USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

Four Bar Mechanism

Figure 4.5 shows a four bar mechanism

 

VB =w1. AB = w0.BO

w0 =w1 AB / BO

Again,                         VC = w2 CD = w0 . CD

w0 = w2 CD /CO

 

From (4.11) and (4.12), we have

w1 . AB / BO = w2 .  CD / CO

w1 /w2 = CD /AB . BO /CO

 

SOLVED EXAMPLES

 

Example 4.1. In Fig. P-4.1,find VB if VA = 5 m/s.

 

Solution. Let ro0 is the angular velocity of AB

VA = w0 x AO

VB = w= x BO

VA / VB = AO / BO = tan 450 =1

VA =VB

VB =VA = 5 m/s

 

Example 4.2. The crank of a reciprocating engine is rotating at 200 rpm. The lengths of the crank and connecting rod are 200 mm and 1000 mm respectively. Find the velocity of the point A (velocity of piston) when crank has turned through an angle 30° with the horizontal as shown in Fig. P-4.2.

 

Solution .                 w (crank) = 2πN  /600 = 2π x 200 /60 = 20.94 rad

θ = 300

r sin θ= L sin ϕ

sin ϕ = r /L  sin θ = 0.2 /1  sin 300 = 0.1

ϕ = 5.740

VA = w[ L sin ϕ + cos θ tan ϕ]

= 20.94 [ 0.1+0.2 cos 300

= 2.46 m/s .

 

Example 4.3. Find the acceleration of point A in the problem 4.2 above.

Solution .                            VA  =w [ L  sin ϕ + r cos θ tan ϕ]

dVA   / dt  +w [ L cos ϕ dϕ / dt +r { cos θ. sec 2 ϕ dϕ /dt + tan ϕ x (- sin θ) dθ /dt) }]

now ,                           r sin θ = L sin ϕ

r cos θ . dθ /dt  =L cos ϕ dϕ/dt

wr cos θ = L cos ϕ . dϕ /dt

dϕ/dt = wr cos θ / ϕ  = 20.94 x 0.2 cos 300  / 1 x cos 5.74

 

equation (1)                 = 3.645 rad / sec

dVA / dt = 20.94 [ 1/cos 5.740  x 3 .645 + 0.2 { cos 300  sec 2 5.740  x 3.645}

+ tan 5.740  (- sin 300  ) x 20. 94}]

aA  = 84.9 rad /sec2