ME 301 THEORY OFD MACHİNES I

SOLVED PROBLEM SET 2

PROBLEM 1. For the given mechanisms, find the degree of freedom F. Show the link numbers

on the figures. Number the links and label the joints on the figure. Write the number of links,

number of joints and types of joints in the mechanisms. Indicate if there is any redundant degree

of freedom.

a) Planar mechanism. Assume that there is no slip at the cam joint.

𝜆 = 3

Disregard the spring and the joints related to the spring.

ℓ = 9, 𝑗 = 12 (8𝑅, 2𝑃, 1𝐶𝑝(𝑛𝑜 𝑠𝑙𝑖𝑝), 1𝐶𝑠)

∑𝑓𝑖 =8 + 2 + 1 + 2 = 13

𝐹 = 3(9 − 12 − 1) + 13 = −12 + 13 = 2

1

2 3

4

5 6

7

8 9

1

R, R

R R

R

P

P

R

𝐶𝑠

𝐶𝑝

R

R

b) Planar mechanism. Assume that the loader is stationary.

PROBLEM 2.

The shaft coupling shown is a spatial mechanism and obeys the general degree of freedom

equation. It is known that its degree of freedom is F=1. As shown in the figure, the joints between

links 1 and 2, and links 1 and 4 are revolute joints, and the joint between links 2 and 3 is prismatic

joint. State the type, degree of freedom and the types of relative motion (i.e. rotation and/or

translation) of the joint between links 3 and 4.

𝜆 = 3

ℓ = 9, 𝑗 = 11 (9𝑅, 2𝑃)

∑𝑓𝑖 =11

𝐹 = 3(9 − 11 − 1) + 11 = 2

The two freedoms are the raising and

turning of the bucket 9, which are

controlled by the motions of the two

pistons 3 and 6.

1

1

1

1

2

1

3

1 4

1

5

1

6

1 7

1 8

1

9

1

R

1 R

1

R

1 R

1 R

1 R

1

R

1

R

1

R

1

P

1

P

1

Solution:

PROBLEM 3.

In the planar mechanism given below 90BED and 0 90A CD and the link lengths are

labelled as 𝐴𝑜𝐴 = 𝑟2, 𝐴𝐵 = 𝑟3, 𝐴𝐸 = 𝑎3, 𝐴𝑜𝐶 = 𝑟1 and 𝐶𝐷 = 𝑎1.

a) Find the degree of freedom, the number of independent loops, and the total number of required

joint variables (position variables).

b) Choose a sufficient number of revolute joints which when disconnected yield an open-loop

system. Indicate those joints. Assign the joint variables and show them clearly.

c) Write the necessary number of independent loop closure equations using vectors described by

directed lines such as SQ⃗⃗ ⃗⃗ , RS⃗⃗ ⃗⃗ , etc.

d) Using complex numbers, re-write these loop closure equations in terms of the joint variables

and the fixed parameters of the mechanism.

Solution:

PROBLEM 4.

For the following double slider mechanism with dimensions |OA| = b1, |OC| = c1, |BD| = b3 where

s14 is the input joint variable.

a) Write down the loop closure equations using point to point vectors and using complex

numbers.

b) Write down the scalar components of the loop closure equations.

c) Solve the unknown joint variables analytically in terms of the link lengths and the input

joint variable.

Solution:

a) LCE via point to point vectors : OA AD DB OC CB

LCE via complex numbers: 12 312 ( )

1 23 3 1 14ib s e b e c is

b) Scalar components of LCE :

Real part : 23 12 3 12 3 1cos cos( )s b c

Imaginary part : 1 23 12 3 12 3 14sin sin( )b s b s

c) Let us rearrange the above equations as below:

23 12 14 1 3 12 3

23 12 1 3 12 3

sin sin( )

cos cos( )

s s b b

s c b

Divide side by side:

14 1 3 12 312

12 1 3 12 3

sin( )sin

cos cos( )

s b b

c b

Cross multiplying gives:

1 12 3 12 12 3 14 1 12 3 12 12 3sin sin cos( ) cos cos sin( )c b s b b

or

1

2

3

4

s23 γ3

12

D

s14

O C

14 1 12 1 12 3 12 3 12 12 3 12cos sin sin( )cos cos( )sin 0s b c b

Since sin sin cos cos sin :

12 12cos sinA B C

where

14 1

1

3 3sin

A s b

B c

C b

If we let

cosA D

sinB D

where 2 2D A B

𝜑 = 𝑎𝑡𝑎𝑛2(𝐵, 𝐴) Then:

𝜃12 = 𝜑 ± 𝑐𝑜𝑠−1 𝐶

𝐷

Once 12 is found from the above expression; 23s can be found from either

1 3 12 323

12

cos( )

cos

c bs

(if 23

0

0s )

or

14 3 12 3 123

12

sin( )

sin

s b bs

(if 23

0

0s )

PROBLEM 5.

In the single d.o.f. mechanism shown, the joint variables are assigned as 𝜃12, 𝜃15, 𝑠54, 𝜃16 and 𝑠63

which correspond to disconnecting the revolute joints at A and B. The link dimensions are given

as 𝐴𝑜𝐴 = 𝑟2 = 17cm, 𝐴𝐵 = 𝑟3 = 24cm, 𝐴𝐸 = 𝑎3 = 17cm, 𝐴𝑜𝐶 = 𝑟1 = 47cm, 𝐶𝐷 = 𝑎1 =

26cm, 90BED and 0 90A CD . The loop closure equations in complex numbers are

𝐿𝐶𝐸 1 ∶ 𝑟2𝑒𝑖𝜃12 = −𝑖𝑟1 + 𝑎1 + 𝑠63𝑒

𝑖𝜃16 + 𝑎3𝑖𝑒𝑖𝜃16

𝐿𝐶𝐸 2 ∶ 𝑠54𝑒𝑖𝜃15 = 𝑎1 + 𝑠63𝑒

𝑖𝜃16 + (𝑟3 + 𝑎3)𝑖𝑒𝑖𝜃16

By solving the Loop Closure Equations analytically, determine the joint variables when 𝜃12 =20𝑜. Find all solutions corresponding to different assembly configurations of the mechanism.

(Assume that there are no physical limitations of the prismatic joints).

Solution: