Hans Welleman 1

Slender Structures

Load carrying principles

Basic systems:

• Serial

• Parallel

v2019-1

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

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Learning objectives

Extend the technique for basic models to systems

Find the ODE for a specific system 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

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Basic Cases

Second order DE

Extension

Shear

Torsion

Cable

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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

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Combined Spring Systems (1 dof)

Serial springs

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1 2

1 1 2 2

1 21 2

; ;

N N F

k u k u F

F Fu u

k k

= =

∆ = ∆ =

∆ = ∆ =

1 21 2

1 2

1 2

1 2

1 1 1

e

e

F Fu u u

k k

u F Fk k k

k kk

k k

= ∆ + ∆ = +

= + =

=+

Combined Spring System (1 dof)

Parallel springs

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1 2

1 2

u u u

N N F

∆ = ∆ =

+ =

( )1 1 2 2

1 2

1 2

e

e

k u k u F

k k u F

k u F

k k k

∆ + ∆ =

+ =

=

= +

constrained rotation

Result

Serial

– Load bearing capacity is equal to weakest element (bothelements take the same load)

– Total deformation is summation of each element

– Equivalent stiffness :

Parallel

– Total load carrying capacity is summation of each element

– Both elements have the same deformation

– Equivalent stiffness :

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1 2

1 1 1

ek k k= +

1 2ek k k= +

Slightly more advanced ….

1 22

1

1

force equilibrium

N N F bN F

moment equilibrium l

N l Fb

+ = = −

=

2 1 1 21 2

1 2

2 1 2 1 2

1 2

2 1 2 1 2 1 22 2

1 2 1 2

22 21 2

2 2 2 22 1 1 2

or

( )

( )

1with : e

e

u u u uu u a u u b

l l

Fk l b kFb

k lk k lk kFbu a

lk k l

Fblk Fk a l b Fabk k al k bl k ab k abu F

l k k l k k

k k la bu F F k

kl k l k a k b k

∆ − ∆ ∆ − ∆ = ∆ + = ∆ +

−−

= +

+ − − + − −= =

= + = = +

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( ) ( )

1 21

1 1 1 2

122

2 2 1 2

N FbkFbu

k lk lk k

F l b Fk l bNu

k lk lk k

∆ = = =

− −∆ = = =

Non-dimensional relations1

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Study this from the notes !

1T.J. Folkerts, A more general form for parallel springs, American Journal of Physics, 70(5):493-494, 2002

Application

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b = spring length in deformed state

l = additional cables (no stress)

a = chord which will be cut

What will happen to the block depending upon the values of a,b

and l.

Source : Johan Blaauwendraad, BD magazine 2, nov 1998.

neglect unloaded spring length

Combine basic cases

Parallel system

– Bending and cable

– Bending and shear

Special “serial” system

– Bending and shear, Timoshenko Beam

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System with cable + bending

Same deflection, therefore P-system

Capacity is sum of elements

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2

2

d

dcable

wH q

x− =

4

4

d

dbending

wEI q

x=

4 2

4 2

d d

d dbending cable

w wEI H q q q

x x− = + =

+

A simple example

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Influence of cable force H on deflection at midspan

General solution ?

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4 2

4 2

4 22 2

4 2

d d( ) ( ) sin

d d

sind d

with :d d

b c o

o

w w xEI H q x q x q

x x l

xq

w w Hl

x x EI EI

π

π

α α

− = + =

− = =

wh(x) = ………………………………………

wp(x) = ………………………………………

reduction factor

Result

only bending

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4

2 4

2

1sin

1

oq l xw

Hl EI l

EI

π

π

π

= ⋅

+

4

4sino

bending only

q l xw

EI l

π

π− =

reduction

factor

2

2

H

EI

l

π

2

2

1

1bending onlyw w

Hl

EIπ

−= ⋅

+

Parallel system cable + bending

Suspension Bridge

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Clifton Suspension Bridge (1864) crossing the Avon at Bristol.

