Fea Chapter155

7/27/2019 Fea Chapter155

1/18

Chapter 4

ANALYSIS OF TRUSSES AND FRAMES

The derivation of element equations for one-dimensional structural elements is considered in

previous chapter. These elements can be used for the analysis of bar type systems like planar

trusses, space trusses, beams, continuous beams, planar frames, grid systems and space

frames. Pin-jointed bar elements are used in the analysis of trusses. In planar truss analysis,

each of the two nodes can have components of displacement parallel to x and y axes. In threedimensional truss analysis, each of the two nodes can have components of displacement

parallel to x, y and z directions. Rigidly jointed bar (beam) elements are used in the analysis

of frames. Thus a frame or a beam element is a bar which can resist not only axial forces also

transverse loads and bending moments. In the analysis of planar frames, each of the two

nodes of an element will have two translational displacement components (parallel to global x

and y axes) and a rotational displacement (in the plane of x y). For a space frame element,

each of the two ends is assumed to have three translational displacement components (parallel

to global x, y and z axes) and three rotational displacement components (one in each of the

three planes xy, yz, zx).

If local axes for a finite element are not parallel to global axes for the whole structure,

rotation-of-axes transformations must be applied to element stiffnesses, nodal displacements,

and nodal loads. Thus, when the elements are assembled, the resulting nodal equilibrium

equations will pertain to the global directions at each node. (It is also possible to apply

translation-of-axes transformations in order to refer all nodal equilibrium equations to the

same origin, but this type of operation is unnecessary.)

4.1 TRUSSES

A truss structure consist only of two-force members. That is, every truss element is in directtension or compression. In a truss, it is required that all loads and reactions are applied only at

the joints, and that all members are connected together at their ends by frictionless pin joints.

Every engineering student has, in a course on statics, analyzed trusses using the method of

joints and the method of sections. These methods, while illustrating the fundamentals of

statics, become tedious when applied to large-scale statically indeterminate truss structures.

Further, joint displacements are not readily obtainable. The finite element method on the other

hand is applicable to statically determinate or indeterminate structures alike. The finite

element method also provides joint deflections. Effects of temperature changes and support

settlements can also be routinely handled.


2/18

56

4.1.2 Plane Trusses

The local coordinate system is the x axis running along the element, since a truss element

simply a two-force member. Consequently, the nodal displacement vector in local coordinates is

{ } T

21 q,qq = (4.1)

The nodal displacement vector in global coordinates is now

{ } T

4321 q,q,q,qq = (4.2)

Figure 4.2 Plane truss element.

Transformation between local and global coordinates can be written as

{ } [ ]{ }qq =

where the transformation matrix [] is given by

[ ]

=

ml00

00ml(4.3)

The element stiffness matrix in global coordinates is given by Eq. which yields

[ ]

=

22

22

22

22

e

ee

mlmmlmlmllml

mlmmlm

lmllml

l

AEk (4.4)

y

x

x

x

q1

q2

q3

q4

q1

q2


3/18

57

Any element load vector, {f}, in the local coordinate system can be transformed into the

global coordinates as

{ } [ ] { }ff T =

For example, the element temperature load in the global coordinates can be obtained as

{ } [ ] { }

=

==

m

l

m

l

TAE1

1TAE

m0

l0

0m

0l

ff eeeeeeeeTT

T (4.5)

The element stresses in global coordinates is given by

{ } [ ][ ][ ]{ }

=

==

4

3

2

1

e

e

4

3

2

1

ee

e

q

q

q

q

mlmlL

E

q

q

q

q

ml00

00ml

L

1

L

1EqBE (4.6)

Now let us obtain consistent mass matrix for a truss element. We have four translations for the

truss element, and the shape function matrix is

[ ] 1 2

1 2

0 0

0 0

N N

N N

=

N (4.7)

where

1 21x x

N and NL L

= = (4.8)

Treating the material density to be constant over the element, we can write the following

formulation for the mass matrix

[ ] [ ] [ ]

1

1 1 2

2 1 2

2

21 1 2

21 1 2

2

0 1 2 22

1 2 2

0

0 0 0

0 0 0

0

0 0

0 0

0 00 0

= =

=

m N Ne e

T

V V

L

e

N

N N NdV dV

N N N

N

N N N

N N NA dx

N N NN N N

(4.9)


4/18

58

The shape functions are given with Eq.4.8. The mass matrix can be found as

[ ]

2 0 1 0

0 2 0 1

1 0 2 06

0 1 0 2

e eA L

=

m

Example 4.1 Consider the four-bar truss shown in Fig. E4.1a. It is given that E = 70 Gpa and

Ae = 200 mm2

for all elements. The system is subjected two loads of 25 kN and 20 kN.

