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5. stress function

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YASWANTH BHAIRAVABHOTLA

This document discusses the stress function approach for solving two-dimensional elasticity problems. It begins by presenting the general equations of elasticity, including stress-strain relationships, strain-displacement equations, and equilibrium equations. It then introduces the stress function method proposed by Airy, where a single function of space coordinates is assumed that satisfies all the elasticity equations. The key steps are: (1) choosing a stress function, (2) confirming it is biharmonic, (3) deriving stresses from its derivatives, (4) using boundary conditions to determine the function, (5) deriving strains, and (6) displacements. Examples of polynomial stress functions are also provided.

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ME 6201: Applied Elasticity and Plasticity STRESS FUNCTION Cartesian coordinates Prof. S.K.Sahoo
Plane Stress         0, 1 1 1 , 1        yzxzxyxy yxyxz xyyyxx E E EE           )( 2 1 0)( 2 1 )( 2 1 ,, x w z u y w z xy u z w y v x u xz yz xy zyx                                   0 yzxzz  Stress-strain relationship strain-displacement equations The equilibrium equations Field equations ),(),,(),,( yxyxyx xyxyyyxx   0 0             Y yx X yx yxy xyx   Plane Strain 0),,(),,(  wyxvvyxuu 0,2 )()( 2)( 2)(     yzxzxyxy yxyxz yyxy xyxx G G G     0 0             Y yx X yx yxy xyx   0 )( 2 1 ,,              yzxzz xy yx xy u yx u      )21)(1( 21 2 1 3 3 2              E G KGK  = Lamé’s constant G = shear modulus xyxy xyy yxx E v v v E v v v E v    )1( ) 1 ( 1 ) 1 ( 1 2 2           0 yzxzz  xyxy xyy yxx v E v v E v v E    )1( )( 1 )( 1 2 2        
Simplified Elasticity Formulations Displacement Formulation Eliminate the stresses and strains from the general system of equations. This generates a system of three equations for the three unknown displacement components. Equations are on u, v, w terms. Stress Formulation Eliminate the displacements and strains from the general system of equations. This generates a system of six equations and for the six unknown stress components. Equations are on terms A practical elasticity problem may requires to solve 15 equations with 15 unknowns. It is very difficult to solve many real problems, so modified formulations have been developed. F(z) G(x,y) z x y Even using displacement or stress formulations three-dimensional problems are difficult to solve. So most solutions are assumed to of two-dimensional problems ],,,,,[ zxyzxyzzyyxx 
Displacement Formulation 0 )1(2 0 )1(2 2 2                               Y y v x u yv E vG X y v x u xv E uG Plane stress + ),(),,( yxvvyxuu bb  (B.C.) Plane strain + ),(),,( yxvvyxuu bb  0)( 0)( 2 2                             Y y v x u y GvG X y v x u x GuG   (B.C.) )21)(1( 21 2 1 3 3 2              E G KGK  = Lamé’s constant G = shear modulus
Stress Formulation yxxy xyyx          2 2 2 2 2 2             y Y x X vyx )1()(2  0 0             Y yx X yx yxy xyx   + or + Plane stress Plane strain 0 0             Y yx X yx yxy xyx                y Y x X v yx )1( 1 )(2  yyxy yxxx mlY mlX     yyxy yxxx mlY mlX     (B.C.) or yxxy xyyx          2 2 2 2 2 2 + +
Stress Function Approach 0 0             Y yx X yx yxy xyx   yxxy xyyx          2 2 2 2 2 2 Solution of plain problems:              y Y x X v yx 1 1 )(2  Plain Strain             y Y x X vyx )1()(2  Plain Stress 0)(2 2 2 2            yx yx  3 different equations with 3 unknowns Airy (An English Mathematician) for 2D problems proposes a single stress function (function of space coordinates) that will satisfy all equations yyxy yxxx mlY mlX     When No body force 0)(2  yx  or Philosophy: Stress = f (strain) Strain = f (displacement) = f (space coordinate) = f (x, y, z) => Stress = f (x, y, z)
