Pergamon 0045-7949(94)E0181-Z

Compurrrs & Srructures Vol. 53. No. 6. pp. 1275-1280. 1994 Copyright 0 1994 Elwier Science Ltd

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EFFECT OF HORIZONTAL DIAPHRAGM FLEXIBILITY ON THE P-DELTA ANALYSIS

M. M. El-Hawary

Department of Civil Engineering, Kuwait University, P.O. Box 5969, 13060 Safat, Kuwait

(Received 22 July 1993)

Abstract-This paper investigates the importance of including the effects of the flexibility of the horizontal diaphragms when using the &delta method of analysis, especially when considering the loads applied to intermediate frames on trusses that are not part of the lateral force resisting system. Analyses were conducted for structural systems with a variable number of stories, number of bays and diaphragm stiffnesses and supported by rigid jointed plane frames or vertical trusses.

INTRODUCTION

The present procedure used to design steel compression members in framed construction-the effective length method-is a first order method using a magnification factor to approximately compensate for the second order effects caused by lateral deflec- tion and axial load. In the first order analysis, the effect of sway deformations on the shear equilibrium equations and the influence of axial force on the stiffness of the members are neglected.

The second order effects can be directly taken into account by using a second order analysis. Here, the shear equilibrium equations are formulated on the deformed structure. This process directly accounts for what is commonly called the P-delta (P-8) moments produced by the vertical loads acting through the sway displacement of the structure. The P-6 moments are classified as secondary effects and hence when included in an analysis or design constitute a second order analysis.

In recent years extensive work has been done to show the superiority of the P-6 method over the first order method when applied to the design of steel (and to a lesser degree concrete) frames [ 1,2]. Much of the work deals with the second order analysis being applied to planar structures having all deflections occurring in the plane of the frame. For a typical rectangular building structure this deflection con- straint is brought about by assuming that the frames of the structure are connected by rigid horizontal diaphragms. That is, the diaphragm deflection in the plane of the diaphragm is assumed to be infinitely small compared to the deflection of the vertical resisting systems.

Although most analyses of multistoried buildings treat the slabs as rigid horizontal diaphragms, some work has been done where this constraint has been relaxed. Here, the flexural stiffness of the slab com- pared to that of the supporting shear walls was

treated: by considering the slab in a way analogous to a ‘Timoshenko beam’ and using the slope deflec- tion method [3]; by considering the slab as a series of beams and using a finite difference analysis [4]; by using the finite element method [5]; or by introducing a modification factor and using rigid diaphragms in the analysis [6].

Recent building codes require that consideration be given to the plywood diaphragm deflections. The plywood diaphragm must be considered a flexible diaphragm [7].

As noted above, the P-6 method is sometimes used to effectively account for one of the principal second order effects in determining the design loads applied to the vertical lateral force resisting systems. This paper shows the importance of including the effects of the flexibility of the horizontal diaphragms when using the P-b method especially when considering the loads applied to intermediate frames or trusses that are not part of the vertical lateral force resisting system.

APPROACH AND PROCEDURE

To illustrate the procedure followed in this study, first the classical P-6 analysis in which the dia- phragms are assumed rigid is considered [8]. For simplicity all the fames in each story are chosen to be the same and subjected to the same story deflection 6. Considering a one story frame, a fictitious shear force V’ will be produced to resist the second order moments caused by the vertical forces acting on the displaced structure. Thus,

V’= PCs/L,, (1)

where P is the total vertical force acting on the frame and L,. is the height of the story. The deflection 6 is determined from a first-order analysis that, in this

1275

1276 M. M. El-Hawary

instance, does not include V’. Thus, it is necessary to do another first-order determination of the lateral deflection based on any originally imposed lateral forces and the additional P-6 forces (V’) to deter- mine a corrected S. This procedure can be repeated until there is an insi~ifi~ant change in the corrected value of S from one iteration to the next.

In a similar manner, the above process can be applied to a multistory structure. In this case the fictitious shear forces, VI, are

where

ca

Vi = additional shear in story i due to the sway forces,

I: Pi = sum of the vertical forces in story i, L, = height of story i,

&_, ,6, = displacements of levels i - 1 and i respect- ively.

