, Splines & Cy al chapter 02 - Shatngs, Keys, Splines & Coupling mG PN ela a PROBLEM 7 ms aa ah has an Inner diameter of 0.035 m and an outer diameter A OF 0.06 shear stress Is not to exceed 120 Mpa. nh oe Ares id C. 4,100 N-m : Bt 300 Nem D, 4,150 Nem Saletan 2 6D * a(o% a") 167 (60 120 = ( ) x|(60)' (3s)'| 0.067 T = 4,500,000 N- mm © thus; T = 4,500N—m ‘2. PAST BOARD PROBLEM. A 2 in. solid shaft is driven by a 36 in. gear and transmit power at 120 rpm. if allonate shearing stress is 12 ksi, what horsepower can be transmitted ? A. 29.89 Hp C. 35.89 Hp B. 38.89 Hp D. 34.89 Hp Solation: 16T g; = 8h oD? 12,000 = 1ST x(2) T = 18,850 in-tb TN _ (18850)(120) 63000 63000 = thus;P = 35.90Hp 3. PAST BOARD PROBLEM A hollow shaft has an inner diameter of 0.035 m and outer diameter of 0.06 ™ Determine the polar moment of inertia of the hollow shaft. A. 1.512% 10% mé C. 1.215 x 108 mt B. 1.152% 10° mé D. 4.125 x 108 mé Je yoo) J = F[10.06)*- (0.035)*] = thus; J 1.125 x 10% m*4 Machine Design Reviewer rent ‘Lrast BOARD PROBLEM ee) t ¢ would a spindle 55 mm in diameter ty wrsnor shaft is 59 N/mm? TONE 8480 rem, Stang ana kW ©. 60.61 kW ™ age eokW D. 39.21 kW a xD° 16T n(58)° T = 1,927,390N-mm = 1.92739 kKN-m 2nTN_ 2x (1.92739) (480) a 60 «thus; P = 96.88 kW . 5, PASTBOARD PROBLEM the diameter of a solid shaft transmitting 75 Hp at 1800 rpm. The nature of joad and the type of service is such that the allowable S, based on pure torsion is 6000 MA 1 TB" 24/16" B 115/16" D.3 1/8" Seltion: N p-a- 63000 asi =, Titan) : 63000 T = 2625 in4b ~ 16T & = >) 6000 = 16(2625 nD? * thus ; D= 1.306 or D= 15/16" 5. PAST BomRD PROBLEM, Nas 10 Arbed diameter diamond saw blade is mounted on a pulley driven steel shaft, va 6 Inches ae Peripheral linear speed of 150 f/sec. Motor drive is 125 hp at 1,200 rpm, fe: s diameter pulley. Determine the shaft rpm to attain blade peripheral speedchapter 02 ~Shaftings, Keys, Splines & Couplings » | m4 i C, 638.3 pm A. 716.21m D, 687.56 rpm B. 639.10'(pm Saloon ve rDN 50 =| |N . 150(60) (3) = 687.551pm ¥ 7. PAST BOARD PROBLEM Compute the polar section modulus of a solid shaft with a diameter of 101.6 mm. A. 209.5 cm? C. 209.5 cm* B. 205.9 cm® D. 205.9.m* x 3 = 210. Fplt0-16) 8. PAST BOARD PROBLEM A 15/16 — in wide key has a depth of 5/8 in. It is 12 inches long and is to be used on a 200 hp, 1160 rpm, squirrel! -cage induction motor. The shaft diameter is 378 inches. The maximum running torque is 200%.of the full-load torque. Compute the maximum torque. A. 17,330 in-lb C. 37,210 in-Ib B. 733,211 in-lb D. 21,733 in-Ib Solution: pea. 63000 T = 21,733 in-lb 9, PAST BOARD PROBLEM A helical-coil spring has a mean coll diameter of inch Compute the value of Bergstrassar factor of the ibaa: orate a A172 c. 4.74 B. 1.217 Biter2 Machine Design Reviewer ‘ 0-15 oe -cmeeremnenanm Dye 1 = Ona 1 2g Crt ie c+2 C-3 4(8)+2 = = = 1.1724 Ke 4(8) -3 70. PAST BOARD PROBLEM A 15/16 — in wide key has a depth of 5/8 in. It is 12 inches long and is to be used ‘on a 200 hp, 1160 rpm, squirrel cage induction motor. The shaft diameter is 3 7/8 inches. The maximum running torque is 200% of the full-load torque, Determine the maximum direct shearing stress on the shaft, considering the effect of the keyway? A. 2850 psi C. 4250 psi B. 3250 psi D. 1820 psi Soletson: The direct shearing stress on the shaft considering the effect of the keyway: T : Sg = ky— > Note: minor shocks ; ki = 1.5 _ (1.5)(21,733 ) 2(37] Ss= 2850 psi 11. PAST BOARD PROBLEM, A centrifugal pump is to be driven by a 15-Hp electric motor at 1750 RPM. Compute the diameter of the pump shaft if it is made of AIS! C1045 as rolled. Consider the load is gradually repeated with S, = 72 ksi and factor of safety of 8. A. 11/16 in. + Cc. 2%in B.1%in D. %in e Solution: P= © T= 45 ftlb 63,000 Ss, 842 St = coke Ny 'd 8 ter 2) Spe ee cod 2 ge = OOO) oa 74in, nD? Sq (9000) D=0.674 in. + use: 11/16.in._-gnaningss Keys, Splines & Couplings 4 Chapter 02 12, PAST BOARD PROBLEM 48,000 Hp tums the shaft at 300 rpm. The Lebiapeberogein to thy Aturbine Seone . 500 ib, A hollow steel! shaft having 20 at - le diameter wn Seo ad Detocri Tho inside diameter of the shaft ifthe maximum shears aoe med on the rao aloe ie oto axond 7°70 A B seein D. 12.56 in. 2 T = 262,500 fib Be eat J =4foyt-]=— 32 Ss (262,500 fi-b)(7 in.)(12 int) Soot 4] 32 7500 tin. Answn: d = 9.59 in. 13. PAST BOARD PROBLEM ‘A shaft supported as a simple beam, 18 in. long, is made of carburized steel ( As 3120) of ultimate strength of 90 ksi, size factor of 0.85 and surface factor of 0.87. The Shaft rotating, a steady load of 2000 Ib is applied midway between the bearings. The surfaces are ground. Indefinite life is desired with N = 1.6 based on endurance strengh What should be its diameter if there are no surface discontinuities? A. 1 5/16 C.23/16 B. 1 3/16 D.3 1/16 Sclation: ‘The maximum moment will occur at the midpoint ( x = 9 in. ): M = (1000 pounds ),( 91in.) = 9,000 in-b The flexural stress (S)): _ Sy _ (0.87)(0.85)(90,000 psi) = — = = 41,600 psi N 1.6 _Mc_M M 3 See == whore: 2= SO 1 We) Zz 32 nD? = M 32 By ao ee]. | 9000 in-b m\Sq x \ 41,600 Ib/in2 Avon D=1.3in. use: D=1 5/16 in.% Machine Design Reviewer 1-17 eevevermunmernsneconnt oe wleortrtheniioree 14, PAST BOARD PROBLEM 6 inches pulley is fastened to a 1 1/4-In shaft by a sot screw, If a net tangential force of 75 Ib is applied to the surface of the pulley, what I Joad is steady and N = 3? pulley , what Is the holding force when the 0 ona rman sna senamaaamanaans A. 1080 1b C. 1045 Ib B. 20301b D. 1201 Ib Soledon = Mo =0 F (5/8) =75(3) F =360 Ib ‘The holding force is : F =3 (360) = 1080 Ib F = 1080 Ib 15, PAST BOARD PROBLEM Aline shaft is 2 15/16 in. in diameter and will transmit 50 hp when turning at 200 rpm at constant rate. This shaft furnishes the power to 10 machines each requiring 5 hp to operate, Each of the 10 pulleys is keyed to the shaft by standard fiat key. If the width and thickness of the key are 3/4 and 1/2 respectively, find the length of the key based on shear considering that the allowable shear stress for commercial shaftiig is 6000 psi A 0.325 in C.0.1342 in B. 0.238 in. D. 0.3182 in Seaton: ar S,bD Lz 2(1575.6 in-lb) (6000 Ibfin.? )(3/4 in.)(2 15/16 in.) L = 0.238 in.Chapter 02 ~ Shi ‘eys, Splines & Couplingy pacts sn ceys, Sf Couplings 4 1-18 16, PAST BOARD PROBLEM Aline shaft is 2 15/16 in. in diameter and will transmit 50 hp when turning at rpm at constant rate. This shaft furnishes the power to 10 machings ta requiring 5 hp to operate. Each of the 10 pulleys is keyed to the shat by flat key.{f the width and thickness of the key are 3/4 and 1/2 respects" the length of the key based on compression considering that the alot compressive stress for commercial shafting is 6 000 psi, A. 0.328 in C. 0.2342 in B. 0.387 in. D. 0.2262 in Salection, 4T L= S,tD 4(1575.6 in-Ib) (6000 Ib/in )(1/2 in.\(2 15/16 in.) L = 0.357 in. 17. PAST BOARD PROBLEM A cast iron pulley transmit 65.5 Hp at 1750 RPM. The 1045 as-rolled shaft to which it is to be keyed is 1 % in. in diameter; key material, cold-drawn 1020 Compute the length (based on shear) of flat and square key needed if the 1 3/4 in diameter has a width and thickness 3/8 and 1/4 in. respectively. Note: S,= 66 ksi (with Sys = 0.5 Sy = 33ksi) and N= 1.5 for smooth loading. A. . 0.328 in C. 0.2342 in B. 0.327 in. D. 0.2252 in Sclition: 5, = 32 s N = 22 ksi = 22 000 psi 2(2359 in-Ib) S,bD (22,000 Ib/in.? )(3/8 in.)(1 3/4 in.) Araws: L = 0.327 in — for flat and square keys1-19 %® Machine Design Reviewer enennanrrs 18, PAST BOARD PROBLEM A set screw Is necessary to fastened Hp and rotates at 150 rpm? Find the vorgue cn ha dat boewer Sere fae A. 105 fttb c.470 nA B. 150 feb Bde ede Solaion: 933,000 H pe, 2nN 33,000 (3) p= SOMO «105 tp 2n(150) Aeewer: 105.0 ft-lb 19. PAST BOARD PROBLEM A set screw is necessary to fastened a pulley to a 2 in. shaft while transmits 3 Hp ‘and rotates at 150 rpm? What is the typical size of set screw in practice? A. 1/2in C. 2/3 in. B. 1/3 D. 3/2 Sclotion: 41 set screw = 75 dena = | — |(2 in) ane daet screw = = in. 2 20. PAST BOARD PROBLEM ‘An eccentric is to be connected to a 3 — in, shaft by a set screw. The center of the aocentrc is 1 1/4 in, rom the center of the shaft when a tensile force of 1000 Ip is applied to the eccentric rod perpendioular tothe line of centers. Whats the holding force necessary if the factor of safety is 6? A. 4500 Ibf C. 1230 Ibf B. 5000 Ibf D. 4300 Ibf1-20 Chapter 02 ~ Shaftings, Keys, Splines & Coupling, Sia blsctiaanlncce ne reece" cman Solotion: F(1.5) = (1000)(1,25) F = 833 Ib Foiting = factor of safety x Force = 6 (833) Fhotaing = 3 (1260) = 5000 Ib 21, PAST BOARD PROBLEM A turbine developing 15,000 Hp turns the shaft at 300 rpm. The Propel attached to this shaft develops a thrust of 150,000 Ib. A hollow steel shaft hee an outside diameter of 14 in. is to be used. Determine the inside diameter on shaft if the maximum shearing stressed based on the torsion alone is | nett | exceed 7500 psi. What is the percentage savings in weight. by] A. 37.3% C. 25.73% } B. 35.7% : D. 