MACHINE DESIGN SOLVED PROBLEMS
1. Determine the estimated weigh to fan A-36 steel plates size ½x4x8.
a. 280 kgs
b. 332 kgs SOLUTION:
c. 301 kgs
From faires p 574: density of
d. 297 kgs
steel = 0.284 lb/in3
Weight = ½ (4x12)(8x12)(0.284)
= 654.336 lbs.
= 297 kg
2. Determine the estimated weigh of A-36 steel plates size 3/16x6’x20’.
a. 919 lbs
b. 1012 lbs
SOLUTION:
c. 829 lbs
From faires p 574: density of steel = 0.284 lb/in3
d. 735 lbs
Weight = (3/16) (6x12)(20x12)(0.284) = 920 lbs.
3. The minimum clearance allowed ro meashing spur gears with a circular pitch of .1571 and
diameter pitch of 20. The spur gear have 25 teeth.
a. .007855
b. .007558SOLUTION:
c. .008578
Clearance = 0.1571 / 20 = 0.007855
d. .007585
4. A cylindrical tank with 10” inside diameter contains oxygen gas at 2500 psi. Calculate the
required wall thickness in (mm) under stress of 28,000 psi .
a. 11.44mm
b. 11.34mm
SOLUTION:
c. 10.6 mm
st = PD / 2t
d. 10.3 mm
28,000 = 2500(10) / 2t
t = 0.4464 in. = 11.34 mm
5. A 3”0 diameter short shaft carrying 2 pulleys close to the bearings transmit how much
horsepower if the shaft makes 280 rpm.
a. 199 Hp
b. 198 Hp
SOLUTION:
c. 200 Hp
From PSME Code p 18:
d. 210 Hp
P = D3N / 38 = 33(280) / 38 = 199 Hp
6. Find the horsepower capacity of a standard A60 size V-Belt for a drive in which the pitch
diameter of a small sheave is 4” and that by the large sheave is 10 inches. The small sheaves
is connected to a drive motor of 3 Hp x 1750 rpm.
a. 1.8326
b. 1.90
c.1.56
d.1.48
SOLUTION:
Reference form PSME code pp 19-23
EP Rating: = X0.91 – (YS/de) – ZS3
Where from Table 3.6:
X = 1.945
S = 3.14 D N / 1000 = (3.14)(4/12) (1750) / 1000
Y = 3.801
S = 1.8326
Z = 0.0136
7. Determine the estimated weigh of A-36 steel plates size 3/16x6’x20’.
a. 920 lbs
b. 1012 lbs
SOLUTION:
c. 829 lbs
From faires p 574: density of steel = 0.284 lb/in3
d. 735 lbs
Weight = (3/16) (6x12)(20x12)(0.284) = 920 lbs.
8. The minimum clearance allowed ro meashing spur gears with a circular pitch of .1571 and
diameter pitch of 20. The spur gear have 25 teeth.
a. .007855
b. .007558SOLUTION:
c. .008578
From Vallance p 262, Table 11-1
d. .007585
c = 0.157/Pd = 0.157/20 = 0.00785
9. A 3”0 diameter short shaft carrying 2 pulleys close to the bearings transmit how much
horsepower if the shaft makes 280 rpm.
a. 199 Hp
b. 198 Hp
SOLUTION:
c. 200 Hp
From PSME Code p 18, for short shaft:
d. 210 Hp
P = D3N / 38 = 33(280) / 38 = 199 Hp
10. What pressured required to punch a hole 2” diameter through a ¼” steel plate?
a. 10 tons
b. 20 tons
SOLUTION:
c. 30 tons
From Machinery’s Handbook p 1924:
d. 40 tons
P = d x t x 80 tons = 2 x ¼ x 80 = 40 tons
11. Compute the working strength of 1” bolt which is screwed up tightly in packed joint when
the allowable working stress is 13,000 psi.
a. 3600 lbs
b. 3950 lbs
SOLUTION:
c. 3900 lbs
From Machinery’s Handbook p 1149:
d. 3800 lbs
W = st (0.55d2 – 0.25d)
= 13,000 [0.55(1)2 – 0.25(1)]
= 3900 lbs
12. What is the working strength of a 2” bolt which is screwed up tightly in a packed joint when
the allowable working stress 12,000 psi?
a. 20,120 lbs
b. 20,100 lbs
SOLUTION:
c. 20,400 lbs
From Machinery Handbook p 1149
d. 20,200 lbs
W = st (0.55d2 – 0.25d)
= 12,000 [0.55(2)2 – 0.25(2)]
= 20,400 lbs
13. Compute the speed of the gear mounted on a 52.5 mm diameter shaft receiving power from a
driving motor with 250 Hp.
a. 2182 rpm
b. 2071 rpm
SOLUTION:
c. 2282 rpm
From PSME Code p 18:
P = D3N / 80
250 = (52.5/25.4)3N / 80 = 2265 rpm
14. The minimum whole depth of spur gear of 14-1/2 deg. Involute type with diametral pitch of
24 and circular pitch of 0.1309.
a. 0.09000
b. 0.08900
SOLUTION:
c. 0.089875
From Vallance p 262, Table 11-1
d. 0.089758
h
= 2.157/Pd = 2.157/24 = 0.089875
d. 2341 rpm
15. What is the frictional HP acting on a collar loaded with 100 kg weight? The collar has an
outside diameter of 100 mm and an internal diameter of 40 mm. The collar rotates at 1000
rpm and the coefficient of friction between the collar and the pivot surface is 0.15?
