STRESS TRANSFORMATION Combined Static Loading

STRESS TRANSFORMATIONCombined Static LoadingSlide No. 1• Introduction– Formulas were developed previously fordetermining normal and shearing stresseson a specific planes in• Axially loaded bars• Circular shafts, and• Beams– For example, the normal stress at a pointon a cross section of a beam can becomputed by the flexural formula.

Combined Static LoadingSlide No. 2• Introduction– The elastic flexural formula for normalstress is given byandσM cr=Imax (1)M yr (2)σ =xICombined Static LoadingSlide No. 3Figure 1• IntroductionDistribution of Normal Stress in a Beam CrossSectionywPNeutral axisCentroidal axisF CcxF TdAy CccydyRV r

Combined Static LoadingSlide No. 4• Introduction– Also, as an example, the shearing stress atthe same point on the cross section of thebeam can be computed using the shearingstress formula:VQ=Itτ (3)Combined Static LoadingSlide No. 5M V• Introduction(a)(b)A BdyC DM + ∆MV + ∆VFigure 2y 1ycyτ xyV Lτ yxτ yx(c)τ xyV Rx∆xtF 1 F 2V H = τ t ∆x

Combined Static LoadingSlide No. 6• Introduction– Another example is the stress on circularshafts due to torsion.– The stress of a point on a circular shaft dueto torsion can be computed fromTcτ =JTρτ = ρJmax(4)Combined Static LoadingSlide No. 7• Introduction• Distribution of Shearing Stress within theCircular Cross Sectionτ τ maxτ τ maxτ = Tcτ mincJcr oρ r iρFigure 3

Combined Static LoadingSlide No. 8• IntroductionFigure 4y Axtτ x y dA cos αy (c)τ n t dAnαασ n dAxτ yx dA sin αyτ xyατ yxτ yxτ xy(b)x(a)Combined Static LoadingSlide No. 9+ ∑ Ft= 0τ dA −τntFrom whichτ = τnt+ ∑ Ft= 0σ dA −τnFrom which• Introduction( dAcosα) sinα−τ( dAsinα)σ = 2τsinαcosα= τ sin 2αnxyxyxyxy( dAcosα) cosα+ τ ( dAsinα)22( cos α − sin α ) = τ cos 2αxyxyyxyxsinα= 0(5)cosα= 0(6)tτ x y dA cos αFigure 5yτ n t dAnαασ n dAτ yx dA sin α

Combined Static LoadingSlide No. 10• Introduction– The method of finding the stress on aspecified inclined plane as illustrated byfree-body diagram of Fig. 5 is not suitablefor determining the maximum normal andmaximum shearing stresses.– A more general approach will bedeveloped to handle this type of analysis.IntroductionSlide No. 11ENES 220 ©Assakkaf• The most general state of stress at a point maybe represented by 6 components,σ x,σ y,σ z normal stressesτ xy,τ yz,τ zx shearing stresses(Note : τ xy = τ yx,τ yz = τ zy,τ zx = τ xz )• Same state of stress is represented by adifferent set of components if axes are rotated.• The first part of the chapter is concerned withhow the components of stress are transformedunder a rotation of the coordinate axes. Thesecond part of the chapter is devoted to asimilar analysis of the transformation of thecomponents of strain.

IntroductionSlide No. 12ENES 220 ©Assakkaf• Plane Stress - state of stress in which two faces ofthe cubic element are free of stress. For theillustrated example, the state of stress is defined byσ , σ , τ and σ τ = τ 0.xyxy z = zx zy =• State of plane stress occurs in a thin plate subjectedto forces acting in the midplane of the plate.• State of plane stress also occurs on the free surfaceof a structural element or machine component, i.e.,at any point of the surface not subjected to anexternal force.Stress at a General Point in anArbitrary Loaded MemberSlide No. 13• State of Stress– Stress at a point in a material body hasbeen defined as a force per unit area.– But this definition is somewhat ambiguoussince it depends upon what area weconsider at the point.– To see this, consider a point O in theinterior of the body shown in Fig. 6a.

