slender structures load carrying principles

Hans Welleman 1

Slender Structures

Load carrying principles

Part 1 : Basic cases:

• Extension, Shear, Torsion, Cable

• Bending (Euler)

v2021-1

Content (5 parts)

Basic cases

– Extension, shear, torsion, cable

– Bending (Euler-Bernoulli) …. extra temperature

Combined systems

- Parallel systems

- Special system – Bending (Timoshenko)

Continuously Elastic Supported (basic) Cases

Cable revisit and Arches

Matrix Method

Hans Welleman 2Part

1, In

troduction

Learning trajectory

Order of topics based on increasing complexity of the governing differential equation (!!! not in line with the notes !!!)

Recap of math partly in class but primarily DIY !

Extensive use of MAPLE, so install MAPLE and get involved in it!

MatrixMethod with Python

Hans Welleman 3Part

1, In

troduction

Learning objectives

Understand the outline of the classical displacement

method for finding the ODE’s for basic load cases

Find the general solution for these loadcases and

define the boundary and/or matching conditions for a

specific application

Solve the ODE’s (by hand and MAPLE)

Investigate consequences/limitations of the model

and check results with limit cases

Hans Welleman 4Part

1, In

troduction

Model

(ordinary) Differential Equation – (O)DE

– Displacement field is the unknown to solve

– Boundary conditions

– Matching conditions Hans Welleman 5

ODE

Part

1, In

troduction

General recipe to find the ODE

(classical displacement1 method)

Kinematic relation

Constitutive relation

Equilibrium condition

Hans Welleman 6

DIFFERENTIAL EQUATION (with its general solution)

+ Conditions at boundaries/interfaces

= The Solution

1 counterpart is the Force Method

displacements

deformations

stresses

loads

Part

1, In

troduction

Extension(prismatic)

Hans Welleman 7

“axial stiffness” EA

external load q

internal generalised stress, normal force N …. BC

axial deformation or strain ε

displacement field (longtitudinal) u …. BC

Maeslantkering

Part

1, E

xte

nsio

n

ODE

Fundamental relations

Kinematic relation

Constitutive relation (Hooke)

Equilibrium

Hans Welleman 8

0

dlim

dx

u u

x xε

∆ →

∆= =

∆

N EAε=

0

0

dlim

dx

N q x N N

N N Nq q

x x x∆ →

− + ∆ + + ∆ = ⇔

∆ ∆= − ⇔ = = −

∆ ∆

2

2

d d d

d d d

u N uN EA EA q

x x x= → = = −�

Check:

• direction of displacement field

• same direction of loading

• equilibrium condition in this direction

Part

1, E

xte

nsio

n

Hans Welleman 9

Example

Solve the ODE using parameters

Write down the boundary conditions

q = 5 kN/m

F = 25 kN

l = 10 m

EA = 2500 kN

Part

1, E

xte

nsio

n

I found as an answer

u(x) = ………………………………………

Boundary Conditions:

……………………………………………..

……………………………………………..

……………………………………………..

……………………………………………..

Hans Welleman 10Part

1, E

xte

nsio

n

Plot the results

Hans Welleman 11

Suggestions for a check ?

Maple> restart;

> ODE:=EA*diff(u(x),x$2)=-q;

> u:=rhs(dsolve(ODE,u(x)));

> eps:=diff(u,x); N:=EA*eps;

kinematic or Dirichlet boundary condition at x=0

> x:=0: eq1:=u=0;

dynamic or Neumann boundary condition at x=L

> x:=L; eq2:=-N+F=0;

> sol:=solve({eq1,eq2},{_C1,_C2}); assign(sol);

> x:='x'; u; N;

> L:=10; F:=25; q:=5; EA:=2500;

> _C1; _C2;

> plot(u,x=0..L,title="displacement u(x)");

> plot(N,x=0..L,title="normal force N(x)");

Part

1, E

xte

nsio

n

What if …


1, E

xte

nsio

n

Assignment : Foundation PileDimensions:

l = 20 m

A = 250 × 250 mm

EA = 5×109 N

Pile capacity:

- point bearing, modulus of subgrade reaction

(Beddingsconstante)

c = 16×108 N/m3

- (positive) friction or adhesion:

τ = 3×104 N/m2

Load:

F = 2500 kN

Find the DE to model this pile and the

boundary conditions. Solve the DE and plot the

normal force distribution in the pile. Find the

displacement of the pile tip.


