Full transcript
0:01Welcome to this video where we will study the step-by-step method applied to the construction of motion sequences in electro-pneumatic circuits . The step-by-step method applied to electro-pneumatics has certain design requirements . The first requirement is to use two-position directional valves ,
0:23and the second is that both positions must be activated by solenoids , which is why these valve types are common . Here we have a three-way , two-position directional valve , and we can see that both positions are activated by solenoids . This 3/2 valve would
0:47typically be used to control a single-acting cylinder . Here we have a five-way , two-position directional valve . And we can also see that both positions are activated by solenoids . This valve would typically be used to
1:07move a double-acting cylinder . From these two requirements , we can say that in summary , bistable solenoid valves must be used in the step-by-step method . If you want to delve deeper into what monostable and bistable valves are , I
1:27remind you that this topic was already covered here on the channel , and I’ll leave the link to that video in the description below or in a card here . Next , we will continue with the steps required to obtain a motion sequence using the step-by-step method . The
1:51first step is to write down the motion sequence . The second step is to separate simultaneous movements into steps . If there are no simultaneous movements , then there would be one step for each movement . The third step is to
2:15write the condition that must be met to perform that step below the motion sequence . Conditions are usually expressed in terms of pushbuttons or limit switches . The fourth step is to write the element that must be
2:36activated to trigger that step above the motion sequence . Usually , these activation elements are solenoids and relays . Here we must keep in mind that in the step-by-step method , one relay is used for each step of the sequence .
2:59The fifth and final step is to install the general structure for each step of the motion sequence and assign names to the elements . What is this general structure ? The general structure is what we have in this figure . This must be installed for each step of the
3:19motion sequence . On the left side , we have the 24V power supply , and on the right side , we have the circuit ground or the 0V line . This general structure , as we can see here , is composed of a normally open contact in series with
3:41another normally open contact , and those two contacts are in parallel with a normally open contact of a relay . If we continue here , after the parallel connection , we would have a normally closed relay contact in series , followed by a relay coil . Finally , on
4:06this third line , we have a normally open relay contact and the solenoid of the directional valve . This relay coil corresponds to the relay that is activated in the current step . That is to say , in the step to which this general structure belongs . And these
4:29two normally open contacts would belong to the current step's relay ; meaning these two normally open contacts would be associated with this coil . Here we have the current step's solenoid , that is , the solenoid in charge of generating the movement we have in that
4:50step of the sequence . Here we find a normally open contact . This normally open contact depends on the step . If it is the first step , it would be a normally open contact of the push button that starts the sequence . But if
5:10it is a step other than the first , then it would be a normally open contact of the previous step's relay . Here we find another normally open contact , which would be the condition that allows me to generate the next step of the sequence and , therefore , would
5:31correspond to the limit switch that tells me the previous movement has finished . And finally , we have this normally closed contact , which would be from the next step's relay . What would be the function of each of these elements that we find in the general
5:51structure ? We have these two normally open contacts in charge of performing a preparation phase , that is , ensuring that the conditions to generate the movement have been met . We can see that if these two contacts close , the energy
6:12would continue along this line , pass through this normally closed contact , and reach the relay coil , energizing it . When this coil is energized , since it has that relay associated with these two normally open contacts , these two contacts would close . The function of
6:35this contact is to hold , that is , to keep the coil energized even though these conditions are no longer met . Then , the function of this second normally open contact associated with this relay is to perform the work phase , since when the relay coil is
6:57energized , this contact closes , allowing energy to reach the valve solenoid . With this , we activate the solenoid , and the switching of the valve position can be performed , which allows me to generate the movement associated with that step of the sequence . Finally , we have this
7:20normally closed contact which functions to cut the power to perform the next step , since this contact would be associated with the next relay . So , when the next relay is energized , this contact that was normally closed opens , interrupting the power to this relay
