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Studying the forces acting on the hydraulic soil digging machines Theory

Conference Paper · January 2012

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 Proceeding of the 12th of the European Regional Conference of the
ISTVS Pretoria, South Africa, Sept. 24-29, 2012 101

Studying the forces acting on the hydraulic soil digging

machines
Theory
Shaker H. Aday
Mech, Dept., College of Agriculture, University of Basrah, Basrah, Iraq
E. mail: shaker.aday @ uobasrah.edu.iq
Abstract:
The forces acting on the hydraulic soil digging machines was studied
theoretically when the machine at static and dynamic state. The studying
included the machines provided with straight and bent arms. Different types of
equations were derived to predict the forces acting on the machine as well as the
moment which can cause stability problem during the machine operation. The
distances between the forces and their components and the supporting point
whether on the machine body or the soil surface when the bucket at rest can be
calculated by derived equations.

Introduction
The soil digging machine is widely used during the last century and still
used widely to carry out many heavy duties work. These machines are used to
open irrigation and drainage channel as well as used to open trenches to lat out
electrical and telephone cables and to lay out the subsoil drains such as the
plastic pipes and the tiles. These machines are provided with bucket to dig the
soil. The bucket was controlled mechanically by using cables but this sort of
method is slow and the bucket can not penetrate the soil Easley because
depends on its weight. But using the hydraulic systems developed the
performance of the machine greatly through controlling the bucket during the
operation as well as the bucket is forced to penetrate the soil by the pressure of
the hydraulic of the cylinders.

However, many forces act on the machine when it is on static and dynamic
states. In the static state the weight of the machine and the weight of their arms
are the main forces. When the machine in the dynamic state many force acting
on it in addition to the previously mention forces. The most important force is
the weight of the bucket when loaded with soil. The weight of the load bucket
 Studying the forces acting on the hydraulic soil digging machines

adds weight to the machine as well as creates moment which would affect the
201
balance of the machine. For better machine performance the force acting on it
should be studied to determine the best method to remain the balance of the
machine and to determine the load of the bucket which give the highest
productivity with least danger of overturning the machine.

The soil digging operation and transferring the load from the digging area
to the dumping place cause overturn moment thereupon some forces appear to
keep the machine balance at least at minimum. For good machine operation and
high balance to reduce the danger of overturn the forces acting on the machine
when at static and dynamic states should be studied. The hydraulic soil digging
machines are in two types. These types are the machines with straight and bent
arms. This paper is to study the forces on the machines of straight and bent
arms.

The machine parts are shown in figure (1). Different forces acting on the
machine some of them come from the soil which the machine operating in and
the other forces are due to the forces of the soil. Most of the forces in the
hydraulic cylinders which are used to move the bucket and the arms appear due
to the weight of the bucket. Because the importance of the bucket the forces
acting on it will be studied first.

Weight of the Bucket

The weight of the bucket includes its weight when empty and the weight of
the material when filled. The materials usually dogged out by the hydraulic
digger are soil sand and gravel. The machine is widely used to dig the soil so
that it will be used as example for calculation.
The specific weight of the soil when the bucket is filled with soil can be
expressed by equation (1).

S   g ............(1)

Where  S = the specific weight of the soil (kN/m3)

ρ= soil bulk density (kg/m3)

g= ground acceleration (9.81m/sec2)

The weight of the soil within the bucket can be expressed by equation (2)
 Studying the forces acting on the hydraulic soil digging machines

201
Figure (1): the forces acting on the hydraulic soil digging machine.
 Studying the forces acting on the hydraulic soil digging machines

WS   S  V ............ (2) 201

Where WS= the soil weight within the bucket (kN).

V= the size of the bucket (m3)

When the soil contain water the weight of the water should be taken in
account, equation (3)

WSW   S  VS   W  VW ............ (3)

Where WSW= weight of the soil and the water (kN)

VS= volume of the soil (m3)

VW= volume of the water (m3)

VW= volume of the water within the soil into the bucket (m3)

γS= specific weight of the water (10kN/m3)

The total weight of the bucket when filled with soil can be expressed by
equation (4).

W  WS  We ............ (4)
Where W= total weight of the bucket when filled with soil (kN)

We= weight of the bucket when empty (kN).

When equation (2) is substitute in equation (4), the total weight of the
bucket can then be expressed by equation (5).

W   S V  We ............ (5)

When the soil contains water the total weight of the bucket can be
expressed by equation (6)

W   S  VS   W  VW  We ............ (6)

The forces acting on the machines.