1, E

xte

nsio

n

Shear

Hans Welleman 14

external load q

internal generalised stress, shear force V …. BC

Shear deformation or shear strain γ

displacement field w …. BC

“shear stiffness” k

Vierendeel shear beam, Hamburg

Part

1, S

hear

ODE


Kinematic relation


Equilibrium

Hans Welleman 15

0

dlim

dx

w w

x xγ

∆ →

∆= =

∆

V k γ=

0

0

dlim

dx

V q x V V

V V Vq q

x x x∆ →

− + ∆ + + ∆ = ⇔

∆ ∆= − ⇔ = = −

∆ ∆

2

2

d d d

d d d

w V wV k k q

x x x= → = = −�

Part

1, S

hear

Conclusion …

Basic Extension and Shear result in an identical ODE

We now know all about this ….

What about the “shear stiffness”?


1, S

hear

Examples of shear beams

Hans Welleman 17

Sketch the deformed situation

Proof it!

Part

1, S

hear

More examples of shear beams

Hans Welleman 18

Sketch the

deformed situation

Proof it!

Part

1, S

hear

Shear stiffness for beams1(in bending)

Hans Welleman 19

assumption in this

model:

- constant V

- constant γ

Shear force V is integrated shear stress

distribution ….

discrete model

1 Work, energy methods & influence lines, appendix A, ISBN 978-90-72830-95-1, J.W. Welleman

Part

1, S

hear

Shear stress due to bendingfor a rectangular cross section

parabolic shear stress distribution

so .. no constant shear stress τ

and no constant shear deformation γHans Welleman 20

fibre model

Part

1, S

hear

Both models should generate the

same amount of strain energy E

Hans Welleman 21

( )2

2

1 11 12 2

2 2124

3

discrete model:

d d with:

Fibremodel: ( prereqisuite knowledge )

6( )d d : ( )

2

h

h

discrete

z

fibre z

E V w V x V k

V h zzE b x z with z

G bh

γ γ

ττ

=

=−

= × = × = ×

−= =

Part

1, S

hear

Elaborate …

Hans Welleman 22

( )

2

2

2

2

21

12

22 21

413

1 1

( )d d d

2

6d

6with: and

5

5with: 1,2

6

h

h

h

h

z

z

z

z

zb x z V x

G

V h zbz V

G bh

VV k A bh

GA

GAk GA

τγ

γ

γ γ

ηη

=

=−

=

=−

= ×

× × − = ×

= = × =

= = =

Part

1, S

hear

Cross section is not a plane ….

Hans Welleman 23

1γ

dx

1d dw xγ=

V

discreteEh

Part

1, S

hear

Shear Stiffness for Frames

Hans Welleman 24

Source: https://gallery.autodesk.com/projects/fusion-360-bow-arch-vierendeel-truss-bridge-wip-17012015

Part

1, S

hear

Shear Stiffness k for Frames

Hans Welleman 25

rigid floors general model

Although the frame is loaded

in bending a shear beam

model can be used based on

the global behaviour.

Part

1, S

hear

Model with rigid floors k = ??

Hans Welleman 26

14

:

4

use virtual work

uM H u

h

M Hh

δδγ δ δγ= ∧ =

=

2 3

2

2

: :

240

3 6 6 24

24:

deformation zero rotation at foundation

u Mh Mh Mh Hh EIu h H

h EI EI EI EI h

EIthus k

h

γ γ− + − = = = = =

=

… try this quickly ..

Part

1, S

hear

General model k = ??

Hans Welleman 27

.. try this yourself ..

:

24

k r

answer

kh b

hEI EI

=

+

Part

1, S

hear

Assignment

Find the displacement at midspan C for both cases

Hans Welleman 28

10 kN;

1000 kN;

5 m;

5

F

k

a

n

=

=

=

=

Part

1, S

hear

So we need an extension …

Add a degree of freedom (rotation)

And add an additional equation (moments)

