Revista Científica Interdisciplinaria Investigación y Saberes
2023, Vol. 13, No. 3 e-ISSN: 1390-8146
Published by: Universidad Técnica Luis Vargas Torres
How to cite this article (APA):
Guerrero, O., Rivera, r., Rivera, R. (2023) Development
of Octave programming to calculate the forces in the structural elements of a scissor lift
table, Revista Científica Interdisciplinaria Investigación y Saberes, 13(3) 95-127
Development of Octave programming to calculate the forces in the
structural elements of a scissor lift table
Desarrollo de la programación en Octave para calcular las fuerzas en los elementos
estructurales de una mesa elevable tipo tijera
Oscar Xavier Guerrero Ferrusola
Mechanical Engineer, Instituto Técnico Superior Simón Bolívar, o_guerrero@istsb.edu.ec
https://orcid.org/0000-0003-0149-0833
Raúl Arnaldo Rivera Obando
Mechanical Engineer, Instituto Técnico Superior Simón Bolívar, ra_rivera@istsb.edu.ec
https://orcid.org/0000-0003-3448-2522
Ruth Verónica Rivera Spain
Bachelor's Degree in Computer Science, Instituto Técnico Superior Simón Bolívar, R_rivera@istsb.edu.ec
https://orcid.org/0009-0009-4554-358X
When building equipment or parts of a machinery, the determination
of the forces acting on the elements is key to dimension them
properly. Newton's laws of motion are used to determine the forces,
in the case of mechanisms whose movement is slow (quasi-static) or
when it does not move, the application of Newton's first law is
relevant. For a table for lifting loads (weights) it is necessary to know
the forces acting on the elements for different positions of the arms
of the table, doing it manually requires repeated calculations and
there is a possibility of placing unintentional errors when performing
arithmetic operations.
Keywords:
Newton of motion, program, arithmetic, arithmetic
Resumen
Cuando se construye un equipo o partes de una maquinaria, la
determinación de las fuerzas que actúan sobre los elementos es clave
para dimensionarlos adecuadamente. Para determinar las fuerzas se
utilizan las leyes de Newton del movimiento, en el caso de
Abstract
Received 2023-02-23
Revised 2023-03-03
Published 2023-09-07
Corresponding Author
Oscar Rivera Ferrusola
o_guerrero@itsb.edu.ec
Pages: 95-127
https://creativecommons.org/lice
nses/by-nc-sa/4.0/
Distributed under
Copyright: © The Author(s)
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
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96
mecanismos cuyo movimiento es lento (cuasi estático) o cuando no
se mueve, la aplicación de la primera ley de Newton es relevante.
Para una mesa elevadora de cargas (pesos) es necesario conocer las
fuerzas que sobre los elementos actúan para diferentes posiciones de
los brazos de la mesa, hacerlo de forma manual exige realizar los
cálculos de forma repetida y hay la posibilidad de colocar errores
involuntarios al hacer las operaciones aritméticas.
Palabras clave:
Newton del movimiento, programa, aritmética
Introduction
It is desired to design a scissor type lifting table with a hydraulic
system that automates the upward or downward movement with a
hydraulic cylinder. To determine the dimensional characteristics of the
arms and the hydraulic equipment, we must obtain the numerical
values of the forces acting on each member as a function of the load
to be lifted and the positional angle of the arms. We try to use a
computer program that allows us to calculate the forces on each
member of the lifting table, and to determine the internal shear and
axial forces and bending moments that allow us to size the element
knowing the material with which the element will be manufactured.
For this purpose we will use the help of a numerical calculation
program to determine the values of forces in different positions of the
table arms, from the low level to the maximum that can be raised.
For the construction of scissor-type folding tables, it is necessary to
know the forces acting on the constituent elements, as well as the
means for lifting the table. The study of the loads (forces) on the joints
and to which the arms of the scissors mechanism are subjected is the
objective of this work.
Among the means we can use for lifting the table, we will mention
that these can be pneumatic, hydraulic, electric or mechanical linear
actuators; For this work we select in advance that the means to lift the
table will be through two hydraulic cylinders, placed at the base of
the table, in horizontal position, (see figure 1) of low speed, no greater
than 0.010 m/s (10 mm/s), which will allow us to analyze the
mechanism practically as a static system since the inertial forces will
be negligible, this is called quasi-static motion system.
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To obtain the parametric formulas of the forces at each point of the
folding table structure we will use the scheme shown in Figure 1, in
which we introduce the relevant variables of distance and support
points of the different elements.
Figure 1.
Simplified schematic of the scissor-type lift table, operated
by a hydraulic cylinder, in its elevation view, for the low level.
