MACHINE DESIGN Spur Gear 1) Find the tooth thickness of a 14 deg. Involute gear having a diametral pitch of 6 . A. 5.33 mm Given Formula/s Solution
C. 8.45 mm
B. 6.65 mm
:DP = 6 in-1
= 0.5236 in
:
0.2618 in x
,
D. 12.36 mm
= 0.2618 in
= x
6.65 mm
2) A gear set having a gear ratio of 3 is to be used at a center distance of 10 inches. If the gear has 60 teeth, what must be the circular pitch? A. 0.7236 in. Given Formula/s
Solution
C. 0.8970 in.
B. 0.7851 in.
: C = 10 in. , T2 = 60 , Gear Ratio
D. 0.7283 in.
, : , since C = 10 in. so :
= 3 ,
By relation,
,
3D1 = D2
By substitution,
then
4D1=20 in
D1 = 5 in , D2=15 in
-1
= 4 in
= 0.7853 in
3) Compute the speed of the gear mounted in a 52.5 mm diameter shaft receiving power from a prime motor with 250 hp. A. 2182 rpm
B. 2081 rpm
Given
: P = 250 hp ,
Formula/s
:
C. 2265 rpm
D. 2341 rpm
Solution
:
4) Find the distance between centers of a pair of gears, one of which has 12 teeth and the other 37 teeth. The diametral pitch is 8. B. 4 in
A. 3 in
Given Formula/s
Solution
-1
:DP = 8 in
C. 5 in
, T1=12
D. 6 in
, T2=37
, : so :
5) Two parallel shafts have an angular velocity ratio of 3 to 1 are connected by gears, the largest of which has 54 teeth. Find the number o f teeth of smaller gear. A. 14 Given Formula/s
Solution
B. 16
C. 12
D. 18
:Gear Ratio = 3 , T1=54
so : :
6) A spur pinion rotates at 1800 r pm and transmits to a mating gear 30 hp. The pitch diameter is 8 inches and the pressure angle is 14 1 /2. Determine the total load in lbs. A. 123.45 lbs
B. 653.45 lbs
Given
: P = 30 HP , N = 1800 rpm
Formula/s
:
Solution
:
C. 271.24 lbs
, D=8 in ,
= 14.5 deg
,
,
D. 327.43 lbs
7) A spur pinion rotates at 1800 rpm and transmits to mating gear 30 HP. The pitch diameter is 4” andthe pressure angle is 14 1/2 . determine the tangential load in lbs. A. 495
C. 535
B. 525
Given
: P = 30 HP , N = 1800 rpm
Formula/s
:
Solution
:
D. 475
, D=4 in
,
8) Two idlers of 28 T and 26 T are introduced between the 24 T pinion with a turning speed of 400 rpm driving a final 96T gear. What would be t he final speed of the driven gear and its direction relative to the driving gear rot ation? A. 120 rpm and opposite direction
C. 80 rpm and same direction
B. 100 rpm and opposite direction
D. 100 rpm and same direction
Given
: T1 = 24 , T2 = 96
Formula/s
:
Solution
:
, N1=400rpm
,
=
9) A spur pinion supported on each side by ball bearings rotates 1 750 rpm and transmit to a mating gear at 25 Hp. The pressure angle is 20 degrees and the pitch diameter is 5. Determine the tangential load in lbs. A. 420 Given Formula/s
Solution
B. 300
C. 360
: P = 25 HP , N = 1750 rpm
, D=5 in
, : :
D. 400
10) The minimum clearance allowed for meshing spur gears with a circular pitch of 0.1571 and diametral pitch of 20. The spur gear has 25 teeth. B. 0.008578
A. 0.007855
Given Formula/s Solution
:PC= 0.1571
C. 0.007553
:
:
,
D. 0.007565
Worm Gear 1) A work rotating at 1150 rpm drives a worm gear. The velocity ratio is 15 to 1. A 10 hp motor is used to supply the worm with worm pitch ge ar diameter of 3 in. Find the tangential force on the worm. A. 365.37 lbs
B. 465.37 lbs
Given
:P= 10 hp
Formula/s
:
Solution
C. 565.37 lbs
, N=1150 rpm ,
D. 665.37 lbs
, Dw=3 in
,
: =
2) A double thread worm gear has a pitch of 1 1/8 and a pitch diameter of 3 in. It has a coefficient of friction of 0.20 and normal angle (pressure angle) of 14.5 degrees . The worm is supplied by 12 hp at 1200 rpm motor. Find the tangential force on the ge ar. The worm is left hand thre ads. A. 597.08 lbs
B. 697.08 lbs
Given
:P= 12 hp , N=1200 rpm
Formula/s
:
C. 797.08 lbs
D. 897.08 lbs
, Dw=3 in , Pw=1.125 in , = 14.5 deg , f = 0.20
,
,
,
Solution
:
=
2(1.125 in.) = 2.25 in
3) A single worm gear has a pitch diameter of 2 in and a pitch of 1 in with coefficient of friction of 0.21. The normal angle is 14.5 degrees w ith tangential force on gear of 100 0 lbs. Find the separation force on gear and worm considering a left hand threads.
