3.6 Cables

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In the first chapter a rope served us well to introduce the basic principles of

the finite element technique. Here we want to add some more details to the

analysis of ropes and cables.

The cable in Fig. 3.29 carries a vertical load p(x) and is prestressed by a

force S. If the angle of the chord is zero, α = 0, the prestressing force S acts

in horizontal direction, S · cos α = H = S · 1, as in chapter one.

In the following the bending moment M(x) is the bending moment in a

beam which carries the same load p and the forces RA and RB in Fig. 3.29

are identical with the support reactions of the beam.

Because the bending stiffness of the cable is neglected, EI = 0, the bending

moment must be zero at any point x (let p(x) = p a uniform load, though it

can be any load)

_x

_

M : RA · x − p · x2

_ _ 2 _

M(x)

−S · sin α · x + S · cos α · y = 0 (3.143)

306 3 Frames

Fig. 3.30. Cable element

or

M(x) − S · sin α · x + S · cos α · y = 0 M(x) +H y = 0 (3.144)

(the second equation applies to a horizontal cable) from which follows the

equation for the shape of the cable

y(x) = tan α · x − M(x)

S · cos α

y(x) = −M(x)

H

. (3.145)

Note that the y-axis in Fig. 3.29 points upward. In this system the differential

equation for the deflection has a positive sign

y(x) = −M(x)

H

⇒ H y

__(x) = −M(x)__ = p(x) (3.146)

while if y points downward as in chapter one the sign is negative, −H y__(x) =

p(x).

The maximum tension in the rope is

SB =

_

(S cos α)2 + (RB + S sin α)2 . (3.147)

With

y

_ = tan α − 1

S cos α

·M

_ = tanα − V

S · cos α

(3.148)

the expression for the length s of the rope becomes

s =

_ l

0

_

1 + (y_)2 dx =

_ l

0

%

1 + (tan α − ( V

S · cos α

)2) dx . (3.149)

The length s is equal to the length s0 of the unstretched cable plus its elastic

elongation

Δs =

_ s

0

ε dx = S · cos α

EA

_ l

0

(1 + (y

_)2) dx

= S · cos α

EA

_ l

0

(1 + (tan α − V

S · cos α

)2) dx (3.150)

3.6 Cables 307

Fig. 3.31. Single lap joint

plus—eventually—thermal effects

s = s0 + Δs + s0 αT ΔT (3.151)

or

s =

_ l

0

%

1 + (tan α − V

S · cos α

)2) dx

= s0 (1 + αT ΔT) + S · cos α

EA

_ l

0

(1 + (tan α − V

S · cos α

)2) dx

_ _ _

Δs

.

This equation allows to determine for any type of load the prestressing force

S and then with (3.145) the shape y of the cable.

A horizontal cable with cross section A assumes under gravity

g(x) = g

cos ϕ

g = γ A, ϕ = arctany

_ (3.152)

the shape of a catenary (the origin, x = 0, y = 0, is the deepest point of the

cable)

y(x) = H

g

&

cosh x · g

H

− 1

'

(3.153)

or if we let g(x) _ g as in a shallow cable the shape of a parabola (x and y as

in Fig. 3.29)

y(x) = − g

2H

(l − x) x . (3.154)

The two parts of the stiffness matrix of a cable element (2-D) (see Fig.

3.30) represent the longitudinal stiffness (EA/l) and the so-called geometric

stiffness (S/l) of the cable, c = cos α, s = sin α,

K = EA

l

⎛

⎜⎜⎝

c2 −c · s −c2 c · s

−c ·s s2 c · s −s2

−c2 c ·s c2 −c · s

c · s −s2 −c ·s s2

⎞

⎟⎟⎠

+ S

l

⎛

⎜⎜⎝

s2 c · s −s2 −c · s

c ·s c2 −c · s −c2

−s2 −c ·s s2 c · s

−c · s −c2 c ·s c2

⎞

⎟⎟⎠

.

(3.155)

308 3 Frames

Fig. 3.32. Linear analysis: normal forces in a) a truss and b) a frame, c) a sway

frame with cables

While a cable element is straight the real cable will sag and create its own

prestress by its own weight. The effect can easily be described by a subdivision

of the cable in different elements and an updated Lagrangian approach.

If we consider for example a horizontal cable with a length of 10.0 m, a

cross section of 84 mm2 and prestressed by a force of 1 kN the displacement

of the straight prestressed rope under gravity load will be about 8 _ of the

length. If we take into account the nonlinear effects of the displacements the

tensile force will increase by a factor of 2 and the deformations and the eigen

frequencies will change considerably, see Table 3.1.

Table 3.1. Linear and nonlinear analysis of a cable

N u f1 f2 f3

linear 1.0 83.36 1.928 3.809 5.596

nonlinear 1.9 43.95 2.374/3.468 4.690 6.890

Of special interest is the doubling of the first eigen value. Introduced by

the deformation we have unsymmetric stiffness and different frequencies for

a movement up and down. As far as we know these effects are not included

automatically in finite element programs. So it is up to the user to detect such

behaviour. There are many examples where those nonlinear effects (which are

favorable in general) have not been included. Even for a simple problem as

in Fig. 3.31 where a single lap joint was modeled with shell elements. The

geometric non linear effects markedly reduced the eccentricity of the load and

so the bending stress was reduced by a factor of 2.

Even if we stick to linear analysis the analysis of cable structures is anything

but simple. Take for example a sway frame with bracings by two diagonal

cables; see Fig. 3.32. The system is defined in general without prestress. In

3.7 Hierarchical elements 309

ment extending from

ϕ1(x), ϕ2(x) and the

ϕ3(x), ϕ4(x), ϕ5(x)

a linear analysis both elements will stay effective. The vertical payload will

shorten the cables and both cables will thus carry compressive forces and

consequently they will be deactivated in a nonlinear analysis. Then we might

have a kinematic system, and it is only an additional horizontal force which

will introduce a tensile stress in one of the cables and thus stabilize the whole

structure. In the construction process the cables will be inserted at a time

when some of the vertical loading has been already applied and for true cables

a small prestress will be applied. To be able to cope with those problems

a series of modifications and assumptions has to be made.

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