✔ME 5463-001 Fracture Mechanics, Homework #2

— David Wagner 2009/09/01 11:50

Determine the resulting stress equations for σx, σy, and τxy given the Airy stress function Phi = a_1 x^4 + a_2 x^3 y + a_3 x^2 y^2 + a_4 xy^3 + a_5 y^5 where the ai are constants. Please determine, if any, the restrictions between the constants.

Stress Equations

{partial Phi}/{partial y} = 5 a_5 y^4+3 a_4 x y^2+2 a_3 x^2 y+a_2 x^3

sigma_x = {partial^2Phi}/{partial y^2} = 20 a_5 y^3+6 a_4 x y+2 a_3 x^2

{partial Phi}/{partial x}= a_4 y^3+2 a_3 x y^2+3 a_2 x^2 y+4 a_1 x^3

sigma_y = {partial^2Phi}/{partial x^2} = 2 a_3 y^2+6 a_2 x y+12 a_1 x^2

tau_xy = {partial^2Phi}/{partial x partial y}

= {partial Phi}/{partial x}( {partial Phi}/{partial y} )

={partial Phi}/{partial x}( 5 a_5 y^4+3 a_4 x y^2+2 a_3 x^2 y+a_2 x^3 )=

tau_xy=a_4 y^3+2 a_3 x y^2+3 a_2 x^2 y+4 a_1 x^3

Constant Restrictions

{partial^3 Phi}/{partial x^3} =6 a_2 y+24 a_1 x

{partial^4 Phi}/{partial x^4} =24a_1

{partial^3 Phi}/{partial y^3} =60 a_5 y^2+6 a_4 x

{partial^4 Phi}/{partial y^4} =120 a_5 y

{partial^4 Phi}/{partial x^2 partial y^2} ={partial^4 Phi}/{partial x^2}({partial^2 Phi}/{partial y^2}) ={partial Phi}/{partial x}({partial Phi}/{partial x}( {partial^2 Phi}/{partial y^2}))

={partial Phi}/{partial x}({partial Phi}/{partial x}( 20 a_5 y^3+6 a_4 x y+2 a_3 x^2))

={partial Phi}/{partial x}( 6 a_4 y+4 a_3 x)

={partial^4 Phi}/{partial x^2 partial y^2} =4 a_3

  • Restrictions among constants come from compatibility constraints.
  • Set the Airy function to zero.

{partial^4 Phi}/{partial x^4} + 2 {partial^4 Phi}/{partial x^2 partial y^2} + {partial^4 Phi}/{partial y^4}= 0 = 24a_1 +8 a_3 +120 a_5 y=0 ∀ x and y ⇒

a_5=0
24a_1 +8 a_3=0 ⇒ a_3 = -3 a_1

Personal Tools