Draft

6.1Theory 125

Hv =

wx

2

2

+C 1 x+C 2 (6.6-b)

andtheconstants of integrationsC 1 andC 2 canbe obtainedfromtheboundaryconditions:

v= 0 atx= 0 andatx=L)C 2 = 0 andC 1 =

wL

2

. Thus

v=

w

2 H

x(Lx) (6.7)

Thisequationgives theshapev(x) in termsof thehorizontalforceH,

5 Sincethemaximumsaghoccursat midspan(x=

L

2

) we cansolve forthehorizontalforce

H=

wL

2

8 h

(6.8)

we notetheanalogywiththemaximummoment in a simplysupporteduniformlyloadedbeam

M=Hh=

wL

2

8

. Furthermore,thisrelationclearlyshowsthatthehorizontalforceis inversely

proportionalto thesagh, ash&H%. Finally, we canrewritethisequationas

r

def

=

h

L

(6.9-a)

wL

H

= 8 r (6.9-b)

6 EliminatingHfromEq.6.7and6.8we obtain

v= 4h

x

2

L

2

• 
  

x

L

!

(6.10)

Thus thecableassumesa parabolicshape (asthemoment diagramof theappliedload).

7 WhereasthehorizontalforceHis constant throughoutthecable,thetensionTis not. The

maximumtensionoccursat thesupportwheretheverticalcomponent is equaltoV =

wL

2

and

thehorizontalonetoH, thus

Tmax=

p

V

2

+H

2

=

s



wL

2



2

+H

2

=H

s

1 +



wL= 2

H

 2

(6.11)

CombiningthiswithEq.6.8we obtain

1

.

Tmax=H

p

1 + 16r

2

H(1 + 8r

2

) (6.12)

8 Hadwe assumeda uniformloadwper lengthof cable(ratherthanhorizontalprojection),

theequationwouldhave beenoneof a catenary

2

.

v=

H

w

cosh



w

H



L

2

x



+h (6.13)

Thecablebetweentransmissiontowersis a good exampleof a catenary.

1

Recallingthat(a+b)

n

=a

n

+na

n 1

b+

n(n1)

2!

a

n 2

b

2

+
or (1 +b)

n

= 1 +nb+

n(n1)b^2

2!

• 
  

n(n1)(n2)b^3

3!

+


;

Thus forb

2

<<1,

p

1 +b= (1 +b)

1

(^2) 1 +
b
2
2
Derivationof thisequationis beyondthescope of thiscourse.