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MACHINE DESIGN SOLVED PROBLEMS
- 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/in^3 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/in^3 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.. b. .007558SOLUTION: c. .008578 Clearance = 0.1571 / 20 = 0. d.. 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 = D^3 N / 38 = 3^3 (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. SOLUTION: Reference form PSME code pp 19- EP Rating: = X0.91^ – (YS/de) – ZS^3 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. Z = 0.
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/in^3 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.. b. .007558SOLUTION: c. .008578 From Vallance p 262, Table 11- d. .007585 c = 0.157/Pd = 0.157/20 = 0. 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 = D^3 N / 38 = 3^3 (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 (0.55d^2 – 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 (0.55d^2 – 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:
Angle of center line (150° arc of contact ….1.0) Normal operating conditions ….. 1. 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 = 2TN 200 = 2T (400/60) T = 4.77465 kN-m L = length = 5m Solving for the polar moment of inertia J: J = /32 (D 04 – Di^4 ) = /32 [(0.140)^4 – (0.124)^4 ] = 1.45042 x 10-5^ m^4 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. 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^ m^4 b. 1.215 x 10-6^ m^4 SOLUTION: c. 1.152 x 10-6^ m^4 J = /32 (D 04 – Di^4 ) d. 1.125 x 10-6^ m^4 J = /32 [(0.06)^4 – (0.035)^4 ] J = 1.125 x 10-6^ m^4 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 800 J= 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 speed of a pulley 55 cm mounted on the shaft.
a. 1796 fpm b. 1766 fpm SOLUTION: c. 1856 fpm For short shaft (PSME Code p 18) d. 2106 fpm P = D^3 N / 38 where D = 61 / 25.4 == 2.4 in. 20 = (2.4)^3 N / 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 10^6 b. 42.300 x 10^6 SOLUTION: c. 41.202 x 10^6 E = stress / strain = (44000)/(0.00105 = 41.905 x 10^6 psi d. 43.101 x 10^6 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. b. 35.89 SOLUTION: c. 38.89 SS = 16t / D^3 = 12,000 = 16T / (2)^3 d. 34.89 T = 18,850 in-lbs = 1,570.8 ft-lbs P = 2TN = [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 mv^2 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. 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. n = 18 – 2 = 16 active coils y = 8FC^3 n / 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?
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. 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 / [(Do^4 – Di^4 )] 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(t 2 – t 1 ) = 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. Clearance = 1.999333 – 1. = 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 10^6 psi b. 41.202 x 10^6 psi SOLUTION: c. 43.101 x 10^6 psi E = Stress / Strain d. 41.905 x 10^6 psi = 44 ,000 / 0.00105 = 41.905 x 10^6 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. b. 3.1831 SOLUTION: c. 0.03141 Tooth thickness = 1.5708/P = 1.5708/5 = 0.31416” d. 0. 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)d^2 )] 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. b. 0.7825 SOLUTION: c. 0.8035 D 1 N 1 = D 2 N 2 d. 0.8085 N 1 / N 2 = D 2 / D 1 = 4 Pd = T/D = T 1 /D 1 = 18/4. D 2 = 4 D 1 = 4. C = (D 1 + D 2 ) / 2 Use std. Pd of 4 10.23 = (D 1 + 4D 1 ) / 2 Pd x Pc = D 1 = 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, b. 14,165 SOLUTION: c. 13,050 L / D = 2.5 A = D x L
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 = 2TN d. 475 30(33,000) = 2T(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. 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. Wahl Factor = 4C-1/4C-4 = 4(13.86) – ¼(13.86) – 4 = 1. 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: T 1 N 1 = T 2 N 2 = T 3 N 3 = T 4 N 4 24(400) = 96(N 4 ) N 4 = 100 rpm opposite direction 52. Compute the lineshaft diameter to transmit 12 HP at 180 rpm with torsional deflection of 0. degrees per ft. length. a. 3 in b. 5 cm SOLUTION: c. 2.35 in P = 2TN d. 62 mm 12(33,000) = 2T(180) T = 350.14 ft-lbs = 4,201.68 in-lbs L = 1 ft = 12 in. J = D^4 / 32 G = 12,000,000 psi for steel = TL /JG 0.08(/180) = 4,201.68(12) / [(D^4 /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. 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. 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 ps i b. 4250 psi SOLUTION: c. 3860 psi Ss = 16T / D^3 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
c. 1.20 x 10^6 P = 2tn d. 2.10 x 10^6 20 = 2(T)(200/60) T = 0.955 kN-m T = 0.955 kN-m x (1000N/kN) x (1000mm/m) T = 0.955x 10^6 N-mm
64. Determine the load in KN on a 25mm diameter x 1200 mm long steel shaft if its maximum elongation exceeds 1 mm. a. 85(ans) b. 125 SOLUTION: c. 103 y = FL/AE d. 93 E for steel = 30,000,000 psi = 206,786 x 10^3 kPa 1 mm x = (F(1200mm) / [(/4)(0.025)^2 m^2 (206,786 x 10^3 )(kN/m^2 )] 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/mm^2. 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, b. 7,100 SOLUTION: c. 8,150 P = 2TN d. 6,300 (ans) 300(33,000) = 2T(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, b. 242,816 SOLUTION: c. 218,182 Stress = kET
d. 196,860 = (0.0000119)(206,786 x 10^3 )(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. b. 0.112 SOLUTION: c. 0.36 (ans) Ff = fN d. 0.38 20 = f(56) f = 0. 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 V 22 = V 12 + 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 = 2TN d. 945 20(33,000) = 2T(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 ft^3 d. 31. 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. b. 11.34 (ans) SOLUTION: c. 10.24 St = PD / 2t d. 12.44 t = 0.446 in = 11.34 mm
= 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 = V^2 / 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 10^6 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 = [(3x10^6 N-mm)(1200mm)//32(92.075)^4 mm^4 (80,000 N/mm^2 )] 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. 3475 lbs b. 4175 lbs SOLUTION: c. 4375 lbs (ans) Working strength = S 1 [0.55d^2 – 0.25d] d. 7513 lbs = 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 = (V 2 – V 1 ) / t = (0 – 30) / 4 = - 7.5 m/s^2 d. 60 (ans) V 22 = V 12 + 2aS = 30^2 + 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 = PL^3 / 12EI b. y = PL^3 / 3EI c. y = PL^3 / 48EI (ans) d. y = PL^3 / 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. b. 18.926 (ans) SOLUTION: c. 25.4 For main power transmitting shaft: d. 30.7 P = D^3 N / 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 in^4 b. 25 in^4 (ans) SOLUTION: c. 30 in^4 For square section with side a J = a^4 / 6 d. 35 in^4 Therefore: J = (3.5)^4 / 6 J = 25 in^4 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
c. 45.45 in^3 Zp = (/32)(Do^4 – Di^4 ) / Do/ d. 51.98 in^3 = (/32)(6^4 – 3^4 ) / 6/2 = 39.76 in^3
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. b. 0.015 SOLUTION: c. 0.0185 (ans) E for steel : 30 x 10^6 psi d. 0.0166 E = FL / AE = [(60,000) (5 x 12)] / [(/4)(2.875)^2 (30 x 10^6 ) = 0.0185 in.
