Wednesday, May 6, 2020
Analyzing Stanley Milgrams The Lost Letter Experiment
The Lost Letter: Stanley Milgrams experiment Stanley Milgram is one of the most influential of the American postwar social scientists. Milgrams reputation lies not so much in his construction of wide, over-arching theories but in his ability to create provocative, strange even controversial experiments that test conventional notions of moral behavior. Although profoundly different, many of Milgrams experiments possess a common theme that of the situation-specific nature of morality. Humans were moral, his experiments suggested, not out of character or innate virtue, but based upon social pressures. Milgrams most famous experiment, rooted his knowledge of Nazi war crimes and groupthink, involved orchestrating an environment in which seemingly ordinary people were encouraged to administer what they believed to be fatal electric shocks to other experimental subjects (who were really Milgrams confederates). The overwhelming majority complied, and roughly 65 per cent of subjects continued to administer shocks up to the maximum of 450 volts despite the apparent screams of pain from their victim, results that have been replicated cross-culturally (Russell 2009). In a similar experiment which replicated Milgrams findings, seminarians were unwilling to help someone in physical distress when instructed to hurry from one building to another to deliver a moral sermon (Russell 2009). Although inspired by the postwar climate, these experiments did not specifically addressShow MoreRelated65 Successful Harvard Business School Application Essays 2nd Edition 147256 Words à |à 190 Pagesopportunity to devise the plan that jump..started their involvement. This experience also highlighted leadership strengths that balance this weakness. Most important, I am a good listener. Openness to others input allows me a broader perspective for analyzing problems and leads to more thorough solutions. Furthermore, I pour my heart into everything I do. My enthusiasm enables me to work well in teams, motivate others, and create a fun and sup.. portive team environment. analysis This essay is evidenceRead MoreStephen P. Robbins Timothy A. Judge (2011) Organizational Behaviour 15th Edition New Jersey: Prentice Hall393164 Words à |à 1573 PagesAbout Organizational Behavior? 4 Myth or Science? ââ¬Å"Most Acts of Workplace Bullying Are Men Attacking Womenâ⬠12 An Ethical Choice Can You Learn from Failure? 24 glOBalization! Does National Culture Affect Organizational Practices? 30 Point/Counterpoint Lost in Translation? 31 Questions for Review 32 Experiential Exercise Workforce Diversity 32 Ethical Dilemma Jekyll and Hyde 33 Case Incident 1 ââ¬Å"Lessons for ââ¬ËUndercoverââ¬â¢ Bossesâ⬠34 Case Incident 2 Era of the Disposable Worker? 35 vii viii CONTENTS Read MoreDeveloping Management Skills404131 Words à |à 1617 PagesTodd Dewett, Wright State University Andrew J. Dubrin, Rochester Institute of Technology Steven Edelson, Temple University Norma Givens, Fort Valley State University Barbara A. Gorski, St. Thomas University David Hampton, San Diego State University Stanley Harris, Auburn University Richard E. Hunt, Rockhurst College Daniel F. Jennings, Baylor University Avis L. Johnson, University of Akron xx PREFACE Jay T. Knippen, University of South Florida Roland Kushner, Lafayette College Roy J. Lewicki, Ohio
Bending of an Aluminum beam free essay sample
ââ¬Å"Beams are long straight members that are subjected to loads perpendicular to their longitudinal axis and are classified according to the way they are supportedâ⬠[1]. When a beam is subjected to an external load there are unseen internal forces within the beam that one must be aware of when implementing it into any design or structure. These internal forces create stress and strain that could result in failure or deformation. This lab looked at how an aluminum cantilevered beam performed under symmetric and unsymmetrical bending as well as the stresses and strains developed as a result. Objectiveââ¬Å"To study the stress