The second order force Hj at level i is the difference between the fictitious shears or

H;= V;+,- V;. (3)

The forces H( are added to the applied lateral loads and the frame reanalysed for the new 6s. The method is repeated until convergence.

Now consider the case in which the diaphragm is assumed rigid but the frames in the story are not the same. The structure is rectangular in plan, one story high and supported by three plane frames. Frames on lines 1 and 3 (Fig. 1) represent rigid frames and serve as the vertical lateral forces resisting systems (LFRS). The frame on line 2 represents an unbraced simple frame which does not participate in resisting the lateral loads. It is assumed that in addition to the vertical loads, which are taken as acting on the

3 2

columns of the frames, there are lateral loads, H, acting parallel to the plane of the frames.

The notations used in sequel are: F = vertical load on each column in the lateral

resisting systems, Q = vertical load on each column in the simple

system, H = lateral load per unit length, A = diaphragm deflection at the location of the

simple frame. The second order moment, mR, due to the sway of the middle frame is

mR = 2QS. (4)

To counteract this moment a fictitious shear force Vm, will be developed, where

VmR = 2QS IL,. (5)

Since the simple frame can not resist this force, the force is transferred to the diaphragm as a concen- trated load. In return, half of the force Vm, will be transferred to each rigid frame. This additional force applied to the end frames will be added to the second order shear force produced by the end frames as required to counteract the second order moment there.

The second order shear force on the end frames is determined in eqn (1) by substituting P = 2F. Thus, the final second order shear force acting on the end frames is

VR = K2F + QYLI~. (6)

This shear force can be added to any previously obtained forces acting at the top of the end frame and will cause a new value of 6. The above process can be repeated until the new value determined is the same as the previous old value of 6. At this point one can proceed to design the structure.

1

I Tdbutsty losd I Tributuy lad Tributary tolinl3 toline

3 2 1

Fig. 1. System consists of horizontal diaphragm, two end LFRS and a middle simple frame.

Effect of horizontal diaphragm flexibility on the P-delta analysis 1277

For a multistory structure with rigid diaphragms but not having all frames in a given story the same, one can use the method just described on to the P-6 analysis. That is, the second order shear at a given story is found to be

v,,=(2~+~Qi)(si_s,+,)/L,i, (7)

where Qi represents the sum of the vertical loads acting on the simple frames in the story. The second order force Hi, at level i is the difference between the fictitious shears or

Hi, = VRi+, - V,,. (8)

The forces Hi, are added to the applied lateral loads and the frame reanalysed for the new 6s. The method is repeated until convergence.

The second order shear can also be determined in a similar fashion when the in-plane flexibility of the diaphragm is taken into account for structures not having all frames in a story the same. The second order moment due to the sway of the middle frame will be

m =2Q(6 +A). (9)

The fictitious shear force due to this moment is

V,,, = 2Q (6 + A)/&. (10)

Half of this force will be added to the second order shear force acting on each rigid frame so the second order shear force, V, acting on the rigid frame will be

V = [2F + Q + Q(A/s)]s /L,. (11)

This force will increase the deflection 6 which will increase the force V,. Note however, that V,,, acts as a lateral load on the diaphragm so the increase in V,,,

will also increase the deflection which will in turn increase the shear force V. After a sufficient number of interations V will converge to a final value.

To simplify the computational work that follows, the two iterations described above were assumed uncoupled so they may be considered as two separate iterations. In the first iteration one iterates for A using eqn (10) assuming 6 equal to zero. The final value of A is then substituted into eqn (11) and iterations performed to obtain the final value for the second order shear V.

The effects of including the flexibility of the dia- phragm on the P-4 second order shears can be measured by the index R where

R = (V’- Vx)/V’, (12)

where V’ is the converged value of the second order shear force considering the diaphragm as flexible [from eqn (1 I)]; Vk is the converged value of the

second order shear force assuming the diaphragm is rigid [from eqn (7)].