A. 171.23 lbs
C. 371.23 lbs
B. 271.23 lbs
Given
:Ft= 1000 lbs , Dw=2 in , Pw=1 in , n = 14.5 deg , f = 0.21
Formula/s
:
D. 471.23 lbs
Solution
:
x =9.04 deg
4) A double thread worm gear has a lead angle of 7.25 degrees and pitch radius of 2 ½ in. Find the pitch of the worm. A. ¼ in
B. ½ in
C. 1 in
Given
: x= 7.25 degrees , Rw=2.5 in, Dw = 2Rw =2 (2.5 in) =5 in
Formula/s
:
Solution
:
D. 1 ½ in
;
5) A triple thread worm gear has a helix angle of 78 degrees. It has a pitch of ¾ in. Find the pitch diameter of the gear. A. 2.37 in
B. 2.77 in
Given
: = 78 degrees , p = 0.75 in
Formula/s
:
Solution
:
C. 3.07 in
D. 3.37 in
;
Helical Gear 1) A 20-tooth helical gear has a pitch diameter of 10 in. Find the diametral pitch of the gear. B. 3
A. 2
C. 4
Given
: D = 10 in ; T = 20
Formula/s
:
Solution
:
D. 5
2) A 28 –tooth helical gear having a pitch diameter o f 7 has a helix angle of 22 degrees. Find the circular pitch in a plane normal to the teeth. A. 0.528
B. 0.628
C. 0.728
Given
: D = 7 in ; T = 28 ; = 22 degrees
Formula/s
:
D. 0.828
;
Solution
:
3) A helical gear having a helix angle of 23 degrees and pressure angle of 20 degrees . Find the pressure angle normal to the teeth. A. 0.235
C. 0.435
B. 0.335
Given
: = 20 degrees ; = 23 degrees
Formula/s
:
Solution
:
D. 0.535
4) A 75 hp motor, running at 4 50 rpm is geared to a pump by means of helical gear having a pinion diameter of 8 in. Find the tangential force of the gear. A. 2,326 lbs Given
B. 2,426 lbs : P = 75 hp ; N = 450 rpm ; D p=8 in
C. 2,526 lbs
D. 2,626 lbs
Formula/s
:
Solution
:
;
5) A turbine at 30,000 rpm is used to drive a reduction gear delivering 3 hp at 3,000 rpm. The gears are 20 degrees involute herringbone gears of 28 pitch and 2 1/8 in effective width. The pinion has 20 teeth with a helix angle of 23 deg. Determine t he load normal to the tooth surface. A. 20.4 lbs
B. 24.4 lbs
C. 28.4 lbs
D. 32.4 lbs
6) A helical gear having 20 teeth and pitch diameter of 5 in. Find the normal diametral pitch if helix angleis 22 degrees. A. 3.31 in
C. 5.31 in
B. 4.31 in
Given
: D = 5 in ; T = 20 ; = 22 degrees
Formula/s
:
Solution
:
D. 6.31 in
Bevel Gear 1) A 20 degrees full depth straight tooth gear has a face width of 3 ¾ in and a pitch diameter o f 12 in with cone pitch angle of 37.5 degrees. Find the mean diameter. A. 6.72 in
B. 7.72 in
Given
:
Formula/s
:
Solution
:
C. 8.72 in
D. 9.72 in
= 37.5 degrees ; Dp = 12 in ; b = 3.75 in
