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2.4.1.G

Ähnliche Aufgabe

2.4.1.D

Gegeben

Symbole:

\[b, h\]

Größen:

\[\begin{split}b &= 5 \, \mathrm{cm} \\ h &= 2 \, \mathrm{cm}\end{split}\]

Zeigen Sie, dass:

\[\begin{split}I_{yy} &\stackrel{\scriptscriptstyle{0{,}1}}{\approx} 1{,}1 \, \mathrm{cm}^4 \\ I_{yz} &\stackrel{\scriptscriptstyle{0{,}1}}{\approx} 1{,}4 \, \mathrm{cm}^4 \\ I_{zz} &\stackrel{\scriptscriptstyle{0{,}1}}{\approx} 6{,}9 \, \mathrm{cm}^4 \\ \\ \max I_{\bar y \bar y} &\stackrel{\scriptscriptstyle{0{,}1}}{\approx} 7{,}3 \, \mathrm{cm}^4 \\ \varphi_1 &\stackrel{\scriptscriptstyle{1{,}0}}{\approx}77^\circ\end{split}\]

Lösung

Für das Dreieck gilt:

\[\begin{split}\xi &= 5\, \mathrm{cm} \\ \xi' &= 5\, \mathrm{cm} \\ \eta &= -2\, \mathrm{cm} \\ \mathrm{sgn}(\eta) &= -1\end{split}\]

und damit:

\[\begin{split}I_{yy} &= \tfrac{\mathrm{sgn}(\eta)\,\xi \eta^{3}}{36}\\ &= \tfrac{-1 \cdot 5 \cdot (-2)^3}{36} \,\mathrm{cm}^4 \\ &= \tfrac{40}{36} \,\mathrm{cm}^4 \\ &= \tfrac{10}{9} \,\mathrm{cm}^4 \\ &= 1{,}\overline{1} \, \mathrm{cm}^4 \\ \\ I_{yz} &= \tfrac{\mathrm{sgn}(\eta)\,\xi \eta^{2} \left(\xi - 2 \xi'\right)}{72} \\ &= \tfrac{-1 \cdot 5 \cdot (-2)^2 \cdot (5 - 2\cdot 5)}{72}\,\mathrm{cm}^4 \\ &= \tfrac{100}{72}\,\mathrm{cm}^4 \\ &= \tfrac{25}{18} \,\mathrm{cm}^4 \\ &= 1{,}3\overline{8} \, \mathrm{cm}^4 \\ \\ I_{zz} &= \tfrac{\mathrm{sgn}(\eta)\,\xi \eta \left(\xi^{2} - \xi \xi' + \xi'^{2}\right)}{36} \\ &= \tfrac{-1 \cdot (-10) \cdot \left( 25 - 25 + 25 \right)}{36}\,\mathrm{cm}^4 \\ &= \tfrac{250}{36}\,\mathrm{cm}^4 \\ &= \tfrac{125}{18}\,\mathrm{cm}^4 \\ &= 6{,}9\overline{4} \, \mathrm{cm}^4 \\\end{split}\]

Mohr'scher Kreis

Tensor Transformation

Programm zur Berechnung der Komponenten des Flächenträgheitstensors

Siehe Polygon und Dreieck für Lösung ohne Integration.

SymPy

Source Code eines Programms dazu:

from sympy.physics.units import kg, m, s
from sympy import pi, S, pprint, symbols, cos, sin, Matrix, Eq, solve, sqrt, exp, simplify
from sympy import atan

# Units:
N         =  kg*m/s**2
k, M, G   =  10**3, 10**6, 10**9
kN        =  k*N
Pa        =  N/m**2
MPa, GPa  =  M*Pa, G*Pa
deg       =  pi/180
half      =  S(1)/2
cm        =  m/100
mm        =  m/1000

# Rounding:
def iso_round(x, pv):
    try:
        x = float(x)
        tmp = round(x, pv)
    except TypeError:
        x = x.evalf()
        from functools import partial
        func = partial(round, ndigits = pv)
        tmp = x.applyfunc(func)
    return tmp

a = 1*cm
prec = 1

# quiz 1:
(b, h) = (5*a, 2*a)
sign = +1

(b2, h2) = (b*b, h*h)
(b3, h3) = (b*b*b, h*h*h)

pprint("\n(Iyy, Iyz, Izz) / cm⁴:")
Iyy = b*h3/36
Iyz = sign * b2*h2/72
Izz = h*b3/36

for tmp in (Iyy, Iyz, Izz):
    tmp /= a**4
    pprint(tmp)
    pprint(iso_round(tmp,prec))
    pprint("\n")

Tyy = Iyy/a**4
Tyz = Iyz/a**4
Tzz = Izz/a**4

pprint("\nTm:")
Tm = (Tyy + Tzz)/2
tmp = Tm
tmp = iso_round(tmp,prec)
pprint(tmp)

pprint("\nTt:")
Tt = (Tyy - Tzz)/2
tmp = Tt
tmp = iso_round(tmp,prec)
pprint(tmp)

pprint("\nr:")
r = sqrt(Tt**2 + Tyz**2)
tmp = r
tmp = iso_round(tmp,prec)
pprint(tmp)

pprint("\nmax Tȳȳ:")
tmp = Tm + r
tmp = iso_round(tmp,prec)
pprint(tmp)

pprint("\nmin Tȳȳ:")
minT = Tm - r
tmp = minT
tmp = iso_round(tmp,prec)
pprint(tmp)

pprint("\nφ₁ / deg:")
p1 = atan(Tyz/(Tyy - minT))
tmp = p1
tmp /= deg
tmp = iso_round(tmp,0)
pprint(tmp)

Programm ausführen? Dazu:

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