lab 4 report shipp

8/14/2019 Lab 4 Report Shipp

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Kyle ShippIME 462 Ergonomics

Lab 4A Energy Expenditure

Job Title

Lead-off operator on the Superior Machine at Plant 75, Dept. 34

Job Description

Transfer clutch plates from pallets to conveyor

Job Data

Pallets of parts: Length 4 feet

Width 4 feetHeight from floor: full 4 feet; empty 4 inches

Conveyor: Width 16 inches

Height 40 inchesDistance between pallets and conveyor 30 inches

Clutch plates: Weight 18.4 pounds

Height 2.44 inches

Time to unload: 15 minutes

Operator Data

Age: 23Height: 71 inches

Weight: 210 poundsSex: Male (1)

Total Energy Expenditure from hand calculations in Appendix A: 160.37 kcal/cycle, 10.69 kcal/min

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Plant: 75, Dept. 34 Worker: Shipp,K. Analyst: Shipp,

Job: Lead-off Operator Depalletize Gender (0=f,1=m): 1 MAC: 78.85 %ile = 17.0 kcal/min

Job # Transfer Clutch Plates from Pallets to Conveyor Age: 23 Work: 25 Rest: 7 Cycle: 31.1762 inutes max

Cycle: 15 minutes Weight (pounds): 210 0 % of all of given gender-age group capable without restTask Energy (kcal) Freq Force V/H1< V/H2 Time Distance Speed Slope

Number Element Description NameType min % /cycle /min /cycle pounds inches inches min feet mph %

18 Total 160.37 10.69

Postures Sit 0 0 0.00 2.21

Stand 10 67 22.89 2.29

Stand Bent 5 33 13.55 2.71

1 Lift Top row part LiftArm 4.39 54 18.4 40.66 52 0 0 0

2 Return LowerArm 2.04 54 0 40.66 52 0 0 0

3 Lift Bottom row part LiftStoop 13.93 54 18.4 11.33 32 0 0 0

4 Return LowerStoop 8.75 54 0 11.33 32 0 0 0

5 Lift Middle row part LiftStoop 4.04 54 18.4 26 32 0 0 0

6 Return LowerStoop 2.54 54 0 26 32 0 0 07 Lift middle/bottom row parts from 32" LiftArm 13.41 108 18.4 32 52 0 0 0

8 Return LowerArm 4.08 108 0 32 52 0 0 0

9 Lower part to conveyor LowerArm 11.13 162 18.4 40 52 0 0 0

10 Return LiftArm 4.08 162 0 40 52 0 0 0

11 Walk Walk 9.08 4.99 1 0 0 0 1.82 480 3 0

12 Move part towards conveyor Lateral ArmTwo180 26.85 162 18.4 0 0 0 0 0

13 Return Lateral ArmTwo180 17.01 162 0 0 0 0 0 0

14 Get Top cardboard LiftArm 0.08 6 0 40.66 42.66 0 0 0

15 Get Middle cardboard LiftStoop 0.28 6 0 26 28 0 0 0

16 Get Bottom cardboard LiftStoop 0.98 6 0 11.33 13.33 0 0 0

17 Grasp and move cardboard Lateral ArmStand90 0.66 18 1 0 0 0 0 0

18 Return Lateral ArmStand90 0.60 18 0 0 0 0 0 0 Figure 1.Energy Expenditure Spreadsheet Results

Total Energy Expenditure from spreadsheet in Figure 1: 160.37 kcal/cycle, 10.69 kcal/min

Work/Rest Cycle Determination

Assumption: Percentile for aerobic capacity corresponds to population percentile for body weight.

50th

Percentile Male Body Weight: 183.4 33.2 pounds

Worker Body Weight Percentile: 78.85 %

Calculation Notes:

Percentile calculated using Excel formula =normdist(x,mean, std_dev, cumulative)Wherex is the subjects measurement, mean is the 50

th percentile value, std_dev is the standard deviation of the

values, and cumulative is a logical operator set to True to return the cumulative distribution function as opposed

to the probability mass function.E = 10.69 kcal/min

MAC = 15.8 kcal/mi

78.85%ile MAC = 17.0 kcal/minfVO2,max = E/MAC = 10.69/17 = 0.6288

T k=(40/ fVO2 ) 39 = (40/0 6288) 39 = 24 61 minutes



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On the chart of expected increase in heart rate in Appendix 1, these WB and DB parameters are in the range for

an increase of 1 bpm/1 C > 25 C. Therefore, there would be an expected increase of (30-25) = 5 bpm.

fVO2,max increase = 0.74%/bpm * 5 bpm = 3.7 %

fVO2,max (Adjusted) = (0.6288)*1.037 = 0.6521

Twork=(40/ fVO2,max)-39 = (40/0.6521)-39 = 22.34 minutesTrest = 8.8 LN(fVO2,max - 0.5) +24.6 = 8.8 LN(0.6521-0.5) + 24.6 = 8.03 minutes

Cycle time = 22.34 + 8.03 =30.37 minutes

Solution to Reduce Energy Expenditure

My solution to reduce energy expenditure is to put the pallet of parts on a lift. This would cause all of the parts

to come from the first tier height and eliminate the drastic vertical changes.

