Esercizi svolti per preparazione esame scritto - metodo 4 momenti e studio sezioni parte 2

19 dic 2017

CN Ve+c≥3m

l > 3 ⇒ 1 volta PER

F = 1/5 ql

Sconnetto a Rotazione

CN: Ve+c≥3m

l+l > 3 . 3

8<9 ⇒ 1 volta L'ABILE

Numero le incognite PER

• x, y, z, ξ

eq. congruenza

• P_BA = P_BC
• P_CB = P_CD
• EQ NODO
• EQ PIANO

EA MODO

• -x-y+9l²/2 = 0
• -x-y = -9L²/2

eq piano

T_BA-T_CD+ qL-¹/₅ R_L

A:∫ x - T_BAL =

= -T_BAL = -^x/₇

T_BA= ^x/₇

D:∫ z-T_CDL=

= -T_CDL = -^z/_-7 = T_CD = ^z/₇

^x/_L - ^z/_L = -9L+¹/₅ qL

9L - ^x/₇ = -⁴/₅θ L ⇒ x-z = -⁹/₅ qL²

∫^L/_6ES {(2x) + y} = ^L/_6ES [(2y+2)]

^L/_6ES [(2z-4y)] = ^L/_6ES [(2z)]

-x-y = -^q/₂ L²

x-z = -⁴/₅ qL²

2x + A = 2y + z

-2θ - y = 2z - A

-x-y = -¹/₂ qL²

x-z = -⁹/₅ qL²

{ 2x-2y-2z+A =

{ x = ¹⁹/₆₀ qL²

{ y = ⁴⁹/₆₀ qL²

{ z = θ qL²

A = ²⁷/₆₀ qL²

¹/₅ qL²

d_min > ³ √_{(204,16)(10⁶)(5,27a)} / (448,95)(200)

= 22,88 + 10% = 25,16 mm

≈ 25 mm

d_min = 25 mm

A_tot = 34(25)² = 2 1250 = 2,12 x 10⁹ mm²

S_xp = 448,15 (25)² = 1 75371093,8 = 1,75 x 10⁸ mm⁴

1) CALCOLO TENSIONI NORMALI ESPRESSA SNE

σz = ^N_max / _Atot + [^M_max / _Sxp] . y

corde

y(a) = (4,73)-(4,73)(e³) = 428,25 mm = 222,4⁹⁸MPa

y(b) = (3,73)-(3,73)(25) = 93,25 mm = 175,129 MPa

y(c) = [0]

= 0 MPa

y(d) = (1,22a) = (1,12)(25) = 106,75 mm = 200,169 7 MPa

y(e) = (-5,27) = (-5,27)(23) = 131,75 mm = 247, 66 6 MPa

2) CALCOLO V₂

∮ = ^N / _{Jx + ∛8 x}

S_x(b) = (b)(h)(G_b . 2) = (5a)(4,123a)²

= 21,15a³ = 3,3 x 10⁵ m³

S_x(c) = (b)(h)(G_c . 2)

(a:3,73:1)(1,8a:c) = 6,93 a³

= 102,82 = 2,55 t/ 314^8x

F = 23Hc3 = 414 a

= 2

Tratto

AB

• O
• -128q_l² / 460
• -74q_l² / 460

BA

• -64q_l² / 460
• -128q_l² / 460
• 74q_l² / 460

BC

• 64q_l² / 460
• 94q_l² / 460
• -220q_l / 460

CB

• 10q_l² / 460
• 74q_l / 460
• -220q_l / 460

CO

• 10q_l² / 460
• 220q_l / 460
• 74q_l / 460

DC

• O
• -240q_l / 460
• 74q_l / 460

q = 10 kN/ml = 5 mG_om = 200 MPa

H_MAX = -64 / 460 (10 kN/m) (5 m)² = -341.78 kN m = 34.78 x 10⁶ N mm

T_MAX = -128 / 460 (10 kN/m) (5 m) = -13.91 kN = -13.91 x 10³ N

N_MAX = -74 / 460 (10 kN/m) (5 m) = -8.04 kN = -8.04 x 10³ N

eq di congruenza

φ_BA = φ_BC

φ_CB = 0

eq xodp

eq piano

• 2_qES{[(2x) - (- ²⁄₃q¹²)] - (-³⁄₃q¹²)]} = (2)_qES[(-2y + z)] + ^A⁄_2L
• 2_qES{(-2x + 9q¹²)} = 2(-2y + z) + ^A⁄₂
• 2_qES{(-2x + 2y) + ^A⁄₂
• x + 4y = ¹⁄₂q¹²
• -y - z = 0
• -4x + 2y + ^A⁄₂ = 0
• x + 4y = ¹⁄₂q¹²
• -4 - z = 0

• x = ¹⁄₂q¹²
• y = 0
• z = 0
• A = 0

1. 2q¹²
2. V
3. 1

1. 1/5
2. 1
3. -2,5

V = 1 - 2,05

1 - 2,05

↑1₃

(+) (-) 2,5

↑φ

∝1

∝3