The Kutzbach’s Principle to find the Degree of Freedom
Lower Pairs and Higher Pairs
Where two links are paired in a way that the links have surface contacts between them, the pair is called a lower pair, or lower
pairs have surface contact between the links. For planar linkages, the revolute and prismatic pairs are basically used.
When the links are connected in a way that there exists ideally a point or a line contact between the links, the pair is called a
higher pair. The pair between the teeth of a pinion and a gear, or the cam and a flat face follower are some examples of higher
pair. The contact in reality can seldom be a point or a line, however, it may be interpreted that the area of contact is negligible
with respect to that in the case of lower pair. As the contact between two bodies is over a very small area, the contact may
admit both rotation about the point of contact and sliding of the point of contact. Thus it admits two motions or degrees of
freedom and restricts one degree of freedom.
The Kutzbach’s Equation: F = 3(n-1) – 2j – h, where, ‘n’ is the total number of links, ‘j’ is the total number of joints or pairs
which restrict two degrees of freedom each, and ‘h’ is the total number of higher pairs in the linkage.
Degree of freedom specifies the number of independent coordinates necessary to represent the configuration of the linkage.
Please note that Kutzbach’s equation specifies a generic idea of the degree of freedom, without taking into consideration the
shape, size, way of assembly of links etc. Therefore the effective degree(s) of freedom may be different from that predicted by
the Kutzbach’s principle if particular conditions are considered. We shall discuss some modifications.
Degree(s) of freedom continued
R
Let us count the links and joints in the mechanism at the instant
shown.
3
Cam
Please note that the cam (link 2) is connected to the oscillating
follower (link 4) through a circular roller, which is link 3 in the
diagram.
In the figure shown,
the total number of links n = 4,
the total number of joints, which take away two degrees of
freedom each, j = 3
the number of higher pair(s), h = 1
Applying the Kutzbach’s formula we get
F = 3*(4-1) - 2*3 – 1*1 = 2
H
2
Circular
Roller
4 Follower
R
R
1
1
Cam roller follower mechanism
This mismatch is due to the fact that Kutzbach’s
equation does not have a provision to consider the
shape of a link. So it is modified as
F = 3 * (n-1) – 2 * j – h – Fr ;
Fr is the redundant degrees of freedom in a linkage.
Is it a mechanism giving the constrained motion? Ideally speaking: No
However experience says that corresponding to an angular orientation of the cam there should be a unique angular orientation of the
follower. The physical degrees of freedom are 1) the angular orientation of the cam and 2) the angular orientation of the roller, which
is redundant as this is a circle, because any orientation of it will not change the angle of the follower.
So, actually the effective degree of freedom = 2 -1 = 1, and the linkage is a mechanism. The result would have been accurate if the
shape of the roller would have been non circular
Degree(s) of freedom continued
Point of
contact
R
3
Cam
H
2
4 Follower
R
R
1
3
Circular
Roller
≡
n = 3, j = 2, h = 1
F = 3 * (3-1) – 2 * 2 – 1*1 = 1
2
1
1
1
Cam roller follower mechanism
The modified mechanism without revolute pair at the centre of the circular roller
The follower is integrally connected to the circular roller.
You may note that at the point of contact both rolling and sliding are permitted, as it is a higher pair at the point of
contact. However, if you choose to arrest sliding or slipping at the point of contact, you may assume the presence of a
revolute pair at the point of contact in the first case and find the degree of freedom. However, if slipping is not permitted
at the point of contact in the case of the modified mechanism, the linkage will lose all degrees of freedom, or F will be
equal to 0 and the linkage will be a structure.
Let us see another example
Consider the 2R-2P 4 bar chain shown on the right.
For this n = 4, j = 4 (2 revolute and 2 prismatic pairs). Note that the
prismatic pairs are in line or have a common axis so that a solid prism can
be passed through both of them.
The degree of freedom is F = 3 * (4-1) – 2 * 4 = 1. So the linkage should
move. However, this is not correct.
The physical degree of freedom is the to and fro motion of the solid prism,
the link 4, which does not cause any angular motion of the links 2 and 3.
Hence, as such the linkage behaves as a structure. The to and fro motion
of link 4 is possible due to a special situation that the prismatic pairs are
placed in line. Had they not been so, the to and fro motion of the link 4
would not have been possible. So the degree of freedom is redundant.
