1Dynamic AnalysisStress 1Geol341/342Reading: Chapter 3Outline• Force, vectors• Units• Normal, shear components• Pressure• Lithostatic Stress• Stress Tensor• Resolving stress on a planeForce• What is it?F= Mass . acceleration• Units?kg .m/s2Newtons• How much is a Newton?A normal person can throw a bowling ball at 90 NAction and reaction2Force and deformationForce is a vectorFy= F . SinAFAFx= F . CosAyxNormalShearDoes applied force alone control deformation?FFLand’O LakesStressStress = force / areaUnits = Pascals= N/m2=kg/(m . s2)3Stress vs. Pressure• Units of pressure = force/area• 33 psi = 33 pounds per square inchWhat is the difference?• Pressure - Scalar quantity• Stress - Tensor quantityHow much is 1 Pascal?• 1 Pa= 0.00014 psi• Car tire ~200,000 Pa • 1 Megapascal (1 MPa) = 1 million Pa= 1 x 106PaCalculate State of Stress at 1000 m depthLithostatic Stress• Similar to hydrostatic pressure• Magnitude of stress components is the same in all directions4Stress TensorOn-in convention:ON the x-planeIN the y-directionσxyStress TensorσxxσxyσxzσyxσyyσyzσzxσzyσzzNormal Stress Components2D Stress Tensorxzσxxσxxσzzσzzσxzσxzσzxσzx52D Stress TensorσxxσxzσzxσzzSign Conventions• Compression pos•Tension neg• Right Lateral Shear pos• Left Lateral Shear negResolving stress on a planeGiven σxxand σzzWhat are σnand σsacting on a given plane?xzσxxσzzσnσsResolving stress on a planeMust first correct for the change in the areaxzσxxσzzaσzzSin aσxx Cos aThen, you can resolve the vectors into normal and shear components6Exampleσxx=45 Mpaσzz=26 Mpax- northz- upWhat are the normal and shear stresses acting on a fault oriented 090/40S?Step 1: Understand the geometryupσxxσzzNDip=40Cross Sectional ViewMap View40NorthfaultSfaultStep 2: Find Magnitude of stress components along x and zzxa= 90-dip = 50 deg.σfx= σxx Cos a= 45 MPa Cos 50= 28.9 Mpaσfz= σzz Sin a= 26 MPa Sin 50= 19.9 MpaσxxσzzaFσfxσfzStep 3: Add the two vectorszxσf2= σfx2+ σfz2σf = sqrt{28.9 2 +19.9 2 } Mpaσf= 35.1 Mpab= arctan (σfz/σfx)b= 34.5 deg.σfxaFσfzσfb7Step 4: Resolve σfinto normal and shear componentszxc = a-b = 50- 34.5 = 15.4σn= σf Cos c= 35.1 Mpa Cos 15.4= 33.8 Mpaσs= σf Sin c= 35.1 Mpa Sin 15.4= 9.32 MPaσnaFσfbσscMagnitude of Normal Force and Normal StressAs a function ofAngle ΘσnMaxσn= σ Cos2ΘMagnitude of Shear Force and Shear StressAs a function ofAngle ΘσsMinσsMinσs= σ ½Sin 2 ΘStress Componentsxzσxxσxxσzzσzzσxzσxzσzxσzx8Principal Stress ComponentsYou always can find an orientation of the reference cube where there won’t be any resolved shear stressesxzσ1σ1σ3σ3Most compressiveLeast compressiveStress componentsStress Ellipse- Normal Stressσ3σ3σ1σ1Maximum Normal StressMinimum Normal StressStress Ellipsoidσ1 > σ2 > σ39Magnitude of Normal and Shear Stressesσ1σ3NormalShearPrincipal Stress Componentsσ1000 σ2000σ3GreatestIntermediateLeast10General Stress Equationsσn= ½ (σ1 + σ3) + ½(σ1- σ3) cos 2Θ(eq. 3.7)σs= ½(σ1- σ3) sin 2Θ (eq. 3.10)Θ= angle between plane and σ3 or between normal to the plane