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 set­ tingup 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 sim­ plification 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 move­ ment. The moments induced at the ends of each member are fixed·end moments. This procedure is similar to that used in the moment distribu- '. tion method in Chapter 13. 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 the·frame in Figure I8.la. This frame is composed, of two members connected by a rigid joint at B. Under the

•

•

•

.•.."' .....

684

Chapter 18

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

p

B

I / !'

t 1 t,

I /' t

Ab

M

+

\1

\1

A

A

MD=Mb+M;)

Milc+MBC

V

~ .. '.~

Mb

MilCB Milc

MCB = Mea + MeB

MeB

MBA = MBA I

AD

I

I •

MBA

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

+ il MAB

/ (a)

(c)

(b)

Figure 18.1: Analysis by the stiffness method. (a) Deflected shape and moment diagrams (bot­

tom of figure) produced by the vertical load at D; (b) loads applied to the restrained structure; imag­ inary 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.

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 rota­ tion 8B , small displacements in the x and y directions. Since these dis­ placements 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 of kinematic 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 stmc­ ture into two fixed-end beams. The analysis of these beams can be read­ ily 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

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 super­ scripted 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 Fig­ ure 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 deter­ mined for the restrained condition. In the next step of the analysis we determine values of joint displace­ ments for which the restraining forces vanish. This is done by first apply­ ing the joint restraining forces, but with the sign reversed, and then solv­ ing 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 struc­ ture stiffness matrix. The term degree of freedom (DOF) refers to the independent joint displacement components that are used in the solution of a particular prob­ lem 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 free­ dom 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' . .

.

... ..... :;;.

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685

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 ele­ mentsof the matrix K. More explicitly, a typical element kij of the struc­ ture stiffness matrix K is defined as follows: kij = force at degree of free­ dom 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

~.iI

The 2 x 2 Rotational Stiffness Matrix for a Flexural Member

In this section we derive the member stiffness matrix for an individual flex­ ural 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 practi­ cal problems, such as continuous beams and braced frames where joint translations are prevented. Furthermore, it is a basic item in the deriva­ tion of the more general 4 X 4 member stiffness matrix to be presented .in Section 18.4 . . Figure 18.2 shows beam of length L with cnd moments M j 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. 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

a

chord

(18.2) Figure 18.2: End rotations produced by member end moments.

where k is the 2 X 2 member rotational stiffness matrix. 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

and

2EI M·J = -(0· L '

687

The 2 x 2 Rotational Stiffness Matrix for a Flexural Member

+ 20.) J

(18.4)

Equations 18.3 and 18.4 can be written in matrix notation as

[~;]= 2:/[~ ~J

[:;] ,

(185)

By comparing Equations 18.2and 18.5 it follows that the.member rota­ tional stiffness matrix

k is - = 2E/[2 k

L

IJ

(18.6)

1 2

We will now illustrate the use of the preceding equations by solving a number of examples. To analyze a structure, it is necessar.y to identify the degree of freedom first. After the degree of freedom has been identi­ fied, the solution process can be conveniently broken down into the fol­ lowing five steps:

1. Analyze the restrained structure and calculate the clamping forces at the joints. 2. Assemble the structure stiffness matrix. 3. Apply the joint clamping forces but with the sign reversed to the original structure, and then calculate the unknown joint displacements using Equation 18.1. 4. Evaluate the effects of joint displacements (for example, deflections, moments, spears). 5. Sum the results of steps 1 and 4 to obtainthe final solution.

EXAMPLE 18.1 Using the direct stiffness method, analyze the frame shown in Figure 18.3a.

The change in length of the members may be neglected. The frame con­

sists of two members of constant flexural rigidity EI connected by a rigid

joint at B. Member Be supports a concentrated load P acting downward

at midspan. Member AB carries a uniform load w acting to the right. The

magnitude of w (in units ofload per unitlength) is equal to 3P/L.

p

B

1 L

W=

J

Solution

With axial deformations neglected, the degree of kinematic indetermi­

nacyequals 1 (this structure is discussed in Sec. 18.1). Figure 18.3b ilhiS­

trates the positive direction (clockwise) selected for the rotational degree

of freedom at joint B.