If any motion of any link is possible without moving any other link in a
linkage, the degree of freedom corresponding to that motion is considered
redundant.
4
2
3
1
1
The Kutzbach’s formula is
modified as
F = 3 * (n-1) – 2 * j – h – Fr ;
Fr is the redundant degrees of
freedom in a linkage.
Degree(s) of freedom continued
A
Let us consider the linkage shown. In this the length of the links
AB, CD and EF is the same or AB = CD = EF. Let us find the
degree of freedom of the linkage.
However according to our common sense the linkage should
move as at any instant the arms ABCD or CDFE will form
parallelograms. So where is the difference?
The formula has no provision to consider the equality of the
links. In this case any one of the arms AB or CD or EF may be
taken to be redundant to get the effective degree of freedom.
In that case n = 4, j = 4 and F = 3 * (4-1) – 2 * 4 = 1.
E
4
5
B
D
F
1
1
2
In this n = 5, j = 6, or
F = 3 * (5 – 1) – 2 * 6 = 0
Kutzbach’s formula predicts that the linkage should not move
or is a structure.
C
3
1
So we modify the Kutzbach’s formula to write
F = 3*(n- nr - 1) – 2*j – h – Fr.
In the above nr is the number of redundant links. In
this example nr = 1.
Degree(s) of freedom continued
3
If in the above parallelogram linkage, one of the links (link 4) is not
parallel to the rest, as shown, the linkage will be a structure, i.e. will not
move and the prediction of Kutzbach’s formula will be true.
Kutzbach’s formula gives a generic prediction of the degree(s) of
freedom.
2
1
4
5
1
1
Degree(s) of freedom continued
Consider the straight shaft supported on three rigid
bearings as shown on the right.
Suppose we want to find out whether it will be able to
spin about its axis as shown, by finding out the degree of
freedom of the shaft in rotation, by utilizing the
Kutzbach’s formula, though it is obvious that the shaft
must rotate.
Bearings
A straight shaft supported on three bearings
For this
This anomaly happens because, placing three revolute
n = 2, the ground and the shaft
j = 3, as each bearing acts as a revolute pair in the plane pairs for the same rotation constrains the linkage unduly.
Again thinking practically, by virtue of rigidity the
of rotation of the shaft.
straight shaft does not need three bearings to maintain
F = 3 * (2 -1) -2 * 3 = 3 -6 = -3.
Kutzbach’s formula predicts that not only it should not straightness and rotate. Any two bearings are redundant.
So we shall modify the Kutzbach’a formula as
rotate but it should be a statically indeterminate
F = 3*(n- nr - 1) – 2*(j – jr) – h – Fr.
structure also. Why is it so?
In this ‘jr’ is the redundant number of joints.
Applying the modified formula we shall get F = 1.
The Kutzbach’s formula and its modifications
F: The degree of freedom, Fr: the redundant degrees of freedom
n: total number of links, nr: total number of redundant links
j: total number of joints each restricting two degrees of freedom, jr: number of redundant joints
h: the number of higher pairs
The original Kutzbach’s equations for planar linkages: F = 3 * (n-1) – 2 * j
When the higher pairs are included: F = 3 * (n-1) – 2 * j – h
When the redundant links are included: F = 3 * (n - nr - 1) – 2 * j – h
When the redundant joints are considered: F = 3 * (n - nr - 1) – 2 * (j – jr) – h
When the redundant degrees of freedom are included: F = 3 * (n - nr - 1) – 2 * (j – jr) – h – Fr
Kutzbach’s formula with the modifications: F = 3 * (n - nr - 1) – 2 * (j – jr) – h - Fr
Let us calculate the degrees of freedom of the
linkage shown.
Lower Pair
5
Let us count the number of links ‘n’ and joints ‘j’
and the higher pair(s) ‘h’.
n = 5, j = 5 (Revolute pairs: 4, Prismatic pair: 1),
h = 1.
From Kutzbach’s formula the degree(s) of
freedom F = 3*(5-1) -2*5 – 1*1 = 1
So F = 1, or the given linkage is a mechanism.
In this case the input is the rotation of the cam
mounted integrally on the cam shaft and the
output is the up and down movement of the valve
inside the valve guide.
