CAD/CAM/CIM

CAD/CAM/CIM Unit III

Analysis of Automated Flow Lines In Analysing the Performance of Automated flowlines two general problem areas must be distinguished. 1. Production Process 2. Flowline Performance 1. Production Process: Consider a Transfer line that performs a series of machining operations; this technology includes proper specification & use of cutting tools, the metallurgy & Machinability of work material, chip control, machining economics, machine tool vibration etc. Many of the problems encountered in the operation of a metal cutting transfer line are directly related to and can be solved by the application of good machining principles. The same is true for other production processes in each area of production by making the best use of the given process technology each individual workstations on the line can be made to operate in an optimal way. 2. Flowline Performance: This Problem area is concerned with the system aspects of designing & running the line. The Problem of Reliability is normally associated with the operation of an automated flowline. Since the line operates as a single mechanism, failure of one component of the mechanism will result in stoppage of the entire line, however there are approaches to this problem that transcend the manufacturing process at individual station. Some of the parameters that affect the performance of the Automated Flowlines are as follows: a. How much Improvement can be obtained by using one or more buffer storage zone? b. What is the effect of component quality on the operation of an automated assembly machine? c. How will the use of manual workstation affect the line? In addition to reliability problem another system design problem is proper balancing of the flow line. In Line Balancing the objective is to spread the total workload as evenly as possible among the station in the line.

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Terminologies & Analysis of Automated Flow Lines Flow line performance is measured by following parameters: 1. Cycle Time 2. Average Production Rate 3. Line Efficiency 4. Cost per piece Assumptions in the analysis of automated flowlines: a) The work part transfer is synchronous type b) The workstation perform processing operations only (Similar to machining operation) no Assembly Operations are involved c) Processing time at each work station is constant (But may not be equal to that of other workstation) d) Parts are loaded at the first workstation and after processing these parts are transported to the subsequent work station for further processing Terminologies: a) Cycle Time (Tc) : The Time consumed from the moment it enters a work station till it reaches the next workstation defines the theoretical or ideal cycle time for a flow line. Thus it comprises different components such as: a. Actual processing time at the longest work station. b. Ideal time spent, if any, at the work station. c. Transfer time.

Therefore, Tc= Process Time + Transfer Time (Tr) + Idle Time (If Any) Note: In An Automated Flowline the the process taking longest time will set the pace of the Flowline & Hence Longest processing time should be considered b) Production Time (Tp): Production Time is always greater than cycle time by an amount equal to the down time. The average production time can be mathematically expressed by the relation, Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III

Where, Tp = Average Production Time Tc= Cycle Time F= Frequency of downtime (Linestops/cycle) Td= Average downtime per line stop

Few Possible Reasons for Downtime are as follows: a) Mechanical Failure: Failure of workstation or the transfer system Eg: Failure of gears, linkages, Jam in the sliders/conveyor mechanism & so on. b) Electrical Failure: Loss of main power supply, Electric Motor Failure, Short Circuit, Overload/Switch gear Tripping etc c) Tool Failure: Toll Wear, Breakage or jam with the work part due to excessive cut/feed or incorrect programming of the feeds/depth of cut. d) Tool Setting: Need for Tool Change, Tool Adjustment etc e) Input/output delays: Stockout of starting Parts f) Quality Related Issues: Poor Quality Part, Poor Quality Tool etc g) Scheduled Maintenance h) Other Reasons: Accidents, No Line worker etc... c) Production Rate: the Production rate is the reciprocal of average production time. Thus it is expressed as

Where, Rp = Production Rate (parts/min or parts/hour) Tp = Production Time (In Mins or Hrs) Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III The Ideal Production Rate is given by

Where, Rc = Ideal Production rate. Tc = Cycle Time (Note: Ideal Production Rate will not have any downtime) d) Line Efficiency: It refers to the proportion of uptime on the flowline & indicative of the reliability (rather than efficiency). It is given by the relation,

Where E = the proportion of uptime on the production line. Tc= Cycle Time Tp= Production Time F= Frequency of Downtime Td= Average Downtime The other measure of performance/reliability is the proportion of the downtime, which is nothing but,

D=1 – E Downtime can also be expressed by the relation,

The sum of proportion of uptime & downtime is always unity i.e.

E + D =1 (100% efficient flowline) e) Cost per Piece (Cpc) : An important economic measure of the performance of an automated production line is the cost of the unit produced.

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III The cost of 1 piece includes the cost of the starting blank that is to processed, the cost of time on the production line & the cost of the tool consumed. The cost per unit can be expressed as the sum of three factors:

Where, Cpc = cost per piece (Rs / pc) Cm = cost of material (Rs / min) Tp = average production time per piece (min / pc) Ct = cost of tooling per piece (Rs / pc) Co = the allocation of capital cost of the equipment over the service life, labour to Operate the line, applicable overheads, maintenance, & other relevant costs all reduced to cost per min

FLOWLINES WITHOUT STORAGE BUFFER

In the analysis of Automated flowlines without buffer two approaches are followed a. Upper Bound Approach b. Lower Bound Approach a) Upper Bound Approach: The upper bound approach provides an upper limit on the frequency on the line stops per cycle. In this approach we assume that the part remains on the line for further processing. It is possible that there will be more than one line stop associated with a given part during its sequence of processing operations. Let Pi = probability or frequency of a failure at station i where i = 1, 2,………. n Since a part is not removed from the line when a station jam occurs it is possible that the part will be associated with a station breakdown at every station. The expected number of lines stops per part passing through the line is obtained by summing the frequencies Pi over the n stations. Since each of the n stations is processing a Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III part of each cycle, then the expected frequency of line stops per cycle is equal to the expected frequency of line stops per part i.e. 𝑛

𝐹=

𝑃𝑖 𝑖=1

where F = expected frequency of line stops per cycle Pi = frequency of station break down per cycle, causing a line Stop n = number of workstations on the line If all the Pi are assumed equal, which is unlikely but useful for computation purposes, then F = n.p Where, all the Pi are equal (P1= P2= P3........Pn = P) b) Lower Bound Approach: The lower bound approach gives an estimate of the lower limit on the expected frequency of line stops per cycle. Here we assume that a station breakdown results in destruction of the part, resulting in removal of the part from the line & preventing its subsequent processing at the remaining workstations. Let Pi = the probability that the work piece will jam at a particular station i. Then considering a given part as it proceeds through the line, Pi = probability that the part will jam at station 1 (1 - Pi) = probability that the part will not jam station 1 & thus will available for processing at subsequent stations. A jam at station 2 is contingent on successfully making it through station 1 & therefore the probability that the same part will jam at station 2 is given by P1 (1 – P2 ) Generalising the quantity Pi (1 – Pi - 1) (1 – Pi - 2) = (1 – P2) (1 – P1) Where i = 1,,2, ……….n

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Is the probability that a given part will jam at any station i. Summing all these probabilities from i = 1 through i = n gives the probability or frequency of line stops per cycle. Probability that the given part will pass through all ‘n’ stations without a line stop is 𝑛

(1 − 𝑃𝑖 ) 𝑖=1

Therefore the frequency of line stops per cycle 𝑛

𝐹 =1−

(1 − 𝑃𝑖 ) 𝑖=1

If all the probabilities, Pi, are equal, Pi = P, then 𝐹 = 1 − (1 − 𝑃)𝑛

Because of parts removal in the lower bound approach, the number of parts coming of the Line is less than the number launched onto the front of the line. If F= frequency of line stops & a part is removed for every line stop, then the proportion of Parts produced are (1 - F). This is the yield of the production line. The production rate Equation then becomes:

Where Rap = average actual production rate of acceptable parts from the line.

