Download Calculating the Applied Load

510E
Calculating the Applied Load
The LM Guide is capable of receiving loads and moments in all directions that are generated due to
the mounting orientation, alignment, gravity center position of a traveling object, thrust position and
cutting resistance.
Reverse radial load
Radial load
Lateral
load
Lateral
load
MA
Moment
in the pitching
direction
MB
Moment
in the yawing
direction
MC
Moment
in the rolling
direction
Fig.1 Directions of the Loads Applied on the LM Guide
Calculating an Applied Load
[Single-Axis Use]
 Moment Equivalence
When the installation space for the LM Guide is limited, you may have to use only one LM block, or
double LM blocks closely contacting with each other. In such a setting, the load distribution is not
uniform and, as a result, an excessive load is applied in localized areas (i.e., both ends) as shown in
Fig.2. Continued use under such conditions may result in flaking in those areas, consequently shortening the service life. In such a case, calculate the actual load by multiplying the moment value by
any one of the equivalent-moment factors specified in Table1 to Table6 A1-43.
Moment load
Moment load
Rows of balls under a load
Rows of balls under a load
Ball displacement line
Maximum ball deflection
Load distribution curve
Maximum applied load on a ball
LM rail
Ball
displacement line
Load distribution curve
Fig.2 Ball Load when a Moment is Applied
An equivalent-load equation applicable when a moment acts on an LM Guide is shown below.
P = K•M
P
K
M
: Equivalent load per LM Guide (N)
: Equivalent moment factor
: Applied moment
(N-mm)
B1-56
510E
Point of Selection
Calculating the Applied Load
 Equivalent Factor
Equivalent Factors for the MA Moment
PR=KAR•MA
Equivalent in the radial direction
PL=KAL•MA
Equivalent in the reverse-radial direction
Fig.3 Equivalent Factors for the MA Moment
Equivalent factors for the MA Moment
Equivalent factor
in the radial direction
KAR=
C0
MA
Equivalent factor in the
reverse radial direction
KAL=
C0L
MA
C0
C0L
=
=1
KAR•MA KAL•MA
Equivalent Factors for the MB Moment
PT=KB•MB
Equivalent in the lateral direction
PT=KB•MB
Equivalent in the lateral direction
Fig.4 Equivalent Factors for the MB Moment
Equivalent factors for the MB Moment
Equivalent factor in
C0T
KB=
the lateral directions
MB
C0T
=1
KB•MB
B1-57
LM Guide
Since the rated load is equivalent to the permissible moment, the equivalent factor to be multiplied
when equalizing the MA, MB and MC moments to the applied load per block is obtained by dividing
the rated loads in the corresponding directions.
With those models other than 4-way equal load types, however, the load ratings in the 4 directions
differ from each other. Therefore, the equivalent factor values for the MA and MC moments also differ
depending on whether the direction is radial or reverse radial.
510E
Equivalent Factors for the MC Moment
PR=KCR•MC
Equivalent in the radial direction
PL=KCL•MC
Equivalent in the reverse-radial direction
Fig.5 Equivalent Factors for the MC Moment
Equivalent factors for the MC Moment
Equivalent factor
in the radial direction
KCR=
C0
MC
Equivalent factor in the
reverse radial direction
KCL=
C0L
MC
C0
C0L
=
=1
KCR•MC KCL•MC
C0
C0L
C0T
PR
PL
PT
:
:
:
:
:
:
B1-58
Basic static load rating (radial direction)
Basic static load rating (reverse radial direction)
Basic static load rating (lateral direction)
Calculated load (radial direction)
Calculated load (reverse radial direction)
Calculated load (lateral direction)
(N)
(N)