2) A straight tooth bevel gear has a face width of 4” and a pitch diameter of 14 in with cone pitch angle of 40 degrees . If the torque on the gear is 8000 in-lb, what is the tangential force on the gear? A. 1400 lbs
B. 1400 lbs
Given
:
Formula/s
:
Solution
:
C. 1400 lbs
D. 1400 lbs
= 40 degrees ; Dp = 14 in ; b = 4 in ; T=8000 lb-in
;
3) A pair of straight tooth bevel ge ar connect a pair of shafts 90 degrees . The velocity ratio is 3 to 1. What is the cone pitch angle of smaller gear? B. 65.34 deg
A. 71.57 deg
Given
:
Formula/s
:
Solution
:
C. 18.43 deg
D. 12.34 deg
4) A spiral bevel pinion with a left hand spiral rotates c lockwise transmits power to a mating gear with speed ratio of 2 to 1. Determine the pitch angle of t he pinion. A. 16.56 deg
Given
B. 20.56 deg
:
C. 26.56 deg
D. 32.56 deg
Formula/s
:
Solution
:
( left hand rotates clockwise )
5) A spiral bevel pinion with a left hand spiral transmits 4 hp at 1200 rpm to a mating gear. The mean diameter of the pinion is 3 in. Find the tangential force at the m ean radius of the gear. A. 110 lbs
B. 120 lbs
C. 130 lbs
Given
: P = 4 hp ; N = 1200 rpm ; Dp avg=3 in
Formula/s
:
Solution
:
D. 140 lbs
;
6) A pair of 4-pitch, 14.5 degree s , involute bevel gear have 2 :1 reduction. The pitch diameter of the driver is 10 in and face width of 2 in. Determine the face angle of pinion. (Shafts at right angle) A. 20.85 deg
B. 23.85 deg
Given
:
Formula/s
:
C. 25.85 deg
;
; P = 4 in
-1
() ---------
;
;
------
Solution
D. 27.85 deg
:
;
;
Belts 1) Find the angle of contact on t he small pulley for a belt drive with center distance of 72 inches if pulley diameters are 6 in. and 12 in., respectively. A. 180.60 deg.
B. 243.40 deg.
C. 203.61 deg.
Given
: C = 72 inches ; D2 = 12 in ; D1 = 6 in
Formula/s
:
Solution
:
D. 175.22 deg.
2) Determine the belt length of an open belt to connect the 6 cm and 12 cm diameter pulley at a center distance of 72 cm. B. 160.39 cm
A. 172.39 cm
C. 184.39 cm
Given
: C = 72 inches ; D2 = 12 in ; D1 = 6 in
Formula/s
:
Solution
:
D. 190.39 cm
3) A 12 cm pulley turning at 600 rpm is driving a 20 cm pulley by means of belt. If total belt slip is 5%, determine the speed of driving gear. A. 360 rpm
B. 342 rpm
Given
: D1 = 12 cm , D2 = 20 cm
Formula/s
:
Solution
:
C. 382 rpm
, N1=600rpm
D. 364 rpm
4) The torque transmitted in a belt connecte d 300 mm diameter pulley is 4 KN-m . Find the power driving the pulley if belt speed is 20 m/sec. A. 358.88 KW Given
