Esami Scienza delle costruzioni 1

ESAME GENNAIO 2014 N. 1

tensione normale?

t(m * ?

Deformazione?

Ac = 4

A₀ = πh²

A_q = πh² = 1,54π² = S⁰²

A_cot = hn - A_q = 2,63π²

* S₀₂ = - (0,641H + 1,54H²)

S₀₁ =

cos(θ)H = (0,9) (q)

(0) (0)

S_F = ∫₀^1/2 ∫_θB ^{H/2 5/2}

ρ²   cos(α

θ_B = 3 ¹ 2 ^π = H³

∫ 0 H^3/2

- (1 - 1) - 1 ^cosH3

S_F = ∫₀^3/2 (3 ¹ 1¹ ϕwϑwdoϕ

= 8^3/4) (icon)

H³ = 0 = 0

S₀₂ = [(9⨯4)

H

S_F

(0) (0)

S₂ = * S₀₂ = (.

(0,9) (1⨯3)

1 ^(0,5⨯4)

(2)

τ_cσ = S₀₁ - (0,5⨯4) = (0)

J

p^o 1,54π² (0)

(0)(0)

τ_cστ = - (0,33) H = (0,33) H

(0) (0)

J₁ = =

8h² h

8J² H 3 π

₁ (.

33 (0) (33 0) (33 (0)

.3

8

(0 h⁴ = J₁

(12 12)

J₀ = 0

J_p = 8h

₁ (5 ⨯ 3)

H¹ = ∫₀^1/9 ∫₀

ρ² cos(2α

θρ

(1 (20 + ξθ

(16) (20 8عی

= 8⨯5 (1 4) 8π =

[q

= 8π (1 (1 - q) - .034]

J₂ = ∫^2π₀ p³ senαdα = ρ₄ |^π/2₀ senα dα

<π/2∫₀ (1/2 - 1/2) = 0

J₂₂ = ∫^2π₀ (r³ senαdα) = ρ₄ H ∫ (28 - sen(28))^π/2₀

< = J₂₁ = 0,94, J₁₁ = 1,93, J₁₂ = 0,641, π/2 = 0

H = J₂₂ = 0,93H⁴

J₂₁ = 0,962 H = 0 , H⁴

J₂ = J₃ = J₁₂ = (0,94 0 H⁴)

0 0,981

J_a = J - A_aa ,

(0,41 0 H⁴ - 2,43H⁴) (0,34 0)

0,9810

ΔI (0,94, 0),

B( 0,84 0 )

C (0,89, 0)

r = 0,25

J_α = 0,64

J_β = 0,89

J_γ = 0,62

J_δ = 0,43

ϕ = 0

A ha ordnate minore rispetto a B

J_v = 1/4 = (0,84 0), 1*4 ,

0 0,04 1_ΰ4 = 229,⁰, 1/4

ε_pq = ( 0 ) H T₂₂ = ( 1 ) H, ρ = γB|| = 2 γ_wH

J_pq = T_pqT₄₅ [ ( 0 0 0 ) ( H₃/3 ( 0 0 ) H₂/2 ) ( 0 0 ) H₃( 0 0 0 ) ( H₂/3 ( 0 0 ) 0 ) ]( 0 0 ) H₂J = T_pqT₄₅ [ ( 0 0 ) 0.033 (H/2 0.85 1.47H) 0 0 ]

T₄₅ ( 0.033 0.84 1.47 )( 0.94 0.37 )T_pq = [ 0.64 0.26 ] H₃( 0.94 0.33 )

J = ∫T₂₂J₃ = (0.8 0.24 ) H₃(0.24 0.8)

J_α = J - A_ccB_cc = [ 0.8 0.26 ( H₃ - γ_sα(1.5) )( 0.62 0.12 ) = ( -0.83 -0.14 ) H₃]( 0.24 0.8 ) ( 0.10 0.12 ) ( 0.94 0.39 )

J_α^-1 = [ (0.93 0.94) 1 (0.52 0.23) 1 (0.94 0.29) ]

( 0.94 0.39 ) H₃ (0.23 0.52) 1 H₃

dt = J_α^-1Q = (0.52 0.93) ( Q √2 / 2 ) + (0.53) Q (0.23 0.52) H₃ [ √2 / 2 ] (0.50) H₃

αNO = 1

▢

T_s (0.53 0.9 ) [ Q (0.35 H - M / 2) 1 (0.53) (Q / 0.35 H - M / 2) ]( 0.53 ) H₃ (0.35) ( 0 ) H (0.53) (0 0.53H)( 0.53 ) (0.35H )

-0.9β (Q) M₂ -0.24β (Q) M₂ -0.24β M²

-Im (H) L→γ₊=0(α)-Im (H)(H→T_y = γ + 1 (α)₅

ESAME GIUGNO 2016 N. 1

A = π * 4H²

= 12,5H²

A₀ = 4H²

Area = A₁ + A₀ = 8,5H²

J₂⁴ = ∫₀^2π ∫₀¹ ρ ^{(2x + sen(θt)}

(2H * π) = 11

J_k = ∫₀ ^2π ∫₀ ¹ ...