Problems on Automated Flowlines without Buffer Storage: 1. A ten-station transfer machine has an ideal cycle time of 30 sec. The frequency of line stops is 0.075 stops per cycle. When a line stop occurs, the average downtime is 4.0 min. Determine (a) average production rate in pc/hr, (b) line efficiency, and (c) proportion downtime. Solution: (a) Tp = 0.5 + 0.075(4) = 0.5 + 0.3 = 0.8 min

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Rp = 1/0.8 = 1.25 pc/min = 75 pc/hr (b) E = 0.5/0.8 = 0.625 = 62.5% (c) D = 0.3/0.8 = 0.375 = 37.5% 2. Cost elements associated with the operation of the ten-station transfer line in Problem 16.4 are as follows: raw workpart cost = Rs 0.55/pc, line operating cost = Rs 42.00/hr and cost of disposable tooling = Rs 0.27/pc. Compute the average cost of a workpiece produced. Solution: Refers to Problem 16.4: Cpc = 0.55 + 42(0.8)/60 + 0.27 = 0.55 + 0.56 + 0.27 = Rs 1.38/pc 3. In Problem 1, the line stop occurrences are due to random mechanical and electrical failures on the line. Suppose that in addition to these reasons for downtime, that the tools at each workstation on the line must be changed and/or reset every 150 cycles. This procedure takes a total of 12.0 min for all ten stations. Include this additional data to determine (a) average production rate in pc/hr, (b) line efficiency, and (c) proportion downtime. Solution: Refers to Problem 16.4: (a) F1Td1 = 0.075(4.0) = 0.3 min F2Td2 = 12.00/150 = 0.08 min Tp = 0.5 + 0.3 + 0.08 = 0.88 min Rp = 1/0.88 = 1.13636 pc/min = 68.2 pc/hr (b) E = 0.5/0.88 = 0.5682 = 56.82%, (c) D = 0.38/0.88 = 0.4318 = 43.18% 4. A 30 station Transfer line is being proposed to machine a certain component currently Produced by conventional methods. The proposal received from the machine tool builder states that the line will operate at a production rate of 100 pc / hr at 100% efficiency. From a similar transfer line it is estimated that breakdowns of all types will occur at a frequency of F = 0.20 breakdowns per cycle & that the average downtime per line stop will be 8.0 minutes. The starting blank that is machined on the line costs Rs. 5.00 per part. The line operates at a cost for 100 parts each & the average cost per tool = Rs. 20 per cutting edge. Compute the following: 1. Production rate 2. Line efficiency 3. Cost per unit piece produced on the line Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Solution: 1. At 100% efficiency, the line produces 100 pc/hr. The reciprocal gives the unit time or ideal cycle time per piece. Tc = 1/100 = 0.010hr / pc = 0.6 mins The average production time per piece is given by: Tp = Tc + FTd = 0.60 + 0.20 (8.0) = 0.60 + 1.60 Tp= 2.2 mins / piece Rp = 1 / 2.2m = 0.45 pc / min = 27 pc / hr Efficiency is the ratio of the ideal cycle time to actual production time E = 0.6 / 2.2= 27 %

Tooling cost per piece

= Rs 6/ Piece

The hourly ratio of Rs 100 / hr to operate the line is equivalent to Rs. 1.66 / min. Cpc = 5 + 1.66 (2.2) + 6 = 5 + 3.65 + 6 Cpc = Rs 14.65 / piece

5. For a 10 Station Transfer line the following data is given: P=0.01 (All Stations have an equal probability of failure), Tc=0.5 min , Td= 5.0 min Using the upper bound approach & Lower Bound Approach, determine a. The Frequency of Line stops b. The avg Production rate c. The Line Efficiency Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Solution: Upper Bound Approach a. Frequency of Line Stops, F Upper Bound Approach assumes that no work part is taken out of the station and the frequency of line stops is given by F= np =10X 0.01 F = 0.1 breakdowns/cycle b. Avg Production rate Rp W.k.t. Rp=1/Tp Tp= Tc +F Td =0.5 + 0.1 X 5 = 1 min/pc Therefore Rp= 1/1= 1 pc/min

c. Line Efficiency, E E= Tc/Tp = 0.5/1 =0.5 E=50%

Solution: Lower Bound Approach a. Frequency of Line Stops, F Lower Bound Approach assumes that work part is taken out of the station and the frequency of line stops is given by 𝑭 = 𝟏 − (𝟏 − 𝒑)𝒏 =1- ( 1- 0.01)10 F = 0.096 Breakdown/cycle b. Avg Production Rate, Rp Rp= (1 – F ) / Tp Tp=Tc + F Td Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III = 0.5 +0.096 X 5 Tp= 0.98 min/pc Rp= (1 – 0.096)/ 0.98 Rp= 0.922 pc/min Rp= 55.35 pc/hr

c. Line Efficiency, E E = Tc/ Tp = 0.5/0.98 =0.51 E=51% Note: Line efficiency is greater with lower bound approach even though production rate is lower. This is because lower bound approach leaves fewer parts remaining on the line to jam. 6. In the operation of a certain 15-station transfer line, the ideal cycle time = 0.58 min. Breakdowns occur at a rate of once every 20 cycles, and the average downtime per breakdown is 9.2 min. The transfer line is located in a plant that works an 8-hr day, 5 days per week. Determine (a) line efficiency, and (b) how many parts will the transfer line produce in a week? Solution: (a) Tp = 0.58 + 9.2/20 = 0.58 + 0.46 = 1.04 min E = 0.58/1.04 = 0.5577 = 55.77% (b) Rp = 60/1.04 = 57.69 pc/hr Weekly production = 40(57.69) = 2307.7 pc/wk.

7. A ten-station rotary indexing machine performs nine machining operations at nine workstations, and the tenth station is used for loading and unloading parts. The longest process time on the line is 1.30 min and the loading/unloading operation can be accomplished in less time than this. It takes 9.0 sec to index the machine between workstations. Stations break down with a frequency of 0.007, which is considered equal for all ten stations. When these stops occur, it takes an average of 10.0 min to diagnose the problem and make repairs. Determine (a) line efficiency and (b) average actual production rate.

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Solution: (a) F = np = 10(0.007) = 0.07 Tc = 1.30 + 0.15 = 1.45 min Tp = 1.45 + 0.07(10) = 1.45 + 0.7 = 2.15 min/pc E = 1.45/2.15 = 0.674 = 67.4% (b) Rp = 1/2.15 = 0.465 pc/min = 27.9 pc/hr

ANALYSIS OF TRANSFER LINES WITH STORAGE BUFFERS: In an automated production line with no internal storage of parts, the workstations are interdependent. When one station breaks down all other stations on the line are affected either immediately or by the end of a few cycles of operation. The other stations will be forced to stop for one or two reasons 1. Starving of Stations: Starving on an automated production line means that a workstation is prevented from performing its cycle because it has no part to work on. When a breakdown occurs at any workstation on the line, the stations downstream from the affected station will either immediately or eventually become starved for parts. 2. Blocking of Stations: Blocking means that a station is prevented from performing its work cycle because it cannot pass the part it just completed to the neighbouring downstream station. When a break down occurs at a station on the line, the stations upstream from the affected station become blocked because the broken down station cannot accept the next part for processing from the neighbouring upstream station. Therefore none of the upstream stations can pass their just completed parts for work.

