Esercizi Iperstatiche Metodo dei Momenti

Y + x - ¹/₂gc² - 1/100 Θ = 0

x - y - ¹/₉gc² - 6/9 Θ = 0

-z + x - ³/₂gc²

z + ³/₂gc² - z + ¹/₅qc² - Θ = 0

+¹⁹/₉Θ + 3/9 gc²

-8 + ^-1/₁₉ Θ = 0

³⁴/₁₉

x = ²⁴/₆₂gc²

Y = ²/₃₇gc²

z = -¹⁵/₃₇gc²

9^qe

2q^c2

q = 10 KN/m

C_s = 4 m

σ_adm = 120 MPa

Sezione critica CD

τ_max = 2q^c2 = 320 kN

T_max = ¼ 9^c - 99,35 KN

N = 0

A_tot = 52 a²

Y'g = Σ₁/_Atot = [2(-12a²)(-5a) + 28a²(a)]/52a² = -2.85a

x = - ²²¹/₁₈₈q^c2   y = ¹²⁹/₁₈₈q^c2   z = - ¹²³/₁₈₈q^c2   K = - ⁴/₄₉q^c2

-²⁰/₄₉qc^̅C;

-²²⁹/₉₂qc^̅A)

SEZIONE CRITICA BC6_vdo=160≥ 160

A = 1000 kNmT = 90 qc^̅ = 191,69 kNN = ²¹⁹/₃₉₆ qc^̅ = 59,91 kN

Area = 8a² + 8a² + 32a² = 48a²Y_i^̅ = ⁵ⁱ/_Dtot = 2·8a·^(6a2)/_48a² + 32e² (^a)

-2,69a- ^{17 σ0x-1}/_YoA^̅ ≤ 6σ _max

→ 1000·16 N/mm → ^1,33/₁₆₀ N/mm → α > 52 mm → + 10% → α = 52 mm

Calcolo τ̅zy

Corda a: τ̅zy = 0

Corda b:

• b̅r = 6a² → Iⁿa 50ⁿ10
• b̅r = 2a 50ⁿ10 20,89⸱10⁻³N
• τ̅zy S× 4,68⸱10⁻⁵mm³
• br 1,385552⸱10⁻⁸mm⁴
• ≈ I⸮y S×
• 2a → τ̅zy 1,50 ᴺ⸮a
• 1/6a → τ̅zy = 0,21 ᴺ⸮a

Corda c:

• A̅r 1/6a⸱a2 = br 1/6a → 8,8⸱10⁻⁵ mm³
• ≈ 20,89⸱10⁻³N
• 1,385552⸱10⁻⁸mm⁴
• br:1/6a → τ̅zy = 0,21 ᴺ⸮a

Corda d:

• A̅r = 6a²
• b̅r 2a50❍20
• τ̅zy =I⸮y S×
• 3a S×=6a²(-2a)=418⸱10⁻⁶mm³

Corda e: τ̅zy = 0

Calcolo τ̅zx

Corda a: τ̅zx = 0

Corda b:

• A̅r = a₁2
• b̅r a₂ S×
• 1a D×
• S× = A₂2⸱0 = 0
• τ̅zx = 0

1. 431,49 0 0 ✓
2. 188,86 -1,5 0 ✓
3. 188,85 -0,21 0 ✓
4. 9,07 -0,25 0 ✓
5. -18,17 0,21 ✓ ✓
6. -18,17 1,5 0 ✓
7. -431,33 0 0 ✓

σ̅₁₀ = √σ̅z² + 3 (τ̅zy² - τ̅zx²) < σᴬᴺᴺ

q=15KN/m

L=5m

σ=1600Pa

ε

V2 ≥ 3 + 5

2+ε≥3+1+1

Una volta carico

x = y + 1/2 9e^2

5x + 1 ≥ 29/

y=-1/10 9e^2 + 9/8 6e

x=11/50 9e^2

c'è un punto di nullo

Struttura 3 volte iperstatica

Applichiamo metodo delle forze

Sconnettendo a rotazione

V_E + 3 + 5 − 2 + 2 + 1/2 + 2 + 1/2 = 15

x₂ + x₃ + x₃ = 0

Per equilibrio alla rotazione del nodo

Eq. Congruenza

• φ_B = 0
• φ_BC = φ_BE
• φ_BD = φ_BD

φ_B = 0

^C/_6EI [2x_n - (−x₂)] − δv/l = 0

φ_BC = φ_BE

^C/_6EI [(−2x₂ − x_n)] + δv/l = ^C/_6EI [−(2x₃ − 0)] + δv/l

φ_BD = φ_BD

^C/_6EI [(−2x₂ − x_n)] − δv/l = −^C/_6EI [−(2x₃ − 0)] + −¹/₁₂ 9q² - 1/12q²

2x₁ + x₂ = ^6EI/_l² δv

x₁ + 2x₂ − 2x₃ = −2 ^6EI/_l²

x_n + 2x₂ − x₃ = −^6EI/_l² δv − ¹/₅ q²

x₂ + x₃ + x₄ = 0

Equilibrio Nodo B:

+ qc² + T_B' + T_B" = 0

T_BC + T_DC + ¹/₂ qc² = 0 ➔ T_BE = -^p₀/_c + ¹/₂ qc²

Eq. Congruenza

φ_BD = φ_BC ➔ ^c/_GE1 (2T_E) = 0 - ^ε/_c L/GE₁

φ_CB = φ_CD ➔ ^c/_GE2 (2T_C - T_C) = ^ε/_c (2T_C)

-T_B' + T_B" + qc² = 0

T_B + T_BC - ¹/₂ qc² = 0

¹/₃ T_BC = ¹/_G T_BC + ¹/_G (T_B - T_C)

-¹/₃ T_BC + ¹/_G T = 0 ➔ -^p₀/_c + ^1/