东南大学：《弹性力学》课程教学课件（英文讲稿）04 Stress and Equilibrium

Stress and Equilibrium

Stress and Equilibrium

OutlineBodyand SurfaceForcesTraction/Stress Vector Stress TensorTraction on Oblique PlanesPrincipal Stresses and DirectionsMohr's Circles of Stresses Octahedral StressesSpherical and Deviatoric StressesConservation of Linear Momentum Conservation of Angular MomentumEquilibrium Equations: Equilibrium Equations in Curvilinear Coordinates2

Outline • Body and Surface Forces • Traction/Stress Vector • Stress Tensor • Traction on Oblique Planes • Principal Stresses and Directions • Mohr’s Circles of Stresses • Octahedral Stresses • Spherical and Deviatoric Stresses • Conservation of Linear Momentum • Conservation of Angular Momentum • Equilibrium Equations • Equilibrium Equations in Curvilinear Coordinates 2

Body and Surface ForcesExternal loads include body and surface forcesP:Body Forces: F(x)++1Surfaceforces. Forces are vectors (unit: N)F= Fe, +Fze2 + Fe, = Fe,: Often interpreted in terms of density: body force densityand surface force densityF, = (l,T"(x)dSF, = JF(x)dV3

Body and Surface Forces • External loads include body and surface forces. Surface forces • Forces are vectors (unit: N) F e e e e     F F F F 1 1 2 2 3 3 i i • Often interpreted in terms of density: body force density and surface force density 3  d b V  V  F F x  d S S  S  n F T x

Traction/Stress VectorAF: Given △F as the force transmitted across △A, a stresstraction vector can be defined asAFT" (x) = limM-0 △AUnits: Pa (N/m2), 1 MPa = 106 Pa, 1 GPa = 109 PaDecomposition of the traction vectorT"(x)=on+tt =on+t't'+t"t"4

Tn σ  Units: Pa (N/m2 ), 1 MPa = 106 Pa, 1 GPa = 109 Pa. • Given ΔF as the force transmitted across ΔA, a stress traction vector can be defined as F A n 4 Traction/Stress Vector   0 lim  A A    n F T x                 n T x n t n t t • Decomposition of the traction vector

Stress TensorXoer+to.eTry+treOeOTT.L1VZ2Z福yxJZ2Xz0.TTZX1ZTy0xvon KarmanNotation5

5 Stress Tensor         x x xy y xz z n T x e e e   yx x y y yz z       n T x e e e   zx x zy y z z       n T x e e e x xy xz ij yx y yz zx zy z                         von Karman Notation

Sign Convention: Normal stress: tension positive / compression negative: Shear stress: product of the surface normal (the firstsubscript) and the stress direction (the second subscript): All stress components shown in the figure are positive.TVtyztxyTzyZXtxza6

6 Sign Convention • Normal stress: tension positive / compression negative • Shear stress: product of the surface normal (the first subscript) and the stress direction (the second subscript) • All stress components shown in the figure are positive

Traction on an Obligue Plane - 2Df0-ZFn[0=ZF,TiT"A= o,AAcos0+twAsin0AATnAA = t..AAcosO+o.Asin0=o.n.+tn.yxXXyx"trn.+o.n>{T" T"}vxTnoBa2D Cauchy's relation=n·o7

    0 0 cos sin cos sin x y x x yx y xy y x x x yx y y xy x y y x xy x y x y yx y F F T A A A T A A A T n n T n n T T n n T n                                                                n n n n n n n n T = n σ 7 Traction on an Oblique Plane - 2D 2D Cauchy’s relation n t A  x  y yx  xy  n T 

Traction on an Obligue Plane - 3D. The state of stress at a point is defined by?nzOx,Ty,TxTyx,Oy,Tye,T-xTay,O,o2 Consider the tetrahedron with unit normal nTxyftxz(nn·ei620cos(n,e,n,fnne,Wzy0 =ZF，00=ZF, 0=ZFtozT"A=o,AAn,+tuAAn,+tAnT"A= txAn,+o,AAn,+t,AAnT"A=t.AAn, +tAn, +o.AAn=n,o3D Cauchy's relationT=n.o8

  Tx n   Ty n   Tz n • The state of stress at a point is defined by: , , , , , , , ,          x xy xz yx y yz zx zy z • Consider the tetrahedron with unit normal n cos ,   0 , 0 , 0 i i i i x y z x x x yx y zx z y xy x y y zy z z xz z yz y z z i j ji n F F F T A An An An T A An An An T A An An An T n                                                  n n n n n n e n e n e T = n σ 3D Cauchy’s relation 8 Traction on an Oblique Plane - 3D

Principal Stresses and Directions: Seeking the solution through an eigenvalue equationT"=n·=·n=o,n→det[o,-O,,=0>-o3 +Io-,o, +I, =0. Three invariants of the stress tensorI, =(o,o,-0,on), I,=det[0]I, =0k,00af00026i0003I,=0+02+03,x2=0,02+020,+03013I, =0,0203.(GeneralCoordinateSystem)(PrincipalCoordinateSystem)

3 2 1 2 3 det 0 0 n ij n ij n n n I I I                        n T = n σ σ n = n 9 Principal Stresses and Directions • Seeking the solution through an eigenvalue equation • Three invariants of the stress tensor 1 2 3   1 , , det . 2 kk ii jj ij ji ij I I I              1 2 3 1 1 2 3 2 1 2 2 3 3 1 3 1 2 3 0 0 0 0 0 0 , , . ij I I I                                  

Stress Transformation - 2DOβ=QayQO '= QQ!sinecosOsincosOTx1-sinecosOo,ll-sinecoseaO+ocos20+txysin20O22Tya.+oax-oycos 20-Tx, sin 2022Da0ainsin 20+tx cos20宁2oCrB72=R2+0E2Mohr's stress circleStress symmetry will be proved shortly

 2 cos sin cos sin sin cos sin cos cos 2 sin 2 2 2 cos 2 sin 2 2 2 sin 2 cos 2 2 T T x y x y x y x xy x y x y y xy x xy xy y x y xy x ave x       Q Q                                                                                                 σ QσQ σ 2 2 2 2 ; 2 2 y x y x y ave xy R R                          10 Stress Transformation - 2D Mohr’s stress circle Stress symmetry will be proved shortly.  