a. 0.8 HP
b. 0.5 HP
SOLUTION:
c. 0.3 HP
T = frictional torque = f W rf
d. 1.2 HP
= f W [(r0+r1)/2] = 0.15(100)(0.00981)[(0.050+0.020)/2]
= 0.00515 kN-m
T = 2TN
= 2 (0.00515)(1000/60)
= 0.723 HP
16. A solid cylindrical shaft 48.2 cm long is used for a transmission of mechanical power at a rate
of 37 KW running at 1760 rpm. The ss is 8.13 Mpa. Calculate the diameter.
a. 30 mm
b. 35 mm
SOLUTION:
c. 40 mm
P = 2TN
d. 50 mm
37 = 2t (1760/60)
T = 0.2 KN-m
Ss = 16T /  D3
8130 = 16(0.2) /  D3
D = 0.050
m = 50 mm
17. An internal gear is set up with a 5-in. diameter opinion and center distance of 18 inches. Find
the diameter of the internal gear.
a. 36”
b. 21.5”
SOLUTION:
c. 26”
C = [D2 – D1]/2
d. 41”
18 = [D2 – 5] / 2
D2 = 41”
18. A 100 Hp engine running at 225 rpm is belted to an electric generator. The band wheel in the
engine is 6 feet in diameter. Assuming an arc of contact on the generator pulley to be 150°,
determine the necessary width for a double-ply oak tanned leather belt 3/8 in. thick.
SOLUTION:
V = belt speed = DN
=  (6)(225) = 4241 ft/min
From Vallance, p 387, Table 16-6
HP Rating = 13.08 hp/in (interpolated)
From Table 16-8, p 389
correction factors:
Angle of center line (150° arc of contact ….1.0)
Normal operating conditions ….. 1.0
Width of Belt = 100hp / 13.08 hp/in = 7.64 in.
From table 16-2, p 383: use standard width of 8 inches.
19. A hallow shaft with outside diameter of 14 cm and wall thickness of 0.8 cm transmits 200
KW at 400 rpm. What must be the angular deflection of the shaft if the length is 5 meters?
The material of the shaft is C4140 steel.
SOLUTION:
Solving for the torque:
P = 2TN
200 = 2T (400/60)
T = 4.77465 kN-m
L = length = 5m
Solving for the polar moment of inertia J:
J =  /32 (D04 – Di4)
=  /32 [(0.140)4 – (0.124)4] = 1.45042 x 10-5 m4
From valance p 31. Table 2-6 for C4140 steel:
G = 12,000,000 psi
= 12,000,000 (101.325/14.7) = 82,714,286 KPa
 = angular deflection = TJ / JG = 4.77465(5) / 1.45042 x 10 -5 (82,714,286)=0.0199
rad = 1.14 deg.
20. A 2.5” diameter by 2 in. long journal bearing is to carry a 5500-lb load at 3600 rpm using
SAE 40 lube oil at 200°F through a single hole at 25 psi. Compute the bearing pressure
a. 1100 psi
b. 900 psi
SOLUTION:
c. 1000 psi
Sb = F / LD = 5500 / (2)(2.5) = 1100 psi
d. 950 psi
21. A hollow shaft has an inner diameter of 0.035 m and outer diameter of 0.06 m. Determine the
polar moment of inertia of the hollow shaft.
a. 1.512 x 10-6 m4
b. 1.215 x 10-6 m4
SOLUTION:
c. 1.152 x 10-6 m4
J =  /32 (D04 – Di4)
-6
4
d. 1.125 x 10 m
J =  /32 [(0.06)4 – (0.035)4]
J = 1.125 x 10-6 m4
22. Determine the thickness of a steel air receiver with 30 inches diameter and pressure load of
120 psi, design stress of 8000 psi.
a. ¼ in
b. 5/8 in
SOLUTION:
c. 3/8 in
st = PD / 2t
d. ½ in 800J= 120 (3) / 2T
t = 0.225 in use ¼ in. std. Thickness (Kent’s p. 6-03)
23. A short 61 mm  shaft transmits 120 Hp. Compute the linear spee dof a pulley 55 cm 
mounted on the shaft.
a.
b.
c.
d.
1796 fpm
1766 fpm
1856 fpm
2106 fpm
SOLUTION:
For short shaft (PSME Code p 18)
P = D3N / 38 where D = 61 / 25.4 == 2.4 in.
20 = (2.4)3N / 38
N = 330 rpm
V = vfn = (0.55)(3.28)(330) = 1870 fpm
24. What is the modulus of elasticity if the stress is 44,000 psi and a unit strain of 0.00105?
a. 41.905 x 106
b. 42.300 x 106
SOLUTION:
6
c. 41.202 x 10
E = stress / strain = (44000)/(0.00105 = 41.905 x 106 psi
d. 43.101 x 106
25. A 2-in solid shaft is driven by a 36-in. gear and transmits power at 20 rpm. If allowable
shearing stress is 12 ksi, what horsepower can be transmitted?
a. 29.89
b. 35.89
SOLUTION:
c. 38.89
SS = 16t / D3 = 12,000 = 16T / (2)3
d. 34.89
T = 18,850 in-lbs = 1,570.8 ft-lbs
P = 2TN = [2(1,570.8)(120/60)] / 550 = 35.89 HP
26. The first derivative with respect to velocity of kinetic energy is:
a. power
b. acceleration
SOLUTION:
c. momentum
KE = ½ mv2
d. none of these Dke / dv = mv (momentum)
27. Using oxyacetylene welding method to weld a 3ft. long seam in a 3/8 in. thick plate at a
consumption rate of 9 cu.ft./ft. of weld for oxygen and 7 cu.ft./ft. for acetylene. What is the
total combined gas consumption in cu. Ft.?
