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Kabel (Freileitung) zwischen zwei Masten

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                                          Kabel (Freileitung) zwischen zwei Masten  

 

                                                Univ.-Prof. Dr.-Ing. habil. Josef  BETTEN  

                                                           RWTH Aachen University  

                                                                    Templergraben 55  

                                                           D-52056  A a c h e n ,  Germany  

 

                                                            <betten@mmw.rwth-aachen.de>  

 

> restart:
 

> Dgl:=diff(y(x),x$2)-a*sqrt(1+(diff(y(x),x))^2)=0;
 

`assign`(Dgl, `+`(diff(y(x), `$`(x, 2)), `-`(`*`(a, `*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2)))))) = 0) (1)
 

Darin ist der Parameter  a  gem??  a:= q / H  definiert, wobei  q = q(x,y)  die auf 

die L?ngeneinheit des Kabels bezogene Kraftkomponente (z.B.: Eigengewicht  + 

Windkraft + Schnee- & Eislast) ist, w?hrend  H  die auf die St?tzen wirkende  

Horizontalkraft bedeutet. Bei gegebenem  q = q(x,y)  kann  H  aus der Stabilit?t 

der St?tzen vorgeschrieben werden, so dass damit der Parameter  a  bestimmt ist, 

der die L?nge des Kabels festlegt. Mithin kann f?r ein  Kabel mit gegebenem 

q = q(x,y)  die erforderliche Kabell?nge allein ?ber den Horizontalzug bestimmt  

werden. Zu ber?cksichtigen ist auch der Temperatureinfluss (Sommer / Winter) 

auf die Kabeldehnung.  

Mit abnehmendem Horizontalzug  H  nimmt der Parameter  a  und damit auch die 

Kabell?nge zu, so dass die vertikale Belastung der St?tzen gr??er wird. 

 

Randbedingungen  #  Aufh?ngung in Hanglage: 

> R[1]:=y(0)=1;   R[2]:=y(2)=2;  #  beispielsweise
 

 

`assign`(R[1], y(0) = 1) (2)
 

`assign`(R[2], y(2) = 2) (2)
 

> dsolve({Dgl,R[1],R[2]},y(x)):
 

> Y(x,a):=simplify(convert(rhs(%),radical)) assuming a>0;
 

`assign`(Y(x, a), `+`(`-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(`-`(`*`(3, `*`(exp(`+`(`*`(3, `*`(a)))), `*`(a)))), `-`(`*`(2, `*`(cosh(`+`(`*`(x, `*`(a)), `-`(a), ln(`+`(`*`(exp(a), `*`(a)), `*`(`^`(`+`(`*`...
`assign`(Y(x, a), `+`(`-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(`-`(`*`(3, `*`(exp(`+`(`*`(3, `*`(a)))), `*`(a)))), `-`(`*`(2, `*`(cosh(`+`(`*`(x, `*`(a)), `-`(a), ln(`+`(`*`(exp(a), `*`(a)), `*`(`^`(`+`(`*`...
`assign`(Y(x, a), `+`(`-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(`-`(`*`(3, `*`(exp(`+`(`*`(3, `*`(a)))), `*`(a)))), `-`(`*`(2, `*`(cosh(`+`(`*`(x, `*`(a)), `-`(a), ln(`+`(`*`(exp(a), `*`(a)), `*`(`^`(`+`(`*`...
(3)
 

 

Diese MAPLE-L?sung kann folgenderma?en ausgedr?ckt werden: 

 

> Y(x,a):=3/2-((1+exp(2*a))/(2*a*(exp(3*a)-exp(a))))* sqrt(1+(a^2-2)*exp(2*a)+exp(4*a))+                            (1/a)*cosh(a*(x-1)+ln((a*exp(a)+ sqrt(1+(a^2-2)*exp(2*a)+exp(4*a)))/(exp(2*a)-1)));
 

