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U. FOREST SERVICE RESEARCH PAPER FPL 43 OCTOBER U. S. DEPARTMENT OF AGRICULTURE FOREST SERVICE FOREST PRODUCTS LABORATORY MADISON, WIS. LATERAL STABILITY OF DEEP BEAMS WITH SHEAR-BEAM SUPPORT

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U. FOREST SERVICE RESEARCH PAPER FPL 43 OCTOBER

U. S. DEPARTMENT OF AGRICULTURE FOREST SERVICE

FOREST PRODUCTS LABORATORY MADISON, WIS.

LATERAL STABILITY OF DEEP BEAMS WITH SHEAR-BEAM SUPPORT

The FOREST SERVICE of the U. S. DEPARTMENT OF AGRICULTURE is dedicated to the principle of tiple use management of the Nation’s forest resources for sustained yields of wood, water, forage, wildlife, and recreation. Through forestry research, cooperation with the States and private forest owner’s, and management of the National Forests and National Grasslands, it – directed by Congress – to provide increasingly greater service to a growing

SUMMARY

The stability of roof and floor systems whose proportions allow

lateral buckling of the supporting beams is analyzed, with particular

attention to the stabilizing influence of the stiffness of the

attached deck. For the class of problems considered, the stabilizing

effect is shown to be mathematically analogous to that of an axial pull

acting on the beams along the line of deck attachment. The model em-

ployed for the deck is a "shear-beam" in which the effects of shear

are primary and those of bending are negligible. Such a beam is a dual

of the usual Euler-Bernoulli beam.

The main result is a pair of linear differential equations govern-

ing lateral displacement and twist in which the influence of vertical

deflection is disregarded. In particular cases, a power-series solution

was assumed and eigenvalue pairs were determined with the aid of a

computer. Numerical results are presented in the form of curves for

four cases where the vertical loading and deck attachment are at the

centroidal axis of the beam.

i

CONTENTS

Page

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

DEFINITIONS OF SYMBOLS USED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

MATHEMATICAL MODEL OF DECK--THE "SHEAR-BEAM" . . . . . . . . . . . . . . . . . . . . 4

THE SUPPORTING BEAM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Equilibium analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Load-Deflection Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

GENERAL DIFFERENTIAL EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

STABILITY CRITERIA FOR PARTICULAR CASES . . . . . . . . . . . . . . . . . . . . . . . . . 13

Case 1.--Pure Bending and Simple Support . . . . . . . . . . . . . . . . . . . . . 14 Case 2.--Uniform Load and Simple Support . . . . . . . . . . . . . . . . . . . . . 16

3.--End Loaded Cantilever . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Case 4.--Uniformly Loaded Cantilever . . . . . . . . . . . . . . . . . . . . . . . . . 21

DISCUSSION AND RECOMMENDATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

LITERATURE CITED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

APPENDIX--DETAILS OF SERIES SOLUTION FOR CASE 2 . . . . . . . . . . . . . . . . . 25

ii

Lateral Stability of Deep Beams with Shear-Beam Support

By JOHN J. ZAHN, Engineer

Forest Products Laboratory,1 Forest Service U.S. Department of Agriculture

INTRODUCTION

In many modern roof structures, a few deep laminated beams have re-

placed a system of many smaller rafters. As a result, the possibility

of failure by lateral buckling has arisen calculating the buckling

load, the only formulas presently available to the designer do not include

the stabilizing effect of the decking attached to the beams, although it

seems likely that this effect has a major influence on the stability of

the system. It is clear that present design practice depends upon the

decking to provide stability, although the margin of safety contained in

these designs is difficult to estimate.

Early papers (4, 5)2,3 contain formulas for the critical loads of

cantilevers and simple beams under various loadings. Timoshenko (7)

summarized these and indicated how an approximate solutition for point

loadings could be obtained by an energy method. Temple and Bickley (6)

improved and clarified the energy approach. Flint (2) and Hooley and

Madsen (3) reviewed these solutions, verified them experimentally , and

1Maintained at Madison, Wis., in cooperation with the University of Wisconsin. 2 Underlined numbers in parentheses refer to Literature Cited at the end of

this report. 3 Prandtl, L. "Kipperscheinungen.'' Ph.D. thesis at Nuremberg. 1899.

urged their adoption in design codes. Flint (1) also used Timoshenko's

energy method to investigate the effects of elasticity of supports and

elastic restraint at an intermediate point on the span.

The purpose of this Research Paper is to assist the designer of roof and

floor systems by theoretically analyzing the effect of deck stiffness on the

lateral stability of the supporting beams. Figure 1 shows the basic beam

and deck system, which is periodic with period S and extends to infinity in

the z direction; that is, any influence of end walls is neglected. The deep

beams are assumed to be supported by sidewalls (not shown) either as canti-

levers or as simple beams whose ends are restrained against axial rotation.

