MECHANICAL DESIGN PROCESS AND REVIEW OF DYNAMICS Lecture #1 Course Name : DESIGN OF MACHINE ELEMENTS...
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Transcript of MECHANICAL DESIGN PROCESS AND REVIEW OF DYNAMICS Lecture #1 Course Name : DESIGN OF MACHINE ELEMENTS...
MECHANICAL DESIGN PROCESS AND REVIEW OF DYNAMICS
Lecture #1
Course Name : DESIGN OF MACHINE ELEMENTSCourse Number: MET 214
Market segments involving mechanical design process:
Consumer Products:Household appliances (Can openers, food processors, mixers, toasters, vacuum cleaners, clothes washers), lawn mowers, chain saws, power tools, garage doors openers, air conditioning systems, and many others.
Manufacturing systems:Material handling devices, conveyers, cranes, transfer devices, industrial robots, machine tools, automated assembly systems, special purpose processing systems, forklift trucks, and packaging equipment.
Construction equipment:Tractors with front-end loaders or backhoes, mobile cranes, power shovels, earth movers, graders, dump trucks, road pavers, concrete mixers, powdered nailers and staplers, compressors, and many others.
Agricultural equipment: Tractors, harvesters (for corn, wheat, tomatoes, cotton, fruit, and many other crops),rakes, hay balers, plows, disc harrows, cultivators, and conveyers.
Transportation equipment:(a) Automobiles, trucks, and buses, which include hundreds of mechanical devices such as suspension components (springs, shock absorbers, and struts); door and window operators; windshield wiper mechanisms; steering systems; hood and trunk latches and hinges; clutch and braking systems; transmissions; drive shafts; seat adjusters; and numerous parts of the engine systems. (b) Aircraft, which include retractable landing gear, flap and rudder actuators, cargo handling devices, seat reclining mechanisms, dozens of latches, structural components, and door operators.
Ships:Winches to haul up the anchor, cargo-handling cranes, rotating radar antennas, rudder steering gear, drive gearing and drive shafts, and the numerous sensors and controls for operating on-board systems.
Space systems:Satellite systems, the space shuttle, the space station, and launch systems, which contain numerous mechanical systems such as devices to deploy antennas, hatches, docking systems, robotic arms, vibration control devices, devices to secure cargo, positioning devices for instruments, actuators for thrusters, and propulsion systems.
Mechanical Design:
Process for producing a product that is practical to manufacture and will satisfy the needs of a customer in a safe, efficient, reliable and economical manner.
The Mechanical Design process:Most Designs evolve through a cycle of activities identified in the figure below.
Comments supporting Figure 1:
Customer requirements:• Key to establishing customer requirements is to identify the end users of the product and envision how the
product will be utilized from the perspective of the end users. • Envisioning how a product will be utilized throughout the life cycle of the product facilitates identifying key
features that need to be designed as part of the product and/ or enables an order of priorities to be established that can guide the product development. Understanding customer requirements is fundamental to identifying the most appropriate design alternative.
Various techniques can be utilized to assist in defining customer requirements including:
• Focus groups• Surveys
Customer requirements are typically classified into categories to assist in performing comprehensive assessments
1. Function – tells what the device must do using general non quantitative statements- generalizing avoids the tendency to incorporate preconceived ideas into product.
2. Design requirements – detailed, usually quantitative statements of expected levels of performance.3. Evaluation criteria – statements of desirable qualitative characteristics of a design that assist the
designer in deciding which design option is to be selected as the best and final configuration.
Concurrent Engineering:Simultaneous consideration of product design with manufacturing process design so that the product development process results in a profitable product. Combining both considerations during the evolution of a product from concept to deliverables guides the design process toward a final design that is capable of being manufactured in a realizable manner consistent with customer requirements at a reasonable profit.Must develop products that can be manufactured at a profit.
Evaluation criteria:• Purpose of this step is to develop acceptance test procedures for use in testing the product. The results
of acceptance tests procedures performed on the product can be used to assist customers in deciding if they are going to accept the product and/or to insure that the manufacturer is manufacturing a product compliant with customer requirements.
• Level of performance achieved with acceptance testing can be used to establish product specifications. • Good designs comply with product requirements and result in a product that can be manufactured at an
acceptable level of profit.
Techniques for generating alternative concepts.1. Reverse engineer competitive designs and/or products2. Perform literature search
a) Patentsb) Journalsc) Vendor application notes
3. Experience/ modifications to existing designs4. Innovate/ product groups surveys/ product gap analysis.
Working several design alternatives in parallel enhances the opportunity for identifying a suitable design.
