UDC 624.072.3

An iterative method of moment distribution for space frames * B. RAWLINGS BSc MEngSc PhD AMlStructE

Synopsis A method is presented for the analysis of rigid space frames using iterative techniques. In no case are matrices larger than 3 x 3 used, so that the procedure may be carried out by manual methods or may, if desired, be programmed for computation using an elec- tronic computer. The method is illustrated by two examples, one nithout sway and one involving sway of the frame.

The elastic analysis of rigid non-orthogonal space frames is a problem of considerable complexity, as structures of this nature cannot be treated by pro- jecting them into planes and carrying out plane frame analyses. Each slope-deflexion relationship between end deformation and end moments, written for any member of the structure, must be transformed by resolution of the moments into the three components and the deformations into their similar components, so that the member characteristics are represented by a 3 X 3 matrix.

The moment distribution method of analysis, first developed for plane framesl, has been extended to include both orthogonal space frames2 and the general case of non-orthogonal space frames 3t4. Problems which include sway of frames can also be analysed by use of the principle of superposition.

The normal moment distribution procedure, as con- ceived by Hardy Cross, has the limitation that it is not iterative, so that errors which may be introduced during the process of computation are not eliminated in the next cycle of the process but are retained. Like- wise, the full degree of precision of calculation must be adopted at the start, as such a method does not permit of an improvement in accuracy being made during the solution of the problem. Furthermore, in the Hardy Cross procedure i t is normal to commence the analysis by assuming all joints to be encastrd, so that i t is im- possible to assume a likely configuration for the structure and to proceed from this point. In the case of plane frames, iterative methods exist which do not possess the disadvantages mentioned above. Amongst these may be mentioned the methods described by Gaspar Kani5 and by Bull and Sved6, who apply iterative

Fig 1

ln

Fig 2

methods to the solution of plane frames of various types, either restrained from, or free to sway.

In the present paper an outline is given of a method of analysis which is similar to those described above, which is here extended to the case of three-dimensional, non-orthogonal frames.

Analysis of Non-orthogonal Frames Consider firstly a prismatic member A B , of length S, and having linear elastic characteristics, which is sub- jected to end rotations and displacements, as shown in Fig 1. It has been shown4 that the moments induced at the ends of the member are given by

--c N, = - (0 n A - 0 n ~ )

S

Fvhere I1 and I, are the second moments of area for bending about the l and m axes and C is the torsional rigidity of the member, established for appropriate warping conditions. If the member axes ( I , m, n) are inclined to the reference set (x, y , z ) which is seIected for the whole structure, as shown in Fig 2, the direction cosines relating ( l , m, n) to (x, y , z ) may be written in an array [M], where

*Reader in Civil Engineering, University of Sydney, Australia.

THE STRUCTURAL ENGINEER SEPTEMBER 1966 No 9 VOLUME 44 295

;l 11 cl The moments

components along the (x, y , x) axes, by the relationship

[ ] = [ M ] [ ] and similarly the rotations may be resolved, if small, by

[ 3 = [ M ] [ 5. ] and the end deflexions by

and

f [ M ] - 1

+ [M] - '

r . I hese are the slope-deflexion equations for the end A , when resolved into the (.Y,J~,Z) directions.

In general this matrix equation can be rewritten in the form l x,,, 3 = L S,, 3 I K, j + I L , % -1 L 013

- -1

+ i ; L h , ; -t L X(dAL3 1 where double suffis notation is now introduced for moment components and Ivhere for the case of a pris- matic member, the stiffness matrix !SAiI3;

has thc \ . due

the induction matrix i I A I , 1 has the value 0 0 U 11

[M]-' [ 0 2 R , 0 ] [M] 2Hl 0 0

0 0 --H,

0 0 0

296 V O L U M E 44 N o 9 SEPTEMBER 1966 THE STRUCTURAL ENGINEER

Table 1

--R

1 R _-

Joint Jo in t B E

_ _ _ _ . _ ~ _

6 . 8 ] [ 0 6 . 8 0

[ ]

6 . 8 0

0 8.4 0 0 8.4 O I

-0.147 0 -0.147 0

-0.147 0 [ 0 0 -0.119 0 -0.147 0

0 -0.1 19

matrices may be derived as shown in Table 1 where B = EI/10. Distribution of moments then proceeds as shown in Table 2 until the values of the components of 101 remain virtually unchanged with further cycles of the process. The moments at the ends of the members are then calculated, using equation l , and are entered at the foot of the Table. The moments, after resolving back to the separate member axes, are shown in Fig 5, where they are represented vectorially in accordance with the right-hand screw rule.