4
1
1
1
3
2
Higher Pair
1
Lower Pair
Please note:
1) The valve guide is a part of the frame. However, it does not
dictate the direction of the valve rod and so, the valve rod
and the guide do not form a prismatic pair. Therefore, no
pair has been assumed between the valve rod and guide. In
fact, the valve rod, as a result of its motion translates and
rotates very slightly. Had the guide been not there, still the
valve will function.
2) The assumption of a prismatic pair between link 4 and 1 is
logical, because, if the pair is not there the follower will not
be able to able to follow the displacement of the cam
An Example of a complex joint
We shall attempt to find out the degrees of freedom of the Peaucelliar
mechanism. It is a mechanism to generate straight line motion of a
point, the point Q. The input link is OPP, which upon rotation will
cause the point Q move ┴ to the line joining points O and OP.
The linkage contains revolute pairs only. Please note that the points O,
A, B and P, at each of which more than two links have met. These
joints are called complex joints.
Question Is how many revolute pairs exist at each of these points?
The number of revolute pairs is:
Number of links meeting at a point -1
Peaucelliar Mechanism
Therefore, for this linkage, n = 8, j = 10 (2 each at points O, A, B, P
and 1 each at points OP and Q)
Therefore F = 3*(8-1) – 2*10 = 1.
ሷ
The Gr𝐮bler’s
equation
The original Kutzbach’s equation is given by F = 3(n-1) – 2j.
ሷ
The Gr𝐮bler’s
equation is obtained, when we try to find out the number of joints necessary for getting a mechanism
for a closed chain, i.e. with all binary links in a linkage.
For a mechanism F = 1, therefore putting the value of F in the equation we get 3(n-1) – 2j = 1 from which we solve j
and find j = (3/2)n – 2. Hence these many joints each restricting two degrees of freedom are necessary to produce
constrained motion in a closed chain.
An important count is the number of joints that exceed the number of links. This is (3/2)n – 2 – n = (n/2) – 2. This
count is necessary to find the highest order of link in a closed chain.
An example:
Say we have a closed 6 bar chain. How many joints do we need to make it behave as a mechanism?
For the closed 6 bar chain to behave as a mechanism we need to have j = (3/2)*6 -2 = 7 joints. In this case the
number of joints required – the number of links = 7 -6 = 1. From this count we shall be able to find out the
maximum order of link in a closed 6 bar chain.
Please note that in an ‘n bar’ closed chain, there are at least n joints, i.e. n links connected by n joints will make all
the links binary.
You have seen that if a six bar chain is connected by 6 pairs, then all the links will be binary, however, the motion
will not be constrained. If we put this extra joint on a binary link, then the maximum order of link in the 6 bar chain
will be ternary.
Another example:
Suppose we have a closed 5 bar chain. How many joints are necessary to obtain constrained motion?
𝟏
ሷ
From Gr𝐮bler’s
criterion, j = 𝟓 𝟐 or we need 5 full joints and a half joint. The half joint may be interpreted as a
higher pair as it restricts half the degrees of freedom compared with a full joint, which restricts 2 degrees of
freedom.
Let us draw the diagrams each of a closed 5 bar and a closed 6 bar chain.
5 bar closed chain with 5
revolute pair. Each link is binary,
F = 3(5-1) -2*5 = 2.
Physical degrees of freedom: the
angles made by the two links
connected with the base with the
base.
5 bar closed
chain, with 5
revolute pairs,
and 1 higher pair.
F = 1.
Elements:
In planar linkages, we use primarily the revolute and the prismatic pairs. Each of these pairs is composed of two
elements, the pin and the hole for revolute pair and the solid and hollow prisms for the prismatic pair are thought as
elements.
1
Pin
Hole
These form the joint
or the revolute pair
2
Front orthographic view of the pin and
hole element for the revolute pair.
We may notice that the total number of elements of all the joints is 2*j, as each joint, revolute or prismatic,
consists of two elements. So the original Kutzbach’s formula may be rewritten as
F = 3(n-1) – e
where ‘e’ is the total number of elements. Using this concept we shall be able to find out different composition
of links in an ‘n’ bar closed chain, where ‘n’ is even, as otherwise, we shall have half joints or higher pairs.
