CHAPTER5 Analysis

7/31/2019 CHAPTER5 Analysis

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ANALYSIS OF VIBRATING-GRIZZLY FEEDER

(1) Gravity Analysis No load applied Only load due to gravity is applied. Boundary conditions are shown by blue colour in below figure.

Fig. 1 Boundary condition

Fig. 2 Equivalent (von-Mises) Stress


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Fig.3 Total deformation

Result achieved:

(i) Total deformation: 0.023588mm(ii) Von-mises stress: 4.44 MPa


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(2)Modal Analysis (No loading condition)Modal Analysis is done to find natural frequency of the system.

Same boundary condition applied as above. Total 10 natural frequencies are found out.

Fig.4 Boundary condition for Modal analysis


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Fig.5 Meshing

Highly fine meshing is done with 15mm element size. Meshing is done with quadrilateral

element for better result. Total nodes generated are 771299 representing 187 bodies active

components. Total elements created are 116174.

Fig.6 First Mode (132.06 Hz)


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Fig.7 Second Mode (136.97 Hz)

Fig. 8 Third Mode (141.22 Hz)

Result of Modal analysis:

Total 10 natural frequencies are found as below:

` Mode Frequency[Hz]

1 1 132.06

2 2 136.97

3 3 141.22

4 4 148.06

5 5 159.86

6 6 171.35

7 7 176.09

8 8 181.58

9 9 182.18

10 10 183.67


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(A)CONDITION AT INSTANT WHEN MATERIAL JUST FALLS ON DECK

(1)MATERIAL IS STORED IN A CHAMBER ABOVE DECK AREA, NO HOPPERIS PROVIDED.

As the material is falling from height of 2 m, we are neglecting hopper effect. So we are

considering direct impact due to material falling. Material is stored at 2 m height. Potential

energy is stored at the height, as lump falls down the Potential energy is converted in to Kinetic

Energy, so we can find the value of falling velocity by formula,

V=

V= Impact velocity

V0= Initial velocity at 2m level (In our case it is assumed to be zero)

g= Gravitational acceleration (9.81 m/)

The impact velocity find is 6.465 m/

Explicit Dynamics is one of the modules of ANSYS, which is used to find deflection of

object due to impact loads. Lump is assumed cubical shape. Here the maximum lump size taken

as 300*300*300 , the total lump are assumed as cover the total plate. Bulk density is takenas 800 kg/. The explicit dynamics analysis is carried out as below to find effect of impactforce.

EXPLICIT DYNAMICS

Explicit dynamics is used to found impact stress and deflection of falling object. The material

falling is of 300mm*300mm*300mm cube of coal. Material falling velocity is 6.465 m/.Boundary conditions are given to feeder. This is shown below.


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Fig. 9 Boundary condition for explicit dynamics

Yellow coloured objects in above figure are coal lump which are impacting on VGF deck area.

Boundary conditions are shown by blue colour in above figure.


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Fig. 10 Directional deformation

Result of directional deformation is 0.27647mm, due to 300*300*300 lump size.


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Fig. 11 Von Mises stress

Von-Mises stress is found: 240.39 MPa due to 300*300*300 lump size.

(2)MATERIAL IS STORED IN HOPPER JUST ABOVE THE DECK PLATE AREAOF FEEDER.

In this condition material is taken falling from hopper just above the deck. By different

configuration of hopper different analysis is carried out. The hoper may conical shaped outlet

or wedge shape outlet. It is then differentiate by inner lining material, the material may be

mild steel or stainless steel. It may further differentiate by surcharge shape, it may be conical

or triangular. Based on it different cases are been taken to analyzed with different hopper

configuration.
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CASE 1: HOPPER OUTLET IS WEDGE SHAPED; HOPPER IS INLINED WITH M.S

MATERIAL, CONICAL MATERIAL SURCHARGE AT TOP OF HOPPER:

i.

Bin

Opening Dimension (B) 2.5 m

Height H 2.0 m

Width D 3.90 m

Surcharge Hs 1.0 m

Height of hopper Hh 1.0 m

Half angle 35

Parallel section of bin mild steel

Hopper section Lined with stainless steel type 304-2B

Length of opening L 1 m

ii. Bulk Solid Type = coal

Effective angle of

internal friction = 50

Angle of friction between

coal and mild steel = 30


coal and stainless

steel n = 18

Bulk density = 0.80 t/m

Loads act on feeder are two types. Initial Load on feeder and flow load. Initial loads are at when

hopper is fully empty, in this time the loads acts on a feeder are higher, while flow loads are less

compare to initial loads. So, for safe design we will analyze for initial load because it is higher.