Airy Stress Function Method Plane Problems with No Body Forces 0 0             yx yx yxy xyx   0)(2  yx Stress Formulation yxxy xyyx                2 2 2 2 2 ,, Airy Representation 02 4 4 4 22 4 4 4          yyxx Biharmonic Governing Equation (Single Equation with Single function  satisfy equilibrium and compatibility equations. The unknown parameters of stress function can be find out by putting boundary conditions. ) 0)(2 2 2 2            yx yx  Let assume a stress function = (x,y) such that: or Putting it on equilibrium equations, it can be checked that both equations are satisfied. Putting on compatibility equation gives
Steps of the Stress Function Approach 1. Choose a stress function,  2. Confirm  is biharmonic (compatibility is satisfied) 3. Derive stresses from second derivatives of  4. Use boundary conditions to identify unknown parameters of  5. Derive strains from stress-strain relationships 6. Integrate strain-displacement relations to get displacements
Example of biharmonic Polynomial Stress Functions       2 0211 2 20011000 0 0 ),( yAxyAxAyAxAAyxAyx m n nm mn terms do not contribute to the stresses and are therefore dropped1 nm terms will automatically satisfy the biharmonic equation3 nm terms require constants Amn to be related in order to satisfy biharmonic equation 3 nm cbyax ,If 22 If, cybxyax  04   ...........,If 2234222322  ykxyjxhxgyfxyyexdxcybxyax c y xx 2002 2       0022 2     a x yy   00 2     b yx xy   Stress distribution can be obtained but all to satisfy biharmonic equation, which requires that there exist some relations among coefficients. ..........2006200200 2 kxgyfxcxx  ...........26120026002 22 kyjxyhxeydxayy  ..........430022000 2 kxyjxfyexbxy  stress function have no meaning, as it gives zero stress in the body. Any quadratic function of x and y will automatically satisfy the biharmonic equation but give constant values of stresses. 04  
Stress functions with body forces (Plane strain) 0         X zyx zxxyxx  0         Y zyx yzyyxy  0         Z zyx zzyzzx  0      X yx xyxx  0      Y yx yyxy  0   z zz We have general equilibrium equation; For plane strain condition; The component of body force X and Y independent Z & Z=0. The body force can be expressed by a potential function Ω, so that; x X    y Y      ,0      yx xy xx     0      yy xy yy  So equilibrium equation becomes, 2 2 y xx      yx xy      2 2 2 x yy      If we now choose a stress function Φ so that, The third equation will be dropped from the discussion for the time being, since its solution only indicates , which value may later be determined as a function of  yxfzz , yyxx  & We can verify that the equilibrium equations are identically satisfied. For complete solution, we must consider the compatibility condition, i.e.              y Y x X v yyxx 1 1 )(2 
)())(( 2 2 2 2 2 2 2 2 2 2 2 2 1 1 yxxyyx                      Stress functions with body forces (Plane strain)…….. xyyyxx  &,Now substituting the in the terms of stress function. This will result a differential equation will be such that any solution to it will satisfy both equilibrium & compatibility condition. Simplifying, It gives, If there is no body force, 0 1 1 22 )()()( 2 2 2 2 22 4 4 4 4 4 2 2 2 2                       yxyxyxyx   0 1 21 2 )()( 2 2 2 2 22 4 4 4 4 4                  yxyxyx    0 1 21 222        0 1 21 24        04   This gives the condition of stress function equation for the solution of problem of plane strain, provided the proper boundary conditions are also satisfied.