One of the most widely used mechanical systems for lifting loads and
people are lifting tables. These are manufactured in various
arrangements or structures to meet different needs of loads or
weights to be lifted. (Navarro Fandos, 2019)
In the work of Maldonado Anrubia, on the design of a scissor lift, he
mentions the UNE-EN 280 classification for lifts intended for the use
of people in work activities, in this he mentions the classification
according to the vertical projection of the center of gravity, being
called group A those whose center of gravity is always within the
overturning lines of the machine and group B those that do not have
their center of gravity within the overturning lines. A second
complementary classification is based on the translation that the
equipment can have, so we can mention type 1, in which the
translation can only be performed if the equipment is in transport
position; type 2, the translation of the equipment is performed when
it is on a chassis; type 3 when the translation of the equipment is
performed together with the platform. Figure 2 shows an example of
the aforementioned classification. (Maldonado Anrubia, 2016)..
Hydraulic piston
Load to be lifted
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Figure 2
Types of lifting tables (Source "Instituto Nacional de
Seguridad e Higiene en el Trabajo" of Spain)
Of the many possible configurations of an elevated table, the scissor
type has been chosen for this work, which can be used in construction,
maintenance and cleaning activities.
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In the thesis of Navarro Fandos (2019), a detailed explanation is made
of the constituent parts of a simple scissor type lift table, which
consists of:
Base. It is the support for the rest of the constituent elements. In
general it must be strong, rigid and stable, most of the times it is built
to be located on the ground or inside a pit. There are also those with
wheels which lock when the elevation is used.
Scissors. They provide vertical movement while supporting the
platform with the load to be lifted. The scissors are connected by
shafts at the center point of the arms, as well as at the top and bottom.
Depending on the number of scissors the table can be single or
multiple scissors with two, three, four or more pairs, depending on
the height to be reached.
Platform. Dimensions can be any, but compatible with the base and
scissor arms. Dimensions cannot be less than the width of the scissors
when the scissors are in the lowest position.
Drive. There is a range of drives to raise and lower the platform with
the scissors, they can be cable or belt mechanisms, which allow high
frequencies and working speeds; with a spindle formed by nut and
screw, which gives a good quality-price ratio; finally there are the
lifting tables by a hydraulic mechanism that allows a long duration and
smoothness of work with minimal maintenance.
It is precisely on this type of drive where the present study will be
focused, for this we will mention that there are different positions in
which the drive cylinder or cylinders are placed for the scissor type lift
table.
Several scientific and technological papers or articles have been
developed around the scissor lift tables, among which it is worth
mentioning some articles in which a study of the calculation of the
forces supported by the structural members of the table is made.
In Hamidi's work on the need for a dynamic analysis of certain lifting
platforms for the Faculty of Mechanical Engineering of the University
of Pristina, he develops his work by creating a model for the
calculation of the lifting mechanism of the platform and then follows
with the stresses of the structural elements by the finite element
method (Hamidi, 2012).
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The importance of using equation parameterization to determine
forces in lifting loads on scissor lifts in a symmetrical structure is the
work of Shanghong He, from the College of Mechanical and
Automotive Engineering of the People's Republic of China, who
conclude that the optimization of the cylinder force acting in lifting a
scissor table is closely related to four position parameters of the
installation and using simulation models they manage to optimize the
position in which the cylinder is placed reducing by 12% the maximum
cylinder load during lifting (He, Ouyang, Gong, & Liu, 2019).
The use of mathematical models has been established for scissor lift
research in the work Liu & Sun. They make a kinematic and kinetic
simulation analysis which was carried out with MATLAB/Simulink. The
relative kinetic relationship between the hydraulic cylinder and other
parts, as well as their change rules have been found. A 3D model of
scissor lift mechanism was established with Pro/Engineer design
software. The mechanism design was optimized in Pro/Mechanical
based on the findings of simulation analysis, which can guide and
improve the subsequent design. The design proved to be scientific
and reasonable and could serve as a theoretical guide and reference
for the scissor lift mechanism design of other uses (Liu & Sun, 2009-
06).
Another study on lifting scissor tables analyzes the applicability of
numerical methods using a variable in the design of lifting devices
using scissor mechanisms. The study comes to the following
conclusions: that the enabling dimensions of the hydraulic cylinder
and the whole system can be selected on purpose and that the
method proposed in the study allows them to establish force
equations at the limit load locations of the system. In this way, design
recommendations can be made for the dimensioning and structure of
the structural elements of the scissor table (Dang & Nguyen-Dinh,
2020-11).
Methodology
Working with parametric expressions gives us the advantage of
having general expressions that can be replaced by numerical values
such as different bar lengths (links), mass placement positions,
position angle of the drive bar of the mechanism, etc. For the case
shown in figure 1, the anglea will be the input parameter for the
Development of Octave programming to calculate the forces in the structural elements of a
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formulas; the total mass M is distributed in two structures in the same
way as shown in figure 1 so each structure will have half of the total
weight of the assigned mass.
Determination of the parametric expression for the reactions at
supports B and F.
Figure 3 shows the free body diagram of the table and the
representation of the force vectors acting on the outside of the scissor
lift table. It is worth mentioning that the force that will move the arms
of the table making it rise or descend is represented by Fp, which is
the force generated by a piston of a hydraulic system.
Figure 3.
Free-body diagram (FWD) of the scissor-type lift table
mechanism.