a. 51
b. 48
SOLUTION:
c. 45
V = (9 + 7) 3 = 48 cu. ft.
d. 55
28. A helical spring having squared and ground ends has a total of 18 coils and its material has
modulus of elasticity in shear of 78.910 GPa. If the spring has an outside diameter of 10.42
cm and a wire diameter of 0.625 cm, compute the maximum deflection that can be produced
in the spring due to a load of 50 kgs.
a. 302 mm
b. 342 mm
SOLUTION:
c. 490 mm
Dm = Do – d = 10.42 – 0.625 = 9.795 cm
d. 322 mm
C = Dm / d = 9.795 / 0.625 = 15.672
n = 18 – 2 = 16 active coils
y = 8FC3n / Gd
= 8(50 x 9.81)(15.672)3(16)
78,910,000,000(0.00625)
= 0.490 m = 490 mm
29. With the electric arc welding rate of 18 in/min, how long will it take to weld a ½ in. thick
plate by 3 ft long seam?
a. 3 min.
b. 2 min.
SOLUTION:
c. 1.5 min
Time = 3(12) / 18 = 2 min
d. 4 min.
30. How long will it take to mill a ¾” by 2” long keyway in a 3” diameter shafting with a 24
tooth cutter turning at 100 rpm and 0.005” feed/tooth?
a. 0.136 min
b. 0.196 min
SOLUTION:
c. 0.166 min
Time = Length of Cut / Cutting Rate
d. 0.106 min
= 2” / [ (24 teeth/rev)x(100 rev/min)x(0.005”/tooth)]
= 0.167 min.
31. How long will it take to saw a rectangular piece of aluminum plate 8 in. wide and 1½ in.
thick if the length of the cut is 8 in., the power hacksaw makes 120 rev/min and average feed
per stroke is 0.0060 in.?
a. 13.11
b. 11.11
SOLUTION:
c. 14.01
Time = Length of Cut / Cutting Rate
d. 12.03
= 8” / [ (120 teeth/rev)x(1cutting stroke/rev)x(0.0060”/cutting stroke)]
= 11.11 min.
32. A cast iron flywheel is rotated at a speed of 1200 rpm and having a mean rim radius of 1 foot.
If the weight of the rim is 30 lbs., what is the centrifugal force? Use factor C=41.
a. 14,800 lbs
b. 7 mt
SOLUTION:
c. 14,860 lbs
V = DN
d. 14,760 lbs
= (2) (1200/60)
= 125.664 ft/sec
Fc = WV2/GR = 30(125.664)2 / 32.174(1)
= 14,724 lbs
33. A steel cylindrical air receiver with 5 feet diameter and pressure load of 180 psi, design stress
of 9500 psi maximum. The pressure vessel is to be provided with 1½” diameter drain valve
installed at the bottom of the vessel and safety pressure relief valve installed either at the top
most or at the side with pop-out rating of 200 psi. Assume a 100% weld joint efficiency. The
lap welding tensile strength is 65,000 psi. Determine the bursting pressure of this air receiver.
a. 1154 psi
b. 1454 psi
SOLUTION:
c. 1354 psi
Solving for the wall thickness of the air receiver:
d. 1254 psi
St = PD/2t
Based on operating pressure:
9500 = (180)(60) / 2t
t = 0.568”
Based on pop-out pressure:
9500 = (200)(60) / 2t
t = 0.631”
A std. Plate thickness of 5/8” (0.625”) would be safe. Solving for the bursting
pressure using a wall thickness of 5/8” (0.625”):
Sut = PD / 2t
65,000 = P(60) / 2(0.625)
P = 1354 psi
34. The tooth thickness of a gear is 0.5 inch and its circular pitch is 1.0 inch. Calculate the
dedendum of the gear.
a. 0.3183
b. 1.250
SOLUTION:
c. 0.3979
P (Pc) = 
d. 0.1114
P=/1=
From valance p 262, Table 11-1:
Dedendum = 1.25 / P = 1.25 /  = 0.3979”
35. A hollow shaft has an inner diameter of 0.035 m and an outer diameter of 0.06 m. Compute
for the torque if the shear stress is not to exceed 120 Mpa in Nm.
a. 4500
b. 4100
SOLUTION:
c. 4300
Ss = 16TDo / [(Do4 – Di4)]
d. 4150
120,000,000 = 16T(0.06) / [[(0.06)4 – (0.035)4]]
T = 4500 N-m
36. It is a problem of expansion and shrinkage of steel material so that the slightly bigger shafting
of 2” diameter can be inserted/fitted to the slightly smaller hole of a steel bushing of 1.999”
diameter with the following process material/data to apply:
Coefficient of expansion of carbon steel – 0.0000068”/”°F
Temperature raised by gas heating = 24.5°F
Cooling media to use dry ice with boiling point of –109.3°F (-78.5°C)
Shrinkage rate below boiling point is 0.00073 in/in
Determine the final clearance between the expanded steel bushing hole against the shrinkage
of the steel shaft.
