A belted transmission system is loaded as shown. The input shaft takes its power from a 50 kW motor that runs at 1200 rpm.

The output shaft drives a certain machinery equipment. Determine the following:

a. torque available at each of shafts A, B, and C.

b. torsional stress induced on the 50mm diameter output shaft

c. length of belt required to connect each pair of pulleys

d. maximum bending moment on shaft B, considering a belt coefficient of friction of 0.25.

Community Answer

(a) Calculate torque available at shaft A using the relation.P=(2piN_(A)T_(A))/(60)Here, P is power transmited through motor, N_(A) is speed of imput shaft at A and T_(A) is torque transmitted by input shaft A.Substitute 50kw for P and 1200rpm for N_(A).{:[50kwxx(10^(3)w)/(1kw)=(2pi xx12008pmxxT_(A))/(60)],[T_(A)=397.887N-m]:}Therefore, Torque available at A is 397.887N-mDetermine Torque arailable at shaft B.Calculate torque torque transmitted by 450mm pulley using the relation.{:[T_(450)N_(450)=T_(A)N_(A)],[T_(450)=T_(A)((N_(A))/(N_(450)))],[(N_(A))/(N_(450))=(D_(450))/(D_(A))],[=(450(mm))/(250(mm))],[(N_(A))/(N_(450))=1.8],[T_(450)=397.887N-mxx1.8],[T_(450)=716.8N-m.]:}pulley f 450mm is lies on the shaft B, there fore same torque is being tranmitted through out the shast.Therefore, Torque arailable at B is T_(B)=716.19N-m Calculate torque arailable at shaft using Using the relation.T_(B)N_(B)=T_(C)*N_(C){:[T_(C)=T_(B)((N_(B))/(N_(C)))],[(N_(B))/(N_(C))=(D_(C))/(D_(B))],[=(5.00(mm))/(200(mm))],[=2.5.],[T_(C)=716.19N-mxx2.5],[T_(C)=1790.475N-m]:}Therefore torque available at c is 1790.475N-m(b) Determine torsional stress induced on the 50mm diameter output shaft using the relation.tau_(c)=(16T_(c))/(pid^(3))Substitute 1790.475N-m for T_(C) and 50mm ford{:[tau_(c)=16 xx1790.475N-mxx(10^(3)(N)*mm)/(1(N)-m)],[pi(50mm)3]:}tau_(c)=72.95(N)/(mm^(2))There fore, torsional Stress at output shaft is72.95N//mm^(2)" or "MPa(C) Determine Length of belt between input shaft pulley and 450mm diameter pulley using the relation{:[L=(pi)/(2)(d_(1)+d_(2))+2x+((d_(1)-d_(2))^(2))/(4x)],[=(pi)/(2)(250mm+450mm)+2xx1000mm+],[=((250(mm)-450(mm))^(2))/(4xx1000(mm))],[L=31090557mm.]:}Therefore length of input bett is is3109^(˙).557mmDetermine the length of output belt using the relation{:[L=(pi)/(2)((D_(B)+D_(C))/(2))+2x+((D_(B)-D_(C))^(2))/(4x)],[=(pi)/(2)(200+500)+2xx1000+((200-500)^(2))/(4xx1000)],[=1099.55+2000+22.5mm],[= ... See the full answer