Stress at a General Point in anArbitrary Loaded Member• State of StressySlide No. 14Figure 6zx•oodF = σ dAxx(a)(b)dF = τ dA xyydF = τ dA xzzStress at a General Point in anArbitrary Loaded Member• State of StressySlide No. 15dF = σ dA yoydF = τ dA yxxxzdF dAz= τyz o dF = τ dAxzx(c)Figure 6dF = σ dA zzdF = τ dA zyy(d)

Stress at a General Point in anArbitrary Loaded MemberSlide No. 16• State of Stress– Let x, y, and z be the usual rectangularcoordinates axes and let us pass a cuttingplane through point O perpendicular to thex axis as shown in Fig. 6b.– If dA is the area, then by definition,dFdFxy= τxy=τdAdAx xz=dFzdAσ (5)Stress at a General Point in anArbitrary Loaded MemberSlide No. 17• State of Stress– In a similar manner, let us pass a cuttingplane through point O perpendicular to they axis as shown in Fig. 6c.– The corresponding components of stressescan be written asdFydFx= τyx=τdAdAy yz=dFzdAσ (6)

Stress at a General Point in anArbitrary Loaded MemberSlide No. 18• State of Stress– Finally, by passing a cutting planeperpendicular to the z axis as in Fig. 6d,this can result indFzdFx= τzx=τdAdAdFydAσz zy=(7)Stress at a General Point in anArbitrary Loaded MemberSlide No. 19• State of Stress– Now it is quite likely that each of the ninestresses provided by Eqs. 5, 6, and 7 willhave a different value.– Therefore, what is the stress at a point?– It seems that we have nine choices!– There is no such a thing as the stress at apoint O, but rather there is a combinationor state of stress at point O.

Stress at a General Point in anArbitrary Loaded MemberSlide No. 20• State of Stress– It is convenient to depict the state of stressby the scheme of Fig. 7, in which thestresses on three mutually perpendicularplanes are labeled in the manner describedabove.– The state of stress shown in the figure iscalled the general or triaxial state of stresswhich can exit at any interior point.Stress at a General Point in anArbitrary Loaded MemberFigure 7• General or Triaxial State of Stressτ yzτ zyσ zσ yτ zxτ yxτ xzτ xyσ xzτττxyyzzxy= τ= τ= τyxzyxzxSlide No. 21(7)

Stress at a General Point in anArbitrary Loaded Member• General or Triaxial State of StressSlide No. 22• The most general state of stress at a point maybe represented by 6 components,σ x,σ y,σ z normal stressesτ xy,τ yz,τ zx shearing stresses(Note : τ xy = τ yx,τ yz = τ zy,τ zx = τ xz )Stress at a General Point in anArbitrary Loaded MemberSlide No. 23• General or Triaxial State of Stress– Sign ConventionsNormal stresses indicated by the symbol σ anda single subscript to indicate the plane (actuallythe outward normal to the plane) on which hestress acts.Normal stresses are positive if they point in thedirection of the outward normal. Thus, normalstresses are positive if tensile and negative ifcompressive.

Stress at a General Point in anArbitrary Loaded MemberSlide No. 24• General or Triaxial State of Stress– Sign Conventions (cont’d)Shearing stresses are denoted by the symbol τfollowed by two subscripts, the first subscriptdesignates the normal to the plane on whichthe stress acts and the second designate thecoordinate axis to which the stress is parallel.A positive shearing stress points in the positivedirection of the coordinate axis of the secondsubscript if it acts on a surface with an outwardnormal in the positive direction.Stress at a General Point in anArbitrary Loaded MemberSlide No. 25• General or Triaxial State of Stress– In dealing with the states of stresses theengineer is confronted with two problems:1. First, how does she/he determine the state ofstress of a point; that is, how does she/hecalculate values for σ x , σ y , τ xy , and so on?2. Second, how does the maximum value of thestress (normal or shear) be determined at apoint? The stresses σ x , σ y , τ xy , and so on maynot be the maximum possible value.

