Chassis and body design and manufacturing

STATICALLY INDETERMINATE STRUCTURES AND FRAMES

In static, a structure is STATICALLY INDETERMINATE if the static equilibrium equations are not sufficient for determining the constraints reaction (thus the "internal forces") and that structure has more than

n^u UNKNOWN REACTIONS n^e EQUILIBRIUM EQUATIONS.

In such a case the structure may be solved by taking into account its STIFFNESS characteristics (thus matrix deformation or numerical methods).

ANALYSIS METHODS: FORCE METHOD VS DISPLACEMENT METHOD

Structural analysis requires that the equations governing the following physical relationships be satisfied:

• EQUILIBRIUM OF FORCES AND MOMENTS
• COMPATIBILITY OF DEFORMATIONS
• CONSTITUTIVE LAW (ε = σ)

FORCE METHOD

DISPLACEMENT METHOD

PRIMARY UNKNOWN: UNKNOWN FORCES

GENERALIZED FORCES

GENERALIZED DISPLACEMENTS IN EQUILIBRIUM

COMPATIBILITY CONDITIONS

CONTINUITY CONDITIONS

EQUILIBRIUM CONDITIONS

1) FORCE METHOD

The force method converts, or converts, the indeterminate structure to a determinate one, by suppressing "sufficient" numbers of, support, a... rd force (unknown)

COMPATIBILITY ENFORCED TO FIND THE UNKNOWNS

In other words, among the o^o (n = number of unknowns) balanced configurations, the only one which also satisfies compatibility is retained.

EXAMPLE

EQUIVAL STRUCTURES

PRIMARY STRUCTURE

Remove the redundant constraint and make the structure in equilibrium

Add force or a moment that create same deformation in the original structure

NB Primary structure arbitrary must be self statically determined

1) EQUILIBRIUM CONDITION

Σx = 0

Σy = 0

ΣP = 0

2) COMPATIBILITY CONDITION

The compatibility conditions make sure point B must not move -> v = 0

Step 1: load P

ΔE = SE Σ

Reaction components X, Y can be obtained by placing unit displacements along direction of components

1. Equilibrium equation

X_A = 0

Y_A + Y_B + F = 0

(1) a · Y_A + b · Y_B = L · E · i

(2)

2) Compatibility conditionY_A = 0

X_A + 1/3 · P · E = 0 → X_A = -1/3 · P · EY_A + 1/5 · P · E = 0 → Y_A = -1/5 · P · E1/4 · P - 1/3 · P · E = 0 → P = 3/4 · P · E

So, that

Correct method for points A, B (P and a)

• Find F_B and, F_A
• Find Y_A and Y_B froma E R = F₂/E₂
• -1 YB=2E
• n_B
• F/29 2 are λ F

3) THE LOAD is OE, F Y, G E, 2 codes Properties of maximum response to P = X/_min

Derivation Method (Displacement method)

The derivation method consists of finding the only codecompatible displacement unknowns.

This method can also be called displacement method.In analysis we can locate particular essence of eachWe start from complete displacement structure and then place external works

φ_AC = 0

φ_CE = -θ_CE + ψ_CE = - F_L ^L / 2EI

ψ_CE = -θ_CE = F_L ^L / 6EI

φ_CE = F_L ^L / 3EI

θ_CE = F_L ^L / 6EI

φ_BC = -θ_BC + ψ_BC = 0

ψ_BC = θ_BC = F_L ^L / 3EI

=> θ_BC = F_L ^L / 3EI

= φ_BC

φ_BE = -θ_BE + ψ_BE =

ψ_BC = φ_BC - θ_BC

= -FL ^L / 2EI =

ψ_BE = -θ_BE

= -FL

ψ_BE = 0

φ_BE = 0

φ_BC = -FL ^L / 2EI

ψ_BC = FL ^L / 3EI

Sol. the equations compatibility 4) φ_AC

φ_AC = -θ_AC + ψ_AC =

(1) φ_AC + ψ_BC =

(3) φ_BC + φ_BE =

φ_BE = -φ_CE

Mech / Rk s

[3 EI] = -[EI][FL / 6EI]

2 EI [F]

FL₃

= L

= FL₂ ² / L_EA

= FL₂

- [FL³

/ 6EI] = EI

Discretization

A complete circle (2508 m). Have to calculate the mass p of a circumference by dividing it into a finite number of isoceles polygon.

Approximation

The approximate solution CONVERGES to the (so that e tends to 0) cost solution. It is an intermediate domain, no greater than a certain number m:

• Discretization error m
• Convergence error e

Convergence

Element

Assembly

Stiffness Matrix of a Finite Element

The approximate solution for the element (E) can be written as one of the following procedures:

1. Direct Formulation
2. Variational Formulation

1) Direct Formulation

Force E, applied to [...]. Homogeneous Poissons ratio of transformations. Geometrical transformations but need of transformations based on transformations adjustment by finite element solution

2) Variational Formulation

Truss element. Truss element is a small element under a certain area section A and Young's module E.

Example of stiffness matrix F: linear elastically medium homogeneous with a constant area variation. The axial element is perfectly to constrain changes based on E and [...].

Example of a truss element: A=1 (bindelet formation (2))

Threaded Joints

Key Dimensions

• d = nominal screw diameter
• d₁ = minimum nut thread diameter
• d₂ = mean thread diameter
• d₃ = minimum screw thread diameter
• d_c = stress diameter

H = basic triangle height

H_c = 5/8 P

P = thread pitch

150 basic thread profile according to ISO 68-1

Bolt = Screw + Nut

Primary Functions

• Bolt = holds together parts on which external forces act which tend to separate them: reciprocal (direction + external forces perpendicular to the seam) force (RT) acts to separate them (direction of force parallel to the axis of the bolt (RN))
• Bolt generates axial preload in clamping load (traction force) RT used to hold together parts upon which external forces act to separate them (RN) generates friction coefficient between parts to the targeted preload to reduce amplitude of fatigue stress.
• Optimum torque T is necessary to induce axial preload

Strength Grades

Class 8 nuts allow preloading of the screw up to its yield point (R_p0.2 = F_x) without stripping the thread out. UTS _{preloading has to be prevented into plastic range, class mismatched}

The most common strength grades for screws are reported by the international standard ISO 888-1. These classes range from 3.6 to 12.9

Screw 8.8 Nut 8

R_F stress under proof load R_m