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/sME357 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                                