To understand the forces acting on the machines will be divided to three
parts.
 Studying the forces acting on the hydraulic soil digging machines

Part 1
201
A reaction force F3 shows up in the hydraulic cylinder which supporting
and moving the bucket. F3 is due to the weight of the bucket WS. When the
bucket is inclined with horizontal line the weight of the bucket is analyzed into
horizontal WSh and vertical WSV components while F3 analyzed into horizontal
F3h and F3V components (Fig 2 (.

(A) the bucket is inclined (B) the bucket is level.

Figure (2): the forces acting on the bucket when inclined and level.

The horizontal forces:

The horizontal component which acting on the bucket anfd the hydraulic
cylinder are equal, equation (7).

F3 h  WS h ............ (7)

The horizontal component (WSh)of the weight WS, equation (8).

WS h  WS  sin  ............ (8)

 = The inclination angle of the bucket with horizontal line (degrees)

When the bucket is level the weight WS acting downward.
The angle can be calculated from triangle BCD, equation (9), Figure (3).
 Studying the forces acting on the hydraulic soil digging machines

201

Figure (3): the inclination angle of the bucket.

hg
  sin 1 ............ (9)
Lb

Lb = the length of the bucket (m).

h g = the height of bucket edge relative the horizontal line (m).

The value of  is between zero and 900. the vertical position of the bucket
(900) is to unload the soil.

When equations (8), (1) and (2) are substituted in equation (7), therefore
F3h is:

F3 h    g  V  sin  ............ (10)

The vertical forces

The vertical component of the bucket weight (WSV) equals to the vertical
component of the force of the hydraulic cylinder (F3V), equation (11)

F3V  WSV ............ (11)

The vertical component of the bucket weight can be calculated by equation

(12)

WSV  WS  cos ............ (12)

When equations (1), (2) and (12) are substituted in equation (11), therefore
F3h can be expressed by equation (13)

WS V   .g.V . cos ............ (13)

 Studying the forces acting on the hydraulic soil digging machines

Part 2 201
Part 2 represents the second arm which is connected with first arm (A1)
from the upper end and with bucket from the lower end. This arm is freely
connected to the first arm and the bucket to move the bucket far or close to the
machine body. Also the position of the bucket can be changed to unload the soil
in it and can be vertically positioned to the soil surface. Many forces act on this
arm among them the force of the hydraulic cylinder (F2) which determine the
position of arm 2 (A2) relative to arm 1 (A 1), the weight of the arm 2 (W A2 )
and the force of the hydraulic cylinder of the bucket F3.

F2 is analyzed to its horizontal and vertical components F 2h and F2v

respectively. WA2 is also analyzed to its horizontal and vertical components
WA2h and WA2V respectively, figure (4).

Figure (4): the forces acting on the arm 2 of the hydraulic machine.

The vertical forces:

The vertical forces acting on arm 2are equal at the static state of the arm
and the bucket. These forces can be expressed by equation (14).

F2V  WA2V  F3V ............ (14)

F2V= the vertical component of the force F2 (kN)

WA2V =the vertical component of the weight of arm2 (kN)

 Studying the forces acting on the hydraulic soil digging machines

F3V= the vertical component of F3 (kN)

201
However, the vertical component of the arm 2 weight can be expressed by
equation (15).

WA2V  WA2 cos ............ (15)

 = the inclination angle of arm 2 relative to the horizontal line (degrees).

 can be calculated from triangle BEG (fig. 5) and equation (16).

hA 2
  sin 1 ............ (16)
LA2
hA2= the vertical distance between the horizontal line passes through the
upper end and the lower end of arm 2.

LA2= the length of the arm (m).

E G

hA2
LA 2
B
Figure (5): the inclination angle of arm2 relative to the horizontal line.
Substitute equations (11), (13) and (15) in equation (14). F2V, therefore
can be expressed by equation (17).

F2V    g V  cos  WA2  cos ............ (17)

The horizontal forces:

The horizontal components acting on arm 2 can be expressed by equation
(18).

