finite element analysis

LECTURE 3 The main topics in this session will include the following:     

Finite Element formulation of trusses Global and local coordinate systems Transformation of vectors Truss element Example of solution of a truss using Direct Stiffness Method

Learning objectives: At the end of this session, you should be:   

3.1

Familiar with FE formulation of a truss element Able to develop the stiffness matrix for a plane truss Able to compute stresses and axial forces in members of a plane truss

DEFINITION OF A TRUSS

The main objective of this session is to introduce the basic concepts in finite element formulation of trusses. A truss is an engineering structure consisting of straight members connected at their ends by means of bolts, welding, etc. Trusses offer practical solutions to many structural problems in engineering, such as bridges, roofs of buildings, and towers. A plane truss is defined as a truss whose members lie in a single plane. The forces acting on such a truss must also lie in the same plane. Members of a truss are generally considered to be two-force members. This means that internal forces act in equal and opposite directions along the members. A typical truss structure consists of a large number of elements connected together. One of the simplest techniques for solving such structures is based on the method of joints (enforcing equilibrium of forces at each of the nodes) for a statically determinate structure. The statically indeterminate trusses can also be solved using the force or displacement methods. The analysis of the truss may also be obtained by considering each of the individual elements in isolation from the other elements. This procedure will provide the stiffness matrix for an individual element, each of which may be combined to provide the overall (global) stiffness matrix for the structure. The deflections of the nodes and the forces in each of the elements due to applied external loading and restraints can then be determined.

3.2

BASIC CHARACTERSTICS OF A TRUSS ELEMENT

2 A truss element is considered to be a tension and compression element, as shown in Figure 3.1:

Figure 3.1 A truss member subjected to a force F

Recall that the average stresses in any two-force member are given by



F A

(3.1)

The average strain of the member can be expressed by



L L

(3.2)

Over the elastic region, the stress and strain are related by Hooke’s Law,

  E

(3.3)

 AE  F   L  L 

(3.4)

Combining Eqs. (3.1), (3.2), and (3.3)

Note that Eq. (3.4) is similar to the equation of a linear spring, F  k . Therefore, a centrally loaded member of uniform section may be modelled as a spring with an equivalent stiffness

keq  3.3

AE L

(3.5)

LOCAL AND GLOBAL COORDINATE SYSTEMS

In many problems, it is convenient to introduce both local and global coordinates. Local coordinates are always chosen to represent the individual element of the structure. Global coordinates are chosen to be convenient for the whole structure.

3 We choose a fixed global coordinate system, xy (1) to represent the location of each node and to keep track of the orientation of each element, using angles such as ; (2) to apply the constraints and the applied loads in terms of their respective global components; and (3) to represent the solution – that is, the displacement of each node in global directions. A local, or an elemental, coordinate system is used to describe the behaviour of individual elements. The relationship between forces in the local and global coordinate systems is shown in Figure 3.2.

f2 y

y

fˆ2 y

yˆ f1 y fˆ1 y

xˆ

xˆ

2

xˆ

fˆ1x

1

fˆ2 x

f2 x



f1x

x Figure 3.2 systems

Relationship between global

( x, y) and local ( xˆ, yˆ ) coordinates

The global forces are related to the local forces according to the following equations:

f1x  fˆ1x cos   fˆ1 y sin  f1 y  fˆ1x sin   fˆ1 y cos  f 2 x  fˆ2 x cos   fˆ2 y sin 

(3.6)

f 2 y  fˆ2 x sin   fˆ2 y cos  or, in matrix form,

 f   T  fˆ  where

(3.7)

