Chapter 7. FEM: Plane Stress and Strain

Byeng D. Youn

System Health & Risk Management Laboratory Department of Mechanical & Aerospace Engineering

Plane Triangular Element

2

FEM in a Plane Stress Problem

3

Fig. 7.1 Plane stress: (a), (b)

(a) (b)

Fig. 7.2 Plane strain: (a), (b)

(a) (b)

• Plane stress and plane strain problems

• Constant-strain triangular element

• Equilibrium equation in 2-D

Plane stress: The stress state when normal stress, which is perpendicular to the plane x-y, and shear stress are both zero.

Plane strain: The strain state when normal strain , which is perpendicular to the plane x-y, and shear strain are both zero.

ε

z

xz, yz

γ γ

Stresses in 2-D Principal stress and its direction

𝜎𝜎 = 𝜎𝜎_𝑥𝑥 𝜎𝜎_𝑦𝑦

𝜏𝜏_{𝑥𝑥𝑦𝑦} 𝜎𝜎₁ = 𝜎𝜎_𝑥𝑥 + 𝜎𝜎_𝑦𝑦

2 + 𝜎𝜎_𝑥𝑥 − 𝜎𝜎_𝑦𝑦 2

2 + 𝜏𝜏_{𝑥𝑥𝑦𝑦}² = 𝜎𝜎_{𝑚𝑚𝑚𝑚𝑥𝑥}

𝜎𝜎₂ = 𝜎𝜎_𝑥𝑥 + 𝜎𝜎_𝑦𝑦

2 − 𝜎𝜎_𝑥𝑥 − 𝜎𝜎_𝑦𝑦 2

2 + 𝜏𝜏_{𝑥𝑥𝑦𝑦}² = 𝜎𝜎_{𝑚𝑚𝑖𝑖𝑖𝑖} tan2𝜃𝜃_𝑝𝑝 = 2𝜏𝜏_{𝑥𝑥𝑦𝑦}

𝜎𝜎_𝑥𝑥 − 𝜎𝜎_𝑦𝑦

Displacement and rotation of plane element ^{x − y}

𝜀𝜀_𝑥𝑥 = 𝜕𝜕𝑢𝑢

𝜕𝜕𝑥𝑥 𝜀𝜀_𝑦𝑦 = 𝜕𝜕𝜐𝜐

𝜕𝜕𝑦𝑦 𝛾𝛾_{𝑥𝑥𝑦𝑦} = 𝜕𝜕𝑢𝑢

𝜕𝜕𝑦𝑦 + 𝜕𝜕𝜐𝜐

𝜕𝜕𝑥𝑥 {𝜀𝜀} = 𝜀𝜀_𝑥𝑥 𝜀𝜀_𝑦𝑦 𝛾𝛾_{𝑥𝑥𝑦𝑦}

Stress-strain matrix(or material composed matrix) of isotropic material for plane stress (𝜎𝜎_𝑧𝑧 = 𝜏𝜏_{𝑥𝑥𝑧𝑧} = 𝜏𝜏_{𝑦𝑦𝑧𝑧} = 0)

Stress-strain matrix(or material composed matrix) of isotropic material for plane deformation (𝜀𝜀_𝑧𝑧 = 𝛾𝛾_{𝑥𝑥𝑧𝑧} = 𝛾𝛾_{𝑦𝑦𝑧𝑧} = 0)

𝐷𝐷 = 𝐸𝐸

1− ν²

1 ν 0

ν 1 0

0 0 1 − ν 2

𝐷𝐷 = 𝐸𝐸

(1 + ν)(1 − 2ν)

1 − ν ν 0

ν 1 − ν 0

0 0 1 − 2ν 2 𝜎𝜎 = 𝐷𝐷 {𝜀𝜀}

• Step 1: Determination of element type

• Step 2: Determination of displacement function

• Step 3: Relation of deformation rate – strain and stress-strain

• Step 4: Derivation of element stiffness and equation

• Step 5: Construction of global system equations and application of boundary conditions

• Step 6: Calculation of nodal displacement

• Step 7: Calculation of force (stress) in an element

Considering a triangular element, the nodes i, j, m are notated in the anti- clockwise direction.

The way to name the nodal members in an entire structure must be devised to avoid negative element area.

Step1: Determination of element type

𝑑𝑑 =

𝑑𝑑_𝑖𝑖 𝑑𝑑_𝑗𝑗 𝑑𝑑_𝑚𝑚 =

𝑢𝑢_𝑖𝑖 𝑣𝑣_𝑖𝑖 𝑢𝑢_𝑗𝑗 𝑣𝑣_𝑗𝑗 𝑢𝑢_𝑚𝑚 𝑣𝑣_𝑚𝑚

Linear function gives a guarantee to satisfy the compatibility.

A general displacement function {𝜓𝜓} containing function 𝑢𝑢 and 𝑣𝑣 can be expressed as below.

Step2: Determination of displacement function

Substitute nodal coordinates to the equation for obtaining the values of a.

