Machine-design-refresherpdf compress PDF

Preview

Summary

Machine Design & Shop Practice Refresh Trivia # 1MACHINE DESIGN AND SHOP PRACTICE (Refreshers Trivia # 1) Prepared by: Jose R. Francisco, PME September 2012 INSTRUCTION : Select the correct answer for each of the following questions. Mark only one answer for each item by shading the box correspo...

Description

Machine Design & Shop Practice

Refresh Trivia # 1

JCSF ENGINEERING REVIEW CENTER 2nd Floor Santos-Causing Building, National Highway Halang, City of Calamba, Laguna MACHINE DESIGN AND SHOP PRACTICE (Refreshers Trivia # 1) Prepared by: Jose R. Francisco, PME September 2012 INSTRUCTION: Select the correct answer for each of the following questions. Mark only one answer for each item by shading the box corresponding to the letter of your choice on the answer sheet provided. STRICTLY NO ERASURES ALLOWED. Use pencil No. 1 only 1.

o

If the angular deformation of a solid shaft should not to exceed 1 in a length of 1.8 m and the allowable shearing stress is 83 MMa, what is the 6 diameter of the shaft? Assume that the shaft material has G = 77 x 10 kPa. a) 222.34 mm b) 234.22 mm c) 23.42 cm d) 24.22 cm

⎛ πD 3s s ⎞ ⎟L 32⎜ ⎜ 16 ⎟ TL 32T L ⎝ ⎠ = 2s sL Solution: θ = = = JG π D4 G DG π D 4G 2.

E=

(

D =3

53.5 P N

=3

)

)

Where,

d) 3.28 inches

53.5(200) = 2.28 inches 900

For Main Power Transmitting Shaft:

For Small, Short Shaft:

P=

P=

D3 N 80

For Line Shaft Carrying Pulleys:

P=

D3 N 53.5

D3 N 38

P = transmitted power, Hp

N = shaft rpm

D = shaft diameter, in inches

A round steel shaft rotates at 200 rpm and is subjected to a torque of 275 N-m and a bending moment of 415 N-m. Determine the equivalent twisting moment. a) 597.84 N-m b) 456.42 N-m c) 546.43 N-m d) 497.85 N-m Solution:

ME =

M2 + T2 =

( 415) 2 + ( 275) 2

= 497.85 N ⋅ m 2

A vertical steel cylinder water tank is 30 m in diameter and 45 m high. The allowable stress of the steel plate is 1224 kg/cm . Without reinforcing angle bars and rods, what is the thickness of the steel plate? a) 55.15 mm b) 51.55 mm c) 65.15 mm d) 61.55 mm Solution:

Where, 6.

(

F 3163.27 kg / cm 2 (2.205 lb / kg )(2.54 cm / in )2 Stress = 42 857142.86 psi = 43 x 10 6 psi = A = δ 0.00105 Strain L

A line shaft is to transmit 200 Hp at 900 rpm. Find the diameter of the shaft. a) 2.18 inches b) 2.28 inches c) 3.18 inches

Note:

5.

2(83 000 )(1.8 ) = 222.34 mm π ⎞ 6 1 ⎜⎜ 77 x 10 ⎟ ⎟ ⎝ 180 o ⎠ o⎛

2

Solution:

4.

2 ss L = θG

What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 from a load producing a unit tensile stress of 3163.27 kg/cm ? 6 6 6 6 a) 40 x 10 psi b) 43 x 10 psi c) 45 x 10 psi d) 46 x 10 psi Solution:

3.