Two-Dimensional StressSlide No. 26• Plane Stress– The analysis of plane stresses of a pointare more simpler than the analysis ofgeneral state of stress.– Plane stress is considered a special caseof the general state of stress of a point.– Two parallel faces of the small elementshown in Fig. 7 and 8 are assumed to befree of stress.Two-Dimensional StressSlide No. 27• Plane Stressσ yFigure 8σ yτ yxτ yzτ xyτ xyτ zyτ zxσ xσ xσ xσ zτ xzτ xyτ yxσ y

Two-Dimensional StressSlide No. 28• Plane Stress• Plane Stress - state of stress in which twofaces of the cubic element are free ofstress. For the illustrated example, thestate of stress is defined byσ , σ , τxyandxy σ z = τ zx = τ zy =0.Two-Dimensional StressSlide No. 29• Plane Stress• State of plane stress occurs in a thin plate subjectedto forces acting in the midplane of the plate.• State of plane stress also occurs on the free surfaceof a structural element or machine component, i.e.,at any point of the surface not subjected to anexternal force.

Two-Dimensional StressSlide No. 30• Plane Stress– For the purpose of a analysis, let thesefaces be perpendicular to the z-axis,therefore,σz= τzx= τzy= 0(8)– As was shown earlier, this also implies thatτ xz= τ yz= 0(9)IntroductionSlide No. 31ENES 220 ©Assakkaf• Plane Stress - state of stress in which two faces ofthe cubic element are free of stress. For theillustrated example, the state of stress is defined byσ , σ , τ and σ τ = τ 0.xyxy z = zx zy =• State of plane stress occurs in a thin plate subjectedto forces acting in the midplane of the plate.• State of plane stress also occurs on the free surfaceof a structural element or machine component, i.e.,at any point of the surface not subjected to anexternal force.

Two-Dimensional StressSlide No. 32• Plane Stress– Components:σyNormal Stress σ xNormal Stress σ yShearing Stress τ xyShearing Stress τ yxσ xAτ xyθτ yxτ xyAσ xτxy= τ yxτ yxσ yFigure 9The Stress TransformationEquations for Plane StressSlide No. 33• Plane Stress Equations– Equations relating the normal and shearingstresses σ n and τ nt on an arbitrary plane(oriented at an angle θ with respect to areference x-axis) through a point and theknown stresses σ x , σ y , and τ xy = τ yx on thereference planes can be developed usingthe free body-diagram method.

σ xThe Stress TransformationEquations for Plane Stress• Plane Stress Equations(a)Aθτ xyτ yxFree-body Diagramσ yσ yτ yxτ xyAtσ xytσ x dA cos θnxτ x y dA cos θyθτ n t dAθSlide No. 34σ n dAxτ yx dA sin θσ y dA sin θFigure 10(b)nThe Stress TransformationEquations for Plane StressSlide No. 35• Plane Stress Equations– Figure 10b is a free-body diagram ofwedge-shaped element in which the areasof the faces are• dA for the inclined face (plane A-A)• dA cos θ for the vertical face, and• dA sin θ for the horizontal face

The Stress TransformationEquations for Plane Stress• Plane Stress EquationsSumming forces in the n-direction gives+ ∑ Fn= σndA−−τyxSinceτ= τ , thereforexyyxσx( dAcosθ) cosθ−σy( dAsinθ)( dAsinθ) cosθ−τ( dAcosθ) sin22σn= σxcos θ + σysin θ + 2τxysinθcosθIn terms of the double angle,σx+ σyσx−σyσn= + cos 2θ+ τxysin 2θ2 2xysinθθ = 0Slide No. 36(10a)(10b)σ xThe Stress Transformation Equationsfor Plane Stress(a)Aθτ xyτ yx• Plane Stress EquationsFree-body Diagramσ yσ yτ yxτ xyAtσ xytσ x dA cos θnxτ x y dA cos θyθτ n t dAθσ n dAxτ yx dA sin θσ y dA sin θSlide No. 37Figure 10(b)n