Hans Welleman 29

d

d

dd

d

w

x

MV M V x

x

γ φ= +

= =

NOTE

Not in the notes, all shear beams in the notes

have zero rotation of the cross sections

Part

1, S

hear

Torsion

Hans Welleman 30

external load m

internal generalised stress, torsional moment Mt …. BC

torsional deformation or specific twist θ

rotation around the axis ϕx …. BC

“torsional

stiffness” GItP

art

1, Tors

ion

ODE


Kinematic relation


Equilibrium

Hans Welleman 31

x x

0

dlim

dx x x

ϕ ϕθ

∆ →

∆= =

∆

t tM GI θ=

0

0

dlim

d

t t t

t t t

x

M m x M M

M M Mm m

x x x∆ →

− + ∆ + + ∆ = ⇔

∆ ∆= − = = −

∆ ∆

2x x

2

dd d

d d d

tt t t

MM GI GI m

x x x

ϕ ϕ= → = = −�

Part

1, Tors

ion

Storebaelt Bridge, Danmark

Cable 1load distributed along the projection of the cable

Hans Welleman 32

external load q

internal generalised stress, vertical component V …. BC

Cable slope tanα

Position of the cable z …. BC

“cable stiffness” H

“cable without

elongation” ??!!

Part

1, C

able

ODE


geometrical relation

Equilibrium of Moments

Equilibrium

Hans Welleman 33

0

dtan lim

dx

z z

x xα

∆ →

∆= =

∆

tanV H α=

0

0

dlim

dx

V q x V V

V V Vq q

x x x∆ →

− + ∆ + + ∆ = ⇔

∆ ∆= − ⇔ = = −

∆ ∆

2

2

d d d

d d d

z V zV H H q

x x x= → = = −�

Part

1, C

able

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 34

( )2

22 2 2 dtan 1

d

zT H V H H H

xα

= + = + = +

Part

1, C

able

35

Alternative

Cable : equilibrium method

F

h

a b

A

Bzk

H

H

vA

vB

In all section the horizontale component of the force in the

cable is equal to H! (cable is a string of hinges)

Part

1, C

able

Hans Welleman

36

Create a section ..

a

A

C

h a

a b

⋅

+

kz

H

vA

( )

( )

( ) 0 total structure 0

0 left part 0

v B

v k C

A a b H h F b T

haA a H z T

a b

+ − ⋅ − ⋅ = =

⋅ − ⋅ + = =

+

F

h

a b

A

B

k

FabH z

a b⋅ =

+

zk

Part

1, C

able

Hans Welleman

37

Conclusion

The funicular line under the line AB is

identical to the shape of the bending moment

distribution for a simply supported beam with

identical load.

The distance zk under the line AB is

proportional to the magnitude of the moment

M. The horizontal component H of the cable

force is the scaling factor to be used.

Part

1, C

able

Hans Welleman

Examples(ODE or equilibrium)

Hans Welleman 38

Find the funicular curve and express f in terms of H, q

and l.

Part

1, C

able

39

Horizontal component H

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=

Part

1, C

able

Hans Welleman

Results

Hans Welleman 40

2 21 14 4

2

2 21 14 4

:

1 1

8

8 8 1 8 1

assume ql F

HH F

F

ql l ff

H l

λ

λ λ

λ λ

λ λ

=

= − = −

= = =− −

,H f

F l

H

F

f

l

ql

Fλ =

Part

1, C

able

Bendingin x-z plane

Hans Welleman 41

external load q

internal generalised stresses M, V …. BC

deformation is the curvature κ

displacement field w, ϕ …. BC

“bending stiffness” EIP

art

1, B

endin

g

A lot of parameters …

Hans Welleman 42

( )

d ( ) d( )

d d

u z z

u zz z z

x x

ϕ

ϕε κ

=

= = =

2 2 2

2 2 2

2 2

2 2

2

d d ( )d

( )d d

h h h

h h h

h h

h h

z z z

z z z

z z

z z

M M z N z z A

M E z z A E z A EI

σ

ε κ κ

= = =

=− =− =−

= =

=− =−

= = =

= = =

Part

1, B

endin

g

Equilibrium

Hans Welleman 43

d

d

d

d

equilibrium of vertical forces

Vq

x

equilibrium of moments

MV

x

= −

=

Proof this in a minute!

Part

1, B

endin

g

ODE


Kinematic relation


Equilibrium

Hans Welleman 44

d d; ;

d d

w

x x

ϕϕ κ= − =

M EI κ=

d d;

d d

V Mq V

x x= − =

Part

1, B

endin

g

Result Bending in x-z plane

Hans Welleman 45

2

2

22

2

2

2

2

d;

d

dd

d

d

d;

d

wM EI

x

wEI

xq

x

Mq

x

= −

=

= −

sufficient to solve static

determinate structures

general applicable

Part

1, B

endin

g

Bending - Euler Bernoulli

Mind the coordinate system used!