D=

t=

(

)

pD 441 297 N / m 2 (30 m ) = = 0.05515 m = 55.15 mm 2 s t E j 2 1224 kg / cm 2 (9.8066 N / kg )(100 cm / m )2

(

)

p = maximum pressure inside the tank, Pa

(

)(

)

p = ρ g H = 1000 kg / m 3 9.8066 m / s 2 (45 m) = 441 297 Pa

st = hoop stress of the tank, Pa The root diameter of a double square thread is 0.55 inch. The screw has a pitch of 0.2 inch. Find the outside diameter and the number of threads per inch. a) 0.75 inch and 5 threads/inch b) 0.50 inch and 5 threads/inch c) 0.75 inch and 4 threads/inch d) 0.50 inch and 4 threads/inch Solution: For the number of threads per inch, p

=

1 Number of Threads per inch

Number of threads per inch = For the major diameter,

Where,

1 1 = =5 p 0.2

Where, p = the pitch

⎛p⎞ D o = D i + 2h = D i + 2⎜ ⎟ = Di + p = 0.55 + 0.2 = 0.75 inch ⎝2⎠

h = height or depth of thread =

p 2

1

→ for square thread

Machine Design & Shop Practice 7.

Refresh Trivia # 1

A flat belt is 6 inches wide and 1/3 inch thick and transmits 15 Hp. The center distance is 8 ft. The driving pulley is 6 inches in diameter and rotates at 2 3 000 rpm such that the loose side of the belt is on top. The driven pulley is 18 inches in diameter. The belt material is 0.035 lb/in and the coefficient of friction is 0.30. Determine the belt net tension. a) 175.5 lb b) 157.5 lb c) 155.7 lb d) 165.7 lb Solution: F = F1 − F2 =

2T 2 ⎛ 63 000 Hp ⎞ ⎛ 2 ⎞ ⎡ 63 000 (15 )⎤ = ⎜ ⎟ = ⎜ ⎟⎢ ⎥ = 157.5 lb D D⎝ n ⎠ ⎝ 6 ⎠ ⎣ 2 000 ⎦

33 000 Hp 33 000(15) ⎛ 6 ⎞(2 000) 3141.59 fpm F F F = 1− 2 = = = 157.56 lb ⎟ = Vm 3141.59 ⎝12 ⎠

Other Solution: Vm = πD n = π⎜ 8.

A right-handed single-thread hardened-steel worm has a catalog rating of 2.25 kW at 650 rpm when meshed with a 48-tooth cast-steel gear. The axial o pitch of the worm is 25 mm, normal pressure angle is 14.5 , and the pitch diameter of the worm is 100 mm. The coefficient of friction is 0.085. Determine the shafts center distance. a) 241 mm b) 142 mm c) 412 mm d) 124 mm Solution:

Speed Ratio, SR = Tanλ =

L πD w

=

Tg Dg cos λ Dg ω w nw = = = = n T D sin λ D w tan λ ωg g w w

p 25 = = 0.07958 πD w π(100)

→ λ = 4.55o

⎛ Tg ⎞ ⎛T ⎞ ⎛ ⎞ ⎛T ⎟D w tan λ = ⎜ g ⎟ D w⎜ p ⎟ = ⎜ g ⎜ πD ⎟ ⎜ T ⎟ ⎜ ⎟ T w ⎠ ⎝ w ⎝ w ⎠ ⎝ w⎠ ⎝ Dw + Dg 100 + 381.97 Center Distance, C = = 241 mm = 2 2

Pitch diameter of the gear, D g = ⎜ ⎜T

9.

⎞ ⎛ p ⎞ ⎛ 48 ⎞⎛ 25 ⎞ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ = 381.97 mm ⎟ ⎝ π ⎠ ⎝ 1 ⎠⎝ π ⎠ ⎠

o

A 20 straight-tooth bevel pinion having 14 teeth and a diametral pitch of 6 teeth/inch drives a 42-tooth gear. The two shafts are at right angles and in the same plane. Find the pitch angle of the pinion. o o o o a) 18.4 b) 20 c) 14.5 d) 20.5

⎛ Tp ⎞ ⎟ = tan −1⎛⎜ 14 ⎞⎟ = 18.4 o ⎜ Tg ⎟ ⎝ 42 ⎠ ⎠ ⎝

1 Solution: γ = tan − ⎜

10. A double-thread worm has a pitch diameter of 3 inches. The wheel has 20 teeth and a pitch diameter of 5 inches. Find the gear helix angle. o o o o a) 4.69 b) 9.46 c) 6.49 d) 6.94 Solution:

⎡⎛ T λ = tan −1 ⎢ ⎜⎜ w ⎢⎣ ⎝ Tg

⎞⎛ D g ⎞⎤ ⎡ 2 ⎛ 5 ⎞⎤ ⎟ ⎟⎟⎜ tan −1 ⎢ ⎜ ⎟ ⎥ = 9.46 o ⎜ D ⎟⎥⎥ = ⎠⎝ w ⎠⎦ ⎣ 20 ⎝ 3 ⎠⎦ o

11. A 36-tooth pinion turning at 300 rpm drives 120-tooth gear of 14.5 involute full depth pressure angle. Determine the rpm of the driven gear. a) 60 rpm b) 45 rpm c) 75 rpm d) 90 rpm

⎛T ⎞

⎛ 36 ⎞ p ⎟ Solution: n g = n p ⎜ = (300 )⎜ ⎟ = 90 rpm ⎜ Tg ⎟ ⎝ 120 ⎠ ⎝

⎠

12. If two parallel shafts are connected by cylinders in pure rolling contact and turning in the same direction, and having a speed ratio of 2.75, what is the Center distance of the two shafts assuming that the diameter of the smaller cylinder is 22 cm? a) 18.25 cm b) 19.25 cm c) 20.25 cm d) 17.25 cm Solution:

Diameter of the bigger cylinder, D 2 = SR (D 1 ) = 2.75 (22 ) = 60.5 cm Center distance, C =

D 2 − D 1 60.5 − 22 = 19.25 cm = 2 2

13. Two extension coil springs are hooked in series that support a single weight of 100 kg. The first spring is rated at 4 kN/m and the other spring is rated at 6 kN/m. Determine the total deflection of the springs. a) 408.6 mm b) 486.0 mm c) 480.6 mm d) 460.8 mm Solution:

δt = δ1 + δ2 =

⎛ k + k2 ⎞ F F ⎡4 + 6⎤ ⎟ = (100 kg)( 9.8066 N / kg)⎢ + = F⎜⎜ 1 ⎥ = 408.6 mm ⎟ k 1 k2 ⎣ 4( 6) ⎦ ⎝ k1 k 2 ⎠

14. A double thread ACME screw driven by a motor at 400 rpm raises the attached load of 900 kg at a speed of 10 m/min. The screw has a pitch diameter of 36 mm; the coefficient of friction on threads is 0.15. The friction torque on the thrust bearing of the motor is taken as 20 % of the total input. Determine the lead angle. o o o o a) 12.465 b) 14.265 c) 15.462 d) 16.452

V 10 = = 0.025 m = 25 mm n 400 −1⎛ L ⎞ − 1⎛ 25 ⎞ o ⎟ For the lead angle, λ = tan ⎜ ⎜ πD ⎟ = tan ⎜⎝ 36 π⎟⎠ = 12.465 m ⎠ ⎝

Solution: For the lead,

L=

15. A cylinder having an internal diameter if 508 mm and external diameter if 914.4 mm is subjected to an internal pressure of 69 MPa and an external pressure of 14 MPa. Determine the hoop stress at the inner surface of the cylinder. a) 90.11 MPa b) 91.10 MPa c) 911.0 MPa d) 19.10 Mpa

2

Machine Design & Shop Practice

Refresh Trivia # 1 2 2 ⎞ ⎛ 508 ⎞ ⎤ ⎛ 914.4 mm⎟ ⎥ − 2(14 MPa)⎜ mm ⎟ +⎜ ⎠ ⎠ ⎝ 2 ⎠ ⎥⎦ ⎝ 2

⎡

Solution:

s ti =

(

)

p i r o2 + ri2 − 2p o r o2 r o2

− r i2

(69 MPa )⎢ ⎛⎜ 914.4 mm⎞⎟ ⎢⎣ ⎝

=

2

2

2

⎞ ⎞ ⎛ 508 ⎛ 914.4 mm ⎟ mm ⎟ − ⎜ ⎜ ⎠ ⎠ ⎝ 2 ⎝ 2

2

sti = 90.11 Mpa sti = maximum tangential or hoop stress at the inside pi = internal pressure, Mpa ro = outside radius, mm Note: For the maximum tangential or hoop stress at the outside, Where,

ri = inside radius, mm po = external pressure, Mpa

16. A flat belt is 6 inches wide and 1/3 inch thick and transmits 15 Hp. The center distance is 8 ft. The driving pulley is 6 inches in diameter and rotates at 2 3 000 rpm such that the loose side of the belt is on top. The driven pulley is 18 inches in diameter. The belt material is 0.035 lb/in and the coefficient of friction is 0.30. Determine the belt net tension. a) 175.5 lb b) 157.5 lb c) 155.7 lb d) 165.7 lb Solution:

F = F1 − F2 =

2T D

2 ⎛ 63 000 Hp ⎞ ⎛ 2 ⎞⎡ 63 000 (15 )⎤ ⎟ = ⎜ ⎟⎢ ⎜ ⎥ = 157.5 lb D⎝ n ⎠ ⎝ 6 ⎠⎣ 2 000 ⎦

=

17. What is the polar section modulus of a solid shaft with a diameter of 101.6 mm? 3 4 3 a) 209.5 cm b) 209.5 cm c) 205.9 cm

4

J 2J π 3 π = = D = (10.16)3 = 205.9 cm3 c D 16 16

Zj =

Solution:

d) 205.9 cm

18. The transmitted torque of a hollow shaft is 3400 N-m at a shearing stress of 55 MPa. If the outside diameter is 1.25 times that of the inside diameter, what is the inside diameter, in mm? a) 64.87 b) 46.87 c) 84.67 d) 74.64

ss =

Solution:

16T

(

πD 1 3

D=3

− β4

)

where, β =

)

=

16 T

(

πs s 1 − β

4

3

Di 1 = = 0.8 D o 1.25

16( 3.4)

[

π(55 000) 1 − (0.8)

4

]

Di =

= 81.092 mm

Do 81.092 = = 64.87 mm 1.25 1.25

19. What is the stress area of a 1-inch diameter bolt that has a TPI (Threads per inch) of 8? 2 2 2 a) 0.565 in b) 0.626 in c) 0.443 in Solution: A = π [ D − 0.9743p]2 : → p = 1 : s

TPI

4

d) 0.606 in

2

→ As = 0.606 in 2

20. Determine the Poisson’s ratio of a material whose modulus of elasticity is 200 GPa and whose modulus of rigidity is 80 GPa. a) 0.33 b) 0.25 c) 0.38 d) 0.22 Solution: G

=

E 2(1 + ν)

E = 200GPa,

∴ν = 0.25

G = 80GPa:

21. A steel has a BHN = 300. What is its approximate ultimate strength in ksi? a) 300 ksi b) 150 ksi c) 75 ksi

d) 200 ksi

Solution: Su ≈ 0.5(BHN), ksi 22. What is the pitch of an American Standard Screw Threads with designation 12-28 UNF? a) 0.083 inch b) 0.0357 inch c) 0.5 inch

d) 0.75 inch

1 1 1 Solution: p = = = = 0.0357 inch n No. of threads per inch 28 23. A thrust washer has an inside diameter of 12.7 mm and an outside diameter of 76.2 mm. For an allowable bearing pressure of 90 psi, determine the axial load that the washer can sustain. a) 618.5 lb b) 537.2 lb c) 702.2 lb d) 871.2 lb

(

)

π ⎛ π⎞ 2 ⎟ D o − D 2i = (90 ) 4 4 ⎝ ⎠

Solution: F = pA = p⎜

⎡⎛ 76.2 ⎞ 2 ⎢⎜ ⎟ − ⎢⎣⎝ 25.4 ⎠

⎛ 12.7 ⎞ ⎜ ⎟ ⎝ 25.4 ⎠

2⎤

⎥ = 618.5 lb ⎥⎦

24. What is the number of threads per mm and the tensile stress area of a standard Metric screw Thread designated by M10 x 1.5? 2 2 2 2 a) 0.555 and 57.99 mm b) 0.667 and 57.99 mm c) 1.5 and 57.99mm d) 1.75 and 57.99 mm Solution: n =

1 1 = = 0.667 threads per mm p 1.5

For Standard Metric Screw Threads, A t = 0.7854 (D − 0.9383p )2 = 0.7854 10 − 0.9383(2.5) 2 = 57.99 mm 2

[

]

2

Where At = tensile stress Area, mm ; D = basic major diameter, mm; p = pitch, mm Note: o M stands for standard metric screw threads; 10 stands for the basic major diameter, which is 10 mm; 1.5 stands for pitch that is 1.5 mm. 2

25. An air cylinder has a bore of 25 mm and is operated with shop air at a pressure of 6.327 kg/cm . Find the push force exerted by the piston rod, in N.

