Appunti esercitazione di Statica e scienza della costruzioni

TEMA D'ESAME 13/11/2017

0. IDENTIFICAZIONE PARTI

1. ASTA DC (2pb)
2. ASTA DE (pb)
3. PANNELLO OCBE (p) (pieno)
4. FORO SEMICERCHIO (-p)
5. FORO 1/4 CERCHIO (-p)

1. CALCOLO BARICENTRI E MASSE COMPONENTI

1) ASTA DC

M₁ = 2pb √(b²+(2b)²) = 2√5 pb²

x_G1 = 0

y_G1 = -b

z_G1 = b/2

2) ASTA DE

M₂ = pb (2b) pi/4 = pb² pi/2

x_G2 = 2b/pi

y_G2 = 0

z_G2 = 2b/pi

3 PANNEUO CBO (r)

M₃ = r · 2b · b = 2rb² x_G3 = b/2 y_G3 = -b z_G3 = 0

4 FORO SEMICERCHIO (-r)

M₄ = -r π (b/2)² · 1/2 = -r π b²/8 x_G4 = 4/3 · b/2 · 1/π = 2/3 · b/π y_G4 = -b/2 z_G4 = 0

5 FORO 1/4 CERCHIO (-r)

M₅ = -r π (b/2)² · 1/4 = -r π b²/16 x_G5 = b - 4/3 · (b/2) · 1/π = b - 2/3 · b/π y_G5 = -2b + 4/3 · (b/2) · 1/π = -2b + 2/3 · b/π z_G5 = 0

PARTE    MASSA    x_Gi    y_Gi    z_Gi

1    2√5 rb²    0    -b    b/2 2    rb² π/2    2b/π    0    2b/π 3    2rb²    b/2    -b    0 4    -rπ b²/8    2/3 · b/π    -b/2    0 5    -rπ b²/16    b - 2/3 · b/π    -2b + 2/3 · b/π    0

2. CALCOLO MASSA TOTALE E MOMENTI STATICI TOTALI

M_tot = M₁ + M₂ + M₃ + M₄ + M₅ = 2√5 ρb²·b²·π/2 + 2ρb² - ρπb²/8 - ρπb²/16 = ρb²(2√5 · π/2 + 2 · π/8 - π/16) =

= ρb²(2√5 · 2 · 5/16 π)

S_{x tot} = Σ M_ix_ai = ρb² π/2 · (2b/π) + 2ρb² (b/2) - ρπb²/8 (2/3 b/π) - ρπb²/16 (b · 2/3 b/π) = ρb³(1 + 1 - 1/12 · π/16 · 1/24) =

= ρb³(67/24 · π/36)

S_{y tot} = Σ M_iy_ai = 2√5 ρb² (-b) + 2ρb² (-b) - ρπb²/8 (-b/2) - ρπb²/16 (-2b - 2/3 b) = ρb³(-2√5 - 2 · π/36 + π/8 - π/24) =

= ρb³(-2√5 - 49/24 · 3π/36)

S_{z tot} = Σ M_iz_ai = 2√5 ρb² b/2 + ρb² π/2 · 2b/π = ρb³(√5 + 1)

3. CALCOLO BARICENTRO

x_{G tot} = S_{x tot} / M_tot = ρb³(67/24 · π/36) / ρb²(2√5 · 2 · 5/16 π) = b (67/24 · π/36) / (2√5 · 2 · 5/16 π)

y_{G tot} = S_{y tot} / M_tot = ρb³(-2√5 - 49/24 · 3π/36) / ρb²(2√5 · 2 · 5/16 π)

z_{G tot} = S_{z tot} / M_tot = ρb³(√5 + 1) / ρb²(2√5 · 2 · 5/16 π)

Tema d'esame 04/02/2021

CBA (asta continua) : p

BEHG : 2p

0. IDENTIFICAZIONE PARTI

1. AB
2. BC
3. BE
4. EH
5. HG

1. CALCOLO BARICENTRI E MASSE COMPONENTI

1 ASTA AB

M₁ : p√2 b

x_G1 = 0

y_G1 = b/2

z_G1 = b/2

2 ASTA CB

M₂ = p b

x_G2 = 0

y_G2 = 0

z_G2 = (b + b/2) = 3b/2

3 ASTA BE

M₃ = 2pb

x_G3 = 0

y_G3 = -b/2

z_G3 = b

4 ASTA EH

M₄ = 2pb

x_G4 = b/2

y_G4 = -b

z_G4 = b

5 ASTA HG

M₅ = 2p pi/2 b = pbx

x_G5 = b + 2b/pi

y_G5 = -2b/pi

z_G5 = b

PARTEI MASSA x_Gi y_Gi z_Gi

1 pV2b 0 b/2 b/2

2 pb 0 0 3b/2

3 2pb 0 - b/2 b

4 2pb b/2 -b b

5 pbx b + 2b/pi -2b/pi b

2. CALCOLO MASSA TOTALE E MOMENTI STATICI TOTALI

M_TOT = M₁ + M₂ + M₃ + M₄ + M₅ = rho sqrt2 b + rho b + 2 rho b + 2 rho b + rho b pi = rho b (sqrt2 + 5 + pi)

