Temi d'esame di Tecnica delle Costruzioni 2

5/07/12

q_t

l_a = 2l_s = 2,86 mm

q_w = 5000 mm

E = 30 GPa = 30⋅10³ N/mm²

EJ₁ = 6,39⋅10³ Nmm²

EJ₂ = 3,195⋅10³ Nmm²

1) AZIONE ASSIALE

• 3/8 qℓ
• 3/8 qℓ⋅2 -

2) Ψ = 1

3) Θ = 1

TELIO A NODI SPOSTABILI

4) MATRICI RIGIDEZZE

• 4EJ₁/l₁ ⋅ Gι_l1 + 3EJ₂/l₂ ⋅ Gι_b - 3EJ₂/l₂² ⋅ Gι_l2
• - 3EJ_l2² ⋅ Gι_b
• 3EJ₁/l₂² ⋅ Gι_vb

l_a = √(3qℓ/4EJ₁) = √(3/I_l) ⋅ √(qℓ/E)

l_b = √(3qℓ/3EJ_l2) = √(3/l_l1) ⋅ √(q⁴E)

√(q²) = l_la

k_ab = l_a ⋅ k_b = l_b ⋅ k_a

k_a = l_a ⋅ k_b

considero inoltre l_b

• 1 ≤ β ≤ 2 → β = -1,5
• 1/β = 0,667 → Gι_b = 0,95
• Gι_vb = 0,81946

det = 4,23⋅10⁴

β = 1,7

_αb / _π = ₁ / β = 0,58823 → G'b = q͠b4₁₂₅

Gvb = -0,1013

det = -1,0643 ⋅ 10¹⁴

β = 2

_αb / _π = ₁ / β = ₁ / 2 → 95

G'b = 98225

Gvb = 0

det = -6,84 ⋅ 10¹³ = -0,837 ⋅ 10¹³ ≠ 0

per β = 2

_αb / _π = ₁ / β → x_b = _π / (b₂β) → √(3₉₀₀) / (0,8 ⋅ ET₁) = _π / (b₂β)

q^b_int = (π² EI₁ 8) / (3ℓ (b₂ β)²) = 4266,56 kN/μ

q^a_int = (π² EI₃ Δ) / (3ℓ (b₄ β)²) = 4261,56 kN/μ

_αla / _π = ₁ / β_a = 4,91 * ₁ / β_b = ₁ / β_a = 4,91 ⋅ 9,5

β_a = ₁ / (4,91 ⋅ 0,5) = 4,047

1)

φ = 1

3E₁ 1 H₂ - G_b

3E₁φ3E₁G₂

3E₁G₁

φ

3E₃GaR

3E_1e

3E₁ G₄

2)

CARICO ESTERNO

R₂²

qe

qe²₂

qe φ qe²₂ φ qe²₂ qe φφ qe²₂ φ qe

(3E₁G_b + 4E₁Ga) φ qe²₂ q_e² = 0

B₁ 5,56x10⁹ q + 9,27x10⁹+ (3,62)(4,523830)Ga φ = 101,1680654

α₁ β₁² 7

3E₁ q⁸3 3E qe₂

- 0,588 φ G_a = 0,7466875

α₁ H⁷ 3E qe¹ G_b = 0,937

G_b = 0,9562275

φ = 101,6666 φ= -0,063q

H)

