Hans Welleman 1

Slender Structures

Load carrying principles

Continuously Elastic Supported (basic) Cases:

• Extension, shear

• Euler-Bernoulli beam (Winkler 1867)

v2019-4

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 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

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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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Extension(prismatic)

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“axial stiffness” EA

external load no

internal generalised stress, normal force N …. BC

axial deformation or strain ε

displacement field (longtitudinal) u …. BC

ODE

Fundamental relations

Kinematic relation

Constitutive relation (Hooke)

Equilibrium

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d

d

u

xε =

N EAε=

d d 0

d

d

N p x N N

Np ku

x

− − + + = ⇔

= =

2

2

d d d

d d d

u N uN EA EA ku

x x x= → = =�

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Example: pull out test

Solve the ODE using parameters

Write down the boundary conditions

concrete

rebar

l not known yet

I found as an answer

u(x) = ………………………………………

Boundary Conditions:

……………………………………………..

……………………………………………..

……………………………………………..

……………………………………………..

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Some values

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2 5 2

3 2 12

10 mm;

400 N/mm ; 2.1 10 N/mm ;

concrete:

30 10 N/mm ;

y s

c c

steel rebar

f E

E k E

φ =

= = ×

= × =

Find some results

for 25 kN load …

Interpretation of results

Slope of u(0) and N(0) intersect x-axis at distance 1/αfrom origin.

1/α is length of the “influence zone”

Find α by experiment: impose F and measure u(0)

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2

(0)e e

e

FF k u k kEA

F

kEA

kk

EA

= = =

=

Shear(prismatic)

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“shear stiffness” GAeff

external load q

internal generalised stress, shear force V …. BC

shear deformation γ

displacement field w …. BC

ODE

2

2

d d d

d d deff eff

w V wN GA GA kw q

x x x= → = = −�

Fundamental relations

Kinematic relation

Constitutive relation (Hooke)

Equilibrium

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d

d

w

xγ =

effV GA γ=

( )d d 0

d

d

V q kw x V V

Vkw q

x

− + − + + =

= −

Veerse Gat 1961

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Photo : Aart Klein, 24 april 1961

Why this shear type of caisson?

Gates are closed after positioning

of all caissons. The walls must

remain parallel in order to operate

the gates.

Result so far

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2 22 2

2 2

2 22 2

2 2

extention (homogeneous):

d d0 0 with:

d d

shear (inhomogeneous):

d dwith:

d deff

eff eff

u u kEA ku u

EAx x

w w q kGA kw q w

GA GAx x

α α

α α

− + = → − = =

− + = → − = − =

1 2

homogeneous solution:

( ) x xhu x C e C e

α α−= +

2

2

1 2

homogeneous solution:

( )

particular solution:

( ) ( )

( ) ( )

( ) sin ( ) sin

x xh

oo p

oo p

ox xo pl l

effl

w x C e C e

qq x q w x

k

qq x q x w x x

k

qq x q w x

GA k

α α

π ππ

−= +

= → =

= → =

= → =+

Assignment : Find improved model for shear beam on elastic foundation

Hint : Adjust for rotation in shear beam

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Beam(prismatic, only bending deformation)

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external load q

internal generalised stress, shear and moment V, M …. BC

bending deformation, curvature κ

rotation and displacement ϕ, w …. BC

k = foundationmodulus [N/m2]

c = modulus of subgrade [N/m3]

(beddingsconstante),

gravel c = 108 N/m3

sand c = 107 N/m3

ODE

4

4

d d d

d d d

V wM EI EI kw q

x x x

ϕ= → = − = −�

Fundamental relations

Kinematic relation

Constitutive relation (Hooke)

Equilibrium

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d d;

d d

w

x x

ϕϕ κ= − =

M EIκ=

( )d d 0

d d;

d d

V q kw x V V

V Mkw q V

x x

− + − + + =

= − =

Solving the ODE

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4

4

44 4

4

d

d

rewrite as:

d4 : 4

d

wEI k w q

x

w q kw with

x EI EIβ β

+ × =

+ = =

move to slide 24 …

or if fan of math continue …

Find homogeneous solution:4

4

4

4 4 4 4

d4 0

d

:

4 0 4

x

ww

x

subst w eλ

β

λ β λ β

+ =

=

+ = = −

… some math …

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( )

( )

14

14

4 4

4 4 4 4

4 2 24 4

4

4

4 ( 1) 4 1

2 1 2 1

2 1 2 1

λ β

λ β λ β

λ λ β β

λ β β

= −

= × − = × −

= = × − = × −

= × − = × −

Complex …

… some background …

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( ) ( )1144

( 2 )

( 2 ) ( / 4 / 2)

1 cos sin

note: plus or min 2 is the same, so:

1 cos( 2 ) sin( 2 )

1 cos( ) sin( )4 2 4 2

four roots for 0,1, 2 en 3

1 1- -1 1-, , ,

2 2 2 2

i

i m

i m i m

i e

m i m e

m me e i

m

i i i i

π

π π

π π π π

π π

π

π π π π

π π π π

+

+ +

− = + =

− = + + + =

− = = = + + +

=

+ + −

… apply this …

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( )14

1,2,3,4

1,2,3,4

1 2

3 4

2 1

12

2

(1 ) (1 )

( 1 ) ( 1 )

i

i i

i i

λ β

λ β

λ β λ β

λ β λ β

= × −

± ± = ×

= × + = × −

= × − + = × − −

Homogeneous solution:

31 2 4( )xx x x

hw x Ae Be Ce Deλλ λ λ= + + +

.. a few more last steps …

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31 2 4( )xx x x

hw x Ae Be Ce Deλλ λ λ= + + +

(1 ) (1 ) (1 ) (1 )( ) i x i x i x i x

hw x Ae Be Ce Deβ β β β+ − − − − += + + +

( ) ( )( ) x ix ix x ix ix

hw x e Ae Be e Ce Deβ β β β β β− − −= + + +

:

cos sin and cos sinix ix

Euler

e x i x e x i xβ ββ β β β−= + = −

what is this complex stuff??

.. almost there …

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{ } { }( )

{ } { }( )

( ) cos sin cos sin

cos sin cos sin

x

h

x

w x e A x i x B x i x

e C x i x D x i x

β

β

β β β β

β β β β−

= + + − +

+ + −

( )

( )

( ) ( ) cos ( )sin

( ) cos ( ) sin

x

h

x

w x e A B x i A B x

e C D x i C D x

β

β

β β

β β−

= + + − +

+ + −

A-B and C-D complex

1 2

3 4

use new constants:

; ( )

; ( )

C A B C i A B real

C C D C i C D real

= + = −

= + = −

complex conjugate - assume:

; ;

2 ; 2

A a ib B a ib

A B a A B ib

= − = +

+ = − = −

.. Okay we are there …

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( ) ( )1 2 3 4( ) cos sin cos sinx x

hw x e C x C x e C x C xβ ββ β β β−= + + +

damping term for x < 0 damping term for x > 0

sinoidal shape (wave)sinoidal shape (wave)

damping or decreased amplitude is governed by β :

4

1 1characteristic length [m]

4

k

EI

β=

Example

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( ) ( )1 2 3 4( ) cos sin cos sinx x

hw x e C x C x e C x C xβ ββ β β β−= + + +

Symmetry

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( )3 4( ) cos sinxw x e C x C xβ β β−= +

only two BC:

(0) 0; (0) ;V Fϕ = = −

Assignment

Solve the example in MAPLE

Use

Find the distributions for

– Displacement

– Rotation

– Moment

– Shear

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414

; 1000 4 ; 10EI Fβ π β= = × =

… one last trick …

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( )3 4( ) cos sinxw x e C x C xβ β β−= +

( )( ) sin cos cos sinxw x A e x xβ ω β ω β−= +

ω

A

4 cosC A ω=

3 sinC A ω=

( )( ) sin cos cos sinxw x e A x A x

β ω β ω β−= +

( )( ) sinxw x Ae xβ β ω−= +( )

math:

sin sin cos cos sina b a b a b+ = +

new integration constants

Why??

