KTU B.Tech S5 Lecture Notes Mechanics 0f Machinery
ME 301 Mechanics of Machinery
Introduction to kinematics and mechanisms – various mechanisms, kinematic diagrams, degree of freedom- Grashof’s criterion, inversions, coupler curves
straight line mechanisms exact, approximate – Ackerman Steering Mechanism – Hooke’s joint – Geneva mechanism – mechanical advantage, transmission angle
Displacement, velocity and acceleration analysis – relative motion – relative velocity – instant centre -Kennedy’s theorem
Scotch Yoke Mechanis
A scotch yoke mechanism converts rotational motion to linear sliding motion, or vice Versa. A pin on a rotating link is engaged in the slot of a sliding yoke. With regards to the input and output motion, the scotch yoke is similar to a slider-crank, but the linear sliding motion is pure sinusoidal. In comparison to the slider-crank, the scotch yoke has the advantage of smaller size and fewer moving parts, but can experience rapid wear in the slot.
A Geneva mechanism converts the motion of a continuous rotating crank to a intermittent motion of the Geneva wheel. For a four-slot wheel, the Geneva wheel rotates 90 degrees for a complete rotation of a crank.
As the crank rotates continuously, the output slider moves intermittently in the linear Geneva mechanism
This saw (in yellow) has reciprocating motion inside the arm (in green), driven by a rotating disk (in orange) with a cur (in grey) . This is a crank slider mechanism. The arm pivots at the center of the disk to feed the saw downward to cut the metal bar (in purple).
Oil field pump
As the motor turns the crank (in orange), the walking beam (in yellow) oscillates. The pumping (sucker) rod, which is immersed in the oil, is connected to the horse -head of the walking beam by a cable. Therefore, the oscillation of the walking beam is converted to the reciprocating motion of the pumping rod to pump oil.
Straight line motion mechanisms
Straight line motion mechanisms are mechanisms, having a point that moves along a straight line, or nearly along a straight line, without being guided by a plane surface.
Condition for exact straight line motion:
If point B (fig.1.40) moves on the circumference of a circle with center O and radius OA, then, point C, which is an extension of AB traces a straight line perpendicular to AO, provided product of AB and AC is constant.
Fig.1.40: Condition for exact straight line motion
Locus of pt.C will be a straight line, ┴ to AE if, is constant
Peaucellier exact straight line motion mechanism:
Fig.1.41: Peaucellier exact straight line motion mechanism
Here, AE is the input link and point E moves along a circular path of radius AE = AB. Also, EC = ED = PC = PD and BC = BD. Point P of the mechanism moves along exact straight line, perpendicular to BA extended.
To prove B, E and P lie on same straight line:
Triangles BCD, ECD and PCD are all isosceles triangles having common base CD and apex points being B, E and P. Therefore points B, E and P always lie on the perpendicular bisector of CD. Hence these three points always lie on the same straight line.
To prove product of BE and BP is constant.
In triangles BFC and PFC,
But since BC and PC are constants, product of BP and BE is constant, which is the condition for exact straight line motion. Thus point P always moves along a straight line perpendicular to BA as shown in the fig.1.41.
Approximate straight line motion mechanism: A few four bar mechanisms with certain modifications provide approximate straight line motions.
Fig.1.42: Robert’s mechanism
This is a four bar mechanism, where, PCD is a single integral link. Also, dimensions AC, BD, CP and PD are all equal. Point P of the mechanism moves very nearly along line AB.
Hooke’s joint (Universal joints)
Hooke’s joins is used to connect two nonparallel but intersecting shafts. In its basic shape, it has two U –shaped yokes ‘a’ and ‘b’ and a center block or cross-shaped piece, C. (fig.1.47(a))
The universal joint can transmit power between two shafts intersecting at around 300 angles (α). However, the angular velocity ratio is not uniform during the cycle of operation. The amount of fluctuation depends on the angle (α) between the two shafts. For uniform transmission of motion, a pair of universal joints should be used (fig.1.47(b)). Intermediate shaft 3 connects input shaft 1 and output shaft 2 with two universal joints. The angle α between 1 and 2 is equal to angle α between 2 and 3. When shaft 1 has uniform rotation, shaft 3 varies in speed; however, this variation is compensated by the universal joint between shafts 2 and 3. One of the important applications of universal joint is in automobiles, where it is used to transmit power from engine to the wheel axle.
Fig.1.47(b): Hooke’s joint
Steering gear mechanism
The steering mechanism is used in automobiles for changing the directions of the wheel axles with reference to the chassis, so as to move the automobile in the desired path. Usually, the two back wheels will have a common axis, which is fixed in direction with reference to the chassis and the steering is done by means of front wheels.