B. 565.88 KW.
D. 433.33 KW
: v = 20 m/sec , d = 300 mm --- r = d/2 =300 mm/2 = 150 mm , T = 4 kN-m
Formula/s
C. 533.33 KW
:
,
Solution
:
5) A 3/8 inch flat belt is 12 inches wide and is used on 24 inches diameter pulley rotating at 600 3
rpm. The specific weight of belt is 0 .035 lb/in . The angle of contact is 150 degrees. If coefficient of friction is 0.3 and stress is 300 psi, how much power can it deliver? A. 65.4 Hp Given
B. 69.5 Hp
C. 60.5 Hp
: t = 3/8 in. ; d = 24 in ; b = 12 in ; f=0.3 ; N = 600 rpm ;
; ;
Formula/s
D. 63.5 Hp
:
;
;
;
------
;
Solution
:
6) A belt connected pulleys has 10 cm diameter and 30 cm diameter. If center distance is 50 cm, determine the angle of contact of smaller pulley. A. 152 deg.
B. 154 deg.
C. 157 deg.
Given
: C = 50 inches ; D2 = 30 cm in ; D 1 = 10 cm
Formula/s
:
Solution
:
D. 159 deg.
Brake 1) A brake has a difference in band tension of 4 KN. The drum diameter is 1 meter and rotating at 300 rpm. Determine the power needed to drive the drum. A. 54.62 KW
B. 56.85 KW
C. 62. 83 KW
D. 64.86 KW
Given
: F1 – F2 = 4kN ; D = 1 m , r = D/2 = 1m/2 = 0.5 m ; N = 300 rpm
Formula/s
:
Solution
:
;
2) In a brake, the tension on tight side is thrice the slack side. If coefficient of friction is 0.25, find the angle of contact of the band. A. 240.61 deg.
B. 251.78 deg.
Given
: f = 0.25
Formula/s
:
Solution
:
C. 286.75 deg.
D. 275.65 deg.
----- ln = 1 so ; ;
3) On a brake drum the difference in tension between the slack side and tight side is 5. If the ratio in band tension is 3, determine the tension in tight side. A. 5.0 KN
B. 5.5 KN
C. 6.5 KN
Given
: F1 – F2 = 5 ;
Solution
: F2 = F1 – 5 , by substitution in
D. 7.5 KN
; -----
-------
4) A steel band have a maximum tensile stress o f 55mpa and thickness of 4 mm. If the tension in tight side is 6 KN, what width of band should be used? A. 25.64 mm
B. 27.27 mm
C. 28.28 mm
D. 29.29 mm
Given
:
Formula/s
:
Solution
:
; t = 4 mm ; F = 6 kN 1
then
5) A band brake has a straight brake arm of 1.5 m and is placed perpendicular to t he diameter bisecting the angle of contact of 27 0 degrees which is 200 mm from the end of slack side. If 20 0 N force is applied at the other end downward of brake arm, determine the tension at slack side. A. 2121.32 N
B. 4646.32 N
Given
:
Formula/s
:
Solution
:
C. 3131.32 N
D. 4141.32 N
; ;
;
(No Answer in Choices)
6) A band brake has a 76 cm diameter drum sustains a load of 1 Mton to a hoisting drum 50 cm in diameter.What is the band tension difference? A. 657.89 kg.m
Given
:
Formula
:
Solution
:
B. 785.98 kg.m
C. 686.86 kg.m
D. 948.71 kg.m
; Load (F)= 1Mton = 1000 kg ;
; ( ) ;
Clutch 1) A cone clutch has an angle of 12 degrees and coefficient of friction of 0.42. Find the axial force required if the capacity of the clutch is 8 KW at 500 rpm. The mean diameter of the active conical sections is 300 mm. Use uniform wear me thod. B. 604.27 N
A. 504.27 N
C. 704.27 N
D. 804.27 N
Given
: P = 8 kW ; N = 500 rpm ;α = 12 deg ;f = 0.42 ; D f = 300 mm -- rf = Df /2 =150mm
Formula/s
:
Solution
:
; ;
