MECH 401 Mechanical Design ApplicationsDr. M. K. OMalley Master NotesSpring 2008Dr. D. M. McStravickRice University

Failure from static loadingTopicsFailures from static loadingDuctile FailuresMaximum Shear StressMaximum Distortion EnergyBrittle FailuresMaximum Normal StressCoulomb-Mohr Modified Mohr Reading --- Chapter 5

What is Failure?Failure any change in a machine part which makes it unable to perform its intended function.(From Spotts M. F. and Shoup T. E.) We will normally use a yield failure criteria for ductile materials. The ductile failure theories presented are based on yield.

Failure TheoriesStatic failureDuctileBrittleStress concentration

RecallDuctileSignificant plastic deformation between yield and fractureBrittleYield ~= fracture

Tensile Test

Linear Stress Strain Plot

Mohrs Circle for Tensile Test

Static Ductile FailureTwo primary theories for static ductile failureVon Mises criterion Maximum Distortion-energy TheoryMDEMaximum Shear Stress criterionMSS

Failure Theory Problem StatementGiven:Stress-strain data for simple uniaxial tension

Find:When failure occurs for general state of stress

Static Ductile FailureMax Shear Stress criterion Material yields (fails) when:

Factor of Safety:

or1)2)

Maximum Shear Stress Criteria

Static Ductile FailureVon Mises criterion Let the Mises stress (se, equivalent stress) be:

Then failure (yield) occurs when:

Factor of Safety:

Typically,Want a margin of error but not completely overdesigned

Which theory to use?Look at a plot of the principal stresses sB vs. sAThe non-zero principal stressesFailure occurs when the principal stresses lie outside the enclosed areaShape of area depends on the failure theoryData points are experimental resultsMSS Slightly more conservativeEasier to calculateMDEMore accurateIf not specified, use this one!

Comparison of MDE, MSS,MNS

Hydrostatic Stress State Diagonal

Ductile failure theory exampleGiven:Bar is AISI 1020 hot-rolled steel A DUCTILE materialF = 0.55 kNP = 8.0 kNT = 30 NmFind:Factor of safety ()Two areas of interest:ATop where max normal stress is seen (bending!)BSide where max shear stress is seen

Element AConsider the types of loading we haveAxial?Yes due to PBending? Recall that bending produces s and t, depending on the element of interestYes due to M (s at A, t at B) Torsion?Yes due to T

Element ACalculate stresses due to each loadAxial:

Bending:

Shear:

Torsion:

Element ALook at a stress elementSum up stresses due to all the loads

sx = 95.5 MPatxz = 19.1 MPa

Element ADraw Mohrs Circle with the stresses that we calculatedsx = 95.5 MPatxz = 19.1 MPax at (sx, txz)(95.5, 19.1)y at (sy, tzx)(sy, -txz)(0, -19.1)Find C

Find radius

Out of Plane Maximum Shear for Biaxial State of StressCase 1s1,2 > 0s3 = 0

Case 2s2,3 < 0s1 = 0 Case 3s1 > 0, s3 < 0s2 = 0

Element AFind principal stressess1 = C + R 99.2 MPas2 = C - R -3.63 MPaThink about 3-D Mohrs Circle!This is Case #3We want s1 > s2 > s3Assign s2 = 0 and s3 = -3.63 MPaNo failure theory was given, so use MDE

Element AFind the von Mises stress (se)

Sy for our material = 331 MPaCalculate the factor of safety For yield

Element BConsider the types of loading we haveAxial?Yes due to PBending? Recall that bending produces s and t, depending on the element of interestYes due to M (s at A, t at B)Torsion?Yes due to T

Element BCalculate stresses due to each loadAxial:

Bending: Use equation for round solid cross-section

Shear:

Torsion:

Element BLook at a stress elementSum up stresses due to all the loads

sx = 25.5 MPatxy = 19.1 MPa Note small contribution of shear stress due to bending

