REVISION TEST 4 (Page 142)
This assignment covers the material contained in chapters 9 to 11. The marks available are
shown at the end of each question.
Problem 1. A beam, simply supported at its ends, is of length 1.4 m. If the beam carries a centrallyplaced downward concentrated load of 50 kN, determine the minimum permissible diameter of the
beam’s cross-section, given that the maximum permissible stress is 40 MPa, and the beam has a
solid circular cross-section.
Marks
Since each support carries
50
1.4
= 25 kN, bending moment, M = 25 kN 
m
2
2
1
= 17.5 kN m
Since

y

M
I
from which,
and
then
40  106 17.5 103

d
 d4
2
64
2
d 4 17.5 103  64

 4.4563 103 m3  4.4563 103 109 mm3
d  2  40 106
d=
3
4.4563 106 = 164.6 mm
3
i.e. the minimum diameter of the beam is 164.6 mm
Total:
6
Problem 2. Determine the force applied tangentially to a bar of a screw-jack at a radius of 60 cm, if
the torque required is 750 N m.
Marks
Torque, T = force  radius,
from which, force =
torque
750 N m

= 1250 N
radius 60 10 2 m
3
Total:
3
28
© John Bird & Carl Ross Published by Taylor and Francis
Problem 3. Calculate the torque developed by a motor whose spindle is rotating at 900 rev/min and
developing a power of 4.20 kW.
Marks
Power P = 2nT, from which, torque, T =
P
Nm
2n
where power, P = 4.20 kW = 4200 W and speed, n = 900/60 rev/s
Thus, torque, T =
P
=
2n
2
4200  60
4200
=
= 44.56 N m
2 900
 900 
2 

 60 
3
5
Total:
Problem 4. A motor connected to a shaft develops a torque of 8 kN m. Determine the number of
revolutions made by the shaft if the work done is 7.2 MJ.
Marks
work done
torque
1
Work done = 7.2 MJ = 7.2  10 6 J and torque = 8 kN m = 8000 N m
1
Work done = T, from which, angular displacement,  =
Hence, angular displacement,  =
7.2 106
= 900 rad
8000
2
2 rad = 1 rev, hence,
the number of revolutions made by the shaft =
900
= 143.2 revs
2
2
Total:
6
Problem 5. Determine the angular acceleration of a shaft which has a moment of inertia of
32 kg m 2 produced by an accelerating torque of 600 N m.
29
© John Bird & Carl Ross Published by Taylor and Francis
Marks
Torque, T = I, from which, angular acceleration,  =
T
, where torque,
I
1
T = 600 N m and moment of inertia I = 32 kg m 2
Hence, angular acceleration,  =
600
= 18.75 rad/s 2
32
4
Total:
5
Problem 6. An electric motor has an efficiency of 72% when running at 1400 rev/min. Determine
the output torque when the power input is 2.50 kW.
Marks
Efficiency =
power output
power output
100% hence 72 =
100
2500
power input
from which, power output =
72
 2500 = 1800 W
100
2
Power output, P = 2nT, from which
torque, T =
Hence,
P
where n = (1400/60) rev/s
2n
output torque =
1800
= 12.28 N m
 1400 
2 

 60 
3
Total:
5
Problem 7. A solid circular section shaft is required to transmit 60 hp at 1000 rpm. If the
maximum permissible shear stress in the shaft is 35 MPa, determine the minimum permissible
diameter of the shaft. Determine the resulting angle of twist of the shaft per metre, assuming that
the modulus of rigidity G = 70 GPa and 1 hp = 745.7 W.
30
© John Bird & Carl Ross Published by Taylor and Francis
Marks
Power =
i.e.
2NT
W
 60 hp  745.7
60
hp
21000  T
 60  745.7
60
from which, torque, T =
3
 T

r J
Now
i.e.
60  745.7  60
= 427.25 N m
2  1000
35  106 427.25

d
 d4
2
32
from which,
d3 
and
d=
427.25  32
 6.2170 105 m3  6.2170 105 109 mm3
6
 35 10  2
3
62170  39.6 mm
Hence, the maximum permissible diameter of the shaft is 39.6 mm
Now
 G

r
L
from which,
N
1m
2
L
m


= 0.02525 rad
G r 70 109 N  39.6 103 m
m2
2
3
35 106
= 0.02525 rad 
360
2 rad
i.e. the resulting angle of twist per metre,  = 1.45
4
Total:
10
TOTAL MARKS FOR REVISION TEST 4: 40
31
© John Bird & Carl Ross Published by Taylor and Francis