20.57 % Weis, [On 6]: [a (OP- O76] 96 Mise FE srg gt Ws HE (Ds Fle Kee 16T J ee fib)(42in/tty 1S (7500) Ib/in2 (0, ? -[(0y? - (a? | (12,886)? -[(14? - (9.59) | (0, ? (12.886) = 12.886 in. % Weavings = Anawer: Yo Neavings = 37.3 % 22. PAST BOARD PROBLEM A 15/16 — in wide key has a depth of 5/8 in. It is 12 inches long and is to be used on @ 200 hp, 1160 rpm, squirrel —cage induction motor. The shaft diameter is 3 78 inches. The maximum running torque is 200% of the full-load torque. Determine the maximum compressive stress of the key. A. 5,990 psi C. 7,290 psi B. 2,990 psi D. 9,920 psiMachine Design Ravina ve 1-21 ~—e Uc ah oe em mene Seletions Sc max = Ao 11,217 Ib = ————___ = 2990 psi 1/2(5/8 in.) (12 in.) Aeseen: 2,990 psi 3 PAST BOARD PROBLEM A 15/16 - in wide key has a depth of 6/8 in. It is 12 inches long and is to be used ‘on @ 200 hp, 1160 rpm, squirrel ~cage induction motor. The shaft diameter is 3 7/8 inches. The maximum running torque is 200% of the full-load torque. Determine the maximum direct shearing stress on the shaft , considering the effect of the keyway. A. 797 psi C. 897 psi B. 997 psi D. 729 psi fg 11,217 Ib (15/16 in,)(12 in.) foowe: 997 psi = 997 psi ,c max 14, PAST BOARD PROBLEM Alever 16 in. long is to be fastened to a 2 ~in. shaft. A load of 40 Ib is to be applied normal to the lever at its end. What is the holding force necessary if the factor of safety is 5? A. 2000 Ib C. 3200 Ib B. 3000 Ib D. 6400 Ib Satin: F(1.0) = (40)(16) F = 640 Ib Fhoiding = factor of safety x Force (640) = 3200 Ib _hollow, tube has an outer diameter of 2 in. and an inner di wy is ten ino, A crank 16 in. long is keyed to one end , and the other end inh iameter of 4 y i Th produce a tonal shee f S000 pa est 2 Pld yt pits 5 mt ts Ss = a or T= 8.) Mis r Px 15in, = 5000 bin? (1.0738 in) (1 in) Calculate the interface Pressure for a solid shaft and collar assembly with both Parts fan where radial interface of 0.0006 in, interface radius (R= 5 cm), radius ovtee (Re= 8 cm) and modulus of elasticity (E = 207 Gpa and 8 = 0.00001 ) A. 12.6 Mpa C. 13.5 Mpa B. 11.7 Mpa D. 10.54 Mpa Solation: p= £5, (R ° _(207x10°)(0.00001) | 4-(0.98 2 © REAR) | 20.05) (os) = 12,627,000 Pa = 12.6 Mpa 27. PAST BOARD PROBLEM. the maximum torque that can be applied without exceeding a shearing stress of 70 MN/m? OF a twist of 2.5 degrees in the 3.5-m length. Use G = 83 GNim?, A. 4kN-m ©. 10.44 kN-m 15m B. 3kN-m D474 kN 5 :‘& Machine Design Reviewer saree onnenannermr drt see 11-23 The Polar moment of Inertia: dy #5 (0.1 -0.07*) = 7.46 x 10° 32 ©1907)" + = (0.07)' =2.36 x 10 4 yp (0.07) ‘The Torque: (without considering the twist angle) te She 70x10" (7.46x10°) ‘a 1Osad kbd * R 0.05 Sk 70x10° (2.96x10*) ahaa nen c 0.035, The Torque (T) in consideration of the twist angle (6): TL o= 256 a ( a ) T(2) a TAS $480) “7.46x10*(63xi0") ” 2.36x10°(63x10") T=4.0 kN-m — chose the least value 124, PAST BOARD PROBLEM ‘Afiexible shaft consists of a 5-mm diameter steel wire encased in a stationary tube ‘that fits closely