next up previous

Next: Boundary Conditions (§ 2.5.0)
Up: Tutorial Contents
Previous: Linear Elastic Behaviour (§ 2.3.0)


Combining Equilibrium, Smooth Deformation, and Linear Elastic Behavior (§ 2.4.0)

The table below summarizes all the equations we have derived in previous sections of this tutorial. If they are still unfamiliar, the numbers correspond to the equation number. Return to the section and review the derivation of the equations before continuing the tutorial.

Equilibrium
δσxx
δx
+ δτxy
δy
= 0
(1)
δτyx
δx
+ δσyy
δy
= 0
(2)
Smooth
Deformation
δ2εxx
δy2
+ δ2εyy
δx2
- δ2γxy
δxδy
= 0
(11)
Linear
Elastic Behavior
εxx = 1
E
xx - νσyy)
(22)
εyy = 1
E
yy - νσxx)
(23)
γxy = τxy
G
(24)
G = E
2 (1 + ν)
(25)

Our first step is to combine the constitutive relations (from § 2.1.3: Linear Elastic Behavior) with the continuity expression (from § 2.1.2: Smooth Deformation) and equilibrium expressions (from § 2.1.1: Equilibrium). Starting with the continuity expression (Equation 11),
δ2εxx
δy2
+ δ2εyy
δx2
- δ2γxy
δxδy
= 0
(26)

and inserting the constitutive relations (Equations 22-25) into Equation 26, we get:
δ2
δy2
(1
E
xx - νσyy)) + δ2
δx2
(1
E
yy - νσxx)) - δ2
δxδy
((2+2ν)
E
τxy) = 0
(27)

Multiplying through by E and expanding the products leads to:
δ2σxx
δy2
+ δ2σyy
δx2
- ν (δ2σyy
δy2
+ δ2σxx
δx2
) - 2δ2τxy
δxδy
- 2 ν δ2τxy
δxδy
= 0
(28)

Our next step is to take the equilibrium expressions (Equations 1-2) and put them in a form where they can be substituted into Equation 28 above. To do this, take the partial derivatives of the equilibrium expressions, as shown below.

  Equilibrium Expression Partial Derivative
x-direction
δσxx
δx
= - δτxy
δy
(29)
δ2σxx
δx2
= - δ2τxy
δxδy
(30)
y-direction
δτyx
δx
= - δσyy
δy
(31)
δ2σyy
δy2
= - δ2τxy
δxδy
(32)

Plugging these new expressions for equilibrium (Equations 29 - 32) into our partial differential equation for smooth deformation in a linear elastic solid (Equation 28) and simplifying leads to the following expression:
δ2σxx
δy2
+ δ2σyy
δx2
-ν( - δ2τxy
δxδy
+ - δ2τxy
δxδy
) - 2δ2 τxy
δxδy
- 2 νδ2τxy
δxδy
= 0
(33)
δ2σxx
δy2
+ δ2σyy
δx2
-2δ2τxy
δxδy
= 0
(34)

This single partial differential equation describes the stresses, σxx, σyy, and τxy in any plane strain problem in a linear elastic solid in which deformations are smooth, satisfying our three assumptions mentioned in the Introduction to this chapter.

Remember that Equation 34 was derived in the same way that most mechanics problems are solved: by combining separate expressions developed from equilibrium, continuity, and material properties. Our next step is to fine-tune the equation to the specific problem setup of fracture mechanics by means of defining the boundary conditions.


Exercise 2: For the following questions, match each value to its variable.
1) Shear stress:
σ
x
ε
τ
2) Young's Modulus:
G
γ
E
σ


next up previous

Next: Boundary Conditions (§ 2.5.0)
Up: Tutorial Contents
Previous: Linear Elastic Behavior (§ 2.3.0)


August 9, 2004