In the calculations that follow, the deflections, A, will be calculated by considering the diaphragms as simple beams spanning between the support provided by the LFRS. As the length-to-width ratio of the diaphragm is not too large, the shear terms in the strain energy expression are included.

The deflections of the LFRS-vertical trusses or rigid frames-are calculated using a locally developed general purpose frame analysis computer pro- gram [9].

The reference frame used in this study is frame 15- 1 used by Korn and Galambos [IO]. The frame has L, = 14 ft, E = 30,000 ksi, F, = 36 ksi and D = 20 ft, is subjected to a wind load of 20 psf and vertical dead and live loads of 150 psf and 100 psf respectively. Some of the member sizes used by Korn and Galambos were not found in the eighth edition (1980) of the AISC Manual so the nearest standard sizes were used (see Table 1). The frame, in general, is 15 stories high but the number of stories considered each time in the present analysis is dependent on the individual case. As shown in Fig. 1 the vertical load contribution to the middle frame is double that applied to each end frame, so in all cases considered, Q =2F.

The lateral load was taken to be 0.1 k/ft along the edge of the diaphragm for the top story and 0.2 k/ft for all the other stories. The typical vertical loadings are shown in Fig. 2. The vertical loads were idealized as if acting only on the junctions and the values of the Qs are shown in Table 2. The deflections of some frames evaluated turned out to be very large. These frames are used only to illustrate the problem and surely would not be used in practice.

ANALYSIS AND RESULTS

Two different forms of the LFRS were used to support the horizontal diaphragms, rigid frames and vertical trusses. Structures supported laterally by rigid jointed frames are usually more flexible than

Table I. Sizes of the reference structure members

Story Column Girder

1 W 8x18 W 8x21 2 W 8x28 W 10 X 26 3 W 10 x 39 W 12 x 30 4 do Wl2x40 5 W 12 x 40 do 6 W 12 x 50 Wl2x45 7 Wl2x58 W 14 x 53 8 W 14 x 61 W 14 x 61 9 w 14 x 74 do

10 w 14 x 90 w 14 x 74 11 w 14 x 120 W 14 x 82 12 do w 14 x 90 13 do W 16 x 89 14 W 14 x 132 Wl6x 100 15 w 14 x 145 W 18 x 97

I278 M. M. El-Hawary

Rmax 1CO

@J .

@J . . .,..

4() .

0 50 loo 150 200 250 300 350 4ao 450 500 550 600

El millions k.ff

Fig. 2. R,,, vs EI for 5 story, 4 bay structure, rigid jointed frames.

those supported by vertical trusses. The effect of considering the diaphragm’s flexibility on the second order P-6 shear forces for a differing number of bays, number of stories and diaphragm stiffness is discussed for each of the LFRS.

To illustrate the typical method of analysis the case of a one story two bay structure consisting of two rigid jointed end frames, a simple interior frame and a diaphragm having a stiffness equal to 8 x IO6 k ft* is discussed in detail. The sizes of the frame members are those of the members of the first story of the reference structure given in Table 1.

First consider the diaphragm as being rigid. Since the simple interior frame cannot resist any lateral load, half of the total lateral load will be carried by each end frame. That is h = 1.7 kips, where h is the lateral load applied to each end frame. Using eqn (6) we get the iteration shown in Table 3. The deflection 6 due to the lateral load was obtained using the frame analysis computer program, the second order shear force VR was calculated using eqn (6) and the lateral load for the second cycle was obtained as the sum V, + h for the first cycle. After three cycles the second order shear force V, converged to Vi = 0.0725 k. In the case of a multi-story structure the same procedure was followed for each story.

