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Introduction... Mechanism types... Mobility... Grashof's Law... Mechanical Advantage... Freudenstein's Equation... Velocity Vectors... Acceleration Vectors...
Linkages include garage door mechanisms, car wiper mechanisms, gear shift mechanisms. They are a very
important part of mechanical engineering which is given very little attention...
Examples of Lower Pair Links with associated Degrees of Freedom
Planar, Spatial and Spherical Mechanisms
A planar mechanism is one in which all particles describe plane curves is space and all
of the planes are co-planar.. The majority of linkages and mechanisms are designed as planer systems. The main
reason for this is that planar systems are more convenient to engineer. Spatial mechanisms are far
more complicated to engineer requiring computer synthesis. Planar mechanisms ultilising only lower
pairs are called planar linkages. Planar linkages only involve the use of revolute and prismatic
An important factor is considering a linkage is the mobility expressed as the number of degrees
of freedom. The mobility of a linkage is the number of input parameters which must be controlled
independently in order to bring the device to a set position. It is possible to determine
this from the number of links and the number and types of joints which connect the links...
DOF = 3 (n-1)
Connecting two links using a joint which has only one degree of freedom adds two constraints. Connecting two links with a joint which has two degrees of freedom include 1 restraint to the systems. The number of 1 DOF joints = say j 1 and the number of joints with 2 DOF's = say j 2.. The Mobility of a system is therefore expressed as
mobility = m = 3 (n-1) - 2 j 1 - j 2
Examples linkages showing the mobility are shown below..
A system with a mobility of 0 is a structure. A system with a mobility of 1 can be fixed
in position by positioning only one link. A system with a mobility of 2 requires two links to be positioned to
fix the linkage position.
When designing a linkage where the input linkage is continuously rotated e.g. driven by
a motor it is important that the input link can freely rotate through complete revolutions.
The arrangement would not work if the linkage locks at any point.
For the four bar linkage Grashof's law provides a simple test for this condition
Referring to the 4 inversions of a four bar linkage shown below ..Grashof's law states that one of the links (generally the shortest link) will be able to rotate continuously if the following condition is met...
b (shortest link ) + c(longest link) < a + d
Note: If the above condition was not met then only rocking motion would be possible
for any link..
Mechanical Advantage of 4 bar linkage
The mechanical advantage of a linkage is the ratio of the output torque exerted
by the driven link to the required input torque at the driver link.
It can be proved that the mechanical advantage is directly proportional to
Sin( β ) the angle between the coupler link(c) and the
driven link(d), and is inversely proportional to sin( α )
the angle between the driver link (b)
and the coupler (c) . These angles are not constant so it is clear that the
mechanical advantage is constantly
The linkage positions shown below with an angle α = 0 o and 180 o
has a near infinite mechanical advantage. These positions are referred to as toggle
positions. These positions allow the 4 bar linkage to be used a clamping tools.
The angle β is called the "transmission angle".
As the value sin(transmission angle) becomes small the mechanical advantage of the linkage approaches zero.
In these region the linkage is very liable to lock up with very small amounts of friction. When using four bar linkages
to transfer torque it is generally considered prudent to avoid transmission angles below 450 and 500.
This equation provides a simple algebraic method of determining the position of
an output lever knowing the four link lengths and the position of the input lever.
The position vector of the links are related as follows
l 1 + l 2 + l 3 + l 4 = 0
Equating horizontal distances
l 1 cos θ 1 + l 2 cos θ 2 + l 3 cos θ 3 + l 4 cos θ 4 = 0
Equating Vertical distances
l 1 sin θ 1 + l 2 sin θ 2 + l 3 sin θ 3 + l 4 sin θ 4 = 0
Assuming θ 1 = 1800 then sin θ 1 = 0 and cosθ 1 = -1 Therefore
- l 1 +
l 2 cosθ 2 +
l 3 cosθ 3 +
l 4 cos θ 4 = 0
Moving all terms except those containing l 3 to the RHS and Squaring both sides
l 32 cos 2 θ 3 = (l 1 - l 2 cos θ 2 - l 4 cos θ 4 ) 2
l 32 sin 2 θ 3 = ( - l 2 sin θ 2 - l 4 sin θ 4) 2
Adding the above 2 equations and using the relationships
Freudenstein's Equation results from this relationship as
K 1 cos θ 2 + K2 cos θ 4 + K 3 = cos ( θ 2 - θ 4 )
K1 = l1 / l4 K2 = l 1 / l 2 K3 = ( l 32 - l 12 - l 22 - l 2 4 ) / 2 l 2 l 4
This equation enables the analytic synthesis of a 4 bar linkage. If three positions of the output lever are required corresponding to the angular position of the input lever at three positions then this equation can be used to determine the appropriate lever lengths using three simultaneous equations...
Velocity Vectors for Links
The velocity of one point on a link must be perpendicular to the axis of the link, otherwise
there would be a change in length of the link.
Considering the four bar arrangement shown below. The velocity vector diagram is built up as follows:
The velocity vector diagram is easily drawn as shown...
Velocity of sliding Block on Rotating Link
Acceleration Vectors for Links
The acceleration of a point on a link relative to another has two components:
The diagram below shows how to to construct a vector diagram for the acceleration components on a single link.
The diagram below shows how to construct an acceleration vector drawing for a four bar linkage.
The location of the acceleration of point p is obtained by proportion bp/bc = BP/BC and the absolute
acceleration of P = ap
The diagram below shows how to construct and acceleration vector diagram for a sliding block on
a rotating link..
The total change in velocity in the radial direction = dv- ω r d θ
Radial acceleration = dv / dt = ω r d θ / dt = a - ω2 r
The total change in velocity in the tangential direction = v dθ + ω dr + r α
Tangential acceleration = v dθ / dt + ω dr/dt + r d ω / dt
The acceleration vector diagram for the block is shown below
Note : The term 2 v ω representing the tangential acceleration of the block relative
to the coincident point on the link is called the coriolis component and results whenever
a block slides along a rotating link and whenever a link slides through a swivelling block
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Last Updated 09/09/2011