= 41 * π = 12 - 12,5

J₀ = (1,3H 41)

(H/3 * sen 2θ) = 0

J = J₂ - J_a = C

A

Esame Aprile 2016 N.2

A = _{2π lH}/² = 6.28Hs

SF = ∫₀² 4H³ cosδ dδ = 0

S_F = ∫₀² aH³ sinω dω = aH² (cosω) | = -aH³ (1 + A) = 812³

S_F = ₈H³ / ²²A = S - (0/8)

J⁴ = ∫₀⁸ aH³ cosδ dδ = 2H₈ (20 + sen(2ω)/2) | = -12.54H³

ESAME GENNAIO 2016 N. 1

H_p = H₀

∆ = ^π/₂ (4H²-H²) = 1.5πH²; 6.₁H²

S¹ = ∫_H^2H∫₀^π ρ²cosαdαdρ - ρ³/₃⎥^π₀⎥^2H_H = 0

S² = ∫_H^2H∫₀^π ρ³senαdαdρ - ρ³/₄⎥(-cosα)⎥^π₀⎥

= (^{3H3 - H3}/₃)⎥ (1 - -)

S= ( ¹/₀ H )³ τ_g = ^S/_A = ( ¹/₀ H )³

( ¹/₀ H ) ( ^3.038/_0.982 )

J¹ = ∫_H^2H∫₀^π ρ³ cosαdαdρ = ^ρ4/₄ ⎥ ( 2α + sen(2α)⎥^π₀⎥)( ^(4H)4/₄ ) ∣ ∣π

J₀ ² = ∫₀^H∫₀^π ρ³senαdαdρ = ^ρ4/₄ ⎥ -sen2 ⎥^π₀⎥ ( ^{(4H + H⟩4}/₄ ) ( 2α - sen(2α⟩

J_c = J - A_area ( ⁵⁸⁹/₅₈₉ ⁰/₀ )^H4/sup> + 0.0 =

r_y = ( ²³⁴⁄₃₆ - ²⁄_H³ - ^5H2⁄₂ ) -

= 2.1³Q + 2.3³Q -

0.34 Q² - . . . - 2.84 Q³ + 2.08 Q³

-1m z = 0 -> r_x = 0.9

-1m z + H -> r_S = . .4 *

r_L (0.8H - H)⁄ (0.82 - 0.8H Q = .9⁴Q

∫0 x - x r_y H_S ( ^{2,8 x q m1 2,3 0m2 ) dm - HS (2,10 Q2 1,8 q2) d l2}

J_e = J_s λσ_c0.029

( 0.088 0 ) ( H₃ + H₃ ) ( 0.25 -0.5 )

( 0 0 ) ( 0.5 1 )

( 0 0 )

J_λ = J_z λ_eq0.029

( 0.25 0.8 ) ( 0.8 0.25 ) ( 0.25 -0.5 )

( 0 0.5 ) ( 0.5 0.25 ) ( 0.5 1 )

( 25 1 )

Z_eq3 ( ( 1 -1 ) ( 1 ) ) ( 1 7Ω3 -1 ) ( 1 ) (ρ = 2Ω6 | H 2.8Ω4)

( 1 ) ( 1 -1 )

J_e = 2.88HS [ ( 1 -1 ) H² ( 1 1 ) H² ( -1 1 ) H_I² ( H² )

( -1 1/3 ) ( -1 3 ) ( 1 -1 ) H_f ( 1 1 ) H_f ( 1 1 ) H₂²

( 2 ) ( -1 1/6 ) ( -1 3 ) ( 1 -1 ) ( -1 6 )

( 1 6 )

2.88HS [ ( 0.8Ω3 -Ω3 ) H² ( 0.8 -Ω2 ) H₂ 0.94 0.94 H_f

( -0Ω3 0Ω3 ) ( 0Ω3 0Ω3 ) 0.44 0.94 ) H²

( -0Ω3 0Ω3 ) ( 0Ω3 0Ω3 )]

= 2,88HS ( ( 0.82 -0Ω2 ) H_f ( 0Ω1 -0ΩΩ2 ) H₃

( 0Ω2 0Ω3 ) ( -0Ω1 0Ω1 )

J_z = 0ΩΩ₂ +ΩΩ+0Ω₃ = ( 3Ω4 -2Ω1 ) H₃ = J_λ

( -2Ω1 3Ω4 )

Δ ( 2Ω4 2Ω1 )

B ( -3Ω4 -2Ω₃ )

C ( 3Ω4 0 )

τ = 0.91

μ = 0.68

J_c = 0,6Ω6

J_t = 0,5Ω4

β₂ = 0,31

φ = ( 1 ) /2 (-0Ω) ≈ -45°