(a)Determine the element stiffness matrix for each element(b)Assemble the structural (global) stiffness matrix [K] for the entire truss.(c)Using the elimination approach, solve for the nodal displacement.(d)Recover the stresses in each element(e)Calculate the reaction forces.

Figure E4.1

Solution:

Lets model the truss with the bar elements.

x

400 mm

300 mm

25 kN

20 kN

y

1

2

3

4

1

23

4

Q1

Q2

Q7

Q6

Q4

Q3

Q5

Q8

25 kN

20 kN x

y


5/18

59

a) It is recommended that a tabular form be used for representing nodal coordinate data and

element information as shown below. The nodal coordinate data is

Node x (mm) y (mm)

1 0 02 400 0

3 400 300

4 0 300

The element connectivity table is

Element 1 2

1 1 2

2 3 2

3 1 34 4 3

Note that the user has a choice in defining element connectivity. For example, the

connectivity of element 2 can be defined as 2-3 instead of 3-2 as above. However,

calculations of the direction cosines will be consistent with adopted connectivity scheme. For

example, the direction cosines of element 3 are obtained as

( ) ( )

( ) ( )

6.08.0

500/0300500/0400

l/yyml/xxl 313313

==

==

==

Similarly, using formulas in Eqs 4.6 and 4.7, together with the nodal coordinate data and

element connectivity information given above, we obtain the direction cosines table:

Element Le l m

1 400 1.0 0.0

2 300 0.0 -1.0

3 500 0.8 0.6

4 400 1.0 0.0

Now, using Eq. 4.13, the element stiffness matrices for the elements can be written as


6/18

60

4 41 2

43

1 0 1 0 0 0 0 0

0 0 0 0 0 1 0 17 10 200 7 10 200

1 0 1 0 0 0 0 0400 300

0 0 0 0 0 1 0 1

0.64 0.48 0.64 0, 48

0.48 0.36 0.48 0.367 10 200

0.64 0.48 0.64 0.48500

0.48

Global serbestlik derecesi

x x x xk k

x xk

= =

=

44

1 0 1 0

0 0 0 07 10 200

1 0 1 0400

0.36 0.48 0.36 0 0 0 0

x xk

=

The global dofs are indicated in the stiffness matrices of elements. These numbering scheme

assist in assembling the various element stiffness matrices.

(b) The structural stiffness matrix [K] is now assembled from the element stiffness matrices.

By adding the element stiffness contributions, noting the element connectivity, we get

[ ]

=

00000000

00.1500.150000

0032.2476.50.20032.476.500.1576.568.220076.568.7

000.2000.20000

000000.1500.15

0032.476.50032.476.5

0076.568.700.1576.568.22

3

10x7K

3

(c)The structural stiffness matrix [K] given above needs to be modified to account for theboundary conditions (Q1 = Q2 = Q4 = Q7 = Q8 = 0). The elimination approach will

be used here. The rows and columns corresponding to dofs 1,2,4,7, and 8, which

correspond to fixed supports, are deleted from the [K] matrix above. The reduced

finite element equations are given as

=

25

0

20

10

Q

Q

Q

32.2476.50

76.568.220

000.15

3

10x7 3

6

5

33

Solution of the above equations yields the displacements

Q3 = 0.571 mm, Q5 = 0.119 mm, Q6 = -0.469 mm,

The nodal displacement vector for the entire structure can therefore be written as

