Documents about Von Mises

  • 14 Pages

    HW6

    Maryland, ENME 000

    Excerpt: ... Time Aigbe ENME470 Sec. 0101 HW6 - Page 1 of 14 6-1. von Mises Stress Plot Displacement Distribution Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 2 of 14 1st Principal Stress Distribution Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 3 of 14 6-2. Sensitivity Study Results: 6-3. von Mises Stress Distribution (Left) and Displacement Distribution (Right) Plots: Time Aigbe ENME470 Sec. 0101 HW6 - Page 4 of 14 6-4. von Mises Stress Distribution Plot: Deformation Pattern Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 5 of 14 6-5. Temperature Distribution Plot: 6-6. Flow Trajectories: Time Aigbe ENME470 Sec. 0101 HW6 - Page 6 of 14 6-7. CASE 1: Camera dropped on rigid surface Displacement Distribution Plot: von Mises Stress Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 7 of 14 CASE 2: Camera dropped on soft surface Displacement Distribution Plot: von Mises Stress Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 8 of 14 CASE 3: Camera ...

  • 14 Pages

    HW11

    Maryland, ENME 470

    Excerpt: ... Time Aigbe ENME470 Sec. 0101 HW6 - Page 1 of 14 6-1. von Mises Stress Plot Displacement Distribution Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 2 of 14 1st Principal Stress Distribution Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 3 of 14 6-2. Sensitivity Study Results: 6-3. von Mises Stress Distribution (Left) and Displacement Distribution (Right) Plots: Time Aigbe ENME470 Sec. 0101 HW6 - Page 4 of 14 6-4. von Mises Stress Distribution Plot: Deformation Pattern Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 5 of 14 6-5. Temperature Distribution Plot: 6-6. Flow Trajectories: Time Aigbe ENME470 Sec. 0101 HW6 - Page 6 of 14 6-7. CASE 1: Camera dropped on rigid surface Displacement Distribution Plot: von Mises Stress Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 7 of 14 CASE 2: Camera dropped on soft surface Displacement Distribution Plot: von Mises Stress Plot: Time Aigbe ENME470 Sec. 0101 HW6 - Page 8 of 14 CASE 3: Camera ...

  • 1 Pages

    SOC 220 Intro

    Purdue, SOC 220

    Excerpt: ... 2008_01_09 Lecture 2: Introduction continued Wednesday, January 09, 2008 1:38 PM Durkheim - Anomie Skeptical of psychology and capitalism Favored looking at macro level Suicide studies Suicide rates w/in countries tended to be stable Variation between countries and regions (southern regions had lower suicide rates) Upward trend across negative dependent variables Anomie High social density = small town, everyone knows each other Low social density = urban area Social norms stronger in high social density - according to Durkheim, this is good, as it reduces anomie, (deviance, crime, suicide) Small, close knit communities => + Social density => +Social norms => -anomie => + anomic suicide C. Wright Mills - The Sociological Imagination Start w/ your own life, mindset, experiences, broaden to friends, examining problems and issues Take these problems to the societal level, looking for societal reasons Ludwig Von Mises - Human action Austrian free-market economist Wrote the book Human Action Arg ...

  • 3 Pages

    ME475.ConnectingLug_SS09

    Michigan State University, ME 475

    Excerpt: ... xamine the results. For each mesh density, record the predicted maximum (tensile) and minimum (compressive) von Mises stresses and maximum 2-component of displacement (magnitude). Once all of these values are recorded, plot the stresses and displacement versus number of nodes in the model. The number of nodes can be found by opening the input deck (the *.inp file) and finding the keyword toward the top that starts with *Node. The next keyword should be *Element, search for this. The line above will be the last node written, and the first value is the number of nodes in your model. This is because Abaqus/CAE starts its node numbering from 1, and numbers sequentially. Not all preprocessors do this. For the most refined mesh (seed of 0.003 m), the largest node number will be 19257. Note our educational license of Abaqus is limited to 20,000 nodes. Model Validation: The connecting lug is not too dissimilar from a cantilevered beam with a rectangular crosssection and similar dimensions. Perform this calculation a ...