Stress functions with body forces (Plane stress) 0         X zyx zxxyxx  0         Y zyx yzyyxy  0         Z zyx zzyzzx  0      X yx xyxx  0      Y yx yyxy  We have general equilibrium equation; For plane stress condition; The component of body force X and Y independent Z & Z=0. The body force can be expressed by a potential function Ω, so that; x X    y Y      ,0      yx xy xx     0      yy xy yy  So equilibrium equation becomes, 2 2 y xx      yx xy      2 2 2 x yy      If we now choose a stress function Φ so that, We can verify that the equilibrium equations are identically satisfied. For complete solution, we must consider the compatibility condition, i.e.             y Y x X vyyxx )1()(2 
)())(( 2 2 2 2 2 2 2 2 2 2 2 2 )1( yxxyyx                    Stress functions with body forces (Plane stress)…….. xyyyxx  &,Now substituting the in the terms of stress function. This will result a differential equation will be such that any solution to it will satisfy both equilibrium & compatibility condition. Simplifying, It gives, If there is no body force, 0)1(22 )()()( 2 2 2 2 22 4 4 4 4 4 2 2 2 2                      yxyxyxyx   0)1(2 )()( 2 2 2 2 22 4 4 4 4 4                yxyxyx  0)1( 222   0)1( 24   04   This gives the condition of stress function equation for the solution of problem of plane stress, provided the proper boundary conditions are also satisfied.
Summary: Solutions to Plane Problems in Cartesian Coordinates yxxy xyyyxx                2 2 2 2 2 ,, Airy Representation for stresses 02 4 4 4 22 4 4 4          yyxx Biharmonic Governing Equation ),(,),( yxfYyxfX yx  Boundary Conditions R S x y 0)1( 24   0 1 21 24        If there is no body force, Plane Stress Plane Strain yyxy yxxx mlY mlX    
Example: )23( :FunctionStressFollowingthemin 2 3 ydxy d F forstressestheeDetr  Appears to Solve the Beam Problem: x y dF )( 6 ,0 )( 6 : 3 2 2 2 32 2 ydy d F yxx ydx d F y Answer xyy x                 
T 1 T +X +Y L L C C O Uniaxial Tension Example: Two-dimensional plane stress of a long rectangular narrow beam by a uniform traction T at each end. Boundary conditions are: 0yy Txx  0xy Cy  at (3) at (2) For all values of yLx  For all values of x 0xy Cy at (4)For all values of x at (1)For all values of yLx  Surface force, T is per unit area
Because constant stresses at boundaries, let choose a second-order stress function: 2 Ay Φ is a valid stress function as 04   Hence, 0,22 2     xyyyxx A y    The first boundary condition implies that , 2 T A  So, the stress field solution is, 0,  xyyyxx T  Let find the displacement field, i.e., u ,v   E T Ex u yyxxxx     1   E T Ey xxyyyy     1v 0 v        Gxy u xy xy  Total shear strain, (5) (6) (7)
Integrating Equations (5) and (6) w.r.t. x & y )(yfx E T u  )(v xgy E T   f(y) is a arbitrary function of y g(x) is a arbitrary function of x Putting in equation (7) cxgyfxgyf  )()(0)()( '''' Integrating, Dcyyf )( Dcxxg )( c, D are arbitrary constants It gives, Dcyx E T u  Dcxy E T  v Displacement boundary conditions are, At x=0, y=0 u=0 & v=0 also v=0 at x=±L for all y It gives, c=0 & D=0 So, final displacement field is, x E T u  y E T v
22 ByAx  3 Ax  224 3 yxxA  Problems: Prove that the followings are Airy’s stress function & examine the stress distribution represented by them: , , 22 ByAx  ,2Ax x    By y 2   A x 22 2    B y 22 2     04 4    x  04 4    y  022 4    yx  00002 4 4 22 4 4 4 4           yyxx   0,2,2 2 2 2 2 2           yx A x B y xyyyxx       Ans: Hence it is a Airy’s stress function. It is a uniform two-dimensional stress function. ,3 2 Ax x    04 4 2 2          yyy  Ax x 62 2     A x 63 3     04 4    x  04   0,2,0  xyyxx Ax  One-dimensional linear stress field. b) 022 4    yx   23 64 xyxA x      yAxyxA y 22 66     2 2 2 612 yxA x       22 2 2 66 AxxA y     Ax x 243 3     04 4    y  A x 244 4     Axy yx 12 2     Ay yx 122 3     A yx 1222 4     0024244  AA 2 6Axxx   22 612 yxAyy  Axyxy 12 , , So it is a Airy’s stress function, and c)