F2 h  F3 h  W A2 h ............ (18)
Substitute equation (10) in equation (18), F2h is:

F2 h    g V  sin   WA2  sin ............ (19)

Third part:
Third part is arm 1which is connected to the machine body from the lower
end and with arm 2 from the upper end. The arm is movable at both ends. A
 Studying the forces acting on the hydraulic soil digging machines

hydraulic cylinder (2) is fixed on the upper end of arm 1 to move arm 2 (A2)
201
relative to arm 1(A1). The arm is also provided with hydraulic cylinder at the
lower end to move the arm relative to the machine body. The forces acting on
the arm 1 are its weight WA1 which analyzed to horizontal and vertical
components WA1h and WA1V respectively and the force of the hydraulic
cylinder (1) F1. The horizontal and vertical components of F1 are F 1h and F1V
respectively (fig. 1).

The horizontal components can be expressed by equation (20)

F1h  F2 h  WA1h ............ (20)

The horizontal component of the weight of arm1 (A1) can be expressed by

equation (21).

WA1h  WA1  sin ............ (21)

WA1= the weight of arm1 (A1) (kN).

 = The inclination angle of arm1 (A1).

The inclination angle of A1 can be calculated from triangle JKM (fig. 6)

and equation (22).

h A1
  sin 1 ............ (22)
L A1
hA1= the height of the upper end of arm (A1) (m)

LA1= the length of arm (A1) (m).

K K LA1 cg
LA1
h A1 h A1
WA1h

M J M
WA1

WA1v J

)B( )A(

Figure (6) (A) the forces acting on arm A1 (B) the inclination angle of A1
 Studying the forces acting on the hydraulic soil digging machines

Substitute equations (19) and (21) in equation (20), therefore F 1h canbe

expressed by equation (23). 220

F1h    g  v  sin   WA2  sin   WA1  sin ............ (23)

Equation (23) can be rearranged as follows, equation (24)

hg hA 2 h
F1h    g V  sin  WA2  sin  WA1  sin A1 ............ (24)
Lb LA 2 LA1

The vertical forces:

The vertical forces acting on arm A1 can be expressed by equation (25).

F1V  WA1h  F2V ............ (25)

The vertical component of the arm A1 can be expressed by equation (26)

WA1h  WA1  cos ............ (26)

Substitute equations (17) and (26) in equation (25), therefore F1V can
expressed by equation (27).

F1V    g  V  cos  WA2  cos  WA1  cos ............ (27)

Equation (26) can be rearranged as follows:

L2b  hg2 L2A2  hA2 2 L2A1  hA21

F1v    g  v  cos  WA2  cos  WA1  cos
Lb LA 2 LA1

............ (28)
The vertical force component F1V increases the weight of the machine
because it includes the weight of the bucket with load. The weights of the arms
are already included in the weight of the machine. The weights of the arms
create moment at the front of the contact the track with soil. The horizontal
force component F1h crates resistance force (R) at the contact area of the track
with soil because tries to pull the machine forward. The rolling resistance (R)
moves the soil reaction (Z) on the machine forward distance equal K. K can be
calculated by equation (29), figure (1).

Rh  Z k ............ (29)

Where K= the distance between the machine center of gravity and soil
 Studying the forces acting on the hydraulic soil digging machines

reaction (Z) (m).

h= the height of the center of gravity from the soil surface (m). 222
Z is soil reaction on the machine track and equal the weight of the machine
(Z= Gp). Substitute Gp in equation (29), K therefore expressed by equation
(30).
R
K h ............ (30)
Gp
The rolling resistance coefficient (CRR), is :
R
CRR  ............ (31)
Gp
Substitute equation (31) in equation (30), K therefore can be expressed by
equation (31).
K  CRR  h ............ (32)

The moment action on the machine:

The forces acting on the machine and causing instability are the weights of
arms A1 and A2 and the weight of the bucket when loading (Ws). The forces
within the hydraulic cylinders are the results of the previous forces. These
forces are not taken in to account when the balance of the machine is calculated.
The machine balance can be calculated by taking moment about point A (fig.7).
The moment can be expressed by equation (33).

Gp  a  WA  L4  WA1V  L1  WA1h  h1  WA2V  L2  WA2 h  h2  WsV  L3  Wsh  h3

...(33)
Where WA=the addition weight (kN).
L1= the distance between the center of gravity of arm (A1) and
point A (m).
L2= the distance between the center of gravity of arm (A2) and
point A (m).
L3= the distance between the center of gravity of the bucket and
point A (m).
L4= the distance between the additional weight and point A (m).
h1 and h2= height of the center of gravity of arms A1 and A2 from
point A respectively (m).
h3= height of the center of gravity of the bucket from point A.
 221
Studying the forces acting on the hydraulic soil digging machines

Figure (7): The force and their components acting on the machine and their distance relative to point A.
 Studying the forces acting on the hydraulic soil digging machines

Arm A1 inclined at angle  relative to the soil surface. L1 and h1 can be

calculated from the length of arm A1 (LA1) and angle  using triangle JNq, figs. 221
7 and 8. H1 and L1 are calculated by equations (34) and (25) respectively.
h1  0.5L A1  sin  t ............ (34)
Where t= the height of the point Aa from A (m).
L1  0.5LA1 cos ............ (35)

Figure (8): the inclination angle of arm A1.

