Report

Young’s Modulus: Statistical Analysis Khalid Alamri Mohammed Alzayer Andres Glasener Mat E 316 April 28th, 2014 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 Procedures1 Dial 1. Radius of rod and span Supports length were measured, and the apparatus was set-up. Hook 2. The dial on the apparatus was calibrated at a weight of zero. 1 Cademartiri, Rebecca. Mat E 215L – Young’s modulus. 2012. Web. Sample Bucket Experimental Procedures1 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 Factors Operator Material Type of factor Blocking factor Levels 1) Khalid 2) Mohammed 3) Andres 1) Steel Experimental variable 2) Glass (Nominal) 3) Plastic 4) Wood List of Factors Factors Units Type of factor Radius (r) m Experimental variable (Continuous) Levels 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 Factors Units Type of factor Span length (L) m Load (P) N Deflection (d) m Experimental variable (Continuous) Levels 1) 0.293 2) 0.285 3) 0.295 Each level corresponds to an operator. 59 Response 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. Modeling Deflection Material Equation Other terms R2 Adj R2 Steel -0.0074 + 0.58*r + 0.020*L + 8.99E-5*P +0.0013*[(L + 0.08*[(r-0.29)(P0.99 0.0023)*(P-10.13)] + 10.13)] 1.44E-6*(P-10.13)2 0.98 Glass +0.0058*[(P -3.93)*(L0.99 0.29)] 0.98 -0.027 + 19.51*r 0.043*L + 6.15E-4*P Modeling Deflection Material Equation Other terms R2 Adj R2 Plastic -0.16 + 3.72*r + 0.50*L + 3.02E-4*P + 0.020*[(r+1.76E-6 0.0045)*(P-19.86)] + *(P0.014*[(L-0.29)*(P19.86)2 19.86)] 0.99 0.98 Wood 3.58E-4*P - 8.33E4*(Andres) - 1.24E- 6.38E-4 4*[(Andres)*(P-11.75)] + 3.64E0.97 1.47E-4*[(Khalid)*(P4*(Khalid) 11.75)] 0.96 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 Original E Predicted Original New R2 (GPa) E (GPa) R2 Material Operator Steel Mohammed 221.116 222.751 0.9847 Khalid 576.696 590.704 0.9727 0.9963 Andres 167.675 168.483 0.9918 0.9966 Mohammed 65.96 66.122 0.9981 1 Khalid 66.004 66.56 0.9882 1 Andres 62.256 63.471 0.9837 1 Glass 0.992 Application of Models Original E Predicted Original New R2 (GPa) E (GPa) R2 Material Operator Plastic Mohammed 4.1277 4.3817 0.9407 0.9986 Khalid 11.876 11.934 0.9903 0.9951 Andres 3.2087 3.27 0.998 0.9987 Mohammed 20.453 20.471 0.9991 1 Khalid 32.562 33.865 0.9615 1 Andres 11.651 11.875 0.9995 1 Wood Stress-Strain Graphs Glass 200 180 160 140 120 100 80 60 40 20 0 52.5 Stress, σ, MPa Stress, σ, MPa Steel 47.5 42.5 37.5 0 32.5 0.0003 0.0002 0.0004 0.0006 0.0008 Strain, ɛ Plastic 160 120 Stress, σ, MPa Stress, σ, MPa 140 100 80 60 40 20 0 0 0.001 0.002 Strain, ɛ 0.0007 Strain, ɛ Wood -0.001 0.0005 0.003 0.004 50 45 40 35 30 25 20 15 10 5 0 0 0.005 Strain, ɛ 0.01 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. Figure: Real steel stress-strain curves. • Steel Example: 2 Operator E (GPa) Mohammed 222.751 Khalid 590.704 Andres 168.483 Callister, William. Materials science and engineering: an introduction. 2010. Print. 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.