`assign`(Y(x, a), `+`(`/`(3, 2), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(1, exp(`+`(`*`(2, `*`(a))))), `*`(`^`(`+`(1, `*`(`+`(`*`(`^`(a, 2)), `-`(2)), `*`(exp(`+`(`*`(2, `*`(a)))))), exp(`+`(`*`(4, `*`(a))...
`assign`(Y(x, a), `+`(`/`(3, 2), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(1, exp(`+`(`*`(2, `*`(a))))), `*`(`^`(`+`(1, `*`(`+`(`*`(`^`(a, 2)), `-`(2)), `*`(exp(`+`(`*`(2, `*`(a)))))), exp(`+`(`*`(4, `*`(a))...
(4)
 

> Ableitung(x,a):=simplify(diff(%,x))  assuming a>0;
 

`assign`(Ableitung(x, a), sinh(`+`(`*`(x, `*`(a)), `-`(a), ln(`+`(`*`(exp(a), `*`(a)), `*`(`^`(`+`(`*`(`^`(a, 2), `*`(exp(`+`(`*`(2, `*`(a)))))), `-`(`*`(2, `*`(exp(`+`(`*`(2, `*`(a))))))), exp(`+`(`*... (5)
 

>  
 

> for i in [1,1.5,2,2.5] do Y(x)[a=i]:=evalf(subs(a=i,Y(x,a)),4) od;
 

`assign`(Y(x)[a = 1], `+`(`-`(.176), cosh(`+`(x, `-`(.5866))))) (6)
 

`assign`(Y(x)[a = 1.5], `+`(`-`(.163), `*`(.6667, `*`(cosh(`+`(`*`(1.5, `*`(x)), `-`(1.155))))))) (6)
 

`assign`(Y(x)[a = 2], `+`(`-`(.452), `*`(.5000, `*`(cosh(`+`(`*`(2., `*`(x)), `-`(1.728))))))) (6)
 

`assign`(Y(x)[a = 2.5], `+`(`-`(1.004), `*`(.4000, `*`(cosh(`+`(`*`(2.5, `*`(x)), `-`(2.295))))))) (6)
 

> for i in [1,1.5,2,2.5] do        Neigung(x)[a=i]:=evalf(subs(a=i,diff(Y(x,a),x))) od:
 

> erforderliche_Kabell?nge:=Int(sqrt(1+(diff(y(x),x))^2),x=0..2)= value(int(sqrt(1+(sinh(a*x-a+ln(a*exp(a)+sqrt(a^2*exp(2*a)-2*exp(2*a)+ exp(4*a)+1))-ln(exp(2*a)-1)))^2),x=0..2))  assuming a>0:
 

> alias(H=Heaviside,th=thickness):
 

> p[1]:=plot(rhs(%%),a=0..2.5,2..5,color=black,th=2):
 

> p[2]:=plot({5,5*H(a-2.5)},a=0..2.5001,color=black,           title="erforderliche Kabell?nge  L(a)"):
 

> plots[display]({p[1],p[2]});
 

Plot_2d  
 

> for i in [1,1.5,2,2.5] do erforderliche_Kabell?nge[a=i]:= Int(sqrt(1+(diff(y(x),x))^2),x=0..2)=     evalf(int(sqrt(1+(Neigung(x)[a=i])^2),x=0..2),4)  od;
 

`assign`(erforderliche_Kabell?nge[a = 1], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 2) = 2.554) (7)
 

`assign`(erforderliche_Kabell?nge[a = 1.5], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 2) = 3.010) (7)
 

`assign`(erforderliche_Kabell?nge[a = 2], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 2) = 3.762) (7)
 

`assign`(erforderliche_Kabell?nge[a = 2.5], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 2) = 4.942) (7)
 

> alias(H=Heaviside,th=thickness):
 

> p[1]:=plot({Y(x)[a=1],Y(x)[a=1.5],Y(x)[a=2],Y(x)[a=2.5]}, x=0..2,-1..2,scaling=constrained,th=2,color=black):
 

> p[2]:=plot(2*H(x-2),x=0..2.001,color=black,                          title="Kabel zwischen zwei Masten  #  Parameter  a  =  [1, 1.5, 2, 2.5]"):
 