𝜓𝜓 = 𝑎𝑎₁ + 𝑎𝑎₂𝑥𝑥 + 𝑎𝑎₃𝑦𝑦

𝑎𝑎₄ + 𝑎𝑎₅𝑥𝑥 + 𝑎𝑎₆𝑦𝑦 = 1 𝑥𝑥 𝑦𝑦

0 0 0 0 0 0 1 𝑥𝑥 𝑦𝑦

𝑎𝑎₁ 𝑎𝑎₂ 𝑎𝑎₃ 𝑎𝑎₄ 𝑎𝑎₅ 𝑎𝑎₆ 𝑢𝑢 𝑥𝑥,𝑦𝑦 = 𝑎𝑎₁ + 𝑎𝑎₂𝑥𝑥 + 𝑎𝑎₃𝑦𝑦

𝑣𝑣 𝑥𝑥,𝑦𝑦 = 𝑎𝑎₄ + 𝑎𝑎₅𝑥𝑥 + 𝑎𝑎₆𝑦𝑦

Calculation of

a

₁

, a

₂

, a

₃

:

Step2: Determination of displacement function (Continued)

Solving a, 𝑎𝑎 = 𝑥𝑥 ⁻¹{𝑢𝑢}

Obtaining the inverse matrix of [x],

where, 2𝐴𝐴 = 𝑥𝑥_𝑖𝑖 𝑦𝑦_𝑗𝑗 − 𝑦𝑦_𝑚𝑚 + 𝑥𝑥_𝑗𝑗(𝑦𝑦_𝑚𝑚 − 𝑦𝑦_𝑖𝑖)+𝑥𝑥_𝑚𝑚(𝑦𝑦_𝑖𝑖 − 𝑦𝑦_𝑗𝑗) : 2 times of triangle area.

or 𝑢𝑢_𝑖𝑖

𝑢𝑢_𝑗𝑗

𝑢𝑢_𝑚𝑚 = 1 𝑥𝑥_𝑖𝑖 𝑦𝑦_𝑖𝑖 1 𝑥𝑥_𝑗𝑗 𝑦𝑦_𝑗𝑗 1 𝑥𝑥_𝑚𝑚 𝑦𝑦_𝑚𝑚

𝑎𝑎₁ 𝑎𝑎₂ 𝑎𝑎₃

𝑥𝑥 ⁻¹ = 1 2𝐴𝐴

𝛼𝛼_𝑖𝑖 𝛼𝛼_𝑗𝑗 𝛼𝛼_𝑚𝑚 𝛽𝛽_𝑖𝑖 𝛽𝛽_𝑗𝑗 𝛽𝛽_𝑚𝑚 𝛾𝛾_𝑖𝑖 𝛾𝛾_𝑗𝑗 𝛾𝛾_𝑚𝑚 𝑢𝑢_𝑖𝑖 = 𝑎𝑎₁ + 𝑎𝑎₂𝑥𝑥_𝑖𝑖 + 𝑎𝑎₃𝑦𝑦_𝑖𝑖

𝑢𝑢_𝑗𝑗 = 𝑎𝑎₁ + 𝑎𝑎₂𝑥𝑥_𝑗𝑗 + 𝑎𝑎₃𝑦𝑦_𝑗𝑗 𝑢𝑢_𝑚𝑚 = 𝑎𝑎₁ + 𝑎𝑎₂𝑥𝑥_𝑚𝑚 + 𝑎𝑎₃𝑦𝑦_𝑚𝑚

𝛼𝛼_𝑖𝑖 = 𝑥𝑥_𝑗𝑗𝑦𝑦_𝑚𝑚 − 𝑦𝑦_𝑗𝑗𝑥𝑥_𝑚𝑚 𝛽𝛽_𝑖𝑖 = 𝑦𝑦_𝑗𝑗 − 𝑦𝑦_𝑚𝑚 𝛾𝛾_𝑖𝑖 = 𝑥𝑥_𝑚𝑚 − 𝑥𝑥_𝑗𝑗

𝛼𝛼_𝑗𝑗 = 𝑦𝑦_𝑖𝑖𝑥𝑥_𝑚𝑚 − 𝑥𝑥_𝑖𝑖𝑦𝑦_𝑚𝑚 𝛽𝛽_𝑗𝑗 = 𝑦𝑦_𝑚𝑚 − 𝑦𝑦_𝑖𝑖 𝛾𝛾_𝑗𝑗 = 𝑥𝑥_𝑖𝑖 − 𝑥𝑥_𝑚𝑚

𝛼𝛼_𝑚𝑚 = 𝑥𝑥_𝑖𝑖𝑦𝑦_𝑗𝑗 − 𝑦𝑦_𝑖𝑖𝑥𝑥_𝑗𝑗 𝛽𝛽_𝑚𝑚 = 𝑦𝑦_𝑖𝑖 − 𝑦𝑦_𝑗𝑗

𝛾𝛾_𝑚𝑚 = 𝑥𝑥_𝑗𝑗 − 𝑥𝑥_𝑖𝑖

Step2: Determination of displacement function (Continued)

Similarly,

Derivation of displacement function u(x, y) (v can also be derived similarly) 𝑎𝑎₁