(N)
(N)
(N)
(N)
510E
Point of Selection
Calculating the Applied Load
Example of calculation
When one LM block is used
Model No.: SSR20XV1
No.3
No.4
No.2
No.1
ℓ1
LM Guide
Gravitational acceleration g=9.8 (m/s2)
Mass m=10 (kg)
ℓ1=200（mm）
ℓ2=100（mm）
ℓ2
m
m
Fig.6 When One LM Block is Used
No.1
No.2
No.3
No.4
P1=mg+KAR1•mg•ℓ1+KCR•mg•ℓ2=98+0.275×98×200+0.129×98×100=6752 (N)
P2=mg–KAL1•mg•ℓ1+KCR•mg•ℓ2=98–0.137×98×200+0.129×98×100=–1323 (N)
P3=mg–KAL1•mg•ℓ1–KCL•mg•ℓ2=98–0.137×98×200–0.0644×98×100=–3218 (N)
P4=mg+KAR1•mg•ℓ1–KCL•mg•ℓ2=98+0.275×98×200–0.0644×98×100=4857 (N)
When two LM blocks are used in close contact with each other
Model No.: SVS25R2
Gravitational acceleration g=9.8 (m/s2)
Mass m=5 (kg)
ℓ1=200（mm）
ℓ2=150（mm）
No.3
No.4
No.2
No.1
ℓ1
ℓ2
m
m
Fig.7 When Two LM Blocks are Used in Close Contact with Each Other
No.1 P1=
mg
49×150
mg•ℓ2 49
+0.0188×49×200+0.0814×
=507.9 (N)
+KAR2•mg•ℓ1+KCR•
=
2
2
2
2
No.2 P2=
mg
49×150
mg•ℓ2 49
–KAL2•mg•ℓ1+KCR•
=
–0.0158×49×200+0.0814×
=168.8 (N)
2
2
2
2
No.3 P3=
49×150
mg
mg•ℓ2 49
–KAL2•mg•ℓ1–KCL•
=
–0.0158×49×200–0.0684×
=–381.7 (N)
2
2
2
2
No.4 P4=
49×150
mg
mg•ℓ2 49
+KAR2•mg•ℓ1–KCL•
=
+0.0188×49×200–0.0684×
=–42.6 (N)
2
2
2
2
Note1) Since an LM Guide used in vertical installation receives only a moment load, there is no need to apply a load force
(mg).
B1-59
510E
[Double-axis Use]
 Setting Conditions
Set the conditions needed to calculate the LM system’s applied load and service life in hours.
The conditions consist of the following items.
(1) Mass: m (kg)
(2) Direction of the working load
(3) Position of the working point (e.g., center of gravity): ℓ2, ℓ3, h1(mm)
(4) Thrust position: ℓ4, h2(mm)
(5) LM system arrangement: ℓ0, ℓ1(mm)
(No. of units and axes)
(6) Velocity diagram
Speed: V (mm/s)
Time constant: tn (s)
Acceleration: n(mm/s2)
V
(αn = tn
)
(7) Duty cycle
Number of reciprocations per minute: N1(min-1)
(8) Stroke length: ℓs(mm)
(9) Average speed: Vm(m/s)
(10) Required service life in hours: Lh(h)
Gravitational acceleration g=9.8 (m/s2)
ℓ3
mg
h2
h 1 ℓ0
Duty cycle
Speed (mm/s)
ℓ1
ℓ2
V
tn
ℓ4
Fig.8 Condition
B1-60
t1
tn
ℓS
Velocity diagram
（s）
（mm）
510E
Point of Selection
Calculating the Applied Load
 Applied Load Equation
The load applied to the LM Guide varies with the external force, such as the position of the gravity
center of an object, thrust position, inertia generated from acceleration/deceleration during start or
stop, and cutting force.
In selecting an LM Guide, it is necessary to obtain the value of the applied load while taking into account these conditions.
LM Guide
Calculate the load applied to the LM Guide in each of the examples 1 to 10 shown below.
m
: Mass
(kg)
ℓn
: Distance
(mm)
: External force
(N)
Fn
: Applied load (radial/reverse radial direction) (N)
Pn
(N)
PnT : Applied load (lateral directions)
g
: Gravitational acceleration
(m/s2)
(g =9.8m/s2)
V
: Speed
(m/s)
: Time constant
(s)
tn
: Acceleration
(m/s2)
n
V
(αn = tn
)
[Example]
Condition
Applied Load Equation
Horizontal mount
(with the block traveling)
Uniform motion or dwell
mg
P2
P4
1
mg
mg•ℓ2
mg•ℓ3
+
–
4
2•ℓ0
2•ℓ1
P2 =
mg
mg•ℓ2
mg•ℓ3
–
–
2•ℓ0
2•ℓ1
4
P3 =
mg
mg•ℓ2
mg•ℓ3
–
+
2•ℓ0
2•ℓ1
4
P4 =
mg
mg•ℓ2
mg•ℓ3
+
+
2•ℓ0
2•ℓ1
4
P1 =
mg
mg•ℓ2
mg•ℓ3
+
+
2•ℓ0
2•ℓ1
4
P2 =
mg
mg•ℓ2
mg•ℓ3
–
+
2•ℓ0
2•ℓ1
4
P3 =
mg
mg•ℓ2
mg•ℓ3
–
–
2•ℓ0
2•ℓ1
4
P4 =
mg
mg•ℓ2
mg•ℓ3
+
–
2•ℓ0
2•ℓ1
4
ℓ1
P3
ℓ2
P1 =
P1
ℓ0
ℓ3
Horizontal mount, overhung
(with the block traveling)
Uniform motion or dwell
P3
ℓ1
P2
P4
2
P1
ℓ0
ℓ3
mg
ℓ2
Note) Load is positive in the direction of the arrow.