> plots[display]({p[1],p[2]});
 

Plot_2d  
 

>  
 

Symmetrische Aufh?ngung im Abstand von  x  =  1: 

restart: 

> Dgl:=diff(y(x),x$2)-a*sqrt(1+(diff(y(x),x))^2)=0;
 

`assign`(Dgl, `+`(diff(y(x), `$`(x, 2)), `-`(`*`(a, `*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2)))))) = 0) (8)
 

> R[1]:=y(0)=0;   R[2]:=y(1)=0;
 

`assign`(R[1], y(0) = 0) (9)
 

`assign`(R[2], y(1) = 0) (9)
 

> dsolve({Dgl,R[1],R[2]},y(x));
 

y(x) = `+`(`/`(`*`(cosh(`+`(`*`(x, `*`(a)), ln(RootOf(`+`(`*`(exp(a), `*`(`^`(_Z, 2))), `-`(1))))))), `*`(a)), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(exp(a), 1), `*`(RootOf(`+`(`*`(exp(a), `*`(`^`(_Z, 2))... (10)
 

> Y(x,a):=convert(rhs(%),radical);
 

`assign`(Y(x, a), `+`(`/`(`*`(cosh(`+`(`*`(x, `*`(a)), ln(`*`(`^`(`/`(1, `*`(exp(a))), `/`(1, 2))))))), `*`(a)), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(exp(a), 1), `*`(`^`(`/`(1, `*`(exp(a))), `/`(1, 2)))... (11)
 

> Y(x,a):=(1/a)*cosh(a*(x-1/2))-(1/2/a)*(exp(a)+1)*sqrt(1/exp(a));
 

`assign`(Y(x, a), `+`(`/`(`*`(cosh(`*`(a, `*`(`+`(x, `-`(`/`(1, 2))))))), `*`(a)), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(exp(a), 1), `*`(`^`(`/`(1, `*`(exp(a))), `/`(1, 2)))), `*`(a))))))) (12)
 

> Ableitung(x,a):=diff(%,x)  assuming a>0;
 

`assign`(Ableitung(x, a), sinh(`*`(a, `*`(`+`(x, `-`(`/`(1, 2))))))) (13)
 

> for i in [1,1.5,2,2.5] do Y(x)[a=i]:=                     evalf(subs(a=i,Y(x,a)),4) od;
 

`assign`(Y(x)[a = 1], `+`(cosh(`+`(x, `-`(.5000))), `-`(1.128))) (14)
 

`assign`(Y(x)[a = 1.5], `+`(`*`(.6667, `*`(cosh(`+`(`*`(1.5, `*`(x)), `-`(.7500))))), `-`(.8632))) (14)
 

`assign`(Y(x)[a = 2], `+`(`*`(.5000, `*`(cosh(`+`(`*`(2., `*`(x)), `-`(1.))))), `-`(.7712))) (14)
 

`assign`(Y(x)[a = 2.5], `+`(`*`(.4000, `*`(cosh(`+`(`*`(2.5, `*`(x)), `-`(1.250))))), `-`(.7552))) (14)
 

> for i in [1,1.5,2,2.5] do Neigung(x)[a=i]:=       evalf(subs(a=i,diff(Y(x,a),x)),4) od;
 

`assign`(Neigung(x)[a = 1], sinh(`+`(x, `-`(.5000)))) (15)
 

`assign`(Neigung(x)[a = 1.5], sinh(`+`(`*`(1.5, `*`(x)), `-`(.7500)))) (15)
 

`assign`(Neigung(x)[a = 2], sinh(`+`(`*`(2., `*`(x)), `-`(1.)))) (15)
 

`assign`(Neigung(x)[a = 2.5], sinh(`+`(`*`(2.5, `*`(x)), `-`(1.250)))) (15)
 

> erforderliche_Kabell?nge:= Int(sqrt(1+(diff(y(x),x))^2),x=0..1)=       value(int(sqrt(1+(sinh(a*x-a/2))^2),x=0..1))  assuming a>0;
 