𝑎𝑎₂

𝑎𝑎₃ = 1 2𝐴𝐴

𝛼𝛼_𝑖𝑖 𝛼𝛼_𝑗𝑗 𝛼𝛼_𝑚𝑚 𝛽𝛽_𝑖𝑖 𝛽𝛽_𝑗𝑗 𝛽𝛽_𝑚𝑚 𝛾𝛾_𝑖𝑖 𝛾𝛾_𝑗𝑗 𝛾𝛾_𝑚𝑚

𝑢𝑢_𝑖𝑖 𝑢𝑢_𝑗𝑗 𝑢𝑢_𝑚𝑚

𝑎𝑎₄ 𝑎𝑎₅

𝑎𝑎₆ = 1 2𝐴𝐴

𝛼𝛼_𝑖𝑖 𝛼𝛼_𝑗𝑗 𝛼𝛼_𝑚𝑚 𝛽𝛽_𝑖𝑖 𝛽𝛽_𝑗𝑗 𝛽𝛽_𝑚𝑚 𝛾𝛾_𝑖𝑖 𝛾𝛾_𝑗𝑗 𝛾𝛾_𝑚𝑚

𝑣𝑣_𝑖𝑖 𝑣𝑣_𝑗𝑗 𝑣𝑣_𝑚𝑚

𝑢𝑢 = 1 𝑥𝑥 𝑦𝑦 𝑎𝑎₁ 𝑎𝑎₂

𝑎𝑎₃ = 1

2𝐴𝐴 1 𝑥𝑥 𝑦𝑦

𝛼𝛼_𝑖𝑖 𝛼𝛼_𝑗𝑗 𝛼𝛼_𝑚𝑚 𝛽𝛽_𝑖𝑖 𝛽𝛽_𝑗𝑗 𝛽𝛽_𝑚𝑚 𝛾𝛾_𝑖𝑖 𝛾𝛾_𝑗𝑗 𝛾𝛾_𝑚𝑚

𝑢𝑢_𝑖𝑖 𝑢𝑢_𝑗𝑗 𝑢𝑢_𝑚𝑚 𝑎𝑎} = 𝑥𝑥 ⁻¹{𝑢𝑢

Step2: Determination of displacement function (Continued)

As the same way,

Simple expression of u and v:

Arranging by deployment:

where 𝑢𝑢(𝑥𝑥,𝑦𝑦) = 1

2𝐴𝐴{(𝛼𝛼_𝑖𝑖 + 𝛽𝛽_𝑖𝑖𝑥𝑥 + 𝛾𝛾_𝑖𝑖𝑦𝑦)𝑢𝑢_𝑖𝑖 + (𝛼𝛼_𝑗𝑗 + 𝛽𝛽_𝑗𝑗𝑥𝑥 + 𝛾𝛾_𝑗𝑗𝑦𝑦)𝑢𝑢_𝑗𝑗 + (𝛼𝛼_𝑚𝑚 + 𝛽𝛽_𝑚𝑚𝑥𝑥 + 𝛾𝛾_𝑚𝑚𝑦𝑦)𝑢𝑢_𝑚𝑚}

𝑣𝑣(𝑥𝑥,𝑦𝑦) = 1

2𝐴𝐴{(𝛼𝛼_𝑖𝑖 + 𝛽𝛽_𝑖𝑖𝑥𝑥 + 𝛾𝛾_𝑖𝑖𝑦𝑦)𝑣𝑣_𝑖𝑖 + (𝛼𝛼_𝑗𝑗 + 𝛽𝛽_𝑗𝑗𝑥𝑥 + 𝛾𝛾_𝑗𝑗𝑦𝑦)𝑣𝑣_𝑗𝑗 + (𝛼𝛼_𝑚𝑚 + 𝛽𝛽_𝑚𝑚𝑥𝑥 + 𝛾𝛾_𝑚𝑚𝑦𝑦)𝑣𝑣_𝑚𝑚}

𝑢𝑢(𝑥𝑥,𝑦𝑦) = 𝑁𝑁_𝑖𝑖𝑢𝑢_𝑖𝑖 + 𝑁𝑁_𝑗𝑗𝑢𝑢_𝑗𝑗 + 𝑁𝑁_𝑚𝑚𝑢𝑢_𝑚𝑚 𝜐𝜐(𝑥𝑥,𝑦𝑦) = 𝑁𝑁_𝑖𝑖𝜐𝜐_𝑖𝑖 + 𝑁𝑁_𝑗𝑗𝜐𝜐_𝑗𝑗 + 𝑁𝑁_𝑚𝑚𝜐𝜐_𝑚𝑚

𝑁𝑁_𝑖𝑖 = 1

2𝐴𝐴(𝛼𝛼_𝑖𝑖 + 𝛽𝛽_𝑖𝑖𝑥𝑥 + 𝛾𝛾_𝑖𝑖𝑦𝑦) 𝑁𝑁_𝑗𝑗 = 1

2𝐴𝐴(𝛼𝛼_𝑗𝑗 + 𝛽𝛽_𝑗𝑗𝑥𝑥 + 𝛾𝛾_𝑗𝑗𝑦𝑦) 𝑁𝑁_𝑚𝑚 = 1

2𝐴𝐴(𝛼𝛼_𝑚𝑚 + 𝛽𝛽_𝑚𝑚𝑥𝑥 + 𝛾𝛾_𝑚𝑚𝑦𝑦)

Step2: Determination of displacement function (Continued)

Making the equation be simple in a form of matrix, 𝜓𝜓 = 𝑁𝑁 {𝑑𝑑}

Arranging by deployment:

The displacement function is represented with shape functions and nodal displacement {d} .