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5. stress function

  • 1. ME 6201: Applied Elasticity and Plasticity STRESS FUNCTION Cartesian coordinates Prof. S.K.Sahoo
  • 2. Plane Stress         0, 1 1 1 , 1        yzxzxyxy yxyxz xyyyxx E E EE           )( 2 1 0)( 2 1 )( 2 1 ,, x w z u y w z xy u z w y v x u xz yz xy zyx                                   0 yzxzz  Stress-strain relationship strain-displacement equations The equilibrium equations Field equations ),(),,(),,( yxyxyx xyxyyyxx   0 0             Y yx X yx yxy xyx   Plane Strain 0),,(),,(  wyxvvyxuu 0,2 )()( 2)( 2)(     yzxzxyxy yxyxz yyxy xyxx G G G     0 0             Y yx X yx yxy xyx   0 )( 2 1 ,,              yzxzz xy yx xy u yx u      )21)(1( 21 2 1 3 3 2              E G KGK  = Lamé’s constant G = shear modulus xyxy xyy yxx E v v v E v v v E v    )1( ) 1 ( 1 ) 1 ( 1 2 2           0 yzxzz  xyxy xyy yxx v E v v E v v E    )1( )( 1 )( 1 2 2        
  • 3. Simplified Elasticity Formulations Displacement Formulation Eliminate the stresses and strains from the general system of equations. This generates a system of three equations for the three unknown displacement components. Equations are on u, v, w terms. Stress Formulation Eliminate the displacements and strains from the general system of equations. This generates a system of six equations and for the six unknown stress components. Equations are on terms A practical elasticity problem may requires to solve 15 equations with 15 unknowns. It is very difficult to solve many real problems, so modified formulations have been developed. F(z) G(x,y) z x y Even using displacement or stress formulations three-dimensional problems are difficult to solve. So most solutions are assumed to of two-dimensional problems ],,,,,[ zxyzxyzzyyxx 
  • 4. Displacement Formulation 0 )1(2 0 )1(2 2 2                               Y y v x u yv E vG X y v x u xv E uG Plane stress + ),(),,( yxvvyxuu bb  (B.C.) Plane strain + ),(),,( yxvvyxuu bb  0)( 0)( 2 2                             Y y v x u y GvG X y v x u x GuG   (B.C.) )21)(1( 21 2 1 3 3 2              E G KGK  = Lamé’s constant G = shear modulus
  • 5. Stress Formulation yxxy xyyx          2 2 2 2 2 2             y Y x X vyx )1()(2  0 0             Y yx X yx yxy xyx   + or + Plane stress Plane strain 0 0             Y yx X yx yxy xyx                y Y x X v yx )1( 1 )(2  yyxy yxxx mlY mlX     yyxy yxxx mlY mlX     (B.C.) or yxxy xyyx          2 2 2 2 2 2 + +
  • 6. Stress Function Approach 0 0             Y yx X yx yxy xyx   yxxy xyyx          2 2 2 2 2 2 Solution of plain problems:              y Y x X v yx 1 1 )(2  Plain Strain             y Y x X vyx )1()(2  Plain Stress 0)(2 2 2 2            yx yx  3 different equations with 3 unknowns Airy (An English Mathematician) for 2D problems proposes a single stress function (function of space coordinates) that will satisfy all equations yyxy yxxx mlY mlX     When No body force 0)(2  yx  or Philosophy: Stress = f (strain) Strain = f (displacement) = f (space coordinate) = f (x, y, z) => Stress = f (x, y, z)
  • 7. Airy Stress Function Method Plane Problems with No Body Forces 0 0             yx yx yxy xyx   0)(2  yx Stress Formulation yxxy xyyx                2 2 2 2 2 ,, Airy Representation 02 4 4 4 22 4 4 4          yyxx Biharmonic Governing Equation (Single Equation with Single function  satisfy equilibrium and compatibility equations. The unknown parameters of stress function can be find out by putting boundary conditions. ) 0)(2 2 2 2            yx yx  Let assume a stress function = (x,y) such that: or Putting it on equilibrium equations, it can be checked that both equations are satisfied. Putting on compatibility equation gives
  • 8. Steps of the Stress Function Approach 1. Choose a stress function,  2. Confirm  is biharmonic (compatibility is satisfied) 3. Derive stresses from second derivatives of  4. Use boundary conditions to identify unknown parameters of  5. Derive strains from stress-strain relationships 6. Integrate strain-displacement relations to get displacements
