Hibbeler R.C. Structural Analysis

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Application of the stiffness method requires subdividing the structure

into a series of discrete finite elements and identifying their end points

as nodes. For truss analysis, the finite elements are represented by each

of the members that compose the truss, and the nodes represent the

joints. The force-displacement properties of each element are

determined and then related to one another using the force equilibrium

equations written at the nodes. These relationships, for the entire

structure, are then grouped together into what is called the structure

stiffness matrix K. Once it is established, the unknown displacements of

the nodes can then be determined for any given loading on the

structure. When these displacements are known, the external and

internal forces in the structure can be calculated using the force-

displacement relations for each member.

Before developing a formal procedure for applying the stiffness method,

it is first necessary to establish some preliminary definitions and concepts.

Member and Node Identification. One of the first steps

when applying the stiffness method is to identify the elements or

members of the structure and their nodes. We will specify each member

by a number enclosed within a square, and use a number enclosed

within a circle to identify the nodes. Also, the “near” and “far” ends of

the member must be identified. This will be done using an arrow

written along the member, with the head of the arrow directed toward

the far end. Examples of member, node, and “direction” identification

for a truss are shown in Fig. 14–1a. These assignments have all been

done arbitrarily.*

Global and Member Coordinates. Since loads and displacements

are vector quantities, it is necessary to establish a coordinate system in

order to specify their correct sense of direction. Here we will use two

different types of coordinate systems. A single global or structure

coordinate system, x, y, will be used to specify the sense of each of the

external force and displacement components at the nodes, Fig. 14–1a.A

local or member coordinate system will be used for each member to

specify the sense of direction of its displacements and internal loadings.

This system will be identified using axes with the origin at the

“near” node and the axis extending toward the “far” node. An

example for truss member 4 is shown in Fig. 14–1b.

x¿

y¿x¿,

540

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

*For large trusses, matrix manipulations using K are actually more efficient using

selective numbering of the members in a wave pattern, that is, starting from top to bottom,

then bottom to top, etc.

14.1 FUNDAMENTALS OF THE STIFFNESS METHOD 541

Kinematic Indeterminacy. As discussed in Sec. 11–1, the

unconstrained degrees of freedom for the truss represent the primary

unknowns of any displacement method, and therefore these must be

identified. As a general rule there are two degrees of freedom, or two

possible displacements, for each joint (node). For application, each degree

of freedom will be specified on the truss using a code number, shown at the

joint or node, and referenced to its positive global coordinate direction

using an associated arrow. For example, the truss in Fig. 14–1a has eight

degrees of freedom, which have been identified by the “code numbers”

1 through 8 as shown. The truss is kinematically indeterminate to the

fifth degree because of these eight possible displacements: 1 through

5 represent unknown or unconstrained degrees of freedom, and 6 through

8 represent constrained degrees of freedom. Due to the constraints, the

displacements here are zero. For later application, the lowest code numbers

will always be used to identify the unknown displacements (unconstrained

degrees of freedom) and the highest code numbers will be used to identify

the known displacements (constrained degrees of freedom). The reason for

choosing this method of identification has to do with the convenience of

later partitioning the structure stiffness matrix, so that the unknown

displacements can be found in the most direct manner.

Once the truss is labeled and the code numbers are specified, the

structure stiffness matrix K can then be determined. To do this we must

first establish a member stiffness matrix for each member of the truss.

This matrix is used to express the member’s load-displacement relations

in terms of the local coordinates. Since all the members of the truss are

not in the same direction, we must develop a means for transforming

these quantities from each member’s local coordinate system to

the structure’s global x, y coordinate system.This can be done using force

and displacement transformation matrices. Once established, the

elements of the member stiffness matrix are transformed from local to

global coordinates and then assembled to create the structure stiffness

matrix. Using K, as stated previously, we can determine the node

displacements first, followed by the support reactions and the member

forces. We will now elaborate on the development of this method.

y¿x¿,

k¿

(a)

y¿

(b)

Fig. 14–1

14.2 Member Stiffness Matrix

In this section we will establish the stiffness matrix for a single truss

member using local coordinates, oriented as shown in Fig. 14–2.

