Presentation on theme: "Khalid Alamri Mohammed Alzayer Andres Glasener Mat E 316 April 28 th, 2014 Young’s Modulus: Statistical Analysis."— Presentation transcript:
Khalid Alamri Mohammed Alzayer Andres Glasener Mat E 316 April 28 th, 2014 Young’s Modulus: Statistical Analysis
Outline Basics Objectives Experiment o Apparatus o Procedures Statistical Analysis: o List of Factors Blocking Factors Experimental Factors Response Variable o Modeling Deflection o Application of Model Conclusion and Suggestions
Basics Stress and strain are related by: σ= E ε σ: Applied stress (Pa) usually in MPa ε: Strain resulted from a load (mm/mm) E: Young’s modulus (Pa) usually in GPa E is the slope of the elastic regime of a stress-strain curve. The steeper the slope, the stiffer the material
Objectives Purpose of experiment? find E for 4 samples of rods. Materials? steel, glass, plastic, wood. How? by measuring deflection resulted from applying increasing loads. Purpose of this study? come up with a well fit model that describes the deflection.
Experimental Procedures 1 1.Radius of rod and span length were measured, and the apparatus was set-up. 2.The dial on the apparatus was calibrated at a weight of zero. Sample Supports Bucket Dial Hook 1 Cademartiri, Rebecca. Mat E 215L – Young’s modulus. 2012. Web.
Experimental Procedures 1 3.Max affordable weight for each rod was calculated. Given max stresses of: Glass: 60 MPa Plastic: 60 MPa Steel: 200 MPa Wood: 100 MPa 4.Weight of bucket was measured. First deflection was recorded with empty bucket. 5.Weight was added to the bucket, in the form of water. Deflection was recorded for 5 different weights. 1 Cademartiri, Rebecca. Mat E 215L – Young’s modulus. 2012. Web.
List of Factors FactorsType of factorLevels Operator Blocking factor 1) Khalid 2) Mohammed 3) Andres Material Experimental variable (Nominal) 1) Steel 2) Glass 3) Plastic 4) Wood
List of Factors FactorsUnitsType of factorLevels Radius (r)m Experimental variable (Continuous) 1) Khalid Steel: 0.002 Glass: 0.002 Plastic: 0.004 Wood: 0.0025 2) Mohammed Steel: 0.002325 Glass: 0.002 Plastic: 0.00475 Wood: 0.003075 3) Andres Steel: 0.0024525 Glass: 0.002025 Plastic: 0.004775 Wood: 0.00396
List of Factors FactorsUnitsType of factorLevels Span length (L) m Experimental variable (Continuous) 1) 0.293 2) 0.285 3) 0.295 Each level corresponds to an operator. Load (P)N59 Deflection (d)mResponse variable N/A
Modeling Deflection One model for the entire data? not possible. Why? not enough data. Materials behave differently. What to do? model each material separately. Advantages? Considers all variables and interactions. Better fit.
Application of Models Stress and strain equations derived from the geometry of the apparatus: Deflections were calculated using the new models. They were plugged into the strain equation. Stresses were also calculated using raw data. Stress versus strain plots were obtained using Excel.
Application of Models MaterialOperator Original E (GPa) Predicted E (GPa) Original R 2 New R 2 Steel Mohammed221.116222.7510.98470.992 Khalid576.696590.7040.97270.9963 Andres167.675168.4830.99180.9966 Glass Mohammed65.9666.1220.99811 Khalid66.00466.560.98821 Andres62.25663.4710.98371
Application of Models MaterialOperator Original E (GPa) Predicted E (GPa) Original R 2 New R 2 Plastic Mohammed4.12774.38170.94070.9986 Khalid11.87611.9340.99030.9951 Andres3.20873.270.9980.9987 Wood Mohammed20.45320.4710.99911 Khalid32.56233.8650.96151 Andres11.65111.8750.99951
Conclusions How accurate are the models? Pretty accurate. Better stress-strain curve fit. Same material, different E, why? some samples were plastically deformed, i.e. cold worked when they were tested. Steel Example: OperatorE (GPa) Mohammed222.751 Khalid590.704 Andres168.483 2 Callister, William. Materials science and engineering: an introduction. 2010. Print. Figure: Real steel stress-strain curves. 2
Conclusions Operator effect? No effect. Wood is an exception according to JMP. Why? It has low degree of structural consistency, due to its composite nature. JMP emphasized effect of operator and operator-load interaction in wood, why? Operator does not really affect results. It’s JMP way to say wood behavior is difficult to predict.
Suggestions 1.Emphasizing the role of interactions: –Big variations in radii and lengths in plastic data makes its model the best. It has more effective interactions than other materials. –Maximize the differences between a factor’s levels for better results in future. 2.Decreasing the magnitude of applied loads: –Apply much smaller loads than allowable to reduce the effects of fatigue and get identical results of E for the same material.