ME 352 - Machine Design I
Name of Student:_
Fall Semester 2012
Lab Section Number:_
EXAM 1. OPEN BOOK AND CLOSED NOTES.
Wednesday, September 26th, 2012
Please use the blank paper provided for your solutions. Write on one side of the paper only.
Where nece
ME 352 - Machine Design I
Fall Semester 2012
EXAM 2. Answers
Problem 1 (25 Points)
(i) X C = 0.5 cm/cm
YC = 0.866 cm/cm
X C = + 0.825 cm/cm2
YC = + 0.390 cm/cm2
(ii) ut = + 0.5 i + 0.866 j
un = 0.866 i + 0.5 j
(iii) C = 1.925 cm
(iv) X CC = 8.33 cm
YCC =
Problem 1 (25 Points)
(i) 3.8 kg-m2
(ii) 3.39 N-m, CCW
Problem 2 (25 Points).
Part I.
Exact: 1 = 477.41 rad/s
and
2 = 823.82 rad/s
(i) 1 = 499.69 rad/s
(ii) 1 = 413.05 rad/s
Part II.
(i) FB = 1.1944 i + 3.5715 j N = 3.766 N 108.5
FA = 4.8424 i 3.9025 j N
Problem 1 (25 Points)
For the mechanism shown in the figure below, the input link 2 rotates with an angular velocity of
2 = 2 rad/s CCW and an angular acceleration of 2 = 5 rad/s 2 CC W. The masses and second moments
of inertia for the two moving links ar
ME 352
Spring 2013
Homework Answers
Homework No. 5
a) R2 (A to O4) + R7 (O4 to B) R3 (A to B) = 0
Using this vector loop: dR2/dt = 0.15 m/sec d2R2/dt2 = 0.10 m/sec2
b) R3 = 34.641 mm
theta3 = 120 degrees
theta7 = 210 degrees
-1
c) R3 = 1.155
theta3 = 14.4
ME 352
Spring 2013
Homework No. 4
4a) Problem 3.29 on p. 162 of the text using instant centers:
Follower angular velocity = 4.38 rad/sec (hence CCW)
Roller angular velocity = -83.0 rad/ sec (hence 83.0 rad/sec CW)
4b) Use instant centers to solve for the
ME 352
Spring 2013
Homework No. 3
3a) Refer to the figure for Problem 2.16 on p. 95 of the text
Position Solution:
theta 3 (vector AB) = 45 degrees
theta 4 (vector CB) = 90 degrees
Kinematic coefficients: theta 3 prime = -0.2
theta 4 prime = 0.0
theta 5 p
ME 352 - Machine Design I
Spring 2013
Homework No. 3a
Name
Lab. Div. Number
Note: this is the 1_st of m problems due
Due: Friday, February 1, 2013
Refer to problem 2.16 on p. 95 of the ME 352 text: Theory of Machines and Mechanisms, and
determine
ME 352
Spring 2013
Answers to Homework No. 2
Configuration #1:
Configuration #2:
theta3 = 8.942 degrees
theta3 = 291.058 degrees
theta4 = 79.471 degrees
theta4 = 220.529 degrees
In part 4) after the first iteration: theta3 = 9.549 degrees and theta4 = 78.
ME 352 - Machine Design I
Name_
Spring 2013
Lab. Div. Number_
Homework No. 2
Due: Parts 1, 2, 3, 4 on Friday, January 25, 2013 (10 pts)
Part 5 in Lab at your first Lab period the week of January 28, 2013 (10 pts)
Also remember the Homework Notes (Will app
ME 352
Spring 2013
Answers to Homework No. 1
1) Problem 1.5 on p. 41 of the text
a) M=1
b) M=1
c) M=0 (exception since mobility is one) d) M=1
2) Problem 1.6 on p. 41 of the text
n=5 J1=5 J2=1
M=1
3) Phi = 30 degrees. There are an infinite number of solut
ME 352 - Machine Design I
Name_
Spring 2013
Lab. Div. Number_
Homework No. 1
Due: Wednesday, January 16, 2013
Homework Notes (Will apply to all homeworks this semester)
(i) The handout sheet must be stapled as the cover sheet to your solution.
(ii) Your n
ME 352 - Machine Design I
Fall Semester 2012
EXAM 1 - Answers
Problem 1.
Part I. (7 Points).
MA 1.86
Part II (18 Points).
m 3(7 1) 2(8) 1 1
A possible set of vectors is shown in figure below.
1
Problem 2 (25 Points)
0.0144 rad / cm
3
R 4 1.154 rad / cm
3
ME 352 - Machine Design I
Name of Student:_
Fall Semester 2012
Lab Section Number:_
EXAM 2. OPEN BOOK AND CLOSED NOTES.
Thursday, November 8th, 2012
Write your solutions on the blank paper provided and write on one side of the paper only. Where
necessary,