7 Module 4: Analysis of Frame Structures

Lecture 2: Analysis of Truss

4.2.1 Element Stiffness of a 3 Node Truss Member

Fig. 4.2.1 3-node truss member

Here, the displacement function using Pascals triangle can be expressed as:

20 1 2u x x x 0

21

2

1 x x

(4.2.1)

Applying boundary conditions: At x= 0, u(0)= u 1 , x=L/2, u(L/2) = u2 and at x=L, u(L) = u3

And solving for 0 , 1 and 210 u , 1 2 31 3 4u u uL

and 1 2 32 22 4 2u u uL

Therefore,

2 2 21 2 32 2 23 2 4 4 21 x x x x x xu x u u u N uL L L L L L

(4.2.2)

Here, N is the shape function of the element and is expressed as:

2 2 22 2 23 2 4 4 21 x x x x x xN L L L L L L

(4.2.3)

Now, the element stiffness matrix can be written as Tk B D B d

(4.2.4)

Where, 2 2 23 4 4 8 1 4d N x x xB dx L L L L L L So, the stiffness matrix will be:

Tk B D B d

0L TB E B Adx

8

3

44

8

1 4

3

4

4

8

1

4

9 16

24 12

40

32 3

16

16

12 40 32 16

64

64 4

24

32

3 16 16 4

24

32 1

8

16

(4.2.5)

After integrating the above equation, the stiffness matrix of the 3-node truss member will become:

7 8 18 16 81 8 7

(4.2.6) 4.2.2 Worked Out Example Analyze the truss shown below by finite element method. Assume the cross sectional area of the inclined member as 1.5 times the area (A) of the horizontal and vertical members. Assume modulus of elasticity is constant for all the members and is E.

Fig. 4.2.2 Plane truss

9

Solution The analysis of truss starts with the numbering of members and joints as shown below:

Fig. 4.2.3 Numbering of members and nodes The member information for the truss is shown in Table 4.2.1. The member and node numbers, modulus of elasticity, cross sectional areas are the necessary input data. From the coordinate of the nodes of the respective members, the length of each member is computed. Here, the angle has been calculated considering anticlockwise direction. The signs of the direction cosines depend on the choice of numbering the nodal connectivity. Now, let assume the coordinate of node 1 as (0, 0). The coordinate and restraint joint information are given in Table 4.2.2. The integer 1 in the restraint list indicates the restraint exists and 0 indicates the restraint at that particular direction does not exist. Thus, in node no. 2, the integer 0 in x and y indicates that the joint is free in x and y directions.

Table 4.2.1 Member Information for Truss

Member No.

Starting Node

Ending Node

Value of

Area Modulus of Elasticity

1 1 2 90 A E 2 2 3 315 1.5A E 3 3 1 180 A E

Table 4.2.2 Nodal Information for Plane Truss

Node No. Coordinates Restraint List x y x y

1 0 0 1 1 2 0 L 0 0 3 L 0 1 1

10

The stiffness matrices of each individual member can be found out from the stiffness matrix equation as shown below.

22

22

22

22

sinsincossinsincos

sincoscossincoscos

sinsincossinsincos

sincoscossincoscos

LAEk

Thus the local stiffness matrices of each member are calculated based on their individual member properties and orientations and written below.

and

5 6 1 25

6

1

2

0000

0101

1000

0101

3 LAEk

3 4 5 6 3

4

5

6

1111

1111

1111

1111

243

2 LAEk

1 2 3 41

2

3

4

1010

0000

1010

0000

1 LAEk

11

Global stiffness matrix can be formed by assembling the local stiffness matrices into globally. Thus the global stiffness matrix are calculated from the above relations and obtained as follows:

243

243

243

24300

2431

243

243

24301

243

243

2431

24310

243

243

243

24300

001010010001

LAEK

The equivalent load vector for the given truss can be written as:

1

1

2

2

3

3

00

2

00

x

y

x

y

x

y

FFF P

FF PFF

Let us assume that u and v are the horizontal and vertical displacements respectively at joints. Thus the displacement vector will be expressed as follows:

00

00

2

2

3

3

2

2

1

1

vu

vuvuvu

d

Therefore, the relationship between the force and the displacement will be:

1 2 3 4 5 61 2 3 4 5 6

12

1

1

2

2

3

3

1 0 0 0 1 00 1 0 1 0 0

03 3 3 30 0 04 2 4 2 4 2 4 22 3 3 3 30 1 1

4 2 4 2 4 2 4 23 3 3 3 01 0 1

4 2 4 2 4 2 4 2 03 3 3 30 0

4 2 4 2 4 2 4 2

x

y

x

y

FF

P uAEP vL

FF

From the above relation, the unknown displacements u2 and v2 can be found out through computer programming. However, as numbers of unknown displacements in this case are only two, the solution can be obtained by manual calculations. The above equation may be rearranged with respect to unknown and known displacements in the following form:

F k k dF k k d

Thus the developed matrices for the truss problem can be rearranged as:

243

24300

243

243

2431

24301

243

243

001010010100

243

243

1024

31243

243

24300

243

243

LAE

. The above relation may be condensed into following

2

2

3 32 4 2 4 2

3 314 2 4 2

uP AEvP L

2P

P

Fx1 Fy1

Fx3

Fy3

u2

v2 0 0

0

0

13

The unknown displacements can be derived from the relationships expressed in the above equation.

1

2

2

3 3 3 312 24 24 2 4 2 4 2 4 2

3 3 3 3314 2 4 2 4 2 4 2

u P PAE Lv P PL AE

Thus the unknown displacement at node 2 of the truss structure will become:

2

2

8 233

3

u PLv AE

Support Reactions: The support reactions {Ps} can be determined from the following relation:

s csP P K d Where, {Pcs} correspond to equivalent loadings at supports. Thus, the support reaction of the present truss structure will be:

0 00 0 1

8 20 33 330 4 2 4 2 3

0 3 34 2 4 2

sAE PLPL AE

PPP

2230

Member End Actions: Now, the member end actions can be obtained from the corresponding member stiffness and the nodal displacements. The member end forces are derived as shown below. Member 1

1

1

2

2

00 0 0 0 0

00 1 0 1 3

8 20 0 0 0 0330 1 0 1 3

3

mx

my

mx

my

FF PAE PLF L AEF P

14

Member 2

2

2

3

3

8 21 1 1 1 2331 1 1 1 23 3

1 1 1 1 24 2 01 1 1 1 2

0

mx

my

mx

my

F PF PAE PLF PAELF P

Member 3

0000

0000

0000010100000101

1

1

3

3

AEPL

LAE

FFFF

my

mx

my

mx

Thus the member forces in all members of the truss will be:

2 23 3

2 2 2 200

m

P P

F P P P

The reaction forces at the supports of the truss structure will be:

032

2

R

PF

PP

Thus the member force diagram will be as shown in Fig. 4.2.4.

Fig. 4.2.4 Member Force Diagram