a. 0.000793”
b. 0.000750”
SOLUTION:
c. 0.000693”
Solving for diameter of bushing hole after heating:
d. 0.000800”
Y = kL(t2 – t1)
= 0.0000068(1.999)(24.5) = 0.000333”
D = 1.999 + 0.000333 = 1.999333”
Solving for diameter of shaft after cooling:
y = -0.00072(2) = -0.00146”
d = 2 – 0.00146 = 1.998540
Clearance = 1.999333 – 1.998540
= 0.000793”
37. What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 m/m
from a load producing a unit ensile stress of 44,000 psi?
a. 42.300 x 106 psi
b. 41.202 x 106 psi
SOLUTION:
c. 43.101 x 106 psi
E = Stress / Strain
d. 41.905 x 106 psi
= 44 ,000 / 0.00105 = 41.905 x 106 psi
38. Compute the pitch angle of a bevel gear given the pinion’s number of teeth of 14 and 42
teeth on the gear.
a. 18.4°
b. 28.4°
SOLUTION:
c. 33.4°
From machinery’s Handbook p 844:
d. 38.4°
Pitch Angle = tan-1 n/N = tan-1 14/42 = 18.4°
39. Compute for the tooth thickness of 14½° spur gear with diametral pitch = 5.
a. 0.3979
b. 3.1831
SOLUTION:
c. 0.03141
Tooth thickness = 1.5708/P = 1.5708/5 = 0.31416”
d. 0.31416
40. If the ultimate shear strength of a steel plate is 42,000 lb/in 2, what force is necessary to
punch a 0.75 inch diameter hole in a 0.625 inch thick plate?
a. 63,008 lbs
b. 68,080 lbs
SOLUTION:
c. 61,800 lbs
Force = Stress x Area
d. 66,800 lbs
= Ss (dt)
= 42,000()(0.75)(0.625)
= 61,850 lbs.
41. A steel tie rod on bridge must be made to withstand a pull of 5000 lbs. Find the diameter of
the rod assuming a factor of safety of 5 and ultimate stress of 64,000 lb/in 2.
a. 0.75 in
b. 0.71 in
SOLUTION:
c. 0.84 in
St = F/A
d. 0.79 in
64,000 / 5 = 5000 / [(/4)d2)]
d = 0.71 in
42. If the weight of 6” diameter by 48” long SAE 1030 shafting is 174.5 kg., then what will be
the weight of chromium SAE 51416 of same size?
a. 305.5 lbs.
b. 426.4 lbs
SOLUTION:
c. 384.6 lbs
The major component of different steels is iron, therefore
d. 465.1 lbs
their densities do not differ much.
Weight = 174.5 kg
= 174.5 (2.205) = 384.8 lbs.
43. Compute the circular pitch (in inch) of a pair of gears having a ratio of 4 and a center distance
of 10.23. Each gear has 72 teeth and pinion has 18 teeth.
a. 0.8095
b. 0.7825
SOLUTION:
c. 0.8035
D1N1 = D2N2
d. 0.8085 N1 / N2 = D2 / D1 = 4
Pd = T/D = T1/D1 = 18/4.092
D2 = 4 D 1
= 4.4
C = (D1 + D2) / 2
Use std. Pd of 4
10.23 = (D1 + 4D1) / 2
Pd x Pc = 
D1 = 4.092 in.
Pc = /4 = 0.7854 in.
44. A journal bearing with diameter of 76.2 mm is subjected to aload of 4900 N while rotating at
200 rpm. If its coefficient of friction is 0.02 and L/D = 2.5, find its projected area in mm 2.
a. 12,090
b. 14,165
SOLUTION:
c. 13,050
L / D = 2.5
A=DxL
d. 14,516
L / 76.5 = 2.5
= 76.2 (190.5)
L = 190.5 mm
= 14,516 mm2
45. A flywheel rotates at 120 rpm or 12.57 rad/sec slowed down to 102 rpm or 10.68 rad/sec
during the punching operation that requires ¾ second of the punching portion of the cycle.
Compute the angular acceleration of the flywheel in rad/sec 2.
a. –2.52 rad/sec2
b. 3.15 rad/sec2
SOLUTION:
c. –2.75 rad/sec2
 = (W2 – W1) / t = (10.68 – 12.57) / 0.75 = -2.52 rad/sec2
2
d. 2.22 rad/sec
46. A drop hammer of 1 ton dead weight capacity is propelled downward by a 12 in. diameter
cylinder. At 100 psi air pressure, what is the impact velocity if the stroke is 28 inches.
a. 63.2 fps
b. 15.8 fps
SOLUTION:
c. 31.6 fps
Solving for the force of the hammer:
d. 47.4 fps
F = pA = 100(/4)(12)2 = 11,310 lbs
Solving for the acceleration of the hammer caused by the
applied force:
F = ma
11,310 = 2000/32.2 a
a = 182.1 ft/s2
Solving for the impact velocity
V22 = V12 + 2as
V22 = 0 + 2(182.1 + 32.2)(28/12)
V2 = 31.6 fps
47. Compute the nominal shear stress at the surface in MPa for a 40 mm diameter shaft that
transmits 750 KW at 1500 rpm. Axial and bending loads are assumed negligible. Torsional
shearing stress is 28 N/mm2.
a. 218
b. 312
SOLUTION:
c. 232
P = 2TN
d. 380
750 = 2T (1500/60)
T = 4.775 kN-m
Ss = 16T /  D3
= 16(4.775) /  (0.04)3
= 379,982 KPa
= 380 Mpa
48. A truck skids to a stop 60 m after the application of the brakes while traveling at 90 km/hr.
What is the acceleration in m/sec2?
a. –5.21 /sec2
b. 6.36
SOLUTION:
c. –7.06
V1 = 90(1000) / 3600 = 25 m/s
d. 5.76
V22 = V12 + 2as
0 = (25)2 + 2 a(60)
a = -5.21 m/sec2
49. A spur pinion rotates at 1800 rpm and transmits to a mating gear 30 HP. The pitch diameter is
4” and the pressure angle is 14½. Determine the tangential load in lbs.
a. 495
b. 525
SOLUTION:
c. 535
P = 2TN
d. 475
30(33,000) = 2T(1800)
T = 87.535 ft-lbs = 1050.4 in-lbs
Force = Torque / Radius
Ft = 1050.4 / 2 = 525 lbs.