2) How much torque can a cone clutch transmit if the angle of the conical elements is 10 degrees. The mean diameter of conical sections is 20 0 mm and an axial force of 600 N is applied. Consider a coefficient of friction of 0.45. A. 135.49 N.m
B. 155.49 N.m
C. 175.49 N.m
D. 195.49 N.m
Given
: Fa= 600N ; f = 0.45 ; α = 10 deg ; D f = 200 mm ---- rf = Df /2 =200mm/2=100mm
Formula/s
:
Solution
:
3) A clutch has an outside diameter of 8 in and inside diameter of 4 in. An ax ial force of 500 lb is used to hold the two parts together. If friction is 0.4, how much t orque can the clutch handle? A. 322.22 in-lb
B. 422.22 in-lb
C. 522.22 in-lb
Given
: Fa= 500 lb ; f = 0.4 ; D = 8 in ; d = 4 in
Formula/s
:
Solution
:
D. 622.22 in-lb
; – –
4) A disc clutch has 6 pairs of co ntacting friction surfaces. The frictional radius is 2 in and the coefficient of friction is 0.30. An axial force of 1 00 lb acts on the clutch. The shaft speed is 400 rpm. What is the power transmitted by the clutch? A. 1.28 HP
C. 3.28 HP
B. 2.28 HP
Given
: Fa= 100 lb ; f = 0.3 ; rf = 2 in ; Nc = 6 ; N = 400 rpm
Formula/s
:
Solution
:
D. 4.28 HP
;
5) A cone clutch has cone elements at an angle of 12 degrees . The clutch transmit 25 HP at a speed of 1200 rpm. The mean diameter of the conical friction section is 16 in and the coefficient of friction is 0.35. Find the axial force needed to engage the clutch. A. 238.04 lbs
B. 248.04 lbs
C. 258.04 lbs
D. 268.04 lbs
Given
: P = 25 kW ; N = 1200 rpm ;α = 12 deg ;f = 0.35 ; Df = 16 in -- rf = Df /2 =8 in
Formula/s
:
Solution
:
; ; ;
Bearing 1)
The main bearing of a one cylinder steam are 152 mm diameter by 280 mm long and support a load of 4400 kg. Find the bearing str ess. B. 517.10 kpa
A. 507.10 kpa
C. 527.10 kpa
D. 537.10 kpa
Given
: m = 4400 kg ; D = 152 mm ; L = 270 mm
Formula/s
:
Solution
: Bearing = 2 for cylinder steam so mass will be distributed in 2
;
2) A bearing 150 mm diameter and 3 00 mm long supports a load of 5000 k g. If coefficient of friction is 0.18, find the torque required to rotate the shaft. A. 331 N-m
C. 873 N-m
B. 662 N-m
Given
: m = 5000 kg ; D = 150 mm ; L = 300 mm ; f b = 0.18
Formula/s
:
Solution
:
D. 1020 N-m
;
3) A bearing journal rotates at 460 rpm is use to support a load of 50 KN. It has a diameter of 20 cm and length of 40 cm. Find the friction loss in kw per bearing. Use f = 0.12. A. 20.45 KW
Given
B. 18.45 KW : N = 460 rpm ; F = 50 kN ;
:
Solution
:
D. 14.45 KW
; f = 0.12 ; ;
L
Formula/s
C. 16.45 KW
;
;
Assume there are two bearings -----
4) A bearing has a per unit load of 5 50 Kpa. The load on bearing is 20 KN and it has a diametral ratio of 0.0012. If diametral clearance is 0.120 mm, find the length of journal. A. 163.63 mm
B. 263.63 mm
Given
:
Formula/s
:
Solution
:
C. 363.63 mm
; F = 20 kN ; Cd = 0.12 mm ;
D. 463.63 mm
;
5) A bearing whose shaft rotates at 50 0 rpm, has a friction loss of 15 KW. The bearing load is 30 KN and friction of 0.14. Find the bearing diameter. B. 146.42 mm
A. 136.42 mm
C. 156.42 mm
D. 166.42 mm
f = 0.14 ; ; D = 2r
Given
: N = 500 rpm ; F = 30 kN ;
Formula/s
:
Solution
:
;
;
----
D = 2(68.21 mm)= 136.42 mm
