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MACHINE DESIGNSpur Gear Find the tooth thickness of a 14 deg. Involute gear having a diametral pitch of 6. A. 5 mm B. 6 mm C. 8 mm D. 12 mm Given :DP = 6 in-Formula/s ,Solution := 0 in= 0 in0 in xx= 6 mm A gear set having a gear ratio of 3 is to be used at a center distance of 10 inches. If the gear...

Description

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 B. 6.65 mm C. 8.45 mm D. 12.36 mm Given

:DP = 6 in-1

Formula/s Solution

, :

= 0.5236 in = 0.2618 in x

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. B. 0.7851 in. C. 0.8970 in. D. 0.7283 in. Given

: C = 10 in. , T2 = 60 , Gear Ratio

Formula/s

:

,

Solution

:

, since C = 10 in. so

,

=3 ,

By relation, By substitution, D1 = 5 in , D2=15 in

3D1 = D2 then

4D1=20 in

= 4 in-1 = 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 C. 2265 rpm D. 2341 rpm Given

: P = 250 hp ,

Formula/s

:

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. A. 3 in B. 4 in C. 5 in D. 6 in Given

:DP = 8 in-1

Formula/s

:

Solution

:

, T1=12

, 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 of teeth of smaller gear. A. 14 B. 16 C. 12 D. 18 Given

:Gear Ratio = 3 , T1=54

Formula/s

:

Solution

:

so

6) A spur pinion rotates at 1800 rpm 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 C. 271.24 lbs D. 327.43 lbs Given

: P = 30 HP , N = 1800 rpm , D=8 in ,

Formula/s

:

Solution

:

,

,

= 14.5 deg

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 B. 525 C. 535 D. 475 Given

: P = 30 HP , N = 1800 rpm , D=4 in

Formula/s

:

Solution

:

,

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 the final speed of the driven gear and its direction relative to the driving gear rotation? 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 Formula/s

: T1 = 24 , T2 = 96 , N1=400rpm :

Solution

:

, =

9) A spur pinion supported on each side by ball bearings rotates 1750 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 B. 300 C. 360 D. 400 Given

: P = 25 HP , N = 1750 rpm , D=5 in

Formula/s

:

Solution

:

,

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. A. 0.007855 B. 0.008578 C. 0.007553 D. 0.007565 Given

:PC= 0.1571

Formula/s

:

Solution

:

,

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 gear diameter of 3 in. Find the tangential force on the worm. A. 365.37 lbs B. 465.37 lbs C. 565.37 lbs D. 665.37 lbs Given

:P= 10 hp

Formula/s

:

Solution

:

, N=1150 rpm

,

, 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 gear. The worm is left hand threads. A. 597.08 lbs B. 697.08 lbs C. 797.08 lbs D. 897.08 lbs Given

:P= 12 hp , N=1200 rpm

Formula/s

:

,

,

, Solution

, Dw=3 in , Pw=1.125 in ,  = 14.5 deg , f = 0.20

 

:

= 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 with tangential force on gear of 1000 lbs. Find the separation force on gear and worm considering a left hand threads.

A. 171.23 lbs

B. 271.23 lbs

C. 371.23 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 D. 1 ½ in Given

: x= 7.25 degrees , Rw=2.5 in, Dw = 2Rw =2 (2.5 in) =5 in

Formula/s

:

Solution

:

;

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 C. 3.07 in D. 3.37 in Given

: = 78 degrees , p = 0.75 in

Formula/s

:

Solution

:





;



Helical Gear 1) A 20-tooth helical gear has a pitch diameter of 10 in. Find the diametral pitch of the gear. A. 2 B. 3 C. 4 D. 5 Given

: D = 10 in ; T = 20

Formula/s

:

Solution

:

2) A 28 –tooth helical gear having a pitch diameter of 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 D. 0.828 Given

: D = 7 in ; T = 28 ;  = 22 degrees

Formula/s

: 

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 B. 0.335 C. 0.435 D. 0.535 Given

:  = 20 degrees ;  = 23 degrees

Formula/s

: 

 

Solution

:

 

 

4) A 75 hp motor, running at 450 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 B. 2,426 lbs C. 2,526 lbs D. 2,626 lbs Given

: P = 75 hp ; N = 450 rpm ; Dp=8 in

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 the 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 B. 4.31 in C. 5.31 in D. 6.31 in Given

: D = 5 in ; T = 20 ;  = 22 degrees

Formula/s

: 

Solution

: 

Bevel Gear 1) A 20 degrees full depth straight tooth gear has a face width of 3 ¾ in and a pitch diameter of 12 in with cone pitch angle of 37.5 degrees. Find the mean diameter. A. 6.72 in B. 7.72 in C. 8.72 in D. 9.72 in Given

:

Formula/s

:

Solution

:

= 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 C. 1400 lbs D. 1400 lbs Given

:

Formula/s

:

Solution

:

= 40 degrees ; Dp = 14 in ; b = 4 in ; T=8000 lb-in ;

3) A pair of straight tooth bevel gear connect a pair of shafts 90 degrees . The velocity ratio is 3 to 1. What is the cone pitch angle of smaller gear? A. 71.57 deg B. 65.34 deg C. 18.43 deg D. 12.34 deg Given

:

Formula/s

:

Solution

:

4) A spiral bevel pinion with a left hand spiral rotates clockwise transmits power to a mating gear with speed ratio of 2 to 1. Determine the pitch angle of the pinion. A. 16.56 deg B. 20.56 deg C. 26.56 deg D. 32.56 deg Given

