# Chapter 18 Matrix Analysis of Beams and Frames by the Direct Stiffness Method

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### Text of Chapter 18 Matrix Analysis of Beams and Frames by the Direct Stiffness Method

• Matrix Analysis of Beams

and Frames by the Direct Stiffness Method

nu., ;~:~'f~~;'~;~1 .. ; ....... .................................................HH + H ~ , u

18:1 Introducti.on In Chapter 17 we discussed the analysis of trusses using the direct stiff ness method. In this chapter we extend the method to structures in which loads. may be applied to joints as well as to members between joints, and induce both axial forces and shears and moments. Whereas in the case of. trusses we had to consider only joint displacements as unknowns in settingup the equilibrium equations, for frarneswemtlsf adrljoint rotations. Consequently, a total of three' equations of equilibrium, two for forees and one for mQm~t, can be written for each joint in a plane frame.

Even thou,'gh the analysis of a plane frame using the direct stiffness method involves three displacement components per joint (8, Il", Il), we can often reduce the number of equations to be solved by neglecting the change in length of the members. In typical beams or frames, this simplification introduces little error in the results.

In the analysis of any structure using the stiffness method, the value of any quantity (for example, shear, moment, or displacement) is obtained from the sum of two parts. The first part is obtained from the analysis of a restrained structure in which all the joints are restrained against movement. The moments induced at the ends of each member are fixedend moments. This procedure is similar to that used in the moment distribu- '. tion method in Chapter13. After the net restraining forces are computed and the signs reversed at each joirtt, these restraining forces are applied to the original structure in the second part of the analysis to determine the effect induced by joint displacements,

The superposition of forces and. displacements from two parts can be explained using as an example theframe in Figure I8.la. This frame is composed, of two members connected by a rigid joint at B. Under the

..."' .....

• I

684 Chapter 18 Matrix Analysis of Beams and Frames by the Direct Stiffness Method .

p

B

I / !'

t 1t,\1 A

MD=Mb+M;)

~

Milc+MBC V ..'.~

MCB = Mea + MeB MBA = MBA

r:.======~::::;::!f

/

(a)

Figure 18.1: Analysis by the stiffness method. (a) Deflected shape and moment diagrams (bottom of figure) produced by the vertical load at D; (b) loads applied to the restrained structure; imaginary clamp at B prevents rotation, producing two fixed-end beams; (c) deflected shape and moment diagrams produced by a moment opposite to that applied by the clamp at B.

I /'M ADAb t II + \1

A

Mb

MilCB Milc MeB

MBA

+

M ilAB

(b) (c)

loading shown, the structure will defonn and develop shears, moments, and axial loads in both members. Because of the changes in length induced by the axial forces, joint B will experience, in addition to a rotation 8B, small displacements in the x and y directions. Since these displacements are small and do not appreciably affect the member forces, we neglect thePl .. With this simplification we can analyze the frame as having only one degree ofkinematic indeterminacy (Le., the rotation of joint B). . In the first part of the analysis, which we. designate as the restrained

condition, .we introduce a rotational restraint (an imaginary c1amp)at joint R (see Fig. 18.1h). The annition of the clamp transfonns the stmcture into two fixed-end beams. The analysis of these beams can be readily carried out using tables (e.g., see Table 12.5). The deflected shape and the corresponding moment diagrams (directly under the sketch of the frame) are shown in Figure 18.lb. Forces and displacements associated with this case are superscripted with a prime.

Since the counterclockwise moment M applied by the clamp at B does not exist in the original structure, we must eliminate its effect. We

'a-,...... _ --

• Section 18.2 . Structure Stiffness Matrix 685

do this in the second part of the analysis by solving for the rQtation OR of joint B produced by an applied moment that is equal in magnitude but opposite in sense to the moment applied by the clamp. The moments and displacements in the members for the second part of analysis are superscripted with a double prime, as shown in Figure 18.1c. The final results, shown in Figure 18.la, follow by direct superposition of the cases in Figure 18.lb and c.