Figure (9): The position of arm A2 when the machine at rest (vertical) and
at work (inclined ).

h2 can be calculated as follows:

EF can be calculated from angle GEF, equation (36), (figs. 9 and 10)

EF  0.5LA2  sin  ............ (36)

 Studying the forces acting on the hydraulic soil digging machines

E 221
 G

0.5LA2
F
Figure(10): triangle GFE (inclination angle of arm A2)

 = inclination angle of arm A2 (degrees).

GF represents half the length of arm A2 and equals to Gk 1 and CH,
equation (37).

Gk1  GF  CH  0.5LA 2 ............ (37)

FK is the height of the center of gravity of arm A1 (K1) from the soil
surface. FK can be expressed by equation (38).

Fk  CH  EF ............ (38)
When equations (36) and (37) are substituted in equation (38), Fk therefore
can be expressed by equation (39).

Fk  0.5LA2  0.5LA2  sin  ............ (39)

When equation (39) is simplified, FK is:

Fk  0.5LA2 1  sin   ............ (40)

h2 can therefore can be expressed by equation (41)

h2  F1k1  k1d  dn ............ (41)

However,

F1 k1  Fk : dn  hb : k1d  0.5L A2 ............ (42)

When equations (40) and (42) are substituted in equation (41), h2 is
expressed by equation (43).

h2  0.5LA 2  0.5LA 2 1  sin    hb ............ (43)

Equation (43) can rearranged as follows, equation (44).

h2  0.5LA 2 2  sin    hb ............ (44)

 Studying the forces acting on the hydraulic soil digging machines

Where hb= depth of the bucket (m).

221
L2 is the distance between the vertical components (WA2V) of the weight
of arm A2 from A which can be expressed by equation (45).

L2  PQ  kk1 ............ (45)

PQ can be calculated from triangle PQG (fig. 11) and fig. (7), equation
(46).

LA1


Q P
LA1 cos

Figure (11): the inclination angle of arm A1.

PQ  LA1  cos ............ (46)

KK1 can be expressed by equation (47), Fig. (9)

kk1  GE  0.5LA 2  cos ............ (47)

Substitute equations (46) and (47) in equation (45), therefore, L2 can be

expressed by equation (48).

L2  LA1  cos  0.5L A 2 cos ............ (48)

L3 is the distance between the vertical component of the bucket weight

from A and can be expressed by equation (49).

L3  L2  EB  XY ............ (49)

EB can be calculated by equation (49), fig. (9).

EB  GB  EB ............ (50)

EB  LA2  cos  0.5LA2 cos  0.5LA2 cos .....(51)

XY can be calculated by equation (52), figs. (9) and (12).
 Studying the forces acting on the hydraulic soil digging machines

X Y
221


0.5 Lb
m
Figure (12) the inclination of the bucket with horizontal line

XY  0.5Lb  cos  ............ (52)

Where Lb= the length of bucket (m).

 = the inclination angle of the bucket (degrees)

Substitute equations (48), (50) and (52) in equation (49), L3 therefore, can
be expressed by equation (53).

L3  LA1  cos  0.5LA2  cos  0.5LA2 cos  0.5Lb  cos ........... (53)
Equation (53) can be simplified to be come, equation (54).

L3  L A1  cos  L A2  cos  0.5Lb  cos ............ (54)

h3 is the distance between the horizontal component (W bh) of the bucket

weight and the soil surface (point A) and represents the height of the center of
gravity of the bucket from the soil surface. It can be calculated by equation (55),
fig. (9).

h3  TR  Ym ............ (55)

TR= the height of the bucket from soil surface (m).

hb= the bucket depth (m).