Basic model for cross sections with one axis

of symmetry

For prismatic beams use:

Hans Welleman 46

4

4

3 2

3 2

d;

d

d d d; ; ;

dd d

wEI q

x

w w wV EI M EI

xx xϕ

=

= − = − = −

Part

1, B

endin

g

Options for Boundary Conditions

kinematic

ϕ and/or w

dynamic

M and or V

Note:

M or ϕ

V or w

Static Indeterminate (SI)

for 3 or more kinematic

BC

Hans Welleman 47

SD

SD

SI

SI

V=0

V=0

V=0

V

V

VV=0 V

Part

1, B

endin

g

Examples

Hans Welleman 48

Find the deflection

line and the force

distribution.

Bending:

Neglect the influence

of possible axial

deformation.

Part

1, B

endin

g

Double Bending

Hans Welleman 49

z

x-axis

z-axis

ϕy

w

u

ϕy

zd

dy

w

xϕ = −

y

x-axis

y-axis

ϕzv

u

ϕzy

d

dz

v

xϕ =

yu zϕ∆ =

zu yϕ∆ = −

cross section at :

( , , ) : ( ), ( ) ( )

x

displacements x y z u x v x and w x

inhomogeneous / unsymmetricalP

art

1, D

ouble

Bendin

g

Complete theory without rotation into principal direction

Kinematic relation for P

Fibre

Model

Horizontal displacement for fibre through P(x,y,z) :

P y

P

( , , )

d( , , )

d

u x y z u z

or

wu x y z u z

x

ϕ= +

= −

z

d

d

y

vy

x

ϕ−

−

d

dy

w

xϕ = −

d

dz

v

xϕ =

z

x-axis

z-axis

ϕy

w

u

ϕy

z

yu zϕ∆ =

y

x-axis

y-axis

ϕzv

ϕzy

u

zu yϕ∆ = −


1, D

ouble

Bendin

g

x-axis

z-axis

x-axis

y-axis

strain

strain

z

y

κz

κy

Strains in cross section

2 2

2 2

d d d( , )

d d d

u v wy z y z

x x xε = − −

P

d d( , , )

d d

v wu x y z u y z

x x= − −

P ( , , )( , )

u x y zy z

xε

∂=

∂

with:2

2

2

2

d; strain of beam axis

d

d; curvature in - plane

d

d; curvature in - plane

d

y

z

u

x

vx y

x

wx z

x

ε

κ

κ

=

= −

= −

y z( , )y z y zε ε κ κ= + +

Hans Welleman 51; ;y zε κ κ

cross sectional deformations:

Part

1, D

ouble

Bendin

g

Strain to Stress

Constitutive relation

for cross section at x

y z

( , ) ( , ) ( , )

( , ) ( , )

y z E y z y z

y z E y z y z

σ ε

σ ε κ κ

=

= + +

Hans Welleman 52

inhomogeneous

Part

1, D

ouble

Bendin

g

Cross Section Forces (N, M )

Static relationsy

z

( , )

( , )

( , )

A

A

A

N y z dA

M y y z dA

M z y z dA

σ

σ

σ

=

=

=


1, D

ouble

Bendin

g

2 2 ; tan ;zy z m

y

MM M M

Mα= + =

plane of loading

Elaborate …( )

( )

( )

y z

y y z

z y z

( , ) ( , )

( , ) ( , )

( , ) ( , )

A A

A A

A A

N y z dA E y z y z dA

M y y z dA E y z y z ydA

M z y z dA E y z y z zdA

σ ε κ κ

σ ε κ κ

σ ε κ κ

= = + +

= = + +

= = + +

y z y y z z

2

y y z y yy y yz z

2

z y z z yz y zz z

( , ) ( , ) ( , )

( , ) ( , ) ( , )

( , ) ( , ) ( , )

