Hans Welleman 1

Slender Structures

Load carrying principles

Cables and arches

v2018-1

Content (preliminary schedule)

� Basic cases

– Extension, shear, torsion, cable

– Bending (Euler-Bernoulli)

� Combined systems

- Parallel systems

- Special system – Bending (Timoshenko)

� Continuously Elastic Supported (basic) Cases

� Cable revisit and Arches

� Matrix Method

Hans Welleman 2

Learning objectives

� Extend the technique for basic cases

� Find the ODE for a specific case and the boundary

conditions for the specific application

� Solve the more advanced ODE’s (by hand and

MAPLE)

� Investigate consequences/limitations of the model

and check results with limit cases

Hans Welleman 3

Basic Cases

Second order DE

� Extension

� Shear

� Torsion

� Cable

Hans Welleman 4

Fourth order DE

� Bending

2

2

2

2

2

2

2

2

d

d

d

d

d

d

d

d

xt

uEA q

x

wk q

x

GI mx

zH q

x

ϕ

− =

− =

− =

− =

4

4

d

d

wEI q

x=

Model

� (ordinary) Differential Equation – (O)DE

– Boundary conditions

– Matching conditions

Hans Welleman 5

Cable

� How to find H in a cable structure

� Catenary solution,load along cable

� Axial deformation of cables

� Susceptive to changes in load

� Horizontal displacements of cables

Hans Welleman 6

Cable 1load distributed along the projection of the cable

Hans Welleman 7

� external load q

� internal generalised stress, vertical component V …. BC

� Cable slope tanα

� displacement field w …. BC

“cable stiffness” H

“cable without

elongation” ??!!

ODE

Fundamental relations

� geometrical relation

� Moment equilibrium

� Equilibrium

Hans Welleman 8

dtan

d

z

xα =

tanV H α=

d d 0

d

d

V q x V V

Vq

x

− + + + = ⇔

= −

2

2

d d d

d d d

z V zV H H q

x x x= → = = −⋯

Remarks� Cable takes no bending

� Cable ODE describes an equilibirum in the deflected state (funicular curve) so this is a non-linear approach! (no superposition)

� H can be regarded as constant only if no horizontal loads are

applied

� This model is not valid for loads distributed along the cable!

� Derivation is strictly based upon equilibrium only!

� Cable force can be expressed in H and z :

Hans Welleman 9

( )2

22 2 2 dtan 1

d

zT H V H H H

xα

= + = + = +

10

How to find H ?

� With specified cable length

� With (external) specified load

11

1st option : with specified length

Example length AB given: L = 11.6344 m

10 kN

60 kN

3.0 m 4.0 m 2.0 m 6.0 m

3.0 m

3.0 m

A

B

C

chord AB

zk1

zk2

2 2

2 2 2 2

1 2 1 2

7 23 ( 1) 4 ( 1) 2

3 3k k k k

L z z z z− −

= + − + + − − − + + +

12

Find position of the cable in terms of

the constant H

10 kN 60 kN

3.0 m 4.0 m 2.0 m

20 kN 50 kN

100 kNm

60 kNm

M-lijn 1kH z

2kH z

12 1

2

60 10

100 6

kk k

k

H zz z

H z= ⇒ =

13

1st option : with specified length

Example for fixed cable length L

A B

q

l

L

f

s∆

x∆

z∆

2

1( )

4 ( )2( )k

qx l xfx l x

z xH l

−−

= =

2d

d 1 dd

zs x

x

= +

2

0 0

dd 1 d

d

x l x l

x x

zL s x

x

= =

= =

= = +

∫ ∫

OEPS ..

14

Continue …

2 2 2 3

2

0 0 0

1 d 1 dd 1 d d

2 d 2 d 24

x l x l x l

x x x

z z q lL s x l x l

x x H

= = =

= = =

= ≈ + ≈ + = +

∫ ∫ ∫

A B

q

l

L

f

overlength ∆ = L – l :2 3

2

24

q lH =

∆2

3 8or

8 3

ff l

l= ∆ ∆ =Check with MAPLE

2

0 0

dd 1 d

d

x l x l

x x

zL s x

x

= =

= =

= = +

∫ ∫

15

A B

l

L

f

F

katrol.

q

F

T = F

Free Body

Diagram of

the block

support

reaction of

the block

force polygon

for the block F

F

VT = F

H

components

of the cable

force T

q

A

B

l

Lf

F

block.

