際際滷

際際滷Share a Scribd company logo

Design of 14 speed gear box

Download as PPTX, PDF
K
kamaraj college of engineeing and technology

This document provides an example problem to design a 14-speed gearbox with output speeds ranging from 125 to 2500 rpm. It involves 4 steps: 1) Calculating the standard step ratio of 1.12 using the minimum and maximum speeds. 2) Selecting the 14 output speeds using multiples of the 1.12 step ratio. 3) Developing the structural formula of 3(1) 3(3) 2(5) to represent the gear stages. 4) Drawing the kinematic arrangement with 4 shafts to connect the gears based on the structural formula.

1 of 8
Downloaded 229 times
By Mr. B.Balavairavan Assistant Professor Mechanical Engineering Kamaraj College of Engg and Tech Virudhunagar PSG Design Data Book DEPARTMENT OF MECHANICAL ENGINEERING Problem on Design of 14 Speed Gear Box
Sample Problem A 14 speed gear boxis required to furnish output speeds in the range of 125 to 2500 rpm. Draw the speed diagram and kinematic arrangement. Given: n = 14 Nmin = 125 rpm Nmax = 2500 rpm
Step - 1 Calculation of Step ratio Referring PSG Data Book P. No : 7.20 the calculated step ratio is not a std. value Nmax Nmin = n-1 2500 125 = 14 -1 = 1.259
Since is not a std. value, Using multiples of std. value the required step ratio is calculated 1.6 - 1.25 - 1.12 - 1.06 - Multiples of 1.12 gives nearest value of 1.258 As 1.12 is multiplied 1 time, so we skip 1 speed Hence std. = 1.12 & R 20 series is selected Cannot be used Cannot be used 1.12 1.06 x 1.12 = 1.254 x 1.06 x 1.06x 1.06 = 1.238
Step - 2 Selection of Speeds The obtained speeds are; 160,200,250,315,400,500,630,800, 1000,1250,1600,2000,2500 100 112 125 140 160 180 200 224 250 280 315 355 400 450 500 560 630 710 800 900 100 0 112 0 125 0 140 0 160 0 180 0 200 0 224 0 250 0 The speeds are withing the range hence no need to check for deviation
Step - 3 Structural formula & Ray Diagram The structural formula for 14 speed gear box is 3 (1) 3 (3) 2 (5)
3 (1) 3 (3) 2 (5) Stage 1 Stage 2 Stage 3 Nmin Ni/p 0.25 Nmax Ni/p 2 125 315 400 315 = 0.4 = 1.27 = = Nmin Ni/p 0.25 Nmax Ni/p 2 315 800 1250 800 = 0.39 = 1.56 = = Nmin Ni/p 0.25 Nmax Ni/p 2 800 1600 1250 1600 = 0.5 = 0.78 = = Stage - 3 Stage - 2 Stage - 1 2500 2000 1600 1250 1000 800 630 500 400 315 250 200 160 125
Step - 4 Kinematic Arrangement Shaft - 1 / Input Shaft - 2 / Intermediate Shaft - 3 / Intermediate 1 3 42 5 6 13 15 8 7 9 10 16 14 Shaft - 4 / Output 11 12

More Related Content

Design of 14 speed gear box

  • 1. By Mr. B.Balavairavan Assistant Professor Mechanical Engineering Kamaraj College of Engg and Tech Virudhunagar PSG Design Data Book DEPARTMENT OF MECHANICAL ENGINEERING Problem on Design of 14 Speed Gear Box
  • 2. Sample Problem A 14 speed gear boxis required to furnish output speeds in the range of 125 to 2500 rpm. Draw the speed diagram and kinematic arrangement. Given: n = 14 Nmin = 125 rpm Nmax = 2500 rpm
  • 3. Step - 1 Calculation of Step ratio Referring PSG Data Book P. No : 7.20 the calculated step ratio is not a std. value Nmax Nmin = n-1 2500 125 = 14 -1 = 1.259
  • 4. Since is not a std. value, Using multiples of std. value the required step ratio is calculated 1.6 - 1.25 - 1.12 - 1.06 - Multiples of 1.12 gives nearest value of 1.258 As 1.12 is multiplied 1 time, so we skip 1 speed Hence std. = 1.12 & R 20 series is selected Cannot be used Cannot be used 1.12 1.06 x 1.12 = 1.254 x 1.06 x 1.06x 1.06 = 1.238
  • 5. Step - 2 Selection of Speeds The obtained speeds are; 160,200,250,315,400,500,630,800, 1000,1250,1600,2000,2500 100 112 125 140 160 180 200 224 250 280 315 355 400 450 500 560 630 710 800 900 100 0 112 0 125 0 140 0 160 0 180 0 200 0 224 0 250 0 The speeds are withing the range hence no need to check for deviation
  • 6. Step - 3 Structural formula & Ray Diagram The structural formula for 14 speed gear box is 3 (1) 3 (3) 2 (5)
  • 7. 3 (1) 3 (3) 2 (5) Stage 1 Stage 2 Stage 3 Nmin Ni/p 0.25 Nmax Ni/p 2 125 315 400 315 = 0.4 = 1.27 = = Nmin Ni/p 0.25 Nmax Ni/p 2 315 800 1250 800 = 0.39 = 1.56 = = Nmin Ni/p 0.25 Nmax Ni/p 2 800 1600 1250 1600 = 0.5 = 0.78 = = Stage - 3 Stage - 2 Stage - 1 2500 2000 1600 1250 1000 800 630 500 400 315 250 200 160 125
  • 8. Step - 4 Kinematic Arrangement Shaft - 1 / Input Shaft - 2 / Intermediate Shaft - 3 / Intermediate 1 3 42 5 6 13 15 8 7 9 10 16 14 Shaft - 4 / Output 11 12