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ME-403 · THEORY OF MACHINES/Unsolved PYQ Paper

ME-403 Theory of Machines - Jun 2022 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

  1. 7 Marks
    aDefine Grashof's law. State how is it helpful in classifying the four-link mechanisms into different types.
    How to answer
    • State Grashof's law: s + l <= p + q for continuous relative rotation
    • Define shortest, longest and other links
    • Explain classification: crank-rocker, double-crank, double-rocker based on fixed link
    • Give condition for Grashof and non-Grashof chains
  2. 7 Marks
    bExplain the phenomena of 'slip' and 'creep' in a belt drive.
    How to answer
    • Define slip: relative motion due to insufficient friction
    • Define creep: differential elongation of tight and slack sides
    • Explain causes and effects on velocity ratio
    • Differentiate between slip and creep
  3. 14 Marks
    Simple quick return mechanism is shown in the figure. The forward to return ratio of the quick return mechanism is $2:1$. If the radius of the crank $O_1P$ is $125$ mm, then the distance 'd' (in mm) between the crank centre to lever pivot centre point should be.
    Crank centre O1 and lever pivot O2 separated vertically by distance d. Crank radius O1P with slider P on slotted lever O2Q. Lever O2Q pivoted at O2, slider at P. Vertical dashed line O1-O2.
    How to answer
    • Use quick-return ratio = cutting angle / return angle = 2:1
    • Relate crank radius O1P = 125 mm to centre distance d via cos of extreme positions
    • Calculate angle: forward 240 deg, return 120 deg
    • Compute d = O1P / cos or d = 2*O1P = 250 mm
  4. 5 Marks
    aAn instantaneous configuration of a four-bar mechanism, whose plane is horizontal, is shown in the figure below. At this instant, the angular velocity and angular acceleration of link $O_2A$ are $\omega = 8$ rad/s and $\alpha = 0$, respectively, and the driving torque $(\tau)$ is zero. The link $O_2A$ is balanced so that its centre of mass falls at $O_2$. a) Which kind of 4-bar mechanism is $O_2ABO_4$?
    Four-bar O2ABO4: O2-O4 horizontal fixed 240 mm. O2A vertical 60 mm with omega=8 rad/s, alpha=0. O4B vertical 160 mm. Coupler AB connects tops of cranks. Both cranks at 90 deg to fixed link.
    How to answer
    • Check Grashof condition with link lengths O2A=60mm, O2O4=240mm, O4B=160mm
    • Identify shortest, longest links and fixed link
    • Classify as crank-rocker / double-rocker etc.
    • Justify type O2ABO4
  5. 5 Marks
    bb) At the instant considered, what is the magnitude of the angular velocity of Q4B?
    How to answer
    • Use relative velocity method or instantaneous centre for four-bar
    • Velocity of A = omega_O2A * O2A
    • Equate component along AB or use rigid body rotation
    • Compute omega_Q4B = v_B / O4B
  6. 4 Marks
    cc) At the same instant, if the component of the force at joint A along AB is $30$ N, then the magnitude of the joint reaction at $O_2$.
    How to answer
    • Draw free-body of links with force at A along AB = 30 N
    • Apply equilibrium / dynamic force analysis with tau=0, alpha=0
    • Take moment about O4 or O2 to find reactions
    • Compute joint reaction magnitude at O2
  7. 7 Marks
    aDiscuss relative merits and demerits of belt and chain drive for transmission of power.
    How to answer
    • Define belt drive vs chain drive power transmission
    • List merits of belt: low cost, shock absorption, long distance
    • List demerits: slip, creep, low efficiency vs chain merits: positive drive, high efficiency
    • Conclude with applications
  8. 7 Marks
    bDescribe with a neat sketch a centrifugal clutch and deduce an equation for the total torque transmitted.
    How to answer
    • Sketch centrifugal clutch: drum, shoes, springs, spider
    • Explain working: centrifugal force engages shoes at speed
    • Derive normal force = m*omega^2*r, friction torque
    • Obtain total torque transmitted = n*mu*F*R
  9. 7 Marks
    aDescribe various inversions of a slider-crank mechanism giving examples, with neat sketch.