6) A shaft revolving at 1740 rpm is supported by bearing with a length of 105 mm and diameter of 64 mm. If the load is light and SAE Oil No. 20 (u = 2.4 x 10 -6 reyns) is used and diametral clearance is 0.136 mm, find the power loss due to friction. A. 164 watts
Given
B. 174 watts
C. 184 watts
D. 194 watts
; ;
: ns = 1740 rpm ;
Formula/s
:
; ; ; ;
;
Solution
:
Roller Chains 1)
A chain and sprocket has 18 t eeth with chain pitch of 1/2 in. Find the pitch diameter of sprocket. A. 0.879 in
B. 1.879 in
Given
:
Formula/s
:
Solution
:
C. 2.879 in
D. 3.879 in
;
2) A chain and sprocket has 24 teet h with chain pitch of ½ in. If the sprocket turns at 600 rpm, find the speed of chain. B. 621.72 fpm
A. 601.72 fpm
Given
:
Formula/s
:
;
Solution
:
;
C. 641.72 fpm
D. 661.72 fpm
; N = 600 rpm
3) A chain and sprocket has a pitch diamete r of 9.56 in and a pitch o f ¾ in. How many teeth are there in sprocket? A. 25
B. 30
Given
:
Formula/s
:
Solution
:
C. 35
D. 40
;
4) A chain and sprocket has a pitch diamete r of 28.654 in and there are 90 teeth available. Find the pitch of the chain. A. ½ in Given
B. ¾ in :
C. 1 in
;
D. 1 ¼ in
Formula/s
:
Solution
:
5) A fan require at least 4.5 hp to deliver 18,000 CFM of air r unning at 320 rpm. For a service factor of 1.15, find the designed horsepower of the sprocket. A. 4.5 hp
B. 4.84 hp
Given
:
Formula/s
:
Solution
:
C. 6.34 hp
D. 5.175 hp
; N = 1..15
6) A 20-tooth driving sprocket that rotates at 600 rpm and pitch chain of ½ in drives a driven sprocket with a speed of 200 rpm. Find the diameter of the driven sprocket. A. 7.55 in
B. 8.55 in
Given
:
Formula/s
:
Solution
:
C. 9.55 in
; N1 = 600 rpm ; N2 = 200 rpm; p = 0.5 in ;
D. 10.55 in
Wire Ropes 1)
A 6 x 19 IPS wire rope with rope diameter of 44.5 mm is to be used for hoisting. Find the weight of the rope if the depth of mine hoist is 100 m. A. 1411.31 lbs
B. 1511.31 lbs
Given
:
Formula/s
:
Solution
:
D. 1711.31 lbs
C. 1611.31 lbs
;
2) A 6 x 19 IPS wire rope has a wire rope diameter of 50.8 mm and sheave diameter of 40 in. Find the bending stress of the rope. A. 96,500 psi Given
B. 97,500 psi :
Formula/s Solution
C. 98,500 psi
D. 99,500 psi
; D = 40 in ;
;
:
:
3) A 6 x 19 IPS wire rope has a wire rope diameter of 2 in and sheave diameter of 50 in. Find the equivalent bending load of the rope. A. 124,800 lbs Given
B. 134,800 lbs :
; D = 50 in ;
C. 144,800 lbs
D. 154,800 lbs
Formula/s
:
Solution
:
;
;
;
4) A mine hoist is to carry a cage loaded with ore at a total load of 8 tons. The depth of mine shaft is 328 ft. A 6 x 19 IPS wire rope with rope diameter of 1.75 in is to be use with sheave diameter of 39 in. The cage loaded is to start from rest and will attain a maximum velocity of 525 fpm in 10 seconds. Find the total load of the rope. A. 12,086 lbs Given
B. 14,086 lbs :
C. 16,086 lbs
D. 18,086 lbs
; ; ---- (for downward) ; d = 1.75 in ;
;
Formula/s
Solution
:
:
(No answer in choices)
5) A wire rope is used with total load of 9000 kg and bending load of 50, 000 kg. If bearing strength of the rope is 110,000 kg, find the factor of safety of the rope. A. 4.67
C. 8.67
B. 6.67
Given
:
Formula/s
:
Solution
:
D. 10.67
;
;
