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Chapter 7
sameerg edited this page May 27, 2011
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- Relative vels in Slider Crank Mechanism
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The velocity of the slider A (i.e. v A) may be determined by relative velocity method as - discussed below :
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1. From any point o, draw vector ob parallel to the direction of v B (or perpendicular to OB) - such that ob = v B = ω.r, to some suitable scale, as shown in Fig. 7.5 (b).
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2. Since A B is a rigid link, therefore the velocity of A relative to B is perpendicular to A B. - Now draw vector ba perpendicular to A B to represent the velocity of A with respect to B i.e. v AB.
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3. From point o, draw vector oa parallel to the path of motion of the slider A (which is along - AO only). The vectors ba and oa intersect at a. Now oa represents the velocity of the slider A i.e. v A,
- to the scale.
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The angular velocity of the connecting rod A B (ωAB) may be determined as follows:
- Rubbing Velocity at a Pin Joint
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The links in a mechanism are mostly connected by means of pin joints. The rubbing velocity
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is defined as the algebraic sum between the angular velocities of the two links which are connected
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by pin joints, multiplied by the radius of the pin.
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- Rubbing velocity at the pin joint O
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= (ω1 – ω2) r, if the links move in the same direction
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= (ω1 + ω2) r, if the links move in the opposite direction
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Example 7.1. In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40
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- mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and
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- AD are of equal length. Find the angular velocity of link CD when angle BAD = 60°.
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Solution. Given : N BA = 120 r.p.m. or ωBA = 2 π × 120/60 = 12.568 rad/s
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Since the length of crank A B = 40 mm = 0.04 m, therefore velocity of B with respect to A or
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- velocity of B, (because A is a fixed point),
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vBA = v B = ωBA × A B = 12.568 × 0.04 = 0.503 m/s
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First of all, draw the space diagram to some suitable scale, as shown in Fig. 7.7 (a). Now the
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- velocity diagram, as shown in Fig. 7.7 (b), is drawn as discussed below :
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1. Since the link A D is fixed, therefore points a and d are taken as one point in the velocity
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- diagram. Draw vector ab perpendicular to B A, to some suitable scale, to represent the velocity of B
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- with respect to A or simply velocity of B (i.e. v BA or v B) such that
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vector ab = v BA = v B = 0.503 m/s
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2. Now from point b, draw vector bc perpendicular to CB to represent the velocity of C with
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- respect to B (i.e. v CB) and from point d, draw vector dc perpendicular to CD to represent the velocity
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- of C with respect to D or simply velocity of C (i.e. v CD or v C). The vectors bc and dc intersect at c.
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By measurement, we find that
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v CD = v C = vector dc = 0.385 m/s
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We know that CD = 80 mm = 0.08 m
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∴ Angular velocity of link CD,
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vCD 0.385
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ωCD = = = 4.8 rad/s (clockwise about D
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Example 7.2. The crank and connecting rod
of a theoretical steam engine are 0.5 m and 2 m long respectively. The crank makes 180 r.p.m. in the clockwise direction. When it has turned 45° from the inner dead centre position, determine : 1. velocity of piston, 2. angular velocity of connecting rod, 3. velocity of point E on the connecting rod 1.5 m from the gudgeon pin, 4. velocities of rubbing at the pins of the crank shaft, crank and crosshead when the diameters of their pins are 50 mm, 60 mm and 30 mm respectively, 5. position and linear velocity of any point G on the connecting rod which has the least velocity relative to crank shaft
* Solution. Given : NBO = 180 r.p.m. or ωBO = 2 π × 180/60 = 18.852 rad/s
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Since the crank length OB = 0.5 m, therefore linear velocity of B with respect to O or velocity - of B (because O is a fixed point),
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vBO = v B = ωBO × OB = 18.852 × 0.5 = 9.426 m/s -
. . . (Perpendicular to BO) -
- Velocity of piston
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First of all draw the space diagram, to some suitable scale, as shown in Fig. 7.8 (a). Now the - velocity diagram, as shown in Fig. 7.8 (b), is drawn as discussed below :
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1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of B - with respect to O or velocity of B such that
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vector ob = v BO = v B = 9.426 m/s -
2. From point b, draw vector bp perpendicular to BP to represent velocity of P with respect - to B (i.e. v PB) and from point o, draw vector op parallel to PO to represent velocity of P with respect
- to O (i.e. v PO or simply v P). The vectors bp and op intersect at point p.
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By measurement, we find that velocity of piston P, -
vP = vector op = 8.15 m/s Ans. -
- Angular velocity of connecting rod
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From the velocity diagram, we find that the velocity of P with respect to B, -
vPB = vector bp = 6.8 m/s -
Since the length of connecting rod PB is 2 m, therefore angular velocity of the connecting rod, -
vPB 6.8 -
ωPB = = = 3.4 rad/s (Anticlockwise) Ans. -
2 -
PB -
- Velocity of point E on the connecting rod
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The velocity of point E on the connecting rod 1.5 m from the gudgeon pin (i.e. PE = 1.5 m) - is determined by dividing the vector bp at e in the same ratio as E divides PB in Fig. 7.8 (a). This is
- done in the similar way as discussed in Art 7.6. Join oe. The vector oe represents the velocity of E. By
- measurement, we find that velocity of point E,
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vE = vector oe = 8.5 m/s Ans. - Note : The point e on the vector bp may also be obtained as follows :
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BE × bp -
BE be -
= or be = -
BP bp BP -
- Velocity of rubbing
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We know that diameter of crank-shaft pin at O, -
dO = 50 mm = 0.05 m -
Diameter of crank-pin at B, -
dB = 60 mm = 0.06 m - and diameter of cross-head pin,
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dC = 30 mm = 0.03 m -
We know that velocity of rubbing at the pin of crank-shaft -
dO 0.05 -
× ωBO = × 18.85 = 0.47 m/s Ans. -
= -
2 2 -
Velocity of rubbing at the pin of crank -
0.06 -
dB -
(ωBO + ωPB ) = (18.85 + 3.4) = 0.6675 m/s Ans. -
= -
2 2 -
...(3 ωBO is clockwise and ωPB is anticlockwise.) - and velocity of rubbing at the pin of cross-head
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0.03 -
dC -
× ωPB = × 3.4 = 0.051 m/s Ans. -
= -
2 2 -
...(3 At the cross-head, the slider does not rotate and only the connecting rod has angular motion.) -
- Position and linear velocity of point G on the connecting rod which has the least velocity
- relative to crank-shaft
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The position of point G on the connecting rod which has the least velocity relative to crank- - shaft is determined by drawing perpendicular from o to vector bp. Since the length of og will be the
- least, therefore the point g represents the required position of G on the connecting rod.
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By measurement, we find that -
vector bg = 5 m/s -
The position of point G on the connecting rod is obtained as follows: -
bg -
bg BG 5 -
or BG = × BP = × 2 = 1.47 m Ans. -
= -
bp -
bp BP 6.8 -
By measurement, we find that the linear velocity of point G, -
vG = vector og = 8 m/s Ans.