By Adding one or more parts storage buffers between workstations production lines can be designed to operate more efficiently. The storage buffer divides the line into stages that can operate independently for a number of cycles. Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III The number depending on the storage capacity of the buffer If one storage buffer is used, the line is divided into two stages. If two storage buffers are used at two different locations along the line, then a three stage line is formed. The upper limit on the number of storage buffers is to have a storage between every pair of adjacent stations. The number of stages will then be equal to the number of workstations. For an ‘n’ stage line, there will be n – 1 storage buffers. This obviously will not include the raw parts inventory at the front of the line or the finished parts inventory that accumulates at the end of the line. Consider a two – stage transfer line, with a storage buffer separating the stages. If we assume that the storage buffer is half full. If the first stage breaks down, the second stage can continue to operate using parts that are in the buffer. And if the second stage breaks down, the first stage can continue to operate because it has the buffer to receive its output. The reasoning for a two stage line can be extended to production lines with more than two stages.

Effectiveness of Buffer Storage Two extreme cases of storage buffer effectiveness can be identified: 1. No buffer storage capacity at all (Zero Buffer Storage Capacity). 2. Infinite capacity storage buffers 1. Zero Buffer Storage Capacity: In the case of no storage capacity, the production line acts as one stage when a station Breaks down the entire line stops. This is the case of a production line with no internal Storage. Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III

The line efficiency of a zero capacity storage buffer:

𝐸𝑜 =

𝑇𝑐 𝑇𝑐 + 𝐹. 𝑇𝑑

2. Infinite capacity storage buffers The opposite extreme is the case where buffer zones of infinite capacity are installed Between every pair of stages. If we assume that each storage buffer is half full, then each stage is independent of the next. The presence of the internal storage buffer means that then no stage will ever be blocked or starved because of a breakdown at some other stage.

An infinite capacity storage buffer cannot be realized in practice. If it could then the Overall line efficiency will be limited by the bottleneck stage. I.e. production in all other stages would ultimately be restricted by the slowest stage. The downstream stages could only process parts at the output rate of the bottle neckstage. Given that the cycle time Tc is the same for all the stages the efficiency for any stage k is given by:

where k is used to identify the stage. The overall line efficiency would be given by:

𝐸∞ = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 {𝐸𝑘 } where the subscript ∞ identifies E∞ as the efficiency of a line whose storage buffers have infinite capacity.The actual value of line efficiency will fall somewhere between these extremes for a given buffer capacity Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III

Eb Indicates the line efficiency with a definite buffer capacity, b.

Analysis of Two Stage Transfer Line The two stage line is divided by a storage buffer of capacity is expressed in terms of the number of work parts that it can store. The buffer receives the output of stage 1 forwards it to stage 2, temporarily storing any parts not immediately needed by stage 2 upto its capacity b. The ideal cycle time Tc is the same for both stages. We assume the downtime distributions of each stage to be the same with mean downtime = Td, let F 1 & F2

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III

Problems on Automated Flowlines with Buffer Storage: 1.

A 30-station transfer line has an ideal cycle time of 0.75 min, an average downtime of 6.0 min per line stop occurrence, and a station failure frequency of 0.01 for all stations. A proposal has been submitted to locate a storage buffer between stations 15 and 16 to improve line efficiency. Determine (a) the current line efficiency and production rate, and (b) the maximum possible line efficiency and production rate that would result from installing the storage buffer. Solution: (a) Tp = 0.75 + 30(0.01)(6.0) = 0.75 + 1.8 = 2.55 min/pc E = 0.75/2.55 = 0.2941 = 29.41% Rp = 1/2.55 = 0.392 pc/min = 23.53 pc/hr (b) Tp1 = Tp2 = 0.75 + 15(0.01)(6.0) = 0.75 + 0.90 = 1.65 min/pc. E = 0.75/1.65 = 0.4545 = 45.45% Rp = 1/1.65 = 0.6061 pc/min = 36.36 pc/hr

2.

Given the data in Problem 1, solve the problem except that (a) the proposal is to divide the line into three stages, that is, with two storage buffers located between stations 10 and 11, and between stations 20 and 21, respectively; and (b) the proposal is to use an asynchronous line with large storage buffers between every pair of stations on the line; that is a total of 29 storage buffers. Solution: (a) Tp1 = Tp2 = Tp3 = 0.75 + 10(0.01)(6.0) = 0.75 + 0.60 = 1.35 min/pc For each stage, E = 0.75/1.65 = 0.5555 = 55.55% Rp = 1/1.35 = 0.7407 pc/min = 44.44 pc/hr (b) Tp1 = Tp2 = . . . = Tp29 = 0.75 + 0.01(6.0) = 0.75 + 0.06 = 0.81 min/pc For each stage, E = 0.75/0.81 = 0.926 = 92.6% Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Rp = 1/0.81 = 1.235 pc/min = 74.1 pc/hr 3.

In Problem 1, if the capacity of the proposed storage buffer is to be 20 parts, determine (a) line efficiency, and (b) production rate of the line. Assume that the downtime (Td = 6.0 min) is a constant. Solution: From Problem 1, Eo = 0.2941 and E = 0.4545 D1' 

Td Tc

15(0.01)(6.0) 2.55

=

= 0.3529,

r=

F1 F2

= 1.0

6.0 = 8.0. If b = 20, then B = 2 and L = 4 0.75

Substituting the values of B & L in the equation h(20) =

2 1  2 4 1   0.75   4   =    = 0.7083 2  1  6.0   (2  1)(2  2)  3 8  12 

E = 0.2941 + 0.3529(0.7083)(0.4545) = 0.4077 = 40.77% (b) Rp = 4.

E Tc

=

0.4077 = 0.5436 pc/min = 32.62 pc/hr 0.75

Solve Problem 3, but assume that the downtime (Td = 6.0 min) follows the geometric repair distribution. Solution: From previous Problem 3, Eo = 0.2941 and E = 0.4545 D’1 =

15(0.01)(6.0) = 0.3529, 2.55

h(20) =

20(0.75 / 6.0) 2  (20  1)(0.75 / 6.0)

r=

=

F1 F2

= 1.0

2.5 = 0.5714 4.375

E = 0.2941 + 0.3529(0.5714)(0.4545) = 0.3858 = 38.58% (b) Rp = 5.

