King Fahd University of Petroleum & Minerals
Mechanical Engineering Department
(Semester 162)
Solution
ME 309 Mechanics of Machines
HW 1
(Assigned: Thursday 16/02/2017
Due: Sunday 26/02/2017)
_________________________________________________________________________________
Problem 1.
Do problems 1.5 and 1.11 in the textbook.
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j1
j2
j1
j1
j1
j1
(a) n = 6, j1 = 7, j2 = 0, and hence mobility = m = 3(n1)  2j1 – j2 = 3(61) – 2(7) – 0 = 1.
(b) n = 8, j1 = 10, j2 = 0, and hence mobility = m = 3(81) – 2(10) – 0 = 1.
(c) n = 7, j1 = 9, j2 = 0, and hence mobility = m = 3(71) – 2(9) – 0 = 0.
(d) n = 4, j1 = 3, j2 = 1, and hence mobility = m = 3(41) – 2(3) – 1 = 2.
7
j1
j1
j1
4
1
j1
6
5
j2
j1
j1
2
3
j1
j1
n = 7, j1 = 8, j2 = 1, and hence mobility = m = 3(71)  2j1 – j2 = 3(71) – 2(8) – 1 = .
This answer is correct, because, indeed the mechanism has a mobility of 1.
Problem 2.
Find the range of values for the unknown length of link 3 (L3) for the four-bar linkage to act as a crankrocker mechanism. Given: L1 = 200 mm, L2 = 50 mm and L4 = 150 mm.
B
2
3
C

4
O2
O4
1
1
s = L2 = 50 mm (for crank-rocker)
If l = L3, then p = L1 = 200 mm and q = L4 = 150 mm, and hence
s + l < p + q  50 + L3 < 200 + 150  L3 < 300 mm
If p = L3, then l = L1 = 200 mm and q = L4 = 150 mm, and hence
s + l < p + q  50 + 200 < L3 + 150  L3 > 100 mm
Therefore, 100 mm < L3 < 300 mm.
Problem 3. For the above problem 2, find L3 so that the minimum transmission angle min = 45o. Then,
what is the maximum transmission angle max for this linkage.
3 min 4
2
3
2
max
4
For min, (L1L2)2 = L32 + L42  2L3L4cosmin,  (20050)2 = L32 + 1502  2L3(150)cos45o
We find L3 = 212 mm. Using this value:
(L1L2)2 = L32 + L42  2L3L4cosmax,  (20050)2 = 2122 + 1502  2(212)(150)cosmax
We find max = 85.5o.
Problem 4. Consider the offset slider-crank mechanism given below. Let L2 = 60 mm, L3 = 120 mm
and offset distance = e = 20 mm. Show the mechanism in its limiting positions (corresponding to
extreme positions of the slider (link 4). Determine the angle through which the crank (link 2) turns as
the slider moves from the extreme left to the extreme right position. Find the stroke of the slider
(distance between limiting positions of the slider).
B
2
3
A
1
Offset=e
C
4
1

 = 180 o   cos 1
e
e
 cos 1
L2  L3
L3  L2

 = 180 + 83.6o  70.5o = 193.1o.
Consequently,  = 360o   = 166.9o.

20
20

  180 o   cos 1
 cos 1

60  120
120  60 


o
Also, Stroke = S =
( L2  L3 ) 2  e 2  ( L3  L2 ) 2  e 2  (60  120) 2  20 2  (120  60) 2  20 2
S = 178.9 – 56.6 = 122.3 mm.