Mechanical Engineering Assignment: Analysis of Manufacturing Systems

This mechanical engineering assignment solution addresses several key topics within the field. The first question explores strategies for managing challenges in manufacturing, including cost reduction, equipment utilization, and process confusion, along with discussions of re-engineering, IT integration, and data collection. It further differentiates between hard and soft enablers in automation. The second question focuses on a bottleneck model within a flexible manufacturing cell, calculating production rates, machine utilization, and server numbers. The third question delves into material handling and storage systems, specifically analyzing an AGVS for a warehouse, determining the number of required trains. Finally, the fourth question covers the sizing of an AS/RS rack structure, calculating dimensions based on pallet load, aisle configuration, and specified allowances.

Running Head: MECHANICAL ENGINEERING
Mechanical Engineering
Student’s Name
Institution

MECHANICAL ENGINEERING 2
Question 1:
a) They are;
 High cost. Managed by increasing the supervisory span.
 Low equipment utilization. Managed by reduction of machine output requirements.
 Confusion. Managed by making more constant schedules in process requirements
(Mohd, et al., 2017).
b) They include:
 Re-engineering of business processes.
 Managing execution and planning tools.
 Designing in assembly /manufacturability.
 Processes involving reorganization.
 Incorporation that is inter-organizational.
 In the improvement of information technology.
 In the collection of point-of-scale data (Wei, 2016).
c) Hard enablers;
 Conveyor system. They move goods from one place to another via the conveyor
belt.
 Lifting cart. They are machines are employed to lift goods when operated by the
man.
Soft enabler;
 Robot. They are automated to make decisions either for lifting or packing.
 Sensors. They detect situations in the environment by the use of logic information.
d) Hard automation
This has its reference to using equipment that has the special purpose of automated fixed-
sequence assembling or processing operations. Every sequenced operation has simple and
plain rotational or linear movement. This automation has relatively higher difficulty in its
product design. An example is the cart lift (Wei, 2016).
Soft automation
This automation involves the use of programs. The instrument used has the designed
capability of changing the sequence of operation to suit the variety of product
configurations. The program controls the operations with the use of instructed codes. The
instrument’s system is able to read and make an interpretation. One example is a robot
(Mohd, et al., 2017).

MECHANICAL ENGINEERING 3
Question 2: Bottleneck Model
A flexible manufacturing cell consists of three machining workstations plus a load/unload station. The
load/unload station is station 1 with one server (human worker). Station 2 consists of two identical CNC
horizontal milling machines. Station 3 has one CNC vertical machine. Station 4 consists of two identical CNC
drill presses. The four stations are connected by a part-handling system that has one work carrier. The mean
transport time is 2.0 min. The FMC produces two parts, A, and B. The part mix fractions and process routings
for the two parts are presented in the table below. The operation frequency fijk = 1.0 for all operations.
Determine (a) maximum production rate of the FMC, (b) corresponding production rates of each product, (c)
utilization of each machine in the system, and (d) number of busy servers at each station. A spreadsheet
calculator is recommended for this problem.
Part j Part mix pj Operation k Description Station i Process time tijk
A 0.2 1 Load 1 3 min
2 Mill 2 20 min
3 Drill 3 12 min
4 Unload 1 2 min
B 0.3 1 Load 1 3 min
2 Mill 2 15 min
3 Drill 3 30 min
4 Unload 1 2 min
C 0.5 1 Load 1 3 min
2 Drill 3 14 min
3 Mill 2 22 min
4 Unload 1 2 min
Solution:
a) WL1 = (3+2)(0.2)(1.0) + (3+2)(0.3)(1.0) + (3+2)(0.5)(1.0) = 5.0 min
WL2 = 20(0.2)(1.0) + 15(0.3)(1.0) + 22(0.5)(1.0) = 19.5 min
WL3 = 12(0.2)(1.0) + 30(0.3)(1.0) + 14(0.5)(1.0) = 18.4 min
nt = 3(0.2)(1.0) + 3(0.3)(1.0) + 3(0.5)(1.0) = 3, WL4 = 3(2.5) = 7.5 min
Bottleneck station has largest WLi/si ratio:
Station WLi/si ratio
1 (load/unload) 5.0/1 = 5.0 min
2 (mill) 19.5/1 = 19.5 min  Bottleneck
3 (drill) 18.4/1 = 18.4 min
4 (material handling) 7.5/1 = 7.5 min
Bottleneck is station 2: Rp* = 1/19.5 = 0.05128 pc/min = 3.077 pc/hr
b) RpA = 0.05128(0.2) = 0.01026 pc/min = 0.6154 pc/hr
RpB = 0.05128(0.3) = 0.01538 pc/min = 0.9231 pc/hr
RpC = 0.05128(0.5) = 0.02564 pc/min = 1.5385 pc/hr

