1

Introduction

Analysis of framed structures may be carried out using stiffness method. However, such

types of structures can also be analyzed by finite element method. A unified formulation will

be demonstrated based on finite element concept in this module for the analysis of framed

structures. A truss structure is composed of members pin jointed together at their ends. Truss

element can resist only axial forces (tension or compression) and can deform only in its axial

direction. Therefore, in case of a planar truss, each node has components of displacements

parallel to X and Y axes. Planar trusses lie in a single plane and are used to support roofs and

bridges. Members of trusses will not be able to carry bending moment. Configuration of few

standard truss structures are shown in Fig.1.

Fig. 1 Configuration of various truss structures

Element Stiffness of a Truss Member

Since, the members of truss carry only axial force, the displacement along its axis only will

be developed due to axial load. The displacement function of truss member for the

development of shape function may be expressed as:

0

0 1

1

1u x x x

(1)

2

Fig. 2 Axial force on the member along X axis

Applying boundary conditions as shown in Fig.2:

At x= 0, u(0)= u1 and at x=L, u(L) = u

2

Thus, 10

u and L

uu12

1

. Therefore,

uNuL

xu

L

xxu

211

(2)

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

L

x

L

xN 1 (3)

So we get the element stiffness matrix as

T

k B D B d

(4)

Where,

LLdx

NdB

11

So, the stiffness matrix will become:

11

1111

1

1

00 L

AEdx

LL

L

LAEAdxBEBLL

T

Thus, the stiffness matrix of the truss member along its member axis will be:

11

11

L

AEk (5)

3

Generalized Stiffness Matrix of a Plane Truss Member

Let us consider a member making an angle ‘’ with X axis as shown in Fig.3. By resolving

the forces along local X and Y direction, the following relations are obtained.

1 1 1

2 2 2

1 1 1

2 2 2

c o s s in

c o s s in

s in c o s

s in c o s

x x y

x x y

y x y

y x y

F F F

F F F

F F F

F F F

(6)

Where, 1x

F and 2x

F are the axial forces along the member axis X . Similarly, 1y

F and 2y

F are

the forces perpendicular to the member axis X .

Fig.3 Inclined truss member

The relationship expressed in eqn. (6) may be rewritten in matrix form as:

11

11

22

22

c o s 0 0

c o s 0 0

0 0 c o s

0 0 c o s

xx

yy

xx

yy

FF s in

FF s in

FF s in

FF s in

(7)

Now, eqn.(7) may be expressed in short as:

FTF (8)

4

Here, [T] is called transformation matrix. This relates the global (𝑋, 𝑌 axes) to member axis

(�̅�, �̅� axes). Similarly, the relations of nodal displacements between two coordinate systems

may be written as:

d T d (9)

Again, the eqn.(5) may be generalized and expressed with respect to the member axes

including force and displacement vector as:

1 1

1 1

2 2

2 2

1 0 1 0

0 0 0 0

1 0 1 0

0 0 0 0

x

y

x

y

F u

F vA E

F uL

F v

(10)

Where, the nodal forces in Y direction are zero. Eqn.(10) may also be expressed in short as:

dkF (11)

The matrices in eqn.(11) are written with respect to the member local axes. Now, eq. (11)

may be rewritten with the use of eqns. (8) and (9) as follows:

dTkFT (12)

Or,

dTkTF1

(13)

Here, the transformation matrix [T] is orthogonal, i.e., [T]-1 is equal to [T]T. Therefore, from

the above relationship, the generalized stiffness matrix can be expressed as:

TkTkT

(14)

Thus,

c o s s in 0 0 1 0 1 0 c o s s in 0 0

s in c o s 0 0 0 0 0 0 s in c o s 0 0

0 0 c o s s in 1 0 1 0 0 0 c o s s in

0 0 s in c o s 0 0 0 0 0 0 s in c o s

A Ek

L

(15)

Or,

5

22

22

22

22

sincossinsincossin

cossincoscossincos

sincossinsincossin

cossincoscossincos

L

AEk (16)

The above stiffness matrix can be used for the analysis of two-dimensional truss problems.