  • 7 Pages

    ANSYS

    RIT, P 08003

    Excerpt: ... Handle Von Mises Stress Handle Deformation Elements Boxtop Von Mises Stress Deformation: Side View: ...

  • 3 Pages

    StaticFailureTheories

    Mines, L 411

    Excerpt: ... MD: Static Failure Theory Norton,Chapter 5, pages 253-316. In[10]:= SetDirectory@ "UsersjsteeleDocumentsEG411MachineDesignLecturesL5StaticFailureTheory"D ut[10]= UsersjsteeleDocumentsEG411MachineDesignLecturesL5StaticFailureTheory Maximum Distortion Energy Theory also called the Octahedral stress or von Mises -Hencky theory In[11]:= Show@fig0503 = Import@"figuresFigure05-03.jpg"D, ImageSize -> 700D; 2 StaticFailureTheories.nb The von Mises effective stress, seff , The plot above used the following equivalence with s2 = 0, ! ! ! ! Sy = s1 2 + s2 2 + s3 2 - s1 s2 - s2 s3 - s3 s1 is defined as the uniaxial tensile stress that would create the same distortion energy as is created by the actual combination of applied stresses. This is the expression for the effective stress using principal stresses. seff = "# s1 s3 - s2 s# s2 - s1 s2 + s2# ...

  • 3 Pages

    MECH320-Lab2-CBT-Manual-2009

    Virgin Islands, MECH 320

    Excerpt: ... tory, we will be evaluating two particular failure criteria, which apply to ductile failure, since we will use ductile metal samples. These two failure criteria are: (a) Maximum Shearing Stress Theory (a.k.a. Tresca Yield Criterion, 1864) (b) Maximum Distortion Energy Theory (a.k.a. von Mises Theory, 1913) For most metals yielding occurs at very small distortional strain, usually less than 10 . Until yield occurs, the relation between stress and strain is essentially linear, that is -2 MECH 320 Lab Manual -1- Hooke's law holds. After yielding occurs the relation between stress and strain is, in general, nonlinear and there isn't a one-to-one relation between stress and strain. (a) Tresca yield condition According to the Tresca yield condition, yielding occurs when the maximum shearing stress reaches a critical value k. In other words, the limit of elasticity occurs when the maximum shearing stress reaches k. We know that the maximum shearing stress is given by: max x - y 2 1 - 2 2 = + xy = ...

  • 30 Pages

    me383_note02

    Alabama, ME 383

    Excerpt: ... .5 15 lf Example 2 = 100,000 0.5 psi Calculate True and Engineering =n UTS at necking n 0.5 = Kn = 1 00,000( 0.5) = 70710 A0 = n = 0.5 Aneck = A0 e -0.5 ln A neck P = A = A0 e -0.5 P = 42,850 A0 UTS P = = 42,850 psi A0 16 End Questions ? 17 Today's Lecture Stress-Strain State: Hooke's Law Yield Criteria: 1) Tresca 2) von Mises Effective Stress and Strain Work of Deformation and Temperature Case Studies 18 Stress State - Triaxial Stress Equilibrium: Principal Stress 1 2 3 M F =0 i i =0 3 1 2 19 Strain State - Triaxial Strain Principal Strain 1 2 3 3 1 2 20 Stress-Strain Relationship in Triaxial State Generalized Hooke's Law 1 1 = [ 1 - ( 2 + 3 ) ] E 1 2 = [ 2 - ( 1 + 3 ) ] E 1 3 = [ 3 - ( 1 + 2 ) ] E Example: In Tension, = = 0 2 3 1 1 = E 1 2 = 3 = - E 21 Yield Criteria Tresca max( 1 - 2 , 2 - 3 , 1 - 3 ) = y von Mises ( 1 - 2 ) 2 + ( 2 - 3 ) + ( 1 - 3 ) = 2 y 2 2 2 Difference < 15% ...