MECHANICAL ENGINEERING 4
c) U1 = (5.0/1)(0.05128) = 0.256 = 25.6%
U2 = (19.5/1)(0.05128) = 1.0 = 100%
U3 = (18.4/1)(0.05128) = 0.944 = 94.4%
U4 = (7.5/1)(0.05128) = 0.385 = 38.5%
d) BS1 = (5.0)(0.05128) = 0.256 servers
BS2 = (19.5)(0.05128) = 1.0 servers
BS3 = (18.4)(0.05128) = 0.944 servers
BS4 = (7.5)(0.05128) = 0.385 servers
Question 3: Material Handling and Storage Systems
A driverless train AGVS is being planned for a warehouse complex. Each train will consist of a towing vehicle
plus four carts. The speed of the trains = 150 ft/min. Only the pulled carts carry loads. Average loaded travel
distance per delivery cycle is 3000 ft and empty travel distance is the same. Assume the travel factor = 0.8,
and availability = 1.0. The load handling time per train per delivery is expected to be 20 min. If the
requirements of the AGVS are 16 cartloads/hr. Determine the number of required trains.
Tc = 20 + [(3000 + 3000) / 150]
= 60 min/delivery cycle
Rdv = [(60)(0.8)] / 60
= 0.8 deliveries / hour per vehicle
Rf = (16 cartloads / hour) /(4 cartloads / train)
= 4 trainloads / hours = 4 deliveries / hour
nc = 4/0.8
= 5 AGV trains
Question 4: Sizing The AS/RS Rack Structure
A unit load AS/RS is being designed to store 1000 pallet loads in a distribution centre located next to the
factory. Pallet dimensions are: x = 1000 mm, y = 1200 mm; and the maximum height of a unit load =
1300mm. The following is specified: (1) the AS/RS will consist of two aisles with one S/R machine per aisle,
(2) length of the structure should be approximately five times its height, and (3) the rack structure will be
built 500 mm above floor level. Using the allowances: a = 150 mm, b = 200 mm, and c = 250 mm, determine
the width, length, and height of the AS/RS rack structure. Assume that the L/H ratio does not include the 500
mm foundation.
1000 pallets / 2 aisles = 500 pallets per aisle
500 pallets/2 sides of aisle = 250 pallets per side of aisle
ny, nz = 250
L = ny (y+b)
= ny (1200 + 200)
= 1400 ny (ny in mm)

MECHANICAL ENGINEERING 5
= 1.4 ny (ny in m)
H = nz (z+c)
= nz (1300 + 250)
= 1550 nz (nz in mm)
= 1.55 nz (nz in m)
L/H = 5
1.40 ny / 1.55 nz
= 0.9032 ny/nz = 5
ny = 5.536 nz
nynz = (5.536 nz) nz = 250
nz2 = 250 / 5.536
= 45.161
nz = 6.72 = 7
ny = 250 / nz
= 250 / 7
ny = 35.7 = 36
W = 3 (1000 + 150)
= 3450 mm
= 3.45 m/aisle
For 2 aisles,
W = 2 x 3.45
W (for 2 aisle) = 6.9 m
L = 1.4 ny
= 1.4(36)
= 50.4 m
H = 1.55 nz
= 1.55 x 7
= 10.85 m
Since the rack is built 500 mm or 0.5 m above the floor level,
H = 10.85 m + 0.5 m
= 11.35 m
Capacity = 2 x 2 x 36 x 7
= 1008 pallets
L / H = 50.4 / 10.85
= 4.645

MECHANICAL ENGINEERING 6
References
Mohd, F., Tee, B., Mohd, A., Mohd, Z., I., R., & Haslinda, M. (2017). Proceedings of Mechanical
Engineering Research Day 2017. Timmins: Centre for Advanced Research on Energy.
Wei, J. (2016). Mechanical Engineering And Control Systems - Proceedings Of The 2016
International Conference On Mechanical Engineering And Control System (Mecs2016). La
Tuque: World Scientific.