and strain induced in an I-beam under symmetric and unsymmetrical bendingâ⬠[2]. Theory: ? ââ¬â Normal stress (Mpa) ? ââ¬â Strain (mm/mm) M ââ¬â Moment (kNâ⬠¢m) I ââ¬â Moment of inertia (mm^6) E ââ¬â Modulus of elasticity (Mpa) G ââ¬â Modulus of elasticity (Mpa) v ââ¬â Poissonââ¬â¢s ratio. L ââ¬â Length (m) *Subscripts x, y, z indicate plane of reference. We will write a custom essay sample on Bending of an Aluminum beam or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page The strain rosettes are orientated so that ? b = 0, ? c = -45, and ? a = 45. The strain gauge equations then simplify to ?x = ? b, ? y= ? c+ ? a- ? b, and ? xy = ? c- ? a Using Hookeââ¬â¢s Law: ?x= ? xE, ? y= -v ? x, ? xy=? xyG This Experiment consisted of symmetric and unsymmetrical bending. For symmetric bending the relevant theory is as follows: Because the moment about the z-axis here is zero the equation equates to: Where: My = PLA. When rotated 45 degrees: My = PLA Cos(45) and Mz = PLA Sin(45) there is compressive stress along the y-x axis The moment of inertia about the y-axis is found by determining the inertia of the shape and subtracting the imaginary parts as shown: The max normal stress with be at the furthest distance from the neutral axis which is h/2 therefore: (? x)max = The strains can be found by implementing Hookeââ¬â¢s Law: Since ? y and ? z are zero in symmetric loading, the two equations simplify to:Because the there is no shear stress in the x-y plane when the normal stress is at maximum the shear strain will also be zero. The vertical displacement of the end of the beam is determined by multiplying the area under the moment diagram and the distance between the end and the centroid of the diagram. This equates to: For unsymmetrical bending the th eory is the same however there is a moment about the y-axis and z-axis. This will affect the calculation of the normal stress and the strain in the x and y plane. Also the moment of inertia in the z-direction will need to be determined.Procedure (a) *Mount the I-beam on to the support frame. Make sure the mounting screws are tight. (b) Measure the dimensions of the I-beam including its components. (c) Mount the magnet bases of the dial gauges at appropriate positions to permit the measurements of the deflections at the free end of the beam in the vertical and the horizontal directions. (d) *Connect properly the wires from the strain gauges to the readout unit. (e) Place weights to the hanger in increments: 4, 6, 10, 26, and 42 kg. (f) Unload the hanger in increments in the reversed order as for loading.(g) For each increment, measured the strain readings at the given locations and the vertical and horizontal deflections at the free end of the beam. (h) Repeat steps (a) to (g) by rotating the beam with the following angles: 45à °. [3] Results *Refer to appendix for sample calculation and calculated results. Part 1: I-beam at 0o Loading Loading (Kg) 4 6 10 26 42 Strain Gauge 1 (? ) 1 2 4 12 20 Strain Gauge 2 (? ) 6 10 16 43 69 Strain Gauge 3 (? ) 3 4 7 18 29 Displacement 1 (mm) 0. 09 0. 15 0. 23 0. 44 0. 5 Displacement 2 (mm) -0. 19 -0. 34 -0. 55 -1. 4 -2. 25 Load (N) 39. 2 58. 5 97. 9 255. 5 413. 1Unloading Loading (kg) 42 26 10 6 4 Strain Gauge 1 (? ) 20 10 -3 -5 -7 Strain Gauge 2 (? ) 69 42 19 11 9 Strain Gauge 3 (? ) 29 18 6 3 2 Displacement 1 (mm) 0. 5 0. 49 0. 25 0. 16 0. 07 Displacement 2 (mm) -2. 25 -1. 46 -0. 59 -0. 37 -0. 23 Load (N) 413. 1 255. 6 96. 4 58. 7 39. 2 Part 2: I-Beam at 45o Loading Loading (kg) 4 6 10 26 42 Strain Gauge 1(? ) 1 2 2 7 13 Strain Gauge 2 (? ) 5 9 14 36 54 Strain Gauge 3 (? ) 1 1 2 8 13 Displacement 1 (mm) -0. 33 -0. 50 -0. 79 -1. 88 -2. 75 Displacement 2 (mm) -0. 66 -1. 02 -1. 69 -4. 23 -6. 40 Load (N) 39. 4 58. 7 98. 2 256. 5 413. 6 Unloading Loading (kg)42 26 10 6 4 Strain Gauge 1 (? ) 13 4 -22 -25 -26 Strain Gauge 2 (? ) 54 38 22 20 17 Strain Gauge 3 (? ) 13 6 2 0 0 Displacement 1 (mm) -2. 75 -1. 95 -0. 92 -0. 62 0. 46 Displacement 2 (mm) -6. 40 -4. 46 -2. 17 -1. 51 -1. 15 Load (N) 413. 6 256. 3 98. 1 58. 7 39. 