The vertical force By and Fy can be determined by applying Newton's
first law for rigid bodies, which states that the sum of forces of bodies
at rest is zero and the sum of moments. Around a point of the system,
it is also zero; doing the summation of moments analysis around the
point F, with the counterclockwise rotation as positive, we obtain the
following expression:
( )
0
cos 0
F
y
M
aW L B
a
+=
-=
å
F
y
C
L cos
a
Development of Octave programming to calculate the forces in the structural elements of a
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ec. (0.1)
Doing the summation of forces, we obtain:
ec. (0.2)
To determine the expression for the horizontal force Bx that would be
the one developed by the piston of the hydraulic cylinder, we must
decompose each element of the structure and place the internal
forces that occur in the joints, for the effect we will analyze the forces
in the table exclusively and then in the arms.
Determination of the parametric equations of the forces on
supports E and D.
Figure 4.
Free body diagram of the table with forces at the E and D
joints.
Application of Newton's first law to rigid bodies:
( ) ( )
cos 2 cos
y
aW aMg
B
LL
aa
==
0
0
y
yy y y
F
BFW FWB
+! =
+-=® =-
å
( ) ( )
2cos 2 2cos
y
aMg M g aMg
FW
LL
aa
=- = -
( )
1
2cos
y
Mg a
F
L
a
æö
=-
ç÷
ç÷
èø
( )
( )
0
cos 0
cos
D
yy
M
aW
aW L E E
L
a
a
+=
-=®=
å
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ec. (0.3)
ec. (0.4)
According to the FCD (free body diagram) there are no forces acting
on the horizontal axis, it follows that Ex and Dx are zero.
Determination of the parametric equations in the BD and EF link
joints.
In Figure 5, the forces on the joints constituting the link or member
BD are shown. The expressions for Dy and By are known and Bx, Cx
and Cy are unknown. In the diagram the force Dy has been drawn in
the opposite direction to that seen in the free body diagram of the
table, Figure 4, since Newton's third law applies when separating the
elements that make up the mechanism.
Figure 5.
Free body diagram of link (bar) BD. Note the change of
direction of the force at joint D.
Applying Newton's first law for rigid bodies, we begin by determining
the sum of moments around point C to determine the force Bx
( )
2cos
y
aMg
E
L
a
=
( )
0
0
22cos
y
yy y y
F
Mg aMg
EDW DWE
L
a
+" =
+-=® =-= -
å
( )
1
2cos
y
Mg a
D
L
a
æö
=-
ç÷
ç÷
èø
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ec. (0.5)
For the reactions at point C, we use the summation of forces on the
vertical and horizontal axes.
ec. (0.6)
ec. (0.7)
For the link or bar EF the values can be determined since the values
of the forces in Cx , Cy , Ey and Fy are known, therefore, it only
remains to find the expression for the force Fx. In Figure 6 is the DCL
of the link and it is worth mentioning that the direction of the forces
at node C change by the application of Newton's third law, in the case
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
0
sin cos cos 0
22 2
sin cos cos 0 sin cos cos 0
2
sin cos 1 cos 0
2 cos 2 cos
sin cos
2 cos
C
xy y
xy y xy y
x
x
M
LL L
BBD
L
BB D BB D
aMg Mg a
B
LL
aMg
B
L
aaa
aaa aaa
aa a
aa
aa
a
+=
-- =
éù
-- =® -- =
ëû
æö
---=
ç÷
ç÷
èø
-
å
( )
( )
( )
cos cos
2 2 cos
Mg aMg
L
aa
a
-+
( ) ( )
0
sin cos 0
2
x
Mg
B
aa
=
-=
( )
( )
( )
cos
1
cot
2 sin 2
x
Mg
BMg
a
a
a
==
0
0
x
xx x x
F
BC C B
+® =
+=® =-
å
( )
1
cot
2
x
CMg
a
=-
( ) ( ) ( ) ( )
0
0
1
2cos2cos22cos2cos
y
yyy y yy
y
F
BCD C DB
Mg a aMg Mg aMg aMg
C
LL LL
aa aa
+" =
+-=® = -
æö
=- - =- -
ç÷
ç÷
èø
å
( ) ( )
22
1
22cos 2 cos
y
Mg aMg Mg a
C
LL
aa
æö
=- = -
ç÷
ç÷
èø
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of Cx the equation 1.6 was obtained with negative sign, indicating
that it points in a negative direction (to the left) of the horizontal
reference of the diagram, therefore, the sign is changed for the
application in the EF link.
Figure 6.
Free body diagram of the EF link.
ec. (0.8)
To quantify the variation of the forces in the joints of the lifting table
mechanism, we will use the following numerical values described in
Table 1. The total mass M to be lifted will be 500 kg, which will include
the mass of the metallic elements of the table and the mass to be
lifted. From the review of catalogs of commercial hydraulic cylinders,
a cylinder of 0.400 m (400 mm) length, 0.20 m (200 mm) stroke and
provisionally 50 mm bore and 30 mm rod diameter is obtained in the
first instance. Therefore, the values for L1 and L2 will be 0.4 m and 0.6
m respectively. The value b was made numerically the same as the
length of the table, which will be 1.7 m (1,700 mm). For the position
where the mass M is placed on the table we assume that the center
of mass coincides with half of the table length, which will be 0.65 m
(650 mm).
