Advanced Strength of Materials: Stresses on Planes and Stress Transformation, Study notes of Civil Engineering

Download Advanced Strength of Materials: Stresses on Planes and Stress Transformation and more Study notes Civil Engineering in PDF only on Docsity! 8/27/2008 1 Advanced Strength of Materials Lecture 3 Review of last lecture l Elementary bending theory l Definition of stress at a point (Art. 2.1) l Stress notation (Art. 2.2) l Stress tensor (Art 2.3.1) l Differential equations of equilibrium (Art. 2.5) l Symmetry of stress components (Art 2.3.1) 8/27/2008 2 Stresses on a differential element Sign Convention for positive stress: Positive axis direction on a positive plane or Negative axis direction on a negative plane Stress Tensor ~ xx xy xz yx yy yz zx zy zz σ σ σ σ σ σ σ σ σ σ    =      Nine independent stress components Moment equilibrium gives: yzσ =xyσ = xzσ = ~ xx xy xz xy yy yz xz yz zz σ σ σ σ σ σ σ σ σ σ    =      Six independent stress components 8/27/2008 5 Stresses on arbitrary planes z y x xyσ xzσxxσ yzσ xzσ zzσ xzσ yzσ yyσ kji xzxyxxx σσσσ ++= kji pzpypxp σσσσ ++= yσ = zσ = Stresses on arbitrary planes xdA lds= 3 22 3 lm n d s= Volume of the tetrahedron- OABC ds dAzdAx dAy z x y O AC B Areas and Volume of Tetrahedron ydA = zdA = 8/27/2008 6 Stresses on arbitrary planes 0xFΣ = 3 22 3 0 x px xx x xy y xz z lmnB ds ds dA dA dA σ σ σ σ   + −     − − = lds mds nds Neglect Stresses on arbitrary planes px xx xy xzl m nσ σ σ σ= + + pyσ = pzσ = 0xFΣ = 0yFΣ = 0zFΣ = , , , 8/27/2008 7 Stresses on arbitrary planes px xx xy xz py xy yy yz xz yz zzpz σ σ σ σ σ σ σ σ σ σ σσ            =              Stress vector on a plane Stress at a point Normal and Shear Stresses on an Oblique Plane p px py pzi j kσ σ σ σ= + + N l i m j n k= + + pNσ = 8/27/2008 10 Stress Transformation for 3D Case x y z xyσ xzσ xxσ yzσ zzσ xzσ yyσ yzσ xyσ           zzyzxz yzyyxy xzxyxx σσσ σσσ σσσ We know stresses at point O with respect to xyz axes: O Stress Transformation for 3D Case           ZZYZXZ YZYYXY XZXYXX σσσ σσσ σσσ We seek stresses at point O with respect to XYZ axes: X Y Z x y z O 8/27/2008 11 Stress Transformation for 3D Case x y z X l1 m1 n1 Y Z e.g. l1= Cos θxX, l2= Cos θxY Define direction cosines Properties of Direction Cosines 2 2 2 1, 1,2,3i i il m n i+ + = = 2 2 2 1 2 3 1l l l+ + = 1 2 1 2 1 2 0l l m m n n+ + = 1 1 2 2 3 3 0l m l m l m+ + = 8/27/2008 12 Stress Transformation for 3D Case 1 1 1XN l i m j n k= + + X Xx Xy Xzi j kσ σ σ σ= + + 1 1 1Xx xx xy xzl m nσ σ σ σ= + + 1 1 1Xy xy yy yzl m nσ σ σ σ= + + 1 1 1Xz xz yz zzl m nσ σ σ σ= + + Normal on X-plane: Stress Transformation for 3D Case XX X X Nσ σ= • ( ) ( )1 1 1Xx Xy Xzi j k l i m j n kσ σ σ= + + • + + 1 1 1X x X y X zl m nσ σ σ= + + ( ) ( ) ( ) 1 1 1 1 1 1 1 1 1 1 1 1 xx xy xz xy yy yz xz yz zz l m n l l m n m l m n n σ σ σ σ σ σ σ σ σ = + + + + + + + + 8/27/2008 15 Principal Stresses and Directions pσ N xz y p Nσ σ= σ = principal stress Principal Stresses and Directions p px py pzi j kσ σ σ σ= + + N l i m j n k= + + px xx xy xzl m nσ σ σ σ= + + py xy yy yzl m nσ σ σ σ= + + pz xz yz zzl m nσ σ σ σ= + + p N li mj nkσ σ σ σ σ= = + + 8/27/2008 16 Principal Stresses and Directions xx xy xz xy yy yz xz yz zz l l m m n n σ σ σ σ σ σ σ σ σ σ            =               Linear Operator Vector Vector Scalar Principal Stresses and Directions When Lx is aligned along x, it is an eigenvalue problem Lx = λx Lx λx 8/27/2008 17 Principal Stresses and Directions This is not an acceptable solution, because Law of Direction Cosines 0 0 0 l m n        =            2 2 2 1l m n+ + = Trivial Solution of Lx = λx: Principal Stresses and Directions l For non-trivial solution 0 0 0 xx xy xz xy yy yz xz yz zz l m n σ σ σ σ σ σ σ σ σ σ σ σ  −         − =         −      0 xx xy xz xy yy yz xz yz zz σ σ σ σ σ σ σ σ σ σ σ σ − − = − Determinant