4 Discussion For both the symmetric and unsymmetrical bending the theoretical stresses and strains were greater than experimentally determined ones. However the experimental displacement was much higher than the theoretical displacement. These two factors can lead one to believe the I-beam has undergone this procedure many times before.Another interesting point to note is that the stresses and strains are higher at equivalent loads when unloading demonstrating that there is residual stress in the I-beam even after it has been fully unloaded. For the most part however the measured and theoretical values are very close. It is to be expected that the theoretical stresses would be higher than the experimental values. The theoretical calculations rely on a ââ¬Ëperfectââ¬â¢ material. The modulus of elasticity and cross-sectional are said to remain the same through the length of the beam which is rarely the case. Minor imperfections in the beam would result in a weaker beam and less stress is required to deflect the beam. This is exactly what has been observed in this experiment. For the symmetric and bending theoretically there would be no horizontal displacement however some horizontal displacement was shown on the readouts. This is most likely due to the slight swaying of the weights. Since the scale of this experiment was relatively small a lot of the sources of error are pretty large. Just by not having the readout computer not calibrated properly or zeroed all the way would cause pretty large discrepancies.Even the measuring or millimeters by eye caused some error. Rounding errors would be relatively small for this experiment. Conclusion In conclusion theoretical and experimental values for stress and strain are very similar to the values observed in experimental conditions. The theoretical and experimental displacements were pretty far off and at larger scales the theoretical values would not be of much use. Closer results could have been obtained by collecting more accurate measurements or by collecting multiple sets of data using a series of strain rosettes. APPENDIX I Sample Calculations Iy= = (Mz)a =(4kg)(9. 81m/s2)(0.77m) =30. 215 Nm (Mz)b =(4kg)(9. 81m/s2)(0. 33m) =12. 95 Nm (? x)a = = = 1. 259 Mpa (? x)b = 0. 5397 Mpa (? b)v = = = -0. 0902 mm ?xy = = = 0. 0398mm (? x)a = = =17. 22*10^-6 (? y)a = -0. 35*(? x)a = 6. 027*10^-6 Experimental Symmetric Mass (Kg) 4 6 10 26 42 26 10 6 4 ?x (E-6) 6 10 16 43 69 42 19 11 9 ?y (E-6) -2 -4 -5 -13 -20 -14 -16 -13 -14 ?xy (Mpa) 2 2 3 6 9 8 9 8 9 (? x) (Mpa) 0. 438 0. 731 1. 17 3. 14 5. 04 3. 07 1. 39 0. 804 0. 657 (? y) (Mpa) -0. 146 -0. 292 -0. 365 -0. 950 -1. 46 -1. 02 -1. 17 -0. 950 -1. 02 ?xy (Mpa) 0. 054 0. 054 0. 081 0. 162 0. 243 0. 216 0. 243 0. 216 0. 243 Theoretical ââ¬â SymmetricMass(Kg) 4 6 10 26 42 (Mz)a (Nâ⬠¢m) 30. 2 45. 3 75. 5 196 317 (Mb)b (Nâ⬠¢m) 12. 9 19. 4 32. 3 84. 1 135 (? x)a (Mpa) 1. 25 1. 88 3. 12 8. 13 13. 1 (? x)b (Mpa) 0. 536 0. 804 1. 34 3. 48 5. 62 ?xy (Mpa) 0. 0398 0. 0598 0. 0996 0. 258 0. 418 (? x)a (E-6) 17. 1 25. 7 42. 8 111 179 (? x)b (E-6) 7. 33 11. 0 18. 3 47. 6 77. 0 (? y)a (E-6) -5. 99 -8. 98 -14. 9 -38. 9 -62. 8 (? y)b (E-6) -2. 57 -3. 85 -6. 41 -16. 6 -26. 9 ?a (mm) 0. 0902 0. 135 0. 225 0. 586 0. 947 ?b (mm) 0. 00710 0. 0106 0. 0177 0. 0461 0. 0745 Experimental ââ¬âUnsymmetrical Bending Mass (Kg) 4 6 10 26 42 26 10 6 4 (? x) (E-6) 5 9 14 36 54 38 22 20 17 (? y) (E-6)-3 -6 -10 -21 -28 -28 -42 -45 -43 ?xy (E-6) 0 -1 0 1 0 2 24 25 26 (? x) (Mpa) 0. 366 0. 658 1. 02 2. 63 3. 95 2. 78 1. 61 1. 46 1. 24 (? y) (Mpa) -0. 219 -0. 439 -0. 731 -1. 54 -2. 05 -2. 05 -3. 07 -3. 29 -3. 14 Theoretical ââ¬â Unsymmetrical Bending Mass (Kg) 4 6 10 26 42 (Mz,y)a (Nâ⬠¢m) 21. 3 32. 0 53. 4 138 224 (Mz,y)b (Nâ⬠¢m) 9. 15 13. 7 22. 9 59. 5 96. 1 (? x) (Mpa) 0. 381 0. 572 0. 954 2. 48 4. 00 (? y) (Mpa) -1. 40 -2. 10 -3. 51 -9. 12 -14. 7 (? x) (E-6) 5. 22 7. 83 13. 1 33. 9 54. 8 (? y) (E-6) 1. 83 2. 74 4. 57 11. 9 19. 2 ?x-y (mm) 0. 0902 0. 135 0. 225 0. 586 0. 946 ?x-z (mm) 0. 391 0. 587 0. 978 2. 54 4. 11
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