Table 1.
Numerical values of the relevant physical parameters that the
elements of the folding table will have.
0
0
x
xx x x
F
CF F C
+® =
+=® =-
å
( )
1
cot
2
x
FMg
a
=-
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Parameter
Magnitude
Units
Description
M
500
kg
Block mass
L
1,3086
m
Link length
L
1
0,400
m
Piston length contracted
L
2
0,600
m
Extended piston length
a
0,650
m
Block position from end B
b
1,700
m
Position of the pivot F from end A
Results
Once all the parametric relationships of the forces on the joints have
been found we can determine the operating range of the anglea . For
this we know the parameters representing the length of the linear
actuator when the stem is retracted, L1 and the length of the actuator
when the stem is fully out, L2. From this we can determine the
parametric equations for the minimum and maximum value ofa and
thereby perform a study of the variation in joint forces.
In Figure 7 and 8, we have the two level ends that the scissor type lift
table will have and will serve to determine the range of motion of the
opening angle alpha (α).
Figure 7.
Minimum position of the scissor lift table.
Figure 8
. Maximum position of the scissor lift table.
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Expression for the minimum angle: eq. (0.9)
Expression for the maximum angle: eq. (0.10)
Using equation 1.9 and replacing the numerical values, we obtain:
Using equation 1.10 and replacing numerical values, we obtain:
Therefore, for the force values obtained in this study, the opening
range of the anglea will be from 6.6° to 32.7° .
To quantify the variation of the forces in the joints of the lifting table
mechanism, we will make use of a computational numerical
calculation program such as OCTAVE (ver. 7.1.0) with free license,
which allows us to calculate equations and obtain graphs of the
results. The program lines and the graphs obtained with the results of
the programming of the equations with OCTAVE are presented
below.
3. Development of the computer program with the free software
OCTAVE, version 7.1.0.
1
arccos
MIN
bL
L
a
-
æö
=
ç÷
èø
2
arccos
MAX
bL
L
a
-
æö
=
ç÷
èø
min
1700 400
arccos 6,57º 6,6º
1308,6
a
æö
-
==»
ç÷
èø
max
1700 600
arccos 32,79º 32,8º
1308,6
a
æö
-
==»
ç÷
èø
Development of Octave programming to calculate the forces in the structural elements of a
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On the program's web page, you will find that this is a scientific
programming language, mathematically oriented with integrated
plotting and visualization tools in 2 and 3 dimensions. The program is
a free software, does not require license purchase, and is developed
for different operating systems such as Linux, macOS, and Microsoft
Windows.
Octave's syntax is compatible with that of another mathematical
program such as Matlab, which is a proprietary program. Information
on the use of Octave can be obtained from the product's web site,
https://wiki.octave.org/Using_Octave
We are going to detail the instruction lines of the script that has been
developed to determine the main reactions in the elements of the
scissor mechanism of the lifting table.
The first lines indicate the title and letter assignment for the
mechanism variables, as shown in Figure 1.
1 % Calculation of forces on elevated platform elements
2 clear % Command to clear previously created variables
3 clc % Command for clearing write window (terminal)
4 M = 500; % Mass of the body to be lifted in kg.
5 L = 1.3086; % Length of table arms in m.
6 L1 = 0.400; % Length of hydraulic cylinder, retracted, in m.
7 L2 = 0.600; % Length of the hydraulic cylinder completely outside,
in m.
8 a = 0.650; % position of the mass from the right edge of the table,
in m.
9 b = 1,700; % Length from cylinder base to joint F, in m.
10 g = 9.81; % Gravity in m/s^2.
11 %
Then we write the equations to determine the anglea of opening of
the table arms, both minimum and maximum, and the number of
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angular divisions with which we will calculate the forces, for this case
it will be 100 angular divisions betweena min anda max.
12 alpha1 = acos((b-L1)/L); % minimum angle, position below the
table
13 alpha2 = acos((b-L2)/L); % max. angle, high table position
14 n_parts = (alpha2-alpha1)/100; % division of the angular path in
100 parts
15 alpha = (alpha1:n_parts:alpha2);
Having done so, we write the parametric equations of the forces at
joints B, F, C, E, and D, both for the horizontal and vertical
components, as well as their resultant.