    How to answer
    • Define slider-crank and its four inversions
    • Give examples: reciprocating engine, Whitworth, crank-slotted, hand pump
    • Explain each inversion with fixed link change
    • Provide neat sketches for each
  10. 7 Marks
    bFor the position of the mechanism shown in Fig.3. Find the velocity of the slider B for the given configuration if the velocity of the slider A is $3$ m/s.
    Two sliders A and B on inclines 20 deg (left) and 30 deg (right) to horizontal, connected by link 450 units. Slider A velocity 3 m/s down the incline. Dimensions 200 (mm) shown.
    How to answer
    • Resolve velocities along and perpendicular to link AB
    • Apply relative velocity equation v_B = v_A + omega_AB x r_AB
    • Use incline angles 20 deg and 30 deg for sliders
    • Solve for magnitude of velocity of slider B
  11. 5 Marks
    aA gear set has an opinion with 20 teeth and a gear with 40 teeth. The pinion runs at 0 rev/s and transmits a power of 20 kW. The teeth are on the $20^{\circ}$ full-depth system and have module of 5 mm. The length of the line of action is 19 mm. a) Find the center distance for the above gear set in mm?
    How to answer
    • Compute pitch diameters: d = m*z, centre distance = (d1+d2)/2
    • Find base pitch and contact ratio = length of action / base pitch
    • Compute tangential force from power and speed, radial = Ft*tan(phi)
    • Resultant = Ft / cos(phi)
  12. 5 Marks
    bb) The contact ratio of the contacting tooth?
    How to answer
    • Use contact ratio = path of contact / base pitch
    • Compute base pitch = pi*m*cos(phi)
    • Substitute length of action 19 mm
    • Obtain ratio value
  13. 4 Marks
    cc) The resultant force on the contacting gear tooth in N?
    How to answer
    • Find torque / tangential effort from power
    • Resolve into tangential and radial components
    • Compute resultant = sqrt(Ft^2+Fr^2)
    • Express in N
  14. 14 Marks
    For the epicyclic gear arrangement shown in the figure, $\omega_2 = 100$ rad/s clockwise (CW) and $\omega_{arm} = 80$ rad/s counter clockwise (CCW). The angular velocity $\omega_5$, (in rad/s) is
    Epicyclic arrangement with shaft axis vertical on right. Arm connects gear 2 (bottom) to gears 3-4. Gear 5 is ring connected to shaft axis. Teeth: N2=20, N3=24, N4=32, N5=80.
    How to answer
    • Apply tabular method for epicyclic gear: arm, gears 2-5
    • Use teeth numbers N2=20, N3=24, N4=32, N5=80
    • Insert omega2=100 CW, omega_arm=80 CCW with sign convention
    • Solve for omega5
  15. 14 Marks
    A cam consists of a circular disc of diameter 75 mm with its centre displaced 25 mm from the camshaft axis. The follower has a flat surface (horizontal) in contact with the cam and the line of action of the follower is vertical and passes through the shaft axis as shown in Fig.5. The mass of the follower is 2.3 kg and is pressed downwards by a spring which has a stiffness of 3.5 N/mm. In the lowest position the spring force is 45 N.
    1. i)Derive an expression for the acceleration of the follower in terms of the angle of rotation from the beginning of the lift.
    2. ii)As the cam shaft speed is gradually increased, a value is reached at which the follower begins to lift from the cam surface. Determine the camshaft speed for this condition.
    Eccentric circular cam diameter 75 mm, centre offset 25 mm above shaft axis. Flat horizontal follower above cam with vertical line of action through shaft axis, loaded by spring.
    How to answer
    • Define eccentric circular cam: lift x = e(1-cos theta)
    • Differentiate to get velocity and acceleration vs rotation angle
    • Include follower mass 2.3 kg, spring stiffness 3.5 N/mm, preload 45 N
    • For jump condition equate inertia force to spring + weight; solve camshaft speed
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