Chapter 3 beams and frames (cont)

1 2 3P 2 3 Shear force diagram for the beam Bending moment diagram for the beam By using the beam theory sign conventions, the shear force V and bending moment M diagrams are shown in t

F FINITE ELEMENT METHOD

Dr NGUYỄN HỒNG ÂN

Head Department of Mechanics of Materials & Structures Tel: 0909.48.58.38

Email: anhnguyen@hcmut.edu.vn

I Beams

1 Definition

A beam is a long, slender structural member

displacement rotation).

02)(0

)(0

)()(

0

2

2

x w

dx M d

V dx

dM or dx

wdx Vdx

M dM M

M m

epuilibriu moment

x w dx

dV or dx

x w dV V V

F m

epuilibriu Force

A y

−

−+

=+

−

=

=

−+

−

=



+

2 Simple beam theory.

Assumptions: (A)member cross section is constant

(B)cross section dimension < overall length (L/ t>10)(C) linear variation of stress and strain.(Small deformation theory)

) /

( :

) ( :

) (

: : :

length force

loading d

distribute x

w

moment bending

M

transverse force

Shear V

rotation or

slape

nt displaceme lateral

v



Equilibrium equation of a differential element of the beam

Beam under distributed load

Differential beam element

For constant EI and only nodal forces and moments, equation becomes

Solution of displacement v(x) is of cubic polynomial function of x

) ( )

2 2

2 2

2

x

w dx

v d EI dx

d dx

: radius of deflected curve.

E: modulus of elasticity.

I: principal moment of inertia about Z-axis

4 3

2 2

3 1

)

3 Beam element stiffness formulation (Direct equilibrium approach)

(1) Beam element (No axial effects are considered.)

moment bending

M

force shear

F

rotation nodal

nts displaceme nodal

v v

j i

j i

: :

: ,

: ,





Sign convection of the beam element:

direction y

positive positiv

v F

ckwise counterclo

positive M

,:

,

,:

Simple beam theory sign convection for positive shear forces and moments

(2) Assume displacement function (Without distributing loading, w(x)=0)

4 3

2 2

3 1

2 1

4 3

2 2

3 1 3 4

2 3

) (

) (

) 0 (

) 0 (

a L a L

a j

dx

L dv

a L a L a L a v

L v

a dx

dv

a v v

j i i

+ +

=

=

+ +

)

(A total D O Fs v i i v j j

) 0 (

)

dx

v d EI E D beam basic

satisfy B

j i j

i j

i j

L v

v L

x L

v

v L

2 )

(

v N

N N N d

:

, ,

2 3

3 3

3 2

2 3

3 2

3 2

3 3

1

1 ,

3 2

1

2

1 ,

3 2

1

x L

Lx L

N L

x

x L

N

xL L

x L

x L

N L

L x

x L

N

−

= +

−

=

In matrix form, we have

(3)Element stiffness Matrix and Equations

Relationships between moment, force and displacement from

elementary beam theory are

Using the nodal and beam theory sign convections for shear forces and

bending moments, we obtain

2

2

dx

v d EI

) (

) (

) 0 (

) 0 (

dx

L v d EI

dx

L v d EI

dx

v d EI dx

v d EI

M V M V

M F M

F

f

j j i i

3

3

dx

v d EI

V =

Beam element Beam theory

Hence, the beam element equation relating nodal forces and nodal

j j i i

v v

L L

L L

L L

L L

L L

L L

L EI M

F M F





2 2

2 2

3

4 6

2 6

6 12

6 12

2 6

4 6

6 12 6

N

d

L Lx L

x L

Lx L

x L

dx

N

d

d dx

N d dx

v d d dx

N d dx

v d d dx

N d

dx

dv

d N

v

6 , 12 , 6 , 12 1

) 2 6

( ), 6 12 ( ), 4 6

( ), 6 12 ( 1

, ,

3 3

3

2 2

3 2

2

3

3 3

3 2

2 2

2

−

=

− +

4 Example:

   

j j i i

M F M F

L L

L L L

L L

L

EI K

3

) 2 ( )