50. A high alloy spring having squared and ground ends and has a total of 16 coils and modulus
of elasticity in shear of 85 GPa. Compute the Wahl factor. The spring outside diameter is 9.66
cm and wire diameter is 0.65 cm.
a. 1.058
b. 1.10
SOLUTION:
c. 1.185
Dm = mean diameter = DO – d = 9.66 – 0.65 = 9.01 cm
d. 1.2
C = spring index = Dm / d = 9.01 / 0.65 = 13.86
Wahl Factor = 4C-1/4C-4 = 4(13.86) – ¼(13.86) – 4 = 1.058
51. Two idlers of 28T and 26 T are introduced between the 24 T pinion with a turning speed of
400 rpm dirving a final 96 T gear. What would be the final speed of the driven gear and its
direction relative to the driving gear rotation?
a. 120 rpm and opposite direction
b. 80 rpm and same direction
c. 100 rpm and opposite direction
d. 100 rpm and same direction
SOLUTION:
T1N1 = T2N2 = T3N3 = T4N4
24(400) = 96(N4)
N4 = 100 rpm opposite direction
52. Compute the lineshaft diameter to transmit 12 HP at 180 rpm with torsional deflection of 0.08
degrees per ft. length.
a. 3 in
b. 5 cm
SOLUTION:
c. 2.35 in
P = 2TN
d. 62 mm
12(33,000) = 2T(180)
T = 350.14 ft-lbs = 4,201.68 in-lbs
L = 1 ft = 12 in.
J = D4 / 32
G = 12,000,000 psi for steel
 = TL /JG
0.08(/180) = 4,201.68(12) / [(D4/32)(12,000,000)
D = 2.35 in.
53. Calculate the rpm for machining a cast iron workpiece 6 inches in diameter. The lowest
cutting speed for cast iron is 50 rpm.
a. 35.3
b. 33.3
SOLUTION:
c. 43.3
V = DN
d. 53.32
50 =  (6/12)N
N = 31.83 rpm
54. A helical gear having a 14½° normal pressure angle and transverse diametral pitch of 2.3622
per cm. The helix angle is at 45° and has 8 teeth. Compute the transverse pressure angle in
degrees.
a. 22.2°
b. 19.3°
SOLUTION:
c. 18.9°
Ref: Vallance p 282
d. 20.1°
tan n = tan  cos x
tan 14½° = tan  cos 45°
 = 20.1°
55. Compute the maximum unit shear in a 3 inch diameter steel shafting that transmits 24000 in
lb. Torque at 99 rpm.
a. 4530 psi
b. 4250 psi
SOLUTION:
c. 3860 psi
Ss = 16T / D3
d. 4930 psi
= 16(24,000) /  (3)3
= 4527 psi
56. To facilitate the milling (roughing) work of a cast steel material using a 1½ inch diameter
cutter choose between the two available stock. Find the speed of the cutter in rpm.
- high speed steel cutter with a cutting speed of 50 rpm
- carbide tipped cutter with a cutting speed of 200 fpm
a. 389
b. 572
SOLUTION:
c. 509
V = DN
d. 412
High speed steel cutter (slower)
50(12) = (1.2)N
n = 127 RPM
Carbide tipped cutter (faster)
200 (12) = (1.2)N
N = 509 rpm
57. A lineshaft runs at 360 rpm. A 18” pulley on the same shaft is belt connected to a 12” pulley
on the countershaft. From a 15” pulley on the countershaft, motion is transmitted to the
machine. Compute/check the required diameter of the pulley on the machine to give a spindle
speed of 660 rpm.
a. 16”
b. 12¼”
SOLUTION:
c. 10½”
DCNI = DCI NC
d. 8½”
18(360) = 12NC
NC = 540 rpm (speed of countershaft)
DC2NC = DM NM
15 (540) = DM (660)
DM = 12.27 say 12¼”
58. Two parallel shafts connected by pure rolling turn in the same direction having a speed ration
of 2.75. What is the distance of the two shafts if the smaller cylinder is 22 cm in diameter?
a. 16.6 cm
b. 30.25 cm
SOLUTION:
c. 25.25 cm
D1N1 = D2 N2
d. 19.25 cm
N1 / N2 = D2 / D1
2.75 = D2 / 22
D2 = 60.5 cm
C = (D2 – D1) / 2 = (60.5-22)/2 = 19.25 cm
59. Determine the torque received by the motor shaft running at 4,250 rpm, transmitting 11Hp
through a 10 in. diameter 20° involute gear. The shaft is supported by ball bearings at both
ends and the gear is fixed at the middle 8” shaft length.
a. 163 in-lb
b. 167 in-lb
SOLUTION:
c. 132 in-lb
P = 2TN
d. 138 in-lb
11(33,000) = 2 T(4,250)
T = 13.594 ft-lbs.
