Computer Aided Engineering Design

16 COMPUTER AIDED ENGINEERING DESIGN

The magnitude and direction (Input) for the crank r 2 are given. The offset r 1 has known magnitude
and direction (–90°). The slider has known direction (0°) but has variable magnitude (r 4 ). The
connecting rod r 3 has known magnitude but variable direction (θ 3 ). In component form:

X : r 2 cos θ 2 +r 3 cos θ 3 – r 4 = 0

Y : r 2 sin θ 2 – r 3 sin θ 3 + r 1 = 0 (1.10)

Differentiating with respect to θ 2 and using the KC’s h

d

d

f

dr

(^3) d
3
2
4
4
2
= , = :
θ
θθ
dX
d
rrhf
θ
θθ
2
: – sin 223334 – sin – = 0
dY
d
Xr r h
θ
θθ
2
: : cos 22333 – cos = 0
⇒ =

• sin – 1


• cos 0

sin

• cos

3

4

33

33

–1

22

22

h

f

r

r

r

r

⎡

⎣

⎢

⎤

⎦

⎥

⎡

⎣

⎢

⎤

⎦

⎥

⎡

⎣

⎢

⎤

⎦

⎥

θ

θ

θ

θ

(1.11)

The second order KC’s can be similarly determined.

dX

d

rrhhf

2

2

2 22333 33

2

: – cos – (sin + cos ) – = 0 4

θ

θθθ′′

dY

d

rrhh

2

2

2 223 33^33

: – sin – (cos – sin^2 ) = 0

θ

θθθ′

⇒ =

• sin –1


• cos 0

cos + cos

sin – sin

3

4

33

33

–1

(^22333)
2
(^22333)
2
′
′
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
h
f
r
r
rrh
rrh
θ
θ
θθ
θθ (1.12)
Following on similar lines as in Example 1.1, the position, velocity and acceleration of other linkages
with respect to the input crank (link 2) are given by
sin =
( + sin )
3 12 2 , = ( cos + cos )
3
θ 42 23 3
θ
θθ
rr
r
rr r
θω ω ̇ 33324 = = hr, = slider velocity = ̇ f 42 ω
θωα ̇ ̇ 33 ̇ ̇ ωα
2
2
32 4 (^42)
2
= hhr′′ + , = slider acceleration = f f + 42 (1.13)
Applying Eqs. (1.10)-(1.13), an algorithm can be developed to determine the velocity and acceleration
of the connecting rod and slider for various orientations of the crank.
Example 1.3 (Design of Helical Compression Springs)
Machine component design is specially suited to computerized solutions. A computer program can
help in

• looking up tables for materials and standard sizes


• iteration using a set of formulas and constraints


• logic and options


• graphical display and plots with changing parameters