16 % Parametric equations of forces in the joints
17 By = 0.5*M*g*a./(L*cos(alpha)); % Normal force at joint B, in N
18 Fp = 0.5*M*g*cot(alpha); % Piston force at joint B, in N.
19 %
20 Cx = -Fp; % Force at joint C, horizontal x-axis.
21 Cy = 0.5*M*g*(1-2*a./(L*cos(alpha))); % Force at C-joint, vertical
axis
22 C = sqrt(Cx.^2+Cy.^2); %Resultant of the force in C
23 %
24 Fx = Cx; % Force at joint F, horizontal axis.
25 Fy = 0.5*M*g*(1-a./(L*cos(alpha)));% Force on joint F, vertical axis
26 F = sqrt(Fx.^2+Fy.^2); %Resultant of the force in F
27 %
28 Dy = Fy; %Force at table support D
29 Ey = By; % Force at table support D
30 %
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The following instructions plot the database with the results of the 100
values calculated for each parametric variable.
31 alpha_degree = alpha*180/pi; % conversion from radians to
sexagesimal degrees
32 %
33 figure(1)
34 plot(alpha_degree,Fp);
35 title('Horizontal force at joint B')
36 xlabel('Angle in degrees')
37 ylabel('Force in N')
38 grid
39 figure(2)
40 plot(alpha_grade,By,'r')
41 title('Vertical force at joint B')
42 xlabel('Angle in degrees')
43 ylabel('Force in N')
44 grid
45 figure(3)
46 plot(alpha_degree,Cx);
47 title('Horizontal force at the C-joint')
48 xlabel('Angle in degrees')
49 ylabel('Forces in N')
50 grid
51 figure(4)
52 plot(alpha_degree,Cy);
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53 title('Vertical force at the C-joint')
54 xlabel('Angle in degrees')
55 ylabel('Forces in N')
56 grid
57 figure(5)
58 plot(alpha_degree,Ey,alpha_degree,Dy);
59 title('Forces in E and D joints');
60 xlabel('Angle in degrees');
61 ylabel('Forces in N')
62 legend('Ey','Dy');
63 grid
64 figure(6)
65 plot(alpha_degree,Fx);
66 title('Horizontal force at joint F');
67 xlabel('Angle in degrees');
68 ylabel('Forces in N')
69 grid
70 figure(7)
71 plot(alpha_degree,Fy);
72 title('Vertical force at joint F');
73 xlabel('Angle in degrees');
74 ylabel('Forces in N')
75 grid
76 %
Change of axes for the forces in the BD link
.
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For the study of the loads and stresses to which the BD link is
subjected, the axes on which the forces are determined must be
changed. During the development of the equations, the conventional
x (horizontal) and y (vertical) axes were used, but it is convenient to
change the axes to a normal or axial axis (parallel to the axis of the
link) and another transverse or tangential axis (perpendicular to the
axis of the link). To perform the change of axes, the equations formed
through trigonometric relations will be taken, as shown in the
attached figure.
The equations that allow the change can be written in an ordered form
obtaining a system of equations of the following form:
ec. (0.11)
If we use the matrix form to write the equations, we would have the
following expression:
ec. (0.12)
The 2 x 2 matrix containing the trigonometric expressions is called the
rotation matrix and can be used to build a computer program for
numerical calculation of the axial and transverse components of the
horizontal and vertical forces of the link. We will again use the
OCTAVE program to generate the matrix equations and plot the
variation of the forces, but in the normal and transverse axes of the
BD link.
The graph of link BD is shown in Figure 14, with its horizontal, x-axis,
and vertical, y-axis, components, with its forces taken to the normal,
n, and transverse, t, axes.
The axis transformation equations for joints B, C and D would be:
ec. (0.13)
( ) ( )
( ) ( )
cos cos
sin cos
nx y
tx y
FF F
FF F
aa
aa
=+
=- +
( ) ( )
( ) ( )
cos sin
sin c os
x
n
y
t
F
F
F
F
aa
aa
éù
éù
éù
=
êú
êú
êú
-
ëû
ëû
ëû
( ) ( )
( ) ( )
cos sin
sin cos
x
n
y
t
B
B
B
B
aa
aa
éù
éù
éù
=
êú
êú
êú
-
ëû
ëû
ëû
Development of Octave programming to calculate the forces in the structural elements of a
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113
ec. (0.14)
ec. (0.15)
Next, the programming lines as a continuation of the previous lines,
to obtain normal and transverse forces and the graphs obtained in the
joints:
77 % Calculation of normal and axial forces on link BD
78 n = length(alpha);
79 forcesB = [Fp;By];
80 forcesC = [Cx;Cy];
81 array_zeros = zeros(1,n);
82 forcesD = [array_zero;-Dy];
83 forceBrot = zeros(2,n);
84 forceCrot = zeros(2,n);
85 forceDrot = zeros(2,n);
86 % loop to determine the normal and axial vectors of link BD
87 for i = 1:n
88 rotation_matrix = [cos(alpha(i)) sin(alpha(i));-sin(alpha(i))
cos(alpha(i))];
89 forceBrot(:,i) = rotation_matrix*forceB(:,i);
90forceCrot (:,i) = rotation_matrix*forceC(:,i);