  • 9. Example of biharmonic Polynomial Stress Functions       2 0211 2 20011000 0 0 ),( yAxyAxAyAxAAyxAyx m n nm mn terms do not contribute to the stresses and are therefore dropped1 nm terms will automatically satisfy the biharmonic equation3 nm terms require constants Amn to be related in order to satisfy biharmonic equation 3 nm cbyax ,If 22 If, cybxyax  04   ...........,If 2234222322  ykxyjxhxgyfxyyexdxcybxyax c y xx 2002 2       0022 2     a x yy   00 2     b yx xy   Stress distribution can be obtained but all to satisfy biharmonic equation, which requires that there exist some relations among coefficients. ..........2006200200 2 kxgyfxcxx  ...........26120026002 22 kyjxyhxeydxayy  ..........430022000 2 kxyjxfyexbxy  stress function have no meaning, as it gives zero stress in the body. Any quadratic function of x and y will automatically satisfy the biharmonic equation but give constant values of stresses. 04  
  • 10. Stress functions with body forces (Plane strain) 0         X zyx zxxyxx  0         Y zyx yzyyxy  0         Z zyx zzyzzx  0      X yx xyxx  0      Y yx yyxy  0   z zz We have general equilibrium equation; For plane strain condition; The component of body force X and Y independent Z & Z=0. The body force can be expressed by a potential function Ω, so that; x X    y Y      ,0      yx xy xx     0      yy xy yy  So equilibrium equation becomes, 2 2 y xx      yx xy      2 2 2 x yy      If we now choose a stress function Φ so that, The third equation will be dropped from the discussion for the time being, since its solution only indicates , which value may later be determined as a function of  yxfzz , yyxx  & We can verify that the equilibrium equations are identically satisfied. For complete solution, we must consider the compatibility condition, i.e.              y Y x X v yyxx 1 1 )(2 
  • 11. )())(( 2 2 2 2 2 2 2 2 2 2 2 2 1 1 yxxyyx                      Stress functions with body forces (Plane strain)…….. xyyyxx  &,Now substituting the in the terms of stress function. This will result a differential equation will be such that any solution to it will satisfy both equilibrium & compatibility condition. Simplifying, It gives, If there is no body force, 0 1 1 22 )()()( 2 2 2 2 22 4 4 4 4 4 2 2 2 2                       yxyxyxyx   0 1 21 2 )()( 2 2 2 2 22 4 4 4 4 4                  yxyxyx    0 1 21 222        0 1 21 24        04   This gives the condition of stress function equation for the solution of problem of plane strain, provided the proper boundary conditions are also satisfied.
  • 12. Stress functions with body forces (Plane stress) 0         X zyx zxxyxx  0         Y zyx yzyyxy  0         Z zyx zzyzzx  0      X yx xyxx  0      Y yx yyxy  We have general equilibrium equation; For plane stress condition; The component of body force X and Y independent Z & Z=0. The body force can be expressed by a potential function Ω, so that; x X    y Y      ,0      yx xy xx     0      yy xy yy  So equilibrium equation becomes, 2 2 y xx      yx xy      2 2 2 x yy      If we now choose a stress function Φ so that, We can verify that the equilibrium equations are identically satisfied. For complete solution, we must consider the compatibility condition, i.e.             y Y x X vyyxx )1()(2 
  • 13. )())(( 2 2 2 2 2 2 2 2 2 2 2 2 )1( yxxyyx                    Stress functions with body forces (Plane stress)…….. xyyyxx  &,Now substituting the in the terms of stress function. This will result a differential equation will be such that any solution to it will satisfy both equilibrium & compatibility condition. Simplifying, It gives, If there is no body force, 0)1(22 )()()( 2 2 2 2 22 4 4 4 4 4 2 2 2 2                      yxyxyxyx   0)1(2 )()( 2 2 2 2 22 4 4 4 4 4                yxyxyx  0)1( 222   0)1( 24   04   This gives the condition of stress function equation for the solution of problem of plane stress, provided the proper boundary conditions are also satisfied.