Element BDraw Mohrs Circle with the stresses that we calculatedsx = 25.5 MPatxy = 19.1 MPax at (sx, txy)(25.5, 19.1)y at (sy, tyx)(sy, -txy)(0, -19.1)Find C

Find radius

Out of Plane Maximum Shear for Biaxial State of StressCase 1s1,2 > 0s3 = 0

Case 2s2,3 < 0s1 = 0 Case 3s1 > 0, s3 < 0s2 = 0

Element BFind principal stressess1 = C + R 35.8 MPas2 = C - R -10.2 MPaThink about 3-D Mohrs Circle!This is Case #3We want s1 > s2 > s3Assign s2 = 0 and s3 = -10.2 MPaNo failure theory was given, so again use MDE

Element BFind the von Mises stress (se)

Sy for our material = 331 MPaCalculate the factor of safety For yield

Example, concludedWe found the factors of safety relative to each element, A and BA 3.28B 7.91A is the limiting factor of safety = 3.3

Static Brittle FailureThree primary theories for static brittle failureMaximum Normal Stress (MNS)Coulomb-Mohr Theory Modified-Mohr Theory

Mohrs Circle for MNS

Static Brittle FailureMaximum Normal Stress (MNS)Oldest failure hypothesis, attributed to RankineFailure occurs whenever one of the three principal stresses equals the yield strengthSay s1 > s2 > s3 (as we typically do)Failure occurs when eithers1 = St or s3 = -Sc Note brittle materials have both a tensile and compressive strength = St / s1 or = -Sc / s3

Plot of sB vs. sA

Static Brittle FailureCoulomb-Mohr Theory (AKA Internal Friction)

SutSucSucSutUse table, or look at load line

The Load Line Coulomb-MohrsB = rsAEquation of line from origin to point (sA, sB)Then,

Note: Strength of a part can be considered the stress necessary tocause failure. To find a parts strength at onset of failure, use h = 1.

Modified Mohr Failure Theory

Static Brittle FailureModified Mohr Theory

|sB| = SutA:B:Which to use? (C-M or Mod-M)In general, Mod-M is more accurate

The Load Line Modified MohrsB = rsAEquation of line from origin to point (sA, sB)Then,

Note: Again, strength of a part can be considered the stress necessary tocause failure. To find a parts strength at onset of failure, use h = 1.

For both Coulomb-Mohr and Modified Mohr, you can use either the table equations or the load line equations.

Brittle Failure exampleGiven: Shaft of ASTM G25 cast iron subject to loading shownFrom Table A-24Sut = 26 kpsiSuc = 97 kpsiFind: For a factor of safety of = 2.8, what should the diameter of the shaft (d) be?

Brittle Failure exampleFirst, we need to find the forces acting on the shaftTorque on shaft from pulley at BTB = (300-50)(4) = 1000 inlbTorque on shaft from pulley at CTC = (360-27)(3) = 1000 inlbShaft is in static equilibriumNote that shaft is free to move along the x-axis (bearings)Draw a FBD Reaction forces at points of attachment to show constrained motion

EquilibriumUse statics to solve for reactions forcesRAy = 222 lbRAz = 106 lbRDy = 127 lbRDz = 281 lbOK, now we know all the forces. The problem gives us a factor of safety, but unlike our last example, we arent told specific places (elements) at which to look for failure!We are going to have to calculate stresses What do we need?Axial forces, bending moments, and torquesWe need to find our moments HOW?Shear-Moment diagrams will give us the forces and moments along the shaft.Failure will likely occur where the max values are seen

Torsion and moment diagramsLets look at torsion and how it varies across the shaftWe calculated the torques at B and C to be 1000 inlb eachPlot that along the shaft and we see that max torque occurs at B and C (and all points between)

Torsion and moment diagramsNow lets look at the momentsWe have a 3-D loadingHow are we going to do the V-M diagrams?Look at one plane at a timeMoment in the x-y planeFrom geometry you can calculate the values of the moment at B and C