91 forceDrot(:,i) = rotation_matrix*forceD(:,i);
92 endfor
93 %
94 figure(8)
( ) ( )
( ) ( )
cos sin
sin cos
x
n
y
t
C
C
C
C
aa
aa
éù
éù
éù
=
êú
êú
êú
-
ëû
ëû
ëû
( ) ( )
( ) ( )
0
cos sin
sin cos
n
y
t
D
D
D
aa
aa
éù
éù
éù
=
êú
êú
êú
-
-
ëû
ëû
ëû
Development of Octave programming to calculate the forces in the structural elements of a
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114
95 plot(alpha_degree,forceBrot(1,:));
96 title('Normal forces at joint B')
97 xlabel('Angle in degrees')
98 ylabel('Force in N')
99 grid
100 figure(9)
101 plot(alpha_degree,forceBrot(2,:),'r');
102 title('Transverse forces at joint B')
103 xlabel('Angle in degrees')
104 ylabel('Force in N')
105 grid
106 figure(10)
107 plot(alpha_degree,forceCrot(1,:));
108 title('Normal forces in the C-joint')
109 xlabel('Angle in degrees')
110 ylabel('Force in N')
111 grid
112 figure(11)
113 plot(alpha_degree,forceCrot(2,:),'r');
114 title('Transverse forces in the C-joint')
115 xlabel('Angle in degrees')
116 ylabel('Force in N')
117 grid
118 figure(12)
119 plot(alpha_degree,forceDrot(1,:));
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
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115
120 hold on
121 plot(alpha_degree,forceDrot(2,:),'r');
122 title('Normal and transverse forces at joint D')
123 xlabel('Angle in degrees')
124 ylabel('Force in N')
125 legend('Dn','Dt');
126 grid
127 %
Determination of the diagram of axial forces, shear forces and
bending moment in the BD link (bar) of the lifting table.
To make the graphs with the shear forces and bending moments of
bar BD we will determine three positions of the table whose angles
will be 6.6° , 20° and 32.7° ; at these positions we will calculate the
values of normal and transverse forces at nodes B, C and D, with the
help of the program developed with the free software OCTAVE for
numerical calculations. The results obtained also apply to link EF since
it presents the same numerical values of forces.
1 % Calculation of forces on elevated platform elements
2 clear
3 clc
4 % Data
5 M = 500; % Mass of the body to be lifted in kg.
6 L = 1.3086; % Length of the bars in m.
7 L1 = 0.400; % Piston rod length retracted, m.
8 L2 = 0.600; % Piston rod length fully out, in m.
9 a = 0.650; % Length of the position from the right edge of the mass,
in m.
10 b = 1.700; % Length from cylinder base to joint F, in m.
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
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116
11 g = 9.81; % Gravity in m/s^2.
12 %
13 alpha = [6.6*pi/180 20*pi/180 32.7*pi/180]; % conversion of
angles in radians
14 %
15 By = 0.5*M*g*a./(L*cos(alpha)); % vertical component at joint B in
N
16 Fp = 0.5*M*g*cot(alpha); % Piston force at joint B in N.
17 Cx = -Fp;
18 Cy = 0.5*M*g*(1-2*a./(L*cos(alpha)));
19 C = sqrt(Cx.^2+Cy.^2);
20 Fx = Cx;
21 Fy = 0.5*M*g*(1-a./(L*cos(alpha)));
22 F=sqrt(Fx.^2+Fy.^2);
23 Dy = Fy;
24 Ey = By;
25 %% Calculation of normal and transverse forces of the BD link
26 forcesB = [Fp;By];
27 forcesC = [Cx;Cy];
28 array_zeros = zeros(1,3);
29 forcesD = [array_zero;-Dy];
30 forceBrot = zeros(2,3);
31 forceCrot = zeros(2,3);
32 forceDrot = zeros(2,3);
33 % loop to determine the normal and transverse vectors of link BD
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
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117
34 for i = 1:3
35 rotation_matrix = [cos(alpha(i)) sin(alpha(i));-sin(alpha(i))
cos(alpha(i))];
36 forceBrot(:,i) = rotation_matrix*forceB(:,i);
37 forceCrot(:,i) = rotation_matrix*forceC(:,i);
38 forceDrot(:,i) = rotation_matrix*forceD(:,i);
39 endfor
40 Bn = forceBrot(1,:)
41 Bt = forceBrot(2,:)
42 Cn = forceCrot(1,:)
43 Ct = forceCrot(2,:)
44 Dn = forceDrot(1,:)
45 Dt = forceDrot(2,:)
46 %
Results displayed in the OCTAVE command window:
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
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118
From the results obtained from the program we made a table to sort
the values appropriately.
To establish the parametric equations that allow us to obtain the
diagrams of shear forces and bending moments in the bar BD, we will
analyze the bar disregarding the self-weight and rotating the normal
and transverse axes so that these are aligned with the horizontal and
vertical axes to facilitate the determination of the equations. In the
diagram of the free body diagram all the forces have been drawn in
positive direction, according to the table the values of Ct are positive
but the values of the forces Bt and Dt are negative therefore these
forces point downwards.
