Sub:-Kinematics of Machines
Topic:- Synthesis and Analysis of Mechanism
Prepared By:- Goriya Vaishali
(130590119037)
Guided by:-Prof. M. U. Parasara
•Generally
the steering is provided on front wheel only.
•Front wheel is mounted on stub axle.
•The mechanism change the position of stub axle about
their pivot point for steering.
•The rear wheel always remains straight.
•During steering the wheel must has a pure rolling
action.
•As
shown in figure when vehicle takes turn, the inner
wheel makes a large angle θ then outer wheel angle ø.
•As
shown in figure
•L=wheel base.
•a=distance between two wheels.
•b=distance between two pivots.
•cotø – cotθ = b/L.
•This equation is the condition of correct steering.
•This
mechanism is mounted on front axle having
four bar chain with four turning pairs.
•tanα=b/2L.
•This equation is devis steering
gear mechanism.
•This
mechanism is simpler than devis steering
gear mechanism. The difference between two
steering mechanism are :1) Construction of devis mechanism is on the
front side of the front axle where as ackermann
is at back side of the front axle.
2) The devis steering gear mechanism consist
four turning pair and two sliding pair where
ackermann steering gear mechanism consists
four turning pairs.
•In
devis steering mechanism link AB is a fixed link.
Link BC and AD are of same length and connected to
both the wheel pivoted at B and A.
•When the vehicle moves on a straight path the link
AB and CD are parallel to each other and link BC and
AD are equally inclined to link AB.
•A hook
joint is used to connect to shaft, which are
intersecting at a small angle.
•The end of each shaft is forked to u type and each
forked provides to bearing for the arm of a cross.
•The arm of the cross are perpendicular to each other.
2
/ω = cosα
/ 1-cos θ*sin2α
•ω1
2
•In
order to have a constant velocity ratio of the driving
and driven shaft, and intermediate shaft with a hooke’s
joint at each.
•This type of joint is known as double hooke’s joints.
•This joint gives a velocity ratio = unity ; if
1) The axis of driving & driven shaft are in the same
plane.
2) The driving & driven shaft make equal angle
with intermediate shaft.
•Consider
figure.
a slider crank mechanism as shown in
•Link
a, b, c, d are connected and this type of chain is
called Four bar chain.
•Consider Four bar chain mechanism as shown in
figure.
Velocity polygon
Mark a fix point of the mechanism as one point. Take
any point o and from it draw oa=V =ω*OA with some
suitable scale perpendicular to link OA as in fig
•The velocity of point B w.r.t point A is perpendicular
to line AB. So from point ‘a’ draw a line perpendicular
to AB representing V .
•
AO
AB
•The
velocity of slider B relative to O is parallel to the
line OB as B is a slider. From O draw a line parallel to
the line of stroke OB to intersect V at point b.
•Measure V = ab and V = ob from the velocity
diagram.
AB
BA
OB
•Acceleration
polygon
f =ω2* OA =V2 /OA
•Crank OA rotates with constant angular velocity. So
the tangential component of acceleration of point A is
zero.
1) Draw vector oa’ = f t parallel to AO opposite
direction with some suitable scale.
2) From point a’ draw vector a’x = f r = V /BA which
is the radial2 component of acceleration of B w.r.t A. it
is parallel to BA but in opposite direction of link VBA.
• rAO
AO
AO
BA
BA
Acceleration polygon
3) From x draw vector xb’ perpendicular to or AB. The
vector xb’ represent the tangential component of
acceleration of B w.r.t A i.e. f . It is known in
direction only and contains b’ .
r
4) Joint a’ to b’ vector a’b’ represents acceleration of B
w.r.t A i.e. f
5) f , f can be found by measurement with
the scale.
BA
BA
B
BA
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Chebyshev spacing of precision points is very useful in
minimizing the structural error.
For points in the range x0 ≤x≤xn+1 , the precision points x,
according to chebyshev spacing are given as,
xj=1/2(xn+1+x0)-1/2(xn+1-x0) * cos(π(2j-1)/2n)
where, j=1, 2, ..... n
θj= θi+kθ(xj-xi)
øj = øi + kø (yj-yi)
kθ=∆θ/∆x
kø=∆ø/∆y
∆x=xj-xi
∆y=yj-yi

k1cosø+ k2cosθ +k3 =cos(θ-ø)

This equation is freudenstein's equation.
Where , k1 = a/b
k2 = -a/d
2
2 2
2
k3 = (a+b-c+d )/2bd

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YOU