FENET Glasgow Oct2004 EDU Katz

8/3/2019 FENET Glasgow Oct2004 EDU Katz

1/29

Beam Elements

Are they precise ?

Dr.-Ing. Casimir Katz

FENet Technology Workshops,

October 2004, Glasgow, Scotland.


2/29

Cable Analysis

Cable

-H w(x) = p(x)

Solution for uniform load is aquadratic parabola or a sinh

H

p

HP1 P2Cable element

Linear between the nodes

Solution space: polygonal


3/29

Cable Analysis

Solution Space

wh(x) = w1j1(x) + w2j2(x) + ... Solution is exact for:

Single forces

Other loadings:

Minimum of energy => equilibrium

Exact solution is obtained in the nodes!Why ?


4/29

Greens Functions

The influence function (Greens function) for

the deformation of the cable at a point is thecable polygon created by a point load P=1 at

the point of interest.

If the FE-System is able to model this solution

exactly, the solution will be exact for this

value. In all other cases, i.e. if the Greens function

is not within the possible solution space,

one has an approximate solution only.


5/29

What are Beam Elements ?

A 3D Continua with a length >> width / height

Simplification of the solution space(Bernoulli-Hypothesis, persistence of shape)

Simplification for manual analysis

(gravity centre, principal axis, shear etc.)

Superstition:

Beam elements are simple

Beam elements are exact


6/29

Problematic continuous beam

Mechanic is not consistent

Plan view

Moments


7/29

Sections

Normal stress

0 y z

y zx x

u u z y

u u E E z y

x x x x

= +

= = = +


8/29

Sections Shear Stress

V is only valid if the normal

force and the section areconstant along the axis

Z is only valid if the section

is not multiple connected The shear stress is not

necessarily constant along

the width b For biaxial bending or non effective parts of

the section, the equation becomes morecomplex (Swains formula)

V Z

I b =


9/29

An other strategy:

a deformation method

2 2

2 2

:

0

xx y

x

x z

x

x y y x z z

wG zy x

w

G yz x

w wG w G

y z x

B o u n d a r y C o n d i t i o n s

n n

=

= +

= + =

+ =


10/29

Unit Warping

SOFiSTiK AG, 85764 Oberschleiheim, Bruckmannring 38, Tel:089/315-878-0UP (V11.05-21) 24.10.2002

M 1 :X

8.00 6.00 4.00 2.00 0.00 -2.00 -4.00 -6.00 -8.00

Querschnitt Nr 12 VollquerschnittVerwlbung 1 CM = 50000.000 [cm2]

-20525.-12353.

12300.20471.

19815.

-9150.

15582.

24686.

33324.

41220.

13931.

-14264.

-25462.

-32880.

-26721.-16045.

16024.

267

00.

32

859.

254

41.

14243.

-13952.-41244. -33349.

-24712.-15608. 9123.

-19869.

12390.

-12446.

-21149.

-11843.

9088.

14050.

-14131.

-9162.

11781.

21093.


11/29

Shear stress from Mt

M 1 : 68X

Z

8.00 6.00 4.00 2.00 0.00 -2.00 -4.00 -6.00 -8.00 m

Querschnitt Nr 12 Vollquerschnitt

Schub aus MT=1000.00 1 CM = 0.05 [MPa]

0.050.0

6

0.06

0.06

0.06

0.05

0.02

0.05

0.05

0.05

0.03

0.01

0.010.01

0.010.01

0.010.030.06

0.060.06

0.060.06

0.060.06

0.03

0.010.01

0.01

0.01 0.01 0.010.03

0.05

0.05

0.05

0.02

0.040.04

0.040.04

0.040.03

0.02

0.02

0.03

0.040.04

0.040.04

0.040.04

0.04

0.03

0.02

0.02

0.03

-0.04-0.03-0.020.020.030.04

-0.02

-0.03-0.03

0.030.03

0.02

-0.05-0.04-0.030.030.040.05

-0.06

-0.05

-0.04

0.04

0.05

0.06

-0.06

-0.05

-0.04

0.04

0.05

0.06

-0.06

-0.05

-0.04

0.030.040.05

-0.06

-0.05

-0.04

0.030.040.05

-0.06

-0.05

-0.04

0.04

0.05

0.06


12/29

M 1 : 68XYZ

Y 8.00 6.00 4.00 2.00 0.00 -2.00 -4.00 -6.00 -8.00 m

Z

2.0

0

Querschnitt Nr 12 Vollquerschnitt

Schub aus VZ=1000.00 1 CM = 0.10 [MPa]

0.15

0.10

0.03

-0.03

-0.10

-0.15

-0.04

-0.2

0

-0.23

-0.22

-0.17-0

.09

-0.07

-0.05

-0.03

0.020.050.060.08

-0.09-0.05-0.020.020.050.09

-0.08-0.06-0.05-0.02

0.03 0.05 0.07 0.09

0.17

0.22

0.23

0.20

0.04

0.07

0.04

-0.04

-0.07

-0.14

-0.21

-0.23

-0.22

-0.19-0.12

-0.06

0.060.12

0.19

0.22

0.23

0.21

0.14

0.210.220.22

0.220.22

0.21

0.130.170.23

0.23

0.17

0.13

0.210.21

0.200.200.21

0.21

-0.08-0.08-0.08

0.13

0.150.14

-0.14

-0.15

-0.13

0.200.210.21-0.08

-0.07

-0.08

0.200.210.21

0.13

0.150.14

Shear stress from Vz


13/29

Shear deformation Theory of Timoshenko/Marguerre (not exakt!)

