Theory of Machines MODEL ANSWER 1st Mech.

Power Engineering
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Model Answer for Mid Term Examination:

Academic year 2014
Answer All The Following Questions:

Q1:-

Fig. 1 shows diagrammatically a compound epicyclic gear. Wheels A, D and E are

free to rotate independently on spindle O, wheel B and C are compound and rotate
together on spindle P, on the end of arm OP. all the wheels have teeth of the same
pitch. A has 12 teeth, B has 30 and C has 14 teeth, the wheels B and C which are cut
externally. Find the number of teeth on wheels D and E which are cut internally. If
the wheel A is driven clockwise at 100 r.p.m, while D is driven counter-clockwise at
500 r.p.m. determine the magnitude and direction of the velocity of the arm OP, and
wheel E. E
P
Solution B
D C
Given : TA = 12 ; TB = 30 ;TC = 14 ; NA = 1 r.p.s. ;
O
ND = 5 r.p.s. A

Number of teeth on wheels D and E

Let TD and TE be the number of teeth on wheels D and E
respectively. Fig. 1
Let DA , DB , DC , DD and DE be the pitch circle diameters of wheels A, B, C, D and E
respectively. From the geometry of the figure,
DE = DA + 2DB and DD= DE – (DB – DC)
Since the number of teeth are proportional to their pitch circle diameters for the same
module, therefore
TE = TA + 2TB = 12 + 2 × 30 = 72 Ans.
and TD = TE – (TB – TC) = 72 – (30 – 14) = 56 Ans.
Magnitude and direction of angular velocities of arm OP and wheel E

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Dr. Mamdouh El-Elamy
Theory of Machines MODEL ANSWER 1st Mech. Power Engineering
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The table of motions is drawn as follows:

Conditions of motion NArm NB,C NA ND NE

1. Add + y revolutions to all elements

y y y
y y
2. Arm fixed-wheel A rotated
through + x revolutions 0 x

Total motion y x+y

Since the wheel A makes 100 r.p.m. clockwise, therefore from the table,

(1)

Also, the wheel D makes 500 r.p.m. counter clockwise, therefore

(2)

From eqns. (1) & (2)

x = 218.18 r.p.m and y = 445.45 r.p.m
 Velocity of arm OP = y = 445.45 r.p.m (counter clockwise) Ans.

Q2:-
The turning moment diagram for a four stroke gas engine may be assumed for
simplicity to be represented by four triangles, the areas of which from the line of zero
pressure are as follows: Suction stroke = 0.45 × 10–3 m2; Compression stroke = 1.7 ×
10–3 m2; Expansion stroke = 6.8 × 10–3 m2; Exhaust stroke = 0.65 × 10–3 m2. Each m2
of area represents 3 MN-m of energy. Assuming the resisting torque to be uniform,
find the mass of the rim of a flywheel required to keep the speed between 202 and
198 r.p.m. The mean radius of the rim is 1.2 m.

Solution
Given: a1 = 0.45 × 10–3 m2 ; a2 = 1.7 × 10–3 m2 ; a3 = 6.8 × 10–3 m2; a4 = 0.65 × 10–3
m2; N1 = 202 r.p.m; N2 = 198 r.p.m.; k = 1.2 m

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Dr. Mamdouh El-Elamy
Theory of Machines MODEL ANSWER 1st Mech. Power Engineering
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The turning moment crank angle diagram for a four stroke engine is shown in Fig. 2.
The areas below the zero line of pressure are taken as negative while the areas above
the zero line of pressure are taken as positive.
Net area = a3 – (a1 + a2 + a4) =
= 6.8 × 10–3 – (0.45 × 10–3 + 1.7 × 10–3 + 0.65 × 10–3) = 4 × 10–3 m2
Since the energy scale is 1 m2 = 3 MN-m = 3 × 106 N-m, therefore,
Net work done per cycle = 4 × 10–3 × 3 ×106 = 12 × 103 N-m (i)
We also know that work done per cycle,
W.D. = Tmean × 4π (ii)
From equations (i) and (ii),
Tmean = 12 × 103/4π = 955 N-m
Work done during expansion stroke
EXP = a3 × Energy scale = 6.8 × 10–3 × 3 × 106 = 20.4 × 103 N-m (iii)
Also, work done during expansion stroke

We know that the maximum fluctuation of energy ΔE equal to the area between
maximum torque and mean torque.
( ) ( )

N = Mean speed of the flywheel

The total fluctuation of speed ks

Mass of the rim of a flywheel

Ans.

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Dr. Mamdouh El-Elamy
Theory of Machines MODEL ANSWER 1st Mech. Power Engineering
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Turning moment

Tmax

ΔE
Tmean Line of zero
pressure

0 θ

π π π π
Suction Compression Expansion Exhaust

Crank angle
Fig. 2

Q3:-
Draw the cam profile with a base radius rb = 2.5 cm to give the shown motion, in Fig.
3 shown, to the radial roller follower with a radius 1 cm. if the cam rotates uniformly
speed at 500 r.p.m. clocks wisely, draw the displacement, velocity and acceleration-
time diagrams for the follower during the rise and return stroke.

y
Parabolic rise S. H. M. return
30 mm

Fig. 3

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Dr. Mamdouh El-Elamy
‫‪Theory of Machines‬‬ ‫‪MODEL ANSWER‬‬ ‫‪1st Mech. Power Engineering‬‬
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‫‪y‬‬
‫‪Parabolic rise‬‬ ‫‪S. H. M. return‬‬
‫‪30 mm‬‬ ‫‪G‬‬ ‫‪H‬‬
‫‪F‬‬ ‫‪I‬‬
‫‪E‬‬
‫‪J‬‬
‫‪D‬‬ ‫‪K‬‬

‫‪C‬‬ ‫‪L‬‬
‫‪D‬‬ ‫‪C‬‬ ‫‪B‬‬ ‫‪M‬‬
‫‪E‬‬ ‫‪B A‬‬ ‫‪A‬‬ ‫‪N‬‬
‫‪θ‬‬
‫‪F‬‬

‫‪G‬‬ ‫‪N‬‬

‫‪Velocity‬‬
‫‪M‬‬
‫‪L‬‬

‫‪K‬‬

‫‪J‬‬

‫‪I‬‬
‫‪Acceleration‬‬

‫‪H‬‬

‫‪Fig. 4‬‬

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‫‪Dr. Mamdouh El-Elamy‬‬