The terms in this matrix will represent the load-displacement relations

for the member.

A truss member can only be displaced along its axis since the

loads are applied along this axis. Two independent displacements are

therefore possible. When a positive displacement is imposed on

the near end of the member while the far end is held pinned, Fig. 14–2a,

the forces developed at the ends of the members are

Note that is negative since for equilibrium it acts in the negative

direction. Likewise, a positive displacement at the far end, keeping

the near end pinned, Fig. 14–2b, results in member forces of

By superposition, Fig. 14–2c, the resultant forces caused by both

displacements are

(14–1)

(14–2)

These load-displacement equations may be written in matrix form* as

(14–3)

where

(14–4)

This matrix, is called the member stiffness matrix, and it is of the same

form for each member of the truss. The four elements that comprise it are

called member stiffness influence coefficients, Physically, representsk

k¿,

1 -1

-11

q = k

1 -1

-11

ﬂ

x¿q

1x¿ axis2

y¿x¿,

542

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

Fig. 14–2

*A review of matrix algebra is given in Appendix A.

(

)

y¿

x¿

q¿

(b)

y¿

x¿

q–

–

ⴙ

x¿

(c)

ⴝ

14.3 DISPLACEMENT AND FORCE TRANSFORMATION MATRICES 543

the force at joint i when a unit displacement is imposed at joint j.For

example, if then is the force at the near joint when the far joint

is held fixed, and the near joint undergoes a displacement of i.e.,

Likewise, the force at the far joint is determined from so

that

These two terms represent the first column of the member stiffness

matrix. In the same manner, the second column of this matrix represents

the forces in the member only when the far end of the member under-

goes a unit displacement.

14.3 Displacement and Force

Transformation Matrices

Since a truss is composed of many members (elements), we will now

develop a method for transforming the member forces q and

displacements d defined in local coordinates to global coordinates. For

the sake of convention, we will consider the global coordinates positive

x to the right and positive y upward. The smallest angles between the

positive x, y global axes and the positive local axis will be defined as

and as shown in Fig. 14–3. The cosines of these angles will be used

in the matrix analysis that follows. These will be identified as

Numerical values for and can easily be

generated by a computer once the x, y coordinates of the near end N

and far end F of the member have been specified. For example, consider

member NF of the truss shown in Fig. 14–4. Here the coordinates of N

and F are and respectively.* Thus,

(14–5)

(14–6)

The algebraic signs in these “generalized” equations will automatically

account for members that are oriented in other quadrants of the x–y plane.

= cos u

- y

21x

- x

+ 1y

- y

= cos u

- x

21x

- x

+ 1y

- y

, y

2,1x

, y

= cos u

x¿

= k

j = 1,i = 2,

= k

= 1,

i = j = 1,

Fig. 14–3

Fig. 14–4

*The origin can be located at any convenient point. Usually, however, it is located

where the x, y coordinates of all the nodes will be positive, as shown in Fig. 14–4.

y¿

x¿

(far)

(near)

Displacement Transformation Matrix. In global coordinates

each end of the member can have two degrees of freedom or

independent displacements; namely, joint N has and Figs. 14–5a

and 14–5b, and joint F has and Figs. 14–5c and 14–5d. We will

now consider each of these displacements separately, in order to

determine its component displacement along the member. When the far

end is held pinned and the near end is given a global displacement

Fig. 14–5a, the corresponding displacement (deformation) along the

member is * Likewise, a displacement will cause the

member to be displaced along the axis, Fig. 14–5b.The

effect of both global displacements causes the member to be displaced

In a similar manner, positive displacements and successively

applied at the far end F, while the near end is held pinned, Figs. 14–5c and

14–5d, will cause the member to be displaced

Letting and represent the direction cosines for

the member, we have

which can be written in matrix form as

(14–7)

