum Wall Friction Loads (Discharging)

versus Effective Angle of Internal Friction

0 10 20 30 40 50 60

w=10

20

30

40

-50

p*h

shaft

p*n hopper

20

10

30

20

40

50

-10

e=30

1

2

1,5

2,5

0 0

0,5

0,6

0,7

0,4

0,3

0,2

0,1

p*

w,max

boundaries

rough wall

0,5

60

4050

p*

w,max

e in deg

p*

w,max

p*

w,max

Maximum Wall Friction Loads (Discharging) versus

Wall Friction Angle

e=3040

50

5040

0

0,5

1

1,5

2

2,5

0

0,5

0,6

0,7

0,4

0,3

0,2

0,1


0 10 20 30 40 50 60

60

60

e=

30

w in deg

p*w shaftp*w hopper

p* = p U

bg A

.. .


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Fig. 5.11



1

2

3

4

5

6

78

p*v,max

9

10

11

w= 10

20

30

40

50

boundary

rough wall

= 30 core flow

=0,725

= 2,414

A

UH

D

shaft

hopper

w=10

2030 40

50

0 10 20 30 40 50 60e in deg

Maximum Vertical Pressures (Filling) versus Effektive

Angle of Internal Friction

1

2

3

4

5

6

7

8

9

10

11

e=30 4050

0 10 20 30 40 50 60

12

13

shaft

hopper

isistatic pressure

boundary

rough wall

p*v,max

60

w in deg

Maximum Vertical Pressures (Filling)


1

2

3

4

5

6

7

8

9

10

11w= 10

boundaries

rough wall

w

=10

2030

4050

0 10 20 30 40 50 60

p*h shaft

p*n hopper

20304050

60

p*h,max,

p*n,max

60

e in deg

Maximum Normal Pressures (Discharging) versusEffective Angle of Internal Friction

1

2

3

4

5

6

7

8

9

10

11

e

=30 40 50

0 10 20 30 40 50 60

e

=30

4050

60

60


p*h,max,

p*n,max

w in deg

p*h shaft

p*n hopper

Maximum Normal Pressures (Discharging)versus Wall Friction Angle

e=30 40 50

50

40e=

30

0 0

Ip*

w,max

0,5

0,6

0,7

0,8

0,9

1,0

0,4

0,3

0,2

0,10,5

1

1,5

2

2,5

3

p*

w,max boundaries

rough wall

0 10 20 30 40 50 60

60

60

w in deg

p*h shaft

p*n hopper



3

w=10 20

30

40

50

20

10

30

2030

40

50

10

e=60

1

2

1,5

2,5

0

0 10 20 30 40 50 600

0,5

0,6

0,7

0,8

0,9

1,0

0,4

0,3

0,2

0,1

Ip*

w,max

boundaries

rough wall

0,5

e in deg

p*

w,max

p*h shaft

p*n hopper


versus Effective Angle of Internal Friction

p* = p U

bg A

.

. .


12/13

Fig. 5.12



Necessary Steel Reinforcement of Concrete ASt

Di,1

pn

ASt

H

Di

Di,2

pn

Di= Di,1+ Di,2 2

(1)

(2)

(3)

(4)

load factors according to TGL 32 274/09

( ) coscc1c14or3j +=

cj= c1 or c3or c4

with c1= 1,2 ... 1,6 c3= 1,7 ... 2,1

c4= 2,1 ... 4,0

hopper

shaft

dischargingcore flow

mass flow

H2cos/DHp0FStAin ==

cos2

cDpA

StF,

jsin

St

=

= iwF

StF,w

js

2

ib

StD

H

tan4exp1tan8

cDg

A

and for the cylindrical shaft:


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