MECHANICAL DESIGN PROCESS AND
REVIEW OF DYNAMICS
Lecture #1
Course Name : DESIGN OF MACHINE ELEMENTS
Course 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 statementsgeneralizing 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 products
2. Perform literature search
a) Patents
b) Journals
c) Vendor application notes
3. Experience/ modifications to existing designs
4. 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/ Centroid
Center 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.
w1
w2
wT  w1  w2
X cg wT   X i wi  X1w1  X 2 w2
wi
X w
i
X cg 
i
wi
w
i

X 1w1  X 2 w2
w1  w2
wi
Note: if
X1  3 & X 2  3 & w1  w2 then X cg  0
If w1 and w2 are made of a homogenous material having a density ρ then note the following.
w1  m1 g  ( vol1 ) g   (h1L1t1 ) g
w2  m2 g  ( vol2 ) g   (h2 L2t2 ) g
If t1  t2 for a homogenous material having a density ρ, then ρ,t and g can be factored out of each term
in the expression for X cg
X cg 
tg[ X 1 (h1L1 )  X 2 (h2 L2 )] X 1 A1  X 2 A2

tg[(h1L1 )  (h2 L2 )]
A1  A2
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, X cg is referred to as the centroid and given the symbol X.
X 
X
i
Ai
Ai
A
i
Note in example above, if A1  A2 then
X  X cg  0
Similar arguments exist for determining the “balance point” in the Y direction.
Y A
i
Y 
i
Ai
A
i
A
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 X  Yi 2ai
ai
IY   X i2ai
I= in4, mm4
ai
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.
Parallel Axis Theorem
I X  I ox  adx2
I X   ( I oxi  adx2 )
ai
Where I ox = moment of inertia of an area a about centroidal axis Ox
I X = moment of inertia of an area a about axis X. Axis X is parallel to Ox and offset from Ox by d x
Similar formulas holds for transferring moment of inertia about an axis offset from I o .
y
IY  I o y  ady2
I Y   ( I o y  ad y2 )
ai
Where
I oy
IY
i
= moment of inertia of an area a about centroidal axis Oy
= moment of inertia of an area a about axis X. Axis X is parallel to Oy and offset from Oy by d y
Given the following geometry, Calculate
X , Y , I X andIY
X 
A2
a1
Y
a2
A1
01
1.5a2  (3  0.5)a1 1.5(3)  3.5(3) 4.5  10.5 15


  2.5
a1  a2
6
6
6
bh3
(3)(1) 3
3



I
12
12
12 x
I x01
Iy
1.5a1  1.5a2 1.5(3)  1.5(3)

 1.5
a1  a2
6
hb 3 (1)(3) 3 27



12
12
12
02
I y 02
(1)(3) 3
27


12
12
(3)(1)3 3


12
12
a1=a2=3in2
b1=3; h1=1;b2=1; h2=3;
3
27
I X  ( I X 01  a1d x21 )  ( I x02  a2 d x22 )  (
 (3)(3.5) 2 )  (
 (3)(1.5) 2 ) = 46
12
12
I Y  ( I y01  a1d y21 )  ( I y02  a2 d y22 )  (
27
3
 (3)(1.5) 2 )  (  (3)(1.5) 2 )  16
12
12
Polar moment of inertia for areas  J 
2
2
2
Note: ri  X i  Yi
r
2
ai
ai
J   ( X i2  Yi 2 )ai
ai
  ( X i2 ai  Yi 2 ai )
ai
  X i2ai  Yi 2ai
ai
ai
J  IY  I X  I X  IY
Note: For circular cross section
I X 0  I Y0
d 4

64
J  I X 0  IY0  2 I X 0
Figure 14
d 4

32
Review of dynamics
Newton’s three laws govern dynamics:
1)
2)
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.
 F  ma a-> acceleration ft 2 or m 2
sec
sec
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.
Fnet  ma   a 
F
m
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.
Recall acceleration may be related to velocity and/or position by the following relationships
dv
dt
dX
v
dt
d2X
a
dt
a
where
ft
m
or
sec2 sec2
ft
m
v  velocity
or
sec sec
X  Position, relative displacement, ft, m
a  acceleration
We may use the relationships existing between acceleration, velocity and position to rewrite
Newton’s second law as follows
F  ma
dv
 F  m dt
d2X
 F  m dt
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)
 (t )   (t0 )  a[t  t0 ]
2) X (t )  X 0   0t 
1 2
at
2
if
 (t0 )  0 @ t0  0 then
if X (t0 )  X 0 @ t0  0
 (t )  0  at
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 time
b) Force as a function of time
c) 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
ft
m
W=mg where g-> acceleration due to gravity 32.2 sec 2 or 9.8 sec 2
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∆θ
Since r is constant
X

r
 r
t
t
  r
where ω = rads/sec angular speed of rotation
∆θ = Change in angle –rads
where γ = dx/dt = linear velocity ft/sec or m/sec
To extend the analysis, perform the following manipulations
d d r  dr
d

 r
 r
dt
dt
dt
dt
Where
dr
d
  0&

dt
dt
d
   angular acceleration (rads/sec2)
dt