In automobiles, the front wheels are placed over the front axles (stub axles), which are pivoted at the points A & B as shown in the fig.1.48. When the vehicle takes a turn, the front wheels, along with the stub axles turn about the pivoted points. The back axle and the back wheels remain straight.
Always there should be absolute rolling contact between the wheels and the road surface. Any sliding motion will cause wear of tyres. When a vehicle is taking turn, absolute rolling motion of the wheels on the road surface is possible, only if all the wheels describe concentric circles. Therefore, the two front wheels must turn about the same instantaneous centre I which lies on the axis of the back wheel.
Condition for perfect steering
The condition for perfect steering is that all the four wheels must turn about the same instantaneous centre. While negotiating a curve, the inner wheel makes a larger turning angle θ than the angle φ subtended by the axis of the outer wheel.
In the fig.1.48, a = wheel track, L = wheel base, w = distance between the pivots of front axles.
Fig.1.48: Condition for perfect steering
From cotθ = and
from cotφ =
. This is the fundamental equation for correct steering. If this condition is satisfied, there will be no skidding of the wheels when the vehicle takes a turn.
Ackermann steering gear mechanism
Fig.1.49: Ackermann steering gear mechanism
fig.1.50: Ackermann steering gear mechanism
Ackerman steering mechanism, RSAB is a four bar chain as shown in fig.1.50. Links RA and SB which are equal in length are integral with the stub axles. These links are connected with each other through track rod AB. When the vehicle is in straight ahead position, links RA and SB make equal angles α with the center line of the vehicle. The dotted lines in fig.1.50 indicate the position of the mechanism when the vehicle is turning left.
Let AB=l, RA=SB=r; and in the turned position, . IE, the stub axles of inner and outer wheels turn by θ and φ angles respectively.
Neglecting the obliquity of the track rod in the turned position, the movements of A and B in the horizontal direction may be taken to be same (x).
Angle α can be determined using the above equation. The values of θ and φ to be taken in this equation are those found for correct steering using the equation . 
This mechanism gives correct steering in only three positions. One, when θ = 0 and other two each corresponding to the turn to right or left (at a fixed turning angle, as determined by equation ).
The correct values of φ, [φc] corresponding to different values of θ, for correct steering can be determined using equation . For the given dimensions of the mechanism, actual values of φ, [φa] can be obtained for different values of θ. T he difference between φc and φawill be very small for small angles of θ, but the difference will be substantial, for larger values of θ. Such a difference will reduce the life of tyres because of greater wear on account of slipping.
But for larger values of θ, the automobile must take a sharp turn; hence is will be moving at a slow speed. At low speeds, wear of the tyres is less. Therefore, the greater difference between φc and φa larger values of θ ill not matter.
As this mechanism employs only turning pairs, friction and wear in the mechanism will be less. Hence its maintenance will be easier and is commonly employed in automobiles.
Geneva Mechanism: its history, function, and weaknesses
The Geneva mechanism is a timing device. According to Vector Mechanics for Engineers for Ferdinand P. Beer and E. Russell Johnston Jr., says, “ [It] is used in many counting instruments and in other applications where an intermittent rotary motion is required.” (945) Essentially, the Geneva mechanism consists of a rotating disk with a pin and another rotating disk with slots (usually four) into which the pin slides (see right).
According to Brittanica.com, the Geneva mechanism was originally invented by a watch maker. The watch maker only put a limited number of slots in one of the rotating disks so that the system could only go through so many rotations. This prevented the spring on the watch from being wound too tight, thus giving the mechanism its other name, the Geneva Stop. The Geneva Stop was incorporated into many of the first film projectors used in theaters.
In Optimum Design of Mechanical Elements, Ray C. Johnson makes many references to the use of the Geneva mechanism to provide an intermittent motion the conveyor belt of a “film recording marching.” (13) He also discusses several weak points in the Geneva mechanism. For instance, for each rotation of the Geneva (slotted) gear the drive shaft must make one complete rotation. Thus for very high speeds, the drive shaft may start to vibrate. Another problem is wear, which is centralized at the drive pin. Finally, the designer has no control over the acceleration the Geneva mechanism will produce.
Also, the Geneva mechanism will always go through a small backlash, which stops the slotted gear. This backlash prevents controlled exact motion. (Picture at left from Optimum Design of Mechanical Elements.)
Below are models of the Geneva mechanism made with Working Model 2d v4.0. The second model shows velocity vectors for the slotted gear and the drive shaft. Velocity is the black arrow and acceleration is the green arrow. Move the mouse over the mechanisms to start them running.
As a mechanism moves over a range of motion its geometry changes. If we are using a mechanisms to transmit torque, or force then we must consider the ratio between the input and output force in various positions.
