Strain Problems [PDF]

Lecture 18: Plane Stress/Strain Problems. ▫. Consider Example 6.1 of §6.2: ▫. Kinematic boundary conditions exist a

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Lecture 18: Plane Stress/Strain Problems. „ „

„

Consider Example 6.1 of §6.2: Kinematic boundary conditions exist at all three node points of the single element. Determine the element principal stresses. „ „

„

Assume the plane stress conditions. Using a CST element we have constant strains and stresses everywhere in the element domain. We can use Mohr’s circle to resolve principal stresses once we have element stresses. MECH 420: Finite Element Applications

E = 30 × 106 psi t = 1 in ν =0.25 1

u1 = 0.0000 in v1 = 0.0025 in u 2 = 0.0012 in v2 = 0.0000 in u 3 = 0.0000 in v3 = 0.0025 in

G d1



Lecture 18: Plane Stress/Strain Problems. A synopsis of the problem solution:

„

G

G

σ = Dε

;

⎡ ⎤ ⎢1 ν 0 ⎥ ⎥ E ⎢ D= ν 1 0 ⎥ 1 − ν 2 ⎢⎢ 1 −ν ⎥ ⎢0 0 ⎥ 2 ⎦ ⎣

∂uˆ dxˆ ∂vˆ εy = dxˆ ∂uˆ ∂vˆ + γ xy = dyˆ dxˆ

Specific to plane stress conditions (εz=0)

εx =

⎧uˆ ⎫ ⎨ ⎬ = [ N1 ⎩ vˆ ⎭

N2

N 3 ] dˆ

G

ε = Bdˆ

G ∴ σ = DBdˆ MECH 420: Finite Element Applications

; B = [ B1

B2

B3 ]3×6

Lecture 18: Plane Stress/Strain Problems. „

Can use any reference frame for the problem since we only need a consistent description of the node locations. „

„

In fact, when using the CST element as we have defined it there is little need to apply rotation transform, T. The choice of reference frame does not complicate the calculation of the B matrix. β i = y j − ym

B = ⎡⎣ Bi

Bj

⎡ ⎡ βi 1 ⎢⎢ Bm ⎤⎦ = 0 ⎢ ⎢ 2A ⎢ ⎢ ⎣⎣γ i

0⎤ γ i ⎥⎥ βi ⎥⎦

⎡β j ⎢ ⎢0 ⎢ ⎣γ j

0 ⎤ ⎡βm ⎥ γ j ⎥ ⎢⎢ 0 β j ⎦⎥ ⎢⎣ γ m

i =1 , j = 2 , m = 3

MECH 420: Finite Element Applications

0 ⎤⎤ ⎥ γ m ⎥⎥ ⎥ β m ⎥⎦ ⎥⎦

γ i = xm − x j β j = ym − yi γ j = xi − xm β m = yi − y j γ m = x j − xi

Lecture 18: Plane Stress/Strain Problems. „

Forming the stress expression directly:

1

β i = −1 γ i = −2 β j = −2 γj =0 β m = −1 γm = 2

⎡ −1 0 2 0 −1 0 ⎤ 1 ⎢ 0 −2 0 0 0 2 ⎥ B= ⎥ 2 ( 0.5 ⋅ 2 ⋅ 2 ) ⎢ ⎢⎣ −2 −1 0 2 2 −1⎥⎦

MECH 420: Finite Element Applications

Lecture 18: Plane Stress/Strain Problems. ⎡ ⎤ ⎢ 1 0.25 0 ⎥ 6 ⎥ 30 × 10 ⎢ D= 0.25 1 0 ⎥ 1 − 0.252 ⎢⎢ 1 − 0.25 ⎥ 0 ⎢ 0 ⎥ 2 ⎦ ⎣ „

T dˆ = {0.0 0.0025 0.0012 0.0 0.0 0.0025}

The stresses become:

⎧ 0.0 ⎫ ⎪0.0025⎪ ⎡ ⎤ ⎪ ⎢ 1 ⎧σ x ⎫ 0.25 0 ⎥ ⎡ −1 0 2 0 −1 0 ⎤ ⎪ 6 ⎞⎢ ⎥⎛ 1 G ⎪ ⎪ 30 × 10 ⎢ ⎥ ⎪0.0012 ⎪ 0.25 1 0 0 2 0 0 0 2 σ = ⎨σ y ⎬ = − ⎥ ⎜⎜ 2 0.5 ⋅ 2 ⋅ 2 ⎟⎟ ⎢ 2 ⎢ ⎥ ⎨ 0.0 ⎬ 1 0.25 − ( ) ⎠ ⎢ −2 −1 0 2 2 −1⎥ ⎪ ⎪τ ⎪ ⎪ ⎢ 1 − 0.25 ⎥ ⎝ ⎣ ⎦ ⎩ xy ⎭ ⎪ 0.0 ⎪ 0 ⎢ 0 ⎥ 2 ⎦ ⎣ ⎪ ⎪ 0.0025 ⎩ ⎭ ⎧ 19200 ⎫ ⎪ ⎪ = ⎨ 4800 ⎬ psi ⎪−15000 ⎪ ⎩ ⎭ MECH 420: Finite Element Applications

Lecture 18: Plane Stress/Strain Problems. „

We can recover any stress-strain information using the given the node displacements, including: „

„ „

Stresses in terms of the original x-y reference frame used to form the element equations (and the N and B matrices). Principle stress values. The FE solution was based on the approximate linear displacement field. „

„ „

„

Simplicity of the displacement field gives us the chance to get a solution. One we have that solution we can extract a lot of information. Always check for convergence. Postprocessors in ANSYS and other packages will blend the constant stresses/strains to give what appears to be much more resolved answers. Such blending procedures are only valid as long as the system has already converged. MECH 420: Finite Element Applications

Lecture 18: Plane Stress/Strain Problems. „

Consider any unit cell of material within the element of example 6.1.

y

σ y = 4.8 ksi

τ xy = −15 ksi

B

x

A

σ x = 19.2 ksi

MECH 420: Finite Element Applications

Lecture 18: Plane Stress/Strain Problems. „

„

§6.4 Explicit expression for the Constant Strain Triangle Stiffness Matrix. The element stiffness matrix kˆ = (t ⋅ A) BT DB has to be evaluated on an element by element basis. „

„

„

We usually form B and D separately and then carry out the matrix product. Pg.# 290 gives the full k matrix for the plane strain case.

Consider ν=0.5 (discussion on pg. #290-291): „

„

The material is totally incompressible – there can be no change in the volume of a material sample. Consider the conditions of plane strain – ν=0.5 gives a singular D matrix (you can experiment with this in ANSYS #3). MECH 420: Finite Element Applications

Lecture 18: Plane Stress/Strain Problems. „

Example 6.1 is an example of a postprocessing problem. „

„

§6.5 Finite Element Solution of a Plane Stress Problem. „

„

„

A displacement field had already been obtained by solving for the nodal displacements ui,vi (i=1..3). Outlines the solution process, using the CST element, behind the acquisition of these nodal displacements. Example problems in §6.5 are helpful for solving problem P. 6.7 of Assignment #3.

§6.3 Treatment of body and surface forces. „

„

We will now revisit example 6.1 and apply a gravitational load to the element and look at how the solution process will be set-up. In Tutorial #7 we will look at an example similar to P. 6.9 that illustrates how to deal with surface tractions. MECH 420: Finite Element Applications

Lecture 18: Plane Stress/Strain Problems. „

Example: treating body fixed loads. Planform area A and constant thickness t.

1

Weight per unit volume of ρg acting uniformly throughout the element volume. MECH 420: Finite Element Applications

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