However, the height of the bucket from the soil surface is equal to the
height of the arm A2 center of gravity which equals FK and its depth h b, fig. (9).
The distance between the joint point of arm A2 with edge of the bucket
(X), TR, can be calculated by equation (56),

TR  Fk  hb ............ (56)
Substitute equation (56) in equation (55), h3 therefore can be expressed by
equation (57).
 Studying the forces acting on the hydraulic soil digging machines

h3  Fk  hb  Ym ............ (57) 221

Ym can be calculated by equation (58), fig. (12)

Ym  0.5Lb  sin  ............ (58)

Substitute equations (40) and (40) in equation (57), h3 can be expressed by
equation (59).

h3  0.5L A1 1  sin    hb  0.5Lb sin  ............ (59)

The hydraulic digging machine with bent arm:

This type of machine is widely used to dig irrigation and drainage
channels, trench to lay out electric and telephone cables and waste water. The
advantage of this machine is highly capable to dig hard soils. This machine is
provided with big bucket so that its productivity is quite high. This type of
bucket can not be used on the machine of long straight arms because the
balance problem. Another advantage of this machine is their arms do not go
high so that the turning over moment due to the weight of the bucket of the
machine.

The parts of the machine are similar to that shown in fig. (1) but the arm
A1 consists from parts Aa and Ab. The angle between Aa and Ab is greater than
900. Part Aa is at angle ϴ with horizontal line. Part Ab is at angle ϴ1 with
horizontal line. The arm A1 is supported by the chasse of the machine and its
angle (ϴ) changed by the hydraulic cylinders (1) which are fixed on both sides
of arm A1. Arm A2 is moving relative to arm A1 by the hydraulic cylinder (2).
Arm A2 is straight. The bucket is fixed at the end of arm A2 and is moving by
the hydraulic cylinder (3).
 Studying the forces acting on the hydraulic soil digging machines

The Forces acting on the hydraulic digging machine when on static

221
state:
The forces acting on the machine when at static state are the weights of the
machine Gp, Arm A1 which includes weights of Aa (WAa), Ab (WAb)and arm A2.
WAa is analyzed to the horizontal component (WAah) and the vertical component
WAav. The weight of arm A2 is acting downward when the arm is vertical and
that occurred when the machine at rest. The weight of the machine (Gp) is
supported by the machine tracks. The soil reaction to Gp is Z. The weights of
the arms and the bucket are carried by the supporting point on the machine body
(A) and the support point at the contact area between the bucket and the soil
surface (O). When the machine is at rest most of the arms and bucket weight is
supported by point O while the small part of the weights is carried by point A.

The weights acting on points O and A can be calculated by taken moments

about point A and O respectively. However, the dimensions between the forces
components and the point O and A should be calculated.

The dimensions of the forces components relative to point A:

L1 is the distance between the vertical component WAav and the point A
can be calculated by equation (60) using triangle AJK and fig. (13). K is the
center of gravity of Aa, therefore, AK=0.5LAa .

Figure (14): triangle AJK

L1 = AJ = 0.5 LAa cosθ ............ (60)

L2 is the distance between WAav and point A and it can be calculated by
equation (61) using fig. (13)

L2 = AB + CH ............)61(
 Studying the forces acting on the hydraulic soil digging machines

Figure (13) the hydraulic soil digging machine of bent arm. 221
 Studying the forces acting on the hydraulic soil digging machines

AB can be calculated by equation (62) using fig. (15).

210

Figure (15): triangle ABC

AB = LAa cosθ ............ (62)

CH can be calculated by equation (63) using triangle CHG.

G is the center of gravity of part Ab, therefore, CG=0.5LAb

Figure (16): triangle CHG

H = 0.5 LAb cosθ1 ............ (63)

Substitute equations (62) and (63) in equation (61), L 2 can therefore be

expressed by equation (64).

L2 = LAa cosθ + 0.5 LAb cosθ1 ............ (64)

L3 is the distance between WAbv and the point A can be calculated by
equation (65).

L3 = AB + CE ............ (65)
CE can be calculated by equation (66) using triangle CED and fig. (17).

CD= length of part Ab

 Studying the forces acting on the hydraulic soil digging machines

212

Figure (17): triangle CED

CE = LAb cosθ1 ............ (66)

Substitute equations (62) and (66) in equation (65), L3 therefore can be
calculated by equation (67).

L3 = LAa cosθ + LAb cosθ1 ............ (67)

h1 is the distance between component WAah and point O, soil surface, can
be calculated by equation (68).

H1 = KJ + t ............ (68)
t= the height of point A from the soil surface (m).

KJ can be calculated by equation (69) using triangle AJK.