Examble 2 In the second example a non-orthogonal rigid frame has been considered, in which loading has been applied at one joint, B , as in Fig 6. I t will be appreciated that under these conditions displacement of the joints of the structure will occur and it is necessary to conduct a sway analysis. I t will again be assumed that the frame is of uniform tubular section, having G = 0 - 4 E , and that the joints at A , B , C, D, E and F are all rigid. Proceeding as in the first example, and adopting member

Fig 4-fifember axes selected

As in other iterative procedures, the process is continued until the rotations remain virtually unchanged after a further cycle, a t which stage the structure may be assumed to be in its equilibrium configuration.

In order to derive the moments in each member, it is then necessary to substitute the values of [e] back into equation l and to resolve the moments back to the

r L 1 L $1 components.

Illustrations of Method Exanzple l Consider the case of the orthogonal steel frame of uniform tubular section, having a modulus of rigidity equal to 0 - 4 X the Young's modulus. The structure is .d tied at the joint E by a pinned member, to prevent sway, as shown in Fig 3. If the frame is loaded as shown, Fig 5-Moments in orthogonal space frame (ton in); the member and frame axes may be selected, for example, sign convention: right-hand screw rule (moments as in Fig 4, and values of the various member and joint are represented by double-headed arrows)

4 7:6

THE STRUCTURAL ENGINEER SEPTEMBER 1966 No 9 VOLUME 44 297

Table 2-Distribution of Moments in Frame

Joint B j- -0.119

[ ] (ton in)

1 B [ -1Y.9 I +44- 1

Joint E

-0.147 0 0 0 -0. 119

0 l 0 -0.4

I [l 2 . 9 1 + 5 0 - 6

I

I [ -1Y.9 1 +51.5

-0 * 565

Joint E

other end

[B I 3

axes as in Fig 7, the direction cosine matrices, induction and stiffness matrices may be written down as in Table 3, where B = EI / s . Consider now the moments developed if the joint B is displaced to some arbitrary extent 8, while keeping the joint B unrotated. As this structure has only a single degree of freedom for displacement, the joint C will move an amount dependent upon the movement of B , as shown in Fig 8. In the particular frame considered, B and C must remain in the same horizontal plane as A and D. Making the normal assumption that members do not change in length, the components of displacement of one end of each member referred to the tangent drawn from the other end and

sufficiently, the [ 5 ] components of moment may be

evaluated as in Table 6.

It is then necessary to scale these so that the corrected moments are in equilibrium with the actual force applied to the structure. Consider the equilibrium of the forces Fx, F, and F , and moments X , Y and 2 acting on the ends of a typical inclined member PQ as shown in Fig 9. From statics, taking moments about x , y and z axes through Q,

the components o r end moment in members thereby X p f- XQ -k F y P (zQ - ZP) - FzP bQ --P) = 0 produced are shown in Table 4, where B = E I / s . - The method of distribution of these moments follows YP + Y Q + F z P (%Q - %P) - FxP(ZQ - ZP) = 0 the same procedure as in the first example and is illus- trated in Table 5. When the process has converged Zp + Z, -1 F , p (ya - y p ) - F Y P ( x a - XP) = 0

298 VOLUME 44 No 9 SEPTEMBER 1966 THE STRUCTURAL ENGINEER

m U

X

n v)

Z o X ' T I U

T n

X U

n

n

g o i I

op2

U

I .

Fig 6-Non-orthogonal space f r a m e

D

E

L.