, ,

i j m

N N N

{ψ} 𝜓𝜓 = 𝑢𝑢(𝑥𝑥,𝑦𝑦)

𝑣𝑣(𝑥𝑥,𝑦𝑦) =

𝑁𝑁_𝑖𝑖𝑢𝑢_𝑖𝑖 + 𝑁𝑁_𝑗𝑗𝑢𝑢_𝑗𝑗 + 𝑁𝑁_𝑚𝑚𝑢𝑢_𝑚𝑚

𝑁𝑁_𝑖𝑖𝑣𝑣_𝑖𝑖 + 𝑁𝑁_𝑗𝑗𝑣𝑣_𝑗𝑗 + 𝑁𝑁_𝑚𝑚𝑣𝑣_𝑚𝑚 = 𝑁𝑁_𝑖𝑖 0 𝑁𝑁_𝑗𝑗

0 𝑁𝑁_𝑖𝑖 0 0 𝑁𝑁_𝑚𝑚 0 𝑁𝑁_𝑗𝑗 𝑥𝑥 𝑁𝑁_𝑚𝑚

𝑢𝑢_𝑖𝑖 𝑣𝑣_𝑖𝑖 𝑢𝑢_𝑗𝑗 𝑣𝑣_𝑗𝑗 𝑢𝑢_𝑚𝑚

𝑣𝑣_𝑚𝑚

where [𝑁𝑁] = 𝑁𝑁_𝑖𝑖 0 𝑁𝑁_𝑗𝑗

0 𝑁𝑁_𝑖𝑖 0 0 𝑁𝑁_𝑚𝑚 0 𝑁𝑁_𝑗𝑗 𝑥𝑥 𝑁𝑁_𝑚𝑚

Step2: Determination of displacement function (Continued) Review of characteristics of shape function:

𝑁𝑁_𝑖𝑖 =1, 𝑁𝑁_𝑗𝑗 =0, 𝑁𝑁_𝑚𝑚 =0 at nodes (𝑥𝑥_𝑖𝑖, 𝑦𝑦_𝑖𝑖)

A change of N_i of general elements across the surface x - y

Step3: Relation of deformation rate – strain and stress-strain

Calculation of partial differential terms Deformation rate: 𝜀𝜀 =

𝜀𝜀_𝑥𝑥 𝜀𝜀_𝑦𝑦 𝛾𝛾_{𝑥𝑥𝑦𝑦} =

𝜕𝜕𝑢𝑢

𝜕𝜕𝑣𝑣𝜕𝜕𝑥𝑥

𝜕𝜕𝑢𝑢𝜕𝜕𝑦𝑦

𝜕𝜕𝑦𝑦 + 𝜕𝜕𝑣𝑣

𝜕𝜕𝑥𝑥

𝜕𝜕𝑢𝑢

𝜕𝜕𝑥𝑥 = 𝑢𝑢_,𝑥𝑥 = 𝜕𝜕

𝜕𝜕𝑥𝑥(𝑁𝑁_𝑖𝑖𝑢𝑢_𝑖𝑖 + 𝑁𝑁_𝑗𝑗𝑢𝑢_𝑗𝑗 + 𝑁𝑁_𝑚𝑚𝑢𝑢_𝑚𝑚) = 𝑁𝑁_{𝑖𝑖,𝑥𝑥}𝑢𝑢_𝑖𝑖 + 𝑁𝑁_{𝑗𝑗,𝑥𝑥}𝑢𝑢_𝑗𝑗 + 𝑁𝑁_{𝑚𝑚,𝑥𝑥}𝑢𝑢_𝑚𝑚 𝑁𝑁_{𝑖𝑖,𝑥𝑥} = 1

2𝐴𝐴

𝜕𝜕

𝜕𝜕𝑥𝑥 𝛼𝛼^𝑖𝑖 + 𝛽𝛽_𝑖𝑖𝑥𝑥 + 𝛾𝛾_𝑖𝑖𝑦𝑦 = 𝛽𝛽_𝑖𝑖

2𝐴𝐴, 𝑁𝑁_{𝑗𝑗,𝑥𝑥} = 𝛽𝛽_𝑗𝑗

2𝐴𝐴, 𝑁𝑁_{𝑚𝑚,𝑥𝑥}= 𝛽𝛽_𝑚𝑚 2𝐴𝐴

∴ 𝜕𝜕𝑢𝑢

𝜕𝜕𝑥𝑥 = 1

2𝐴𝐴(𝛽𝛽_𝑖𝑖𝑢𝑢_𝑖𝑖 + 𝛽𝛽_𝑗𝑗𝑢𝑢_𝑗𝑗 + 𝛽𝛽_𝑚𝑚𝑢𝑢_𝑚𝑚)