KJ = 0.5 LAa sinθ ............ (69)

Substitute equation (69) in equation (68), h1 is expressed by equation (70).

H1 = 0.5 LAa sinθ + t ............ (70)

h2 is the distance between WAbh and point O and can be calculated by
equation (71).

H2 = BC + GH + t ............ (71)

BC is calculated by equation (72) using triangle ABC.

BC = LAa sinθ ............ (72)
GH is calculated by equation (73) using triangle CHG.

GH = 0.5LAb sinθ1 ............ (73)

Substitute equations (72) and (73) in equation (71), h2 can therefore, be
calculated by equation (74).

H2 = LAa sinθ + 0.5 LAb sinθ1 + t ............ (74)

When parts Aa and Ab are equals, there average equals LA.

 Studying the forces acting on the hydraulic soil digging machines

LA = LAa = LAb
211

Equation 75 becomes as follows:

h2 = 0.5 LA (2sinθ + sinθ1) + t ............ (75)
To calculate the soil reaction (PS2) at point O a moment about point (A)
taken. However, the pressure at the hydraulic cylinder (3) should be zero,
equation (76).

PS2 L3 = WAav L1 + WAbv L2 + WAah h1 + WAbh h2 ............ (76)

............ (77)

When the machine is at static state the weight of arm A2 is added to PS2.

............ (78)

To calculate the reaction force at point A a moment about point O should

be taken, equation (79).

PS1 L3 = WAav (L3 – L1) + WAbv (L3 – L2) + WAah h1 + WAbh h

............ (79)
PS1 can be calculated by equation (80).

............ (80)

The forces acting on the hydraulic soil digging machines with bent arm:
When the hydraulic digging machines at work the forces acting on the
machines arms and the bucket create reaction force at supporting point A, fig.
18. The reaction force at point (A) transferred to the machine as additional
weight and a force try to pull the machine forward which leads to creation of
rolling resistance force at the machine tracks (R). The rolling resistance Pulls
the soil reaction on the machine Z forward distance K.

The forces acting on the machine arms:

In additional to the forces acting on the arms there is the weight of the
bucket and the load (WB). WB adds weight to the machine and causes moment
about point A which causing front tipping moment.
 Studying the forces acting on the hydraulic soil digging machines

To calculate the reaction force at point A the distance between the

211
components of the force acting on the arms and the bucket should be calculated.

The distance between the force components and the supporting force (A):

L1can be calculated by equation (81) using triangle APQ and fig. (19)

Figure (19): triangle APQ

L1 = AP = 0.5 LAa cosθ ............ (81)

Where θ= the inclination angle of part Aa of arm A1 relative to the

horizontal line.

L2 can be calculated by equation (82) using triangles ABC and CXY.

Figure (20): triangles ABC and CXY

L2 = AB + CX ............ (82)
AB and CX can be calculated by equations (83) and (84).

AB = LAa cosθ ............ (83)

 211
Studying the forces acting on the hydraulic soil digging machines

Figure(18): The forces acting on the hydraulic soil digging machine of bent arm.
 Studying the forces acting on the hydraulic soil digging machines

CX = 0.5 LAb cosθ1 ............ (84)

211
Substitute equations (83) and (84) in equation (82), L2 can therefore be
calculated using equation (85).

L2 = LAa cosθ + 0.5LAb cosθ1 ............ (85)

L3 can be calculated by equation (86) using triangles ABC, CED and
DGH.

Figure (21): the triangles CED and DGH

L3 = AB + CE + DG ............ (86)
CE and DG can be calculated by equations (87) and (88) respectively.

CE = LAb cosθ1 ............ (87)

DG = 0.5 LA2 cosω ............ (88)

Substitutes equations (83), (87) and (88) in equation (86), L3 can therefore
calculated by equation (89).

L3 = LAa cosθ + LAb cosθ1 + 0.5 LA2 cosω ............ (89)

L4 is calculated by equation (90).

L4 = AB + CE + DG + VT – 0.5 Lb ............ (90)

VT can be calculated by equation (91) using triangle HTV.
 Studying the forces acting on the hydraulic soil digging machines

211

Figure (22): triangle HVT

VT = 0.5 LA2 cosω ............ (91)

Substitute equations (83), (87), (88) and (90) in equation (91).

L4 = LAa cosθ + LAb cosθ1 + 0.5 LA2 cosω + 0.5 LA2 cosω – 0.5Lb
............ (92)
Substitute equation (89) in equation (92), L4 can therefore be calculated by
equation (93).