Transmission angle is the angle between the coupling member and the output member in a mechanism. As this angle approaches ±90°, the mechanical advantage of the mechanism typically increases.
Toggle positions occur when the input crank has near infinite mechanical advantage. Note: this also applies that the follower has no mechanical advantage on the crank.
Consider the example below, [prob. 1-3 from Shigley & Uicker],
Displacement, Velocity and Acceleration Analysis of Plane Mechanisms
Velocity Analysis in Mechanism
- Let a rigid link OA, of length r rotate about a fixed point O with a uniform angular velocity rad/s in a counter-clockwise direction OA turns through a small angle δθ in a small interval of time δt. Then, A will travel along the arcAA’ as shown in figure.
∴ Velocity of A relative to O
∴ In the limits, when
Thus, velocity of A is ωr and is perpendicular to OA.
Velocity of Intermediate Point
- If represent the velocity of B with respect to O, then
i.e., b divides the velocity vector in the same ratio as B divides the link. The magnitude of the linear velocity of a point on the rotating body at a particular instant is proportional to its distance fromt the axis of rotation.
Velocity Images of Four Link Mechanism
- Figure shows a four link mechanism (quadric cycle mechanism) ABCD in which AD is fixed link and BC is the coupler. AB is the driver rotating at an angular speed of ω rad/s in the clockwise direction if it is a crank or moving at angular velocity ω at this instant if it is rocker.
Velocity Images of Slider-Crank Mechanism
- Consider a slider-crank mechanism in which OA is the crank moving with uniform angular velocity ω rad/s in the clockwise direction. At point B, a slider moves on the fixed guide G.
From the given configuration, the coupler AB has angular velocity in the counter-clockwise direction. The magnitude being.
Velocity of Rubbing
- Let us take two links of a turning pair, a pin is fixed to one of the links whereas a hole is provided in the other to fit the pin. When joined the surface of the hole of one link will rub on the surface of pin of the other link. The velocity of rubbing of the two surfaces will depend upon the angular velocity of a link relative to the other.
Pin at A
- The pin at A joins links AD and AB. AD being fixed, the velocity of rubbing will depend upon the angular velocity of AB only.
- Velocity of rubbing = raω
where, ra = radius of pin at A
Pin at B
ωba = ωab = ω (clockwise)
- rb = Radius of pin at B
Velocity of rubbing = rb(ωab + ωbc)
Pin at C
ωbc = ωcb (counter-clockwise)
ωdc = ωcd (clockwise)
rc = Radius of pin at C
Velocity of rubbing = rc(ωbc + ωdc)
Pin at D
where, rd = radius of pin at D
Velocity of rubbing = rdωcd
Instantaneous Centre of Velocity (I-centre)
- The instantaneous centre of velocity can be defined as a point which has no velocity with respect to the fixed link.
- Suppose there are two link 1 and link 2
- Link 1 may not be fixed. Rigid body 2 is shown to be in plane motion with respect to the link 1.
- In case of fixed link, (link 2) velocity of the point A and B are proportional toPA and PB respectively. So, instantaneously, the rigid body can be thought of as being momentarily in pure rotation about the point P. The velocity of any point C on the body at this instant is given by in a direction perpendicular to PC. This point P is called the instantaneously centre of velocity and its instantaneously velocity is zero.
- If both links 1 and 2 are in motion, we can define a relative instantaneous centre P12 to be a point on 2 having zero relative velocity with respect to a coincident point on 1. Consequently, the relative motion of 2 with respect to 1 be appears to be pure rotation about P12. So P21 and P12 are identical.
- Instantaneous centre is also called centro. So, two coincident points belonging to two bodies having relative motion with the properties.
- They have the same velocities.
- They form a point in one of the rigid bodies about which the other rotates and vice-versa. Which is perhaps true for only an instant.
Primary Centro One which can be easily located by a mere observation of the mechanism.
Secondary Centro Centros that cannot be easily located.
Instantaneous Centre of Acceleration
- It is defined as a point on a link having zero relative acceleration with respect to a coincident point on the other link and is different from the instantaneous centre of velocity.
Aronhold-Kennedy Theorem of Three Centre
- It state that if three bodies are in relative motion with respect to one another, the three relative instantaneous centers of velocity ar collinear.
P12 – Instantaneous centre of fixed ground 1 and body 2.
P13 – Instantaneous centre of fixed ground 1 and body 3.
P23 – Instantaneous centre of body 2 and body 3.
