上海交通大学：《Design & Manufacturing II and Project》课程教学资源（作业）03_Belts Solutions

ME357 Belt Drive Solutions 1. A nylon core flat belt has an elastomer envelope, is 200mm wide,transmits P=60kW,at V=25m/s, mass/length =2kg/m,crossed config d=300mm D=900mm,C=6m a)Calculate belt length wrap angles Sol. Crossed: 0-42m2 0.9m+0.3m =π+2sin =3.34rad=191.5° 2*6m i=[4c-o+aj门°+2o+a) -[46mj-02mj+3409m+03m =13.9m b)Compute belt tensions if f=0.38 P=(F-F)V Sol. F-F,=7 P E=M:_2g+25m1sy=1250N=125N R-F=el0=F-F P rearranging E-F R-F-F -e-()-1w-() 60kW 25m/s =4.59kW 5=斤-=459v-607 =2.19kW 25m/s

ME357 Belt Drive Solutions 1. A nylon core flat belt has an elastomer envelope, is 200mm wide, transmits P=60kW, at V=25m/s, mass/length =2kg/m, crossed config d=300mm D=900mm, C=6m a) Calculate belt length & wrap angles Sol. Crossed: 1 1 2sin ( ) 2 0.9 0.3 2sin 3.34 191.5 2*6 D d C m m rad m                           0.5 2 2 0.5 2 2 4 2 3.34 4*(6 ) 1.2 (0.9 0.3 ) 2 13.9 L C D d D d m m m m m                    b) Compute belt tensions if f=0.38 Sol. 1 2 1 2 P (F F )V P F F V     2 2 2 *(25 / ) 1250 1.25 c kg F MV m s N kN m     1 1 2 1 c f c c c F F F F e F F P F F V         rearranging 0.38*3.34 1 0.38*3.34 2 1 60 1.25 1 1 25 / 4.59 60 4.59 2.19 25 / f f e P e kW F Fc kN e V e m s kN P kW F F kN kN V m s                                

2. A polyamide type A-3 belt is 10 in.wide,connects d=16in.cast iron driving pulley to D=36in.driven pulley in open config, C=15 ft,V=3600ft/min, What's max power transmitted?Use Ks=1.3 Sol. H=(F-F)v 33000K, Table 4-2:polyamide A-3:allowtension=100 width in Cp=0.94 Table 4-4:polyamide A-3:d=16in,so Cr=1.0 (E)。=bFCC,=10×100×0.94×1=9401b What are resulting belt tensions? t=0.13in f=0.8 Sol.Take 4-2 polyamide A-3: 0=0.042 3 7=o*1*widh=0042X0.13m10m=06s56/f F.=T*v2=(0.651b1h)3600*miny Ibf 73.31bf min 60sec 32.21b*fi/sec? 0,=7-2sim-D-4)=7-2sin-(36-l6,12=303ad 2C 2*1512in F=ef(E-F)+F.=es*30(E,-73.3bf)+73.3Ibf F=9401bf →F=1501bf H=G-E)'_940-1501f×36o0f/mim-=66.3p 33000K, 33000×1.3

2. A polyamide type A-3 belt is 10 in. wide, connects d=16in. cast iron driving pulley to D=36in. driven pulley in open config, C=15 ft, V=3600ft/min, What’s max power transmitted? Use Ks=1.3 Sol. 1 2 ( ) 33000 s F F V H K   Table 4-2: polyamide A-3: 100 a allowtension lb F width in   Table 4-4: polyamide A-3: d=16in, so 0.94 1.0 P V C C   1 ( ) 10 100 0.94 1 940 F a a p V  bF C C      lb What are resulting belt tensions? Sol. Take 4-2 polyamide A-3: 3 0.13 0.8 0.042 t in f lb in     3 12 * * (0.042 )(0.13 )(10 )( ) 0.655 / lb in T t width in in lb ft in ft       2 2 2 min * 0.655 / (3600 * ) 73.3 min 60sec 32.2 * /sec c ft lbf F T V lb ft lbf lb ft         1 1 36 16 1 2sin ( ) 2sin ( * ) 3.03 2 2*15 12 d D d ft rad C in             0.8*3.03 1 2 2 1 2 ( ) ( 73.3 ) 73.3 940 150 f F c c e F F F e F lbf lbf F lbf F lbf            1 2  (940 150) (3600 / ) 66.3 33000 33000 1.3 s F F V lbf ft min H hp K       

3. Tractor transmission with 1 V-belt 5hp engine,60%transmitted d=6.2 in.D=12 in.L close as possible to 92 in. n=3100rev/min9。=0。=180°=π Select a belt V=πdn=π(6.2in)3100rev/min)1fi/12in) =5032fi/min H=0.60*5hp=3hp Table4-13:0=180°=K,=1.0 Table 4-15:use K,=1.3 Table4-9:d=-6.2in,H=3hp→H。=K,·H。=l3×3=3.9 B belt Table 4-12:B belt;d=6.2in:V=5032 fi/minH=4.00hp Table 4-14:B belt,L =92in=K2 =1.00 Number of belts:N=101 belt KK2H,1.0*1.0*4hp Table 4-10:90 or 93in Std circum, Choose B90 Table 4-11:L size+1.8 (B Belt) Ln=90+1.8=91.8

3. Tractor transmission with 1 V-belt 5hp engine, 60% transmitted d=6.2 in. D=12 in. L p close as possible to 92 in. n=3100 rev/min 180  d  D     Select a belt (6.2 )(3100 / min)(1 /12 ) 5032 / min V dn in rev ft in ft     H  0.60*5hp  3hp Table 4-13: 1  180  K 1.0  Table 4-15: use 1.3 Ks  Table 4-9: d  6.2in,H  3hp     1.3 3  3.9 HD Ks H p B belt Table 4-12: B belt; 6.2 ; 5032 / min 4.00 r d  in V  ft  H  hp Table 4-14: B belt, 2 92 1.00 L p  in  K  Number of belts: 1 2 1.3*3 0.98 1.0*1.0*4 s r K H hp N K K H hp     1 belt Table 4-10: 90 or 93in Std circum, Choose B90 Table 4-11: 1.8 L p  size  (B Belt) 90 1.8 91.8 L p   