The Stress Transformation Equationsfor Plane Stress• Plane Stress EquationsSumming forces in the t-direction gives+ ∑ Ft= τntdA+ σ−τSinceτ= τ , thereforexyτntyx= −In terms of the double angle,x( dAcosθ) sinθ−σy( dAsinθ)( dAcosθ) cosθ+ τ ( dAsinθ)2 2( σ −σ) sinθcosθ+ τ ( cos θ − sin θ )xyxyσx−σyτnt= sin 2θ+ τxycos 2θ2xyxycosθsinθ= 0(11b)Slide No. 38(11a)Transformation of Plane StressSlide No. 39ENES 220 ©Assakkaf• Consider the conditions for equilibrium of aprismatic element with faces perpendicular tothe x, y, and x’ axes.∑ Fx′= 0 = σ x′∆A−σx( ∆Acosθ) cosθ−τxy( ∆Acosθ) sinθ−σy ( ∆Asinθ) sinθ−τxy( ∆Asinθ) cosθ∑ Fy′= 0 = τ x′y′∆A+ σ x( ∆Acosθ) sinθ−τxy ( ∆Acosθ) cosθ−σy ( ∆Asinθ) cosθ+ τ xy ( ∆Asinθ) sinθ• The equations may be rewritten to yieldσ x + σ y σ x −σyσ x′= + cos 2θ+ τ xy sin 2θ2 2σ x + σ y σ x −σyσ y′= − cos 2θ−τxy sin 2θ2 2σ x −σyτ x′y′= − sin 2θ+ τ xy cos2θ2

The Stress TransformationEquations for Plane StressSlide No. 40• Plane Stress Transformation Equations22σn= σxcos θ + σysin θ +τσ xntAθτ xyτ yxσ yσ yτ yxτ xyAσ xτ x ydA cos θσ x dA cos θ2τxysinθcosθ(12)2 2θ − sin θ( σ −σ) sinθcosθ+ τ ( cos )= −x yxyyθτ n tdAθσ n dAxτ yx dA sin θσ y dA sin θnThe Stress TransformationEquations for Plane StressSlide No. 41• Plane Stress Transformation Equations• Double Angle Form yσ xAθτ xyτ yxσ yτ yxτ xyσ xτ x ydA cos θσ x dA cos θθτ n tdAθσ n dAAτ yxxσ dA sin θyσx−σyσ y dA sin θτnt= − sin 2θ+ τxycos 2θ2σx+ σyσx−σyσn= + cos 2θ+ τxysin 2θ2 2n(13)

The Stress TransformationEquations for Plane StressSlide No. 42• Sign Conventions1. Tensile normal stresses are positive;compressive normal stresses arenegative.2. A shearing stress is positive if it points inthe positive direction of the ordinate axisof the second subscript when it is actingon a surface whose outward normal is ina positive direction.The Stress TransformationEquations for Plane StressSlide No. 43• Sign Conventions (cont’d)3. An angle measured counterclockwisefrom the reference x-axis is positive.Conversely, angles measured clockwisefrom the reference x-axis are negative.4. The n, t, z-axes have the same order asthe x, y, z-axes. Both sets of axesconstitute a right-hand coordinatesystem.

The Stress TransformationEquations for Plane StressSlide No. 44• Example 1– The stresses shown in the figure act at apoint on the free surface of a stressedbody. Determine the normal and shearingstresses at this point on the inclined planeAB shown in the figure.The Stress TransformationEquations for Plane StressSlide No. 45• Example 1 (cont’d)t70MPaτ n28 0 Figure 11xAB06240MPa10MPaσ nn

The Stress TransformationEquations for Plane Stress• Example 1 (cont’d)We have,σx= −10 MPaσ = −70 MPaτyxyθ == + 40 MPa0-28A70MPaB06240MPaSlide No. 4610MPaThe Stress TransformationEquations for Plane Stress• Example 1 (cont’d)Applying Eq. 12 for the given values22σ = σ cos θ + σ sin θ + 2τnx= −10cos2yxysinθcosθSlide No. 472( − 28) − 70sin ( − 28) + 2(40)sin( − 28) cos( − 28)σ = −56.39 MPanτ = −ntnt= −2 2( σx−σy) sinθcosθ+ τxy( cos θ − sin θ )22( −10− ( −70)) sin( −28)cos(−28)+ 40( cos ( − 28) − sin ( − 28))τ = 47.24 MPa

The Stress TransformationEquations for Plane Stress• Example 1 (cont’d)Slide No. 4870MPatAB06240MPa10MPa47.2MPax28 0 56.4MPan