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Solution

i. Static Surcharge Factor qiAssuming material surcharge on top of the bin is of conical shape, for conical shape material

is taken as 1.

=

= 0.333 m (1)

R =

= 0.975 m (2)

=

[1-

] +

(3)

= 0.4 (for incompressible material this value is taken o.4, coal is incompressible)= *g = 0.8*9.81= 7.848 ton/

R= 0.975 m.

= tan = tan 30 = 0.577

H = 2 m.

Put the all above values in equation (3)

We found, =14.13 KPa.

Initial non dimensional surcharge factor ,

= *

[

+

- 1] (4)

In this case material is coming out from wedge shaped outlet, so material flow will be plane flow

and for plane flow the value ofm in above equation is taken as 0.

= hopper half angle= 35tan = 0.70

m=0

by putting all values in equation (4) we get, = 0.861

Initial load acting on feeder for this case is found by,


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W= * * * (5)

m= 0 (for wedge shape outlet)

m=1 (for axial flow or conical outlet)

In this case outlet is wedge shaped, so m=0, put all values in equation (5)

We get load on hopper outlet in this case,

W= 42.257 KN

This force will applied on all four deck plates, so it will converted into pressure to apply on deck

plate in ANSYS 14 workbench.

Pressure, P = 16653 Pa.

Fig 12 Von-Mises stress (Case 1)

Fig 13 Total deformation (Case 1)


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Results found for Case 1:

Total Deformation: 0.1076 mm

Von- Mises stress: 15.83 MPa

CASE 2: HOPPER OUTLET IS WEDGE SHAPED; HOPPER IS INLINED WITH S.S


ii. BinOpening Dimension (B) 2.5 m

Height H 2.0 m

Width D 3.90 m

Surcharge Hs 1.0 m


Half angle 35

Parallel section of bin stainless steel




Effective angle of





coal and stainless

steel n = 18



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Solution

ii. Static Surcharge Factor qiAssuming material surcharge on top of the bin is of conical shape, for conical shape material

is taken as 1.

=

= 0.333 m (1)

R =

= 0.975 m (2)

=

[1-

] +

(3)

= 0.4 (for incompressible material this value is taken o.4, coal is incompressible)= *g = 0.8*9.81= 7.848 ton/

R= 0.975 m.

= tan = tan 18 = 0.3249

H = 2 m.


We found, =15.78 KPa.


= *

[

+

- 1] (4)




m=0




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W= * * * (5)





W= 67.55 KN




Fig 14 Boundary condition (Case 2)


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Fig 15 Total Deformation (Case 2)

Fig 16 Von-Mises stress


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CASE 3: HOPPER OUTLET IS WEDGE SHAPED; HOPPER IS INLINED WITH M.S

MATERIAL, TRIANGULAR SURCHARGE AT TOP OF HOPPER:

iii. BinOpening Dimension (B) 2.5 m

Height H 2.0 m

Width D 3.90 m

Surcharge Hs 1.0 m


Half angle 35





Effective angle of





coal and stainless

steel n = 18


Solution

iii. Static Surcharge Factor qi


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Assuming material surcharge on top of the bin is of triangular shape, for triangular shape

material is taken as 0.

=

= 0.5 m (1)

R =

= 1.95 m (2)

=

[1-

] +

(3)= 0.4 (for incompressible material this value is taken o.4, coal is incompressible)

= *g = 0.8*9.81= 7.848 ton/

R= 0.975 m.

= tan = tan 30 = 0.577

H = 2 m.


We found,

= 13.97 KPa.


= *

[

+

- 1] (4)




m=0



W= * * * (5)



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W= 41.98 KN




Fig 17 Boundary conditions (Case 3)


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Fig 19 Von Mises stress (Case 4)


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CASE 4: HOPPER OUTLET IS WEDGE SHAPED; HOPPER IS INLINED WITH S.S

MATERIAL, TRIANGULAR MATERIAL SURCHARGE AT TOP OF HOPPER:

iv. BinOpening Dimension (B) 2.5 m

Height H 2.0 m

Width D 3.90 m

Surcharge Hs 1.0 m


Half angle 35





Effective angle of





coal and stainless

steel n = 18


Solution

iv. Static Surcharge Factor qi


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Assuming material surcharge on top of the bin is of conical shape, for conical shape material is taken as 1.

