Transcript Screw, Fasteners and the Design of Nonpermanent Joint

Chapter 1
Mathematical Modeling and
Engineering Problem Solving
Lecture Notes
Dr. Rakhmad Arief Siregar
Universiti Malaysia Perlis
Applied Numerical Method
for Engineers
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Background


Over years and years observation and
experiment, engineers and scientists have
noticed that certain aspects of their empirical
studies occur repeatedly.
To understand this behavior, fundamental
knowledge is required to develop a
mathematical model.
2

Engineering Problemsolving process
3
Research data and
Mathematical model
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A Simple Mathematical Model


A mathematical model can be broadly defined as a
formulation or equation that expresses the essential
features of a physical system or process in
mathematical term
In general, it can be represented as a functional
relationship of the form
Dependent
variable
=
f(
independent
variable,
parameters,
Forcing
functions
)
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A falling Parachutist Problem


The net force:
where:



F  FD  FU
FD = mg
FU = -cv
For free falling body near the earth’s
surface, the Newton’s second law
can be used:
F
a
m
or
dv F

dt m
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A falling Parachutist Problem

Then the relation of acceleration
of a falling object to forces is:
dv mg  cv

dt
m

dv
c
g v
dt
m
By using advanced techniques in
calculus solves: (v=0 at t=0)
gm
( c / m ) t
v(t ) 
(1  e
)
c
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Ex. 1.1 Analytical solution

A parachutist of mass 68.1 kg jumps out of stationary hot air
balloon. Use Eq. (1.10) to compute velocity prior to opening
the chute. The drag coefficient is equal to 12.5 kg/s
gm
v(t ) 
(1  e ( c / m )t )
c
(9.8)(68.1)
v(t ) 
(1  e ((12.5) /(68.1)t )
12.5
v(t )  53.39(1  e 0.18355t )
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Ex. 1.1 Analytical solution

After computing, the result is:
v(t )  53.39(1  e 0.18355t )
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Ex. 1.1 Analytical solution
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Numerical Methods

In these methods, the mathematical problem is reformulated
so it can be solved by arithmetic operation.
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Numerical Methods

The change of velocity can be
approximated by:
dv v v(ti 1 )  v(ti )


dt t
ti 1  ti

This is called as a finite divided
difference approximation
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Numerical Methods

Then substitute the net force and
it will yield:
v(ti 1 )  v(ti )
c
 g  v(ti )
ti 1  ti
m

Be rearranged to yield
c


v(ti 1 )  v(ti )   g  v(ti ) (ti 1  ti )
m


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Ex. 1.2 Numerical Solution

Perform the same computation
as in Example 1.1 but use Eq.
(1.12) to compute the velocity.
c


v(ti 1 )  v(ti )   g  v(ti ) (ti 1  ti )
m



12.5 

(0) (1)  9.80
t=1 v  0  9.8 
68.1 


12.5


v

9
.
80

9
.
8

(
9
.
80
)
(1)  17.80
t=2


68.1

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Ex. 1.2 Numerical Solution
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Conservation Laws
and Engineering


Aside from Newton’s second law, there are other
major organizing principles in engineering
The basics one:
Change  increases  decreases
Flow in  Flow out
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For exercise


Compute Questions 1.1, 1.2, 1.3, and 1.4
Question 1.11 will be discussed in the lab if
time permitted
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