{ } [ ]T00469.0119,00571.000Q = mm

1 2 3 4

1

2

3

4

1 2 5 6

1

2

5

6

7 8 5 6

7

8

5

6

5 6 3 4

5

6

3

4

1 2 3 4 5 6 7 8

1

2

3

4

56

7

8


7/18

61

(d)The stress in each element can now be determined from Eq. , as shown below. Theconnectivity of element 1 is 1 2. Consequently, the nodal displacement vector for

element 1 given by { } T

0571.000q = , and Eq. yields

[ ] (MPa)N/mm925.99

0

571.0

0

0

0101400

10x7 24

1 =

=

The stress in member 2 is given by

[ ] (MPa)N/mm43.109

0

571.0

469.0

119.0

1010300

10x7 24

2 =

=

Following similar steps, we get

3 = -26,068 MPa

4 = 20,825 MPa

(e)The final step is to determine the support reactions. We need to determine the reactionforces along dofs 1, 2, 4, 7, and 8, which correspond to fixed supports. These are

obtained by substituting for {Q} into the original finite element equation

{ } [ ]{ } { }RR FQKR = . In this substitution only those rows of [K] corresponding to thesupport dofs are needed, and {F} ={0} for these dofs. Thus we have

N

0

4165

21887

3128

15814

0

0

469.0

119.0

0

571.0

0

0

00000000

00.1500.150000

000.2000.20000

004,32-5.76-004.325.76

005.76-7.68-015.0-5.7622.68

3

10x7

R

R

R

R

R

3

8

7

4

2

1

=

=


8/18

62

Figure 4.1 Three dimensional truss element.

4.1.1 Three-Dimensional Trusses

The local coordinate system is the x axis running along the element, since a truss element

simply a two-force member. Consequently, the nodal displacement vector in local coordinates

is

{ } T

21 q,qq = (4.7)

The nodal displacement vector in global coordinates is now

{ }

T

654321q,q,q,q,q,qq = (4.8)

Transformation between local and global coordinates can be written as

{ } [ ]{ }qq = (4.9)

where the transformation matrix [] is given by

[ ]

=

xoxoxo

xoxoxo

nml000

000nml(4.10)

or we simply say

y

xz

xq1

q3

q2 q6

q4

q1

q5q2


9/18

63

lox = l mox = m nox = n

[ ]

=

nml000

000nml(4.11)

The element stiffness matrix in global coordinates can be obtained using the formula

[ ] [ ] [ ][ ]= kk T (4.12)

If we substitute Eqs. (3.13) and (4.5) into Eq. (4.6), we obtain

[ ]

2 2

2 2

2 2

2 2

2 2

2 2

ln ln

ln ln

ln ln

ln ln

e e

e

l lm l lm

lm m mn lm m mn

mn n mn nE Ak

l l lm l lm

lm m mn lm m mn

mn n mn n

=

(4.13)

Any element load vector, {f}, in the local coordinate system can be transformed into the

global coordinates as

{ } [ ] { }ffT

= (4.14)

The element stresses in global coordinates is given by

{ } [ ][ ]{ } [ ][ ][ ]{ }qBEqBE == (4.15)

4.2 FRAMES

Frames, like trusses, are skeletal structures composed of slender members. However,

unlike trusses, the members of a frame transmit shear and bending, as well as axial loads.

Therefore, the individual frame elements behave like a beam with superimposed axial load.

The joints of a frame are usually considered to be rigid, althoug pin joints may be found in the

structure. Moments can be transfered from one member to another member to another accross

a rigid joint, but not accross a pin.

4.2.1 SPACE FRAME ELEMENT

The 12 x 12 stiffness matrix given in Eq. (3.60) is with respect to the local xyz

coordinate system. Since the nodal displacements in the local and global systems are relatedby the relation


10/18

64

=

12

11

10

9

8

7

6

5

4

3

2

1

zozozo

yoyoyo

xoxoxo

zozozo

yoyoyo

xoxoxo

zozozo

yoyoyo

xoxoxo

zozozo

yoyoyo

xoxoxo

12

11

10

9

8

7

6

5

4

3

2

1

q

q

q

q

q

q

q

qq

q

q

q

nml000000000

nml000000000

nml000000000

000nml000000

000nml000000

000nml000000

000000nml000

000000nml000000000nml000

000000000nml

000000000nml

000000000nml

q

q

q

q

q

q

q

qq

q

q

q

(4.16)

The transformation matrix, [], can be identified to be

[ ]