E Tc

=

0.3858 = 0.5143 pc/min = 30.86 pc/hr 0.75

In the transfer line of Problems 3 and 5, suppose it is more technically feasible to locate the storage buffer between stations 11 and 12, rather than between stations 15 and 16. Determine (a) the maximum possible line efficiency and production rate that would result from installing the storage buffer, and (b) the line efficiency and production rate for a storage buffer with a capacity of 20 parts. Assume that downtime (Td = 6.0 min) is a constant. Solution: F1 = 11(0.01) = 0.11, Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Tp1 = 0.75 + 0.11(6.0) = 0.75 + 0.66 = 1.41 min/pc F2 = 19(0.01) = 0.19, Tp2 = 0.75 + 0.19(6.0) = 0.75 + 1.14 = 1.89 min/pc E = Min{E1, E2 } = Min{(0.75/1.41), (0.75/1.89)} = Min{0.5319, 0.3968} = 0.3968 Rp = 0.3968/0.75 = 0.5291 pc/min = 31.74 pc/hr (b) Eo = 0.75/2.55 = 0.2941, D’1 = 0.66/2.55 = 0.2588, E2 = 0.75/1.89 = 0.3968 r= Td Tc

F1 F2

=

=

0.11 = 0.5789, 0.19

6.0 = 8.0, 0.75

If b = 20, then B = 2 and L = 4 By Substituting the values of B & L in the Equation h(20) = 0.5789

1  0.5789 2 1  0.5789 3

0.75 + 4 

0.5789 3 (1  0.5789)2

 6.0  (1  0.5789 3 )(1  0.5789 4 )

= 0.4776 + 0.0241 = 0.5017

E = 0.2941 + 0.2588(0.5017)(0.4545) = 0.3531 = 35.31% Rp = 0.3531/0.75 = 0.4708 pc/min = 28.25 pc/hr

PARTIAL AUTOMATION Many assembly lines in industry contain a combination of automated & manual work stations. These cases of partially automated production lines occur for two main reasons: 1. Automation is introduced gradually on an existing manual line. Suppose that demand for the product made on a manually operated line increases, & it is desired to increase production & reduce labour costs by automating some or all of the stations. The simpler Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III operations are automated first, & the transition toward a fully automated line is accomplished over a long period of time. Meanwhile, the line operates as a partially automated system. 2. Certain manual operations are too difficult or too costly to automate. Therefore, when the sequence of workstations is planned for the line, certain stations are designed to be automated, whereas the others are designed as manual stations. Examples of operations that might be too difficult to automate are assembly procedures or processing steps involving alignment, adjustment, or fine-tuning of the work unit. These operations often require special human skills and/or senses to carry out. Many inspection procedures also fall into this category. Defects in a product or a part that can be easily perceived by a human inspector are sometimes extremely difficult to identify by an automated inspection device. Another problem is that the automated inspection device can only check for the defects for which it was designed, whereas a human inspector is capable of sensing a variety of unanticipated imperfections & problems. To analyze the performance of a partially automated production line, we build on our previous analysis & make the following assumptions: 1. Workstations perform either processing or assembly operations; 2. Processing & assembly times at automated stations are constant, though not necessarily equal at all stations; 3. Synchronous transfer of parts; 4. No internal buffer storage; 5. The upper bound approach is applicable & 6. Station breakdowns occur only at automated stations. The ideal cycle time Tc is determined by the slowest stations on the line, which is generally one of the manual stations. If the cycle time is in fact determined by a manual station, then Tc will exhibit a certain degree of variability simply because there is a random variation in any repetitive human activity. However, we assume that the average Tc remains constant over time. Given our assumption that breakdowns occur only at automated stations,

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III Let, na = the number of automated stations Td = average downtime per occurrence. pi = the probability (frequency) of breakdowns per cycle; qi = the defect rate mi = probability that the defect will cause station i to stop. The Average Production time is given by,

Where, pi = mi qi For Special Case, When pi = p, mi = m, & qi = q , Then Production time is given by

Where, p = m q Cost Calculations In Partial Automated Flow line Cas = Cost to operate an automatic station Cms = Cost to operate an manual station Cat = Cost to operate transfer lines Cm = Cost of the raw material Ct = Cost of Tooling Pap = Proportion of good assemblies/ product (i.e. 100%=1, 80%=0.8) Then Operating cost of the partially automated transfer line is,

Total cost per unit produced on the line is Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III

Problems on Partial Automation 1) A partially automated production line has a mixture of three mechanized and three manual workstations. There are a total of six stations, and the ideal cycle time of 1.0 min, which includes a transfer time of 6 sec. Data on the six stations are listed in the accompanying table. Cost of the transfer mechanism Cat = $0.10/min, cost to run each automated station Cas = $0.12/min, and labor cost to operate each manual station Cw = $0.17/min. It has been proposed to substitute an automated station in place of station 5. The cost of this station is estimated at Cas5 = $0.25/min and its breakdown rate p5 = 0.02, but its process time would be only 30 sec, thus reducing the overall cycle time of the line from 1.0 min to 36 sec. Average downtime per breakdown of the current line, as well as for the proposed configuration, is 3.5 min. Determine the following for the current line and the proposed line: (a) production rate, (b) proportion uptime, and (c) cost per unit. Assume the line operates without storage buffers, so when an automated station stops, the whole line stops, including the manual stations. Also, in computing costs, neglect material and tooling costs. Station

Type

Process time pi

1

Manual

36 sec

0

2

Automatic

15 sec

0.01

3

Automatic

20 sec

0.02

4

Automatic

25 sec

0.01

5

Manual

54 sec

0

6

Manual

33 sec

0

Solution: For the current line, (a) Tc = 1.0 min,

F = 0.01 + 0.02 + 0.01 = 0.04

Tp = 1.0 + 0.04(3.5) = 1.0 + 0.14 = 1.14 min/unit, Rp = 1/1.14 = 0.877 units/min = 52.6 units/hr (b) E = 1.0/1.14 = 0.877 = 87.7% (c) Co = 0.10 + 3(0.12) + 3(0.17) = $0.97/min.

Cpc = (0.97)(1.14) = $1.106/unit.

For the proposed line in which station 5 is automated, Lecture Notes, Prashanth N, Dept of Mechanical Engg |

22

CAD/CAM/CIM Unit III (a) Tc = 36 sec = 0.6 min

F = 0.01 + 0.02 + 0.01 + 0.02 = 0.06

Tp = 0.6 + 0.06(3.5) = 0.6 + 0.21 = 0.81 min/unit, Rp = 1/0.81 = 1.235 units/min = 74.1 units/hr (b) E = 0.6/0.81 = 0.7407 = 74.1% (c) Co = 0.10 + 3(0.12) + 0.25 + 2(0.17) = $1.05/min Cpc = (1.05)(0.81) = $0.851/unit 2. A manual assembly line has six stations. The assembly time at each manual station is 60 sec. Parts are transferred by hand from one station to the next, and the lack of discipline in this method adds 12 sec (Tr = 12 sec) to the cycle time. Hence, the current cycle time is 72 sec. The following two proposals have been made: (1) Install a mechanized transfer system to pace the line; and (2) automate one or more of the manual stations using robots that would perform the same tasks as humans only faster. The second proposal requires the mechanized transfer system of the first proposal and would result in a partially or fully automated assembly line. The transfer system would have a transfer time of 6 sec, thus reducing the cycle time on the manual line to 66 sec. Regarding the second proposal, all six stations are candidates for automation. Each automated station would have an assembly time of 30 sec. Thus if all six stations were automated the cycle time for the line would be 36 sec. There are differences in the quality of parts added at the stations; these data are given in the accompanying table for each station (q = fraction defect rate, m = probability that a defect will jam the station). Average downtime per station jam at the automated stations is 3.0 min. Assume that the manual stations do not experience line stops due to defective components. Cost data: Cat = $0.10/min; Cw = $0.20/min; and Cas = $0.15/min. Determine if either or both of the proposals should be accepted. If the second proposal is accepted, how many stations should be automated and which ones? Use cost per piece as the criterion of your decision. Assume for all cases considered that the line operates without storage buffers, so when an automated station stops, the whole line stops, including the manual stations.

