Finite Element Analysis Structurs

LECTURE 4 The main topics in this session will include the following:     

Beam element stiffness matrix Beam analysis using the direct stiffness method Formal procedure for stiffness matrix determination 2D plane frame element 3D general beam in a general-purpose FE program

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

4.1

Familiar with FE formulation of a plane frame element Able to develop the stiffness matrix for a beam and a plane frame structure Able to appreciate the difference in finite element formulation of a truss element, a simple beam, a 2D plane beam, and a 3D beam element.

INTRODUCTION

We begin this session by developing the stiffness matrix for the bending of a beam element, the most common of all structural elements as evidenced by its application in buildings, bridges, towers, and many other structures. The beam element is considered to be straight and to have constant cross-sectional area. We will first derive the beam element stiffness matrix by using the principles developed for simple beam theory. We will then present simple examples to illustrate the solution of beam problems by direct stiffness method. The solution of a beam problem shows that the degrees of freedom associated with a node are a transverse displacement and a rotation. We will include the nodal shear forces and bending moments as part of the total solution. We begin with a plane beam element that can resist only in-plane bending and transverse shear force. The element requires only four DoF (two DoF per node) and will be called a simple plane beam element. A plane beam element that also resists axial force requires two additional DoF. It will be described later and will be called a 2D beam element. Finally, a beam element in space that resists all components of nodal force and moment requires six DoF per node and will be called a 3D beam element. We will discuss this element last. 4.2

BEAM STIFFNESS

In this section, we will derive the stiffness matrix for a simple plane beam element. A beam is characterised as a long, slender structural member subjected to transverse loading that produces bending effects as opposed to only axial effects for the truss

2 element. This bending deformation is measured as a transverse displacement and a rotation. Hence, the degrees of freedom considered per node are a transverse displacement and a rotation (as opposed to only an axial displacement for the truss element). Consider the beam element shown in Figure 4.1. The beam is prismatic, with elastic modulus E and centroidal moment of inertia I of its cross-sectional area. The beam is of length L with axial coordinate x and transverse coordinate y. Nodal degrees of freedom (DoF) consist of lateral translations d1y and d2y and rotations 1 and 2 about the z axis (normal to the paper). The nodal forces associated with the DoF are given by f1y and f2y and the bending moments by m1 and m2 as shown. We initially neglect all axial effects. At all nodes, the following sign conventions are used: 1. 2. 3. 4.

Moments are positive in the counterclockwise direction, Rotations are positive in the counterclockwise direction. Forces are positive in the positive y direction. Displacements are positive in the positive y direction.

y

1, m1

2, m2 E, I

1

2

x

L f1y, d1y

Figure 4.1

f2y, d2y

Simple plane beam element with its nodal DoF and nodal loads associated with the DoF

We will now review the standard beam deflection and rotation formulas based on simple beam theory in application to a cantilever beam shown in Figure 4.2. These formulas will be used for derivation of the beam stiffness matrix later on.

3

Figure 4.2

Cantilever beam element with nodal loads and associated nodal deformations

Flexural stress in the beam is computed as

x 

My I

(4.1)

Deflections and rotations due to an applied force F can be calculated by the formulas:

F L3 dy  3EI

F L2 and z  2 EI

(4.2)

Deflections and rotations due to an applied nodal moment M can be calculated by the formulas:

M L2 dy  2 EI

and z 

ML EI

(4.3)

We now want to relate the element nodal forces to the nodal displacements shown in Figure 4.1 for a plane beam element. This relationship will yield the beam element stiffness matrix. That is, we want to find a matrix  k  such that

 f1 y   k11 k12  m  k  1   21 k22    f 2 y   k31 k32  m2   k41 k42

k13 k23 k33 k43

k14  k24  k34   k44 

 d1 y     1   d 2 y   2 

(4.4)

4 To derive the elements of the stiffness matrix  k  , we apply a unit displacement to only one degree of freedom (say, d1 y  1, 1  0, d2 y  0, and 2  0 in Figure 4.1):

 f1 y   k11 k12  m  k  1   21 k22    f 2 y   k31 k32  m2   k41 k42

k13 k23 k33 k43

k14  k24  k34   k44 

1    0    0  0

(4.5)