To find a mathematical expression for Bt and Dt as a function of Ct,
we work with summation of moments around one of the extremes, if
this is D, we will have:
ec. (0.16)
For an angular positiona to obtain the shear and bending moment
diagrams, we will work with the method of sections to develop the
parametric formulas.
Section BC:
ec. (0.17)
ec. (0.18)
0
()0 ()
D
tt t t
M
LB L r C LB L r C
+=
--- =® =--
å
tt
Lr
BC
L
-
æö
=-
ç÷
èø
1
t
r
VC
L
æö
=- -
ç÷
èø
0
0
o
tf
ft
M
xB M
MxB
+=
-+ =
=
å
1
ft
r
Mx C
L
æö
=- -
ç÷
èø
0
0
y
ttt
F
Lr
BV V B C
L
+! =
-
æö
-= ® = =-
ç÷
èø
å
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
Revista Científica Interdisciplinaria Investigación y Saberes , / 2023/ , Vol. 13, No. 3
119
Note: since the vertical force at point C is in the middle of the bar
length, the force is equally distributed at points B and D, the
parametric expression can be used in general when point C is not in
the middle of the bar.
CD Section:
ec. (0.19)
ec. (0.20)
Summarizing, the parametric equations of the BD member would be
as follows:
0
0
11
y
tt tt
tt t
F
BCV VBC
r
VCCC
L
+! =
+-=®=+
æö
=- - + = -
ç÷
èø
å
1
r
L
++
éù
êú
ëû
t
r
VC
L
=
0
() 0
1() 0
o
ttf
ttf
t
M
xB x r C M
r
xCxrCM
L
xC
+=
--- + =
æö
---+=
ç÷
èø
å
tt
r
xC xC
L
-- 0
10
tf
tf
rC M
x
rC M
L
++ =
æö
-+ =
ç÷
èø
1
ft
x
M rC
L
æö
=- -
ç÷
èø
1
0
1
t
ft
r
VC
L
xr
r
Mx C
L
ü
æö
=- -
ç÷
ï
èø ï
£<
ý
æö
ï
=- -
ç÷
ï
èø
þ
Development of Octave programming to calculate the forces in the structural elements of a
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120
Using the free OCTAVE software to perform the programming, we will
obtain the shear force and bending moment plots on the member.
Below are the program lines and the obtained plots:
1 % Diagram of shear force and bending moment in link BD
2 % for the hydraulic folding table.
3 clear
4 clc
5 %Data:
6 Ct = 2436.1; % Ct force value in newtons
7 L = 1.4086; % Length of bar BD in meters
8 r = 0.7043; % Length of force application Ct in meters
9 %
10 % Shear and bending moment equations
11 x = linspace(0,L,2000);
12 for i = 1:2000
13 if x(i)<=r
14 V(i) = -(1-r/L)*Ct;
15 M(i) = -(1-r/L)*Ct*x(i);
16 else
17 V(i) = r/L*Ct;
18 M(i) = -r*Ct*(1-x(i)/L);
19 endif
1
t
ft
r
VC
L
rxL
x
MrC
L
ü
=
ï
ï
££
ý
æö
ï
=- -
ç÷
ï
èø
þ
Development of Octave programming to calculate the forces in the structural elements of a
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Revista Científica Interdisciplinaria Investigación y Saberes , / 2023/ , Vol. 13, No. 3
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20 endfor
21 % graphing of shear forces and bending moments versus x
22 figure(1)
23 plot(x,V)
24 title('Shear force in link BD')
25 xlabel('x in meters')
26 ylabel('Shear in N')
27 axis([0,1.4086,-1500,1500])
28 grid
29 figure(2)
30 plot(x,M)
31 title('Bending moment at link BD')
32 xlabel('x in meters')
33 ylabel('Moment in N.m')
34 axis([0,1.4086,-1000,0])
35 grid
36 %
37 V_max = max(abs(V))
38 [M_max,pos] = max(abs(M));
39 x_max = L*(pos-1)/2000, M_max = M_max
40 %
Then, the maximum values of shear force and bending moment,
which are displayed in the software command window:
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
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From the internet sites
https://matmatch.com/es/materials/minfc37594-sae-j403-grade-
1008, and https://acerosluchriher.com/acero-sae-1008-y-1010/ ,
mechanical properties for SAE J 403 1008 steel are obtained, which
are summarized in Table 3.
Table 2.
Some physical and mechanical properties of the SAE J 403
1008 steel of the structural rectangular pipes of the DIPAC company,
located in Manta-Ecuador.
Property
Value
Unit
Density
7 800 a 7 900
kg/m3
Elastic modulus
200 a 215
GPa
Shear modulus
82
GPa
Poisson's ratio
0,29
Yield Limit
286
MPa
Maximum tensile stress
338
MPa
Maximum elongation
20
%
Area reduction
45
%
Hardness
56
HRB
Since the mechanism will be moving in different positions, the loads
will be fluctuating, therefore a conservative value of the factor of
safety for the allowable stress of the material will be used (Mott, 2009).
