Appunti del corso Chassis and Body Design and Manufacturing

Name: Appunti del corso Chassis and Body Design and Manufacturing
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Appunti personali del corso Chassis and Body Design and Manufacturing di Massimiliano De Agostinis per studenti di ingegneria meccanica LM e del MUNER. Gli appunti sono stati rielaborati dalle …

Esame Chassis and body design and manufacturing

Facoltà Ingegneria

Dal corso del Prof. De Agostinis Massimiliano

Università Università degli Studi di Bologna

A.A. 2022-2023

108 pagine

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Estratto del documento

Statically indeterminate structures and frames

The number of unknown reactions is bigger than the number of equilibrium equations.

Force method

Consider the structure to determine, one by suppressing constraints by replacing them with a generalized force and by imposing compatibility on displacements.

X_A = 0
X_B, Y_B, F = 0
M_A = -F⋅a - Y_B⋅(a+b) = 0
v_a = 0

Deformation method

Castigliano's theorem, the deformation configuration among compatible configurations. For each beam element:

ΔG = Σn ^{cosα_n + _mn⋅sinα_n}
X = L_N - L_N,0 (Small angle assumptions)
Σ X_N + Σ Y_N = 0
Horizontal = P⋅sinβ = 0

Statically indeterminate structures and trusses

The number of unknown reactions is bigger than the number of equilibrium equations.

Force method

Consider the structure is determined once by suppressing constraints by replacing them with a generalized force and by imposing compatibility on displacements.

X_A = 0
Y_A - F = 0
M_A = F · a - Y_B (a + b) = 0
V_A = 0

Deformation method

Castigliano: the deflection configuration energy contributes a configuration for each. Hence, assume:

A_x = Σ P_x a_x = 0, s_in a ⁱ
AD_x = Σ f_i cos (a_i) = 0, T= M_t s_in ^bi mg_x_i
Σ x_i = 0, Σ y_i = 0
Σ M_i = 0, V_A = 0
Σ F = 0, P = s_in f ^bAF

M₀FBCDEl/2Fl/2FxM₀l/2ABxV_dEquilibriumV_A + V_C = F = 0V_A = V_C = F/2

χ_E = 0completelycompatibility + constitutive law

Moment is positive if bends the fiber below

M_Bq(x η³) = 0

le se estendo a uno superposizione di effetti:g(ξ) = 0 perche ξ non γ ≠ [γ hide space] 1η^′0xξ^′0γε x = λ = σ + b - cχ = Cδε

ν(δε) = δε αντιθετο ν(δε)δε μ(|δε|) = E (|δε|)|δε| = |δε|δε λδεi - aδε ν(δε)γnot connected ax x = k = 0x ≠ γ α RTS β (δε δε) δε

Via Fratelli Lumiere, 46/A - 48124 Ravenna

Via Giovan Battista Bonucoli - 47121 Forli

x_ACxEIO = 0

X = F/2V(x₁) = C₁V(x) = C₁ x + C₂

V(l)V\(\tilde{x}\)(x₀) = \( \frac{F(l - x)^{2}}{3EI} \) = \( \frac{F l^{2}}{6EI} \)

Reazione a sxReazioniC₁α da angoli diversi

q(0, x₀) = \( \frac{F l^{2}}{6} \)

V\(\tilde{x}\)(x₀) = \( \frac{F x^{2}}{3EI} \)

V\(\tilde{x}\) = \( \frac{F l}{6EI} x \)

V\(\tilde{x}\)(l) = \( \frac{F l^{3}}{6EI} \)

V(x₀) = \( \frac{5}{8} \frac{F l^{2}}{48 EI} \)da cm

U_C = \(\frac{1}{0.6EI}\)\( \frac{5 F}{0.6EI} \)

Y₀ = \(\frac{5F}{16}\)FRAMES

Frames

23/09 In frames, pieces automatically freeze one inside another. They are designed to stand structures. By converting each continuous node into a pinned node (can move), we define a multiply stationary structure.

Rotating node frame

The primary structure is stable.

(1) For extreme forces. The structure is stable.
3 beams 9 D.O.F.
2 extreme forces 6 D.O.F.
1 internal force between 3 elements = 6 + 6 D.O.F.
(2 - m) = 4 ^d.o.f
(0 - m) = 4 _d.o.f

To solve, we lift the moments once in the time method, applying them to the beams: To ensure equilibrium in models

H _CB + H _AC + M _CD - Y _C = 0
X _CA - P _C = P _CC = 0
M _CC = 0

ine rotations if, we can assume that i.e. is positive, be negative some bonus from rotations:

coordinates system by imposing ∑Me+∑Mb+∑Mc=0 ∑Fe=∑Fa ∑Fb+∑Ca ∑Bc=0 we obtain the reaction mome

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