=

= 0.5 m (1)

R =

= 1.95 m (2)

=

[1-

] +


= *g = 0.8*9.81= 7.848 ton/

R= 1.95 m.

= tan = tan 18 = 0.3249

H = 2 m.


We found,

= 18.15 KPa.


= *

[

+

- 1] (4)




m=0



W= * * * (5)



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W= 72.29 KN




Fig 20 Boundary Condition (Case 4)


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CASE 5: HOPPER OUTLET IS CONICAL SHAPED; HOPPER IS INLINED WITH M.S


v. BinOpening Dimension (B) 1 m

Height H 2.0 m

Width D 2.0 m

Surcharge Hs 1.0 m


Half angle 25





Effective angle of





coal and stainless

steel n = 18


Solution

v. Static Surcharge Factor qi


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=

= 0.333 m (1)

R =

= 0.5 m (2)

=

[1-

] +


= *g = 0.8*9.81= 7.848 ton/

R= 0.5 m.

= tan = tan 30 = 0.577

H = 2 m.


We found,

= 11.28 KPa.


= *

[

+

- 1] (4)

In this case material is coming out from conical shaped outlet, so material flow will be axial flow

and for axial flow the value ofm in above equation is taken as 1.


m=1



W= * * * (5)



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In this case outlet is conical shaped, so m=1, put all values in equation (5)


W= 11.038 KN




Only single hopper above the deck area:



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Two hoppers above the deck area:

Fig 26 Boundary condition (Two Hopper)

Fig 27 Total Deformation (Two Hopper)


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Fig 28 Von Mises stress (Two hopper)


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CASE 6: HOPPER OUTLET IS CONICAL SHAPED; HOPPER IS INLINED WITH S.S


vi. BinOpening Dimension (B) 1.0 m

Height H 2.0 m

Width D 2.0 m

Surcharge Hs 1.0 m


Half angle 25





Effective angle of





coal and stainless

steel n = 18


Solution

vi. Static Surcharge Factor qi


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=

= 0.333 m (1)

R =

= 0.5 m (2)

=

[1-

] +


= *g = 0.8*9.81= 7.848 ton/

R= 0.5 m.

= tan = tan 18 = 0.3249

H = 2 m.


We found,

= 13.79 KPa.


= *

[

+

- 1] (4)




m=1



W= * * * (5)



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W= 17.25 KN




Fig 29 Boundary condition (Case 6)



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Two hoppers above deck:

Fig 32 Boundary Conditions (Two hoppers)


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Fig 33 Total Deformation (Two hoppers)

Fig 34 Von Mises stress (Two hoppers)


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CASE 7: HOPPER OUTLET IS CONICAL SHAPED; HOPPER IS INLINED WITH M.S

MATERIAL, TRIANGULAR SURCHARGE AT TOP OF HOPPER:

vii. BinOpening Dimension (B) 1.0 m

Height H 2.0 m

Width D 2.0 m

Surcharge Hs 1.0 m


Half angle 25





Effective angle of





coal and stainless

steel n = 18


Solution

vii. Static Surcharge Factor qi


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Assuming material surcharge on top of the bin is of triangular shape, for triangular shape

material is taken as 0.

=

= 0.5 m (1)

R =

= 1.0 m (2)

=

[1-

] +


= *g = 0.8*9.81= 7.848 ton/

R= 1.0 m.

= tan = tan 30 = 0.577

H = 2 m.


We found,

= 15.04 KPa.


= *

[

+

- 1] (4)




m=1



W= * * * (5)



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W= 12.51 KN







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Two hopper outlet:

Fig 38 Boundary Condition (Two hoppers)


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Fig 39 Total Deformation (Two hoppers)

Fig 40 Von Mises stress (Two hoppers)


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CASE 8: HOPPER OUTLET IS TRIANGULAR SHAPED; HOPPER IS INLINED WITH

S.S MATERIAL, TRIANGULAR MATERIAL SURCHARGE AT TOP OF HOPPER:

viii. BinOpening Dimension (B) 1.0 m

Height H 2.0 m

Width D 2.0 m

Surcharge Hs 1.0 m


Half angle 25





Effective angle of





coal and stainless

steel n = 18


Solution

viii. Static Surcharge Factor qi


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=

= 0.5 m (1)

R =

= 1.0 m (2)

=

[1-

] +


= *g = 0.8*9.81= 7.848 ton/

R= 1.0 m.