Number of Centros in a Mechanism
- For a mechanism of n links, the number of centros (Instantaneous centre) Nis
Number of Lines of Centros
- The number of lines of centros L for a mechanism with n links is
Acceleration Analysis in Mechanism
- The rate of change of velocity with respect to time is known as acceleration and acts in the direction of the change in velocity. Velocity can changed by only changing its magnitude or its direction. Let a link OA, of length r, rotate in a circular path in the clockwise direction as shown in figure. It has an instantaneously angular velocity ω and an angular acceleration α in the same direction i.e., the angular velocity increases in the clockwise direction.
Tangential acceleration of A relative O is defined as
Centripetal or radial acceleration of A relative to O is defined as
Total acceleration (net acceleration)
There are three cases occurred in the net acceleration as given below
Case I When α = 0 ⇒ ω = constant
So, net acceleration
Case II When ω = 0 ⇒ A has linear motion as
Case III: When α is negative or the link OA decelerates, tangential acceleration will be negative or its direction will be as shown in figure.
Corial’s Acceleration Component
Consider a link AR rotates about a fixed point A on it. P is a point on a slider on the link.
Here, ω = Angular velocity of the link
α = Angular acceleration of the link
v = Linear velocity of the slider on the link
f =Linear acceleration of the slider on the link
r = Radial distance of point P on the slider.
Direction of coriol’s acceleration component (2ω.v) is perpendicular to AR Coriol’s component is positive if
- The link AR rotate clockwise and the slider moves radially outward.
- The link rotate counter clockwise and the slider moves radially inwards. Acceleration of slider (f) is positive if
- Slider has a deceleration while moving in the inward direction.
- Slider has acceleration while moving in the outward direction.
- Acceleration of P | | to AR= acceleration of slider – centripetal acceleration
- Acceleration of P ⊥ to AR.= Coriol’s acceleration + tangential acceleration
- Let Q be a point on the link AR immediately beneath the point P at the instant, then
- Acceleration of P = acceleration of P | | to AR + acceleration of P ⊥ to AR
= acceleration of P relative to Q + Acceleration of Q relative to A + Coriols acceleration component
|Instant centre -Kennedy’s theorem|
The transformation of motion from one shaft to another shaft requires three bodies: a frame (the position of each shaft is fixed in the frame) and two gears. Consider a general case when two disks with arbitrary profiles shown in figure 1, represent gears 2 and 3. Also assume that disk 2 rotates with the constant angular velocity ω2. The motion is transferred through the direct contact at point P (note that P2 and P3 are the same point P, but the first point is associated with disk 2 whereas the second point is associated with disk 3). We need to find out whether or not the angular velocity ω3 will remain constant and if not, what is required to make it remain constant. The answer is given by Kennedy’s theorem.
Kennedy’s theorem identifies the fundamental property of three rigid bodies in motion.
First, recall about that the instantaneous center of velocity, it is defined as the instantaneous location of a pair of coincident points of two different rigid bodies for which the absolute velocities of two points are equal. If one considers body 2 and the frame (represented by point O2) in figure 1, then the instantaneous center of these two bodies is point O2, which belongs to the frame and to disk 2. The absolute velocities of both bodies at point O2 are zero. The same is valid for disk 3 and the frame represented by point O3. For the three bodies in motion there are three instantaneous centers: for all combinations of pairs. Thus, there is an instantaneous center between the two disks.
Figure 1: Kennedy’s theorem illustration
Since point P is a common point for two disks in figure 1, for each disk the velocity component at this point directed along the common normal is the same and equal to Vp. One can move this velocity vector along the common normal until it intersects the line connecting the two instantaneous centers O2 and O3 at point C. According to Kennedy’s theorem, this point (P) is the instantaneous center of velocity for the two disks. Indeed, the velocity Vp, the only instantaneous common velocity for the two disks (disk 2, disk 3) will be equal to
Triangles AO2C and BO3C are similar, it follows that
From the above relationship, the ratio of angular velocities is equal to
Where lengths l2 and l3 denote O2C and O3C, respectively (see the above figure 1).
Velocity Vc in figure 1 is a common velocity for the two disks since
Figure 2: Illustration of the involute profile generation
Velocity Vc is also the absolute velocity of each body at this point because there cannot be another velocity component along the line O2O3 (since the distance between the frame points O2 and O3 is not changing and bodies 2 and 3 are assumed to be rigid).
Note that the relative velocity of two disks at point C is zero, whereas relative velocity at point P is not zero.
The above equation 3 gives the transformation of angular velocities from disk 2 to disk 3. It can be seen from the above relation, in order to remain kinematic ratio (ω2/ω3) constant, legthss l2 and l3 must not change. It is clear from figure 1, however, that for arbitrary profile shapes the common normal changes its direction during the motion, and thus point C moves along line O2O3. Thus, the problem of meeting the constant ratio requirement is to find such disk profiles that the kinematic ratio remains constant. It will be shown in the following that if the disk profile is described by the involute function, the common normal does not change its direction.