(14–8)

where

(14–9)

From the above derivation, T transforms the four global x, y displace-

ments D into the two local displacements d. Hence, T is referred to as

the displacement transformation matrix.

x¿

T = c

00l

d = TD

d= c

00l

= D

+ D

= D

+ D

= cos u

= D

cos u

+ D

cos u

= D

cos u

+ D

cos u

x¿D

cos u

544 CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

*The change in or will be neglected, since it is very small.u

Fig. 14–5

(a)

x¿

cos u

(b)

cos u

x¿

(

)

x¿

cos u

(d)

cos u

x¿

Force Transformation Matrix. Consider now application of the

force to the near end of the member, the far end held pinned,

Fig. 14–6a. Here the global force components of at N are

Likewise, if is applied to the bar, Fig. 14–6b, the global force components

at F are

Using the direction cosines these equations

become

which can be written in matrix form as

(14–10)

(14–11)

where

(14–12)

In this case transforms the two local forces q acting at the ends

of the member into the four global (x, y) force components Q.By

comparison, this force transformation matrix is the transpose of the

displacement transformation matrix, Eq. 14–9.

1x¿2T

= D

0 l

Q = T

T= D

0 l

= q

= cos u

= q

cos u

= q

cos u

= q

cos u

= q

cos u

x¿

(a)

x¿

(b)

Fig. 14–6

14.3 D

ISPLACEMENT AND FORCE TRANSFORMATION MATRICES 545

14.4 Member Global Stiffness Matrix

We will now combine the results of the preceding sections and determine

the stiffness matrix for a member which relates the member’s global force

components Q to its global displacements D. If we substitute Eq. 14–8

into Eq. 14–3 we can determine the member’s

forces q in terms of the global displacements D at its end points, namely,

(14–13)

Substituting this equation into Eq. 14–11, yields the final

result,

(14–14)

where

(14–15)

The matrix k is the member stiffness matrix in global coordinates. Since

T, and are known, we have

Performing the matrix operations yields

(14–16)

k =

-l

k = D

0 l

1 -1

-11

00l

k¿T

k = T

Q = kD

Q = T

k¿TD

Q = T

q = k¿TD

1q = k¿d2,1d = TD2

546

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

14.5 TRUSS STIFFNESS MATRIX 547

The location of each element in this symmetric matrix is

referenced with each global degree of freedom associated with the

near end N, followed by the far end F. This is indicated by the code

number notation along the rows and columns, that is,

Here k represents the force-displacement relations for the member

when the components of force and displacement at the ends of the

member are in the global or x, y directions. Each of the terms in the matrix

is therefore a stiffness influence coefficient which denotes the x or y

force component at i needed to cause an associated unit x or y displacement

component at j. As a result, each identified column of the matrix

represents the four force components developed at the ends of the

member when the identified end undergoes a unit displacement related

to its matrix column. For example, a unit displacement will

create the four force components on the member shown in the first

column of the matrix.

14.5 Truss Stiffness Matrix

Once all the member stiffness matrices are formed in global coordinates,

it becomes necessary to assemble them in the proper order so that the

stiffness matrix K for the entire truss can be found. This process of

combining the member matrices depends on careful identification of the

elements in each member matrix. As discussed in the previous section, this

is done by designating the rows and columns of the matrix by the four

code numbers used to identify the two global degrees of

freedom that can occur at each end of the member (see Eq. 14–16). The

structure stiffness matrix will then have an order that will be equal to the

highest code number assigned to the truss, since this represents the total

number of degrees of freedom for the structure. When the k matrices are

assembled, each element in k will then be placed in its same row and

column designation in the structure stiffness matrix K. In particular,

when two or more members are connected to the same joint or node,

then some of the elements of each member’s k matrix will be assigned

to the same position in the K matrix. When this occurs, the elements

assigned to the common location must be added together algebraically.