Station

qi

mi

Station

qi

mi

1

0.005

1.0

4

0.020

1.0

2

0.010

1.0

5

0.025

1.0

3

0.015

1.0

6

0.030

1.0

Solution: Proposal 1: Current operation: Tc = 1.2 min Lecture Notes, Prashanth N, Dept of Mechanical Engg |

23

CAD/CAM/CIM Unit III Co = 6(0.20) = $1.20/min Cpc = 1.20(1.2) = $1.44/unit. Proposal: Tc = 1.1 min Co = 0.10 + 6(0.20) = 1.30/min Cpc = 1.30(1.1) = $1.43/unit. Conclusion: Accept Proposal 1.

Proposal 2: Tc = 36 sec = 0.6 min if all six stations are automated. F = 0.005(1.0) + 0.01(1.0) + 0.015(1.0) + 0.02(1.0) + 0.025(1.0) + 0.03(1.0) = 0.105 Tp = 0.6 + 0.105(3.0) = 0.6 + 0.315 = 0.915 min/unit Co = 0.10 + 6(0.15) = 1.00/min Cpc = 1.00(0.915) = $0.915/unit Conclusion: Accept Proposal 2.

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

24

CAD/CAM/CIM Unit III

Assembly Systems and Line Balancing Assembly Line An Assembly Line is a production line that consists of a series of workstations where assembly operations are performed. In an assembly line either the components are assembled to form a sub assembly (In Most cases) that becomes an input component for further assembly or a complete assembly (as the final product). Assembly operations involve bringing the mating parts together, Mechanical fastening & Joining Operations. Thus in assembly operations bolting, reveting, welding, soldering, brazing & bonding are common. Assembly operations can be performed using: 1. Manual Single Station 2. Manual Assembly Line 3. Automated Assembly Line Manual Single Station assembly is used for very low volume smaller assemblies or sub assemblies. This is performed in a single work station, involving one or more operators. Eg: Aircraft, Special purpose machine tools, & prototype models of new products, which are produced in low quantities. Manual Assembly Line Manual Assembly line consists of series of workstations where assembly operations are performed by the human operators. Eg: Televison sets, Electric & Electronic Appliances, Fan, Audio Equipment etc....

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

25

CAD/CAM/CIM Unit III Automated Assembly line In this work stations in series are linked by work transport/ transfer system, automatic work/tool handling, assembling & inspection facilities. It is suitable for mass production. Work Transport Mechanism In an assembly line, the transportation of the work part from one work station to the next workstation for further operation, is an important requirement. The Transportation can be performed by: a. Manual Method: Workpart is carried manually to next station

b. Mixed Mode (Manual plus Mechanized also called as Partial Automated Assembly Lines) c. Mechanized Method (Conveyors): In this automated work part transport mechanism is adopted Eg: Walking Beam, Roller/Belt Conveyors etc. In Mechanized Workpart Transport system, the Workpart can be transferred in different methods (These Methods are already discussed in II Unit). i.

ii.

Continuously moving conveyor: operates at constant velocity  Work units are fixed to the conveyor  The product is large and heavy  Worker moves along with the product  Work units are removable from the conveyor  Work units are small and light  Workers are more flexible compared to synchronous lines, less flexible than asynchronous lines Synchronous transport (intermittent transport – stop-and-go line): all work units are moved simultaneously between stations. Problem:  Task must be completed within a certain time limit. Otherwise the line produces incomplete units; Lecture Notes, Prashanth N, Dept of Mechanical Engg |

26

CAD/CAM/CIM Unit III

iii.

 Eexcessive stress on the assembly worker.  Not common for manual lines (variability), but often ideal for automated production lines Asynchronous transport: a work unit leaves a given station when the assigned task is completed.  Work units move independently, rather than synchronously (most flexible one).  Variations in worker task times  Small queues in front of each station.

Assembly Line Models Manual Assembly Line can be distinguished in three categories: a. Single Model Line b. Batch Model Line c. Mixed Model Line a) Single Model Line (No Product Variety): This is a manual assembly line in which only one variety of products is assembled. The Work stations are not designed to take any other assembly operation; such lines are used for the manufacture of products in large quantities. b) Batch Model Line (Hard product Variety): A Batch model assembly line is designed to produce two or more varieties of products. However, each product of desired quantity is produced in batches. c) Mixed Model Line (Soft Product Variety): in a mixed model line, the assembly line are designed to produce more than one product variety and not necessarily in batches. That means 2 or 3 varieties of products can be manufactured simultaneously on the same assembly line.

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

27

CAD/CAM/CIM Unit III LINE BALANCING The line balancing problem is to arrange the individual processing & assembly tasks at the workstations so that the total time required at each workstation is approximately the same. If the work elements can be grouped so that all the station times are exactly equal, we have perfect balance on the line & we can expect the production to flow smoothly. In most practical situations it is very difficult to achieve perfect balance. When the workstations times are unequal, the slowest station determines the overall production rate of the line. Terminologies Used in Line balancing 1. Precedence Constraints Flow line production there are many separate & distinct processing & assembly operations to be performed on the product. Invariably, the sequence of processing or assembly steps is restricted, at least to some extent, in terms of the order in which the operations can be carried out. For example, a threaded hole must be drilled before it can be tapped. In mechanical fastening, the washer must be placed over the bolt before the nut can be turned & tightened. These restrictions are called precedence constraints in the language of line balancing. 2. Minimum Rational Work Element: In order to spread the job to be done on the line among its stations, the job must be divided Into its component tasks. The minimum rational work elements are the smallest practical indivisible tasks into which the job can be divided. These work elements cannot be subdivided further. For example, the drilling of a hole would normally be considered as a minimum rational work element. In manual assembly, when two components are fastened together with a screw & nut, it would be reasonable for these activities to be taken together. Hence, this assembly task would constitute a minimum rational work element. We can symbolize the time required to carry out this minimum rational work element Tej, where j is used to identify the element out of the ne elements that make up the total work or job.