7:40and causing these two contacts to become de-energized , thereby cutting the power that was reaching the solenoid of the current step . To better understand all of this , we are going to do an example in which we will apply the step-by-step methodology to build a motion sequence . In this example , we
8:02want to perform the motion sequence A + , A - , B + , B - by pressing a start button in this pneumatic circuit . This pneumatic circuit is composed of two double-acting cylinders , each one controlled by a five-way , two-position directional valve . And we can see that
8:22both directional valves receive pressurized air from the same source . This would be cylinder A. We can see that the cylinder has two limit switches , one at the start of its movement , which would be limit switch FCA1 , and one at the end of its movement , meaning when it is fully
8:44extended . And this limit switch would be called FCA2 . Cylinder B also has two limit switches . FCB1 at the start of the movement and FCB2 at the end of its travel . Regarding the directional valves , for the valve that controls
9:05cylinder A , we have solenoid S1 for the left position and solenoid S2 for the right position . And in the case of the valve that controls cylinder B , we have solenoid S3 for the left position and solenoid S4 for the right position . Next , we are going to apply all the
9:25steps of the step-by-step methodology to obtain the motion sequence . The first step is to write down the motion sequence . So , we would have A + , A - , B + , B - . The second step is to separate the movements that are performed simultaneously into steps . In the case
9:49of this motion sequence , we do not have simultaneous movements , since first cylinder A extends , then cylinder A returns , then cylinder B extends , and finally cylinder B returns . Each movement is unique . So , they would remain separated in this way . This would be the first step , this would be
10:10the second step , this would be the third step , and this would be the fourth step . Step three of the methodology is to write the condition that must be met to perform each step at the bottom of the sequence . And let's remember that conditions are expressed in terms of push buttons or
10:32limit switches . And the fourth step is to write at the top of the movement sequence the element that must be activated to generate the step . By elements , we refer to solenoids and relays . And we must keep in mind that one relay is used for each step of the sequence . So , since we have four steps
10:55here , it means that to carry out this movement sequence , we will require four relays . We are going to apply steps three and four simultaneously . We have that to perform the first movement , a condition must be met , which is
11:16pressing the start button . To generate this A + movement , we have to activate an element , which would be a relay . And since it is the first relay we are going to introduce in this circuit , we will call it R1 . This relay should
11:38activate one of these four solenoids to generate that A + movement . And we can see that , to generate A + , solenoid S1 should be activated , since , by activating this solenoid , this position is switched , allowing pressurized air to pass from port one to port four ,
11:59continuing through this line to fill this chamber and produce the extension , while this other chamber is evacuated to exhaust 3 . So , we would have that to produce A + , we have to energize solenoid S1 , which would be done via relay R1 . With this , we would have the
12:22extension movement of cylinder A. How do we know that the A + movement has finished ? We would know because cylinder A would hit the limit switch FCA2 . So , that would be the condition to move from this first step to the second step . It would be that cylinder
12:45A hits the limit switch FCA2 . To produce this second step , we will require a second relay , which we will call relay R2 . To produce A - with this relay R2 , we must energize one of these four solenoids . The solenoid in charge
13:10of producing the retraction movement of cylinder A would be solenoid S2 , since , by switching solenoid S2 , we would have this valve position in which pressurized air would pass from port one to port two to continue through this line and fill this chamber , producing the retraction , while the air
13:31trapped in this other chamber would escape through this line in connection with port . four to exhaust 5 . How do we know that the A - movement has finished ? We would know because cylinder A would return completely and hit the limit switch FCA1 . So , FCA1 would be the
13:55condition to move from the second step to the third step . To generate this third step , we have to activate a third relay that we will call R3 . And with this relay , we will generate the B + movement ; for which we would have to energize solenoid S3 , since , by
14:15energizing this solenoid , this position is switched , which allows air to pass from port one to port four , continuing along this line to fill this chamber , causing the cylinder to extend , while the air trapped in this other chamber is discharged into the atmosphere via the connection from port 2 to exhaust 3
14:35. How do we know that this B + movement has finished ? Well , it would be because cylinder B touches the limit switch FCB2 , and with that , we would have the transition from the third step to the fourth step . For this fourth step , we