:

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 mean radius of the gear. A. 110 lbs B. 120 lbs C. 130 lbs D. 140 lbs Given

: P = 4 hp ; N = 1200 rpm ; Dp avg=3 in

Formula/s

:

Solution

:

;

6) A pair of 4-pitch, 14.5 degrees , 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 C. 25.85 deg D. 27.85 deg Given

:

Formula/s

:

;

; P = 4 in-1

--------;

;

------

Solution

:

(

(

)

)

; (

)

;

Belts 1) Find the angle of contact on the 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. D. 175.22 deg. Given

: C = 72 inches ; D2 = 12 in ; D1 = 6 in

Formula/s

:

Solution

:

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. A. 172.39 cm B. 160.39 cm C. 184.39 cm D. 190.39 cm Given

: C = 72 inches ; D2 = 12 in ; D1 = 6 in

Formula/s

:

Solution

:

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 C. 382 rpm D. 364 rpm Given

: D1 = 12 cm , D2 = 20 cm , N1=600rpm

Formula/s

:

Solution

:

4) The torque transmitted in a belt connected 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 B. 565.88 KW. C. 533.33 KW D. 433.33 KW Given Formula/s

: v = 20 m/sec , d = 300 mm --- r = d/2 =300 mm/2 = 150 mm , T = 4 kN-m : ,

Solution

:

(

)

(

)

5) A 3/8 inch flat belt is 12 inches wide and is used on 24 inches diameter pulley rotating at 600 rpm. The specific weight of belt is 0.035 lb/in3 . 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 B. 69.5 Hp C. 60.5 Hp D. 63.5 Hp Given

: t = 3/8 in. ; d = 24 in ; b = 12 in ; f=0.3 ; N = 600 rpm ;

;

; Formula/s

( 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. D. 159 deg. Given

: C = 50 inches ; D2 = 30 cm in ; D1 = 10 cm

Formula/s

:

Solution

:

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 Formula/s

: F1 – F2 = 4kN ; D = 1 m , r = D/2 = 1m/2 = 0.5 m ; N = 300 rpm : ;

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. C. 286.75 deg. D. 275.65 deg. Given

: f = 0.25

Formula/s

:

Solution

:

;

----- 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 D. 7.5 KN Given

: F1 – F2 = 5 ;

Solution

: F2 = F1 – 5 , by substitution in

;

-----------

4) A steel band have a maximum tensile stress of 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 ; F1= 6 kN

then

5) A band brake has a straight brake arm of 1.5 m and is placed perpendicular to the diameter bisecting the angle of contact of 270 degrees which is 200 mm from the end of slack side. If 200 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 C. 3131.32 N D. 4141.32 N Given

:

Formula/s

:

Solution

:

;

;

; (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 B. 785.98 kg.m C. 686.86 kg.m D. 948.71 kg.m Given

:

Formula

:

Solution

:

; 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 method. A. 504.27 N B. 604.27 N C. 704.27 N D. 804.27 N Given

: P = 8 kW ; N = 500 rpm ;α = 12 deg ;f = 0.42 ; Df = 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 200 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 ; Df = 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 axial force of 500 lb is used to hold the two parts together. If friction is 0.4, how much torque can the clutch handle? A. 322.22 in-lb B. 422.22 in-lb C. 522.22 in-lb D. 622.22 in-lb Given

: Fa= 500 lb ; f = 0.4 ; D = 8 in ; d = 4 in

Formula/s

:

Solution

:

; –

–

4) A disc clutch has 6 pairs of contacting friction surfaces. The frictional radius is 2 in and the coefficient of friction is 0.30. An axial force of 100 lb acts on the clutch. The shaft speed is 400 rpm. What is the power transmitted by the clutch? A. 1.28 HP B. 2.28 HP C. 3.28 HP D. 4.28 HP Given

: Fa= 100 lb ; f = 0.3 ; rf = 2 in ; Nc = 6 ; N = 400 rpm

Formula/s

:

Solution

:

; (

)

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 stress. A. 507.10 kpa B. 517.10 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 300 mm long supports a load of 5000 kg. If coefficient of friction is 0.18, find the torque required to rotate the shaft. A. 331 N-m B. 662 N-m C. 873 N-m D. 1020 N-m Given

: m = 5000 kg ; D = 150 mm ; L = 300 mm ; fb = 0.18

Formula/s

:

Solution

:

; )

(

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 B. 18.45 KW C. 16.45 KW D. 14.45 KW Given

: N = 460 rpm ; F = 50 kN ;

; f = 0.12

L

Formula/s

:

Solution

:

;

;

;

;

(

)

Assume there are two bearings ----4) A bearing has a per unit load of 550 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 C. 363.63 mm D. 463.63 mm Given

:

Formula/s

:

Solution

:

; F = 20 kN ; Cd = 0.12 mm ;

;

)

(

5) A bearing whose shaft rotates at 500 rpm, has a friction loss of 15 KW. The bearing load is 30 KN and friction of 0.14. Find the bearing diameter. A. 136.42 mm B. 146.42 mm C. 156.42 mm D. 166.42 mm Given

: N = 500 rpm ; F = 30 kN ;

Formula/s

:

Solution

:

;

;

; f = 0.14

; D = 2r

----

D = 2(68.21 mm)=136.42 mm 6) A shaft revolving at 1740 rpm is ...