Designed by Isambard Kingdom Brunel [source: The Telegraph.co.uk]

Model

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Same deflection, therefore P-system

Capacity is sum of elements

Assumption

Self weight p is taken by cable

only during construction

After completion total load p

and q is taken by the system

2

2

d

d

zH p

x− =

4 2

4 2

( )d w d z wEI H q p

dx dx

+− = +

4 2

4 2

d w d wEI H q

dx dx− =

Some Math

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4 2

4 2

d w d wEI H q

dx dx− =

General solution

Homogeneous solution

Particular solution, assume constant q

I found as an answer

wh(x) = ………………………………………

wp(x) = ………………………………………

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Examples

Wheel load on cables with bending stiffness,

Constrained bending with tension,

Offshore risers,

High rise buildings

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1A.L. Bouma, Mechanica van Constructies, Elasto-static van slanke structuren, VSSD, ISBN 9789040712784, 2000

1

Wheel load

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Model and boundary conditions?

Influence of H on deflection?

Influence of dimension of wheel?

Constrained bending with tension

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Model and boundary conditions?

Influence of H on deflection?

Influence of H on moment ditribution?

Riser

Offshore, pipe from well

Deep water

In tidal water with currents

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6

11 2

3

300 m

12 10 N

3.6 m

45 mm

2 10 N/m

1 m/s (1.9 knots)

1.8 10 N/m

l

H

Diam

thickness

E

current

q

=

= ×

=

=

= ×

=

= ×

334.4 10 N/m (submerged tube weight)p = ×

Model

Assume constant H

Define boundary conditions

Find deformed position h answer 6.75 m

What if H is not constant …

Let’s use MAPLE ….

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Result

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linear H

constant H

High Rise

Shear frame combined with bending element

(elevator or staircase tower)

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Model?

Assume rigid links

Same deflection, therefore P-system

Summation of capacity

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Solve this problem

ODE

Boundary Conditions

Solution and interpretation

Let’s use MAPLE ….

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Displacement w(x)

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primarily bending

shear

Bending

moment M(x)

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total moment

moment in bending element

change of sign!!

Shear V(x)

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total shear Vtot

VbVs

Help!!

Remarkable !

??

Results

Fight between shear and bending

Concentrated force in the top link

All shear load to foundation of the tower

Change of sign in moment distribution

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Beams with bending and shear

deformation, Timoshenko beam

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Displacement w(x) is result of:

Deformation due to shear

Deformation due to bending

Therefore “looks like” a serial system

'

'

shear

bending

w

w

γ

ϕ

=

= −

+d

d

w

xγ ϕ= −

Combined system

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Constitutive relation

Kinematic relation

Equilibrium

d(1)

d

d

d

w

x

x

γ ϕ

ϕκ

= +

=

d

d

d

d

eff eff

wV GA GA

x

M EI EIx

γ ϕ

ϕκ

= = +

= =

d

d

d

d

Vq

x

MV

x

= −

=

Static determinate beams

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2

2

2

2

d

d

d d d 1 d

d d d d

d

d

with: eff

w

x

w V M

x x x k x EI

w q M

x k EI

k GA

γ ϕ

γ ϕ

= − ⇔

= − = −

= − −

=

Rewrite:- use KR and CR for V

- use vertical EQ

k Euler Bernoulli beam

EI Shear beam

→ ∞

→ ∞

Example

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2

2

d

d

w q M

x k EI= − −

Moment distribution ?2

2

d( )

d

Mq x

x= −

Result

This makes sense ….

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22

2 2

2 4

2 4

sin sind

d

sin sin

o o

o o

x xq q l

w l l

kx EI

q l q lx xw

l lk EI

π π

π

π π

π π

= − −

= +

shear bending

Check this out

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use MAPLE !

What about static

indeterminate

systems ??

Take a closer look …

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Lets check system again ..