  • 5 Pages

    Lecture10

    University of Michigan, MECHENG 382

    Excerpt: ... ME 382 Lecture 10 Safety factor on stress = Ratio uniaxial yield stress to effective normal stress Safety factor on pressure = Ratio of pressure that would cause yield / pressure Safety factor on thickness = Ratio of thickness / thickness that would yield Sf > 1 (for design against failure) In this example: Safety factor against yield ! S = 500 MPa & Y = 760 MPa Safety factor against yield: Sf = Y/ ! S =760/500 = 1.5 Example: Plane stress 3 = 0 - conditions for yield with different values of 1 and 2? (i) If 1 & 2 > 0 or if 1 & 2 < 0 (a) (b) If ! 2 < ! 1 yield if ! 1 " ! y If ! 2 > ! 1 yield if ! 2 " ! y (ii) If 1 > 0 & 2 < 0 or if 1 < 0 & 2 > 0 Yield if ! 1 " ! 2 # ! y We can draw a yield surface for 3= 0 (Can also do this in 3-D if 3 0) Experiments indicate that an ellipse describes yielding better 26/ix/07 3 ME 382 Lecture 10 This observation leads to von Mises yield criterion Von Mises yield criterion (Octahedral shear stress criterion ...

  • 2 Pages

    ME475.ConnectingLug

    Michigan State University, ME 475

    Excerpt: ... meshes. You can do this by writing out each model with a different job name, and then executing each one in series from the command line. This will also ensure your results are not overwritten each time. After running the analysis with the four meshes of varying density, examine the results. Choose two locations somewhere on the structure. One location should be on the surface of the hole, another on the surface aligned with the X-Y plane, but not right next to the surface where the boundary conditions were applied. At both locations, record the predicted von Mises stress and 2-component of displacement. To do this, query the element or node closest to the locations chosen (Tools Query, choose element or node, then select the element or node nearest each of the locations chosen). For stress, the values at all of the integration points in the element will be listed in the output screen toward the bottom. Average these to get an approximate value of the stress at the location. For displacement, the value wi ...

  • 3 Pages

    plasticity

    N.C. State, MAT 450

    Excerpt: ... Plasticity (Ch. 3 sections : 3-1 to 3-6; 3-8) Multiaxial Loading - Tresca and von-Mises yield criteria - Plastic flow under uniaxial loading : o is the uniaxial yield strength 3-1 to 3-3 : true vs : note during plastic flow volume is conserved ( = 0) : = V = 1+2+3 = x+y+z = 0 V (Eq. 3-5) -note contrast with elasticity- recall from Hookes law : = x+y+z = 1 2 ( x + y + z ) so for =0, =0.5 here. E A l - constancy of volume implies Al = Aolo so that = ln( ) = ln( o ) lo A 3.4 yield criteria multiaxial loading (ij) : (a) von Mises (distortion criteria) define eff = 1 {(1 2 ) 2 + ( 2 3 ) 2 + (3 1 ) 2 }1/ 2 2 1 (Eq. 3-12) if eff = o, yield occurs these are principal stresses, and in terms of all 6 components Eq. 3-13 : eff = 2 {( x y ) 2 + ( y z ) 2 + ( z x ) 2 + 6( 2xy + yz + 2 )}1/ 2 zx 2 - distortion energy, leading to change of shape, reaches ...

  • 1 Pages

    announce0608

    McGill, MECH 577

    Excerpt: ... ng the maximum von Mises stress, while being able to satisfy its structural functions. Further information is available at http:/www.mcgill.ca/cden/courses/ You may also direct your queries to Prof. J. Angeles, McConnell 452: angeles@cim.mcgill.ca ...

  • 1 Pages

    coverpage

    Michigan State University, ME 471

    Excerpt: ... Cover Page Finite Element Project BRACKET DESIGN AUTHOR: _STUDENT ID _ RESULTS Problem 1: Reference Design Coarse Mesh Amount of material (in ) Number of Elements Number of Nodes Disp of point X (in) 3 Fine Mesh Max Von Mises (psi) Problem 2: Design Variation Coarse Mesh Amount of material (mm ) Number of Elements Number of Nodes Disp of point X (in) 3 Fine Mesh Max Von Mises (psi) Checklist Original Plots Mesh Displacement Von Mises stress Title set to yourname Original work performed independently by _ signature ...