enough to impose a frictional torque of 2 N m/m. Determine the maximum length of the shaft if the shearing stress is not to exceed 140 Mpa. A 2715m C.3.21m B. 1.718m * D. 1.23 m Section: The Torque: = 16T sdoxto® = 16 (2L) (0.005) L=1.718m 29, PAST BOARD PROBLEM Afexible shaft consists of a 5-mm diameter steel wire encased in a stationary tube that fits closely enough to impose a frictional torque of 2 N mim. Determine the ‘maximum length of the shaft if the shearing stress is not to exceed 140 Mpa. What will be the angular rotation of one end relative to the other end? Use G = 83 Gpa. A. 2.715 m and 32.2 deg C. 1.718 mand 23.4 deg 8. 1.718 m and 33.2 deg D. 1.23 m and 12.43 deg1-24 Chapter 02 ~ Shaftings, Keys, Splines & Coupy; . voter dacanbee lomnsticoneeainetyse eee ’ enema ‘ Soletion: The angle of twist (0): ‘The Torque: ‘5 i _ (2d 2 st Jo! IG “IG «16 (2L) 6 Te 0.88 fd 140x10" = 0.008)" (0.008) (83x10") L=4.718m 6 oH 30, PAST BOARD PROBLEM A round steel shaft 3 m long tapers uniformly from a 60-mm diameter at ‘one end. 30-mm diameter at the other end, Assuming that there is no discontinuity resuits 5? each infinitesimal length, compute the angular twist for the entire length when the shaft is transmitting a torque of 170 N-m. Use G = 83 Gpa A 1.72deg C. 1.23 deg B. 1.29 deg D. 1.13 deg Solution: ‘an xe x d= 0.01x z . ==5(0.01x)" =3.1. 10° xx! J =F (0.01%)* =3.125 x 10°%ax The twist angle (6): = pth #Tdx _¢ 170dx 180 62 fig" Re Fl 3.125 x ase (S) 8 =1.29° 31. PAST BOARD PROBLEM A flange bolt coupling consists of eight steel 20-mm diameter bolts spaced evenly around a bolt circle 300 mm in diameter. Determine the torque capacity of the Coupling if the allowable shearing stress in the bolts is 40 MNim?, A. 1.51kN-m C. 5.21 kN-m B. 15.1 kN-m D, 10.51 KN-m% Machine Design Revie Sletins The Torque(T) T= Zesjen = (002) (aoxto)(220 ay T = 15,079.64 N-m 432, PAST BOARD PROBLEM A flange bolt coupling is used to connect a solid shaft 90 mm in diarneter to a hollow shaft 100 mm in outside diameter and 90 mm in inside diameter. If the allowable shearing stress in the shafts and the bolts is 60 MN/m?, how many 10-mm-diameter steel bolts must be used on a 200-mm diameter bolt circle so that the coupling vill be ‘as strong as the weaker shaft? A. Sbolts C. 9 bolts B. 6 bolts D.12 bolts Solations The Torques (T) for solid and hollow Shafts: (60x10°)1(0.09)" = 8588.33 N- a 8 -n ai corso %)(o20t -0.09') ‘ighéesi eat 8 = 4051.48 N-m —> use this value 2 Use the Small Value of the Toque (T) : T= Z¢s,Rn 4 x 2 0.2 (0.01) (eox108)(22)p n= 8.6 bolts = 9 bolts 4051.48 |. PAST BOARD PROBLEM ‘A fiange bolt coupling consists of six 10-mm-diameter steel bolts on a bolt circle 300 mm in diameter, and four 10-mm-diameter steel bolts. on a concentric bolt circle 200 mm in diameter. What torque can be applied without exceeding a shearing stress of 60 Mpa in the bolts? A. 5.5kN-m C. 3.25 kN-m B, 4.8kN-m D. 6.5 KN-m Solation:atings, Keys, Sunes & Comping 4 11-26 Chapter 02 ~ 8h ‘34, PAST BOARD PROBLEM What is the nominal shear stress at the surface for a 50 mm diameter shaft th, subjected to a torque of 0.48 kN-m. A. 16.95 C. 21.65 BL 19.56 D. 25.12 Soleton: 5-1. 