No.000149 Lifting and Carrying Mechanism Using Eccentric Shaft

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Application Overview

Purpose

• Purpose
  • To perform accurate pitch feed of cylinder-shaped workpiece.
• Operation
  • As appearance inspection and NG product ejection are performed at the final position, workpieces move from one positioning groove to the next positioning groove one by one.

Target workpiece

• Shape: aluminum cylinder part
• Size: O20 x 200mm
• Weight: 0.15kg

Design Specifications

Operating Conditions or Design Requirements

• Movement amount and movement pitch: 150mm
• Movement speed: 1sec. / pitch
• Outer dimensions: W970 x D306 x H306mm

Selection Criteria for Main Components

• To allow a movement amount of 150mm, the misalignment amount from the rotation center to the rotation edge is set to 75mm.
• Timing belt is used to drive two eccentric shafts.
• Selected a gear ratio that enables the speed of 1 pitch/second.

Design Evaluation

Verification of main components

• Select motor based on load torque and moment of inertia.
  • Here, calculation for selection of AC motor is explained.
  • Select gear ratio based on workpiece movement time.
  • Conditional value: movement time per stroke = approx. 1.0sec., motor rotation speed = 1,300rpm, timing pulley ratio = 50 : 25 = 2 : 1
  • 60 x gear ratio x 2 / 1.0 ? motor rotation speed Therefore, from gear ratio ? (1,300 x 1.0) / (2 x 60) = 10.8, select 1 : 10 for the gear ratio (i) of gearhead.
  • Calculation of load torque
  • Conditional value: g = 9.807 (m/s2), m = 2 (kg), η = 0.81, r = 0.075m
  • From T'm = F･r/η and F = m･g,
    T'm = 2 x 9.807 x 0.075/0.81 = 1.816N･m
    Considering safety factor Sf = 2 and pulley ratio, the load torque Tm is:
    Tm = T'm x Sf / 2 = 1.816 x 2 / 2 = 1.816N･m
  • Check of loading moment of inertia
    Moment of inertia applied to large belt pulley: J1 = 1.8162 x 10-2kg･m2
    Moment of inertia of small belt pulley: J2 = 1.3312 x 10-?kg･m2
    From moment of inertia applied to small belt pulley: Jd = J2 + J1 (b / a)2 (b / a = pulley ratio), Jd = 1.3312 x 10-? + 1.8162 x 10-2 x (25/50)2?= 4.554 x 10-3kg･m2
    Moment of inertia in motor shaft is, from Jm = Jd･(1 / i)2,
    Jm = 4.554 x 10-3 x (1 / 10)2 = 4.554 x 10-? (kg･m2) = 0.455kg･cm2
  • From allowable moment of inertia of PACMB90-40W-100V = 0.735 (kg･ｃm2) and allowable torque of ACMGX90-10 = 2.25 (N･m), 0.735 (kg･cm2) > 0.455 (kg･cm2), 2.25 (N･m) > 1.816 (N･m)?
    -> Specifications are satisfied.

Other Design Consideration

• The movement amount accuracy depends on the machining accuracy of the workpiece receiving part. Therefore, the required accuracy is the workpiece receiving part manufacturing accuracy.
• The cover of the timing belt is omitted from this drawing for illustrative reasons, but in the actual machine, the cover is mounted for safety and dust prevention purposes.
• The maximum movement value in the specifications represents the movement amount due to misalignment amount, but by changing the workpiece receiving position, the movement amount can be changed accordingly.