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

=

000

000

000

000

1212x (4.17)

Figure 4.3 Three dimensional frame element.

y

x

z

x

q1

q2

q3q4

q5

q6

q7

q8

q9

q10

q11

q12

q3

q2

q1

q4q5

q6

z

y

q7

q8

q9

q10

q11

q12


11/18

65

where

[ ]

=

zozozo

yoyoyo

xoxoxo

3x3

nml

nml

nml

(4.18)

and

[ ]

=

000

000

000

0 33x (4.19)

Here lox, mox, nox denote the direction cosines of x-axis (line ij in Fig. 5.14); loy, moy, noy

represent the direction cosines of y-axis; and loz, moz, noz indicate the direction cosines of z-

axis with respect to global x, y, z axes. It can be seen that finding the direction cosines of x-

axis is a straight forward computation since

L

xxl

ij

xo

= ,

L

yym

ij

xo

= ,

L

zzn

ij

xo

= (4.20)

where xk, yk, zk indicate the coordinates of node k (k= i, j) in the global system. However, the

computation of the direction cosines of y- and z-axes requires some special effort. Finally

the stiffness matrix of the element with reference to the global coordinate system can be

obtained as

[ ] [ ] [ ][ ]= kk T (4.21)

y

x

x

x

y

q1

q2

q4

q5

q3

q6

q1

q2

q3

q4

q5

q6


12/18

66

Figure 4.4 Planar frame element.

4.2.2 Plane Frame Element

In the case of two-dimensional (planar) frame analysis, we need to use an elementhaving six degrees of freedom as shown in Fig. 5.17. This element is assumed to lie in xy

plane and has two axial and four bending degrees of freedom. By using a linear interpolation

model for axial displacement and a cubic model for the transverse displacement, and

superimposing the resulting two stiffness matrices, the following stiffness matrix can be

obtained (the vector {q} is taken as { } 654321T

q,q,q,q,q,qq = )

[ ]

=

22

zz

2

zz

2

2

zz

2

3

zz

L4L60L2L60

120L6120

I

AL00

I

AL

L4L60

Symmetric120

I

AL

L

EI

k (4.22)

It is to be noted that the bending and axial deformation effects are uncoupled while deriving

Eq. (4.21). Equation (4.21 can also be obtained as a special case of Eq. (3.60) by deleting

rows and columns 2, 4, 5, 8, 10, 11.)

In this case the stiffness matrix of the element in the global xy coordinate system can befound as

[ ] [ ] [ ][ ]= kk T

where

[ ]

=

100000

0ml000

0ml000

000100

0000ml

0000ml

yoyo

xoxo

yoyo

xoxo

(4.23)

with (lox , mox) denoting the direction cosines of x-axis and (loy , moy) indicating the

direction cosines of y-axis with respect to the global xy system.

Example: The frame shown in Figure is made from St 37 steel and subjected to a distributed

load of 20 kN/m and a concentrated load of 40 kN.

(a)Model the frame with two beam elements.(b)Calculate the element stiffness matrices and load vectors and then assemble them to

obtain the stiffness matrix and load vector of the frame.


13/18

67

(c)Solve the LAES for the nodal displacements and rotations. Plot the undeformed anddeformed structure.

(d)Obtain the element stresses.(e)Calculate the reaction forces.

Cross-section: DIN 1025 St37 IPB 100

Solution:

a) The nodal coordinate table is

Node x (m) y (m)

1 0 0

2 2 0

3 3.732 -0.5

The element connectivity table is

Element 1 2

1 1 22 2 3

2

Q2

Q1

Q3

Q4

Q5

Q6

Q7

Q8

Q9

1

2

x

y

1

3

2 m

30z

x

y

2 m0.8 m

40 kN

20 kN/m


14/18

68

The direction cosines table is

Element Le l m

1 2 1.0 0.0

2 2 0.866 -0.5

Properties of IPB 100 cross-section are

Area = 0.24800x10-2

m2

Iyy = 0.43227 x10-5

m4

Izz = 0.16681 x10-5

m4

The element stiffness matrices can be calculated as

2 2

5

29 51

3 2 2 2 2

5 5

2 2

0.248 10 2

0.43227 10

0 12

0 6 2 4 2200 10 0.43227 10

2 0.248 10 2 0.248 10 20 0

0.43227 10 0.43227 10

0 12 6 2 0 12

0 6 2 2 2 0 6 2 4 2

2294.860 12

108067.5

x x

x

Symmetric

x xx x xk

x x x x

x x

x

x x x x

S

=

=0 12 16

2294.86 0 0 2294.86

0 12 12 0 12

0 12 8 0 12 16

ymmetric

The all properties of the element 2 are same with those of the element 1 in local axes, the