1 (

4

612

264

6126

− +

− +

2

2 2

2

2 2

3 3 3 2 2 1 1

4

6 12

2 6 4

4

6 12 6

6 12 12

0 0 2

6 4

0 0 6

12 6

L L

L L L

L

L L

L

L L

L

L L

L EI

M F M F M F

y y y

0 ,

0 ,

: , 0 :v2 =v3 = 3 = loads F1 = −P M1= M2 =

− +

− +

4

6 12

2 6 4

4

6 12 6

6 12 12

0 0 2

6 4

0 0 6

12 6

2

2 2

2

2 2

3 3 3

L L L

L

L L

L

L L

L

L L

L EI

M F F P

y y

The final set of equations is

2 2

2 2 3

8 2 6

2 4 6

6 6 12 0

L L L

L L L

EI P

The transverse displacement at node 1 and rotations at node1 and 2 are

EI

PL EI

PL EI

PL v

4 4

3 12

2 2

1 3

1 = −  =  =

where the minus sign indicates that displacement at node 1 and the positive signs indicate

counterclockwise rotations at node1 and 2

By substituting the known global nodal displacements and rotations into the system equation, we

can determine the global nodal forces The resulting equations are

− +

− +

0 4 3 12 7

4

6 12

2 6 4

4

6 12 6

6 12 12

0 0 2

6 4

0 0 6

12 6

12

2

2 3

2

2 2

2

2 2

3 3 3 2 2 1 1

EI PL

EI PL EI PL

L L

L L L

L

L L

L

L L

L

L L

L EI

M F M F M F

y y y

The global nodal forces and moments are

PL M

P F

M P F

M P

2

1 ,

2

3 ,

0 ,

2

5 ,

0

Local nodal force for each element (used for stress analysis of the entire structure)

2 3

3 3 2

2 ) 2 (

3 3 2 2

2

PL P PL P

y

y

v

v K

m f m f

P

v

v k

1 ) 1 (

2 2 1 1

Free body diagrams for element 1 and 2 are shown as follows.

According the results of the global nodal forces and moments,

the free body diagram for the whole beam is given as shown.

P

2 5

P

1 2 3

P

2 3

Shear force diagram for the beam

Bending moment diagram for the beam

By using the beam theory sign conventions, the shear force V and bending moment M

diagrams are shown in the following figures.

5 Distributed loading

Equivalent force system:

Replace the distributed load by concentrated nodal forces and moments tending to

have the same effect on the beam as the actual distributing load based on the concept

of fixed-end reactions from structural analysis theory

Fixed-end reactionsare those reactions at the ends of an element if the ends of the

element are assumed to be fixed

Beam subjected to a distributed load Fixed-end reactions for the beam

Consider the cantilever beam subjected to the uniform load w Find

the right-end vertical displacement and rotation, and nodal forces

equivalent nodal force system for uniform load w

2 2

3 2

2 1 1

4

6 12

2 6 4

6 12 6

12

 n

 n

L L

L L L

L L

L EI M

F M F

2 6

3

3 2 6

3 2

2 2

2

w

wL EI

L wL

wL

L

L L EI

L

 n

Applying the nodal forces and the boundary

, 2

2 2 2

wL M

2

4 6

6 12 12

2



v L L

L L

EI wL

wL

e

Solving the above equation for the displacements, we have

8 2

6

6 12

2 6

6 12

2 2

2

2 2 3

1 1

wL

wL w

wL L

L L

v L L

L L

EI M

F

e e



The equivalent nodal forces are

2/

12/

2/

2

2 0

wL wL wL

wL F

Hence, the effective global nodal forces are

EI wL L

L

L L L

L L

L EI wL

wL wL wL

M F M F F

e e e e e

6 /

8 / 0 0

4

6 12

2 6 4

6 12 6

12

12 /

2 /

12 / 5

2 /

3 4 2

2 2

3 2

2

2 2 1 1

Thus, the correct global nodal forces

2/

2 0

wL

wL F

12 =

−

By comparing the two equivalent system given in the previous page, we have relationships

among the correct nodal forces, the effective nodal forces and the equivalent nodal forces

Uniaxial element with tension, compression, torsion, and bending capabilities The more general Beam element is often used instead of this element The figure, at the end of this section, defines both element types For some analysis programs, MSC/N4W translates both types to the same element type

Application

Used to model general beam/frame structures

Shape

Line, connecting two nodes A third node can be specified to orient the element Y axis

Element Coordinate System

The element X axis goes from the first node to the second.The element Y axis is perpendicular to the

element X axis It points from the first node toward the orientation (or third) node If you use an

orientation vector, the Y axis points from the first node in the direction of the orientation vector The

element Z axis is determined from the cross product of the element X and Y axes

Third Node, or

Orientation Vector

Plane 1 (XY)

Offset AOffset B

Plane 2 (XY)

Line, connecting two nodes A third node can be specified to orient the element Y axis.

