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(MECH103)mech103_midterm_sol_08.pdf
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MECH103 Fall08
Solution to MECH 103 Midterm Examination
Fall Semester 2008

4:30 pm C 6:00 pm
Venue: Lecture Theatre B (LTB)
16 Oct 2007
Student Name:_________________

Student ID:___________________
Student Signature:______________
1 Kinematics of Bio-Machinery (20pts)
(1) How many degrees of freedom (DOF) do you have in your wrist and hand combined?
Ans: Each finger (and thumb) of a human being can rotate up and down and side-to-side about the first joint. In addition, each finger can rotate about each of the two remaining joints for a total of 4 degrees of freedom for each finger. However, the thumb has only 3 degrees of freedom.
Wrist: 2 DOF Thumb: 3 DOF Finger 44 = 16 DOF
Total DOF = 21
(2) How many DOF do the following joints have?
a. Your knee, b. Your ankle, c. Your shoulder, d. Your knuckle
Ans: a) knee: rotate about an axis parallel to the ground = 1 DOF b) ankle: three rotations about mutually perpendicular axes c) shoulder: three rotations about mutually perpendicular axes
d) knuckle: two rotations about mutually perpendicular axes

p.16, A. Tozeren, Human Body Dynamics, Springer, 2000.
2 Kinematics of Rotary Engine (30pts)
This is a 3-cylinder, rotary, internal combustion engine as illustrated in Fig. 1. The pistons (sliders) at C, D, and E drive the output
crank (2) through piston rods (couplers 3, 4, and 5) BC, BD, and BE. There are 3 full joints at B. The cross-hatched crank-shaft at
A is supported by the ground link (1) through bearings. Please calculate the mobility M.


MECH103 Fall08
Fig. 1 Fig. 2
Ans: Number of links, N = 8 Number of J1 joints = 10, Number of J2 joint = 0 Mobility M = 3(N-1)-2J1 CJ2 = 3(7)-2(10) = 21-20= 1
3 Kinematics of Four-bar Linkage (30pts)
The link lengths (in cm) of a four-bar linkage as shown in Fig. 2 are defined in as: a = 5cm, b= 8cm, c= 6cm, d=8cm, 2 = 75.

(1)
Determine the Grashof condition. If the fourbar is Grashof, identify if it is crank-crank, crank-rocker or double rocker.

(2)
Determine the output angles 3 and 4 using the vector loop method.


Ans:
(1)
L + S = 8+5 = 13, P+Q= 6 + 8 = 14, L + S < P + Q

Because the ground link d is adjacent to the shortest link a, this is a Grashof crank-rocker mechanism.

(2)
a = 5, b =8, c = 6, d = 8, 2 = 75 = 1.309 rad, K1 = d/a = 1.6 K2 = 1.333, K3 = 1.017, K4 = 1, K5 = -1.462


A = -0.670, B = -1.932, C = 2.013,
K4 =d/b, K5 = (c2 - d2- a2 -b2)/(2a b)= -1.462

D = -2.545, E = -1.932, F = 0.138

2 .
.
.1 . B
B . 4AC
.
(4 )1,2 = 2tan ..
=.149.735, 78.212
2A .
..
2 .

.1 . E
E . 4DF

.
( ) = 2tan .. =.79.02, 7.497
3 1,2
. 2D .
..
41 and 31 form an open solution, 42 and 32 form a closed solution.

4 Orbital Mechanics (30pts)
A satellite is given an initial velocity v0 = 6000 m/s at a distance r0 = 3RE from the center of the earth, as shown in the figure. Assume the radius of the earth RE is 6