Appunti completi Advanced Machine Design

ADVANCE MACHINE DESIGN

CODE COURSE: 161001 CREDITI FORMATIVI: 6CFU (Corso Integrato)

DOCENTI:

MATERIALE DEL CORSO:

On Teams Channel in section “File” + during the lesson in writing mode.

BOOKS:

• Roark’s formulas for stress and strain
• Mechanical engineering design
• Fundamental for machine component design
• Theory of plates and shells
• Mechanics of materials
• Principles of composite material mechanics
• Theory of elasticity
• Mechanical reliability.

MODALITA’ D’ESAME:

Written exam about theoretical aspects and application - no calculation: presence of three open questions - 1h30min

The result of the exam will be done through e-mail.

For the verbalization of the integrated course, it is necessary send an e-mail to S.Baragetti or E.V.Arcieri in which has to be indicated the date of each module and the results; it is necessary to attach a proof of the claims.

Composite Mechanic Materials

Amplifying material mechanics - Design of structures composed of composite materials.

• Composite in nature

• Wood and wood laminates
• Bone collagen fibers + hard Ca matrix

Composite Classification

• Anisotropic materials
• Design optimizations
• Composite can consist of two or more separate materials

• Matrix - Ductile (Al alloy 6061) - Heat-treated
• Fiber reinforcements
• Particle reinforcements

Example: Fiber glass with σ_f ≥ 3400 MPa

• Higher even higher values with higher σ_f
• T_m: 130 MPa

Example: Direction of the fiber

• Short fibers

Al allocation in matrix

Elastic Symmetry Plane

Considering a cube as reported in the image, X-Y plane is the symmetry plane, which means that a vector normal one XY plane, remains unaltered (same size and direction) after a transformation. There is no symmetry along z axis.

Essentially it means that the deformation respect to XY plane is 0, meaning that deformation along X1 is not possible.

With t_xy = t_yx

This is not enough, in real composite materials T_xz, T_yz != 0, it must be approximately 0. Therefore, in order to understand behavior when T_xz, T_yz ≠ 0 it must be tested.

In composite material testing, symmetry is very first criteria to mention. We have 6 strain deformations, and we have to understand the relations between σ₁ and σ₂.

• σ₁
• σ₁
• σ₂
• σ₃
• σ₄
• σ₅
• σ₆

According to results of the previous mechanics material formula, we can see that the XZ, YZ are not doing in any demonstration to follow the relation from solution. Each moment was treated separately.

r₁ = q₁₁σ₁ + q₁₂σ₂ + q₁₃σ₃ + q₁₄σ₄ + q₁₅σ₅ + q₁₆σ₆

σ_xz = 0 ; σ_yz = 0

We have to impose in order to check how to obtain a symmetry respect to XY plane, meaning θ₁ = 0 and trying to understand how many coefficient can be affected. So in order to have real symmetric composite material:

t_yz = 0 : q₁₁ = 0 ; q₁₂ = 0 ; q₁₃ = 0 ; q₁₄ = 0

t_xz = 0 : σ₁₁ + σ₅σ₇ + σ₁₃t₃ = 0

q₉ = 0 ; q₈ = 0 ; q₇ = 0

We will have restrictions, typically composite must have same symmetry plane.

This means that ₃₆ - ₂₁ = ₁₅ coefficient is different from ₂₂ for 1 material with one symmetry plane.

The same can be done for the shear stress.

FOR Y2 2L_Ry = 1

FOR X2 R = A_y² = M_x or known now

FOR X1 2N_Rx = 1

The lamina assumes a new behavior as it becomes transversely isotropic.

TSIA-HILL CRITERION (1968)

If σ₁ = 0, τ₁₂ = 0, τ_xy = 0 σ_x τ_x = G◦ + σ_xy = 1/2U/ε◦ + 1

This is the HILL CRITERION of the lamina

If the lamina is transversely isotropic, it means that if we consider the lamina where F = the direction of failure and had 0 and A_y⁴ , that for the lamina is correct, for each stress case or periodic case, (other one) the same #TRANSVERSELY ISOTROPIC#

Tsai had demonstrated that:

σ_y² = σ₂ = σ_cy τ₃² = σ_xyc3 τ_x

R_y A_y³ R_xy² = 1

HOFFMAN CRITERION

Which are the ways in which a lamina can break?

Hoffman is a criteria that considers no symmetric material so the lamina inside the laminate may break in different ways – Tension and Compression stress.

RUPTURE IN TENSION (σ , F_t)

If we put a lamina in tension, it may break not the fibers but the matrix can, is important to understand that if the matrix breaks, in an attempt parallel agrees environment inside the material. Obviously, we will not be able to see macroscopic cracks but from microscopical point of view yes.

A huge problem can be when the fiber breaks it will be able to find a solution for propagation crack investing material but not for fibers.

1. Load are ⊥ to middle plane
2. Thickness is constant
3. Small deflection

So p_z = φ_x

φ_x = 0 (partial derivative)

∂w = ∂w_∂x

So M_x = z _∂w/_∂x

In the other plane,

V_x = -∂w/_∂y

DEFORMATION

Already talked about plane stress for an isotropic material. Plane stress theory means neglect of deformation which aris perpendicular in the plane. Already done calculation for beam, so have recognized two equations that linearize the equation.

From basic courses of structural mechanics we have that

ε_x = ε_y = ε_xy = ε.

ε_x = ∂w/_∂x - ∂²w/∂²_z

ε_y = -∂v/_dy - ∂²w/∂²_z

γ_xy = -2 ∂³w/∂²y ∂x + (-2 ∂²w/_∂y dx)

These two derivatives are continuous (Strokes Theorem).

So: -2∇²w/_∂x∂y

So the result that we had for M_x

M_x = [ (E/h₂) (∂²w / ∂x²) + (E/12) d²w ]

But we don't have one only D (lines) but we have to talk about D_x and D_y

M_x = [ D_x ∂²w / ∂x² + D' ∂²w / ∂y² ]

and should remember that M_x for isotropic material was equal to:

M_x = [ -D_x ∂²w / ∂x² + ∂²w ]

M_x = - [ D_x ∂²w + D' ∂²w ]

M_xy = -2 D_xy ∂²w / ∂x ∂y

Some of isotropic plates

Equilibrium is the same of isotropic plates but the results will be different

D_x ∂⁴w / ∂x⁴ + 2 (2 D_xy + D') ∂⁴w / ∂x² ∂y² + D_y ∂⁴w / ∂y⁴ = p

So solving this equation, we have the stress state in the lamina of the equation it is not so simple to solve it just FEM

The step of the escalator there is an improvement & we have different stiffness in x and y direction the problem is the derivation of the stiffness

So D₁ ≥ D₂ D_2y D_x D"

So what can we do?

can't contribute to the stiffness on x direction no ε

y_x = the truss loops like a reinforcement

Considering a piece of plate