1----- L - - - i

(a)

Figure 18.3: (a) Details of frame;

Step 1: Analysis of the Restrained Structure

With the rotation

at joint B restrained by a temporary. clamp, the structure is transformed [continues on next page]

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....... ..- -

•

•

..-.-.:i.. .......

688

Chapter 18

Matrix Analysis of Beams and Frames by the Direct Stiffness Method

Example lB. 1 continues . ..

0.125 PL

degree of freedom

Ih~_

0.125 PL

c

E 18 ,

\

I

\

I

I

M

I

,I

(

!

A

A (b)

_ 1

I I

.)0.25PL

0.125 PL

~)0.125PL

~ 0.25 PL

(d)

(c)

Figure 18.3: (b) CUFed arrow indicates posi­ tive sense of joint rotation at B; (c) fixed-end moments' in restrained structure produced by "applied loads (loads omitted from sketch forclar­ ity); the clamp at B applies moment Ml to the ~tructure (see detail in lower right corner of fig­ ure); (d) moment diagrams for restrained struc­ ltu'e:

into two fixed-end beams (Fig. 18.3c). The fixed-end moments (see Fig.

12.Sd) for member AB are

~~2 ;; _

3: (~~) ;;

,

,PL

MBA = -MAB =

PL 4

(18.7)

4

... '(18.8)'

and for member Be (see Fig. 12.Sa),

M' = _PL Be 8 ,

MeB

(18.9)

,PL

= -MBc = 8

(18.10)

Figure 18.3c shows the fixed-end moments and the deflected shape of the restrained frame. To illustrate the calculation of the restraining moment Mt> a free-body diagram of joint B.is also shown in the lower right comer of Figure I8.3c. For clarity, shears acting on the joint are omitted. From the requirement of rotational eqUilibrium of the joint ('2:MB = 0) we obtain . PL PL M

+-+

48

1

o

from which we compute _ PL M I-

(18.11)

S In this I-degree of freedom problem, the value of M t with its sign reversed is the only element in the restraining force vector F (see Eq.

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Section 18.3

689

The 2 X 2 Rotational Stiffness Matrix for a Flexural Member

18.1). Figure 18.3d shows the moment diagrams for the members in the restrained structure.

Step 2: Assembly of the Structure Stiffness Matrix To assem­ ble the stiffness matrix, we introduce a unit rotation at joint B and calcu­ late the moment required to maintain the deformed configuration. The deflected shape of the frame produced by a unit rotation at joint B is shown in Figure 18.3e. Substituting OA = Oe 0 and OB = 1 rad into Equation 18.5, we compute the moments at the ends of members AB and Be as

[MAB] MBA

2El[2 L 1

2El[2 L 1

'i

+ ~I~J= +

,

(e)

[4EIJ

L,-:-=~~-..;::-...,......,""'c

2EI

L

These moments are shown on the sketch of the deformed structure in Figure 18.3e. The moment required at joint B to satisfy equilibrium can be easily determined from the free-body diagram shown in the lower right corner of Figure 18.3e. Summing moments at joint B, we compute the stiffness coefficient K" as

Kl1

4El L

=-+

4EI 8EI =L L

'I

\ .

a~J~ [Ul]

and

[M BC ] MeB

4EI I L I

2EI

L

(f)

(18.12)

In this problem the value given by Equation 18.12 is the only element of the stiffness matrix K. The moment diagrams for the members corre­ sponding to the condition OB = 1 rad are shown in Figure 18.3/

Figure 18.3: (e) Moments produced by a unit rotation of joint B; the stiffness coefficient KII represents the moment required to produce the unit rotation; (f) moment diagrams produced by the unit rotation of joint B;

Step 3: Solution of Equation 1S.1

Becausetbis problem has only 1 degree offreedom, Equation 18.1 is a simple algebraic equation. Suhsti­ tuting the previously calculated values of F and K given by Equatiqns 18.11 and 18.12, respectively, yields .. . (18.1)

Ka =F

8EL 0 __ PL L B 8

(18.13)

Solving for OR yields (18.14)

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[continues on next page]

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690

Chapter 18

Matrix Analysis of Beams, and Frames by the Direct Stiffness Method

Example 1B.1 continues . ..