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( ) ( )

( ) ( )( )

( ) ( )( )1 14 4

dsin cos

d

sin cos

2 sin cos cos sin

x x

x

x

wAe x Ae x

x

Ae x x

Ae x x

β β

β

β

β β ω β β ω

β β ω β ω

β β ω π β ω π

− −

−

−

= − + + + =

= − + − +

= − + − +

( )

math:

sin sin cos cos sina b a b a b− = −

( )14

d2 sin

d

xwAe x

x

ββ β ω π−= − + −

( )( ) sinxw x Ae xβ β ω−= +

Differentiating becomes

Change sign

Multiply with

Reduce phase with

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2β14

π

( )

( )

( )

( )

14

22 1

22

33 3

43

sin

d2 sin

d

d2 sin

d

d2 2 sin

d

x

x

x

x

w Ae x

wAe x

x

wAe x

x

wAe x

x

β

β

β

β

β ω

β β ω π

β β ω π

β β ω π

−

−

−

−

= +

= − + −

= + −

= − + −“old school”

Study the graphs

…and the notes …

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Classification of beams according

to stiffness ... loadcase 1

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( ) ?w x =

Classification

Wave length of the load

Wave length λ of the beam (no load)

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2l

( )

( )

1 2

3 4

( ) cos sin

cos sin

x

x

w x e C x C x

e C x C x

α

α

β β

β β

−= + +

+

22

πβλ π λ

β= → =

Simplified model

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, 0 , 0

( ) ? ( )

( ) ?bending k bending k

w x w xfind

w x w= =

=

=

, 0 , 0

( ) ? ( )

( ) ?cont reaction EI cont reaction EI

w x w xfind

w x w− = − =

=

=

Influence of k , EI and l

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Find the meaning of this result by studying the extreme values of the

parameters … (also see the assignments in de notes)

Influence of k , EI and l

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Find the meaning of this result by studying the extreme values of the

parameters … (also see the assignments in de notes)

Classification of beams according

to stiffness ... loadcase 2

Also known as Hetényi problemDr. Miklos Hetényi, University of Technical Sciences, Budapest, Hungary, 1924-30; Diploma in Civil Engineering, 1931; Graduate work with H.M. Westerguard, Univ. of Illinois, 1934-35 and with S.P. Timoshenko, Univ. of Michigan, 1935-36; PhD in Eng. Mechanics, 1936

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Model based

on symmetry

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( ) ( )

4

4

1 2 3 4

12

d0

d

( ) cos sin cos sin

0; ; 0;

; 0; 0;

x x

wEI kw

x

w x e C x C x e C x C x

x V F

x l V M

β ββ β β β

ϕ

−

+ =

= + + +

= = − =

= = =

Ok, just solve this ……

Stiffness of slabs

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lβ

12

( )

(0)

w l

w

4

π π

short beams, can be regarded as rigid, neglect bending4

need accurate computations, load on one side has effect on the other end4

long beams, acting force at one end has negligible effect

l

l

l

πβ

πβ π

β π

≤

≤ ≤

≥ on other end

Assignment

Find min and max stress in the continuous support

Find the maximum normal stress in the beam

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rail (beam)

Assignment

Building phases

– A original situation

– B add supports A and B with full load prior to excavation

– C excavate AB under full load

Derive a model and find the moment at B in the beam

Find the moment and shear distribution in terms of qo and l

Find the support reactions at A and B

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4

:

324

assume

kl

EI=

Advanced models:Winkler - Pasternak model

Add shear deformation in the elastic support to create horizontallinkage between the vertical Winkler springs. Stiffnessparameters ks and k will model the elastic support (so-called 2-

parameter model)

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4 2

4 2

d d

d d

!

s

w wEI k k w q

x x

proof this

− + × =