Dynamics of Machines Solved MCQs

Multiple Choice Questions 128 Pages
University Anna University
Course Mechanical Engineering
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    ME8594 - DYNAMICS OF MACHINES
    Multiple choice questions with answers
    UNIT I FORCE ANALYSIS
    1) Which of the following is incorrect regarding inertia force?
    a) Imaginary force
    b) Acts upon a rigid body
    c) Brings the body to equilibrium
    d) Same direction as of accelerating force
    Answer: d
    Explanation: The inertia force is an imaginary force, which on acting upon a rigid body has a
    tendency to bring it in an equilibrium position. Numerically it is equal to the magnitude of
    accelerating force, but the direction of this force is opposite.
    2) D-Alembert’s principle is used for which of the following?
    a) Change static problem into a dynamic problem
    b) Change dynamic problem to static problem
    c) To calculate moment of inertia of rigid bodies
    d) To calculate angular momentum of a system of masses
    Answer: b
    Explanation: D-Alembert’s principle states that the resultant force acting on a body together
    with the reversed effective force (or inertia force), are in equilibrium. This principle is used to
    reduce a dynamic problem into an equivalent static problem.
    3) In the expression F – m.a = 0, the term – m.a is called _______
    a) Reversed effective force
    b) Net force
    c) Coriolis force
    d) Resultant force
    Answer: a
    Explanation: If the quantity m.a is treated as another force with same line of action as the net
    force, then the body could be assumed to be in static equilibrium. This force is known as
    Reversed effective force.
    4) Why the inertia torque acts in the opposite direction to the accelerating couple?
    a) Bring the body in equilibrium
    b) To reduce the accelerating torque
    c) Acts as a constraint torque
    d) Increase the linear acceleration
    Answer: a
    Explanation: The inertia torque is an imaginary torque, which when applied upon the rigid body,
    brings it in equilibrium position. It is equal to the accelerating couple in magnitude but opposite
    in direction.
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    5) A body remains in equilibrium if ________
    a) Inertia force is applied in the same direction to the resultant force
    b) Inertia force is applied in the direction opposite to the resultant force
    c) Inertia force is applied in the direction Perpendicular to the resultant force
    d) Inertia force is applied in the direction Parallel to the resultant force
    Answer: b
    Explanation: Inertia force is an imaginary force which tends to act in the direction opposite to
    the resultant force to bring the body in equilibrium. Numerically it is equal to the magnitude of
    accelerating force.
    6) Inertia force and the reversed effective force are the same.
    a) True
    b) False
    Answer: a
    Explanation: The net force m.a is taken as another force acting in the opposite direction to the
    applied resultant force and is known as inertia force or reversed effective force.
    7) Force which does not act on the connecting rod is ______
    a) Weight of connecting rod
    b) Inertia force of connecting rod
    c) Radial force
    d) Coriolis force
    Answer: d
    Explanation: Since there is no accelerated frame of reference having different velocities,
    therefore coriolis force does not act on the connecting rod.
    8) Inertia forces on the reciprocating parts acts along the line of stroke.
    a) True
    b) False
    Answer: a
    Explanation: The reciprocating parts of the engine experience an Inertia force of magnitude
    m.ω
    2
    .r(cosθ + cos2θ/n), this inertia force acts along the line of stroke.
    9) When mass of the reciprocating parts is neglected then the inertia force is _____
    a) Maximum
    b) Minimum
    c) 0
    d) Not defined
    Answer: c
    Explanation: Inertia force for neglected mass of reciprocating parts is 0 as it depends on the
    mass, since it has a value, it is defined.
    10) For a steam engine, the following data is given:
    Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m. calculate inertia force at θ=30 degrees from IDC.
    a) 19000 N
    b) 19064 N
    c) 19032 N
    d) 20064 N
    Answer: b
    Explanation: l
    1
    = l – G.C = 1.5 – 0.5 = 1m
    Fi = (Mr + G.C,Mc/l).ω
    2
    .r(cosθ + cos2θ/n)
    Mr=300 Kg
    Mc=250 kg
    ω=13.1 rad/s r=0.3m
    substituting these values will give Fi = 19064 N.
    11) Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m
    Find the equivalent length L of a simple pendulum swung about an axis.
    a) 1.35 m
    b) 1.42 m
    c) 1.48 m
    d) 1.50 m
    Answer: b
    Explanation: We know that equivalent length L is given by the expression
    L = (Kg
    2
    + l
    1
    2
    )/l
    1
    Kg = 0.65m l
    1
    = 1m
    therefore L = 1.42 m.
    12) From the data given:
    Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m
    Find the correcting couple in N-m?
    a) 52.7
    b) 49.5
    c) 59.5
    d)56.5
    Answer: c
    Explanation: The correction couple depends on equivalent length, l
    1
    mass of connecting rod and
    angular position and velocity
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    Tc = Mc.l
    1
    .(l-L).(ω
    2
    .sin2θ/2n
    2
    )
    substituting the values into the equation results in
    Tc = 59.5 N-m.