Now, accounting for the diaphragm flexibility, the diaphragm will act as a simple beam spanning be- tween supports provided by the end frames. The diaphragm deflection at the location of the simple interior frame is given as

A = (5HL4 + 8P’L3)/384EI

+ [k(HL* + 2P’L)]/8GA, (13)

where A is the cross-sectional area of the diaphragm, L is the span between the adjacent LFRS, k is the

Table 2. Vertical loads Q

Story Vertical load Q”

I 21 2 63 3 105 4 147 5 189

Table 3. Second order shear force for a one story two bay structure supported by rigid jointed frames assuming the diaphragm is

rigid

Cycle h’ 6in V”R

I 1.7 0.278 1 0.0695 2 1.7695 0.2895 0.0724 3 I .7724 0.2899 0.0725

I .7725

shape factor associated with shear deflections, I is the centroidal moment of inertia of the diaphragm, E is

Young’s modulus and G is the shear modulus. For illustrative purposes, the diaphragm is taken as

1 in thick, k as 1.2, Poisson’s ratio as 0.3 and each bay as 17 ft long. Note that eqn (13) is the deflection of the diaphragm subjected to a uniformly distributed load, H, and a single concentrated load, P’, at the midspan distance between the end frames. If there are more than one simply framed interior frames, eqn (13) should be rewritten to account for the different displacements of the resulting concentrated load(s) associated with the interior frames.

Substituting the values into eqn (13) one obtains

A = 0.0049 + 0.0026P’.

Starting with P’ = 0.0 one gets A = 0.0049 in. For the next cycle P’ = 2QA/L, which will produce a new A which will produce a new P and so on. The iteration is shown in Table 4 and it is seen that the change in A is negligible in this case. Substituting with the final value of A in eqn (11) we get the iteration shown in Table 5. The iteration proceeds as before except we include the effect of diaphragm flexibility in the second order shear equation. The iteration is shown to converge to a value V’ = 0.073 1 k. Using eqn (12) one finds that the ratio of the effect of the consider- ation of the diaphragm’s flexibility on the second order shear force to the second order shear force considering diaphragm flexibility gives R = 0.82%.

The same method was used for different numbers of stories and bays, and for different diaphragm stiffnesses. The ratio R was determined for each diaphragm and the largest one is reported as R,,, . In all cases R,,, pertained to the deflection of the uppermost diaphragm.

The variation of R,,, with number of bays for EI = 8 x lo6 k ft* is shown in Fig. 3 and the variation with EI for a five story, four bay structure is shown in Fig. 2.

Table 4. Variation of A with P for the one story two bay structure with

EI=8x 106kft’

P A

0.0 0.0049 0.0012 0.0049

Effect of horizontal diaphragm flexibility on the P-delta analysis 1279

Table 5. Second order shear force for a one story two bay structure supported by rigid jointed frames

and EI=8x 106kft2

h’ &in v”

1.7 0.278 1 0.0701 1.7701 0.2896 0.0730 I .7730 0.2900 0.073 I 1.773 1

The previous results pertained to the diaphragm being supported by a rigid frame LFRS. The same calculations were done for the diaphragms supported by the LFRS consisting of vertical trusses. The trusses had the same member sizes as the reference frame (Table 1) except they were braced by Wg x 18 members.

The same method of analysis discussed above was used for different combinations of number of bays, number of stories and diaphragm’s stiffness; the results are shown in Figs 4 and 5.

DISCUSSION AND CONCLUSIONS

The effect of the consideration of the diaphragm’s flexibility on the second order P-6 shear forces depends on the relative flexibility of the diaphragms compared to the supporting system. The effect of the consideration of the diaphragm’s flexibility is cap- tured by the parameter R,,, and is seen to increase with the decrease in the Elof the diaphragm and with the increase in the number of bays. Further, one observes that for the same EI, number of stories and number of bays, R,,, is different for the two LFRS considered here, being greater for the stiffer vertical truss.

The interaction between the diaphragm stiffness and the LFRS stiffness is quite complex, depending, in part, on the number of stories in the structure. The P-S calculations generally show that as the number of stories increases so does the deflection A due to the P-6 effect. However, the flexibility of the LFRS also increases with the story height, so the net results on R,,, could be an increase or decrease, depending on the individual structure. For example in Fig. 4 R,,, is seen to first increase and then decrease with the

Rmsx 100

Rmax

80

M)

4()

El in k.ft*

-EI=8E6

‘EI=5E7

*EI=1.456EB

* El = 4.8EB

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

No. of bays No. of bays

Fig. 3. R,,, versus no. of bays, EI = 8E6 k ft’, rigid jointed Fig. 5. R,,, frame.

versus no. of bays for a five story system supported by vertical trusses.