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

28

CAD/CAM/CIM Unit III 3. Total Work Content Time: This time is the sum of all the work element times & represented by T wc. It is Expressed as, 𝑛

𝑇𝑤𝑐 =

𝑇𝑠𝑖 𝑖=1

Where, Ts = Station Time 4. Workstation Process time An assembly line comprises a number of workstations that perform a series of assembly operations in a specified sequence. As indicated, one or more work stations are required to execute one work element. Thus the time required to process at each work station is the sum of all the times all the work elements performed at a particular work station. 5. Cycle Time Tc This refers to the total time taken to complete the assembly process in an assembly line. It is expressed by the relation, 60 𝐸 𝑇𝑐 = 𝑅𝑝 Where, Tc= Cycle time, min/cycle E= Line Efficiency Rp= Production rate, pc/hr 6. Line Efficiency The Ideal cycle rate for a line is given by 60 𝑅𝑐 = 𝑇𝑐 Where Rc= Ideal production rate of the line. Therefore the line efficiency, which is a measure of the performance, is given by, 𝑅𝑝 𝑇𝑐 𝐸= = 𝑅𝑐 𝑇𝑝 Where Tp= The average Production cycle time. 7. Balance Efficiency: This results in line imbalance & measured by the quantity line balance efficiency, simply termed as balance efficiency. It is give by relation, 𝐸𝑏 =

𝑇𝑤𝑐 𝑤 𝑇𝑠

Where, Eb= Balance Efficiency Twc= Total Work Content Time Ts= Max available station time or process time on the line i.e. Max {T si} w= Number of stations or workers Lecture Notes, Prashanth N, Dept of Mechanical Engg |

29

CAD/CAM/CIM Unit III 8. Balance Delay This is referred to as balancing loss. It is a measure of the inefficiency of the assembly line. 𝐷𝑏 =

𝑤𝑇𝑠 − 𝑇𝑤𝑐 𝑤𝑇𝑠

9. Production Rate The production rate is usually expressed as units/hr, depends on the annual market demand and the working hrs (assuming the required capacity is available on the line). It is given by the relation 𝑅𝑝 =

𝐷𝑎 𝑢𝑛𝑖𝑡𝑠/ℎ𝑟 𝑛𝑤 𝑆𝑤 𝐻𝑠ℎ

Where, Rp= Avg production rate, pc/hr Da = Annual demand for the product, pc/year nw= No. of working weeks per year Sw= No. of Shifts per week Hsh=No. of Working Hrs per shift 10. Number of workstations (or Workers), w It is used to calculate total number of workstation or workers required to complete an assembly of a product considering the total workload & available time. It is expressed as follows: 𝑊𝐿 𝑤= 𝐴𝑇 Where, w= the number of workstations (workers) WL= Work Load to be finished in a given time of total work content time, hrs = R pTwc Or WL=

60 𝐸 𝑇𝑤𝑐 𝑇𝑐

AT= Available Time = 60 E (Line Efficiency) 11. Repositioning Loss In assembly some portion of the time is required to reposition the work part or the operator himself. Thus the time available for each operation to perform the actual assembly operation is less than the cycle time Tc. This loss is referred to as repositioning loss. It is expressed as Efficiency factor or Repositioning Efficiency & is given by 𝑇𝑠 𝑇𝑐 − 𝑇𝑟 𝐸𝑟 = = 𝑇𝑐 𝑇𝑐 Where, Ts= Station time or Process time Lecture Notes, Prashanth N, Dept of Mechanical Engg |

30

CAD/CAM/CIM Unit III Tc= Cycle time Tr= Repositioning Time 12. Smoothness Index (SI) Smoothness index is a measure of the relative smoothness of an assembly line. A smoothness value of 0 indicates a perfect balance. That is nothing but the square root of sum of all idle times. It is given by the relation, 𝑤

(𝑇𝑠𝑚𝑎𝑥 − 𝑇𝑠𝑖 )2

𝑆𝐼 = 𝑖=1

Where, w= Total number of stations. Methods of Line Balancing 1. Largest Candidate Rule: It is the easiest method to understand. The work elements are selected to station simply on the basis of the size of their estimated time ‘T e’ values. Procedure: a) List all elements in descending order of ‘Te’ value i.e. Largest ‘Te” at the top of the list. b) To assign elements to the first workstation start at the top of the list & workdown selecting the first feasible element for placement at the station. A feasible element is one that satisfies the precedence requirements & does not cause the sum of ‘Te’ values at the station to exceed the cycle time ‘Tc’. c) Continue the process of assigning work elements to the stations as in step 2 until no further elements can be added without exceeding ‘Tc’. d) Repeat steps 2 & 3 for other stations in the line until all the elements have been assigned. 2. Kilbridge & Wester Method In this, numbers are assigned to each operation indicating the number of predecessors to it. Task with the lowest number of predecessor are assigned first to the wok station. Procedure: a) Draw the precedence diagram for the work elements. First of all, mark in column I all the work elements that do not have any predecessors. Next Mark Column II, with work elements that must follow those in column I. Mark other columns similarly, for all the work elements. Lecture Notes, Prashanth N, Dept of Mechanical Engg |

31

CAD/CAM/CIM Unit III b) Compute the cycle time & station time. c) Prepare a table indicating the work element, number of predecessor for each element and the time for each work element. d) Assign work elements to the station. While, assigning work elements to the station, take care that sum of elemental time does not exceed the station time. Proceed assigning from column I to II & so on. Do not consider the work elements once they are assigned to the station, Repeat step C. Any station, whose sum of elemental times exceeds the station time, it should be assigned to the next station. Similarly assign all work elements to work stations. e) Prepare a Table showing the work station assignments. f) Calculate the required parameters like Line Efficiency, balance Delay & Smoothness Index. 3. Ranked Positional Weight (RPW) or Helgeson & Bernie Method Procedure: a) Draw the precedence diagram ( As explained Previously) b) Find the positional weight for each work element. A Positional Weight of a work element refers to the sum of all the Tek values that fall in all its possible paths from the beginning of the work element through the rest of the diagram. c) Based on the positional weights obtained above, Rank the Work Elements (Arrange the work Elements based on RPW value in Descending Order), i.e Work elemnt with Highest RPW value should be in the first position. d) Next Assign the work elements to the work stations, by assigning the elements of the highest positional weight & rank Priority. e) If the sum of the elemental times for an assigned work station is less than the station time, then assign the next succeeding ranked operation to the work station. See that the assignment maintains precedence relationships & also the total operation time does not exceed the station time. f) Repeat the Steps‘d’ & ‘e’ until all elements are assigned to the work stations. LINE BALANCING PROBLEMS Problem 1: A small component assembly unit has a market demand of 1,20,000 units/year. The line will operate 50 weeks/year, 6 shifts per week and 8 hrs per shift. The company works 8 hrs per shift. The company has a single model assembly line & the assembly work contains the following work elements. Work Element 1 2 3

Duration Tek (Min) 0.52 0.28 0.46

Preceded By 1 2 Lecture Notes, Prashanth N, Dept of Mechanical Engg |

32

CAD/CAM/CIM Unit III 4 5 6 7 8 9 10 11 12

0.3 0.58 0.48 0.22 0.62 0.1 0.38 0.36 0.7

1 4 3,5 6 7 6 6 10 8. 9. 11

Each station can be assigned with one operator, the expected uptime of the line is 95% and the repositioning loss is 0.04min per cycle. Determine: a. Total Work Content Time, Twc b. Production Rate per hour, Rp c. Cycle Time Tc d. Theoretical minimum number of Operators e. Station or Service time, Ts Solution: Da= 1,20,000 units/yr nw=50 weeks Hsh=8 hr/shift E= 95% Tr= 0.04min Sw= 6 Shifts/week a. Total Work Content time, Twc