`assign`(erforderliche_Kabell?nge, Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 1) = `/`(`*`(exp(`+`(`-`(`/`(`*`(a), `*`(2))))), `*`(`+`(`-`(1), exp(a)))), `*`(a))) (16)
 

> alias(H=Heaviside,th=thickness):
 

> p[1]:=plot((exp(-a/2))*(-1+exp(a))/a,a=0..3,1..1.421,          color=black,th=2):
 

> p[2]:=plot({1.42,1.42*H(a-3)},a=0..3.001,color=black,        title="erforderliche Kabell?nde  L(a)"):
 

> plots[display]({p[1],p[2]});
 

Plot_2d  
 

> for i in [1,1.5,2,2.5] do erforderliche_Kabell?nge[a=i]:= Int(sqrt(1+(diff(y(x),x))^2),x=0..1)=      evalf(int(sqrt(1+(Neigung(x)[a=i])^2),x=0..1),4) od;
 

`assign`(erforderliche_Kabell?nge[a = 1], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 1) = 1.042) (17)
 

`assign`(erforderliche_Kabell?nge[a = 1.5], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 1) = 1.096) (17)
 

`assign`(erforderliche_Kabell?nge[a = 2], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 1) = 1.175) (17)
 

`assign`(erforderliche_Kabell?nge[a = 2.5], Int(`*`(`^`(`+`(1, `*`(`^`(diff(y(x), x), 2))), `/`(1, 2))), x = 0 .. 1) = 1.282) (17)
 

> alias(H=Heaviside,th=thickness):
 

> p[1]:=plot({Y(x)[a=1],Y(x)[a=1.5],Y(x)[a=2],Y(x)[a=2.5]}, x=0..1,scaling=constrained,th=2,color=black):
 

> p[2]:=plot({-0.5,-0.5*H(x-1)},x=0..1.001,color=black,                      title="Parameter  a  =  [1,  1.5,  2,  2.5]"):
 

> plots[display]({p[1],p[2]});
 

Plot_2d  
 

> Y(1/2,a):=subs(x=1/2,Y(x,a));
 

`assign`(Y(`/`(1, 2), a), `+`(`/`(`*`(cosh(0)), `*`(a)), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(exp(a), 1), `*`(`^`(`/`(1, `*`(exp(a))), `/`(1, 2)))), `*`(a))))))) (18)
 

> Y(1/2,a):=subs(cosh(0)=1,%);
 

`assign`(Y(`/`(1, 2), a), `+`(`/`(1, `*`(a)), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(exp(a), 1), `*`(`^`(`/`(1, `*`(exp(a))), `/`(1, 2)))), `*`(a))))))) (19)
 

> for i in [1,1.5,2,2.5] do maximaler_Durchhang[a=i]:=                              evalf(subs(a=i,Y(1/2,a)),4)  od;
 

`assign`(maximaler_Durchhang[a = 1], -.128) (20)
 

`assign`(maximaler_Durchhang[a = 1.5], -.1965) (20)
 

`assign`(maximaler_Durchhang[a = 2], -.2712) (20)
 

`assign`(maximaler_Durchhang[a = 2.5], -.3552) (20)
 

> maximaler_Durchhang[a=0]:=Limit(Y(1/2,a),a=0);
 

`assign`(maximaler_Durchhang[a = 0], Limit(`+`(`/`(1, `*`(a)), `-`(`*`(`/`(1, 2), `*`(`/`(`*`(`+`(exp(a), 1), `*`(`^`(`/`(1, `*`(exp(a))), `/`(1, 2)))), `*`(a)))))), a = 0)) (21)
 

> maximaler_Durchhang[a=0]:=value(%);
 

`assign`(maximaler_Durchhang[a = 0], 0) (22)
 

F?r  a = 0  verschwindet der Durchhang. Dann wird der Horizontalzug    

wegen  a  =  q / H  unendlich gro?. 

 

>