    13) Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m
    Find angular acceleration of connecting rod in rad/s
    2
    .
    a) 16.782
    b) 17.824
    c) 15.142
    d) 17.161
    Answer: b
    Explanation: α = -ω
    2
    .sinθ/n
    = 13.1
    2
    .sin3θ/5
    = 17.161 rad/s
    2
    .
    14) Torque due to weight of the connecting rod affects the torque due to connecting rod.
    a) True
    b) False
    Answer: b
    Explanation: The torque due to connecting rod remains same irrespective of the torque caused
    by the weight of the connecting rod.
    15) The net force acting on the crosshead pin is known as __________
    a) Crank pin effort
    b) Crank effort
    c) Piston effort
    d) Shaft effort
    Answer: c
    Explanation: Crank pin effort is the force acting along the direction perpendicular to the crank.
    Crank effort is the torque acting on the crank, piston effort is the force acting on the piston or
    the crosshead pin.
    16) Piston effort acts along the line of stroke.
    a) True
    b) False
    Answer: a
    Explanation: Piston effort is the force acting along the line of action of the stroke on the
    crosshead pin or the piston.
    17) In a horizontal engine, reciprocating parts are accelerated when the piston moves from _______
    a) TDC to BDC
    b) BDC to TDC
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    c) Midway to TDC
    d) BDC to midway
    Answer: a
    Explanation: The reciprocating parts of a horizontal engine are accelerated during the period
    when the piston moves from inner dead center to outer dead center.
    18) In a horizontal engine, reciprocating parts are retarded when the piston moves from _________
    a) TDC to BDC
    b) BDC to TDC
    c) Midway to TDC
    d) BDC to midway
    Answer: b
    Explanation: The reciprocating parts of a horizontal engine are accelerated during the latter half
    of the stroke, i.e when the piston moves from inner dead center to outer dead center and
    retards when the piston moves from outer dead center to inner dead center.
    19) When the piston is accelerated, the piston effort is given by which of the following the
    equation?
    a) F(L) – F(I)
    b) F(L) + F(I)
    c) F(L) ± F(I)
    d) F(L) – F(I) + R(f)
    Answer: b
    Explanation: During acceleration, piston effort is the sum total of the net load on piston and
    inertia forcer. The presence of resistance force also effects the expression.
    20) In the presence of frictional resistance, the expression for piston effort is _________
    a) F(L) – F(I)
    b) F(L) + F(I)
    c) F(L) ± F(I) – R(f)
    d) F(L) – F(I) + R(f)
    Answer: c
    Explanation: Piston effort depends on the net piston load and inertia force, in presence of
    frictional resistance and additional term of R(f) is subtracted from the overall expression.
    21) Crank effort is the product of crank pin radius and _______
    a) Thrust on sides
    b) Crankpin effort
    c) Force acting along connecting rod
    d) Piston effort
    Answer: b
    Explanation: Crank effort also known as turning moment or torque on the crank shaft is the
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    product of the crankpin effort a.k.a FT and the crank pin radius a.k.a r. It is applied at crank pin
    perpendicular to the crank.
    22) In a horizontal engine, the weight of the reciprocating parts also add/subtract to the piston
    effort.
    a) True
    b) False
    Answer: b
    Explanation: The weight of the reciprocating parts of the engine add or subtract to the piston
    effort only in a vertical engine and it depends on the motion of the piston.
    23) For the given data of an Internal combustion engine : Mass of parts = 180 kg bore = 175 mm,
    length of stroke = 200 mm, engine speed = 500 r.p.m., length of connecting rod = 400 mm and
    crank angle = 60° from T.D.C, find the inertia force.
    a) 17.56 N
    b) 19.2 N
    c) 18.53 N
    d) 18.00 N
    Answer: c
    Explanation: Ratio of length of connecting rod and crank n = l/r = 2*200/(200÷2) = 4,
    We know that inertia force is m.ω
    2
    .r(cosθ + cos2θ/ n)
    inserting values will give F = 18.53 N.
    24) The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts
    is 250 kg. When the crank has travelled 30° from T.D.C., the difference between the driving and
    the back pressures is 0.45 N/mm
    2
    . The connecting rod length between centres is 1.2 m and the
    cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is
    neglected, calculate the net load on piston.
    a) 88000 N
    b) 90560 N
    c) 78036 N
    d) 88357 N
    Answer: d
    Explanation: Net piston load is given by F(l) = (p
    1
    -p
    2
    )πD
    2
    ÷4
    p
    1
    -p
    2
    = 0.45 N/mm
    2
    D = 500 mm
    Therefore F(l) = 88357 N.
    25) From the data given:
    crank-pin circle radius = 300mm
    mass of the reciprocating parts = 250kg
    difference between the driving and the back pressures is 0.45 N/mm
    2
    The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m.
    engine runs at 250 r.p.m & 30° from T.D.C.
    Find the piston effort.