Step3: Relation of deformation rate – strain and stress-strain

Summarizing the deformation rate equation, Likewise,

where, 𝜀𝜀 = 1

2𝐴𝐴

𝛽𝛽_𝑖𝑖 0 𝛽𝛽_𝑗𝑗 0 𝛾𝛾_𝑖𝑖 0 𝛾𝛾_𝑖𝑖 𝛽𝛽_𝑖𝑖 𝛾𝛾_𝑗𝑗

0 𝛽𝛽_𝑚𝑚 0 𝛾𝛾_𝑗𝑗 0 𝛾𝛾_𝑚𝑚 𝛽𝛽_𝑗𝑗 𝛾𝛾_𝑚𝑚 𝛽𝛽_𝑚𝑚

𝑢𝑢_𝑖𝑖 𝑣𝑣_𝑖𝑖 𝑢𝑢_𝑗𝑗 𝑣𝑣_𝑗𝑗 𝑢𝑢_𝑚𝑚 𝑣𝑣_𝑚𝑚

= 𝐵𝐵 𝑑𝑑 = 𝐵𝐵_𝑖𝑖 𝐵𝐵_𝑗𝑗 𝐵𝐵_𝑚𝑚 𝑑𝑑_𝑖𝑖 𝑑𝑑_𝑗𝑗 𝑑𝑑_𝑚𝑚

[𝐵𝐵_𝑖𝑖] = 1 2𝐴𝐴

𝛽𝛽_𝑖𝑖 0 0 𝛾𝛾_𝑖𝑖

𝛾𝛾_𝑖𝑖 𝛽𝛽_𝑖𝑖 [𝐵𝐵_𝑗𝑗] = 1 2𝐴𝐴

𝛽𝛽_𝑗𝑗 0 0 𝛾𝛾_𝑗𝑗

𝛾𝛾_𝑗𝑗 𝛽𝛽_𝑗𝑗 [𝐵𝐵_𝑚𝑚] = 1 2𝐴𝐴

𝛽𝛽_𝑚𝑚 0 0 𝛾𝛾_𝑚𝑚 𝛾𝛾_𝑚𝑚 𝛽𝛽_𝑚𝑚

𝜕𝜕𝜐𝜐

𝜕𝜕𝑦𝑦 = 1

2𝐴𝐴(𝛾𝛾_𝑖𝑖𝜐𝜐_𝑖𝑖 +𝛾𝛾_𝑗𝑗𝜐𝜐_𝑗𝑗 +𝛾𝛾_𝑚𝑚𝜐𝜐_𝑚𝑚)

𝜕𝜕𝑢𝑢

𝜕𝜕𝑦𝑦 + 𝜕𝜕𝜐𝜐

𝜕𝜕𝑥𝑥 = 1

2𝐴𝐴(𝛾𝛾_𝑖𝑖𝑢𝑢_𝑖𝑖 +𝛽𝛽_𝑖𝑖𝜐𝜐_𝑖𝑖 +𝛾𝛾_𝑗𝑗𝑢𝑢_𝑗𝑗 +𝛽𝛽_𝑗𝑗𝜐𝜐_𝑗𝑗 +𝛾𝛾_𝑚𝑚𝑢𝑢_𝑚𝑚 +𝛽𝛽_𝑚𝑚𝜐𝜐_𝑚𝑚

Step3: Relation of deformation rate – strain and stress-strain (Continued)

 CST: Constant – Strain Triangle

Strain is constant in an element, for matrix 𝐵𝐵 regardless of x and y coordinates, and is influenced by only nodal coordinates in an element.

Relation of stress - strain 𝜎𝜎_𝑥𝑥

𝜎𝜎_𝑦𝑦

𝜏𝜏_{𝑥𝑥𝑦𝑦} = [𝐷𝐷]

𝜀𝜀_𝑥𝑥 𝜀𝜀_𝑦𝑦

𝛾𝛾_{𝑥𝑥𝑦𝑦}  𝜎𝜎} = [𝐷𝐷][𝐵𝐵]{𝑑𝑑

Step4: Derivation of element stiffness matrix and equation

Total potential energy 𝜋𝜋_𝑝𝑝 = 𝜋𝜋_𝑝𝑝(𝑢𝑢_𝑖𝑖,𝜐𝜐_𝑖𝑖,𝑢𝑢_𝑗𝑗, . . . ,𝜐𝜐_𝑚𝑚) = 𝑈𝑈 + 𝛺𝛺_𝑏𝑏 + 𝛺𝛺_𝑝𝑝 + 𝛺𝛺_𝑠𝑠 Using minimum potential energy principle.

Strain energy 𝑈𝑈 = 1 2�

𝑉𝑉

𝜀𝜀 ^𝑇𝑇 𝜎𝜎 𝑑𝑑𝑑𝑑 = �

𝑉𝑉

𝜀𝜀 ^𝑇𝑇[𝐷𝐷] 𝜀𝜀 𝑑𝑑𝑑𝑑

Potential energy due to body force Ω_𝑏𝑏 = − �

𝑉𝑉

𝜓𝜓 ^𝑇𝑇 𝑋𝑋 𝑑𝑑𝑑𝑑

Potential energy due to concentrated load Ω_𝑝𝑝 = − 𝑑𝑑 ^𝑇𝑇{𝑃𝑃}

Potential energy due to distributed load (or surface force) Ω_𝑆𝑆 = − �

𝑆𝑆

𝜓𝜓 ^𝑇𝑇 𝑇𝑇 𝑑𝑑𝑆𝑆

Step4: Derivation of element stiffness matrix and equation (Continued)