stiffness matrices equal

2 1k k =

Since the local axes of element one are parallel to global axes, there is no need the

transformations for the stiffness matrix and load vector of element 1.

1 1k k =

The stiffness matrix and load vector of element 2 should be transformed into global axes. The

transformation matrix can be calculated as

lox = cos 30 = 0.866 mox = cos 120 = -0.5, loy = cos -60 = 0.5 moy = cos 30 = 0.866,


15/18


16/18

70

{ }

1

1 1

2 22

2

3 3

4 4 0.80.8

0

40000

40000

0

40000

40000

x a

y b b

y b b

x a

y b b

y b b xx

F N

F N N

F N N

F N

F N N

F N N==

= =

bf

The values of shape functions at the specified point on the element are obtained as

( ) 3 2 3 3 2 31 3 30.8 0.81 1

2 3 2(0.8) 3(0.8) 2 2 0.6482

b x xN x x L L

L= = = + = + =

( ) 3 2 2 3 3 2 2 32 3 30.8 0.81 1

2 (0.8) 2 2(0.8) 2 (0.8)2 0.2882x x

N x L x L xLL= =

= + = + =

( ) 3 2 3 23 3 30.8 0.81 12 3 2(0.8) 3(0.8) 2 0.7362x xN x x L

L= = = + = + =

( ) 3 2 2 3 2 24 3 30.8 0.81 1

(0.8) 2 (0.8) 2 0.2562x x

N x L x LL= =

= = =

and the 2nd

element load vector in local coordinates yields as follows

{ }2

0

25920

11520

0

29440

10240

=

bf

A transformation for the first element is not necessary since the local and global axes are in

same directions. The load vector of element 2 in the global axes can be obtained as

{ } [ ] { }2 2

0.866 0.5 0 0 0 0 0 12960

0.5 0.866 0 0 0 0 25920 22447

0 0 1 0 0 0 11520 11520

0 0 0 0.866 0.5 0 0 14720

0 0 0 0.5 0.866 0 29440 25495

0 0 0 0 0 1 10240 10240

T

= = =

b bf f

The load vector of entire structure is

{ } 0 20000 6668 12960 2447 18188 14720 25495 10240T

= F

Now, we can solve the LAES applying the boundary conditions as Q1 = Q2 = Q3 = Q7 = Q8 =

Q9 = 0.


17/18

71

4

9

5

6

1191.90 492.853 0.03112 12960

10 492.853 284.671 0.00834 2447

0.03112 0.00834 0.13140 18188

Q

Q

Q

=

Q4 = 0.00004396 m, Q5 = 0.00008877 m, Q6 = 0.13841 rad

4.2.3 Beam Element

A beam element is one-dimensional element which can resist bending moments in its plane. If

the beam lies in the xy plane, the corresponding degrees of freedom are shown in Fig. 4.4. By

assuming a cubic interpolation polynomial for the transverse deflection of the beam (v), the

stiffness matrix can be derived as

[ ]

=22

22

3

z

L4L6L2L6

L612L612

L2L6L4L6

L612L612

L

EI

k

(4.24)

Figure 4.5 Beam element.

If the beam element has an arbitrary orientation in xy plane as shown in Fig. 4.5, the stiffness

matrix of the beam element in the global xy system is given by

[ ] [ ] [ ] [ ]6 6 6 4 4 4 4 6

T

x x x x=k k (4.25)

where the transformation matrix [], is given by

[ ]

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0 0 1

oy oy

oy oy

l m

l m

=

(4.26)

L

x

y

q1

q2

q3

q4

xy


18/18

72

where (loz, moz) are the direction cosines of z-axis with respect to xy system.

Documents

Fea Chapter155