Homework 1 Solutions
1. (a)-(b)
(c)
(d)
Homework 1 Solutions
3.
S=7.489
L=12.969
P=9.17
Q=9.57
S+L > P+Q, therefore, the linkage is non-Grashof, triple-rocker.
Homework 1 Solutions
Initial Setup
In Motion:
a. The door (the red link) opens down just as you
Homework 2 Solutions
1.
Step 1: Draw link AB in its three design positions A1B1, A2B2, A3B3
Step 2: Draw the perpendicular bisectors of A1A2 and A2A3, and call the intersection of these lines
O2. Repeat for B and call the intersection O4.
Step 3: Connect
Homework 4
09/23, 2011
Due: 09/30, 2011
1. Find all the instant centers of the linkages shown below.
(a)
(b)
2. Find the degrees of freedom (mobility) for the mechanism below. Use the complex variable
method to express the position, velocity and accelerat
3.
a cos 2 b cos 3 c cos 4 d
f 3 , 4
( We want this to be zero)
a sin 2 b sin 3 c sin 4
The iteration scheme is of the form:
3
3
3
3
1
J N f 3 , 4 N
4 N 1 4 N 4 N 4 N
where we have
r1
3
J 3 , 4
r2
3
r1
4 b sin 3
r2 b cos 3
4
c sin
1) Findalltheinstantcentersofthelinkage sshownbelow
2) Findthedegreesoffreedom(mobility)forthemechanismbelow
3) Buildthelinkagefortheproblem
a 9 , b 3 , c 8 , d 7 , 2 85 0 , 2 12 deg/ sec , 2 5 deg/ sec 2 , R pa 9 , and
3 25 deg for the fourbar linkage b