T

HHV V

support reaction

of the block

1

2V ql=

2

2 1

2H F ql

= −

2

2

2 18

2

qlf

F ql

=

−

2

8

qlf

H=

2nd option : with (external) specified load

Example (see also part 1)

Cable 2load distributed along the cable

Hans Welleman 16

� external load q

� internal generalised stress, vertical component V …. BC

� Cable slope tanα

� displacement field w …. BC

“cable stiffness” H

“cable without

elongation due to

cable force” !

ODE

Fundamental relations

� geometrical relation

� Moment equilibrium

� Equilibrium

Hans Welleman 17

dtan

d

z

xα =

tanV H α=

2

d d 0

d d d1

d d d

V q s V V

V s zq q

x x x

− + + + = ⇔

= − = − +

22

2

d d d d1

d d dd

z V z zV H H q

x x xx

= → = = − +

⋯

Catenary solution:

Hans Welleman 18

22

2

d d1

dd

z zH q

xx

= − +

d: sinh (1)

d

zsubsitute

xς=

2d(sinh )1 sinh

dH q

x

ςς= − +

d dcosh cosh

d dH q H q

x x

ς ςς ς= − ⇔ = −

1

qxC

Hς = − + 1

dwith (1): sinh sinh

d

z qxC

x Hς

= = − +

1 2coshH qx

z C Cq H

= − − + +

Catenary versus parabola

Hans Welleman 19

f

l

H

qlβ =

f Hversus

l qlβ =

Parabola

Catenary

Axial cable deformation (strain)

20

( )Tl

l straight cableEA

∆ =

2

2 2 2 2

2

d

d

d1 ( )

d

zT H V H H

x

zT H curved

x

= + = +

= +

2d

d 1 dd

zs x

x

= +

2 2

0 0 0

d d d1 d d

d d

s L x l x l

s x x

T s H z Hl H zL x x

EA EA x EA EA x

= = =

= = =

∆ = = + = +

∫ ∫ ∫

Curved cable:

Small for cables with modest sag

Oostende (B)

Hans Welleman 21

Nelson Mandelabrug in the Maria-Hendrika Park

http://office360.be/catalogus/025-voet-en-fietsbrug-oostende/

Hans Welleman 22

In de natuurlijke omgeving van het Maria Hendrikapark wordt

een sober en evident ogend kunstwerk gerealiseerd dat een

belangrijke schakel vormt in een bovenlokale wandel- en

fietsroute van de Stad Oostende. [a nice bridge in a park]

De brug wordt gereduceerd tot zijn essentie: loopvlak dat zich

uitstrekt van oever tot oever. De structuur wordt in het loopvlak

geïntegreerd, in de vorm van opgespannen kabels. Het loopvlak

neemt de elegante vorm aan van de natuurlijke kettinglijn en

overspant zo met een dikte van 28 cm een afstand van 72 m. De

brug won in 2006 de Staalbouw wedstrijd. [an elegant integrated

cable structure following the shape of a catenary with deck

thickness of 28 cm and a span of 72 m, Steel price winner 2006]

Mandela bridge, some numbers ..

Hans Welleman 23

span l = 72.00 m deck width b = 2.50 m

sag f = 1.30 m thick t = 0.28 m

p = 19.6 kN/m q = 5 kN/m2

EA = 21e6 kN 1 cable force due to dead weight,

2 cable force at full load,

3 sag at full load

24

Variation in H due variation in load

A B

L

L

1q q p= +

2q q p= −

q

12l 1

2l

2 3

224

q lL l

H= +

full load q:

( ) ( )

( ) ( )

221 1

1 1 1 1 222

222 1

2 2 2 3 422

d1

d

d2

d

zfield H H q H H z q x C C x

x

zfield H H q H H z q x C C x

x

− + ∆ = ⇒ − + ∆ = + +

− + ∆ = ⇒ − + ∆ = + +

Variation in load:

25

12

12

2 22 3

1 21 12 22

0

d d1 d 1 d

24 d d

x l x l

x x l

z zq lL l x x

H x x

= =

= =

= + = + + +

∫ ∫

Result …

( ) ( )

( ) ( )

221 1

1 1 1 1 222

222 1

2 2 2 3 422

d1

d

d2

d

zfield H H q H H z q x C C x

x

zfield H H q H H z q x C C x

x

− + ∆ = ⇒ − + ∆ = + +

− + ∆ = ⇒ − + ∆ = + +

Variation in load:

Total cable length remains unchanged (assume no elongation):

some work to do

2

21 1

4

pH H

q

∆ = + −

Suppose:

25% variation in q

0,78% variation in H

7,5% variation in z

26

Horizontal displacements

A B

L

L

u

1q q p= +

2q q p= −

q

12l 1

2l

27

More detail …

αA

z

x

α

α

B

w

u

A′

B′

du

dw

dz

dx

dd tan d with: tan

d

zu w

xα α= − ⋅ =

d d d dd d

d d d d

z u z wu w

x x x x= − ⇔ = − 1

d dd

d d

z wu x C

x x= − +∫

0

d d( ) (0) d 0

d d

x l

x

z wu l u x

x x

=

=

− = − =∫returns for arches

Alternative derivation

Cable length L is constant. Compare length due to load

q with length due to q+p : (axial deformation not included)

Hans Welleman 28

0

d dd 0

d d

x l

x

z wx

x x

=

=

=∫

2 2

1 12 2

0 0

d d( )1 d 1 d

d d

x l x l

x x

z z wx x

x x

= =

= =

+ + = +

∫ ∫

due to q due to q+p

Cable and beam (Euler-Bernoulli)

29

Model with constant H

30

2

2

d

d

zH p

x− =construction : all permanent load on cable:

( )2

2

4

4

d

d

d

d

cable

beam

z wH q

x

wEI q

x

+− =

=

After completion : beam + cable as P-system:

Load:

p = permanent

q = additional load

rigid links

EIbeam

z(x) cable

l

h

Simplified Model (constant H)

31

Final situation : beam + cable as P-system:

cable beamq p q q+ = +

( )24

4 2

dd

d d

z wwEI H q p

x x

+− = +

( )24 2

4 2 2

dd d

d d d

z ww zEI H H q

x x x

+− = − +

4 2 2

4 2 2

d d dwith:

d d d

w w zEI H q H p

x x x− = − =

Refined Model with changing Hnot part of the examination

32

( )( )24

4 2

dd

d d

z wwEI H H q p

x x

+− + ∆ = +

2

2

d

d

zH p

x− =Permanent load:

( )4 2 2

4 2 2

d d dwith:

d d d

w w H zEI H H q p H p

x x H x

∆− + ∆ = − − =

Two unknowns, additional equation required:

( )

0 0

d d d dd 0 or d

d d d d

x l x l

x x

H lz w z wx x

x x x x EA

= =

= =

∆= ≈∫ ∫

no axial deformation with axial deformation

axial deformation

compensated

Example

Hans Welleman 33

cable 0,6 kN/m

beam 9, 2 kN/m

2,0 kN/m

p

p

q

3

2

cable 1650,0 10 kN

beam 800 kNm

EA

EI

×

Question:

Find the deflection w at midspan due to

additional load q?

Assignment (use MAPLE and exact formulation for the length)

� Find length L of cable for z(a+b) = 0.8 m

� Find z(a+b) for change of length of +2.5% and –2.5%

Hans Welleman 34

:

2; 2; 2

5 kN/m; 15 kN;

given

a b c

q F

EA

= = =

= =

= ∞

H = 29,1667 kN, L = 6,25580 m

97.5% H = 47,26377 kN, z(a+b) = 0.49368 m

102.5% H = 22,75174 kN, z(a+b) = 1.02556 m

35

Arches

� Primarily compression

� Shape retained (not like a cable) due to bending stiffness of the arch

� Susceptible to buckling

F

w

q

xz

36

Cable versus Arch

q

l

A BH H

x

z

q

l

A BH H

x

z

EA, EI

arch no moments in the

arch

EA

37

Cable up side down ?

Only if the line of force due to the load corresponds to

the shape of the arch

ANTONI

GAUDI

(1852-1926)

38

Line of Force

8 kN/m

4,0 m

6,0 m 3,0 m

S

A B

C D

Me

N=

Centre of force in a cross section is the point

of intersection of the resultant force of all

normal stresses and the cross section.