14:56are going to use a fourth relay that we will call R4 . to produce B - with relay R4 . we would have to energize solenoid S4 . Since by energizing this solenoid , we switch this right position of the valve , in which the pressurized air passes from port one to port two . It
15:19passes through this line , fills this chamber , produces the retraction , and this other chamber of the cylinder is discharged into the atmosphere via the connection from port four to exhaust 5 . So , we would have relay R4 activating solenoid S4 . And how do we know the movement sequence ends ? How do we know
15:40this B - movement ends ? We would know because the cylinder , upon fully returning , would touch the limit switch FCB1 . Up to this point , we would have already applied steps three and four . In the lower part of the movement sequence , we would have the conditions
16:02that must be met to perform each step , and in the upper part , we would have the elements that must be activated to generate each step , the relays and solenoids . We only need to apply step five , which is to install the general structure for each step of the sequence
16:23and assign the names of the elements . Since we have four steps , it means that for each step we require a general structure . So , we would have four general structures . On the left side , we would have the 24V line , and on the
16:45right side , the ground or 0V line . Having the four general structures , we are going to start naming the different elements . We start with the relays . This would be the relay coil that allows me to generate the first step , which would be relay R1 . Here we would
17:09have the relay coil associated with the second general structure , which is associated with the second step . So this coil would be for relay R2 . Continuing in this way , we would have this as the coil for relay R3 for the third step , and this would be the coil
17:31for relay R4 for the fourth step . As we saw previously , these two normally open contacts are associated with the current relay . So , these two normally open contacts would be contacts of
17:51relay R1 . Likewise , in that second general structure , these two normally open contacts would be for the relay of this current step , that is , they would be contacts of relay R2 . These two normally open contacts would be contacts of relay R3 and these two normally open contacts would be from
18:14relay R4 . Now we are going to name the different solenoids . We have that this relay R1 , according to the methodology we apply here , is in charge of activating solenoid S1 . So , this solenoid of the first general structure would be solenoid S1 . Relay R2 is in
18:39charge of activating solenoid S2 . So this solenoid of the second structure would be S2 . Relay R3 is in charge of activating solenoid S3 . So , this solenoid that we find in the third general structure would be S3 . And
19:00finally , we have that R4 is in charge of activating S4 . So this solenoid of the fourth general structure would be S4 . Now we are going to assign names to the normally open contacts that we find in the first line of each of the general structures . This normally open
19:24contact can be one of two things . It could be from the push button that starts the sequence or it could be a normally open contact from the relay of the previous step . Whenever we are working in the first general structure , that normally open contact would be
19:46from the start button . And in the other general structures , different from the general structure associated with the first step , they would be normally open contacts of the relay from the previous step . So , the symbol would change , it would no longer be this symbol of a normally open contact , but this symbol
20:08of a normally open relay contact . And since it is from the previous step and here we are in the second step which is associated with R2 , then this normally open contact would be of R1 . Here we would then have a normally open contact from the previous step which would be R2 . And here we would have a normally
20:29open contact from the previous step which would be R3 . Now we are going to continue naming the second normally open contact that we find in the first line of each of the general structures . These normally open contacts will be associated with the limit switch that
20:51allows the condition to be satisfied in order to perform each of those steps . Here , I like to always start with the second general structure . So we have that in order to generate the second step , which is A - , a condition must be met , which is that the limit switch
21:11FCA2 must have been triggered . So this normally open contact would be for FCA2 . To be able to generate the third step , which is B + , we have to meet a condition , which is that the limit switch FCA1 must have been triggered . So , this normally open contact would be
21:32from FCA1 . To generate the fourth step , B minus , we must meet a condition : the limit switch FCB2 must have been triggered . So , this normally open contact would be from FCB2 , and to finish the movement sequence , the FCB1
21:53contact must be pressed . So , this normally open contact in the first general structure would be this last one , FCB1 . With this , we only need to name the normally closed contacts found in each of the general structures .