16(0.48) xD° (0.050) = 19,556.96 kPa = 19.56 - MPa ‘35. PAST BOARD PROBLEM A hollow iron pipe to be designed as a column has an outside diameter of 240 mm and is subjected to a force of 80 KN. Compute the pipe thickness considering that te compressive stress is limited to 16 MPa. A. 585mm C. 6.85 mm 8B. 7.85mm D. 885mm ‘ Seaton: D, D, = 0.2263 m Thickness of the Pipe:# Machine Design Reviewer meneabiilbdetemamh ne obabi 36, PAST BOARD PROBLEM A 1-inch diameter shaft has a single disc weighing 76 lb mounted midway between two bearings 20" apart. Compute the | lowest criti ing that the modulus of elasticity is 30 x 10° psi, Neglect he we re ft fn coneidering thi A. 2038 rpm ©. 2 308 rpm B. 2380 rpm D. 2803 on Solon: Moment of Inertia: a xDt W=75b 64 ey i Solving for the lowest critical speed: __ [BT6EIg 26 = N, = 20% (576)(30 x 10°)(0.0491)(32.2) We 30(213:39) N, = 2037.68 pm 37, PAST BOARD PROBLEM The outside diameter of a hollow shaft is 1.6 times the inside diameter. When transmitting 420 W of power at 180 rev/ min the angle of twist is one degree over a length of 40 times the inside diameter. Taking G for the material as 85 GN/m?, calculate the maximum shear stress. A. 29.67 Mpa C. 33.45 MPa B. 25.44 MPa D. 36.78 MPa Sdotions where = 420(60) = 22.28 KN-m 2n(180) is oe 9” 300 L = 40d; , Splines & Coup} nae ___ Chapter 02 —Shattings Oe = enn = Zeta" Hore’) D=1.6d ‘Substituting relations to 1 and 2 03.2 mm D= 165.1 mm thus, S, = 29.67 MPa ‘38, PAST BOARD PROBLEM An automobile weighing 3,000 Ib oe fiom fo) ei oer ns distance “4 the drive-shaft. Rear wheel V mete i Ratio ‘of differential is 40:11. Calculate extreme fiber stress in dq), shafi, neglecting tire, bearing, and gear losses. C. 24,900 psi A. 29,400 psi : B. 20,490 psi ' D. 22,409 psi 5775 in-Ib then: _ 16(575) a = 29,400psi 39. PAST BOARD PROBLEM A pulley is keyed to a 24 mm steel shaft with a metric M6x20 square key. If the hb length is 20 mm and the allowable shearing stress for the key material is 100 MPa, how much torque can be transmitted? A. 0.133 KN-m C. 0.144 kN-m B. 0.155 kKN-m D. 0.166 KN-m 100 a at 24(6)(20) T=144,000 KN-m = 0.144 KN-ma Machine Design Reviewer aeeeeeeeeer seen nranewretinaanviramnnewi sr atarniieneeceeaRN I What would be the diameter of the shi if the shaft is to make {Sovran po mnt aft fo transmit 10 horsepower A. 1.53 in C. 3.641 B. 1.35 in D. 348in Solotions D'N Poaas 9 - D150) 53.5 D=1.53in. 41, PAST BOARD PROBLEM, What horsepower would be transmitted by a short shaft, 2 inches in diameter carrying two pulleys closed to the béarings, if the shaft makes 300 revolutions per minute. A. 36 hp C. 63 hp B. 45hp D. 54hp Seletion: oN “38 _ 2300) 4) 38 P=63hp 42, PAST BOARD PROBLEM What should be the diameter of a power-transmitting shaft to transmit 150 kW at 500 ‘pm? A. 71mm Cc. 65mm 8. 