Multiplying the two matrices in the right-hand side of Eq. (4.5), we get

k11  f1 y

k21  m1 k31  f 2 y

k41  m2

(4.6)

Equations (4.6) contain all elements in the first column of matrix  k  . In addition, they show that these elements, k11 , k21 , k31 , k41 , are the values of the nodal forces and moments that must be applied to sustain a deformation state in which the first DoF has unit value and all other DoF are zero. In a similar manner, the second column in  k  represents the value of forces and moments required to maintain the displacement state 1  0 and all other nodal displacement components equal to zero. We should now have a better understanding of the meaning of stiffness coefficients. We obtain the deflection equations for node 1 by stating that end displacements and rotations produced by force k11 and bending moment k21 are added together to produce unit deflection and zero rotation at node 1 and using standard beam deflection formulas Eqs. (4.2) and (4.3):

f1 y L3

m1 L2 k11 L3 k21L2 d1 y  1     3EI 2 EI 3EI 2 EI f1 y L2 m1 L k11 L2 k21 L 1  0     2 EI EI 2 EI EI

(4.7)

Solving Eqs. (4.7) for two unknowns k11 and k21 results in

k11 

12 EI 6 EI and k21  2 3 L L

(4.8)

The last two coefficients are determined from equations of equilibrium and yield k31 and k41 when k11 and k21 are known:

5

 (forces)  0  (moments)

0  k11  k31

y

node 2

k31  

0

0  k21  k41  k11 L

(4.9)

12 EI 6 EI and k  41 L3 L2

Assuming the remaining three deformations states with a unit deformation in the direction of the degrees of freedom such as (d1 y  0, 1  1, d2 y  0, and 2  0) , (d1 y  0, 1  0, d2 y  1, and 2  0) and (d1 y  0, 1  0, d2 y  0, and 2  1) will result

in the element stiffness matrix:

6 L 12 6 L   12  2 2  EI  6 L 4 L 6 L 2 L   k   3 12 6 L 12 6 L  L  2 2   6 L 2 L 6 L 4 L 

(4.10)

which relates the nodal forces to the nodal displacements in the simple plane beam element:

6 L 12 6 L   d1 y   f1 y   12 m    2 2    1  EI  6 L 4 L 6 L 2 L   1    3    f 2 y  L  12 6 L 12 6 L  d 2 y   2 2   m2   6 L 2 L 6 L 4 L   2 

(4.11)

Equation (4.11) indicates that  k  relates transverse forces and bending moments to transverse displacements and rotations, whereas axial effects have been neglected.

4.3

EXAMPLE OF ASSEMBLAGE OF BEAM STIFFNESS MATRICES

Consider the beam shown below. Assume EI to be constant throughout the beam. A force of 1000 N and a moment of 1000 N-m are applied to the beam at mid-span. The left end is a fixed support and the right end is a pin support. y 1000 N-m 1 1 L

2

2 1000 N

L

3

x

6 First, we discretise the beam into two elements with nodes 1-3. We must include a node at midlength because applied force and moment exist at midlength and, at this time, loads are assumed to be applied only at nodes. Using Eq. (4.10), we find that the global stiffness matrices for the two elements are now given by d1 y

1

d2 y

2

6 L 12 6 L   12  2 2  EI  6 L 4 L 6 L 2 L   k12   3 12 6 L 12 6 L  L  2 2   6 L 2 L 6 L 4 L  d2 y

2

d3 y

(4.12)

3

6 L 12 6 L   12  2 2  EI  6 L 4 L 6 L 2 L   k23   3  12 6 L 12 6 L  L  2 2   6 L 2 L 6 L 4 L 

(4.13)

where the degrees of freedom associated with each beam element are indicated by the usual labels above the columns. Note that here the local coordinate axes for each element coincide with the global x and y axes of the whole beam. The total stiffness matrix can now be assembled for the beam by using the direct stiffness method. When the global stiffness matrix of the beam has been assembled, the external global nodal forces are related to the global nodal displacements. Using the direct stiffness method and Eqs. (4.12) and (4.13), the global stiffness equations for the beam are thus given by