= 13.594(12) = 163.128 in-lbs
60. A drop hammer of 1 ton dead weight capacity is propelled downward by a 12 in. diameter
cylinder. At 100 psi air pressure, what is theimpact velocity if the stroke is 28 inces?
a. 63.2 fps
b. 15.8 fps
SOLUTION:
c. 31.6 fps
Solving for the force on the hammer:
d. 47.4 fps
F = pA =100 (/4)(12)3 = 11,310 lbs.
61. 1 kw is equivalent to
a. 57 Btu/min
b. 764 hp
c. 41 kcal/sec
d. 44,200 ft-lb/sec
SOLUTION:
1kw = 1(kJ/sec) x (Btu/1.055kJ) x (60 sec/min) = 57 Btu / min
62. Compute for the torsional
subjected to twist moment
N/mm2.
a. 0.27
b. 0.31
c. 0.20(ans)
d. 0.24
deflection in degrees of a 110 mm diameter, 1.4 m long shaft
of 3 x 10 6 N-mm. The torsional modulus of elasticity is 83000
SOLUTION:
 = TL / JG
= [(3X106N-mm)(1400mm) / (x/32)(110)4(83,000 N/mm2)]x(180 deg/ rad)
= 0.2 deg
63. Find the torsional moment (in Newton-mm) developed when the shaft delivers 20 kw at 200
rpm.
a. 0.85 x 106
b. 1.0 x 106(ans)
SOLUTION:
6
c. 1.20 x 10
P = 2tn
d. 2.10 x 106
20 = 2(T)(200/60)
T = 0.955 kN-m
T = 0.955 kN-m x (1000N/kN) x (1000mm/m)
T = 0.955x 106 N-mm
64. Determine the load in KN
elongation exceeds 1 mm.
a. 85(ans)
b. 125
c. 103
d. 93
on a 25mm diameter x 1200 mm long steel shaft if its maximum
SOLUTION:
y = FL/AE
E for steel = 30,000,000 psi = 206,786 x 103 kPa
1 mm x = (F(1200mm) / [(/4)(0.025)2m2(206,786 x 103)(kN/m2)]
F = 84.6 kN
65. Find the thickness of a metal needed is the semi-spherical end of cylindrical vessel 0.70 meter
in diameter subjected to an internal pressure of 2.7 N/mm 2. The material is mild carbon steel
and tensile stress is 69 N/mm2.
a. 5.2 mm
b. 6.02 mm
SOLUTION:
c. 7.53 mm
St = PD / 4t
d. 6.84 mm(ans)
69 = 2.7(0.7) / 4t
t = 0.00684 m = 6.84 mm
66. A 102 mm diameter shaft is driven at 3000 rpm by 300 HP prime mover. The shaft drives a
121 cm diameter chain sprocket having 85% output efficiency. Compute the torque n in-lb
develop in the shaft.
a. 5,600
b. 7,100
SOLUTION:
c. 8,150
P = 2TN
d. 6,300 (ans)
300(33,000) = 2T(3000)
T = 525.21 ft.-lb = 6302.5 in-lb.
67. Determine the average time to cut by automatic oxy-acetylene (machine) crosswise a 4 ft x 8
ft. x 4 in thick steel plate. (Oxy acetylene cutting speed = 9 inches per minute).
a. 6.85 min
b. 318 sec (ans)
SOLUTION:
c. 10 min
Length of cut = cutting rate x time
d. 360 sec
4(12) in = 9in/min x min/60sec x time
time = 320 sec.
68. Computes for the induced/compressive stress, in Kpa, of a steel solid shafting of 50 mm
diameter and 800 mm in length that is subjected to an increase of temperature by 80 deg. C.
Use k for steel = 0.0000119 m/m-°C. (E=30,000,000 psi = 206,768 Mpa).
a. 181,816
b. 242,816
SOLUTION:
c. 218,182
Stress = kET
d. 196,860
= (0.0000119)(206,786 x 103)(80)
= 196,860 kPa
69. A mass weighing 56 lbs. Rests on horizontal surface. The force needed to move along the
surface is 20 lbs. Determine the coefficient of friction.
a. 0.0
b. 0.112
SOLUTION:
c. 0.36 (ans)
Ff = fN
d. 0.38
20 = f(56)
f = 0.357
70. Determine the time in seconds, to saw a rectangular magnesium bar 5 in. wide and 2 in. thick
if the length of cut is 5 in. The power hacksaw does 120 strokes/min and the feed/stroke is
0.127 mm.
a. 189
b. 500 (ans)
SOLUTION:
c. 99
Time = 5in / [ (120 strokes/min)x(0.127mm/mm)x(1in/25.4mm)x(1min/60sec)]
d. 20
= 500 sec
71. A car travels with an initial velocity of 10 m/sec or 36 km/hr. If it is decelerating at a rate of 3
m/sec-sec, how far, in meters, does it travel before stopping?
a. 17 (ans)
b. 21
SOLUTION:
c. 19
V22 = V12 + 2aS
d. 15
0 = (10)2 + 2(-3)S
S = 16.67 m
72. Compute for the twisting moment in in-lb developed when the shaft delviers 20 Hp at 1200
rpm.
a. 1166
b. 915
SOLUTION:
c. 1050 (ans)
Power = 2TN
d. 945
20(33,000) = 2T(1200)
T = 87.535 ft-lb = 1050.4 in-lb
73. In an Oxy-Acetylene manual welding method, to weld a 3½ ft. long seam in a 0.375” thick
steel plate at a consumption rate of 9 cu.ft/ft/ for oxygen and 7 cu.ft./ft for acetylene.