∴ 𝜋𝜋_𝑝𝑝

= 1 2�

𝑉𝑉

𝑑𝑑 ^𝑇𝑇 𝐵𝐵 ^𝑇𝑇 𝐷𝐷 [𝐵𝐵] 𝑑𝑑 𝑑𝑑𝑑𝑑 − �

𝑉𝑉

𝑑𝑑 ^𝑇𝑇 𝑁𝑁 ^𝑇𝑇 𝑋𝑋 𝑑𝑑𝑑𝑑 − 𝑑𝑑 ^𝑇𝑇 𝑃𝑃 − �

𝑆𝑆

𝑑𝑑 ^𝑇𝑇 𝑁𝑁 ^𝑇𝑇 𝑇𝑇 𝑑𝑑𝑆𝑆

= 1

2 𝑑𝑑 ^𝑇𝑇 �

𝑉𝑉

𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑑𝑑 𝑑𝑑 − 𝑑𝑑 ^𝑇𝑇 �

𝑉𝑉

𝑁𝑁 ^𝑇𝑇 𝑋𝑋 𝑑𝑑𝑑𝑑 − 𝑑𝑑 ^𝑇𝑇 𝑃𝑃 − 𝑑𝑑 ^𝑇𝑇 �

𝑆𝑆

𝑁𝑁 ^𝑇𝑇 𝑇𝑇 𝑑𝑑𝑆𝑆

= 1

2 𝑑𝑑 ^𝑇𝑇 �

𝑉𝑉

𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑑𝑑 𝑑𝑑 − − 𝑑𝑑 ^𝑇𝑇 𝑓𝑓

Condition having the minimum is

where 𝑓𝑓 = �

𝑉𝑉

𝑁𝑁 ^𝑇𝑇 𝑋𝑋 𝑑𝑑𝑑𝑑 + 𝑃𝑃 + �

𝑆𝑆

𝑁𝑁 ^𝑇𝑇 𝑇𝑇 𝑑𝑑𝑆𝑆

𝜕𝜕𝜋𝜋_𝑝𝑝

𝜕𝜕{𝑑𝑑} = �

𝑉𝑉

𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑑𝑑 𝑑𝑑 − 𝑓𝑓 = 0

Step4: Derivation of element stiffness matrix and equation (Continued) Condition having the minimum is

So, the element stiffness matrix is (Case of an element having constant thickness t)

𝑘𝑘 = �

𝑉𝑉

𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑑𝑑 = 𝑡𝑡 �

𝐴𝐴

𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦 = 𝑡𝑡𝐴𝐴 𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵

𝜕𝜕𝜋𝜋_𝑝𝑝

𝜕𝜕{𝑑𝑑} = �

𝑉𝑉

𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑑𝑑 𝑑𝑑 − 𝑓𝑓 = 0 �

𝑉𝑉

𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑑𝑑 𝑑𝑑 = {𝑓𝑓}



𝑓𝑓_1𝑥𝑥 𝑓𝑓_1𝑦𝑦 𝑓𝑓_2𝑥𝑥 𝑓𝑓_2𝑦𝑦 𝑓𝑓_3𝑥𝑥 𝑓𝑓_3𝑦𝑦

=

𝑘𝑘₁₁ 𝑘𝑘₁₂

𝑘𝑘₂₁ 𝑘𝑘₂₂ ⋯ 𝑘𝑘𝑘𝑘¹⁶₂₆

⋮ ⋱ ⋮

𝑘𝑘₆₁ 𝑘𝑘₆₂ ⋯ 𝑘𝑘₆₆

𝑢𝑢₁ 𝑣𝑣₁ 𝑢𝑢₂ 𝑣𝑣₂ 𝑢𝑢₃ 𝑣𝑣₃

Matrix [k] is a 6x6 matrix, and the element equation is as below

Step5: Introduction a combination of element equation and boundary conditions for obtaining a global coordinate system of equation

Transformation from the global coordinate system to the local coordinate system: (See Ch. 3)

[ ]K [k^{( )e} ]

e

= N

∑

= 1

{ }F {f ^{( )e} }

e

= N

∑

= 1

{ }F =[ ]{ }K d

and

Step6: Calculation of nodal displacement

Step7: Calculation of force(stress) in an element

  

d =Td f = T f k = T k T^T

Constant-strain triangle(CST) has 6 degrees of freedom.

A triangular element with local coordinates system not along to the global coordinate system.

where

�𝑇𝑇 =

𝐶𝐶 𝑆𝑆 0 0 0 0

−𝑆𝑆 𝐶𝐶 0 0 0 0

0 0 𝐶𝐶 𝑆𝑆 0 0

0 0 −𝑆𝑆 𝐶𝐶 0 0

0 0 0 0 𝐶𝐶 𝑆𝑆

0 0 0 0 −𝑆𝑆 𝐶𝐶

𝐶𝐶 = cos𝜃𝜃 𝑆𝑆 = sin𝜃𝜃

below figure) under surface force.

thickness t = 1 in, E = 30 x 106 psi, 𝜈𝜈 =0.30

The global system of the governing equation is

where [K] is a 5x5 matrix.

or

𝐹𝐹_1𝑥𝑥 𝐹𝐹_1𝑦𝑦 𝐹𝐹_2𝑥𝑥 𝐹𝐹_2𝑦𝑦 𝐹𝐹_3𝑥𝑥 𝐹𝐹_3𝑦𝑦 𝐹𝐹_4𝑥𝑥 𝐹𝐹_4𝑦𝑦

=

𝑅𝑅_1𝑥𝑥 𝑅𝑅_1𝑦𝑦 𝑅𝑅_2𝑥𝑥 𝑅𝑅_2𝑦𝑦 50000 50000

= 𝐾𝐾

𝑑𝑑_1𝑥𝑥 𝑑𝑑_1𝑦𝑦 𝑑𝑑_2𝑥𝑥 𝑑𝑑_2𝑦𝑦 𝑑𝑑_3𝑥𝑥 𝑑𝑑_3𝑦𝑦 𝑑𝑑_4𝑥𝑥 𝑑𝑑_4𝑦𝑦

= [𝐾𝐾]