B1-61
510E
Condition
Applied Load Equation
Vertical mount
Uniform motion or dwell
ℓ2
P4
P1T
3
P2 = P3 =
P1
ℓ0
mg•ℓ2
2•ℓ0
P1 = P4 = –
mg
mg•ℓ2
2•ℓ0
F
mg•ℓ3
2•ℓ0
P1T = P4T =
ℓ3
P2
P2T = P3T = –
P2T
mg•ℓ3
2•ℓ0
ℓ1
E.g.: Vertical axis of industrial
robot, automatic coating
machine, lifter
Wall mount
Uniform motion or dwell
ℓ0
P2T
ℓ3
ℓ2
P1T
P2
ℓ1
P1 = P2 = –
P3 = P4 =
P1
mg•ℓ3
2•ℓ1
4
P1T = P4T =
mg
mg•ℓ2
+
4
2•ℓ0
P2T = P3T =
mg
mg•ℓ2
–
2•ℓ0
4
P3
P3T
mg
P4
P4T
E.g.: Travel axis of cross-rail loader
Note) Load is positive in the direction of the arrow.
B1-62
mg•ℓ3
2•ℓ1
510E
Point of Selection
Calculating the Applied Load
Condition
Applied Load Equation
With the LM rails movable
Horizontal mount
P3
P1
5
P4
P1 to P4 (max) =
mg
mg•ℓ1
+
4
2•ℓ0
P1 to P4 (min) =
mg
–
4
ℓ0
LM Guide
ℓ1
–ℓ1
P2
mg
mg•ℓ1
2•ℓ0
ℓ2
E.g.: XY table
sliding fork
Laterally tilt mount
P1 = +
–
P1T=
mg
–
P1
P3
P2T=
6
P1T
P2
ℓ3
ℓ1 θ
P2T
ℓ2
+
P3T=
+
P4T=
mg•cosθ •ℓ3
mg•sinθ •h1
+
2•ℓ1
2•ℓ1
mg•cosθ
mg•cosθ •ℓ2
–
2•ℓ0
4
mg•cosθ •ℓ3
mg•sinθ •h1
–
2•ℓ1
2•ℓ1
mg•sinθ
mg•sinθ •ℓ2
–
2•ℓ0
4
P4 = +
E.g.: NC lathe
Carriage
mg•cosθ
mg•cosθ •ℓ2
–
2•ℓ0
4
mg•sinθ
mg•sinθ •ℓ2
–
2•ℓ0
4
P3 = +
ℓ0
mg•cosθ •ℓ3
mg•sinθ•h1
+
2•ℓ1
2•ℓ1
mg•sinθ
mg•sinθ •ℓ2
+
2•ℓ0
4
P2 = +
h1
mg•cosθ
mg•cosθ •ℓ2
+
2•ℓ0
4
mg•cosθ
mg•cosθ •ℓ2
+
2•ℓ0
4
mg•cosθ •ℓ3
mg•sinθ •h1
–
2•ℓ1
2•ℓ1
mg•sinθ
mg•sinθ •ℓ2
+
2•ℓ0
4
Note) Load is positive in the direction of the arrow.