M-line and N-line required …

cf

cross section

39

Line of force48

48

48

4836

36

24

48

M-line [kNm]

32 16

12

N-line [kN]

8 kN/m

4,0 m

6,0 m 3,0 m

S

A B

C D

12 kN

32 kN16 kN

12 kN

M, e

x

1( ) 8 (6 ) 8 (6 ) 0,0 6,0

2M x x x x x= − ⋅ − + ⋅ ⋅ ⋅ − ≤ ≤

2 813 3

( ) 4e x x x= − + line of force (line of

pressure) A-S

Result

8 kN/m

4,0

m

6,0 m 3,0 m

A B

C

D 1,3

3 m

M, e

xS2 S3

change the

position of the

hinge to influence

the shape and

compressive force

and see ….

Perfect Arch ….

� “flipped” cable

� Geometry (shape) is identical to themoment distribution of a replacing beambut scaled by a constant H ….

� Try this !

Bridge with multiple spans

Bridge crossing the Ulvsund near Kalvehave in Denmark, photograph © Hans Welleman

15,0 m 25,0 m 10,0 m

5 kN/m

7,5 kN/m

5 kN/m

EI = 10000 kNm2

H = 100 kNH = 100 kN

Beam model with M-line ?

15,0 m 25,0 m 10,0 m

5 kN/m

7,5 kN/m

5 kN/m

Flip the M-line and scale the result by H

M [kNm]

x [m]

44

Force distribution in arches

� Line of pressure method … clumsy

� Classical method based on the force method for specific type of arches

� Differential equation for arches with a wider range of application but with limited deformations

45

Classical approach - based on the

force methodq

l

A BH H

x

EA, EI

( )b

z x

H H

z

q

lA

B

x

EA, EI

h

( )bz x

Condition:

Both ends with hinged

supports

46

Force Method

� Reduce structure to a statically determinate (principal) system:

curved beam , find the M-line

� Denote the Statically Indeterminates (redundants):

H

� Define the deformation condition:

h = 0 …. horizontal displacement

47

Horizontal displacementsFind h due to a moment distribution M in a

curved beam:

dd d

d

M Mx

x EI EI

ϕκ ϕ= = ⇒ =Beam (straight) :

horizontal displacement = rotation times vertical distance to centre of rotation

vertical displacement = rotation times horizontal distance to centre of rotation

d d dh a zϕ ϕ= = −

dd ds

ds

M M

EI EI

ϕκ ϕ= = ⇒ =Beam (curved) :

darch

Mzh s

EI= −∫

A

B dh

dϕ

a

z

48

Moment in principal system due

to H

H

HMH

a

0H H

H

M H a M H a

a z M H z

+ ⋅ = ⇒ = − ⋅

= − ⇒ = ⋅

x

z

49

Deformation condition

0

with:

: due to the load on principal system

: due to the redundant

: axial deformation of the arch

a H

a

H

h h h

h

h H

h

∆

∆

+ + =

daa

arch

M z sh

EI= −∫

2d ( ) d dHH

arch arch arch

M z s H z z s H z sh

EI EI EI

⋅ ⋅ ⋅ ⋅= − = − = −∫ ∫ ∫

Hlh

EA

∆ = −

2d d0

a

arch arch

M z s Hz s Hl

EI EI EA− − − =∫ ∫

50

Result classical approach

2

dx

dx

a

arch

arch

M z

EIH

z l

EI EA

= −

+

∫

∫

aM M Hz= +

Moment distribution in the arch is the

superposition of the distribution of the

statically determinate (principal) system

and the distribution due to the statically

indeterminate (redundant) H

51

Exampleq

l

A BH H

x

EA, EI

( ) sinb

xz x f

l

π= −

Given:

l = 100 m

f = 15 m

q = 4,0 kN/m

EA = infinite

1( )

2

aM qx l x= −sin

xz f

l

π = −

2

dx

dx

a

arch

arch

M z

EIH

z l

EI EA

= −

+

∫

∫

aM M Hz= +

52

Result

> restart;

> l:=100; q:=4; f:=15;

> z:=-f*sin(Pi*x/l);

> plot(-z,x=0..l,-50..50,title="boog");

moment Ma in het statisch bepaalde hoofdsysteem> Ma:=(1/2)*q*x*(l-x);

> plot(-Ma,x=0..l,title="Momentenverdeling Ma");

los H op met de klassieke methode:> H:=-int(Ma*z,x=0..l)/int(z^2,x=0..l);

> M:=simplify(Ma+H*z);

> plot(-M,x=0..l,title="Momentenverdeling in de boog");

Moment distribution in the arch

M [kNm]

x [m]