Plant: 75, Dept. 34 Worker: Shipp,K. Analyst: Shipp,

Job: Lead-off Operator Depalletize Gender (0=f,1=m): 1 MAC: 78.85 %ile = 17.0 kcal/min

Job # Transfer Clutch Plates from Pallets to Conveyor Age: 23 Work: cont Rest: 0 Cycle: cont

Cycle: 15 minutes Weight (pounds): 210 27 % of all of given gender-age group capable without rest

Task Energy (kcal) Freq Force V/H1< V/H2 Time Distance Speed Slope

Number E lement Description NameType min % /cycle /min /cycle pounds inches inches min feet mph %

10 Total 125.38 8.36

Postures Sit 0 0 0.00 2.21

Stand 10 67 22.89 2.29

Stand Bent 5 33 13.55 2.71

1 Lift Top row part LiftArm 13.18 162 18.4 40.66 52 0 0 0

2 Return LowerArm 6.12 162 0 40.66 52 0 0 0

3 Lower part to conveyor LowerArm 11.13 162 18.4 40 52 0 0 0

4 Return LiftArm 4.08 162 0 40 52 0 0 0

5 Walk Walk 9.08 4.99 1 0 0 0 1.82 480 3 0

6 Move part towards conveyor Lateral ArmTwo180 26.85 162 18.4 0 0 0 0 0

7 Return Lateral ArmTwo180 17.01 162 0 0 0 0 0 08 Get Top cardboard LiftArm 0.24 18 0 40.66 42.66 0 0 0

9 Grasp and move cardboard Lateral ArmStand90 0.66 18 1 0 0 0 0 0 0

10 Return Lateral ArmStand90 0.60 18 0 0 0 0 0 0 0 Figure 2.Adjusted Energy Expenditure Spreadsheet Results

Total Energy Expenditure from spreadsheet in Figure 2: 125.38 kcal/cycle, 8.36 kcal/minPercent Improvement = (New-Old)/Old = (|125.38-160.37|)/160.37 = 21.8% improvement



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Work/Rest Cycle Determination

Assumption: Percentile for aerobic capacity corresponds to population percentile for body weight.Worker Body Weight Percentile: 78.85 %

E = 8.36 kcal/min

MAC = 15.8 kcal/mi78.85%ile MAC = 17.0 kcal/min

fVO2,max = E/MAC = 8.36/17 = 0.492

fVO2,max < 0.56109 so it is possible to work continuously

Change in Work/Rest Cycle for Temperature

Dry Bulb Temperature: 30 CWet Bulb Temperature: 25 C

On the chart of expected increase in heart rate in Appendix 1, these WB and DB parameters are in the range foran increase of 1 bpm/1 C > 25 C. Therefore, there would be an expected increase of (30-25) = 5 bpm.

fVO2,max increase = 0.74%/bpm * 5 bpm = 3.7 %

fVO2,max (Adjusted) = (0.492)*1.037 = 0.510fVO2,max < 0.56109 so it is possible to work continuously



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Kyle ShippIME 462 Ergonomics

Lab 4B 2-D and 3-D Biomechanical Analysis

2D vs. 3D Static Strength Analysis Comparison

For the purposes of these evaluations, I am assuming the 3D model provides the most correct data because we

can model the position more exactly. By devolving the 3D positions to 2D positions, we lose some of the data if

the position is asymmetric. Axial rotation and lateral bending are also not represented in the 2D analysis. The

original data sheet with the angles used for the 2D and 3D analysis is located in Appendix A.Areas where the 2D Analysis has a large variance from the 3D Analysis are highlighted in yellow in the

following tables.

Force Strength Force Strength

Description (N) % Capable Description (N) % Capable

Elbow (each) -64.8 99 Elbow (each) -62.8 100Shoulder (each) -93.8 99 Shoulder (each) -90.0 99

L5/S1 (Hip-L5-T4) -527.2 99 L5/S1 (Hip-L5-T4) -570.2 94

Hip (each) -353.0 85 Hip (each) -69.5 73

Knee (each) -450.7 71 Knee (each) -188.4 39

Ankle (each) -493.7 62 Ankle (each) -228.5 30

L5/S1 Comp. 3029.9 L5/S1 Comp. 3091.5

2D

Analysis

3D

Analysis

Low Near Position

Low Near 3D Low Near 2D



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Description (N) % Capable Description (N) % CapableElbow (each) -64.8 99 Elbow (each) -62.8 99

Shoulder (each) -93.8 99 Shoulder (each) -90.0 99

L5/S1 (Hip-L5-T4) -527.2 99 L5/S1 (Hip-L5-T4) -570.2 79

Hip (each) -353.0 85 Hip (each) 3.1 95



L5/S1 Comp. 3023.9 L5/S1 Comp. 4565.9

2D

Analysis

3D

Analysis

Low Far Position

Low Far 3D Low Far 2D

For the low far position, there are major differences in the Force and percent of persons who are strength

capable. Looking at the graphic depictions of the 2D and 3D models it is obvious there are distinctdisadvantages to a 2D modeling approach. His arm position is almost symmetrical so it makes sense the values

for Elbow and Shoulder would be similar. The right foot is far ahead of the left foot so there is actually more

balance in the 3D model than there appears to be in the 2D model. That is why the values for Hip, Knee, andAnkle force are much more negative in the 2D model. The interesting thing to note about the L5/S1 joint is the

force numbers do not appear to be much different but there is a difference of 20% in strength percent capable.