  • 3 Pages

    strength_theories

    Washington, ME 354

    Excerpt: ... Strength Theories The majority of material strength data is based on uniaxial tensile test results. Usually, all that you have to work with is the yield strength Sy and/or the ultimate tensile strength Su. This is fine if you only have the one normal stress component present : this is true for simple tension or compression members and for parts loaded only in bending. 1 1 = x In this case, failure (defined as the onset of plastic deformation) occurs when x = 1 =Sy/n `n' is the factor of safety. In many loading cases, we have more than just one normal stress component. E.g. in torsion, we have a single shear stress component: xy xy Or, combined bending and torsion in a shaft: xy x xy These cases can all be reduced to a simple biaxial case by finding the principal stresses, 1 and 2 2 x 1 1 2 Now when does failure occur? For ductile materials there are two commonly used strength theories - the Maximum Shear Stress (MSS) or Tresca theory and the von Mises or Distortion Energy theory. Strength Theories 1. ...

  • 8 Pages

    project

    Grand Valley State, EGR 309

    Excerpt: ... Grand Valley State University Padnos School of Engineering EGR 309 Machine Design I Summer 2000 Jeff Willner Kent Fannin Jayme Dood Combined Loading Final Project August 8, 2000 Executive Summary A model of a road sign pole was analyzed by using a finite element analysis, FEA, by calculating the internal stresses of the pole, and by experimentally determining the stress on the pole by applying a strain rosette. A sign pole undergoes bending stress and torsional shear stress due to the force of the wind blowing. The force of the wind was modeled in our experiment by using a range of weights hanging over a pulley and determining the strain for each weight. A 12.65 mm aluminum bar was used for the model sign and the strain rosette was placed close to the bottom of the bar. The Von Mises stress was determined for the FEA, calculated values, and for the experimental values. The results are shown in Table 1. Table 1 - Von Mises Stresses Von Mises Stress (MPa) Weight (g) Experimental Calculated Ansys 500 8.72 ...

  • 8 Pages

    lecture_21

    Stanford, ME 111

    Excerpt: ... ME111 Instructor: Peter Pinsky Class #21 November 13, 2000 Today's Topics 2. Consider two designs of a lug wrench for an automobile: (a) single ended, (b) double ended. The distance between points A and B is 12 in. and the handle diameter is 0.625 in. What is the maximum force possible before yielding the handle if Sy = 45 kpsi? Failure of ductile materials under static loading. The von Mises yield criterion. 3. A storage rack is to be designed to hold a roll of industrial paper. The weight of the roll is 53.9 kN, and the length of the mandrel is 1.615 m. Determine suitable dimensions for a and b to provide a factor of safety of 1.5 if: (a) The beam is a ductile material with S y = 300 Mpa (b) The beam is a brittle material with Sut = 150 Mpa, S uc = 570 Mpa. The maximum shear stress criterion. Reading Assignment Juvinall, 6.5 6.8 4. For the problem in Example 19.1 (torsion-bar spring), what diameter d will provide a factor of safety of N = 3 against yielding based on von Mises with S y = 150 M ...

  • 1 Pages

    ch7figs5pm

    Virginia Tech, ETD 04192000

    Excerpt: ... (a) (b) (c) (d) FIGURE 7-7. Elastic stress (ksi) distribution for case IIB: (a) column panel zone ( von Mises ); (b) doubler plate ( von Mises ); (c) doubler plate (horizontal shear stress); (d) doubler plate (vertical shear stress). 103 ...

  • 1 Pages

    ch7figs3pm

    Virginia Tech, ETD 04192000

    Excerpt: ... Yielding (a) Yielding (b) (c) (d) FIGURE 7-5. Elastic stress (ksi) distribution for case IB: (a) column panel zone ( von Mises ); (b) doubler plate ( von Mises ); (c) doubler plate (horizontal shear stress); (d) doubler plate (vertical shear stress). 101 ...