Пресс-релиз популярных книг
.
Авторы: 111 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я
Книги: 164 А Б В Г Д Е Ж З И Й К Л М Н О П Р С Т У Ф Х Ц Ч Ш Щ Э Ю Я
На сайте 111 авторов, 92 книг, 72 статей, 5913 глав.
7.3 Green’s functions
To solve the equation
3 · x = 12 ⇒ x =
1
3
· 12 (7.99)
the right-hand side is multiplied by the “Green’s function” xG = 1/3, which
is the solution of 3 · x = 1.
The Green’s functions are the solutions of the adjoint equations. Consider
for example the system Ku = f and the identity
B(u,ˆu) = ˆuT Ku− uT KT ˆu = 0, (7.100)
where KT is the adjoint (= transpose) of the matrix K. Clearly if gi is a
solution of KT gi = ei then ui = gTi
f.
In linear structural mechanics the equations are self-adjoint (or symmetric
K = KT ) so that the Green’s functions are the solutions of the same
equations, EI GIV
0 = δ0, as in the original problem, EI wIV = p.
The complement of the Green’s function is Green’s second identity (Betti’s
theorem), which in the case of the Laplacian reads
B(u, ˆu) =
_
Ω
−Δu ˆudΩ +
_
Γ
∂u
∂n
ˆuds −
_
Γ
u
∂ˆu
∂n
ds −
_
Ω
u (−Δˆu) dΩ = 0.
(7.101)
From this equation we can see what boundary conditions must be imposed on
the Green’s functions; see Fig. 7.4. In a Dirichlet problem
− Δu = p , u = g on Γ (7.102)
things are easy:
− ΔG0 = δ0, G0 = 0 on Γ (7.103)
7.3 Green’s functions 517
Fig. 7.4. Influence function for a beam: a) when a support is displaced, b) Green’s
function, c) theoretically no Green’s function exists for a beam with no supports
and so
u =
_
Ω
G0 pdΩy −
_
Γ
∂G0
∂ n
g ds. (7.104)
In a mixed problem
− Δu = p , u = g on ΓD ,
∂u
∂n
= t on ΓN (7.105)
we require that
− ΔG0 = δ0, G0 = 0 on ΓD ,
∂G0
∂n
= 0 on ΓN (7.106)
and so
u =
_
Ω
G0 pdΩy −
_
ΓD
∂G0
∂ n
g ds +
_
ΓN
G0 t ds . (7.107)
The support conditions of the beam in Fig. 7.4 a are of such a mixed type,
because geometric, w(0) = w_(0) = w(l), as well as static boundary conditions,
M(l) = 0, are prescribed. Hence if the Green’s function of the beam in Fig.
7.4 a solves the boundary value problem
EI GIV
0 = δ0(y − x), G0(0) = G0(l) = G
_
0(0) = −EI G
__
0 (l) = 0,(7.108)
then Betti’s theorem yields
518 7 Theoretical details
B(w,G0) =
_ l
0
pG0 dy + [V G0 −MG
_
0]l
0
− [V0 w −M0 w
_]l
0
−
_ l
0
wδ0 dy =
_ l
0
p G0 dy + B0(x) δ − w(x) = 0 (7.109)
where −V0(l)w(l) = B0(x) δ with the sign convention in Fig. 7.4.
But influence functions for the solution of Neumann problems do not exist,
because one cannot place a force δ0 on Ω and require at the same time that
all the tractions on the boundary vanish
− ΔG0 = δ0
∂G0
∂n
= 0 on Γ ? (7.110)
The reason is that the solution of a Neumann problem is only unique up to a
constant uc as for example in the case of the beam in Fig. 7.4 c, that gives the
impression of a beam on an elastic foundation, EIwIV + cw = p, but it is a
standard beam EIwIV . It is only that the sum of the distributed load on both
sides of the beam happens to be the same, so that no supports are necessary.
Of course Green’s functions for beams on an elastic foundation exist.
Naturally all these problems go away if the solution is made unique by
specifying single values of the solution as w(0) = w(l) = 0, or in terms of
structural mechanics, by adding supports to a structure.
Elastic supports
To be complete let us also discuss the case that the structure rests on an
elastic support. Imagine that the hinged support in Fig. 7.4 is replaced by
a spring with stiffness k. The decisive term in Betti’s theorem (7.109) is the
work term
V0(l)w(l) = kG0(l, x)w(l) = G0(l, x) Vp =
1
k
V0 Vp (7.111)
which encapsulates the interaction between the compression G0(l, x) of the
spring due to the point load P = 1 and the support reaction Vp in the load
case p or—vice versa—the interaction between the support reaction V0 due
to P = 1 and the compression w(l) of the spring in the load case p, so that
the influence function becomes
w(x) =
_ l
0
G0(y, x) p(y) dy + V0(l)w(l) (7.112)
which for the special case w(l) = δ is identical with (7.109). Note that
V0(l)w(l) comes from the “boundary integral” [...]. Such edge contributions
always appear in the influence functions if the structure rests on soft supports.
7.4 Generalized Green’s functions 519
Популярные книги
- Старинные занимательные задачи
- Медоносные растения
- Математика Древнего Китая
- Algebratic geometry
- Workbook in Higher Algebra
- Finite element analysis
- Fields and galois theory
- Пчеловодство
- Mathematics and art
- Black Holes
Популярные статьи
- Higher-Order Finite Element Methods
- Электровакуумные приборы
- Riemann zeta functionS
- Универсальная открытая архитектурно-строительная система зданий серии Б1.020.1-71
- Complex Analysis 2002-2003
- Пример расчета прочности елементов, стыков и узлов несущего каркаса здания
- Составы, вещества и материалы для огнезащитыметаллических консрукций и изделий
- CMOS Technology
- Рекомендации по расчету и конструированию сборных железобетонных колонн каркасов зданий серии Б1.020.1-7 с плоскими стыками ВИНСТ
- Советы старого пчеловода