Torsion and moment diagramsMoment in the x-z planeFailure is going to occur at either B or C, since these are locations where maximum moments are seenBut we have moments in both planesTo find the max bending stresses, we must find the total maximum momentJust as we would vectorally add the two force components to find the force magnitude, we can vectorally add the two moment components to find the moment magnitude

Calculate the max momentWe found the following:MB x-y = 1780 inlbMB x-z = 848 inlbMC x-y = 762 inlbMC x-z = 1690 inlbCalculating the magnitudes with MB = 1971.7 inlbMC = 1853.8 inlbSince the overall max moment is at B, we will expect failure there, and use MB in our stress calculationsIf we had been told the location of interest, we would essentially start here.

Calculate the stresses at BBending stress (s and t)We know from experience that s is the predominant stress, so essentially we will look for failure at an element at the top of the shaftM = 1971Plug in known valuessmax = (20x103)/d3Torsional stressT = 1000t = (5.1x103)/d3

Mohrs CircleLets look at our stress elementNow construct Mohrs circleC at (10 x 103)/d3R = (11.2 x 103)/d3s1 = (21.2 x 103)/d3s3 = (-1.2 x 103)/d3Use Coulomb-Mohr theory for brittle failure

If making a design recommendation, you would recommend the next largest standard dimension (16ths)d = 1.375 in

Stress concentrationA stress concentration is any geometric discontinuity in an element that is subjected to stressAside from reducing the cross-sectional area, these stress concentrations do not significantly affect static ductile failureStress concentrations DO, however, have a significant influence on brittle failure

Analytical approach to stress concentrationssmax = ktsnomtmax = ktstnomkt, kts are stress concentration (SC) factorssnom, tnom are nominal stressesNominal those stresses that are calculated before taking the SCs into accountSC factors are given in the text on page 982-988Equations for the nominal stresses (taking into account geometry change due to the SCs) are given in the same charts

Stresses at a Hole in an Infinite Plate

Hoop Stress at a Hole in an Infinite Plate

Radial Stress at a Hole in an Infinite Plate

Stress Concentration in Ductile Material

Residual Stresses in Ductile Material

Brittle failure exampleGiven:ASTM 30 cast ironSut = 31 ksiSuc = 109 ksiFind:How much torque before failure with and without the stress concentration?Note asked to find failure (not given a safety factor)Use = 1 to find onset of failure

Brittle failure exampleWithout the SC

t = (5.1)TsA = 5.1 T, sB = -5.1 TUse Coulomb-Mohr (easier)

T 4730 inlb

Brittle failure exampleWith the SCRefer to figure A-15-15, pg. 986Picture shows us the loading and geometryEquation is given to calculate the nominal stress considering the geometry with the SCAxis and data labels tell us the quantities we need to calculate (using the figure as a guide)

Brittle failure exampleWith the SCd = D 2r = 1 2(.025) = 0.95D/d = 1/0.95 = 1.05r/d = 0.025/0.95 = 0.026kt ~ 1.8tmax = kttnom

Brittle failure exampletmax = kttnom

tmax = kttnom = (1.8)(5.94T) = 10.69 TConstruct stress element and Mohrs Circle as beforeUse Coulomb-Mohr theory

T 2258 inlbAbout half the load that could be withstood in the absence of the SC!

Brittle failure exampleWhat if we consider a solid shaft (no SCs) with a diameter of d (0.95) ?

Again, use Coulomb-Mohr

Note, this is a greater amount of torque than a shaft with larger diameter but with a SC

Design to avoid stress concentrationsAvoid sudden changes in cross-sectionAvoid sharp inside cornersForce-flow analogyImagine flow of incompressible fluid through partSudden curvature in streamlinesHigh stress concentration!

Design to avoid stress concentrations

Design to avoid stress concentrations