= tan = tan 18 = 0.3249

H = 2 m.


We found,

= 16.84 KPa.


= *

[

+

- 1] (4)




m=1



W= * * * (5)



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W= 13.22 KN







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Fig 44 Boundary Condition (Two hoppers)


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Fig 45 Total Deformation (Two Hoppers)

Fig 46 Von Mises stress (Two Hoppers)


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(B)CONDITION WHEN MATERIAL STARTS SCREENINGIn this situation three conditions are taken:

(1)If the material falls are over sized with 300*300*300 lump size(2)If the material falls are over sized with 100*100*100

lump size

(3) If the material falls average sized with 30*30*30 lump size(1)If the material falls are over sized with 300*300*300 lump size:

At the instant when material is start screening.

Here it is assumed that all lumps are of 300*300*300 size. The conveying is done byvibratory motion, lump are move in parabolic curve to move ahead. Here the amplitude of this

curve not high, so material will not create impact on deck as well as screen. Therefore, It is

assumed that material is moving with sliding motion. So, at any instant the load is equal to

material retained at on particular deck. So direct material load is acting on deck and screen by

W=mg.

Case 1: It is assumed that all lumps are oversized and screening is not done. The material that

falling on deck , leaves in same quantity without screening. So the load act will be same for all

plates.

Load on each row of plates*= width of total 4 plate x length of plate x height of material lump x

density

= 5974.29 N ( This the load acting on each row of plate)

*= each row contains four plates

The total load is converted in to pressure and applied on each plate. The results found are as

below.


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Fig 47 Boundary condition

Fig. 48 Total Deformation


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Fig 49 Von-Mises Stress

Fig 50 First natural frequency (119.34 Hz)

Fig 51 Second natural frequency (126.56 Hz)


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Result found:

(1)Total deformation: 0.041mm(2)Von Mises Stress: 7.6384MPa(3)1st mode: 119.34 Hz(4)2nd mode: 126.56 Hz(5)3rd mode: 128.08 Hz

Case 2: It is assumed that material falling are having some large size lump and some that

can be screened:

Some assumptions made for screening:

(1)First row of screen, screens the material is 35% of total material(2)Second row of screen, screens the material is 25% of total material(3)Third row of screen, screens the material is 20% of total material(4)Fourth row of screen, screen the material is 10% of total material(5)Rest of 10% material passed through last plate, which is oversized.

As above assumption, the pressure on each row of plate is found out:

(1)For first row: 2365 Pa(2)For second row: 1536 Pa(3)For third row: 945 Pa(4)

For forth row: 472 Pa(5)For last plate: 1454 Pa

Fig 50 Boundary and Loading Conditions


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Fig 53 Total Deformation

Fig 54 Von Mises stress

Fig 55 First modal frequency (119.34 Hz)


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Fig 56 Second Modal frequency (126.56 Hz)

Result found:

(1)Total Deformation: 0.040mm(2)Von Mises stress: 7.099 MPa(3)1st mode: 119.34 Hz(4)2nd mode: 126.56 Hz(5)3rd mode: 128.08 Hz

(2)If the material falls are over sized with 100*100*100 lump size:Case 1: It is assumed that all lumps are oversized and screening is not done. The material that

falling on deck , leaves in same quantity without screening. So the load act will be same for all

plates.

Load on each row of plates*= width of total 4 plate x length of plate x height of material lump x

density

Load= 1991.43 N


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Fig 57 Loading and Boundary conditions

Fig 58 Total deformation



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Fig 60 First Modal frequency (126.56 Hz)


Result found:

(1)Total Deformation: 0.028 mm(2)Von Mises stress: 5.5096 MPa(3)1st mode: 119.34 Hz(4)2nd mode: 126.56 Hz(5)3rd mode: 128.09 Hz


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Case 2: It is assumed that material falling are having some large size lump and some that

can be screened:

Same assumptions made for screening as above case:

(1)For first row: 788 Pa

(2)For second row: 512 Pa(3)For third row: 315 Pa(4)For forth row: 157 Pa(5)For last plate: 484 Pa

Fig 62 Boundary and loading conditions


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Result found:



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(6)If the material falls average sized with 30*30*30 lump sizeHere the average material size 30*30*30 , the material screening percentages are as above.

Fig 67 Loading and boundary conditions



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Result found:


CHAPTER5 Analysis

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