B1-63
510E
Condition
Applied Load Equation
Longitudinally tilt mount
P1 = +
–
P1T = +
P3
h1
P2 = +
P2
mg
P2T
–
P4
ℓ2
P1
7
P1T
ℓ3
ℓ1
mg•cosθ
mg•cosθ •ℓ2
+
2•ℓ0
4
mg•cosθ •ℓ3
mg•sinθ •h1
+
2•ℓ1
2•ℓ0
mg•sinθ •ℓ3
2•ℓ0
mg•cosθ
mg•cosθ •ℓ2
–
2•ℓ0
4
mg•cosθ •ℓ3
mg•sinθ •h1
–
2•ℓ1
2•ℓ0
mg•sinθ •ℓ3
2•ℓ0
P2T = –
ℓ0
θ
P3 = +
+
mg•cosθ
mg•cosθ •ℓ2
–
2•ℓ0
4
mg•cosθ •ℓ3
mg•sinθ •h1
–
2•ℓ1
2•ℓ0
mg•sinθ •ℓ3
2•ℓ0
P3T = –
P4 = +
E.g.: NC lathe
Tool rest
+
P4T = +
Horizontal mount with inertia
mg•cosθ
mg•cosθ •ℓ2
+
2•ℓ0
4
mg•cosθ •ℓ3
mg•sinθ •h1
+
2•ℓ1
2•ℓ0
mg•sinθ •ℓ3
2•ℓ0
During acceleration
mg
P1 = P4 =
mg
–
4
m•α 1•ℓ2
2•ℓ0
P2 = P3 =
mg
+
4
m•α 1•ℓ2
2•ℓ0
P1T = P4T =
ℓ3
P1
ℓ2
P2T = P3T = –
P3
P4
P3T
ℓ0
P4T
Speed V (m/s)
ℓ1
t2
t3 Time (s)
Velocity diagram
P1 to P4 =
P1 = P4 =
mg
+
4
m•α 3•ℓ2
2•ℓ0
P2 = P3 =
mg
–
4
m•α 3•ℓ2
2•ℓ0
P1T = P4T = –
E.g.: Conveyance truck
P2T = P3T =
Note) Load is positive in the direction of the arrow.
B1-64
mg
4
During deceleration
V
αn =
tn
t1
m•α 1•ℓ3
2•ℓ0
During uniform motion
8
F
m•α 1•ℓ3
2•ℓ0
m•α 3•ℓ3
2•ℓ0
m•α 3•ℓ3
2•ℓ0
510E
Point of Selection
Calculating the Applied Load
Condition
Vertical mount
with inertia
Applied Load Equation
P4
mg
P1T
ℓ0
P1
F
ℓ3
P2
αn =
P2T
9
V
tn
Speed V (m/s)
ℓ1
t1
t2
t3 Time (s)
Velocity diagram
E.g.: Conveyance lift
Horizontal mount with external force
LM Guide
During acceleration
m (g+α 1) ℓ2
P1 = P4 = –
2•ℓ0
m (g+α 1) ℓ2
P2 = P3 =
2•ℓ0
m (g+α 1) ℓ3
P1T = P4T =
2•ℓ0
m (g+α 1) ℓ3
P2T = P3T = –
2•ℓ0
ℓ2
During uniform motion
mg•ℓ2
2•ℓ0
mg•ℓ2
P2 = P3 =
2•ℓ0
mg•ℓ3
P1T = P4T =
2•ℓ0
mg•ℓ3
P2T = P3T = –
2•ℓ0
P1 = P4 = –
During deceleration
m (g – α 3) ℓ2
2•ℓ0
m (g – α 3) ℓ2
P2 = P3 =
2•ℓ0
m (g – α 3) ℓ3
P1T = P4T =
2•ℓ0
m (g – α 3) ℓ3
P2T = P3T = –
2•ℓ0
P1 = P4 = –
Under force F1
F1•ℓ5
2•ℓ0
F1•ℓ5
P2 = P3 =
2•ℓ0
F1•ℓ4
P1T = P4T =
2•ℓ0
F1•ℓ4
P2T = P3T = –
2•ℓ0
P1 = P4 = –
ℓ2
ℓ4
P3
F2
ℓ5
F1
10
F
ℓ1
F3
P3T
P4
P4T
ℓ3
ℓ0
E.g.: Drill unit,
Milling machine,
Lathe,
Machining center
and other cutting machine
Under force F2
F2
P1 = P4 =
+
4
F2
–
P2 = P3 =
4
F2•ℓ2
2•ℓ0
F2•ℓ2
2•ℓ0
Under force F3
F3•ℓ3
P1 = P2 =
2•ℓ1
F3•ℓ3
P3 = P4 = –
2•ℓ1
F3
P1T = P4T = –
–
4
P2T = P3T = –
F3
+
4
F3•ℓ2
2•ℓ0
F3•ℓ2
2•ℓ0
Note) Load is positive in the direction of the arrow.
B1-65