𝑛

𝑇𝑤𝑐 =

𝑇𝑠𝑖 𝑖=1

Twc = 0.52+0.28+0.46+0.3+0.58+0.48+0.22+0.62+0.1+0.38+0.36+0.7 Twc = 5.0 min b. Production Rate, Rp The hourly Production rate is given by 𝐷𝑎 𝑅𝑝 = 𝑢𝑛𝑖𝑡𝑠/ℎ𝑟 𝑛𝑤 𝑆𝑤 𝐻𝑠ℎ Rp = 120000/ (50 X 6 X 8) Rp = 50 Units/hr

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

33

CAD/CAM/CIM Unit III c. Cycle Time, Tc Cycle Time is given by 60 𝐸 𝑅𝑝 𝑇𝑐 = (60𝑋 0.95) /50 Tc= 1.14min 𝑇𝑐 =

d. Theoretical minimum number of work stations (Workers) w’ = min integer ≥ (Twc / Tc) = 5/1.14 = 4.4 w’=5 Operators e. Station or Service time, Ts

Ts= Tc-Tr = 1.14-0.04

Ts= 1.1 min Problem 2: Solve Problem 1 by applying the largest candidate rule method. Determine the line efficiency, balance delay & Smoothness Index Solution: We have Ts= 1.1 min

a) Precedence Diagram:

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

34

CAD/CAM/CIM Unit III b) Arrange elements in descending order of element time value Work Element 12 8 5 1 6 3 10 11 4 2 7 9

Duration Tek (Min) 0.7 0.62 0.58 0.52 0.48 0.46 0.38 0.36 0.3 0.28 0.22 0.1

Preceded By 8. 9. 11 7 4 3,5 2 6 10 1 1 6 6

c. Assigning Work Content to Stations 2

3

4

1 5

The Work elements assigned to station #1 to #5 is shown in table below:

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

35

CAD/CAM/CIM Unit III Station #

1

2

3

4 5

Work Element 1 4 2 5 3 6 10 7 8 11 12 9

Duration Tek (Min) 0.52 0.3 0.28 0.58 0.46 0.48 0.38 0.22 0.62 0.36 0.7 0.1

Station Time ∑Tek (min)

Idle Time, min

1.1

0

1.04

0.06

1.08

0.02

0.98

0.12

0.8

0.3

a) Line

1. Balance Efficiency, Eb It is given by relation, 𝐸𝑏 =

𝑇𝑤𝑐 𝑤 𝑇𝑠

Here, Ts should be = ∑TekMax at any station. ∑TekMax is at station #1=1.1 min. Eb = 5 / (5 X 1.1) = 90.9 Eb = 90.9%

2. Balance Delay, Db 𝐷𝑏 =

𝑤𝑇𝑠 − 𝑇𝑤𝑐 𝑤𝑇𝑠

= (5 X 1.1 -5)/ (5 X 1.1) Db= 0.091 or 9.1%

Or Db = 1-Eb=1-0.909 Db=0.091

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

36

CAD/CAM/CIM Unit III c. Smoothness Index SI 𝑤

(𝑇𝑠𝑚𝑎𝑥 − 𝑇𝑠𝑖 )2

𝑆𝐼 = 𝑖=1

=

02 + 0.062 + 0.022 + 0.122 + 0.32

SI = 0.33 min Problem 3: Solve the problem 1 using Kilbridge Wester Method. Find Smoothness Index. Solution: We have already computed Ts=1.1min a) Precedence diagram First of all, construct the precedence diagram, keeping the work element nodes of identical precedence in vertical columns, as shown below Col 1

Col 2

Col 3

Col 4

Col 5

Col 6

Col 7

b) Prepare the list of elements and Column Now prepare the a table indicating the work element, column numbers, element time & Sum of column times.

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

37

CAD/CAM/CIM Unit III Work Element, i

Column #

Duration Tek (Min)

1 2 4 3 5 6 7 9 10 8 11 12

I II II III III IV V V, VI V VI VI VII

0.52 0.28 0.3 0.46 0.58 0.48 0.22 0.1 0.38 0.62 0.36 0.7

Column ∑ Tek (min) 0.52 0.58 1.04 0.48 0.7

0.98 0.7

c. Assign elements to stations Now starting with column I, assign elements to station #1. Follow in the order of column number for selecting elements to assign to stations, till sum of T ek does not exceed Ts (1.1 min) Station # 1

2

3

4 5

Work Element, i 1 2 4 3 5 6 7 10 8 9 11 12

Column # I II II III III IV V V VI VI VI VII

Duration Tek (Min) 0.52 0.28 0.3 0.46 0.58 0.48 0.22 0.38 0.62 0.1 0.36 0.7

Station ∑Tek (min)

∑Tek-Ts

1.1

0

1.04

0.06

1.08

0.02

1.08

0.02

0.7

0.4

a) Smoothness Index SI 𝑤

(𝑇𝑠𝑚𝑎𝑥 − 𝑇𝑠𝑖 )2

𝑆𝐼 = 𝑖=1

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

38

CAD/CAM/CIM Unit III =

02 + 0.062 + 0.022 + 0.022 + 0.42

SI = 0.41 min Problem 4: Solve the Problem 1 using ranked positional weight method. Solution: We have computed Ts=1.1min & we have to ensure that Tek does not exceed Ts (1.1 min) while assigning elements to stations. a) Precedence Diagram

b) Compute the Ranked Positional Weight (RPW) Now for each element, compute the RPW & Tabulate them. RPW is the sum of all the Tek values, that fall in all its possible paths. For example, for element 1 the RPW is the sum of all the elements listed, for element 2 the RPW is the sum of Tek of all elements except element 1,4 & 5. The RPW for entire elements are given below Work Element, i 1 2 3 4 5 6 7 8 9 10

RPW 5 3.6 3.32 3.74 3.44 2.86 1.54 1.32 0.8 1.44

Duration Tek (Min) 0.52 0.28 0.46 0.3 0.58 0.48 0.22 0.62 0.1 0.38 Lecture Notes, Prashanth N, Dept of Mechanical Engg |

39

CAD/CAM/CIM Unit III 11 12

1.06 0.7

0.36 0.7

Now Arrange the RPW values in descending order (Largest to smallest) Work Element, i

RPW

Duration Tek (Min)

1 4 2 5 3 6 7 10 8 11 9 12

5

0.52 0.3 0.28 0.58 0.46 0.48 0.22 0.38 0.62 0.36 0.1 0.7

3.74 3.6 3.44 3.32 2.86 1.54 1.44 1.32 1.06 0.8 0.7

Preceded By 1 1 4 2 3,5 6 6 7 10 6 8,9,11

c) Assign to workstations Now assign the elements to stations as per the RPW (Move from Top to Down), Make sure that ∑Tek< Ts Station # 1

2

3

4 5

Work Element, i 1 2 4 3 5 6 7 10 8 9 11 12

Duration Tek (Min) 0.52 0.28 0.3 0.46 0.58 0.48 0.22 0.38 0.62 0.1 0.36 0.7

Station ∑Tek (min)

∑Tek-Ts

1.1

0

1.04

0.06

1.08

0.02

1.08

0.02

0.7

0.4

We can see that this solution is same as the Kilbridge-Wester Method. But for different Assembly line Different type of Line balancing Method will yield

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

40

CAD/CAM/CIM Unit III different results, hence the type of Assembly line will determine which type of Line Balance method to adopt.