53

ODE for Arches

� For all boundary conditions

� Bending and “flipped” cable

q

l

A Bx

EA, EI

z

( )b

z x

2 4

2 4

d dand

d dcable bending

z wH q EI q

x x− = =

geometry z

After completion of the arch the

load q is taken by the arch +

bending. Displacements w with

respect to z will occur and a

constant horizotal component H of

the compressive force in the arch

will occur

2 4

2 4

d dand

d darch bending

z wH q EI q

x x= =

54

Load carrying capacity

� Arch + Bending

2

2

4

4

d

d

d

d

arch

bending

zH q

x

wEI q

x

=

=

4 2

4 2

d d ( )

d darch bending

w z wEI H q q q

x x

++ = + =

4 2

4 2

d d

d d

w zEI q H

x x= −

55

Solution of the ODE

General Solution =

homogeneous solution + particuliar solution

Formulate four boundary conditions

solve the integration constants

Result : The solution in terms of the unknown H

MAPLE

56

Extra condition to find H

0

d dd 0

d d

x l

x

z wx

x x

=

=

=∫

no horizontal displacement at the supports:

0

d dd

d d

x l

x

Hl z wx

EA x x

=

=

= − ∫

Including axial deformation of the arch:

0

d 0

x l

x

w x

=

=

=∫

Parabolic arch:

2

0

8d

x l

x

Hl fw x

EA l

=

=

= ∫

Parabolic arch:

57

Fun ….

2

2

00 0

2

2

00 0

d d d dd d 0

d d d d

or

d d dz dd d 0

d d d d

x lx l x l

xx x

x lx l x l

xx x

z w w wx z z x

x x x x

z w zx w w x

x x x x

== =

== =

== =

== =

− = − + =

− = − + =

∫ ∫

∫ ∫

2

2

0 0 0

dd d 0 d 0

d

x l x l x l

x x x

zw x C w x w x

x

= = =

= = =

= = ⇒ =∫ ∫ ∫

constant for

parabola …

58

Example( )

bz x

q

l

A B x

EA, EI

z Given:

q = 5 kN/m

l = 80 m

f = 15 m

EI = 150000 kNm2

parabolic arch

2

2

d0; 0; 0

d

d; 0; 0

d

wx w

x

wx l w M EI

x

ϕ= = = − =

= = = − =

boundary conditions:

load:

12

5,0 ( )q Heaviside x l= ⋅ −

2

4 ( )fx l xz

l

−= −

arch:

DEMO with MAPLE

59

Conclusion

� Simple parametric model for arches to be used with MAPLE for design purposes

� Results of M-line etc is with respect to

the horizontal axis (x-axis) and not

perpendicular to the beam axis as in FEM programs

??

60

61

ODE result versus FEM result

3

3

3

3

d dcos sin

d d

d dsin cos

d d

w zN H EI H

x x

w zV H EI H

x x

α α

α α

= − + − −

= + − −

local (beam) x-axis

shear force V

shear force Vnormal force N

normal force N

FEM

Vbending Varch

z(x)

α

H

Varch+ Vbending

α

α

ODE

One last detail …

Complete ODE

� Do not neglect w with respect to z

62

4 2 2

4 2 2

d d d

d d d

w w zEI H q H

x x x+ = −

General solution

� use MAPLE

63

( )( )

4

2

1 1 3 422 2 2

2

2sin

4( ) sin cos

4 4

with:

o

xq l

f EIlw x x D x D x D x D

l HEI l H

H

EI

π

β βπ π

β

= − − + + +

−

=

� Buckling length (in plane) = half the span

� Buckling out of plane also possible!

� Snap-through problems Courses on CIE5144 Stabilityand CIE5142 Nonlinear FEM

Assignment Arch-1

Hans Welleman 64

� Find the horizontal component H of the force in the arch

� Find the maximum moment in the arch

( in terms of F and l )25 7

max128 128;

FlH M Fl

f= =

Assignment Arch-2

Hans Welleman 65

3 2

1 2

1 2

1 2

:

100 10 kNm ;

15 m; 25 m;

3 m; 5 m;

150 kN; 30 kN;

( ) sin 0ii i i i

i

Given

EI

l l

f f

F F

xz x f x l

l

π

= ×

= =

= =

= =

= − < <

� Find the moment distribution for a rigid connection betweenthe arches at the midspan support

� Find the moment distribution for a hinged connection betweenthe arches at the midspan support

� Discuss the results