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( , ) ( )

( , ) ( )

u x z z x

w x z w x

ϕ=

=

displacements and shear deformation :

very nice !

theory of

elasticity

1 12 2

d d d2 2

d d dxz

u w w

z x xγ ε ϕ

= = + = +

Differential equations

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2

2

d dand

d d

d d(2)

d d

eff eff

eff

V wq V GA GA

x x

wGA q

x x

γ ϕ

ϕ

= − = = +

+ = −

2

2

d

d dand

d d

d

d d0 (1)

dd

eff eff

eff

wV GA GA

M xV

xM EI

x

wEI GA

xx

γ ϕ

ϕ

ϕϕ

= = + =

=

− + =

unknown displacement

fields:

w(x) and ϕ(x)

System of coupled DE

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2

2

2

2

d d0 (1)

d d

d d(2)

d d

eff

eff

wEI GA

x x

wGA q

x x

ϕϕ

ϕ

− + =

+ = −

What about the

boundary conditions ?

Example

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l = 8 m; q = 10 kN/m

GAeff = 500..50000 kN/m

EI = 10000 kNm2

Conclusion?

MAPLE

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> restart;

Aug 2017, © Hans Welleman

Timoshenko beam, shear and bending, example during lecture

> DV1:=EI*diff(phi(x),x$2)-GA*(diff(w(x),x)+phi(x))=0;

> DV2:=GA*(diff(w(x),x$2)+diff(phi(x),x))=-q;

> sol1:=dsolve({DV1,DV2},{w(x),phi(x)}): assign(sol1):

> w:=(w(x)); phi:=(phi(x));

standard relations for rotation, shear deformation and curvature:

> Gamma:=diff(w,x)+phi: kappa:=diff(phi,x):

> alpha_bending:=phi: alpha_shear:=diff(w,x):

sectional forces:

> V:=GA*Gamma: M:=EI*kappa:

boundary condition, clamped at the left, simply supported at the right:

> x:=0: eq1:=w=0: eq2:=phi=0:

> x:=L: eq3:=w=0: eq4:=M=0:

> sol2:=solve({eq1,eq2,eq3,eq4},{_C1,_C2,_C3,_C4}): assign(sol2); x:='x':

> L:=10: q:=8: EI:=10000: GA:=500:

> plot(-w,x=0..L,title="deflection w(x)");

> plot(-M,x=0..L,title="M");

> x:=0: evalf(M); evalf((-1/8)*q*L^2);

> x:='x': plot([Gamma,alpha_shear,alpha_bending],x=0..L,title="gamma,

dw/dx_shear,

dw/dx_bending",legend=["gamma","dw/dx_shear","dw/dx_bending"]);

Numerical verification

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COORDINATES 101

GENE 101 NODE 1 X 0.0 STEP 0.1 Y 0.0 STEP 0.0 Z 0.0

STEP 0.0

ELEMENTS 100

GENE 100 BEAM 1 NNODE 2 1 STEP 1 2 STEP 1

BEAM 1

1 E 2.0 IZ 5000 AX 100 AY 500 rho 0 NU 0.0 FREE

ELLOAD 100

GENE 100 STARTELE 1 STEP 1 DIRE YL FIXED Q 8

SUPPORTS 4

NODE 1 X FIXED

NODE 1 Z FIXED

NODE 1 PHIY FIXED

NODE 101 Z FIXED

EXECUTE 1

STATIC LINEAR

PRINT 4

DISPLACEMENTS ALL

SUPPORTREACTIONS ALL

LOADING ALL

STRESSES ALL

end

+-------------------------------------------------------------------+

| CALCULATION RESULTS |

+-------------------------------------------------------------------+

---------------------------------------------------------------------

DISPLACEMENTS

---------------------------------------------------------------------

dof/node-dir displacements

---------------------------------------------------------------------

NODE 51 X +0.000000e+000

NODE 51 Z +2.651042e-001

NODE 51 PHIY +9.895833e-003

---------------------------------------------------------------------

SUPPORT REACTIONS

---------------------------------------------------------------------

dof/node-dir support reactions

---------------------------------------------------------------------

NODE 1 X +0.000000e+000

NODE 1 Z -4.625000e+001

NODE 1 PHIY +6.250000e+001

NODE 101 Z -3.375000e+001

---------------------------------------------------------------------