We note that not only are the final moments obtained by adding the values in the restrained case to those produced by the joint rotation OR' but also any other force or displacement can be obtained in the same manner. For example, the deflection directly under the load IlD equals the sum of the corresponding deflections at D in Figure 18.1 band c, that is,

IlD = Ilh + 1l'D

..:1i;2J)lrSt;~ct~~~ ..stiff~;~;M~t~i~.........................................................

In the analysis of a structure using the direct stiffness method, we start by introducing sufficient restraints (i.e., clamps) to prevent movement of all unrestrained joints. We then calculate the forces in the restraints as the sum of fixed-end forces for the members meeting at a joint. The internal forces at other locations of interest along the elements are also determined for the restrained condition.

In the next step of the analysis we determine values of joint displacements for which the restraining forces vanish. This is done by first applying the joint restraining forces, but with the sign reversed, and then solving a set of eqUilibrium equations that relate forces and displacements at the joints. In matrix form we have

KA = F (18.1) where F is the column matrix or vector of forces (including moments) in the fictitious restraints but with the sign reversed, A is the column vector .. of joint displacements selected as degrees of freedom, and K is the structure stiffness matrix.

The term degree of freedom (DOF) refers to the independent joint displacement components that are used in the solution of a particular problem by the direct stiffness method. The number of degrees of freedom may equal the number of all possible joint displacement components (for example,3 times the number of free joints in planar frames) or may be smaller if simplifying assumptions (such as neglecting axial deformations of members) are introduced. In all cases, the number of degrees of freedom and the degree of kinematic indeterminacy are identical.

Once the joint displacements Il are calculated, the member actions (i.e., the moments, shears, and axial loads produced by these displacements) can be readily calculated. The final solution follows by adding these resuits to those from the restr~inf'.rl r~"lf' . .

....:;;. ..... -

• 686 Chapter 18 Matrix Analysis of Beams and Frames by the Direct Stiffness Method

The individual elements of the structure stiffness matrix K can be computed by introducing successively unit displacements that correspond to one of the degrees of freedom while all other degrees of freedom are restrained. The external forces at the location of the degrees of freedom required to satisfy equilibrium of the deformed configuration are the elementsof the matrix K. More explicitly, a typical element kij of the structure stiffness matrix K is defined as follows: kij = force at degree of freedom i due to a unit displacement of degree of freedomj; when degree of freedomj is given a unit displacement, all others are restrained.

18.3 The 2 x 2 Rotational Stiffness Matrix ~.iI

for a Flexural Member In this section we derive the member stiffness matrix for an individual flexural element using only joint rotations as degrees of freedom. The 2 X 2 matrix that relates moments and rotations at the ends of the member is important beca,Qse it can be used directly in the solution of many practical problems, such as continuous beams and braced frames where joint translations are prevented. Furthermore, it is a basic item in the derivation of the more general 4 X 4 member stiffness matrix to be presented .in Section 18.4 .

. Figure 18.2 shows abeam of length L with cnd moments Mj and Mj . As a sign convention the end rotations OJ and OJ are positive when clock

wise and negative when counterclockwise. Similarly, clockwise end moments are also positive, and counterclockwise moments are negative.

chord To highlight the fact that the derivation to follow is independent of the member orientation, the axis of the element is drawn with an arbitrary inclination a. .

In matrix notation, the relationship between the end moments and the resulting end rotations can be written as

(18.2)

Figure 18.2: End rotations produced by member where k is the 2 X 2 member rotational stiffness matrix. end moments. To determine the elements of this matrix, we use the slope-deflection

equation to relate end moments and rotations (see Eqs. 12.14 and 12.15). The sign convention and the notation in this formulation are identical to those used in the original derivation of the slope-deflection equation in Chapter 12. Since no loads are applied along the member's axis and no chord rotation t/J occurs (both t/J and the FEM equal zero), the end moments can be expressed as

(18.3)

........... - .......... - ....;. .....

• Section 18.3 The 2 x 2 Rotational Stiffness Matrix for a Flexural Member 687

2EIand M = -(0 + 20.) (18.4)

J L ' J Equations 18.3 and 18.4 can be written in matrix notation as

(185)[~;]= 2:/[~ ~J [:;] , By comparing Equations 18.2and 18.5 it follows that the.member rotational stiffness matrix k is

- = 2E/[2 IJ (18.6)k L 1 2 We will now illustrate t

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