22:17Remember that these normally closed contacts are associated with the relay of the following step . So , this first general structure is for the first step , that is , the step associated with R1 , and the next step is associated with R2 . Therefore , this normally closed
22:37contact would be from the next relay , R2 . This normally closed contact would be for the next step , that is , for R3 . This normally closed contact would be for the next step , that is , for R4 . And this last normally closed contact would be for the next step . Since we don't
23:01have an R5 step here , but rather the sequence repeats , it would be a normally closed contact of R1 . To finish , we would have to join the different connections of the general structures so they are all powered at 24V and connected to ground . And with
23:23this , we would have the electrical circuit that allows us to perform the movement sequence A + , A - , B + , B - using the step-by-step methodology . In the step-by-step method , you can have steps that involve the simultaneous movement of two or more cylinders . Or we can
23:44also have steps that involve meeting two or more conditions . And for these cases , the general structure we reviewed earlier is modified . First , let's review how the general structure would be modified when we want to execute simultaneous movement of two or more cylinders . In this case , we have
24:08to add one of these lines to the previous general structure for each additional cylinder . These first three lines correspond to the general structure seen previously . We have the two normally open contacts in series installed in parallel with the second line , where we find the relay's
24:29normally open contact . Continuing with that parallel , we had a normally closed contact and then the relay coil of the current step . And in the third line , to trigger a cylinder's movement , we had a normally open contact of the current step's relay that energized the valve
24:52solenoid causing the cylinder movement . So , if we have the simultaneous movement of two cylinders in one step , we would need to have two of these lines . So , this other line would be added , in which we have a normally open contact in series with the valve solenoid that causes that cylinder's
25:13movement . And for every additional simultaneous movement , we would add one of these lines . Secondly , we have the modification of the general structure when two or more activation conditions must be added , that is , two or more limit switches . In this case , the
25:38modification is made to the first line of the general structure . We have the usual two normally open contacts , and a normally open contact is added for each additional condition or each additional limit switch that must be activated to trigger the sequence step . To
26:02understand this better , let's do an example . In this example , we want to execute the motion sequence A + , B + , A - B - simultaneously when a start button is pressed . Here we have the pneumatic
26:22circuit for which we want to perform the motion sequence , and we find two double-acting cylinders , each controlled by a five-way , two-position directional valve . We see that the two valves are connected to the same pressurized air source . This would be cylinder A. This would be cylinder B.
26:43We can see that both cylinders have limit switches at the beginning and end of their travel . And we have that each valve has two solenoids , S1 and S2 for the cylinder A valve and S3 and S4 for the cylinder B valve . With this movement sequence , we want cylinder A
27:05to extend first , then cylinder B to extend , and then both cylinders to retract simultaneously . Let's apply the methodology steps . First , we have to write the motion sequence . We would have A + , B + , A - B - , separating the
27:25steps performed simultaneously . We would have a first step that would be A + , a second step that would be B + , and a third step that would be the simultaneous retraction movement of the two cylinders . A - B - . We are going to execute steps 3 and 4 simultaneously .
27:48In step 3 , we write at the bottom of the sequence the condition that must be met to perform the step in terms of buttons or limit switches . And in step 4 , we write at the top of the motion sequence the element that must be activated to generate the step , keeping in mind that those elements would be
28:07solenoids and relays , and that one relay must be used for each step of the sequence . The condition that must be met to generate the A + movement is that the start button is pressed . To be able to produce this first step , we will need a relay . Since it is the
28:27first relay , we will call it R1 . And with R1 , we are going to activate solenoid S1 . By energizing this solenoid , this position switches , allowing pressurized air to pass from port one to port four , continue along this line , fill this cylinder chamber , produce the extension , and have this
28:48other chamber exhausted to the atmosphere by connecting port 2 to exhaust 3 . To know that this A plus movement finished , a condition must be met , which is that cylinder A has touched limit switch FCA2 . So , that is the condition that allows us to make
29:09the transition from the first step to the second step . To be able to execute this second step , we will require a second relay called R2 . And with this relay R2 , we are going to activate solenoid S3 . Because by energizing this solenoid , this position switches . The
29:30pressurized air can pass from port one to port 4 , filling this cylinder chamber and producing the extension movement of cylinder B , while this other chamber is discharged to the atmosphere by connecting port 2 to exhaust 3 . To be able to know that this B plus movement is finished , we would