  • 14. Summary: Solutions to Plane Problems in Cartesian Coordinates yxxy xyyyxx                2 2 2 2 2 ,, Airy Representation for stresses 02 4 4 4 22 4 4 4          yyxx Biharmonic Governing Equation ),(,),( yxfYyxfX yx  Boundary Conditions R S x y 0)1( 24   0 1 21 24        If there is no body force, Plane Stress Plane Strain yyxy yxxx mlY mlX    
  • 15. Example: )23( :FunctionStressFollowingthemin 2 3 ydxy d F forstressestheeDetr  Appears to Solve the Beam Problem: x y dF )( 6 ,0 )( 6 : 3 2 2 2 32 2 ydy d F yxx ydx d F y Answer xyy x                 
  • 16. T 1 T +X +Y L L C C O Uniaxial Tension Example: Two-dimensional plane stress of a long rectangular narrow beam by a uniform traction T at each end. Boundary conditions are: 0yy Txx  0xy Cy  at (3) at (2) For all values of yLx  For all values of x 0xy Cy at (4)For all values of x at (1)For all values of yLx  Surface force, T is per unit area
  • 17. Because constant stresses at boundaries, let choose a second-order stress function: 2 Ay Φ is a valid stress function as 04   Hence, 0,22 2     xyyyxx A y    The first boundary condition implies that , 2 T A  So, the stress field solution is, 0,  xyyyxx T  Let find the displacement field, i.e., u ,v   E T Ex u yyxxxx     1   E T Ey xxyyyy     1v 0 v        Gxy u xy xy  Total shear strain, (5) (6) (7)
  • 18. Integrating Equations (5) and (6) w.r.t. x & y )(yfx E T u  )(v xgy E T   f(y) is a arbitrary function of y g(x) is a arbitrary function of x Putting in equation (7) cxgyfxgyf  )()(0)()( '''' Integrating, Dcyyf )( Dcxxg )( c, D are arbitrary constants It gives, Dcyx E T u  Dcxy E T  v Displacement boundary conditions are, At x=0, y=0 u=0 & v=0 also v=0 at x=±L for all y It gives, c=0 & D=0 So, final displacement field is, x E T u  y E T v
  • 19. 22 ByAx  3 Ax  224 3 yxxA  Problems: Prove that the followings are Airy’s stress function & examine the stress distribution represented by them: , , 22 ByAx  ,2Ax x    By y 2   A x 22 2    B y 22 2     04 4    x  04 4    y  022 4    yx  00002 4 4 22 4 4 4 4           yyxx   0,2,2 2 2 2 2 2           yx A x B y xyyyxx       Ans: Hence it is a Airy’s stress function. It is a uniform two-dimensional stress function. ,3 2 Ax x    04 4 2 2          yyy  Ax x 62 2     A x 63 3     04 4    x  04   0,2,0  xyyxx Ax  One-dimensional linear stress field. b) 022 4    yx   23 64 xyxA x      yAxyxA y 22 66     2 2 2 612 yxA x       22 2 2 66 AxxA y     Ax x 243 3     04 4    y  A x 244 4     Axy yx 12 2     Ay yx 122 3     A yx 1222 4     0024244  AA 2 6Axxx   22 612 yxAyy  Axyxy 12 , , So it is a Airy’s stress function, and c)