00 00 𝑑𝑑_3𝑥𝑥 𝑑𝑑_3𝑦𝑦 𝑑𝑑_4𝑥𝑥 𝑑𝑑_4𝑦𝑦 𝐹𝐹 = 1

2 (1000psi)(1in.× 10in.) 𝐹𝐹 = 5000lb

𝐹𝐹 = 1 2𝑇𝑇𝐴𝐴

𝐹𝐹} = [𝐾𝐾]{𝑑𝑑

- Element 1

• Calculation of matrix [B]

and where

[𝐵𝐵] = 1 2𝐴𝐴

𝛽𝛽_𝑖𝑖 0 𝛽𝛽_𝑗𝑗 0 𝛾𝛾_𝑖𝑖 0 𝛾𝛾_𝑖𝑖 𝛽𝛽_𝑖𝑖 𝛾𝛾_𝑗𝑗

0 𝛽𝛽_𝑚𝑚 0 𝛾𝛾_𝑗𝑗 0 𝛾𝛾_𝑚𝑚 𝛽𝛽_𝑗𝑗 𝛾𝛾_𝑚𝑚 𝛽𝛽_𝑚𝑚

𝐴𝐴 = 1 2 𝑏𝑏𝑏

= 1

2 20 10 = 100 𝑖𝑖𝑖𝑖.

²

𝛽𝛽_𝑖𝑖 = 𝑦𝑦_𝑗𝑗 − 𝑦𝑦_𝑚𝑚 = 10− 10 = 0 𝛽𝛽_𝑗𝑗 = 𝑦𝑦_𝑚𝑚 − 𝑦𝑦_𝑖𝑖 = 10 −0 = 10 𝛽𝛽_𝑚𝑚 = 𝑦𝑦_𝑖𝑖 − 𝑦𝑦_𝑗𝑗 = 0 −10 = −10

𝛾𝛾_𝑖𝑖 = 𝑥𝑥_𝑚𝑚 − 𝑥𝑥_𝑗𝑗 = 0 −20 = −20 𝛾𝛾_𝑗𝑗 = 𝑥𝑥_𝑖𝑖 − 𝑥𝑥_𝑚𝑚 = 0−0 = 0 𝛾𝛾_𝑚𝑚 = 𝑥𝑥_𝑗𝑗 − 𝑥𝑥_𝑖𝑖 = 20−0 = 20

• Matrix [D] (Plane stress)

• Calculation of stiffness matrix [𝐵𝐵] = 1

200

0 0 10

0 −20 0

−20 0 0

0 −10 0

0 0 20

10 20 −10

𝐷𝐷 = 𝐸𝐸

1 − ν²

1 ν 0

ν 1 0

0 0 1 − ν 2

= 30(10⁶) 0.91

1 0.3 0 0.3 1 0

0 0 0.35

𝑘𝑘 = 𝑡𝑡𝐴𝐴 𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 = 75,000 0.91

140 0 0 −70 −140 70

0 400 −60 0 60 −400

0 −60 100 0 −100 60

−70 0 0 35 70 −35

−140 60 −100 70 240 −130

70 −400 60 −35 −130 435

i = 1 j = 3 m = 2

• Calculation of matrix [B]

and where

[𝐵𝐵] = 1 2𝐴𝐴

𝛽𝛽_𝑖𝑖 0 𝛽𝛽_𝑗𝑗 0 𝛾𝛾_𝑖𝑖 0 𝛾𝛾_𝑖𝑖 𝛽𝛽_𝑖𝑖 𝛾𝛾_𝑗𝑗

0 𝛽𝛽_𝑚𝑚 0 𝛾𝛾_𝑗𝑗 0 𝛾𝛾_𝑚𝑚 𝛽𝛽_𝑗𝑗 𝛾𝛾_𝑚𝑚 𝛽𝛽_𝑚𝑚

𝐴𝐴 = 1 2 𝑏𝑏𝑏

= 1

2 20 10 = 100 𝑖𝑖𝑖𝑖.

²

𝛽𝛽_𝑖𝑖 = 𝑦𝑦_𝑗𝑗 − 𝑦𝑦_𝑚𝑚 = 0− 10 = −10 𝛽𝛽_𝑗𝑗 = 𝑦𝑦_𝑚𝑚 − 𝑦𝑦_𝑖𝑖 = 10 −0 = 10

𝛽𝛽_𝑚𝑚 = 𝑦𝑦_𝑖𝑖 − 𝑦𝑦_𝑗𝑗 = 0−0 = 0 𝛾𝛾_𝑖𝑖 = 𝑥𝑥_𝑚𝑚 − 𝑥𝑥_𝑗𝑗 = 20 −20 = 0 𝛾𝛾_𝑗𝑗 = 𝑥𝑥_𝑖𝑖 − 𝑥𝑥_𝑚𝑚 = 0−20 = −20