S_xTOT = sum M_ix_ai = 2 rho b ^b/₂ + rho b pi (b - ^2b/_pi) = rho b² (3 + pi)

S_yTOT = sum M_iy_ai = rho sqrt2 b (b/2) + 2 rho b (-b/2) + 2 rho b (-b) + rho b pi (-2b/pi) = rho b² (sqrt2/2 - 5)

S_zTOT = sum M_iz_ai = rho sqrt2 b · b/2 + rho b · 3b/2 + 2 rho b · b + 2 rho b · b + rho b pi · b = rho b² (sqrt2/2 + 3/2 + 2 + 2 + pi) = rho b² (sqrt2/2 + 11/2 + pi)

3. CALCOLO BARICENTRO

x_{G TOT} = S_xTOT / M_TOT = b (3 + pi) / (sqrt2 + 5 + pi)

y_{G TOT} = S_yTOT / M_TOT = b (sqrt2/2 - 5) / (sqrt2 + 5 + pi)

z_{G TOT} = S_zTOT / M_TOT = b (sqrt2/2 + pi + 11/2) / (sqrt2 + 5 + pi)

TEMA D'ESAME 16/01/2025

0. IDENTIFICAZIONE PARTI

1. Pannello ABCD
2. Foro
3. Asta DE

1. CALCOLO BARICENTRI E MASSE COMPONENTI

1 Pannello quadrato

M1 = rho b²

xG1 = - b/2

yG1 = - b

zG1 = - b/2

2 Foro

M2 = - rho pi (b/4)² = - rho pi b²/16

xG2 = - b/2

yG2 = - b

zG2 = - b/2

3) ASTA DE

M₃ = rho b sqrt(2) . sqrt(b² . b²) = 2 rho b²

x_G3 = - b/2

y_G3 = - b/2

z_G3 = 0

PARTE 1 MASSA x_Gi y_Gi z_Gi

1 rho b² - b/2 - b - b/2

2 - rho b² pi/16 - b/2 - b - b/2

3 2 rho b² - b/2 - b/2 0

2. CALCOLO MASSA TOTALE E MOMENTI STATICI TOTALI

M_tot = M₁ + M₂ + M₃ = rho b² - rho b² pi/16 + 2 rho b² = rho b² (3 - pi/16)

S_{x tot} = Sum M_i x_Gi = rho b² (-b/2) - rho b² pi/16 (-b/2) + 2 rho b² (b/2) = rho b³ (-3/2 - pi/32)

S_{y tot} = Sum M_i y_Gi = rho b³ (-b) - rho b² pi/16 (-b) + 2 rho b² (-b/2) = rho b³ (-1 - pi/16 - 1) = rho b³ (-2 - pi/16)

S_{z tot} = Sum M_i z_Gi = rho b² (-b/2) - rho b² pi/16 (-b/2) = rho b³ (-1/2 - pi/32)

3. CALCOLO BARICENTRO

x_{G tot} = S_{x tot} / M_tot = - b/2

y_{G tot} = S_{y tot} / M_tot = b (-2 - pi/16) / (3 - pi/16)

z_{G tot} = S_{z tot} / M_tot = b (-1/2 - pi/32) / (3 - pi/16)

TEMA D'ESAME 4/7/2024

pb

PARTE

PANNELLO

MASSA

2pb²

x_Gi

-b

y_Gi

-b/2

z_Gi

0

ASTA FE

pb²

-2b

0

b/2

ASTA AF

pb²p/2

-2b

-2b/p

2b/p

M_tot = 2pb² + pb² + pb²p/2 = (3 + p/2)pb²

S_x = 2pb²(-b) + pb²(-2b) + pb²p/2(-2b) = pb³(-4-p)

S_y = 2pb²(-b/2) + pb²p/2(-2b/p) = -2pb³

S_z = pb²b/2 + pb²p/2(2b/p) = 3/2 pb³

x_G = S_x / M_tot = (-4-p)/(3+p/2) b

y_G = S_y / M = -2/(3+p/2) b

z_G = S_z / M_tot = (3/2)/(3+p/2) b