Temi d'esame Scienza delle costruzioni

03-09-18

4 ASTE = 12 grl13 gdl

ASTA 4 = Biella cinameticamente equivalente ad un cavallo interno

ASTA 1, 2, 5 formano un anello chiuso isostatico

ASTA 3 e il CR1 formano un anello chiuso isostatico

→ CR1 e CR2 in quanto si risvolo: non hanno C.L.R comune

CR1 + CR2 = CR costantemente simcolato a Terra in B e in E.

Σ₁ = 0 → -V_B + V_B - 2qb = 0

V_B = 0

ΣX = 0 → -H_B - V_BC - V_BC + H_C + H_E = 0

H_C = -qb

ΣX = 0 → H_E = -qb

ΣH_A = 0 → H_B = 0

ΣH_E = 0 → V_B + 10^-6 + 2qb² + ½ B^-6 = 0

V_B = 2qb

ΣY = 0 → -2qb + V_E + 2qb = 0

V_E = 0

ΣH_A = 0 → V_C = 0

ΣM_A = 0 → qb² - N_FG b = 0

N_FG = -qb

A_BA: N = T₅ = M = 0

C₃ =: N = -qb , T = h = 0

FAN = ^√2⁄&sub2;b

T = ^√7⁄&sub2;b²

M = ^{½ ex + qb2}

H(0) = ^½b²

H(2b) = 0

BEN = 0

T = 2x²p

H = ^½x₂ - 2p₂b

H(0) = 0

H(b) = -^½3 b²

ECN = -qb

T = -^1₂b

H = ^3₁₂xb

H(0) = ^½x + ^{3₂ ⁄ 7₁ b2}

C² 1b

H(2b) = ^½12

GD

N = -qb

T = -^7₁x

H(0) = 0H(b) = ^½12

CBN = -qb

T -^2₁b

H = ¹1²b

C² = 0H(0) = -^½b²

H(b) = ^½₁b

ΣX = 0 ➔ H_B = 0

3

ΣY = 0 ➔ V₁ - 2φb = 0 ➔ V₁ = 2φb

ΣM_E = 0 ➔ -2qb² - ¹/₂qb² + H_Cb = 0

H_F = -₃φb

ΣX = 0 ➔ H_F + H_B = 0

H_C = ⁵/₂ + 1 ➔ H_T = ⁷/₂

ΣX_A = 0 ➔ -H_F + H_F - qb - H_B + H_C = 0

+ H_F = (⁵/₂ - 1 + ³/₂)φ

H_F = 0

ΣM_B=0 ➔ -H_b + V_B . ¹/₂ - 2 ¹/₂b - H_F2b - V_F2b = 0

- ₁/²V_B² + ⁵/₂V_F2b

2V_F = -qb V_F = -¹/₂qb

Π^F₁ ➔ V_F– V₁ + V_B =0 ➔ -¹/₂b + 2b + V_B =0

V_B = ³/₂qb

N = 0 T = ¹/₂γ x

H = ¹/₂qx² - ¹/₂b x

H(0) = 0

H(2b) = 0

N = -³/₂qb

T = ⁵/₂px

H₅ = ⁵/₂b

M(0)=0 M(b) = ⁵/₂

N = -³/₂b

T = ³/₂b

H = 2qb x - ¹/₂b

H(0) = -1

M(b) = 1

N = -³/₂qb T = ⁵/₂pb

H₅ = ⁵/₂b

M(0) = 0 M(b) = ⁵/₂

H₅ = ⁵/₂b

H(0) = 0

H(2b) = 0

N = ³/₂qb

T = ⁵/₂pb

M(0)=0 N(b) = ³/₂

M = ³/₂b - ³/₂ b x

ΣH_G = 0

ΣH_E = 0

ΣH_B = 0 ⟹ V_cb + H_cb = 0 ⟹ H_c = ¹/_b

ΣH_D = 0 ⟹ 1 - V_cb = 0 ⟹ V_c = ¹/_b

ΣΨ = 0 ⟹ V_B = ¹/_b

ΣX = 0 ⟹ H_B = ¹/_b

G~E N*

E~C

C~D N* = ¹/_b

T* = ¹/_b

M* = - ^x/_b

M*(0) = 0

M*(b) = 1

E~A T* = ¹/_b

N* = ^-1/_b

M* = - ^x/_b

M*(0) = 0

M*(b) = -1

F~A N* = ^-1/_b

T* = ¹/_b

M* = - ^x/_b

M*(0) = 0

M*(b) = 1

Molla:

d_l = √(Xb + √b)

d_t = cedimento

d_ca = 1/EI ∫₀^b (-2 + X_b/b (½πx - ½)² b + 2 X (- ½X_b + X/b)) dx =

= - (3x² b + x³/4X - 2X X_b + 3/2 πx X_b - 2X x + X x²/(3b²))

= 1/(EI) [3/2 b + ∫ (x³ + ⅓ X_b) - 2 X x + X x³/b - 3/4 X_b^1/3 + ⅓ X_b]

= 1/EI [∫ (3/2 x ² + 3x) x b⁶ + (7/3) X_b ]

= 1/EI [ 17/4 qb ³ + 7/3 X_b ]

Ł_ba = 1/EI ∫₀^b (X/(b) ( - X_b/b)) dx = 1/EI ∫₀^b x²/b dx = 1/EI ⋅ 1/3 X_b

17/4qb³ + 8/3 X_b = -1/4 X₀ + 2qh³

(17/4 + 1/2) qb³ = -1/4 X⁰

25/41 qb³ = 25/64 X_b

- 15/7 b³ = X

DC

• N^* = 0
• T^* = -1/b
• H^* = -x/b

M(0) = 0

M(b) = -1

AC

• N^* = -1/b
• T^* = 0
• M^* = -1

BA

• N^* = 0
• T^* = 1/b
• H^* = x/b

M(0) = 0

M(b) = -1

ΣX = 0 → H_B^* = 0

ΣΦ = 0 → V_B^* = 0

1 ΣX = 0 → H_B = 0

Tot ΣX = 0 → H_F^* = 0

ΣH = 0 → V_D^* = 0

3

ΣΦ = 0 → V_E^* = 0

2

Tot ΣH_A = 0 → V_C = 0

Tot ΣΨ = 0 → V_A^* = 0

BA^* N^* = T^* = M^* = 0

N^* = -1

T = M = 0

N^* = 0

T^* = -1

M^* = -1 x

M(0) = 0

M(b) = -b

N^* = 1

T = 0

H = -b

N = 0

T = 1

M = x - b

M(0) = -b

M(b) = 0

y''₁(x₁) = -¹/_EI [-³/₂ qbx₁ · X₁ + X_x1²/₂ + x₁] + A

y'₁(x₁) = -¹/_EI [-³/₄ qx₁² X_x1 + 1Xx₁²/₂ + x₁] + A

y₁(x₁) = -¹/_EI [¹/₄ qbx₁³ -^{1/₂ Xx₁2 + Xx₁3/_6b] + Ax₁ + B}

y''₂(x₂) = -¹/_EI [- qbx -¹/₂ qx₂²]

y'₂(x₂) = -¹/_EI [-¹/₂ qbx₂ -¹/₆ qx₂³] + C

y₂(x₂) = -¹/_EI [-¹/₆ qbx₂ -¹/₂₄ qx₂⁴] + Cx₂ + D

y''₃(x₃) = -¹/_EI [qbx - qb²/₃]

y'₃(x₃) = -¹/_EI [¹/₂ b x₃² - qbx₃] + G

y₃(x₃) = -¹/_EI [-¹/₆ qbx₃ + ¹/₂ qb x₃²/₃] + Gx₃ + H

Condizioni al Contorno

1. y(x₁ = 0) = 0
2. y'(x₁ = 0) = 0
3. y(x₃ = 0) = 0 (Carrello)
4. y(x₃ = b) = y(x₁ = b) b
5. y(x₁ = b) = j ^{q b}/_EI ^Lb/_EI
6. y'(x₂ = b) = -¹/₆ q (Carrello)
7. y (x₁ = b) = y (x₂ = b) (Continuità)

Risolvendo il sistema si ottiene

X = ⁹/₄ l b²