Compute for the total combined gas consumption in cu.ft.
a. 48
b. 24.5
SOLUTION:
c. 56 (ans)
Total Volume = 93.5) + 7(3.5) = 56 ft3
d. 31.5
74. A cylindrical tank with 10 inch inside diameter contain oxygen gas at 2500 psi. Calculate the
required wall thickness in millimeter under stress of 28000 psi.
a. 10.54
b. 11.34 (ans)
SOLUTION:
c. 10.24
St = PD / 2t
d. 12.44
t = 0.446 in = 11.34 mm
75. Compute for the load in KN on a 3 cm diameter, 100 cm long steel rod if its maximum
elongation exceed 0.12 cm.
a. 176 (ans)
b. 196
SOLUTION:
c. 148
y = FL / AE
d. 287
0.12 = F(100) / [ (/4)(3 x 10-2)2 (206,786 x 103)
F = 175.4 kN
76. A baseball is thrown straight upward with a velocity of 20 m/s. Compute for the time elapsed
for the baseball to return. Assume for a zero drag.
a. 1.84
b. 2.21
SOLUTION:
c. 4.08 (ans)
a = (V2 – V1) / t
d. 2.40
-9.81 = (0 –20) / tl
tl = 2.04 sec
total time elapse = 2 x 2.04 = 4.08 sec
77. Compute for the polar section modulus of an SAE 1060 shafting having a diameter of 3
inches. Use a factor of safety of 2 and design stress at 8000 psi.
a. 4.7
b. 6.1
SOLUTION:
c. 4.2
Polar section modulus, Zp = (D3) / 16
d. 5.3 (ans)
= (3)3 / 16
= 5.3 in3
78. Compute the working strength of a bolt having a diameter of 1½ inch under a tensile stress of
8000 psi.
a. 3060 lbs
b. 4560 lbs
SOLUTION:
c. 6900 lbs (ans)
W = St (0.55d2 – 0.25d)
d. 7800 lbs
= 800[0.55(1.5)2 – 0.25(1.5)]
= 6900 lbs.
79. Determine the diameter in inches of a steel countershaft that delivers 13.31 hp at a speed of
15.7 rad/sec, the allowable material shear stress is 8.5 ksi.
a. 1 in
b. 1½ in (ans)
SOLUTION:
c. 2 in
N = 15.7 rad/sec x rev/2rad x 60sec/min = 149.9 rpm
d. 1 ¼ in
HP = D3N / 38 for countershaft
13.31 = D3(149.9) / 38
D = 1.5 in
80. An elevator that moves upward carrying 10 persons with a total weight of 1500 lbs up to the
3rd floor with the height of 30 ft in 20 seconds. What is the power in hp of the motor driving
the elevator?
a. 3.5
b. 5.1
SOLUTION:
c. 2
Power = Energy / Time
d. 4 (ans)
= (1500lbs)(30ft) / 20 sec
= 4.1 HP
81. What angle should the road be banked to prevent a car to skid when traveling at 25 m/s at a
radius of 200m?
a. 21.1 deg
b. 17.67 deg (ans)
SOLUTION:
c. 19.8 deg
tan  = V2 / gr
d. 14.55 deg
= (25)2 / (9.81)(200)
 = 17.67 deg
82. How long, in minutes, to drill a hole of 5 inches thick. The conical height of the drill is ½ in,
feed is 0.001 in/rev, and the speed of the drill is 100 rpm.
a. 50 (ans)
b. 10
SOLUTION:
c. 25
Time = 5in / [(0.001 in/rev)(100 rev/min)]
d. 100
= 50 min
83. Compute for the torsional deflection in degrees of a 3 5/8 inches diameter, 1.2 m long shaft
subjected to a twist moment of 3 x 106 N-mm. The torsional modulus of elasticity is 80,000 Mpa.
a. 0.365 (ans)
b. 0.985
SOLUTION:
c. 0.653
D = 3 5/8 in -= 92.075 mm
d. 1.025
 = TL / JG
 = [(3x106N-mm)(1200mm)//32(92.075)4mm4(80,000 N/mm2)] x (180/ )
 = 0.365 deg
84. Compute the tooth thickness of 14½ degree spur gear with diametral pitch of 7:
a. 5.7 mm (ans)
b. 6.8 mm
SOLUTION:
c. 7.5 mm
t = 1.5708 / Pd = 1.5708 / 7
d. 9.8 mm
t = 0.2244 in = 5.6997 mm
85. Determine the working strength of a 1.25 in bolt/screwed up tightly with a tensile stress of
8000 psi.
a.
b.
c.
d.