Evaluate each proposed alternative:The system configuration, i.e. how sub system components are arranged to implement a product design dictates how system level performance requirements are translated into interface requirements for the subsystem components. For complex systems, during the initial stages of the design process, it is difficult for the designer to envision the demands that are going to placed upon the various subsystem components due to the lack of a detailed system definition and/or the complicated nature of the interactions existing in a complex system. Working several design options in parallel provides an opportunity for designers to avoid adopting a system configuration which places unrealistic demands on the subsystem components. Such considerations emphasize the need to properly translate system level performance requirements into interface requirements for subsystems components.
Typically, motion requirements associated with an application will be utilized to establish a system requirement for horsepower which is subsequently translated into torque and/or force interface requirements for subsystem components. Once a force and/or torque analysis is performed on the subsystem components using the laws of mechanics, stress levels associated with subsystem components can be investigated and components designed to accommodate the stress levels anticipated during operation. Designs of subsystem components are then integrated to produce a best and final design for the product.
An example of material handling device:
• identify how supply chain management strategies evolve to system requirements involving:– Product dimensions– Product weight– Travel time– Travel distance– Velocity and acceleration profiles are derived from product shipping and/or handling
requirements– Discuss how mass moment of inertia of the system components can be reflected to drive shaft
of motor to assist in establishing drive requirements for the conveyor system, i.e. motor horsepower.
More Design Examples:DESIGN OF ENGINE-PUMP SHAFT
The FRF Fire Engine Pump Company manufactures a complete line of engine driven centrifugal pumps. One model under design consists of a 10-hp single cylinder engine direct-connected to the rotor of a centrifugal pump by means of a steel shaft 36 in. long. The drawing below shows the initial dimensions of the drive.
Comments: Requirements of pump translate into requirements for engine and shaft. To design a system, the designer needs to be able to translate pump requirements into torque and/or force requirements for the components. Knowing how to translate horsepower requirements for the pump into torque and force requirements for the components enables a shaft to be designed that can resist the loads incurred during operation of the pump. The shaft must have a diameter consistent with the level of torque to be transmitted to the pump and the level of normal stresses existing in the shaft attributable to operation of the rotor.
Shown in the figures provided below are two alternatives for a saw drive.
Q) Why isn’t the motor connected directly to the shaft of the saw blade, i.e. what purposes or features does the use of a belt and/or chain drive provide that would justify their inclusion in the system?
Q) How does a belt drive differ from a chain drive and how does such differences effect the other components in the system, such as pillow blocks, the diameter of the shafts used to support the sheaves and/or sprocket?
Center of Gravity/ CentroidCenter of Gravity:
The location where the total weight of a body must be placed so that the moment produced by the total weight of the body will be equal to the moment generated by summing together the moments due to each of the individual component weights contained in the body. See the equations below.
Note: if then
If and are made of a homogenous material having a density ρ then note the following.
21 wwwT
2211 wXwXwXwXiw
iiTcg
21
2211
ww
wXwX
w
wX
X
i
i
wi
wii
cg
2121 &3&3 wwXX 0cgX
1w 2w
gtLhgvolgmw )()( 111111
gtLhgvolgmw )()( 222222
1w 2w
If for a homogenous material having a density ρ, then ρ,t and g can be factored out of each term in the expression for
when working with thin uniform distributions of weight, it is an option to rearrange and to work with only the cross sectional planar areas of the distribution instead of the weight of the distribution. In this case, is referred to as the centroid and given the symbol .
Note in example above, if then
Similar arguments exist for determining the “balance point” in the Y direction.
21 tt cgX
21
2211
2211
222111
)]()[(
)]()([
AA
AXAX
LhLhtg
LhXLhXtgX cg
cgX X
i
Aii
A
AX
X i
21 AA 0 cgXX
Ai
Aii
A
AY
Y i
Consider the following distributions of planar areas
Q) What quantitative measures do the two distributions have in common
Q) What quantitative measure indicates that there is a difference in the distributions?
Note: The spread of the distributions are different. The moment of inertia for areas helps to differentiate distribution #1 from distribution #2 on this basis.