Element Coordinate System

The element X axis goes from the first node to the second The element Y axis is perpendicular to the

element X axis It points from the first node toward the orientation (or third) node If you use an orientation vector, the Y axis points from the first node in the direction of the orientation vector The element Z axis is determined from the cross product of the element X and Y axes

Properties

Area, Moments of Inertia (I1, I2, I12), Torsional Constant, Shear Areas (Y, Z), Nonstructural Mass/Length, Stress Recovery Locations, Neutral Axis Offsets (Nay, Naz, Nby and Nbz) If the beam is tapered, you can specify different properties at each end of the element

Beam Element

Third Node, or Orientation Vector

Plane 1 (XY)

Offset A

Offset B

Plane 2 (XY)

If there are no offsets,both the neutral axis and shear center lie directly between the nodes

Stress Recovery Locations define positions in the elemental YZ plane (element cross-section) where you want the analysis program to calculate stresses

Specifying moments of inertia for Beam (and Bar) elements can sometimes be confusing In MSC/N4W, I1 the moment of inertia about the elemental Z axis It resists bending in the outer Y fibers of the beam It is the moment of inertia in plane 1 Similarly, I2 is the moment of inertia about the elemental Y axis If you are familiar with one of the analysis program conventions, the following table may help you convert to

MSC/N4W's convention

MSC/pal & CDA/Sprint Iww Ivv

ALGOR, mTAB & SAP I3 I2

:

,

min :

:

V X product Cross

axis Z

Element

plane Y

X in lies which

V vector by

ed Deter

plane Y

X Element

B End and

A End between

line with

coincident Always

axis X

Element

e

e e

Element coordinate system ( orientation of a beam element)

Interpretation of Element Output

．Bar element internal forces and moments

II Frames and Grids

1 Rigid plane frame :

(1) Definition

A frame consists of a series of beam elements rigidly connected to each other.

(i) joint angles between elements remain unchanged after deformation

(ii) moment continuity exists at the rigid joint

(iii) element centroids and applied loads lie in the pane of the structure

(2) Two Dimensional Beam element

Vectors transform law in 2D

For the beam element, we have (use the 2nd eq of above relation)

Global stiffness matrix for 2D Beam element

where

Here, the global stiffness matrix for a beam element including shear and

bending effect is given as.

(3) 2D Beam element including axial force effect

Axial force effect

Combined with shear force and bending moment effects

or

and relate the local to the global displacement by

The global stiffness matrix for beam element including axial, shear and bending effects

where

The analysis of a rigid plane frame can be performed by using above stiffness

matrix.

Bending moment of Element:

X Z

Z

Z Z

m1 , 1 m1x, 1x m2x, 2x m2Z, 2Z

1

1 ,

x x

x x

Torsional bar element stiffness matrix

x x

x

L

GJ m

1 1





Fig Nodal and element torque sign conventions The relationship between torque and twist is

We assume the shear loading to go through the shear center of these open cross

sections to prevent twisting of the cross section.

By Combining the torsional effect with the shear and bending effects, the local

stiffness matrix equation for a grid element is written as:

z x

b

b b

b b

b

b b

b b

k L k

Lk k

k L Lk

k L

k k

Lk k

Lk k

2 t

2 2

t t

6 0

12

2 0

6 4

0 0

0

6 0

12 6

0 12

1 1

L GJ k

i X Z

_ X j

S C

C S

S C

TG

0 0 0 0

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 1

L

x x

Assume: q, a, EI = constant

- Determine the displacements of nodes.

- Draw the shear and bending moment diagrams for the frame.

Example

Assume: q, a, EI = constant

- Determine the displacements of nodes.

- Draw the shear and bending moment diagrams for the frame.

Example