The L5/S1 compression force is much greater in the 3D model. The drastically different foot position willincrease the compression on the L5/S1 joint.



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L5/S1 (Hip-L5-T4) -527.2 99 L5/S1 (Hip-L5-T4) -90.0 99

Hip (each) -353.0 99 Hip (each) -570.2 98



L5/S1 Comp. 1526.1 L5/S1 Comp. 1033.5

High Near Position

2D

Analysis

3D

Analysis

High Near 3D High Near 3D

The strength percent capable values for the high near position are almost identical. The only differences in the

data are in L5/S1 force, L5/S1 compression and hip force. Looking at the graphic of the 3D high near positionwe can see he is laterally bending toward the left side or conveyor. There is also an asymmetry in the arm

position. Each of these slight changes would make a difference on the hip/L5/S1 area.



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L5/S1 (Hip-L5-T4) -527.2 99 L5/S1 (Hip-L5-T4) -570.2 98

Hip (each) -353.0 97 Hip (each) 171.9 95

Knee (each) -450.7 99 Knee (each) 53.0 97

Ankle (each) -493.7 99 Ankle (each) 13.4 94

L5/S1 Comp. 1879.5 L5/S1 Comp. 4031.4

High Far Position

2D

Analysis

3D

Analysis

High Far 3D High Far 2D

The high far position demonstrates the inability of 2D methods to produce an accurate model for certain

postures. In this position, the legs are mostly symmetrical but from the hips up, he is leaning far to the right side

with his hand outstretched. This posture is impossible to model in 2D from the side view. A better approach formodeling this posture in 2D would be to do it from a front view. I think another difference is the 2D model has

a blue shirt on while the 3D model has a green shirt.

It is interesting to note that the strength percent capable values are almost the same for all joints except theshoulder. I believe this is strictly a coincidence that shows why it is important to look at the specific value and

not only the percent capable. The L5/S1 compression is much greater in the 3D analysis than the 2D because

there is so much compression on the right side from the lateral bending.



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Static Muscle Fatigue for Low Far Posture

For this analysis I am assuming the percentile for muscle strength corresponds to the population percentile forstanding height.

50th %ile male standing height = 68.7 2.6 inches

Operator standing height = 71.0 inchesOperator standing height percentile = 81.2%

Calculation Notes:

Percentile calculated using Excel formula =normdist(x,mean, std_dev, cumulative)Wherex is the subjects measurement, mean is the 50

th percentile value, std_dev is the standard deviation of the

values, and cumulative is a logical operator set to True to return the cumulative distribution function as opposed

to the probability mass function.

Torso Extension Population Mean = 315 n-m

Standard Deviation = 99 n-m

81.2%ile 90 Back Extension MVC = 403 n-m

Torso Extension force from 3DSSPP = 234 n-mfMVC = 234/403 = 0.58

Static Endurance Time = twork= (1.25/fMVC) 1.35 = (1.25/0.58) 1.35 = 0.81 minutes

For the static rest allowance, I am assuming the contraction time (t) for each hold is 2 seconds or 0.033 minutes.

Static rest allowance = RA = 18(t/T)1.4(fMVC-0.15)0.5 = 18(0.033/0.81)1.4(0.58-0.15)0.5 = 0.13

Static Rest Time = trest = RA(t) = 0.13*0.033 = 0.00429 minutes

Minimum Cycle Time < twork+ trest = 0.81 + 0.00429 = 0.81429 minutes

The rest time is very low compared to the contraction time the worker should have the appropriate amount of

rest while he returns to the pickup position. An analysis would have to be done of him moving without a load

in his hand to determine if the worker would be able to work at this cycle time.



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Reduction of Muscle Strength Requirements

By placing the pallet on a lift so it would always be in the high position I estimate we could reduce the musclestrength requirements for this task and therefore increase the static endurance time. This would eliminate thetwo low positions and only leave the high positions for analysis. For comparison purposes, I will calculate the

static work time for the high far position to see the improvement.

Torso Extension Population Mean = 231 n-m

Standard Deviation = 73 n-m

81.2%ile 90 Back Extension MVC = 296 n-m

Torso Extension force from 3DSSPP = 45 n-m

fMVC = 45/296 = 0.15

Because fMVC = 0.15 we can assume the work is able to be done continuously.

The ability to put the pallet on a lift will greatly decrease the fMVC for this task. This would improve his workability and should decrease his injuries.



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APPENDIX A

Hand Calculations for Part A1

Charts for Part A3

Hand-written data collection sheet for Part A

Hand-written data collection sheet for Part B



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