29:52have to meet a condition , which is that cylinder B has touched limit switch FCB2 . So , that will be the transition from step two to step three . To be able to execute this step three , we will require a relay . Although we have the movement of two cylinders , we only
30:14require one relay , since step four of the method tells us that one relay is used for each step of the sequence . It does not matter if we have a single cylinder movement or simultaneous movement of several cylinders . So , this simultaneous A-B movement only requires one relay , and as it is the third relay
30:36, it would be R3 . With this relay , we are going to activate the solenoids that allow me to do A - and B - . To produce A - , we would have to energize solenoid S2 . So that this position switches and pressurized air can flow from port one to port two to fill this
30:59chamber , produce the retraction , and have this other chamber discharged to the atmosphere by connecting port 4 to exhaust 5 . And to produce B - , we would have to energize solenoid S4 , which would allow switching this position in which pressurized air can pass from port one to port 2 to fill this chamber
31:19, produce the retraction , and have this other chamber discharged to the atmosphere by connecting port 4 to exhaust 5 . So R3 will be in charge of energizing two solenoids , S2 and S4 . To know that the movement sequence is
31:40finished , we must meet two conditions : both cylinder A and cylinder B must be fully retracted , meaning these two cylinders must touch the limit switches FCA1 and FCB1 . So , we would have two conditions for completing the sequence : FCA1 and FCB1 . With this in place , we
32:05can move to the fifth step , which is installing the general structure for each step and assigning the component names . Since we have three steps , it means we require three general structures . Here we have the first structure , second structure , and third
32:26structure . Here on the left side , we have the 24V power supply , and on the right side , we have the ground connection . We are going to start naming the different elements that we have in these general structures . We begin with the relays . In the first general structure , we have to activate
32:46relay R1 . So , this would be the coil for relay R1 . In the second general structure corresponding to the second step , we have to activate relay R2 . So , this would be the coil for relay R2 . And in the third general structure corresponding to the third step , we have to activate relay R3 . So , this
33:07would be the coil for relay R3 . These two normally open contacts that we find in the first structure would belong to relay R1 . These two contacts would be for relay R2 , and these three contacts would be for relay R3 . Why three
33:29contacts ? Because remember that since we have simultaneous movement of two cylinders here , we would have to add a line with a normally open contact and the valve solenoid to execute that additional movement . So , compared to the two structures we had before , a
33:49fourth line with those two elements is added . Now let's name the solenoids . With the first general structure , we have to activate AS1 . With the second general structure , we have to activate S3 . And with the third general
34:10structure , we have to activate S2 and S4 . So , this line would activate S2 , and this line would activate S4 . Let's continue with the normally open contacts found on the first line of each of the general structures . In the
34:30first , this normally open contact would be for the start button . In the second line , we replace that normally open contact with a normally open contact from a relay of the previous step . So , here we would have to place a normally open contact from R1 . And the same here
34:53in the third general structure , it would be a normally open contact from the previous step , which was R2 . Now we continue with the second normally open contacts of each of the general structures . With these , I always like to start with the second line , since
35:14this normally open contact would be from the limit switch that allows the transition from step one to step two ; that is , it would be a normally open contact of FCA2 . The next normally open contact would be the one that allows the transition from step two to step
35:34three ; that is , it would be a normally open contact of the limit switch FCB2 . And finally , we go back to the first line . Here we would need to place normally open contacts for this last condition that allows finishing the motion sequence . And since we have two
35:55conditions , two normally open contacts must be added . The first would be from FCA1 and the second from FCB1 . This is the second modification we saw previously . Modification when we have two or more conditions to meet to execute a step . Lastly , we have to name
36:15the normally closed contacts for each general structure . Remember that these contacts are from the next relay , from the relay of the following step . Since this is the first step , this normally closed contact would be from the next one , that is , R2 . This contact would be
36:38from the next step , that is , R3 . And this would be the next step , but since we have an R4 , it would be from the first one ; it would be a normally closed contact of R1 . To finish , we would need to connect all the lines together , so that all the general
36:58structures are powered at 24V and connected to ground . With that , we would have the electrical circuit that allows us to execute the A + B + A - B - motion sequence . With this , we have finished this video on the step-by-step methodology for electropneumatic circuits . If this video was helpful , I
37:20invite you to subscribe to the channel and like this video . In upcoming videos , we will be performing different motion sequences using the step-by-step method and simulating them in FluidSIM .