The reason for this becomes clear if one realizes that each element of the

k matrix represents the resistance of the member to an applied force at

its end. In this way, adding these resistances in the x or y direction when

forming the K matrix determines the total resistance of each joint to a

unit displacement in the x or y direction.

This method of assembling the member matrices to form the structure

stiffness matrix will now be demonstrated by two numerical examples.

Although this process is somewhat tedious when done by hand, it is

rather easy to program on a computer.

= 1

4 * 4

Determine the structure stiffness matrix for the two-member truss

shown in Fig. 14–7a. AE is constant.

548

CHAPTER 14 TRUSS ANALYSIS U SING THE STIFFNESS METHOD

EXAMPLE 14.1

SOLUTION

By inspection, ② will have two unknown displacement components,

whereas joints ① and ③ are constrained from displacement.

Consequently, the displacement components at joint ② are code

numbered first, followed by those at joints ③ and ①, Fig. 14–7b.The

origin of the global coordinate system can be located at any point. For

convenience, we will choose joint ② as shown. The members are

identified arbitrarily and arrows are written along the two members to

identify the near and far ends of each member. The direction cosines

and the stiffness matrix for each member can now be determined.

Member 1. Since ② is the near end and ③ is the far end, then by

Eqs. 14–5 and 14–6, we have

Using Eq. 14–16, dividing each element by we have

The calculations can be checked in part by noting that is symmetric.

Note that the rows and columns in are identified by the x, y degrees

of freedom at the near end, followed by the far end, that is, 1, 2, 3, 4,

respectively, for member 1, Fig. 14–7b. This is done in order to identify

the elements for later assembly into the K matrix.

= AE

12 34

0.333 0 -0.333 0

0000

-0.333 0 0.333 0

0000

L = 3 ft,

3 - 0

= 1 l

0 - 0

= 0

Fig. 14–7

3 ft

4 ft

(a)

3 ft

4 ft

(b)

Member 2. Since ② is the near end and ① is the far end, we have

Thus Eq. 14–16 with becomes

Here the rows and columns are identified as 1, 2, 5, 6, since these

numbers represent, respectively, the x, y degrees of freedom at the

near and far ends of member 2.

Structure Stiffness Matrix. This matrix has an order of since

there are six designated degrees of freedom for the truss, Fig. 14–7b.

Corresponding elements of the above two matrices are added

algebraically to form the structure stiffness matrix. Perhaps the assembly

process is easier to see if the missing numerical columns and rows in

and are expanded with zeros to form two matrices. Then

K = k

+ k

6 * 6k

6 * 6

= AE

1 2 5 6

0.072 0.096 -0.072 -0.096

0.096 0.128 -0.096 -0.128

-0.072 -0.096 0.072 0.096

-0.096 -0.128 0.096 0.128

L = 5 ft

3 - 0

= 0.6

4 - 0

= 0.8

K = AE F

0.405 0.096 -0.333 0 -0.072 -0.096

0.096 0.128 0 0 -0.096 -0.128

-0.333 0 0.333 0 0 0

000000

-0.072 -0.096 0 0 0.072 0.096

-0.096 -0.128 0 0 0.096 0.128

123456

0.072 0.096 0 0 -0.072 -0.096

0.096 0.128 0 0 -0.096 -0.128

000000

-0.072 -0.096 0 0 0.072 0.096

-0.096 -0.128 0 0 0.096 0.128

K = AE

12 3456

0.333 0 -0.333000

000000

-0.333 0 0.333000

000000

+ AE

If a computer is used for this operation, generally one starts with K

having all zero elements; then as the member global stiffness matrices

are generated, they are placed directly into their respective element

positions in the K matrix, rather than developing the member stiffness

matrices, storing them, then assembling them.

14.5 TRUSS STIFFNESS MATRIX 549