1

2.

The number of unknowns a1, a2, a3, a4 equals the number of degrees of freedom of the element

We have assumed that displacement u at coordinate x is described by a third order polynomial function:

P

Why?

1.

2

Transverse displacement u along the beam element is assumed to have the form of 3 rd order polynomial:

This function must satisfy boundary conditions:

bending moment M

shear force F

3

The relation between nodal forces and nodal displacements:

1D beam element stiffness matrix

x =

4

The relation between nodal forces and nodal displacements:

x =

Shape function is defined on nodal displacements so once we know nodal displacements we can calculate displacement anywhere along the element

5

L= 500

D= 10

I= 490.87

E= 2.00E+05

beams A2.xls

6

F= 1000N

600mm

200mm

y

x1 2

3 4

k

Element 1

Element 2

7

k

k

-k

-k

8

9

10

11

Fixed support Fixed

supportRigid connection

12

Fixed support

Fixed support

Hinge connection

Fixed support

Fixed support

F1 ? F2 ?

F

1 2 3 4

13

y

1 2 3 4Element 1 Element 2

Fixed

support Hinge

supportHinge joint

Moment100Nm

14

Beam Beam

Simplify?

1 2 3

4

15

Beam Truss

Simplify?

1 2 3

4

16

Truss

Simplify?

Truss

1 2 3

4

17

1D Beam

1 D Beam

Truss

What is wrong?

1 2

3 4

18

1D Beam

Truss

What is wrong?

1 2

3

19

Beam

Beam

What is wrong?

1 2

3 4

20beam 011A.SLDPRT

fixed immovable

5mm thickness 20Nm

21

Solid elements

Shell elements

Beam elements

beam 011A.SLDPRT

22beam 011B.SLDPRT

23beam 012.SLDPRT