(Mott, 2009).
ec. (0.21)
Replacing values:
Applying eq. 1.22 and clearing the section modulus:
8
cedencia
permisible
s
s
=
286
35,8 MPa
8
permisible
s
==
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From the firm Metal Hierro, the rectangular tube has the following
characteristics:
33
2
857 440 N mm
23 984,3 mm 24 ,0 cm
N
35 ,75
mm
f
permisible
M
S
s
×
== = »
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The value closest to that obtained by the above calculation is that of
the rectangular structural tube 50 mm x 100 mm and 4 mm thick,
which has a section modulus S (W in the catalog) of 26.85 cm3 and a
distributed weight of 84.3 N/m (8.59 kgf/m).
To determine the section resistance for axial loads, we determined
that the maximum value of compressive force is concentrated in the
BC section of the link (bar) which, as can be seen in Figure 25, occurs
when the angle α of the table arms are at their minimum value of 6.6°
and numerically is 21 197 N in compression.
The safety factor for this effort is:
22
N
19, 4 o 19,4MPa en compresión
1095mm mm
21197 NF
A
s
== =
Development of Octave programming to calculate the forces in the structural elements of a
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125
which indicates that it will have no problems
in supporting these loads.
For the case of shear loads the bar is subjected to simple shear and
the shear stress will be:
The allowable shear stress for SAE J 403 1008 steel is obtained from
the following relationship:
Therefore, the safety factor is:
which indicates that there will be no problems due
to shear stresses.
However, the combination of both axial forces, generated by pure
compression and bending, and shear forces must be considered,
mainly in the area where the C-joint is placed, since an orifice in that
area produces a stress concentrator and a reduction in area.
6. Calculation of the pin diameter at the C-joint.
A single shear load is assumed, therefore, the minimum pin diameter
will be:
ec. (0.22)
If the steel selected for the pin is AISI/SAE 1045, its shear strength is
half the yield stress, therefore:
35,8
1,85
19, 4
MPa
FS
MPa
==
22
1218N N
1, 5 1, 5 1, 6 7 o 1, 67 MPa.
1095mm mm
F
A
t
æö
æö
== =
ç÷
ç÷
èø
èø
( )
286
17,9 MPa
28 16
cedencia
permisible
s
t
===
17,9
10,7
1, 67
FS ==
4
permisible
F
D
pt
=
392, 4
196, 2 MPa
22
cedencia
s
t
===
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126
The allowable shear stress, for repetitive loads, shall be
Replacing in equation 1.22, we obtain:
Therefore, the minimum diameter for the pins in joint C will be 35 mm
with AISI 1045 steel (without heat treatment).
Conclusions
This work shows us the usefulness of open source scientific
programming language programs to perform numerical calculations,
such is the case of OCTAVE whose scripts can be compatible with
those of MATLAB and can be executed in operating systems such as
Microsoft Windows and GNU/LINUX. We have used well known
physical principles of physics such as Newton's first and third laws for
rigid bodies, which allows us to parameterize the equations that will
give us the numerical values of the forces on the supports of the links
(bars) that make up the mechanism of the scissors table.
Using numerical data allows us to test the performance of the
equation program in the calculations and instead of obtaining a
numerical table of the results we have preferred to plot these values
for different positional angles of the table arms, angle alpha (α). We
have also developed the programs (scripts) for the determination of
the internal forces of the EF and BD arms that allow us to have the
maximum values of the axial forces, shear forces and bending
moments that are the basis for the determination of the dimensions
of the structural elements, whether it is a rectangular tube, plates,
structural angles or bars to be placed as arms of the table.
Reference
Dang, A.-T., & Nguyen-Dinh, N. (2020-11). A Study of Scissor Lifts
Using Parameter Design. doi:DOI: 10.1007/978-3-030-64719-
3_10.
196, 2
24,5 MPa
8
=
( )
( )
421196,5
33, 2 m m
24,53
D
p
==
Development of Octave programming to calculate the forces in the structural elements of a
scissor lift table.
Revista Científica Interdisciplinaria Investigación y Saberes , / 2023/ , Vol. 13, No. 3
127
Hamidi, B. (2012). Design and calculation of the scissors-type
elevating platforms. Open Journal of Safety Science and
Technology, 8-15.
He, S., Ouyang, M., Gong, J., & Liu, G. (2019). Mechanical simulation
and installation position optimisation of a lifting cylinder of a
scissors aerial work platform. The Journal of Engineering, 74-78.
Liu, T., & Sun, J. (2009-06). Simulative calculation and optimal design
of scissor lifting mechanism. 2009 Chinese Control and Decision
Conference (pp. 2079-2082). IEEE Xplore.
Maldonado Anrubia, M. (2016). Design of a scissor lift platform.
Generation of its virtual prototype and mechanical simulation.
Valenica, Spain: Escuela Técnica Superior Ingenieros
Industriales Valencia.
Mott, R. L. (2009). Strength of Materials. Fifth Edition. Mexico: Pearson
Education.
Navarro Fandos, A. (2019). Analysis and calculation of single scissor
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Zaragoza.