I= in4, mm4
Moment of inertia calculations for shapes commonly encountered in strength of materials and machine elements have been calculated and can be found in the appendix of strength of materials text books.The formulas provided to calculate moment of inertias calculate the moment of inertia with respect to the centroidal axes. To determine the moment of inertia with respect to other coordinate systems, the parallel axis transfer formula must be used.
ia
iiX aYI 2 ia
iiY aXI 2
Parallel Axis Theorem
Where = moment of inertia of an area a about centroidal axis = moment of inertia of an area a about axis X. Axis X is parallel to and offset from by
Similar formulas holds for transferring moment of inertia about an axis offset from .
Where = moment of inertia of an area a about centroidal axis = moment of inertia of an area a about axis X. Axis X is parallel to and offset from by
2xoX adII
x
i
ixa
xoX adII )( 2
xoI
xO
XI xO xO
yoI
2yoY adII
y
i
ia
yyoYadII )( 2
yoI
YIyO
yO yO
xd
yd
Given the following geometry, Calculate
a1=a2=3in2 b1=3; h1=1;b2=1; h2=3;
= 46
5.16
)3(5.1)3(5.15.15.1
21
21
aa
aaX
A1
A2
5.26
15
6
5.105.4
6
)3(5.3)3(5.1)5.03(5.1
21
12
aa
aaY
12
3
12
)1)(3(
12
33
10
bhI x
12
27
12
)3)(1(
12
33
10
hbIy
12
27
12
)3)(1( 3
20
xI
12
3
12
)1)(3( 3
20yI
))5.1)(3(12
27())5.3)(3(
12
3()()( 222
22
1 220110 xxxXX daIdaII
16))5.1)(3(12
3())5.1)(3(
12
27()()( 222
22
1 220110 yyyyY daIdaII
YXandIIYX ,,
1a
2a
Polar moment of inertia for areasNote:
Note: For circular cross section
Figure 14
ia
iarJ 2
222iii YXr
ia
iii aYXJ )( 22
ia
iiii aYaX )( 22
ii a
iia
ii aYaX 22
YXXY IIIIJ
322
644
4
000
00
dIIIJ
dII
XYX
YX
Review of dynamics
Newton’s three laws govern dynamics:
1) Every body persists in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.
2) a-> acceleration
m-> Mass; slugs, Kilograms F -> Force; Lbf, Newtons
3) To every action there is always opposed or equal reaction, or the mutual action of two bodies upon each other are always equal, and directed to contrary parts.
Q) What is mass? To identify how mass influences motion, let us rearrange Newton’s second law.
Note: For the same amount of force F, as m increases then acceleration a decreases.Accordingly : Mass is resistance to motion. When the same force is applied to two different bodies, the body with the larger mass has the lower acceleration. Mass is resistance to motion.
maF22 secsec
mor
ft
m
FamaFnet
Recall acceleration may be related to velocity and/or position by the following relationships
where
Position, relative displacement, ft, m
We may use the relationships existing between acceleration, velocity and position to rewrite Newton’s second law as follows
If we are given a force acting on a body of mass m, then we can determine how the acceleration, velocity and/or position changes with time by rearranging the equations listed above and performing integration.
1) if then
2) if
dt
Xda
dt
dXv
dt
dva
2
X
mor
ftvelocityv
mor
ftonacceleratia
secsec
secsec 22
dt
XdmF
dt
dvmF
maF
2
][)()( 00 ttatt 0@)( 000 tt att 0)(
200 2
1)( attXtX 0@)( 000 tXtX
Example: Given the following velocity profile for transporting a mass of 2 slugs in a horizontal plane, determine the following
a) Acceleration as a function of timeb) Force as a function of timec) Position as a function of time
When dealing with the forces, due to the gravitational attraction to the earth, the forces are referred to as weight and the acceleration due to gravity has a very specific value depending on the mass of the earth, and the distance the object is from the center of the earth. Accordingly, Newton’s second law is reformatted for objects having a mass m as follows
W=mg where g-> acceleration due to gravity m-> mass of the object experiencing gravitational attraction to earth W-> Weight of the object Force due to gravitational attraction to the earth lbs,
Newton To develop Newton’s 2nd law for rotational systems, let us consider the following situations.
Note: ∆X=r∆θSince r is constant where ω = rads/sec angular speed of rotation
∆θ = Change in angle –radswhere γ = dx/dt = linear velocity ft/sec or m/sec
22 sec8.9
sec2.32
mor
ft
r
tr
t
X
r
To extend the analysis, perform the following manipulations
Where
angular acceleration (rads/sec2)
r
dt
dr
dt
dr
dt
rd
dt
d
dt
d
dt
dr&0
dt
d