1 / 19

IE 553 Engineering of Human Work Dr. Andris Freivalds Class #28

IE 553 Engineering of Human Work Dr. Andris Freivalds Class #28. IE 553 Engineering of Human Work Dr. Andris Freivalds Class #28. Cycling Work. Lifting 8.25 lbs, 20 in., 6/min E ave = 1.5 kcal/min, E r = 1.3 kcal/min E w = [E ave (T w +T r ) - E r ×T r ]/T w.

Download Presentation

IE 553 Engineering of Human Work Dr. Andris Freivalds Class #28

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. IE 553 Engineering of Human Work Dr. Andris Freivalds Class #28 IE 553

  2. IE 553 Engineering of Human Work Dr. Andris Freivalds Class #28 IE 553

  3. Cycling Work • Lifting 8.25 lbs, 20 in., 6/min • Eave = 1.5 kcal/min, Er = 1.3 kcal/min • Ew = [Eave(Tw+Tr) - Er×Tr]/Tw IE 553

  4. Metabolic Energy Prediction/Modeling • Need: actual measurements on job: • Difficult, interferes with task • Costly, can’t measure new task • Basis: • Table values (Passmore & Durnin, 1955): • Crude, inaccurate • Many new jobs not found • Predictive models: Ǻberg (1968), Garg (1978) IE 553

  5. Garg Model • Main assumptions (similar to MTM) • Ėjob = function of physical characteristics of job • Job broken down into simpler tasks • Ėjob = ∑ Ėtasks • Two components: • Maintenance of body posture = sitting (Eq. 6.1), standing (Eq. 6.2), standing bent (Eq. 6.3) • Net increase due to tasks Ėjob = [ ∑Ėposture × time + ∑ΔEtasks] / time IE 553

  6. Tasks - 1 • Vertical lift: • Stoop lift (V<32", Eq. 6.4) • Squat lift (V<32", Eq. 6.5) • Arm lift (V>32", Eq. 6.7) • Vertical lower: • Stoop lower (V<32", Eq. 6.8) • Squat lower (V<32", Eq. 6.9) • Arm lower (V>32", Eq. 6.10) IE 553

  7. Tasks - 2 • Walk/carry/hold: • Walk (Eq. 6.11) • Carry arms length (Eq. 6.12) • Carry at waist (Eq. 6.13) • Hold arms length (Eq. 6.14) • Hold at waist (Eq. 6.15) • Hold out in one hand (Eq. 6.16) • Push/pull: • 32" bench height (Eq. 6.17) • 60" height (Eq. 6.18) IE 553

  8. Tasks - 3 • Lateral arm work: • Both hands 180° (Eq. 6.19) • One hand 180° (Eq. 6.20) • Stand 90° (Eq. 6.21) • Sit both hands 90° (Eq. 6.22) • Sit one hand 90° (Eq. 6.23) • Horizontal arm work: • Stand (Eq. 6.24) • Sit (Eq. 6.25) IE 553

  9. Major Variables/Modifiers • Individual characteristics: • Gender • Weight • Working style: posture, speed • Task characteristics: • Frequency • Load • Vertical range • Grade of surface • Horizontal distance moved IE 553

  10. Comments - 1 • Two regions for lift/lower: • 0-32“ (lift body weight) • >32“ (lift arms only) • Freestyle lift = average stoop & squat lifts • Lift and lowers could have horizontal component (in addition to vertical) IE 553

  11. Comments - 2 • Estimate speed: • 2 mph = 88 steps/min • 3 mph = 108 steps/min • Climbing stairs: • ΔE(per step) = 0.1 + 0.0014 (BW+L) • ACU: lbs, in, mph, % IE 553

  12. Ex. #1: Box Lifting • 160 lb male, lifting 20 lb boxes, 10/min, from 30" bench to 40" conveyor • Required tasks? IE 553

  13. Garg’s Model IE 553

  14. Ex. #1: Box Lifting 160 lb male, lifting 20 lb boxes, 10/min, from 30" bench to 40" conveyor • Stoop lift 30"→ 32" ΔE = 10-4[0.375BW(32-h1) + (1.63L + 0.875 S×L)(h2-h1)] ΔE = 10-4[0.375(160)(32-30) + (1.63(20) + 0.875(20))(32-30)] = 0.0235 kcal/lift 2) Arm lift 32"→ 40" ΔE = 10-4[0.06BW(h2-32) + (3.22L - 0.53 S×L)(h2-h1)] ΔE = 10-4[0.06(160)(40-32) + (3.22(20) - 0.53 (20))(40-32)] = 0.0507 kcal/lift IE 553

  15. Ex. #1: Box Lifting – con’t 3) Arm lower 40"→ 32" (no load) ΔE = 10-4[0.09BW(h2-32) + (1.03L + 0.37 S×L)(h2-h1)] ΔE = 10-4[0.09(160)(40-32)] = 0.0115 kcal/lower 4) Stoop lower 32" → 30" (no load) ΔE = 10-4[0.31BW(32-h1) + 0.78L(h2-h1) + 13.25 S(32-h1)] ΔE = 10-4[0.31(160)(32-30) + 13.25(2)] = 0.0125 kcal/lower Standing posture Ė = 0.0109BW = 0.0109(160) = 1.744 kcal/min IE 553

  16. Ex. #1: Box Lifting – Total Ė • Ėjob = [ ∑Ėposture × time + ∑ΔEtasks] / time • ∑ΔEtasks=10×[0.0235 + 0.0507 + 0.0115 + 0.0125] • = 0.982 kcal • Ėjob  = 1.744(1) + 0.982 = 2.73 kcal/min IE 553

  17. Validation • Stepwise multiple regression, with 3 males and 3 females, r2 = 0.98 • Literature: 20 tasks, difference of 2.9 to 38%, average = 13.8% • Field work: 16 tasks, average difference = 5.6% • Garg found overprediction • Beginners typically underpredict • Best feature = relative comparison for redesign (careful with absolute values) IE 553

  18. Garg’s Model IE 553

  19. IE 553

More Related