Scienza delle costruzioni

A

c.s.

B

t e

+ ( )

e

A3

scarica

Bx

K c

eff

V+

m

o

n

4

T: BK

A

A:

A = -

B = scarica

(c troviamo)

L/2

L/2

A

c.t.l.

A B

B C

f

trave

⌀ de l e ci t f

S_F

S₀

S₀

S₁

effe cto

snervato

BE: scarica

A B

e A

x(A)

a:

a:

l/2

1_eff

m:

A 2

B: e

sdsdsdsdsdsdsdsd

AJ

AB

JB

BE

KM

KC

x(a)

x(b)

m_b

Q

A

q

A B

A B C

l/2

9

1 trave

q d.l

q l

x

M(A)= - 9qe l ²/2 q B ₂ -q B ₂ = q e l

q l/2 q l/2

x

AB

(T = 0) q e (2 q B ₂) -

B e l l ² P

T(A) = 0

C x(A) 2 ² p

x(A) =

(T(B)) = P

(T(B)). e

M (B) = x(B)

(B) = 0

q (A) = q

A B

- N e

(B)2

(l : T = 0

M

(C A

(. )

- (q 2.e²)

c ₂ l ^{]( )} M ⁿ

l

M = q e² c ³

c ₃

x

q e² x

- c ³ q e³

- q e³/2 l

- q e² x 3

x q

c ³ l ^{.. l c}

F

M(B) FS - F (z)+ M_A = 0

e_A = Eε

Δ(0) : T = F λ :

M_A = - Fξ₂

M(λ) = - Fξ₂

e(λ) = 0

Δ(1) : 0 :

M(1) = - Fξ₂

0 : M(1) = 0

(x(A))= x (B) = 0

M. l - 2 = 0

M (2)

ω = - 1

Υ=1

0 T : M_A − l = 0

M. l - 2 = 0

AK

KB

BE

• x
• M₀
• 1. P+Fξ₂
• l
• R – l/2

• R – l/2
• M_X
• e_A
• M_B2
• e_M0
• M_X

- F

• 0
• 5Fξ³

M _A = - Fξ₂

M ₂ = - Fξ₂

e_A2

• 5Fξ³

M₁₂

AK :

• - Fξ₂ + F
• - (R − ξ)
• e(M(λ) )
• ξ(F(R) )
• - t + M(x)

- e

M₀ l ctext

⁢

- Fξ₂

5F E²

↘

• F
• 5F E²

5F

• M(x)
• 5 FE^-32

e = Fλ

∆_A (B)

• M₅ = F(R + E)

M_AK

• – AK
• - M_E (R)

M(x) − M(B) + 5F/16 + x

e(λ) = - 5Fξ/48

A₁₂ x (R-ξ) /2

• M(x)
• 5x F + 0 − 5
• e 5 F
• 5

9

1 volta

₂

φ_{1 2}

M(U_A): 1 - y_eL - 0

y_e = x_e

3x_e

φ₂

1^φ

null: _y^x = 0

3q_eL q_e - 2

M(2-2/φ^x) = 0

φ_M e_- A _*

M(U_A): 0

q_{e t} L

q_z 3q_e

M(B): -1

M(C): 0

M₀

M₁

M₁M₀ = 0

M₁² = 1

M₁M₀ = q_z + q₂

M₁² = 2

m₁^φ

e_U1 e_U1 + 1

X ≤ \frac{q_0^2}{q_0}4π \textgreater \frac{q_0^2}{16}

lvl N = x H_1

AB: \frac{q_0^2}{96} \text{-} \frac{q_0^2}{96}

BE: \text{-} \frac{q_0\xi}{2} \frac{q_0\xi}{2} \frac{q_0^2}{16} \nu (\xiC) = \text{-}\frac{q_0^2}{2}

ED: 2 q_0^2 + q_0^2 \frac{16}{2} q_0^2 + q_0^2 \frac{-q_0^2}{2} \frac{q_0^2}{2} \longrightarrow \nu (\xiD) = \text{-}\frac{q_0^2}{2} \nu (\xiD) = 0

Usiamo la linea elastica

Voglio trovare \phi_3

\begin{cases} \begin{array} V(\xiA) = 0 T + 0 \text{-} (\xi A) = 0 \\ \nu (\xiA) = 0 M\xi 0 = \text{-} \frac{q_0^2}{16} \end{array} \end{cases}

V(\xi) = q_4 + \frac{q_0}{2} \xi^2 + \frac{M_0\xi}{2\xi_1} + \frac{T_0\xi^3}{6\xi_1} + \frac{q_0\xi^3}{24\xi_1}

T = \frac{E_l \nu^\prime}{l}

σ = \frac{E_l \nu}{l}

ψ = \text{-} ν \textgreater \phi_1 = \text{-}\nu^"

\phi_B + \phi_B + \phi_B

\xi_1 + E_l^{-1} + \frac{\xi_1}{E}

[Blocco]

V^\prime (\xi) = \frac{q_0}{\rho_A} + \frac{q_0\xi)}{2}3 \xi_2 E_l + \frac{4\xi_3}{6\xi_1} + \frac{q_0\xi^3}{24\xi_1}

\frac{q_0\xi}{\phi_1} \text{-} \nu = \phi_1 ξ

\phi_{\sigma=1} = \frac{q_0\xi}{4\xi_1} ≠ \frac{q_0 \xi}{6\xi_1}

\phi_{\nu=C} = \frac{q_0\xi}{16\xi_1}\frac{4\xi_1}{E_l}

[Equazione]

V(\xi_B) = q_2 + \frac{M_0 \xi}{\xi_1} + \frac{T_0\xi^3}{6\xi_1} = \text{-} \frac{q_0\xi}{2} + (\text{-} q\xi^2_{16} \frac{\xi_2 q_0 \xi}{2})

\nu (\xi) = 0

V(\xi_D) = \frac{M_0 \xi}{24\xi_1} + \frac{M_0}{3ξ} T\xi\xi_1\xi = 0

H_{B_comm_i} = \frac{q_0}{4\xi}q_0 \frac{q_0^2}{32} + \frac{q_0^2}{4} + \frac{q_0^2}{2\xi}

MC = \frac{E_1l \phi}{q_0^3} \nu^\prime

\frac{H_{BE}}{2} = 1 l l σ \phi / [MC]

\xi (\xi_A) = Σ \frac{2}{\xi} q_0\xi_2 + \frac{q_0^2\xi^2}{\xi_2} = \frac{2}{\xi_9} (\frac{q_0^2}{16} \phi) + \frac{\phi_{BE}}{8} + \Phi_B

\phi_C = \frac{q_0^3}{16\xi_1} - \frac{9\phi_1}{32} ξ = \frac{-5}{8\xi^2}

ΔT > 0

ΔT = 0

ν

ν

ν

ν

x_u = v_u • t

φ_u = ν_u • q

x_ = p_

ν_ = p_' = ξ

x = φ_

O = x_e + ξ

x)

L^f.f.

n_iⁱ

T: -1

0 := η_L z₀ ΔT_F

η_L

M_L z_L

0 = H, z₀ < z

ν'(L L) = k + e_x

α(O&) ≡ 0

0 = m_χ + m_L z

η_z,L:

ηz,L

η_L =

∫

z₀

z₀

2

+ z₂α

z

==

R_i²

∫2

Ρ_auxL

+ (R > z)&

z

L

2L

3 Aste

10 g_elR

c : 1

E + x

E 1

x