esercizi seconda parte

Calcoli strutturali e inerzia

Ic₁ = 4_b b = 5 cm F = 20 kN

Gc = 2b - b = l_o; 84₁ = 2b 2₁ = 2b - 10 = l_o; 10A₁ = 4b 4₁ = 320 cm² 4₂ = 2b 2₂ = 2b - 10 = l_o A₂ = 2b 2₁ = 80 cm²

I_z = 16.6 x 3 = 10666.67 cm⁴

I_yx = 8.10^-3 = 666.67 cm⁴

I_yg = 10000 cm⁴

A = A₁ + A₂ = 240 cm²

I_xy = 0

I_z1 = 20.16 ³ = 6826.67 cm⁴

I_yg2 = 8₃ = ^-2 = 426.67 cm⁴

I_yg = 6400 cm 4

N_z = F = 20 kN

M_x = F.g = 20.8 = 160 kN cm

M_y = F E q = 20.10 = 200 kN cm

g_z1 = N₃ = 20000 - 2 0.83 MP

84⁸ = - E = f.88maxfy = - E - 6400 -8 8 = 2 MP

8₄8min N₂ M_x8mx MyM_y(P) = 2000008maxf_g

Analisi delle forze

1. N_z M_x 8 M_g ^I- ^z y = 0-83.33 + 25.y -20 =

y - 83.33 - 20.

²⁵

1. 4; = ; ; 2

N₂ M_x 1 - 2 M_y 4- 83,32 - 50 - 80 = 2,12 MP₈0

I_y = 0

N_z = F = 20 kN

M_x = F · q = 20 · 8 = 160 kN cm

M_y = F · y_G = 20 · 10 = 200 kN cm

σ_z =N_z = 20000 = 0,83 MPa240

σ_maxfx = M_x y_G = 16000 · 8 = 2 MPa

I_x 6400 400

σ_min =N_z M_x y M_y x = 0A I_x I_y-83,33 + 25 · y – 20 x = 0

y = -83,33 20 x = -0,98 x 4,3325Q = (4 ; -2)σ = -83,33 – 50 – 80 = 2,13 MPa

Ipotesi di calcolo

Mensola L = 4 m q = 12 cm F = 15 kN

Calcoli sezione

A_I = a (a + 3/2) = 12 * 18 = 216 cm²

A_II = a * 3 = 4 * 12 = 48 cm²

Centro di gravità

S_xI = A_I * y_{G I} = 1944 cm³

S_xII = A_II * y_{G II} = 288 cm³

S_x = S_xI + S_xII = 1944 + 288

A = A_I + A_II = 216 + 48

y_G = S_x / A

y_G = 9,86 cm

Momento d'inerzia

I_x = I_xI + I_xII = 5991.75 + 1291.18 = 4700.57 cm⁴

I_xI = Bh³/12

I_xII = a³/12 + A_II * (y_{G II} - y_G)² = 12.48³/12

I_y = I_yI + I_yII = 2582.64 + 2528 cm⁴

E = a - y_G = 12 - 9.86 = 2.14

Momento totale

M_x = F * (l - y_G)

M_x = 15000 * (12 - 9.86) = 32100 N cm = 321000 N mm

σ_t = M_x * e / I_x

σ_t = 0.89 MPa

Esempio di calcolo

Maggio 2021

Q = 20 B = 30 h = 47,34 E = 210000

HPBG₁ = (9/2 ; 59+h) (10 ; 60) G₂ = (a / 2 ; 5q + h - q/2) = (30 ; 47,66)

G₃ = a = 9 5q h = q/2

A_I = B (59 q) = 2400 cm²

A_II = 9 q = 400 cm²

A_III = q q = 200 cm²

A = 3000 cm²

y_G = (Σx_I x^I + Σx_II x^II + Σx_III x^III) / A = (2900 · 60) + (400 · 42,66) + (200 · 10) = 61,02 cm

x_G = (Σy_I y^I + Σy_II y^II + Σy_III y^III) / A = (2900 · 10) + (400 · 30,25) + (200 · 25) = 13,6 cm

G (13,6 ; 61,02)

I_x^G = (Sx_I Σx^I + Sx_II Σx^II + Sx_III Σx^III) / 8 = 288966,96 cm⁴+ (400 · (61,02 - 42,66)²) = 98369,17 cm⁴+ 200 · (61,02 - 40)² = 38874,75 cm⁴

= Ix = 357190,88 cm⁴

I_y^G = 2900 (6,6)² = 111000 cm⁴+ 1200 · 20,3 = 120951 cm⁴+ 400 · 20,3 = 121097,1 cm⁴+ 1575 cm = Iy = 159679,83 cm⁴

I_x1 = y₀ - y_G = 240928,8 cm⁴

M_x = 1000 KN cm

y₁ = y_G

5q+q = q₁ = q₂ = 61,02

I_x = 357190,88

M_x = 10001/q₁ = 58,98

I_x = 57qGx = Gq 9,5 = 127623,61 ± 718976p2e = G_x/E = 0 = 0

D_P =

1. 1,9 x 10³

Gennaio 2021

a = 10 cm   E = 260000 MPa N = 8 kN   V = 0,2

A₁ = a² = 324 cm²

Ō₁ = (0; a)^T = (0; 9) Ō₂ = (0; a/3)^T = (0; 15) Ō₃ = (a/3; a)^T = (9; 9)

A = 3/9 = 364 cm² A_T = a/3 = 324 cm²

S_y^' = (S₁y + A₁*(ṳ₁-45)² + A₂*(ṳ₂-45)² + A₃*(ṳ₃-15)²)/288

I_x^' = Ix⊂'+ A₁/₂ = 18*10³/12*324(8/25*-9-15)² + 83*0*25

S_x^' = (S₁x + A₁*(ṳ₁-45)² + A₂*(ṳ₂-45)² + A_I_xx = 74*82 cm⁴ 18*10³/12*8.748 cm⁴ 3x = 324 cm⁴/3 = 3480 =