Subject: Kinematics of Machinery

Topic: Kinetics of Rigid Bodies

Difficulty: High

$n= 3(l-1)-2j-h$

Where $\quad$ l= no. of links

$\quad\quad\quad\quad$ j=no. of binary joints

This equation is called kutzbach criterion for determining the number of degrees of freedom of movability (n) of a plane mechanism.

The no. of degree of freedom or movability (n) for some simple mechanisms having no higher pair (i.e. h=0) as shown in fig. are determined as follows;

1. The mechanism as shown in fig.(a) has three links and three binary joints i.e. l=3, j=3 $\quad\quad\quad\quad$$n=3(3-1)-2\times3=0$ 2. The mechanism as shown in fig. (b) has four links and four binary joints i.e. l=4, j=4 $\quad\quad\quad\quad$ $n=3(4-1)-2\times4=1$

2. The mechanism as shown in fig. (c) has five links and five binary joints i.e. l=5, j=5

$\quad\quad\quad\quad$ $n=3(5-1)-2\times5=2$

1. The mechanism as shown in fig. (d) has five links and six equivalent binary joints (because there are two binary joints at B and D, and two ternary joints at A and C) i.e. l=5, j=6

$\quad\quad\quad\quad$ $n=3(5-1)-2\times6=0$

1. The mechanism as shown in fig. (e) has six links and eight equivalent binary joints(because three are four ternary joints at A, B, C, and D) i.e. l=6, j=8

$\quad\quad\quad\quad$ $n=3(6-1)-2\times8=-1$

Grubler’s Criterion for plane mechanism:

The Grubler’s criterion applies to the mechanism to the only single degree of freedom joints where the overall movability of the mechanism is unity. Substituting n=1 and h=0 in kutzbach equation, we have,

$\quad\quad\quad\quad$ $1=3(l-1)-2j$ or $3l-2j-4=0$

This equation is known as the Grubler’s criterion for mechanisms with constrained motion.

A little consideration will show that a plane mechanism with a movability of 1 and only single degree of freedom joints can not have odd number of links. The simplest possible mechanism of this type are a four bar mechanism and a slider-crank mechanism in which l=4 and j=4.