L4 = L3 + 0.5 LA2 cosω – 0.5 Lb ............ (93)

h1 can be calculated by equation (94).

h1 = QP + t ............ (94)
Qp can be calculated by equation (95) using triangle APQ.

QP = 0.5 LAa sinθ ............ (95)

Substitute equation (95) in equation (94), h1 is therefore, expressed by
equation (96).

h1 = 0.5 LAa sinθ + t ............ (96)

h2 can be calculated by equation (97).

h2 = YX + CB +t ............ (97)
CB and YX can be calculated by equations (98) and (99) using triangles
ABC and CXY respectively.

CB = LAa sinθ ............ (98)

YX = 0.5 LAb sinθ1 ............ (99)

Substitute equations (98) and (99) in equation (97).
 Studying the forces acting on the hydraulic soil digging machines

h2 = 0.5 LAb sinθ1 + LAa sinθ + t ............ (100)

211
h3 can be calculated by equation (101) using fig. (18).

h3 = JM + MN + Lb ............ (101)
JM can be expressed by equation (102).

JM = DM – DJ ............ (102)
DJ can be expressed by equation (103) using triangle DJH. H is the center
of gravity of arm A2.

Figure (23): triangle DJH

DJ = 0.5 LA2 sinω ............ (93)

DM can be expressed by equation (94)

DM = MN = 0.5 LA2 ............ (94)

Substitute equations (93) and (94) in equation (92), JM therefore, is:

JM = 0.5 LA2 – 0.5 LA2 sinω = 0.5 LA2 (1 – sinω) ............ (95)
Substitute equations (94) and (95) in equation (92).

h3 = 0.5 LA2 (1 – sinω) + 0.5 LA2 + Lb ............ (96)

Simplified equation (96), therefore

h3 = LA2 (1 – 0.5sinω) + Lb ............ (97)

h4 can be expressed by equation (98).

h4 = h3 – HT – 0.5 hb ............ (98)

The center of gravity of the bucket is at the middle.

HT can be expressed by equation (99) using triangle HTV.

 Studying the forces acting on the hydraulic soil digging machines

211

Figure (24): triangle HTV

HT = 0.5LA2 sinω ............ (99)

Substitute equations (97) and (99) in equation (98).

h4 = LA2 (1 – 0.5sinω) + Lb – 0.5hb ............ (100)

The supporting force at the hydraulic cylinder of arm A1.

Figure(25): the force of the supporting cylinder of arm A1

To calculate the supporting force within the hydraulic cylinder due to the
forces of arms A1 and A2 and The weight of the bucket when filled with soil a
moment should be taker about point A, fig (18).

Fv L = WAa sinθ L1 + WAa cosθ (h1 – t) + WAb sinθ1 L2 + WAb cosθ1 (h2 – t)

+ WA2 sinω L3 + WA2 cosω (h3 –t) + WS L4 ............ (101)

FV= the vertical component of the force F of the hydraulic cylinder (kN).

L= the distance between supporting point A and the base of the hydraulic
cylinder (1) which is constant for any machine (m), fig. (25).
Studying the forces acting on the hydraulic soil digging machines

Simplified equation (101) to become as follows:

211
Fv L = WAa (L1 tanθ + (h1 – t)) + WAb (L2 tanθ1 + (h2 – t))

+ WA2 (L3 tanω + (h3 – t)) + WS L4 ............ (102)

............ (103)

The total force in the hydraulic cylinder F can be calculated as follows:

............ (104)

Where Ø= the inclination angle of the hydraulic cylinder (1)

The horizontal component (Fh) of F can be expressed by equation (105)

............ (105)

References:
(1) Hannah, J. and M. J. Hillier (1977): Applied Mechanics. The Pitman
press, Bath. first edition.
(2) Hannah, J. and M. J. Hillier (1978): Mechanical Engineering science.
The Pitman press, Bath. first edition.
(3)Shigematsa,T.; T. Muro; N. Terao; N. Oda and T. Hanaoka (2008):
Development of a hard rock excavator by using an edge excavatot.
Proceeding of the 16th Int. conf. of ISTVS Turin Nov. 25-28. pp276-
280. Italy.
(4) Murn, T.; T. Iwatat and K. Kohno (2005): Development of shaft
excavator due to edge excavation system for a rock mass. Proc. Of the
34th symposium on rock mechanics, JSCE, 327-332.
(5)Muck, D.; M. Dinko; V. Koroman and S. Klak (2008): Design of a
forklift with telescopic boom. 16th Int. conf. of ISTVS. Turin Nov. 25-
28, Italy.
 ‫‪Studying the forces acting on the hydraulic soil digging machines‬‬

