{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Let ds dp and de be the pitch circle diameters of

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 revolution. Therefore from the fourth row of the table, y = 1, and x+y=5 or x=5–y=5–1=4 Since the gear E is stationary, therefore from the fourth row of the table, y –x× ∴ TS =0 TE or 1–4× TS =0 TE or TS 1 = TE 4 T E = 4T S Since the minimum number of teeth on any wheel is 16, therefore let us take the number of teeth on sunwheel, T S = 16 ∴ T E = 4 T S = 64 Ans. Let dS, dP and dE be the pitch circle diameters of wheels S , P and E respectively. Now from the geometry of Fig. 13.27, dS + 2 dP = dE 466 l Theory of Machines Assuming the module of all the gears to be same, the number of teeth are proportional to their pitch circle diameters. TS + 2 T P = TE or 16 + 2 T P = 64 or T P = 24 Ans. 2. Torque necessary to keep the internal gear stationary We know that Torque on S × Angular speed of S = Torque on C × Angular speed of C 100 × ωS = Torque on C × ωC ∴ Torque on C = 100 × ωS N = 100 × S = 100 × 5 = 500 N-m ωC NC ∴ Torque necessary to keep the internal gear stationary = 500 – 100 = 400 N-m Ans. Example 13.21. In the epicyclic gear train, as shown in Fig. 13.28, the driving gear A rotating in clockwise direction has 14 teeth and the fixed annular gear C has 100 teeth. The ratio of teeth in gears E and D is 98 : 41. If 1.85 kW is supplied to the gear A rotating at 1200 r.p.m., find : 1. the speed and direction of rotation of gear E, and 2. the fixing torque required at C, assuming 100 per cent efficiency throughout and that all teeth have the same pitch. Solution. Given : T A = 14 ; T C = 100 ; T E / T D = 98 / 41 ; PA = 1.85 kW = 1850 W ; N A = 1200 r.p.m. Fig. 13.28 Let dA, dB and dC be the pitch circle diameters of gears A , B and C respectively. From Fig. 13.28, dA + 2 dB = dC Gears are extensively used in trains for power transmission. 467 l Chapter 13 : Gear Trains Since teeth of all gears have the same pitch and the number of teeth are proportional to their pitch circle diameters, therefore T – TA 100 – 14 TB = C = = 43 T A + 2T B = T C or 2 2 The table of motions is now drawn as below : Table 13.23. Table of motions. Revolutions of elements Step No. Conditions of motion Arm Gear A 1. Arm fixed-Gear A rotated through – 1 revolution (i.e. 1 revolution clockwise) 0 –1 2. 3. 4. Arm fixed-Gear A rotated through – x...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online