A A A

A A A

A A A

N E y z dA E y z ydA E y z zdA EA ES ES

M E y z ydA E y z y dA E y z yzdA ES EI EI

M E y z zdA E y z yzdA E y z z dA ES EI EI

ε κ κ ε κ κ

ε κ κ ε κ κ

ε κ κ ε κ κ

= + + = + +

= + + = + +

= + + = + +

approach with “double-letter” symbols

Hans Welleman 54

Constitutive Relation

=

z

y

zzzyz

yzyyy

zy

z

y

κ

κ

ε

EIEIES

EIEIES

ESESEA

M

M

N

at cross sectional level

INDEPENDENT OF THE

ORIGIN OF THE

COORDINATE SYSTEM

SPECIAL LOCATION OF THE ORIGIN OF THE COORDINATE

SYSTEM TO UNCOUPLE BENDING AND AXIAL LOADING …. NC


1, D

ouble

Bendin

g

Coordinate system taken at the

NORMAL force CENTRE (NC) Bending and axial loading are uncoupled:

=

z

y

zzzy

yzyy

z

y

0

0

00

κ

κ

ε

EIEI

EIEI

EA

M

M

N Axial loading

Bending

y z 0ES ES= =definition NC :

if N = 0 then zero strain ε at the NC and the n.a.

runs through the NC

Hans Welleman 56

A normal force N at the NC will not cause bending

Part

1, D

ouble

Bendin

g

Location NCy

z

NC NCy

NCz

y

z

dAzyE ),(NCNCzzzyyy +=+=

NCzNC

z

NCyNC

y

),(),(

),(

),(),(

),(

zEAESdAzyEzzdAzyE

dAzzyEES

yEAESdAzyEyydAzyE

dAyzyEES

AA

A

AA

A

×+=+×

=×=

×+=+×

=×=

y

NC

zNC

ESy

EA

ESz

EA

=

=

Part

1, D

ouble

Bendin

g

with:

( , ) ;A

EA E y z dA= Hans Welleman 57

Equillibrium

conditions

2

2

2

2

d d0

d d

d d d0 and 0

d d d

d d d0 and 0

d d d

x x

y y y

y y y

z z zz z z

N Nq q

x x

V M Mq V q

x x x

V M Mq V q

x x x

+ = = −

+ = − = = −

+ = − = = −


1, D

ouble

Bendin

g

2 2

2 2

;

; tan ;

y z

zy z m

y

V V V

MM M M

Mα

= +

= + =

plane of loading

Model with ODE’s

d

d

u

xε =

2

2

d

dy

v

xκ = −

2

2

d

dz

w

xκ = −

Kinematics:

Constitutive relations:

y yy yz y yz zy

z zy zz z

0 0

0 :

0

N EA ε

M EI EI κ with EI EI

M EI EI κ

= =

d

dx

Nq

x= −

2

2

d

d

y

y

Mq

x= −

2

2

d

d

zz

Mq

x= −

Equilibrium relations:

subst

ituteaxial loading

bending

2

2

4 4

4 4

4 4

4 4

d

d

d d

d d

d d

d d

x

yy yz y

yz zz z

uEA q

x

v wEI EI q

x x

v wEI EI q

x x

− =

+ =

+ =


1, D

ouble

Bendin

g

Last step : “cleaning-up”

Hans Welleman 60

2

2

4

4 2

4

4 2

d

d

d

d

d

d

x

zz yy y yz yy z

yy

yy zz yz

yz zz y yy zz z

zz

yy zz yz

uEA q

x

EI EI q EI EI qvEI

x EI EI EI

EI EI q EI EI qwEI

x EI EI EI

− =

−=

−

− +=

−

double bending

extension

2 2

with:

( , ) ;

( , ) ; ( , ) ; ( , ) ;

A

yy yz zz

A A A

EA E y z dA

EI E y z y dA EI E y z yzdA EI E y z z dA

=

= = =

Coordinate system through NC

pseudo load

Wrap-up Basic Cases

2nd order DE Extension

Shear

Torsion

Cable

Hans Welleman 61

4th 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

4 4

4 4

4 4

4 4

d

d

d d

d d

inhomogeneous / unsymmetrical d d

d d

yy yz y

zy zz z

wone axis of symmetry EI q

x

v wEI EI q

double bending x x

v wEI EI q

x x

=

+ =

+ =

Part

1, W

rap-u

p B

asic

Cases

Assignment

Hans Welleman 62

21000 kN; 1500 kNm ; 25 kN; 8 kN/m;k EI F q= = = =

Compare force distributions and deflections

Part

1, A

ssig

nm

ent

slender structures load carrying principles

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