I have shared the hw and it is almost done you just need the calculations for d1,d2,d3,d4,d5. I HAVE MARKED THEM IN THE SECOND FILE d1 is a pulley d3 and d4 are both gears d5 is a bearing So the calculations for the diameter (d) of each is different./n Below is a Project #1 depiction of a CNC gearbox shaft. Given the provided data, determine the diameters of the Shaft in each respective section. 300mm Rotation (direc direction C.W. Pulley 50m (тор) Meshing to another gear Gear I ૧૦ 40"m 50% m Gear 2 ножи Bo Bearing 20 200 ww & Meshing to another! gear from bottom! Bearing 180mm di + Given: Belt FP کم Аз Idu ds * Assume the above design * * Input Power on pulley : 30 hp. Shaft rotational Speed = 380 rpm * Output: (gear 1 uses 15 HP Periodically & 15. hp every 30 Sec gear 2 consumes 15hp uniformly + Sur = 800mpa, Sys 580 MPa, ng=3, Ground Surface, Washr=0.25W+ * G₁ & G₂ are helical gears, Assume: Wa=W₁ = 0.25W+ Wa direction for G₁| Wa direction For G2.4 × те RAZ A Wai P = Tw₁ => T == P 2 NU Wri Watt Wr (30x736) 380x27 60 rody Bec = 3700 N Wrz Gu Waz WT₂ 555 N.m RBZ B Fp= тр 555 rp ✓2 300 x 10-3 То 277.5 TM-m + TG2 277.5 TG₁ = 1/2 Tp = 555/2 = 277.5 N.m мал Ты 0 Min 30 30 → (sec) N.M I (N·m) 555 it (sec) 277.5 ++ (sec) TG₂ = 1/2 Tp=277.5 N.m =To₂m * This torque is applied to the shaft from point P to G * From G₂ to B: T=0 From Gi to G₂T= 277.5 (2) b Constant N.m 555-277.5 Tas Between points P to Gg. : 2 Tm= 555+277.5 2 ≈13865 P N-M 416 N.m PORT C₂ To B & Ta=Tm=0, Point G₁ to Gz: Tm = 2775 N-m VG W+1= TC ล 277.5 0.09 = 3082N Wr25 کے TG₂ 277.5 VG₂ s 0.09 Given: Was Wr= 0.25W+ 3082N N Wal = Wr₁ = 0.25W+₁ = 1770.5 Waz Wr₂ = 0.25 W+z: 1770.5 N MG Waix VG₁ = 770.5x0.09 = 69.3 N.m 5) C.C.W. 1 MG25 WALK VG₂ = 770.5x0.09 = 69.3 N.m "C.C.W. (3)

Fig: 1