3475 lbs
4175 lbs
4375 lbs (ans)
7513 lbs
SOLUTION:
Working strength = S1[0.55d2 – 0.25d]
= 8000[0.55(1.25)2 – 0.25(1.25)]
= 4375 lbs
86. Determine the displacement, in meter, of a jeepney skidding in asphalt road with a velocity of
30 m/s before it stops when the stopping period is 4 sec.
a. 80
b. 99
SOLUTION:
c. 100
a = (V2 – V1) / t = (0 – 30) / 4 = - 7.5 m/s2
d. 60 (ans)
V22 = V12 + 2aS
 = 302 + 2(-7.5)S
S = 60m
87. Find the formula of maximum shear in Figure No. 8.
a. V = P /2 (ans)
b. V = P/4
SOLUTION:
c. V = P
V=P/2
d. V = PD / 2
88. What is the equation of the deflection in Figure No. 8?
a. y = PL3 / 12EI
b. y = PL3 / 3EI
c. y = PL3 / 48EI (ans)
d. y = PL3 / 24EI
89. If the weight of 6” diameter by 48” long SAE 1030 shafting is 174.5 kg, then what will be
the weight of chromium SAE 51416 of same size?
a. 288.5 lbs
b. 325.6 lbs
SOLUTION:
c. 384.8 lbs (ans) Different classes of steel have almost the same density
d. 425.8 lbs
because the main component is iron.
Therefore: Weight = 174.5 kg
= 174.5(2.205)
= 384.8 lbs.
90. Find the tooth thickness of a gear with diametral pitch of 4.
a. 0.3178”
b. 0.3925” (ans)
SOLUTION:
c. 0.4523”
Tooth thickness = 1.57 / Pd
d. 0.5293”
= 1.57 / 4
= 0.3925”
91. Find the power in watts transmitted by a main power transmitting shaft with diameter of 53
mm and speed of 200 mm.
a. 15.45
b. 18.926 (ans)
SOLUTION:
c. 25.4
For main power transmitting shaft:
d. 30.7
P = D3N / 80
where: D = 55/25.4 = 2.165”
P = [(2.165)3(200)] / 80
P = 25.37 hp
P = 25.37 (746)
P 18.926 watts
92. Find the polar moment of inertia of a square 3½” by 3½”.
a. 20 in4
b. 25 in4 (ans)
SOLUTION:
c. 30 in4
For square section with side a J = a4 / 6
4
d. 35 in
Therefore:
J = (3.5)4 / 6
J = 25 in4
93. Find the shearing stress of a rectangular steel tank with dimensions L = 800 mm, W =
100mm, H = 400 MM. A force 2000 kN is applied at the edge of the which is welded at the
bottom.
a. 15,000 kPa
b. 20,000 kPa
SOLUTION:
c. 25,000 kPa (ans)
S=F/A
S = (2000) / (800)(100)(400)
d. 30,000 kPa
S = 25,000 kPa
94. Find the working pressure required to punch 4” diameter hole on ¼” thick steel plate.
a. 60 tons
b. 70 tons
SOLUTION:
c. 75 tons
Using derived formula from Machinery’s handbook:
d. 80 tons (ans)
P = d x t x 80
P = 4(¼)(80) P = 80 tons
95. Compute the diameter (in inches) of a SAE 1030 steel shaft to transmit 12 hp at 120 rpm with
torsional deflection below 0.08 degree/foot length as required.
a. 2 7/8
b. 2 5/8 (ans)
SOLUTION:
c. 2 ¼
Solving for the torque: T,
d. 2 3/8
P = 2TN
P = P / 2N
P = 12(550 lb-ft/sec/hp) / 2(120/60)
T = 525.21 lb-ft.
Solving for the diameter, using G = 12x106 psi (for steel)
 = TL / JG
(0.08)(TL / 180) = (525.2 lb-ft.) / [(/32)(D4)(12X106lb/in2)(144)
D = 0.217 ft
D = 2.604 in say 2 5/8 in.
96. Find the center distance between gears with 50 teeth and 25 teeth if the diametral pitch
is 5.
a. 9.5”
b. 6.5”
SOLUTION:
c. 8.5”
Pa = diametral pitch = (no. of teeth / pitch diameter)
d. 7.5” (ans)
P=T/D
D1 = 50 / 5 = 10”
D2 = 25 / 5 = 5”
C = (D2 + D1) / 2
C = (10+5) / 2 = 7.5”
97. Find the tooth thickness of a gear given diametral pitch of 5.
a. 0.2188 in
b. 0.31416 in (ans)
SOLUTION:
c. 0.41367 in
d. 0.57894 in
Tooth thickness = 1.57 / PD = 1.57/ 3 = 0.314 in
98. Compute for the load in kN on a 3 cm diameter, 100 cm long steel rod if its maximum
elongation exceed 0.12 cm
a. 178 (ans)
b. 196
SOLUTION:
c. 148
 = FL / AF
d. 287
where  = 0.12 cm = 0.0012 m
L = 100cm = 1 m
A =  /4 (0.03)2 = 0.000707 m2
E for steel = 30 x 106 psi
= 30 x 106 x (101.325) / 14.7 = 207 x 106 kPa
99. Find the polar section modulus of a hollow shaft with OD = 6” and ID = 3”
a. 28.97 in3
b. 39.76 in3 (ans)
SOLUTION:
3
c. 45.45 in
Zp = (/32)(Do4 – Di4) / Do/2
3
d. 51.98 in
= (/32)(64 – 34) / 6/2 = 39.76 in3
100.
Calculate the elongation of a solid steel shaft of SAE 1030 carbon steel 2 7/8 inches in
diameter by 5 feet in length supporting a direct tension load of 60,000 lbs. The factor of
safety is set at 4, stress concentration factor k – 1 and yield strength of 40,000 psi.
a. 0.020
b. 0.015
SOLUTION:
c. 0.0185 (ans)
E for steel : 30 x 106 psi
d. 0.0166
E = FL / AE
= [(60,000) (5 x 12)] / [(/4)(2.875)2(30 x 106)
= 0.0185 in.