,I:~..~:.:~~

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

-Ei=BEB

+El=E.E7

* El = 1.456E8

* El = 4.8EB

Number of stories

Fig. 4. R,., versus no. of stories for a four bay system supported by vertical trusses.

increase in the number of stories. In Fig. 3, however, R,,, monotonically increases with the increase in the number of stories.

The case where EZ = 8 x IO6 k ft” is near the stiff- ness of a wooden diaphragm. In this case the effect of the consideration of the diaphragm’s flexibility on the second order forces is always considerable for any number of stories and bays as shown in Figs 4 and 5 where the values of R,,, approach 85% for the two-story four-bay structure supported by vertical trusses.

For concrete diaphragms, which are generally con- sidered to be rigid, the effect of the consideration of diaphragm flexibility on the second order shear forces may also be considerable depending on the stiffness of the LFRS. For example in Fig. 4 the largest value of R,,, is 24% for a two-story four-bay structure supported by vertical trusses and a diaphragm stiff- ness equal 1.456 x lo8 k ft2 which is representative of a 4 in thick concrete diaphragm. However, when the same consideration is made for the more flexible rigid jointed frames of this study Fig. 2 shows R,,,,, to be just slightly over 1% for a five-story four-bay struc- ture. In this instance it is clear that the inclusion of the diaphragm’s flexibility in the calculation would be unnecessary.

In the P--6 analysis the flexibility of the horizontal diaphragms should be included in the analysis for wooden diaphragms since the diaphragms are always quite flexible when compared to the flexibility of the LFRS. On the other hand for concrete and horizontal truss type steel diaphragms the flexibility of the

1280 M. M. El-Hawary

diaphragms should be included in the P-6 analysis 4. if the diaphragm deflections are of the same order as the deflections of the vertical supporting system. 5, If the deflections of the diaphragms are very small compared to the deflections of the vertical supporting systems then a standard P-b analysis will suffice.

6.

REFERENCES

Ergin Atimtay, Lateral load analysis of tall buildings with flexible floor slabs. Inr. ConJ Engng for Protection From Natural Disasters, Asian Institute of Technology, Bangkok (January, 1980).

I. B. R. Wood, D. Beaulieu and P. F. Adams, Column 7. W. H. Bower, Lateral analysis of plywood diaphragms. design by the P delta method. J. struct. Div., ASCE 102 J. slrucf. Div., ASCE lOO(ST4), 759-772 (April 1974). (ST2), Proc. Paper 11936, 41 l-427, (February, 1976). 8. B. G. Johnston, Guide to Stability Design Criteria for

2. B. R. Wood, D. Beaulieu and P. F. Adams, Further Mefal Structures, John Wiley, New York (1976). aspects of design by P. delta method. J. strucf. Div., 9. K. M. Romstad, BIGSOV User Manual. Unpublished ASCE 102(ST3), 487-500 (March, 1976). Report, Civil Engineering Department, University of

3. J. E. Goldberg, Analysis of multistory building consid- California, Davis, California (1979). ering shear wall and floor deformations. Tall Buildings, 10. A. Korn and T. V. Galambos, Behavior of elastic plastic Proc. Symp. University of Southampton, April 1966, frames. J. sfruct. Div., ASCE 94(STS), 1119~1140 pp. 349-375, Pergamon Press, Oxford (1967). (May, 1968).

Wai Tso and Adel A. Mohamaud, Effective width of coupling slabs in shear wall buildings. J. sfrucl. Div. ASCE 103, 573-586 (March, 1977). Y. C. Wong and A. Coull, Structural behavior of floor slabs in shear wall buildings, Advances in Concrete Slab Technology, Proc. Inr. Conf. Slabs, Dundee University, April, 1979, pp. 301-312. Pergamon Press, Oxford (1980).