6L 12 6L 0 0   d1 y   F1 y   12 M   6 L 4 L2 2 6 L 2L 0 0   1   1   F2 y  EI  12 6 L 12  12 6 L  6 L 12 6 L  d 2 y    3   2 2 2 2   M 2  L  6 L 2 L 6 L  6 L 4 L  4 L 6 L 2 L   2   F3 y   0 0 12 6 L 12 6 L   d3 y       0 6L 2 L2 6 L 4 L2   3   0  M 3 

(4.14)

Now considering the boundary conditions of the fixed support at node 1 and the hinge support at node 3, we have

1  0 d1 y  0 d3 y  0

(4.15)

7 This leaves us with only the third, fourth, and sixth equations corresponding to the rows with unknown degrees of freedom and using Eqs. (4.15), we obtain

1000   24 0   EI  2  1000   3  0 8L  0  L 6 L 2 L2   

6 L  d 2 y    2 L2   2  4 L2   3 

(4.16)

where F2 y  1000 N , M 2  1000 Nm , and M 3  0 have been substituted into the reduced set of equations. Equations (4.16) can now be solved simultaneously for the unknown nodal displacement d2y and the unknown nodal rotations 2 and 3. 4.4

FORMAL PROCEDURE

The “direct stiffness method” can produce a stiffness matrix only for simple elements, where formulas from mechanics of materials provide relations between nodal displacements and associated nodal loads. For most elements, a general formula for [k] must be used instead. We now take a first look at this formula, and manipulations it requires, by applying it to the truss and beam elements. The general formula is

 k     B  E  B dV T

(4.17)

where  B  is the strain-displacement matrix,  E  is the material property matrix, and dV is an increment of the element volume V. For a truss element, it is easy to demonstrate that because the axial strain is calculated as

d d L d 2  d1    1  2 , or in matrix form L L L L  1 1   d1   x      L L  d 2 

x 

(4.18)

the strain-displacement matrix [B] is the 1 by 2row vector

 B  

1  L

1 L 

(4.19)

Finally, for the truss problem, matrix [E] is simply the elastic modulus E, a scalar, and dV is A dx. Equation (4.17) becomes

1/ L 

 k     1/ L  E  L

0

1   L

1 AE  1 Adx   L L  1

1  1 

(4.20)

8 which agrees with the stiffness matrix for the truss element derived earlier. The special form of Eq. (4.17) applicable to a beam element is L

 k     B EI  B dx T

(4.21)

0

where strain-displacement matrix [B] is the 1 by 4 row vector

 B  

6 12 x  3 2 L  L



4 6x  L L2

6 12 x  L2 L3



2 6x   L L2 

(4.22)

After substitution of Eq. (4.22) into Eq. (4.21), and rather tedious multiplication and integration, Eq. (4.10) again results.

4.5

STRESS CALCULATION

Flexural stress is computed as  x  My / I , and bending moment, M is computed form curvature of the beam element, which in turn depends on nodal displacements d  :

M  EI

d 2v  EI  B d  dx 2

(4.23)

Equation (4.23) shows that bending moment M caused by displacements d  varies linearly with x in each element.

4.6

2D BEAM ELEMENT

A 2D beam element might also be termed a plane frame element. It is a combination of a truss element and a simple plane beam element. It resists axial stretching, transverse shear force, and bending in one plane. By combination of Eqs. (4.20) and (4.10), the stiffness matrix of a 2D beam element that lies along the x axis is

9

0  AE / L  0 12 EI / L3   0 6 EI / L2 ˆ k       AE / L 0   0 12 EI / L3  6 EI / L2  0

0 6 EI / L2 4 EI / L 0 6 EI / L2 2 EI / L

 AE / L 0 0 12 EI / L3 0 6 EI / L2 AE / L 0 0 0

12 EI / L3 6 EI / L2

      6 EI / L2   4 EI / L  0 6 EI / L2 2 EI / L 0

d1 x d1 y

1 d2 x

(4.24)

d2 y

2

where the symbols on the right are appended to show DoF on which the stiffness matrix operates. Note that at node 1 and at node 2 the element has three DoF, namely, two displacements and one rotation, for a total of 6 DoF per element.

