R2B2.B
Berechnen Sie:
die Verschiebung \(w_1\) von Knoten 1
die Wärmedehnung \(\epsilon\), sodass \(w_1=0\).
Lösung "Freischneiden"
Lösung "Gemeinsame Freiheitsgrade"
Source Code eines Programms dazu: Programm im nachfolgenden Frame ausführen? Dazu: Copy: Source Code in die Zwischenablage kopieren. Paste: Source Code ins Eingabefeld hinter Play: Knopf SymPy
from sympy.physics.units import *
from sympy import *
deg = pi/180
# ---
a, q, A, I, E, eT = var("a, q, A, I, E, ϵ")
EA, EI = E*A, E*I
a2 = a*a
a3 = a*a*a
pprint("Using \"FBD\":")
S, psi1, w1 = var("S, psi1, w1")
eq1 = Eq(S, - EA/a * w1 - EA * eT )
K11 = 4*a2
K12 = -6*a
K21 = K12
K22 = 12
K = Matrix([
[K11, K12 ],
[K21, K22 ],
])
K *= EI/a3
u = Matrix([psi1, w1])
f = Matrix([0, S]) + q*Matrix([-a2/12, a/2])
eq2 = Eq(K*u, f)
eqns = [eq1, eq2]
unks = [S, psi1, w1]
sol = solve(eqns,unks)
psi1, w1 = sol[psi1], sol[w1]
pprint("\nψ₁:")
pprint(psi1)
pprint("\nw₁:")
pprint(w1)
pprint("\neT:")
sol = solve( Eq(w1,0), eT)
tmp = sol[0]
pprint(tmp)
pprint("\nUsing \"Shared Dofs\":")
psi1, w1 = var("psi1, w1")
K11 = 4*I
K12 = -6*I/a
K21 = K12
K22 = 12*I/a2 + A
K = Matrix([
[K11, K12 ],
[K21, K22 ],
])
K *= E/a
u = Matrix([psi1, w1])
f = Matrix([0, -EA*eT]) + q*Matrix([-a2/12, a/2])
eq = Eq(K*u, f)
unks = [psi1, w1]
sol = solve(eq,unks)
psi1, w1 = sol[psi1], sol[w1]
pprint("\nψ₁:")
pprint(psi1)
pprint("\nw₁:")
pprint(w1)
[ ]:
einfügen.▶
drücken.