H₁ = 0

H_1c = 4E_1qe φ - qe²₁₂ = -821106933,3 = -821,6t kN ml

N_2c = 3E₁: G₆0 c = 216,9+89 3E₁ K N ml = 214, 9503 kN ml

N_3cH = - 3E₁ G_a = - 205,215509 kN ml = -205,5 kN ml

qe²₂

1250 — 214, 91 = 214 + 821 + 205 - 1250

1261 8333 1250 ≤ 0

H₃ = 2E₁ϕ G = - 95,4275

q₁ q₂₈

3 x 12, 5 kN ml

β = 1,3

L_ph = 1 ÷ 9 ÷ 4

└──────┘

4,3

L_al = 0,8165 ÷ 0,9 ÷ 963 => G_ra = 0,60285

└──────┘

σ

det = 0,203 · 10¹¹

β = 1,28

L_ph = 0,78

└─────┘

σ

Gb = 0,4818

Gvlb = 0,8053

Gvlo = 0,3848

L_ph = 1,64

└──────┘

σ

G_ra = 0,6551

det = 0,885 · 10¹¹

β = 1,28

L_ph = 0,745

└─────┘

σ

Gub = 0,7848

Grb = 0,8867

Gvb = 0,402625

L_al = 2,633

└──────┘

σ

Gvan = 0,60285

det = 0,107 · 10¹¹

4,28 < β < 1,28

β = 1,285

L_ph = 0,748

└─────┘

σ

L_al = 0,633

└────┘

σ

Gub = 0,7823

Gvlib = 0,8860

Gvib = 0,33872685

Gvla = 0,628375

det = 0,00506 · 10¹⁵

abete

Teoria d'esame 02/03/15

e = 6000 mml₁ = 3000 mml₂ = 5000 mml₃ = 4000 mmI₁ = 616·10^-6I₂ = 1,361·10^-6q = 60 kN/m² = 60 N/mmE = 210 GPa = 210·10³ N/mm²EI₁ = 1,2956·10²EI₂ = 4,074·10²

1. Analisi Cinematica
2. Scheda Risolutivo (Fig) qℓ² / 18
3. Ipotesi Azione Assiale
4. φ = 1
5. φ₀ = 1
6. Carico Appicato
7. Sistema Risolvente

(3EI₁ / e + 4EI₃ / l₃) φ - 6EI₁ / l₃ Gvi_a qℓ / 8 + qℓ² / 18 = 0

- 6EI₁ / l₃ Gvi_a φ + (3EI₁ / l₁ Gv_c + 3EI₁ / l₂ Gv_b + 12EI₃ / l₃ Gv_a) φ₀ = 0

1) SISTEMA RISOLVENTE

(3ε₂ + 3ε₂ G_p + 4ε₂₁ (G_ia) φ = 3ε₂ G_p y + qε² = qε² / 32

- 2ε₁ / ε₂² φ + 3ε₂ G_p y + qε² = 0

- (1,561·10⁹ y + 2,16·10⁶ G_p y + 1,624·10⁶ (G_ia) φ - 272300 G_p φ = - 2,184·10⁷

- 2ε·320/G_p + (30,16⁻¹ G_p −4,2)φ = 0

_dy_p / ^dy₁ = qε² / 32 = 0,155 → G_p = 0,9840

G_p = 0,98475

α_p H / π √ qε / 8ε₁ = 0,22 → Cia = 0,9880

Cvia = 0,9980

Gia = 1,0081

φ_p = 0,00° + 15 rad

I / l = 23,316 mm

H)

18,15

14,73

0,5

41,34

5,83

0,27

Y)

21,6

7,2

18,21

21,6

10,8

21,6

42,3

5,34

24,86

tewa d'esame 06/09/16

e₁ = 6000 mmh₁ = 3000 mmh_a2 = 3000 mmh₃ = 4000 mmI₁ = 6,16·10⁶ mm⁴I₂ = 1,36·10⁶ mm⁴E = 210 GPa = 210·10³ N/mm²q = 30 N/mm

{Disegno}

EI₁ = 1,2936·10¹² Nmm²EI₂ = 4,074·10¹² Nmm²

1. ANALISI CINEMATICA
2. SCHEMA RISOLUTIVO

TELAIO A NODI SPOSTABILI (SIMMETRICA)

3. (IPOTESI AZIONE ASSIALE)
4. φ = 1
5. θ₂=d

[ _{3EI1 Gc c} + _{6EI1 G2} ][ ^{3EI₂ h₂ G₂ b} ]

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