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    a) 32.4 kN
    b) 35.2 kN
    c) 37.3 kN
    d) 40.2 N
    Answer: c
    Explanation: Net piston load is given by F(l) = (p
    1
    -p
    2
    )πD
    2
    ÷4
    Fp = Fl – Fi
    Fi = m.ω
    2
    .r(cosθ + cos2θ/n)
    = 51020 N
    Therefore, Fp = 37.3 kN.
    26) Correction couple is applied when masses are placed arbitrarily and to maintain _________
    a) Static equilibrium
    b) Dynamic equilibrium
    c) Stable equilibrium
    d) Unstable equilibrium
    Answer: b
    Explanation: Difference between the torques required to accelerate the two-mass system and
    the torque required to accelerate the rigid body is called correction couple and this couple must
    be applied, when the masses are placed arbitrarily to make the system dynamical equivalent.
    27) Force which does not act on the connecting rod is ______
    a) Weight of connecting rod
    b) Inertia force of connecting rod
    c) Radial force
    d) Coriolis force
    Answer: d
    Explanation: Since there is no accelerated frame of reference having different velocities,
    therefore coriolis force does not act on the connecting rod.
    28) Inertia forces on the reciprocating parts acts along the line of stroke.
    a) True
    b) False
    Answer: a
    Explanation: The reciprocating parts of the engine experience an Inertia force of magnitude
    m.ω
    2
    .r(cosθ + cos2θ/n), this inertia force acts along the line of stroke.
    29) When mass of the reciprocating parts is neglected then the inertia force is _____
    a) Maximum
    b) Minimum
    c) 0
    d) Not defined
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    Answer: c
    Explanation: Inertia force for neglected mass of reciprocating parts is 0 as it depends on the
    mass, since it has a value, it is defined.
    30) For a steam engine, the following data is given:
    Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m
    calculate inertia force at θ=30 degrees from IDC.
    a) 19000 N
    b) 19064 N
    c) 19032 N
    d) 20064 N
    Answer: b
    Explanation: l
    1
    = l – G.C = 1.5 – 0.5 = 1m
    Fi = (Mr + G.C,Mc/l).ω
    2
    .r(cosθ + cos2θ/n)
    Mr=300 Kg
    Mc=250 kg
    ω=13.1 rad/s r=0.3m
    substituting these values will give Fi = 19064 N.
    31) Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m
    Find the equivalent length L of a simple pendulum swung about an axis.
    a) 1.35 m
    b) 1.42 m
    c) 1.48 m
    d) 1.50 m
    Answer: b
    Explanation: We know that equivalent length L is given by the expression
    L = (Kg
    2
    + l
    1
    2
    )/l
    1
    Kg = 0.65m l
    1
    = 1m
    therefore L = 1.42 m.
    32) From the data given:
    Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m
    Find the correcting couple in N-m?
    a) 52.7
    b) 49.5
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    c) 59.5
    d)56.5
    Answer: c
    Explanation: The correction couple depends on equivalent length, l
    1
    mass of connecting rod and
    angular position and velocity
    Tc = Mc.l
    1
    .(l-L).(ω
    2
    .sin2θ/2n
    2
    )
    substituting the values into the equation results in
    Tc = 59.5 N-m.
    33) Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of
    reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m;
    centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an
    axis through the centre of gravity = 0.65 m
    Find angular acceleration of connecting rod in rad/s
    2
    .
    a) 16.782
    b) 17.824
    c) 15.142
    d) 17.161
    Answer: b
    Explanation: α = -ω
    2
    .sinθ/n
    = 13.1
    2
    .sin3θ/5
    = 17.161 rad/s
    2
    .
    34) Torque due to weight of the connecting rod affects the torque due to connecting rod.
    a) True
    b) False
    Answer: b
    Explanation: The torque due to connecting rod remains same irrespective of the torque caused
    by the weight of the connecting rod.
    35) Turning moment is maximum when the crank angle is 90 degrees.
    a) True
    b) False
    Answer: a
    Explanation: Referring the turning moment diagram for a single cylinder double acting steam
    engine, the minimum values occur at 0, 180 and 360 degrees and maximum values occur at 90
    and 270 degrees.
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  • DHIRAJLAL GANDHI COLLEGE OF TECHNOLOGY
    DEPARTMENT OF MECHANICAL ENGINEERING
    36) In the figure given below, the quantity represented by the arrow is known as ___________
    a) Maximum Torque
    b) Minimum Torque
    c) Maximum Force
    d) Mean resisting torque
    Answer: d
    Explanation: In a turning moment diagram given above, the line pointed out by the arrow is the
    value of mean torque and hence it is called mean resisting torque.
    37) When engine torque is more than mean resisting torque, then the flywheel _______
    a) Has uniform velocity
    b) Has 0 velocity
    c) Has acceleration
    d) Has retardation
    Answer: c
    Explanation: When the difference between T-Tmean is positive the work is done by the steam
    and the crankshaft accelerated which results in acceleration of flywheel.
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