‫دراسة ميكانيكية القوى المؤثرة عمى ألة الحفر الهيدروليكية‬ ‫‪210‬‬

‫دراسة نظرية‬

‫شاكر حنتوش عداي‬

‫قسم المكائن واالالت الزراعية‪ /‬كمية الزراعة ‪/‬جامعة البصرة‪ /‬البصرة‪ /‬العراق‪1022/‬‬

‫الخالصة‪:‬‬
‫درس تتل وى تتي وىرت ت رت ت ر عيت ت أى تتل وى ا تتر وىيكدريىاتا تتل ول وا ر وىسس تتر اسل يوا ر وىسة ةا تتل‪.‬‬
‫عةتدس روتيا وا ر سستتر اسل يوىوياتل ا رةتل اتتوا أياوا وىت روعكا يوىوياتتل ريىتد التي ات واليتاة وىيدريىاتاتتل‬
‫وىر ت ر تترذ وا ر يوىوكيتتل‪ .‬ق ت ة وى تتي ر ستتر عةتتد ة اتتل أررو ت ا وى ت ور واي عي ت واىتتل أاا ق ت ة وى تتي‬
‫كتتي كتتدمر ياا وىوياتتل‬ ‫وى تتي‬ ‫ارسست ل ىةىتتل س ت ور ر تيواا‪ .‬أا أا أستترمدول واىتتل ى اتتر ستتيل رمريت‬
‫يوى سر يق و ك د وى ايت دة وى تي وىست رة عيت وا ر يا ىرت ى رد وىادتر عةتد ة اتل ورروت ا وىت ور واي‬
‫رد ادتر وىررىتل وىت واست ل اسستل وى تية وىررىتل وىرت‬ ‫عي واىل يق و اسس ل ظيير عتال كت د وىت وةتدا‬
‫ايوات‬ ‫ر ر عي وىسراركا يعةد ساح وارض‪ .‬ةركلل عتدل ات ل وا ر ات سيالتح يو تد يرسكتر ستح واات‬
‫أاد دق عتا ة اتل أرروت ا وىت ور‬ ‫اوا وى ي وىس رة عي وا ر رر ير وى سرتا ري يق ة وىسرتا ل رمري‬
‫واي ‪ .‬ق ت ة وااد ت د استتتا س ت ي ستتا سد ت دال رتتل أ تتر الي ستتا أاد ت د وا ر يايوا ت ساةةي ت ستتا وىم ت‬
‫ىتتل‬ ‫واا ت يأمرياتتل ق ت ة وىسدتتدال ىيسرتا ت ل واا اتتل عةي ت ىيسرتا ت ل وىدسيداتتل‪ .‬عةتتدس روتتيا أىتتل ا ت‬
‫وىستيا ات ا وىوياتل ري تح ا تتر عستيدى يعيت أستر سل وىت ور وى ت ة يقت و أدى وىت أمترةل وىسدت دال‬
‫وىر رسرمدل ى س ل أاد د سرتا ل وى ي وى رة عي وا ر ‪.‬‬

‫ستتا كتتي وىدتتدد‬ ‫أس ت عةتتدس روتتيا أ ر واىتتل ستتا وىةتتي وىسة ة ت اتتوا وى تتي وى ت رة عيكي ت رمري ت‬
‫رد وىادر عةد ة ال أررو ا وى ور واي عي وىل‪ .‬أس وىسد دال وىر رل أ ر الي‬ ‫يوىسيالح يوىرو كر ييمري‬
‫عا ريتذ وىرت رتل أ تر الي ىةىتل ول وا ر‬ ‫ى س ل أاد د سرتا ل وى ي وىس رة عي وا ر اي رمري‬
‫وىسسر اسل تس الاسل رد وىادر عةد ة ال أررو ا وى ور وىي أو ر سس اسسل ر دل رد ادر وىررىل عي واىل‬
‫وىت واست ل ا تتيرة أو تتر‪ .‬وى تتي وىست رة عيت واىتتل أو تتر اسستتل ت تتر لتتل تكيريت يرتتايد يت و وى لتتل ستتا‬
‫ر أ رعي ‪.‬‬ ‫وىواةل ى‬

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