Principal Axis of shear deformation

zy

z

y y

VG A

w

x

=

= +

=

z

y

zyz

yzy

z

y

V

V

GAGA

GAGA*

11

11


14/29

Shear in haunched beams

h =100 cm

p = 10 kN/m

L = 10.0 m

h = 50 cm

h =100 cm


15/29

FE-Analysis

tau-xy


16/29

FE-Analysis

tau-xy in vertical section

10.62

57.90

52.10

53.96

0

0

0


17/29

Shear in haunches

sig-x

tau-xySchnitt senkrecht

zur SchwerachseSchnitt senkrecht

zur Referenzachse


18/29

FE-Analysis

tau-nq in perpendicular section

-28.

-53.

-32.


19/29

Shear in haunched beams Example for a shear reduction along the

provisions of design codes Shear force V 40.0 kN

Yielding a faulty shear stress 63.2 kN/m2

Reduction of shear by M/d* tan 5.6 kN Reduced shear stress 54.9 kN/m 2

FE-shear stress 53.0 kN/m 2

Although the engineering approach has a

completely different view, the maximum stress

is quite good, distribution is faulty however.


20/29

FE of a haunched beam


21/29

Reference of forces ? Gravity axis with N + V

Results suited for

design Gravity axis with D + T

Similar as for 2nd

order Theory

Superposition offorces

General reference axis

Superposition offorces if acomposite sectionchanges by in situconcrete ortendons.

N Q

D

T

D

T


22/29

Excentricities at the end points

Interpolation u0 linear, v,w,x,y cubic

Displacements within section

Strains from derivative (position of gravity centre is not constant!)

Principles of the FE-Beam

u u z yu u j z y

i i yi i zi i

j yj j zj j

0

0= + = +

( ) ( )u u z z y y y s z s= + 0

( ) ( )

x o y s z s

y s z s

u

xu z z y y

z y

= = +

+


23/29

Example of a haunched beam

h =100 cm

p = 10 kN/m

L = 10.0 m

h = 50 cm

h =100 cm


24/29

Results for haunched beam

w[mm] Ne[kN] Nm[kN] Mye[kNm] Mym[kNm]

Inclined axis

C-Beam (1 element) 0,397 -80,50 -78,00 -73.58 31,91

C-Beam (8 elements) 0.208 -46,30 -43,80 -94,87 19,17

FE-Beam (1 element) 0,172 -39,80 -37,30 -93,65 22,00

Fe-Beam (8 elements) 0.206 -45,80 -43,30 -95,02 19,14

Horiz. reference axis

FE-Beam (1 element) 0.168 -37,90 -37,90 -93.01 22,52

FE-Beam (8 elements) 0.204 -44,20 -44,20 -94.85 19,10

- 3 7 . 8 8 - 3 7 . 8 8

- 9 3 . 0 1


25/29

The precision of the

FE beam (bending)

19.287615.430119.2876Cent. deflection

41.14741.14741.147End Rotations

281.25281.25281.25Max. Moment

2 Elements1 ElementExact


26/29

Bending precision

The theoretical solution is a polynom of 4th order

The shape functions are cubical splines The highest

symmetric ansatz function is the quadratic parabula

As the influence function for the nodal displacementsare cubic functions, the deformations and rotations in

the nodes are exact. The moments and shear forces are also exact.

The only values which are not are the displacements

between the nodes (which are of less interest) For a buckling analysis there are the same principles,

but know the buckling force (which has a high

importance!) is not ok for a simple element.


27/29

The precision of the

FE beam (Buckling analysis)

52879

53249

53550

-

Euler IV

Numerical

132134 Elements

132193 Elements

133122 Elements

5284816065133121 Element

Euler IV

Exact

Euler II

Numerical

Euler II

Exact


28/29

Warping Torsion and Shape

deformation

> 6. Degrees of freedom per node

More forces and moments:secondary torsional moment Mt2 etc.

Pure bending = 84.3 N/mm2

Warping / 2nd Order Torsion = 136.1 N/mm2

240 kN2.625 kN/m

l = 16 m


29/29

Conclusion

Beam elements are not simple in theory

Beam elements are easy to use

There are a lot of modelling errors possible

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