Design for Automated Assembly Line Recommendations and principles that can be applied in product design to facilitate Automated assembly 1. Reduce the amount of assembly required: This principle can be realized during design by combining functions within the same part that were previously accomplished by separate components in the product. The use of plastic molded parts to substitute for sheet metal parts is an example of this principle. A more complex geometry molded into a plastic part might replace several metal parts. Although the plastic part may seem to be more costly, the savings-in assembly time probably justify the substitution in many cases. 2. Use modular design: In automated assembly, increasing the number of separate assembly steps that are done by a single automated system will result in an increase in the downtime of the system. To reduce this effect, Riley suggests that the design of the product be modular, with perhaps each module requiring a maximum of 12 or 13 parts to be assembled on a single assembly system. Also, the subassembly should be designed around a base part to which other components are added. 3. Reduce the number of fasteners required: Instead of using separate screws and nuts, and similar fasteners, design the fastening mechanism into the component design using snap fits and similar features. Also, design the product modules so that several components are fastened simultaneously rather than each component fastened separately. 4. Reduce the need for multiple components to lie handled at once: The preferred practice in automated assembly machine design is to separate the operations at different stations rather than to handle and fasten multiple components simultaneously at the same workstation. (It should be noted that robotics technology is causing a rethinking of this practice since robots can be programmed to perform more complex assembly tasks than a single station in a mechanized assembly system. 5. Limit the required directions of access: This principle simply means that the number of directions in which new components are added to the existing subassembly should be minimized. If all of the components can be added vertically from above, this is the ideal situation. Obviously, the design of the subassembly module determines this. 6. Require high quality in components: High performance of the automated assembly system requires consistently good quality of the components that are added at each

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

41

CAD/CAM/CIM Unit III workstation. Poor-quality components cause jams in the feeding and assembly mechanisms which cause downtime in the automated system. 7. Implement hopperability: This is a term that is used to identify the ease with which a given component can be fed and oriented reliably for delivery from the parts hopper to the assembly workhead. TYPES OF AUTOMATED ASSEMBLY SYSTEMS Based on the type of work transfer system that is used in the assembly system: • Continuous transfer system • Synchronous transfer system • Asynchronous transfer system • Stationary base part system The first three types involve the same methods of work part transport described in automated flow line. In the stationary base part system, the base part to which the other components are added is placed in a fixed location, where it remains during the assembly work. Based on physical configuration: • Rotary type or Dial-type assembly machine • Linear or In-line assembly machine • Carousel assembly system • Single-station assembly machine a) Rotary Type or The dial-type machine, the base partare indexed around a circular table or dial. The workstations are stationary and usually located around the outside periphery of the dial. The parts ride on the rotating table and arc registered or positioned, in turn, at each station a new component is added to base part. This type of equipment is often referred to as an indexing machine or dial index machine and the configuration is shown in Figure below

Lecture Notes, Prashanth N, Dept of Mechanical Engg |

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CAD/CAM/CIM Unit III b) Linear or In-line type configuration The in-line configuration assembly system consists of a sequence of workstations in a more-or-less straight-line arrangement as shown in figure below. The in-line assembly machine consists of a series of automatic workstations located along an in-line transfer system. It is the automated version of the manual assembly line. Continuous, synchronous, or asynchronous transfer systems can be used with the in-line configuration.

c) Carousel assembly system It represents a hybrid between the circular flow of work provided by the dial assembly Machine and straight work flow of the in-line. It is as shown in the figure below. In this system, two segments of assembly lines in parallel and facing one another are connected with a semi rotary end as illustrated in below figure, to make one integrated assembly system. The Base part and the final products are fed and received at the same end. The conveyor system feeds and moves the work part in a rectangular manner along the workstation, thus taking a U turn at the end of the line, and starts moving towards the starting end. Such assembly systems are suitable for large size & complex products involving a large number of components and assembly operations. Compared to the simple in-line configuration, the carousel assembly system utilises the shop floor space in a more efficient manner.

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CAD/CAM/CIM Unit III d) Single Station Assembly System In the single-station assembly machine, the assembly operations are performed at a single location (stationary base part system) as shown in figure below. The typical operation involves the placement of the base part at the workstation where various components are added to the base. The components are delivered to the station by feeding mechanisms, and one or more workheads perform the various assembly and fastening operations.

PARTS FEEDING DEVICES In each of the configurations described above, a means of delivering the components to The assembly workhead must be designed. In this section we discuss these devices and their operation. Elements of the parts delivery system The hardware system that delivers components to the workhead in an automated assembly system typically consists of the following elements as shown in figure below

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CAD/CAM/CIM Unit III a) Hopper: This is the container into which the components are loaded at the workstation. A separate hopper is used for each component type. The components are usually loaded into the hopper in bulk. This means that the parts are randomly oriented initially in the hopper. b) Parts feeder: This is a mechanism that removes the components from the hopper one at a time for delivery to the assembly workhead. The hopper and parts feeder are often combined into one operating mechanism. The vibratory bowl feeder, pictured in Figure below, is a very common example of the hopper-feeder combination.

c) Selector and/or orienteer: These elements of the delivery system establish the proper orientation of the components for the assembly workhead. A selector is a device that acts as a filter, permitting only parts that are in the correct orientation to pass through. Components that are not properly oriented are rejected back into the hopper.

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CAD/CAM/CIM Unit III An orientor is a device that allows properly oriented pans to pass through but provides a reorientation of components that are not properly oriented initially. Several selector and orientor schemes are illustrated in Figure below. Selector and orientor devices are often combined and incorporated into one hopper-feeder system. d) Feed track: The preceding elements of the delivery system are usually located some distance from the assembly workhead. A feed track is used to transfer the components from the hopper and parts feeder to the location of the assembly workhead, maintaining proper orientation of the parts during the transfer. There are two general categories of feed tracks: gravity and powered. The gravity feed track is most common. In this type the hopper and parts feeder are located at an elevation that is above the elevation of the workhead. The force of gravity is used to deliver the components to the workhead. The powered feed track uses vibratory action, air pressure, or other means to force the parts to travel along the feed track toward the assembly workhead. e) Escapement and placement device: The purpose of the escapement device is to remove components from the feed track at time intervals that are consistent with the cycle time of the assembly workhead. The placement device physically places the component in the correct location at the workstation for the assembly operation by the workhead. Several types of escapement and placement devices are shown in Figure below.

(a) Horizontal device

(b) Vertical device for placement of parts onto dial indexing table.

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CAD/CAM/CIM Unit III

(c) Escapement of rivet-shaped parts actuated by work carriers,

(d) Pick & Place Robot

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Analysis of Multistation Assembly Machines In the analysis of multi-station assembly lines, the following assumptions are made: a) The work elements performed at different workstations have constant element times. b) The element times need not be equal c) Synchronous transfer mechanism is used (In-line, Dial-Indexing & Carousel Systems are used). d) There is no storage buffer in between stations. e) Line stopping due to electrical & mechanical failures of the line are not accounted f) Stopping of line will only be due to defective parts added during the assembly

ANALYSIS In any production system, defective parts are bound to be manufactured, and the fraction of defective rate can be considered as, q (0