An Example
Let us try to find out the condition of obtaining constrained motion of an ‘n’ bar closed planar chain, where ‘n’ is
even and each pair has two elements.
Let the chain have N2 binary, N3 ternary, N4 quaternary links and so on in the linkage. We shall have two conditions given
by two equations
Since the number of links in the linkage is n, so N2 + N3 + N4 + N5 + ……. = n … (1)
From the Kutzbach’s equation we have 3(n-1) – e = 3(n-1) – 2N2 - 3N3 - 4N4 - 5N5 - …. = 1… (2)
Substituting (1) in (2) we get
3 (N2 + N3 + N4 + N5 + ……. -1) - 2N2 - 3N3 - 4N4 - 5N5 - …. = 1… (3)
Or
N2 – N4 – 2 N5 - …… = 4 …(4)
this is the condition for obtaining constrained motion.
We observe from (4) that the number of binary links should exceed the quantity (N4 + 2 N5 +……) by 4
Therefore, we understand that the minimum number of binary links in the case when all higher order links are absent is 4.
This means that the minimum number of binary links in an ‘n’ bar closed planar chain, where all the links are
connected by pairs having two elements is 4.
An Example:
Suppose we want to construct six bar closed chain, that gives constrained motion. What is the composition of the chain?
Here, n = 6. So from Grubler’s
ሷ
criterion, the number of joints should be j = (3/2)*6 – 2 = 7, which is 1 more than the
number of links. Therefore the maximum order of the link is ternary. So the closed chain consists of binary and ternary link.
Therefore, let there be N2 binary links and N3 ternary links in the chain. We need to find out the number of binary and
ternary links.
We get two equations
N2 + N3 = 6 … (1) as there are 6 links, and
2N2 + 3N3 = 2*7 = 14 …(2) as each binary link will have two elements and ternary link has three elements.
Solving we get N2 = 4 and N3 = 2. There are two equations and two unknowns, and so N2 and N3 will have unique
solutions.
There are two ways to connect the ternary links to obtain two different six bar chains. The chain, in which the ternary links
are directly connected is called the Watt’s chain, and where the ternary links are connected through a binary link is called
the Stephenson’s chain.
Watt’s chain, kinematic diagram
Watt’s Mechanism
Kinematic diagram of Stephenson’s chain
Let us find different compositions of a closed 8 bar chain to give constrained motion.
Since this is an eight-bar chain, so n = 8.
From the Grubler’s
ሷ
criterion, we get the number of joints necessary will be j = (3/2) * 8 – 2 = 10.
The number of joints exceeding the number of links is 10 – 8 = 2
This means that the highest order of link in the linkage is quaternary. So the linkage may have binary, ternary and
quaternary links. Let us see how many of each type will be present. Let there be N2 binary, N3 ternary N4 quaternary
links.
The first equation is N2 + N3 + N4 = 8 …(1) as this is an eight-bar chain
The second equation 2 N2 + 3 N3 + 4 N4 = 2 * 10 = 20 …(2) from the count of elements
These equations have multiple solutions, as the number of equations is two but the unknowns are three.
Let us choose that N4 = 0, this gives N2 = 4, N3 = 4 and N4 = 0
Let us choose that N4 = 1, this gives N2 = 5, N3 = 2 and N4 = 1
Let us choose that N4 = 2, this gives N2 = 6, N3 = 0 and N4 = 2
Any other solution does not exist, as for N4 > 2, negative values of some links will appear.
Please note that it is possible to prove analytically that the minimum number of links to form a closed chain is 4 with
the concept of elements proposed. Therefore, you may also start the solution by choosing N2 = 4
Activities
1) Identify all the linkages with higher pairs from Tutorial sheet -1 and try to find the equivalent mechanisms
with lower pairs.
2) Identify all the six-bar linkages from the Tutorial sheet -1 and determine their types.
3) You should try to draw these conceptual kinematic diagrams of the different formations of the eight-bar
chain.
4) Find the example of an eight-bar linkage from the Tutorial sheet -1 and draw its kinematic diagram. With
which conceptual kinematic diagram does it match?
5) Using the same principle stated, try to find the formations of a ten-bar constrained planar linkage connected
by pairs, each with two elements.