embed_youtube("PkUh9-hy6bo")Automated Systems Notes
Automated Production Lines
Manufacturing systems with fixed routing. They can have multiple workstations linked together by material handling systems that transfer the parts from one station to another.
The slower workstation sets the pace of the line.
Transfer of the line can be palletized, to hold and move the work parts between stations, or free: where the part geometry permits transfer without pallets.
Linear Transfer Systems
Work part transport mechanisms can be divided into two categories: (1) linear transport systems for in-line and segmented in-line systems, and (2) rotary indexing mechanisms for dial-indexing machines. Some of the linear transport systems provide synchronous movement, while others provide asynchronous motion. The rotary indexing mechanisms all provide synchronous motion. (Mikell P. Groover 2018)
Analysis of transfer lines
The analysis of transfer lines considers three issues:
Line balancing
Process technology used at the work center (subtractive manufacturing, additive manufacturing, etc.)
System reliability
Line balancing has already been covered in the course. Process technology includes materials and machining process technologies used in the work center, this is out of the scope of this course. As for reliability in an automated line, the failure of one or more workstations has an almost immediate effect on the performance of all the line. The analysis that follows considers the following assumptions:
Workstations do not include assembly operations
Processing times at all stations are constant, but not necessarily equal
The work transport used is synchronous
Let \(T_c\) be the ideal processing time of the production line be defined as the sum of processing and transfer time of the slowest workstation:
\[ T_c=MAX (T_{si}) +T_{ri} \]
Where \(T_{si}\) and \(T_{ri}\) are the service and re-positioning times respectively.
If we include breakdowns and stoppages we can calculate the average production time \(T_p\) as:
\[T_p=T_c+FT_d\]
where \(F\) is the downtime frequency \([=] \frac{stops}{cycle}\) and \(T_d\) is the average downtime1 in, say, \(\frac{min}{stop}\). The value \(FT_d\) is downtime time on a per cycle basis.
In general, \(F\) is the count or frequency of failures over all workstations:
\[ F=\sum_i^n f_i \]
where \(n\) is the number of workstations and \(f_i\) is the count of failures at workstation \(i\).
The actual average production rate \(R_p\) is easy to calculate as:
\[ R_p=\frac{1}{T_p} \]
Note that the “ideal” production rate \(R_c\) can be estimated using the ideal cycle time:
\[ R_c=\frac{1}{T_c} \]
We can estimate the proportion of up-time \(E\) of the line with the ratio:
\[ E=\frac{T_c}{T_p}=\frac{T_c}{T_c+FT_d} \]
If we define \(D\) as the proportion of downtime, then:
\[ D=\frac{FT_d}{T_p}=\frac{FT_d}{T_c+FT_d} \]
\(E\) and \(D\) are related: \(E+D=1\).
If we want to estimate the cost of unit produced, we can use:
\[ C_{pc}=C_m+C_oT_p+C_t \]
where \(C_{pc}\) is the cost per unit produced, \(C_m\) the cost of material in $ per pc, \(C_o\) the cost of operating the line in $ per time, and \(C_t\) is the cost of tooling in $ per pc.
Geneva Mechanism
The Geneva drive or Maltese cross is a gear mechanism that translates a continuous rotation movement into intermittent rotary motion.
embed_youtube("3nv2wRo6dEY")Calculation of service and re-positioning time in a Geneva mechanism
Consider a Geneva mechanism with six slots for a six-station dial-indexing table. Each turn of the driver results in a 1/6-th rotation of the worktable, or 60°. The driver only causes motion of the table through a portion of its own rotation. For a six-slotted Geneva, 120° of driver rotation is used to index the table. The remaining 240° of driver rotation is dwell time for the table, during which the operation must be completed on the work unit.
For a general approach, call \(\theta\) the angle of rotation of the work table in degrees. then, the number of equally spaced slots in the Geneva \(n_s\) is related to \(\theta\) by the following equation:
\[ \theta =\frac{360}{n_s} \tag{1}\]
The angle of driver rotation during indexing is \(2\theta\) . The angle of the driver when the table experiences dwell time is \(360-2\theta\) .
In a Geneva mechanism, the number of slots is usually 4,5,6 or 8. Each slot can be considered a work station position.
Thus, the total cycle time \(T_c\) is related to the rotational speed \(N\) of the driver by the equation:
\[ T_c=\frac{1}{N} \tag{2}\]
As part of the total cycle time, the service time –or dwell time– is obtained with the following relation:
\[ T_s= \frac{180+ \theta}{360N} \tag{3}\]
And the indexing, or re-positioning time is:
\[ T_r= \frac{180- \theta}{360N} \tag{4}\]
Walking Beam Transfer System
In this system, the parts are synchronously lifted up from their respective stations by a transfer beam and moved one position ahead, to the next station. The transfer beam lowers the parts into nests that position them for processing at their stations
embed_youtube("rIMa8pkLHBw")Automated Assembly Lines
embed_youtube("iR5anYJxXhM")Custom fully automated assembly systems
embed_youtube("hkPeOxM0hCI")Multi-Station vs Single Station assembly lines
Multi-station assembly lines have the following characteristics:
Faster cycle rates
High production quantities
More operations
More components per assembly
Single-Station assembly lines are usually:
Best suited for robotic assembly
Intended for lower production quantities.
Analysis of assembly lines
General considerations:
Parts delivered at each station should be feed at a net rate greater or equal to the cycle rate of the assembly work-head. If not, the performance of the assembly system is limited by the parts delivery system.
Quality and defect rates of component impacts the performance of the assembly system. Poor quality can generate jams and assembly of defective products.
If the number of assembly stations increases, up time efficiency and production rate are greatly affected by parts quality and reliability.
The slowest station determines the cycle time of a multi-station assembly system.
If a single-station assembly cell has the same number of assembly tasks than a multi-station assembly system:
The production rate of the single-station system is lower than the multi-station system
The up time efficiency of the single-station system will be higher than the multi-station system.
Multi-station systems are better suited for high production applications and long production runs.
Single-station assembly cells tend to have longer cycle times and are appropriate for mid range production runs.
Storage buffers can be used to isolate station s from breakdowns from the automated stations.
Parts delivery systems at workstations
embed_youtube("AZg746y_eos")Let \(\theta\) be the proportion of components of the delivery system that are correctly oriented. Let \(f\) be the component feed rate. Then, the effective feed rate of components is \(f\theta\) . The remaining proportion \(1-\theta\) is recirculated into the hopper. \(f\theta\) should be enough to keep up with the rate of the production machine \(R_c\) .
If \(f\theta>R_c\) we need to limit the queue in the feed track. Usually this is done through a limit switch or optical sensor to turn off the feeder when the feed track is full. This is called the high level sensor. Also, its location defines the active length \(L_{f2}\) of the feed track. If \(L_c\) is the length of the component on the feed track, then the number of parts, or capacity, the track is \(n_{f2}=\frac{L_{f2}}{L_c}\). 2
Another sensor, on the feed track and at some distance from the first sensor, can be used to restart the feeder. This is called the “low level sensor.” Let the location of this sensor be \(L_{f1}\) . Then, the number of components in the feed track is \(n_{f1}=\frac{L_{f1}}{L_c}\).
The rate of reduction of parts in the feed track is \(R_c\), the cycle rate of the automated assembly work head. On average, the rate of increase of parts when the feeder is turn on is \(f\theta-R_c\) . Because this last rate is not uniform, we must have \(n_{f1}\) large enough to avoid stock outs.
Example
A feeder selector device at a workstation of an assembly machine has a feed rate of 70 components per minute and provides a throughput of 2 parts in 8. The ideal cycle time of the assembly machine is 10 seconds the low level sensor is set at 10 components and the high level sensor at 40.
How long will it take for the supply of components to be depleted from the high level sensor to the low level sensor once the feeder selector device is turned off?
Hoe long will it take for the components to be resupplied from the low level sensor to the high level sensor, on average, after the feeder selector device is turned on?
What are the time proportions that the feeder selector device is turned on and off?
References
Mikell P. Groover. 2018. Automation, Production Systems, and Computer-Integrated Manufacturing. Fifth Edition. Pearson.