56mm D. 81mm, Splines & Couplings 11-30 Chapter 02 - Shaftings) Key. oP i enn at \ 43, PAST BOARD PROBLEM 400 rprn? , transmit at What power would a short shaft, 60 mim ih diameter, yr) re able Seletion: DN e P= Taaxi0" __ 50°(400) © 0,83x10° P= 60 kw a 44, PAST BOARD PROBLEM po 6 48 i Find the torsional deflection of a solid steel shaft 4 inches in diameter and 48 inches jong subjected to a twisting moment of 24,000 inch-pounds. C. 0.42 deg A. 0.23 deg B. 0.35 deg D. 0.18 deg Solution: _584TL “DG _ 584(24000)(48) ~“44(1,500,000) =0.23deg. 8 45. PAST BOARD PROBLEM Find the torsional deflection of the solid steel shaft, 100 mm in diameter and 1300 mmloy subjected to a twisting moment of 3 x 10° N-mm. The torsional modulus of elasticity & 80,000 N/mm?. A. 0.285 deg C. 0.258 deg B. 0.379 deg D. 0.397 deg Soletion: } 584( 3 x10° ( 1300 ) (100 (0,000 ) = 0.285 deg, 46, PAST BOARD PROBLEM, Find the diameter of a steel lineshaft to tra torsional deflection not exceeding 0.08 dex nsmit 10 Hp at 150 revolutions per minute wit! gree per foot of length, i A. 3.52 in. B. 2.35 in. C. 3.25 in, D. 2.53 in,% Machine Design Reviewor at Saloon IP 10 o=46fF ae [40 N 150 = 2.35 in, ‘47, PAST BOARD PROBLEM \What power (in Hp) can a 1-inch diameter short shaft transmit at 380 rpm? a te c.10H : p B. 20Hp Bit te Selon: ‘ ND? 38 (1° (980) Fe 3a 710 Hp 48, PAST BOARD PROBLEM AT5-mm diameter shaft is transmitting 350 kW at 650 rpm. A flange coupling is used and has 6 bolts, each 18 mm in diameter. Compute the required diameter of the bolts Circle considering an average shearing stress of 27.5 MPa. A. 245mm C. 325mm B. 300mm D. 425mm Solution 8T _ 60(P) _ 30P. 50m, 7) GaN TaN 8(30)P 7ESNn, 8(30)(350) 7? (0.018)' (27500) (650)(6) =0.245 m=245 mm 49. PAST BOARD PROBLEM What length of a square key is required for a 4-in diameter shaft transmitting 1000 hp” at 1000 rpm? The allowable shear and compressive stresses in the key are 15 ksi and 30 ksi, respectively.ings, Keys, Splines & Couplings, go earner 1-32 ES Chapter 02 - Shaft entreetaernnemaegi oe ~ A. 2.1 inches C. 3.4 inches B. 5.2inches D. 4.2 Inches Solution: 63000HP N = 83000(1000) 63,000 ind 71000 = 1/4 of Diameter) Note: b = io _+ usual proportion (with ‘The Length of the Key (L): (Considering the Shearin vee? (63,000) Spd — 15,000(14) L =2.1 inches The Lenath of the Key(L): (Considering the Compressive Stress: Sc ) iste 4(63,000) “52D 30,000(1)(4) L=2.1 inches 1g Stress: Ss) 50. PAST BOARD PROBLEM meter is to be replaced with a hollow shaft of equal torsiond A solid shaft of 76 mm diat ESnath. Considering that the outside diameter of the hollow shaft is 100 mm, compas the inside diameter and percentage weight saved . A. 86.55 mm, 56.55% * C. 72.54 mm , 56.55 % B. 86.55 mm, 60.25% D. 54.45 mm, 45.45% s Alternate Solution: | 16TD p o'-d? Swotow = aay > Seots = Weave 4 x(D*-" mt Lay the same torsional strength (T): 2 priate ~ [tees (100)(76)? = (100)* ~ d* fs Se () : d= 86,55.mm Eee) Wag = Zales Mtn) _ [8 ~ (40? 4)] PIous (a) note: same density & lena” = 261 - (1007 ~ 86.55") 76 YWerea = x100 = 56.56%