𝛾𝛾_𝑚𝑚 = 𝑥𝑥_𝑗𝑗 − 𝑥𝑥_𝑖𝑖 = 20 −0 = 20

• Matrix [D] (Plane stress)

• Calculation of stiffness matrix [𝐵𝐵] = 1

200

−10 0 10

0 0 0

0 −10 −20

0 0 0

−20 0 20 10 20 0

𝐷𝐷 = 30(10⁶) 0.91

1 0.3 0 0.3 1 0

0 0 0.35

𝑘𝑘 = 75,000 0.91

100 0 −100 60 0 −60

0 35 70 −35 −70 0

−100 70 240 −130 −140 60

60 −35 −130 435 70 −400

0 −70 −140 70 140 0

−60 0 60 −400 0 400

i = 1 j = 4 m = 3

Element 1:

Element 2:

[𝑘𝑘] = 375,000 0.91

28 0 −28 14 0 −14 0 0

0 80 12 −80 −12 0 0 0

−28 12 48 −26 −20 14 0 0

14 −80 −26 87 12 −7 0 0

0 −12 −20 12 20 0 0 0

−14 0 14 −7 0 7 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

1 2 3 4

[𝑘𝑘] = 375,000 0.91

20 0 0 0 0 −12 −20 12

0 7 0 0 −14 0 14 −7

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 −14 0 0 28 0 −28 14

−12 0 0 0 0 80 12 −80

−20 14 0 0 −28 12 48 −26

12 −7 0 0 14 −80 −26 87

global system of stiffness matrix is obtained as below.

1 2 3 4

[𝐾𝐾] = 375,000 0.91

48 0 −28 14 0 −26 −20 12

0 87 12 −80 −26 0 14 −7

−28 12 48 −26 −20 14 0 0

14 −80 −26 87 12 −7 0 0

0 −26 −20 12 48 0 −28 14

−26 0 14 −7 0 87 12 −80

−20 14 0 0 −28 12 48 −26

12 −7 0 0 14 −80 −26 87

Applying given boundary conditions with elimination of columns and rows.

𝑅𝑅_1𝑥𝑥 𝑅𝑅_1𝑦𝑦 𝑅𝑅_2𝑥𝑥 𝑅𝑅_2𝑦𝑦 50000 50000

= 375,000 0.91

48 0 −28 14 0 −26 −20 12

0 87 12 −80 −26 0 14 −7

−28 12 48 −26 −20 14 0 0

14 −80 −26 87 12 −7 0 0

0 −26 −20 12 48 0 −28 14

−26 0 14 −7 0 87 12 −80

−20 14 0 0 −28 12 48 −26

12 −7 0 0 14 −80 −26 87

00 00 𝑑𝑑_3𝑥𝑥 𝑑𝑑_3𝑦𝑦 𝑑𝑑_4𝑥𝑥 𝑑𝑑_4𝑦𝑦

50000 50000

= 375,000 0.91

48 0 −28 14

0 87 12 −80

−28 12 48 −26

14 −80 −26 87

𝑑𝑑_3𝑥𝑥 𝑑𝑑_3𝑦𝑦 𝑑𝑑_4𝑥𝑥 𝑑𝑑_4𝑦𝑦

Therefore, x-component of the displacement at nodes in the equation

∭_𝑉𝑉 𝐵𝐵 ^𝑇𝑇 𝐷𝐷 𝐵𝐵 𝑑𝑑𝑑𝑑 𝑑𝑑 = {𝑓𝑓} of 2-D plane is quite accurate when considering the coarse grids.

The solution of 1-D beam under tension force is 𝑑𝑑_3𝑥𝑥

𝑑𝑑_3𝑦𝑦 𝑑𝑑_4𝑥𝑥 𝑑𝑑_4𝑦𝑦

= 0.91 375,000

48 0 −28 14

0 87 12 −80

−28 12 48 −26

14 −80 −26 87

−1 5000

50000 0

=

609.6 663.74.2 104.1

× 10⁻⁶in.

𝛿𝛿 = 𝑃𝑃𝑃𝑃

𝐴𝐴𝐸𝐸 = (10,000)20

10(30 × 10⁶) = 670 × 10⁻⁶in.

Calculating, Element 1

{𝜎𝜎} = 𝐸𝐸 1 − ν²

1 ν 0

ν 1 0

0 0 1 − ν 2

× 1 2𝐴𝐴

𝛽𝛽₁ 0 𝛽𝛽₄ 0 𝛾𝛾₁ 0 𝛾𝛾₁ 𝛽𝛽₁ 𝛾𝛾₄

0 𝛽𝛽₃ 0 𝛾𝛾₄ 0 𝛾𝛾₃ 𝛽𝛽₄ 𝛾𝛾₃ 𝛽𝛽₃

𝑑𝑑_1𝑥𝑥 𝑑𝑑_1𝑦𝑦 𝑑𝑑_4𝑥𝑥 𝑑𝑑_4𝑦𝑦 𝑑𝑑_3𝑥𝑥 𝑑𝑑_3𝑦𝑦

𝜎𝜎_𝑥𝑥 𝜎𝜎_𝑦𝑦

𝜏𝜏_{𝑥𝑥𝑦𝑦} = 1005 3012.4 psi

FOR LISTENING