4

 f1x  cos  f   sin   1y   f    f  , T    0  2x    f 2 y   0

 sin  cos  0

0 0 cos

0

sin 

0  0  ,  sin    cos  

 fˆ1x    fˆ ˆf   1 y  ˆ (3.8)  f2 x  ˆ   f 2 y 



 f  and  fˆ  represent the components of forces acting at nodes 1 and 2 with ˆˆ coordinates, respectively. T  is the respect to global xy and the local xy transformation matrix that allows for the transfer of local forces and deformations to their respective global values. In a similar way, the local and global displacements may be related according to the equations

d1x  dˆ1x cos   dˆ1 y sin  d1 y  dˆ1x sin   dˆ1 y cos  d 2 x  dˆ2 x cos   dˆ2 y sin 

(3.9)

d 2 y  dˆ2 x sin   dˆ2 y cos  If we write Eqs. (3.9) in matrix form, we have

d   T dˆ

(3.10)

where

 d1x  d  d    d1 y   2x  d 2 y 

and

 dˆ1x    ˆ  d1 y  dˆ    ˆ d2 x  ˆ  d 2 y 



(3.11)

d  and dˆ represent the displacements of nodes 1 and 2 with respect to global (x,y) and the local ( xˆ, yˆ ) coordinate systems, respectively. The transformation matrix [T] is identical to that shown in Eqs. (3.8). We define the angle  to be positive when measured counterclockwise from 3.4

x to xˆ .

STIFFNESS MATRIX FOR A TRUSS ELEMENT IN LOCAL COORDINATES

We will now consider the derivation of the stiffness matrix for the linear elastic truss element shown in Figure 3.3. The derivation here will be directly applicable to the solution of pin-connected trusses. The bar is subjected to tensile forces T directed along the local axis of the bar and applied at nodes 1 and 2. Here we have introduced

5 two coordinate systems: a local one ( xˆ, yˆ ) with xˆ directed along the length of the bar and a global one ( x, y ) assumed to be best suited with respect to the total structure. The truss element is assumed to have constant cross-sectional area A, modulus of elasticity, E, and initial length, L. The nodal degrees of freedom are local axial displacements represented by dˆ1x and Figure 3.3.

dˆ2 x at the ends of the element as shown in

y xˆ

2 dˆ2 x , fˆ2 x

L

 1 dˆ1x , fˆ1x

x Figure 3.3

Truss element; positive nodal displacements and forces in the local coordinate system

It has been shown in Section 3.2 that a truss member of uniform section may be modelled as a spring with equivalent spring stiffness AE/L. Therefore, the truss element stiffness equation becomes  ˆ   f1x    ˆ  f2x   

Now, because

  AE  1 1   dˆ1x    L  1 1  dˆ   2x 

 fˆ  [kˆ]dˆ , we have from Eq. (3.12),

(3.12)

6

[ kˆ] 

AE  1 L  1

1  1 

(3.13)

Equation (3.13) represents the stiffness matrix for a truss element in local coordinates. For structures composed of more than one element, the global stiffness and force matrices and global equations must be assembled using the direct stiffness method described in Lecture Notes 2: N

 K   [k ](e)

N

and

e 1

where now all local element stiffness matrices element stiffness matrices by Eq. (3.14).

F     f 

(e)

(3.14)

e 1

[kˆ] must be transformed to global

[ k ] before the direct stiffness method is applied as indicated

The following several steps of the finite element method will be identical to the procedure discussed for the spring element. The nodal displacements will be determined by imposing boundary conditions and simultaneously solving a system of equations, F   K d  . Finally, the element forces will be determined by backsubstitution of the displacements into the local stiffness equations for each element.

