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MIT 3 11 - Stress Transformations

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LECTURE #16 :3.11 MECHANICS OF MATERIALS F03INSTRUCTOR : Professor Christine OrtizOFFICE : 13-4022 PHONE : 452-3084WWW : http://web.mit.edu/cortiz/www• REVIEW : STRESS TRANFORMATIONSII• SAMPLE PROBLEMS• INTRODUCION TO MOHR’S CIRCLE1. Plane Stress Rotations : Given a state of plane stress, what is the equivalent stress state on an element rotated by an arbitrary angle, θθθθ + CCW defined relative to x-axis• original coordinate axis : x/y, stresses in original coordinate axis : σx, σy, τxy,θ• new coordinate axis : x’/y’, stresses in new coordinate axis : σx’, σy’, τx’y’ ,θ• Consider a free body diagram of a “wedge” where a cut is made along an inclined plane where the transformed stresses are desired :Review Lecture #14 : Stress Transformations Ix•Oσσσσxττττxyσσσσyyxy•Ox’y’σσσσx’ττττx’y’σσσσy’θθθθθθθθx•Oσσσσxττττxyσσσσyyxy•Ox’y’σσσσx’ττττx’y’σσσσy’θθθθθθθθxy•Ox’y’σσσσx’ττττx’y’σσσσy’θθθθθθθθxy•Ox’y’θθσx Aoτxy Aoσy Aotanθτyx Aotanθσx’ Ao/cosθτx’y’ Ao/cosθxy•Ox’y’θθσx Aoτxy Aoσy Aotanθτyx Aotanθσx’ Ao/cosθτx’y’ Ao/cosθEquations of Static Equilibrium, Geometry, Trig ⇒STRESS TRANSFORMATION EQUATIONS :•stresses vary continuously as the axis is rotated :xy xyx' xyxy xyy' xyxyx'y' xy()cos(2)sin(2 )22()cos(2)(sin(2 )22()sin(2)cos(2 )2σσ σσ θστθσσ σσ θστθσσ θττθ+−+−−=+ +=− −=− +Principal and Maximum Shear :xy•Ox’y’σ2 =σx’=σminσ1=σy’=σmaxθp2θp1xy•Ox’y’σx’σy’θs1τx’y’=τmaxxy•Ox’y’σx’σy’θs1τx’y’=τmax2xy xy21, 2xy12xypxyp1 1 p2 2max22maximum normal stressminimum normal stress2tan(2 )Principal Stresses: Principal Angles / Planes : , Maximum Shear Stresses: σσ σσστσστθσσθσθστ+−−=± +===⇒⇒2xy2,minxymaxminxysxys1 max s2 mins2maximum ( ) shear stressmin imum ( ) shear stress()tan(2 )2 Planes / Angles of Maximum Shear :, Relationships Between Principal and Shear :σστττσσθτθτθτθ−−=± +=+=−−=⇒⇒opop1 p2os1 s212max45and 90 apart (mutually perpendicular planes)and 90 apart (mutually perpendicular planes)()2 θθθθθσστ−=±=Derivation of General Equation for Principal Stressesxy xyx' xyxy xyy' xyxypxy()cos(2)sin(2 )22()cos(2)(sin(2 )222tan(2 )σσ σσ θστθσσ σσ θστθτθσσ+−+−−=+ +=− −=Sample Problem #1 : Stresses on an Inclined ElementAn element in plane stress is subjected to :σσσσx=-16,000 psiσσσσy=6000 psiττττxy=4000 psiFind and draw the stresses acting on an element inclined at an angle θ=45o.xy•Oz-face or x/y planeSTRESS TRANSFORMATION EQUATIONS :xy xyx' xyxy xyy' xyxyx'y' xy()cos(2)sin(2 )22()cos(2)(sin(2 )22()sin(2)cos(2 )2σσ σσ θστθσσ σσ θστθσσ θττθ+−+−−=+ +=− −=− +Sample Problem #1 : Stresses on an Inclined ElementAn element in plane stress is subjected to :σσσσx=-16,000 psiσσσσy=6000 psiττττxy=4000 psiFind and draw the stresses acting on an element inclined at an angle θ=45o.STRESS TRANSFORMATION EQUATIONS :xy xyx' xyxy xyy' xyxyx'y' xy()cos(2)sin(2 )22()cos(2)(sin(2 )22()sin(2)cos(2 )2σσ σσ θστθσσ σσ θστθσσ θττθ+−+−−=+ +=− −=− +Sample Problem #1 : Stresses on an Inclined Elementxy•Ox’y’Sample Problem 2 : Principal and Maximum Shear StressesAn element in plane stress is subjected to :σσσσx=12300 psiσσσσy=-4200 psiττττxy=-4700 psi(a) Determine the principal stresses and show them on a sketch of a properly oriented element(b) Determine the maximum shear stresses and show them on a sketch of a properly oriented elementxy•Oz-face or x/y planeSample Problem 2 : Principal and Maximum Shear StressesAn element in plane stress is subjected to :σσσσx=12300 psiσσσσy=-4200 psiττττxy=-4700 psi(a) Determine the principal stresses and show them on a sketch of a properly oriented element(b) Determine the maximum shear stresses and show them on a sketch of a properly oriented elementxy xyx' xy()cos(2)sin(2 )22σσ σσ θστθ+−=+ +Sample Problem 2 : Principal Stressesxy•Ox’y’Sample Problem 2 : Principal and Maximum Shear StressesAn element in plane stress is subjected to :σσσσx=12300 psiσσσσy=-4200 psiττττxy=-4700 psi(a) Determine the principal stresses and show them on a sketch of a properly oriented element(b) Determine the maximum shear stresses and show them on a sketch of a properly oriented elementxysxy()tan(2 )2σσθτ−−=2xy2max, minxy2σσττ−=± +Sample Problem 2 : Maximum Shear Stressesxy•Ox’y’xy xyx' xyxy xyy' xyxyx'y' xy()cos(2)sin(2 )22()cos(2)(sin(2 )22()sin(2)cos(2 )2σσ σσ θστθσσ σσ θστθσσ θττθ+−+−−=+ +=− −=− +MOHRS CIRCLExy xyx' xyxy xyy' xyxyx'y' xy()cos(2)sin(2 )22()cos(2)(sin(2 )22()sin(2)cos(2 )2(1)(2) (3)Stress Transformation Equations :σσ σσ θστθσσ σσ θστθσσ θττθ+−+−−=+ +=− −=− +MOHRS CIRCLE22 2x' avex'y'xyave2xy2xy() R()2R2σσ τσσσσστ−+−+===+CONSTRUCTION OF MOHR’S CIRCLE•C•Oxy•Oσxτxyσyz-face or x/y planex’y’θGiven stress state :22 2x' avex'y'xyave2xy2xy() R()2R2σσ


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MIT 3 11 - Stress Transformations

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