Figure 4.3

4.7

Deformations of a plane frame element

2D BEAM ELEMENT ARBITRARILY ORIENTED IN A SPACE

Many structures, such as buildings and bridges, are composed of frames and grids with arbitrarily oriented beam elements. We will develop the stiffness matrix for an arbitrarily oriented beam element, thus making it possible to analyse plane frames. We can derive the stiffness matrix for an arbitrarily oriented beam element in a manner similar to that used for the truss element. The local axes xˆ and yˆ are located along the beam element and transverse to the beam element, respectively, and the global axes x and y are located to be convenient for the total structure. Recall that we can relate global displacements to local displacements by using the following relationship

10

 d x  C  S   dˆx     ˆ  d S C y   d y   

(4.25)

In case of a beam element that has three degrees of freedom per node, as discussed in Section 4.6), we now relate the global to the local displacements by

 d1x  C d   S  1y    1   0   d2 x   0 d 2 y   0     2   0

S

0

0

C 0

0 1

0 0

0

0 C

0

0

S

0

0

0

ˆ  0   d1x  ˆ 0 0   d1 y    0 0   ˆ1     S 0   dˆ2 x  C 0   dˆ    2y  0 1   ˆ   2  0

(4.26)

where the transformation matrix [T] has now been expanded to include local axial and shear force effects, as well as principal bending moment effects:

S C 0 0

C S  0 T    0  0   0

0 0

0 0 0 0 1 0 0 C

0 0 0 S

0 0

C 0

S 0

0 0  0  0 0  1 

(4.27)

Substituting [T] from Eq. (4.27) and [k] from Eq. (4.24) into Eq. (3.21), we obtain the general transformed global stiffness matrix for a beam element that includes axial force, shear force, and bending moment effects as follows:

E  L  2 12 I 2  AC  L2 S               Symmetry

k  

12 I    A  2  CS L   12 I AS 2  2 C 2 L



6I S L

6I C L 4I

12 I  12 I  6I      AC 2  2 S 2    A  2  CS  S L L  L     12 I  12 I    6I    A  2  CS   AS 2  2 C 2  C  L  L L      6I 6I S  C 2I  L L  12 I 12 I 6 I    AC 2  2 S 2 S   A  2  CS L L  L   12 I 2 6I  2 AS  2 C  C L L  4 I  (4.28)

11

The analysis of a rigid plane frame can now be undertaken by applying stiffness matrix Eq. (4.28). A rigid plane frame is defined as a series of beam elements rigidly connected to each other. From Eq. (4.28), we observe that the element stiffness of a frame are functions of E, A, L, I, and the angle of orientation  of the element with respect to the global coordinate axes. Note that computer programs for frame analysis often refer to the frame element as a beam element, with the understanding that the program is using the stiffness matrix in Eq. (4.28) for plane frame analysis.

4.8

3D BEAM ELEMENT

A beam element in a general-purpose FE program has three-dimensional capability. For explanation, we introduce global coordinate axes XYZ and let the element lie along a local x axis (Figure 4.4). Local coordinate axes xyz may arbitrarily be oriented in global XYZ space. The local x axis is defined by the coordinates of nodes 1 and 2. The web of the beam lies in the xy plane, which contains nodes 1, 2, and 3. Node 3 is either an extra node or another node of the structure, whose coordinate serve to orient the xy plane in XYZ space. No degrees of freedom of the element are associated with node 3. At node 1 and node 2 the element has six DoF, namely, three displacements and three rotations, for a total of 12 DoF per element.

Figure 4.4 3D beam element arbitrarily oriented in global coordinates XYZ

Figure 4.5 Degrees of freedom of the beam element in global coordinates

Question about support conditions: if a 3D beam element is used to model a cantilever beam along the x axis, what DoF must be supported? Answer: All six of them, including rotations 1x and 2x (see Figure 4.4) to prevent the beam from being free to spin about its own axis (even if no torque is applied).