3.5

GLOBAL STIFFNESS MATRIX

We will now use the transformation relationship Eq. (3.10) to obtain the global stiffness matrix for a truss element. We need the global matrix of each element to assemble the total global stiffness matrix of the structure. The following assumptions are used in deriving the truss element stiffness matrix: The bar cannot sustain shear force; that is, fˆ1 y  0 and fˆ2 y  0 . Any effect of transverse displacement is ignored. No intermediate applied loads. We have already shown that for a truss element in the local coordinate system,  ˆ   f1x     fˆ2 x   

  AE  1 1  dˆ1x    L  1 1  dˆ   2x 

(3.15)

Taking into considerations the assumption (1) and setting these terms equal to zero, the local internal forces and displacements can be related through the stiffness matrix

7

 fˆ1x  1    ˆ  f1 y  AE  0  ˆ   f 2 x  L  1  ˆ  0 f  2 y 

ˆ 0 1 0   d1x  0 0 0   dˆ1 y    0 1 0   dˆ2 x   0 0 0   dˆ   2 y 

(3.16)

In Eq. (3.16), because fˆ1 y  0 and fˆ2 y  0 , rows of zeros corresponding to the row numbers fˆ1 y and fˆ2 y appear in  kˆ  . Using matrix form, Eq. (3.16) can be written as  

 fˆ   kˆ dˆ

(3.17)

Now, using Eqs. (3.7) and (3.10) after substituting for

 fˆ and dˆ in terms of  f  and

d  , we have

1 1 T   f   kˆ T  d 

(3.18)

where T  is the inverse of the transformation matrix T  and is 1

 cos    sin  1 T    0   0

sin  cos  0 0

0 0 cos   sin 

0  0  sin    cos 

(3.19)

Pre-multiplying both sides of Eq. (3.19) by T  , we have

 f   T  kˆ  T  d  1

(3.20)

Equations (3.20) express the relationship between the global element nodal forces

f

and the global nodal displacements d  , where the global stiffness matrix for an element is

k   T  kˆ T 

1

(3.21)

Substituting Eq. (3.8) for T  , we obtain the global element stiffness matrix in explicit form by

8

 C2  AE  CS k    2 L C   CS

CS

C 2

S2 CS

CS C2

S 2

CS

CS   S 2  CS   S 2 

(3.22)

where C  cos and S  sin  . The next few steps involve assembling the elemental stiffness matrices, applying boundary conditions and external loads, solving for displacements, and obtaining other information, such as normal stresses. These steps are best illustrated through an example problem.

3.6

SOLUTION OF A PLANE TRUSS

We will now illustrate the use of equations developed in previous sections, along with the direct stiffness method of assembling the total stiffness matrix and equations, to solve the following plane truss example problem. A plane truss is a structure composed of bar (truss) elements that lie in a common plane and are connected by frictionless pins. The plane truss also must have loads acting only in the common plane. Example. For the two-bar truss shown below, determine the displacement in the y direction of node 1 and the axial force in each element. A force of P = 1000 kN is applied at node 1 in the positive y direction while node 1 settles an amount  = 50 mm in the negative x direction. Let the modulus of elasticity E = 210 GPa and the cross-sectional area A = 6.0 x 10-4 m2 for each element. The lengths of the elements are shown in the figure. 2

x

1 3m

3

2

y

1

P = 1000 kN  = 50 mm

4m

9 3.7

COMPUTATION OF STRESS FOR A TRUSS ELEMENT

We will now consider the determination of the stress in a truss element. For a bar, the local forces are related to the local displacements by the following equation:

  fˆ1x   AE  1   ˆ  L  1 f   2x 

1    dˆ1x      ˆ 1  d 2 x  

(3.23)

The usual definition of axial tensile stress is axial force divided by cross-sectional area. Therefore, axial stress is

fˆ2 x x  A

(3.24)

where fˆ2 x is used because it is the axial force that pulls on the bar as shown in Figure 3.3. By Eq. (3.23),

 dˆ1x   ˆf  AE  1 1    2x L dˆ2 x   

(3.25)

Therefore, combining Eqs. (3.24) and (3.25) yields

x 



E  1 1 dˆ L

(3.26)



Now using the transformation matrix to transfer the local displacements dˆ into global coordinates, it can be shown that the stress in the member may be calculated as

x 

E  C L

S C

 d1x  d   1y  S  d2 x  d 2 y 

(3.27)

where vector d  includes the global displacements at nodes 1 and 2 of the truss element.