3

Machine Design & Shop Practice a) 127.57

Refresh Trivia # 1

(

b) 70.42

c) 402.75

)

d) 304.57

⎛ π⎞ 2 ⎛π⎞ 2 2 Solution: F = pA = p⎜ ⎟D = 6.327 kg / cm (9.8066 N / kg )⎜ ⎟ (2.5 cm ) = 304.57 N ⎝ 4⎠ ⎝4⎠ 26. A single plate clutch has an outside diameter of 250mm and a coefficient of friction of 0.3 between the surfaces. Using the uniform wear method, what is its maximum power transmitting capacity at 1000 rpm if the lining pressure is not to exceed 0.4 MPa? a) 38.1 kW b) 27.9 kW c) 29.7 kW d) 31.8 kW Solution:

⎡( D + d ) ⎤ T = Pf ⎢ ⎥ N fs : ⎣ 4 ⎦ To maximize T, take

d (D − d ) 2 D → thus, d = 3

→ P = πp max dT = 0 dd

Then P = 9583N; → T =283411N −mm :

→ d = 144.34mm

→Power =

T (n ) =29.7kW 9.549x10 6

27. Calculate the resultant bending load on a shaft that carries a 200mm-diameter, 20º full depth pinion. The pinion transmits 10 kW at 1750 rpm. a) 258.0 N b) 685.0 N c) 850.2 N d) 580.2 N

⎡ 10 ⎤ 2⎢ x9.549x106 ⎥ 2 T 1750 ⎣ ⎦ : Solution: Ft = = D 200

→ Ft = 546.7 N

→ Fr = Ft tan φ = 198.6 N :

→ FR = Ft 2 + Fr2 = 580.2 N o

28. A pulley 600 mm in diameter transmits 40 kW at 500 rpm. The arc of contact between the belt and pulley is 144 , the coefficient of friction between belt and pulley is 0.35 and the safe working stress of the belt is 2.1 MPa. Determine the belt tensions ratio, neglecting the effect of centrifugal force. a) 2.41 b) 2.14 c) 1.24 d) 4.12 ⎛ π ⎞ F1 0.35( 144)⎜ θ ⎟ = ef = ( e) ⎝ 180 ⎠ = 2.41 F2

Solution:

29. A roller chain and sprocket is to drive vertical centrifugal discharge bucket elevator. The pitch of chain connecting sprockets is 1.75”. The driving sprocket is rotating at 120 rpm and has 11 teeth while the driven sprocket is rotating at 38 rpm. Determine the number of teeth of driven sprocket. a) 33 teeth b) 35 teeth c) 30 teeth d) 34 teeth Solution:

⎞ 120 ⎞ ⎟ = (11)⎛⎜ ⎟ = 34.74 ≈ 35 teeth ⎟ ⎝ 38 ⎠ ⎠

⎛n T2 = T1⎜⎜ 1 ⎝ n2

30. A 20-tooth motor sprocket, running at 1200 rpm, drives a blower at a speed ratio of 4:1. Using the largest permissible chain size and the largest permissible center distance of 80 pitches, what length of chain in pitches is required to connect the sprockets? a) 200 pitches b) 212 pitches c) 216 pitches d) 220 pitches 2

Solution: o

⎡ 900 ⎤ 3 p max = ⎢ ⎥ = 0.825in. : largest RC no. is RC 60 ⎣ n1 ⎦

N t1 + N t 2 ( N − N t1) = 212.pitches + 2Cp + t 2 40Cp 2 2

Lc =

31. A 20 involute spur gear has a tooth whole depth of 16.95 mm, a tooth thickness of 13.2 mm, and a pitch of 3. Determine the circular pitch of the gear. a) 26.6 mm b) 16.6 mm c) 25.6 mm d) 24.6 mm Solution:

Pc =

π π = = 1.0472 inches = 26.6 mm Pd 3 th

32. Which of the following is the 4 most commonly used metal in the world? a) Zinc b) Steel c) Aluminum d) Copper 33. These springs are made from one or more flat strips of brass, bronze, steel or other materials loaded as cantilevers or simple beam. a) Torsion springs b) Leaf springs c) Garter springs d) Drawbar springs 34. This refers to the space between adjacent coils when the spring is compressed to its operating length. a) Coil clearance b) Pitch c) Lead d) Deflection 35. This material is the most popular alloy spring steel for conditions involving higher stresses than can be used with the high-carbon steels and for use where fatigue resistance and long endurance are needed; this is also good for shock and impact loads. a) Chrome silicon b) Chrome vanadium c) hard-drawn wire d) Oil-tempered wire 36. For an American Standard Screw Threads, what does 6-32 UNC designate? a) Size 6, 32 threads per inch, coarse thread b) 6 inches basic diameter, 32 threads per inch, coarse thread c) Size 6, 32 threads per inch, fine thread d) 32 inches basic diameter, 6 threads per inch, coarse thread • Note: 6 stands for the designated size, 32 stands for the number of threads per inch, UNC stands for Coarse threads 37. How do you call the process of producing the residual compressive stress of machine parts, which is performed by directing the a high velocity stream of hardened balls or pellets at the surface to be treated. a) Nitriding b) Shot blasting c) Peening d) Tempering 38. It is a process that produces residual compressive stress on the machine part, which uses a series of hammer blows on the surface. a) Nitriding b) Shot blasting c) Peening d) Tempering o 39. It is a surface-hardening process for alloy steels in which the material is heated to 950 F in a nitrogen atmosphere, typically ammonia gas, followed by slow cooling. a) Quenching b) Nitriding c) Shot blasting d) Peening 40. Which of the following gases is typically used in nitriding process of surface hardening? a) Nitrogen gas b) Carbon dioxide c) Ammonia gas d) Hydrogen gas 41. Which of the following is the benefit in using nitriding as a surface-hardening process for alloy steels? a) Improvement of endurance strength, 50 % or more b) Improvement of endurance strength, less than 50 %

4

Machine Design & Shop Practice

Refresh Trivia # 1

c) Improvement of endurance strength, more than 50 % d) 80 % improvement on the endurance strength 42. In estimating the actual endurance strength of steel parts, one of the factors to be considered is the material factor, which of the following is the recommended material factor for cast steel? a) 0.70 b) 0.80 c) 0.75 d) 1.0 43. How do you call the level of stress that the part will be permitted to see under operating conditions? a) Yield stress b) Endurance stress c) Design stress d) Ultimate stress 44. Which of the following column formulas is applicable to cast iron columns? a) Euler’s formula b) J.B.Johnson’s formula d) Secant formula d) Straight line formula o

Ans. D. Straight line formula. Cast iron columns are usually designed on the basis of

P ⎡L ⎤ = 9000 − 40 ⎢ e ⎥ A ⎣k ⎦ o

→ a Straight line formula

Where the slenderness ratio Le/k should not exceed 70.

45. Which of the following ferrous metals has the lowest carbon content? a) Carbon steel b) Wrought iron c) Cast iron d) SAE 4140 o Wrought iron usually contains less than 0.04% C; steel usually has less than 2.5% C; cast iron has more than 1.7% C; SAE 4140 has approximately 0.4% C. 46. If stiffness is the main criterion in selecting a material, which of the following is the most economical choice? a) SAE 3130 b) SAE 1020 c) SAE 6150 d) AISI 301, ¼ hard stainless steel o Ans. B. SAE 1020- plain carbon steel. All of the above materials are steel with practically equal modulus of elasticity. The three other materials are alloy steels that are relatively more expensive. 47. Which of the following materials can easily be machined? a) AISI C1020 b) AISI C1112 c) ...