6

Example:

Analyze the truss shown in Fig.4 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 Plane truss

Solution

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

Fig. 5 Numbering of members and nodes

The member information for the truss is shown in Table.1. The member and node numbers,

modulus of elasticity, cross sectional areas are the necessary input data. From the coordinate

7

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 us assume the coordinate of node 1 as (0, 0). The coordinate and restraint joint

information are given in Table.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.

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

L

AEk

Thus the local stiffness matrices of each member are calculated based on their individual

member properties and orientations and written below.

Table.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.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

8

and

Global stiffness matrix can be formed by assembling the local stiffness matrices into global

reference. Thus the global stiffness matrix is calculated from the above relations and obtained

as follows:

24

3

24

3

24

3

24

300

24

31

24

3

24

3

24

301

24

3

24

3

24

31

24

310

24

3

24

3

24

3

24

300

001010

010001

L

AEK

1 2 3 4

1

2

3

4

3 4 5 6

3

4

5

6

5 6 1 2

5

6

1

2

1 2 3 4 5 6

1

2

3

4

5

6

9

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

1

1

2

2

3

3

0

0

2

0

0

x

y

x

y

x

y

F

F

F PF

F P

F

F

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:

0

0

0

0

2

2

3

3

2

2

1

1

v

u

v

u

v

u

v

u

d

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

1

1

2

2

3

3

1 0 0 0 1 0

0 1 0 1 0 0

03 3 3 30 0

04 2 4 2 4 2 4 2

2 3 3 3 30 1 1

4 2 4 2 4 2 4 2

3 3 3 3 01 0 1

4 2 4 2 4 2 4 2 0

3 3 3 30 0

4 2 4 2 4 2 4 2

x

y

x

y

F

F

P uA E

P vL

F

F

From the above relation, the unknown displacements u2 and v2 can be found out using

computer program. However, as numbers of unknown displacements in this case are only

two, the solution may 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 d

F k k d

10

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

24

3

24

300

24

3

24

3

24

31

24

301

24

3

24

3

001010

010100

24

3

24

310

24

31

24

3

24

3

24

300

24

3

24

3

L

AE

.

The above relation may be condensed into following

2

2

3 3

2 4 2 4 2

3 31

4 2 4 2

uP A E

vP L

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

equation. 1

2

2

3 3 3 31

2 24 24 2 4 2 4 2 4 2

3 3 3 331

4 2 4 2 4 2 4 2

u P PA E L

v P PL A E

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

2

2

8 23

3

3

u P L

v A E

Support Reactions:

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

s c sP P K d

2P

P

Fx1

Fy1

Fx3

Fy3

u2

v2

0

0

0

0

11

Where, {Pcs} correspond to equivalent loadings at supports. Thus, the support reaction of the

present truss structure will be:

0 0

0 0 18 2

0 33 33

0 4 2 4 23

0 3 3

4 2 4 2

s

A E P LP

L A E

P

P

P

2

2

3

0

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 03

30 1 0 1 3

3

m x

m y

m x

m y

F

F PA E P L

F L A E

F P

Member – 2

2

2

3

3

8 21 1 1 1 23

31 1 1 1 23

31 1 1 1 24 2

01 1 1 1 2

0

m x

m y

m x

m y

F P

F PA E P L

F PA EL

F P

Member –3

0

0

0

0

0

0

0

0

0000

0101

0000

0101

1

1

3

3

AE

PL

L

AE

F

F

F

F

my

mx

my

mx

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

12

2 2

3 3

2 2 2 2

00

m

P P

F P P P

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

0

3

2

2

R

PF

P

P

Thus the member force diagram will be as shown in Fig. 6.

Fig. 6 Member Force Diagram