The minus sign indicates that the rotation of joint B is counterclockwise, that is, opposite in sense to the direction defined as positive in Figure 18.3b.

Step 4: Evaluation .of the Effects of Joint Displacements Since the moments produced by a unit rotation of joint B are known from step 2 (see Fig. 18.3j), the moments produced by the actual joint rota­ tion are readily obtained by mUltiplying the forces in Figure 18.3fby e R given by Equation 18.14; proceeding, we find /I

_

MAB -

/I

2EI Q L UB

-

4EI

_

__

2EI L

MeB I/

e

B

(18.15)

PL

MBA = TeB M'l _ 4EI Q BC L uB

PL 32

-16

(18.16)

PL 16

(18.17)

__

-

PL

(18.18)

= - 32

The double prime indicates that these moments are associated with the joint displacement condition.

Step 5: Calculation of Final Results The final results are obtained by adding the values from the restrained condition (step 1) with those produced by the joint dispiacements (step 4).

= MAE + MAE = - PL 4 I

MAB

/I

+

(PL)

- 32

9PL =-3'2 3PL

=-­

16

0.142PL

r""

I

+ M Bc =

I

+ MCB

M Bc

=

M CB

= MCB

M Bc

."

O.S3L

L

~""'-'--">.

9PL

/I

=

-

PL 8

PL 8

_ 3PL + (PL) -16 --16 (PL)'"

+ -

32

=

3PL 32

32

(g)

Figure 18.3: (g) Final moment diagrams pro­ duced by superimposing moments in Cd) with those in (I) multiplied by Bg.

The member mom.ent diagrams can also be evaluated by combining the diagrams from the restrained case with those corresponding to the joint displacements. Once the end moments are known, however, it is much easier to construct the individual moment diagrams using basic princi­ ples of statics. The final results are shown in Figure 18.3g.

'"

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Section 18.3

691

The 2 X 2 Rotational Stiffness Matrix for a Flexural Member

EXAMPLE 18.2

Construct the. bending moment diagram for the three~span continuous ibeam shown in Figure 18.4a. The beam, which has a constant flexural lrigidity EI, supports a 20~kip c.oncentrated load acting at the .center of ispan Be. In addition, a uniformly distributed load of 4.5 kips/ft acts over . ..... . Ithe length of span CD.

,Solution inspection of the structure indicates that the degree of kinematic 'indeterminacy is 3. The positive directions selected for the 3 degrees of freedom (rotations at joints B, C, and D) are shown with curved arrows in Figure 18.4b.

! An

Step 1: Analysis of the Restrained Structure

The fixed-end moments induced in the restrained structure by the applied loads are cal­ culated using the formulas in Figure 12.5. Figure 18.4c shows the moment diagram for the restrained condition and the free-body diagrams of the joints that are used to calculate the forces in the restraints. Considering moment eqUilibrium, we compute the restraining moments as follows: Joint B: Joint C: Joint D:

+ 100

0

M[

150

0

M2

\.-150+M3

0

M3

Ml

-100

+ M2 +

Figure 18.4: (a) Details of continuous beam;

(b) curved arrows indicate the positive direction

of the unknown joint rotations at B, C, and D; (c) moments induced in the restrained structure

by the applied loads; bottom figures show the

moments acting on free-body dia.~rams of the

clamped joints (shears and reactions omitted for

clarity);

= -100 kip·ft = -'-50 kip·ft . 150 kip·ft

1

20 kips

Reversing the sign of these restraining moments, we construct the force vector F: :~ A

F

100] . 50 kip·ft [ -150

4.5 kips/ft

~IIW ,,___ / . . . c~ _/~~~ .

. _. . . . . ­

(18.19)

J?

"iW~ 150

150

150

20'

B

-..I..- 20' --I

I'C,M2 1'[/