Y

Fig 7-Member axes selected

Fig 8

THE STRUCTURAL ENGINEER SEPTEMBER 1966 N o 9 VOLUME 44 299

Table 3

Member A B

0 $1

l + - 0 42

1 + 7 2 0

l 2

l

- -

+.\/z

+ - 2

' 0 + l 0 '

0 0 + l

$1 0 0 .

l + y 0 + l 0

1 - - 0 \ , 2 0

1 -- I -

' 43 l +z 1 2 + - 1 -1. -

2// 2 -

-0 35

-0 -35 43 $0.65

+0*35dT 4-0.35 1 $ 0 - 3 -0.3545

-0.35y'Z $0.65 i J o i n t U

--. -. .- . _ _

-B L -1- 4 097 -f- 0 * 566 $0.566 $8.028 -1- l a531 -0.566

I I Joint C

.. ___ ____. __ . .

+4.097 -0-566 -- - 1 -531 -0.566 +8*028 -0.566

T 7 -297 -- 1 a 5 3 1 -0.566 -1-7.297

0.2690 -0.0231 -0.0582 -0.0231 -O*I272 -0.0147

-0 - 1504 -0 .0582 -0.0147 -0.1504

-2 1 - 2 s I-' -0.2690 -t-O*0231

-0.0231 -0.12'72 -+ 0.0582 -0-0147

Fig 9 Fig 9

8 41

Fig 10-Moments i n fyanze (assuming F , = 100 units)

Table 4-Components of End Deflexion, and End Moment Developed by Sway, Prior to Distribution

Member E B

1-0 .75 J

Member B C

Member C F

[ + % l

Joint B Joint C

hTo\v, for all members connected to the joint B,

0 x F,, - F 2 FYI3 -

- - 0.254 F

0.437F

c F,, - - 0 where the summations extend over the three members A B , E B and CB. Substitution of the results from the moment distribution into these equations gives the scaling factor, for F , = 100, of 2 -30.

The resultant scaled moments are shown in Table 7, together with the components referred to the individual member axes, obtained by resolution using the equation

These moments are depicted pictorially in Fig 10 and the support reactions in Fig 1 1 .

0-0051 F

0.254 F

Y

Fig 1 l -React ions on frame

THE STRUCTURAL ENGINEER SEPTEMBER 1966 No 9 VOLUME 44 30 1

Table 5-Distribution of Moments in Frame

J o i n t C'

I .l-

-0 * 42

R6 + 0 - 5 5 -3 .50 ] ---0*15

-0 -254

-1- 0 15 l +[ +0-465 j - - -_

'[ -3.495 j +0*546

-0 4-18

3 02 VOLUME 44 No 9 SEPTEMBER 1966 THE STRUCTURAL ENGINEER

L

U X X

U

nn X 4

nn X 4

I

n 0 m m

m o o + l I 0 - 0 - . . .

U

n

c'"? O N O b b m

m o o I i l U

+ + I 1 1 U

l i ir-1 - m m mma;

- 0 0 ?C??

i I + m c c m 0

b o b 3 - 3 . . . l + i

+

U

cq U

0 0 0 - . . . + + l II

+ 1 -t I ++ U- - "

n

? m m

O W 0 L' Q

l I + 1 + 1

I 1 - m m m m a ;

b o o OT-?

I l l

n

??C? + i I

m m - b b

m o o

I 1 1 1

I l I

I I -

O N O - +++ II

-NN m o c + + l I l +

n

2 m I

m o m o m m a - m - m b

m o w o m 0 oao b b m l I + O N O - . . .

I

U I + + II

I

References 1. Cross, Hardy, ' Analysis of continuous frames by distributing

fixed-ended moments ', Trans. A S C E , Vol. 96, 1932. 2. Matheson, J . A . L., ' Moment distribution applied to rectangu-

lar rigid space frames ', Jour. I C E , Vol. 29, No. 3, 1948. 3. Catuncanu, C., Anastasescu, D. and Munteanu, I . , ' The

generalization of the moment distribution method for the analysis of space frames I , Buletinul Znstitutului Politehnic, Timisoara, Vol. 3, No. 17, 1958.

4. Rawlings, B., ' The general moment distribution analysis of space frames ', Slructural Engineer, Vol. XXXVIII No. 6, Junc 1960.

5. Kani, G., Analysis of Multi-Storey Frames, London, Crosby Lockwood, 1957.

6. Bull, F. B. and Sved, G., Moment Distribution in Theory and Practice, T-ondon, Pergamon, 1964.

THE STRUCTURAL ENGINEER SEPTEMBER 1966 No 9 VOLUME 44 303