Problem Statement: Design a power transmission (Figure 1) for an industrial saw

Question

Problem Statement: Design a power transmission (Figure 1) for an industrial saw that will be used to cut tubing for vehicle exhaust pipes to length prior to the forming process. The saw will receive 25 hp from the shaft of an electic motor rotating at 1750 rpm. The drive shaft for the saw should rotate at 500 rpm. 3.500 in Output Shaft Saw Drive ' I Motor Flexible Coupling Flexible Coupling Bearings (4) nion Input Shaft 2.500 in -2.500 in 3.500 in Figure 1: General layout of saw drive with single-stage gear reducer. Design Assumptions: 1. The pinion and gear will not be part of this design project so use the following properties -20° Pressure angle, Number of teeth on pinion, Np 28 Number of teeth on gear, No = 98 Diameter of pinion, Diameter of gear Face width, DP 3.500 in DG 12.250 in F 2.00 in

Explanation / Answer

T1=63000*25/1750=900 Ib-Inch.

GB Power =Motor Hp * FoS=25*4(Given in Question)

RPM=500(Given)

T1=63000*100/500=12600 Ib-Inch.

Ft :Tangential force=2000 T(Kgf –Mt)/Dia of pinion

              No. Teeth=28

              Diametral Pitch (DP)=2.01

              Pitch Diameter = N / DP.=13.904 inch

              OD = PCD + 2 * Corrected Addendum=14.925 inch=0.379 mtr.

              Corr. Add. = (1 ± kp) / DP=0.4975 inch

             Ftp=2000*10 (kgf-m)/0.379 =52770.4 Kgf

               Tangential Calculation for Gear

               No. Teeth=98

             Diametral Pitch (DP)=2.01

              Pitch Diameter = N / DP.=48.762 inch

              OD = PCD + 2 * Corrected Addendum=49.752 inch=1.263 mtr.

              Corr. Add. = (1 ± kp) / DP=0.4975 inch

               Ftp=2000*145.1 (kgf-m)/1.263 =229770.3 Kgf.

Fr=Ftp* tan @=52770.4*0.363(Angle = 20 Degree)

                       =19155.655 Kgf.

Fr=Ftp* tan @=229770.3*0.363(Angle = 20 Degree)

                       =83406.6Kgf.

FrP=face width* Tangetial force /Distance=2.5*19155.655/5   (Converted Inch to mtr)

      = 9577.8 Kgf .(Equal Distance so same forces on Both bearing.)

Frg= face width* Tangetial force /Distance=2.5*83406.6Kgf /5   (Converted Inch to mtr)

      =41703.3 kgf .(Equal Distance so same forces on Both bearing.)

Deflection of Concentreted load at Shaft Mid point on Pinion shaft =Wl3/48EI== 52770.4*127^3/(48*2.8*180 *10^9*30625)

Moment of Interia= (0.5 Mr2)=30625 kg mm2

=0.145 mm on pinion Shaft.

Deflection of Concentrated load at Shaft Mid-point on Gear shaft =Wl3/48EI== 229770.3 *127^3/(48*2.8*180 *10^9*30625)

=0.631 mm

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