Under these assumptions, the degree of lateral support which a deck system

offer the supporting beams is determined by the shear stiffness of the

deck. To analyze the effect of this stiffness on the lateral buckling load

of the beams, the deck system is mathematically modeled as a "shear-beam,"

that is, a beam in the horizontal plane whose deflections are entirely due

to shear and whose bending can be neglected. The stabilizing effect of the

deck on the lateral stability of the supporting beams turns out to be mathe-

matically analogous to the stabilizing effect of an axial pull on the beams

along the line of deck attachment, and when both influences are present

their effects are additive. The results are therefore presented in terms

of a single dimensionless parameter η, which represents the sum of an axial

pull and the shear stiffness of the attached deck.

When η is negative, the buckling load is found to be reduced and this

is interpreted as buckling under combined axial compression and vertical

loading in the plane of greatest flexural rigidity. The b u c k l e d s h a p e

is a combination of twist and lateral displacement in the plane of least

flexural rigidity. The theoretical equations are quite general, but

numerical results are presented only for the case where the deck is

FPL 43 2

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a t t a c h e d a t t h e c e n t r o i d a l a x i s o f t h e beam and v e r t i c a l l o a d i n g i s

th rough t h e deck . Th i s s o l u t i o n shou ld n e x t b e “ p e r t u r b e d ” by a s m a l l

p a r a m e t e r r e p r e s e n t i n g t h e h e i g h t o f t h e deck a t t a c h m e n t .

DEFINITION OF SYMBOLS USED

An, Bn, Cn, Dn Series coefficients. Refer to equations (A1) and 4mn, Bmn, Cmn, Dmn

} (A6) in Appendix.

c

EI 2

F

Fx

JG

k

L

Mx , My , Mz

P

P

q

u, v, w

Vx, Vy

wD

Distance from centroidal axis to line of deck attachment,

Refer to figure 4.

Least flexural rigidity of deep beam.

Shear force in deck.

Axial force in deep beam.

Torsional rigidity of deep beam.

Shear of deck, in pounds. Refer to figure 3

and equation (2) .

Representative length in x direction.

Internal moments in deep beam. Refer to figure 4,

Refer to equations (17).

Distributed vertical load. Refer to figure 4.

Representative vertical force on deep beam.

Distributed load in z direction transmitted by deck

attachment.

Displacements of centroidal axis of deep beam in x,

y, and z directions. We neglect u and v.

Shear forces in deep beam. Refer to figure 4.

Refer to equations (17).

Displacement of deck in z direction.

3

Refer to equations (17).

Coordinate axes. to figure 4.

Rotation of cross section of deep beam due to twist.

Refer to figure 4.

Refer to equations (17)

MATHEMATICAL MODEL OF DECK--THE “SHEAR-BEAM”

Consider a roof system consisting of regularly spaced deep beams

whose top edges are bridged by deck planks (fig. 1).

Figure 1.--Basic beam and deck system used for analysis in this study.

FPL 43 4

For convenience, the planks are considered to be infinitely long and any

influence from their end support is overlooked. When the beams buckle

laterally, the planks are displaced. Although the deck offers no resistance

to a rigid-body displacement, it is assumed that differential displacement

of planks is resisted by internailing between planks so that the deck has

shear stiffness; that is, a deck is considered which has two kinds of

nailing: one which joins adjacent deckboards to each other, and another

which attaches the deck to the supporting The attachment to the

beams is assumed to be such that the can transmit only a lateral

force to the beams. This rather loose attachment renders our system

slightly less stable than a real one.

Since the system is periodic, our attention may be restricted to

one period. Figure 2 isolates a single deckboard with a length of one

period, such as the one shown shaded in figure 1.

M 125 245

Figure 2.--Deckboard considered as an element of a "shear-beam."

In this diagram, F is the force in the nails between deckboards,

qx is the force at the attachment to the beam, and x is the board width.

Lateral forces between boards are not shown and deformations associated

5

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with them will be neglected. Thus, the board in figure 2 is like an

element of a "shear-beam," whose deflection is due entirely to shear.

Figure 3 shows the relation between shear force and slope for such a

beam, where w is the "shear-beam" displacement and k is a shear stiffness

which has the dimension of force:

(1)

M 129 246

Figure 3.--Deck modeled as a "shear-beam" of depth S.

For convenience, the deck is further idealized by permitting x to

approach zero. The individual attachment forces qx then become infini-

tesimal and a model is provided with continuous deck attachment that

transmits a distributed force of q pounds per foot to the support beams.

Then,

For equilibrium of vertical forces:

FPL 43 6

(2)

(3)

From (2) and (3) the load-deflection relationship of a shear-beam is

(4)

Notice that the application of a concentrated load to a "shear-beam"

requires a jump discontinuity in the shear force F and, according to (1) ,

in the slope as well. Since in our system the "shear-beam" is attached to

an Euler-Bernoulli beam where the slope of the line of attachment must be

continuous, there can be no concentrated force transmitted between the

deck and the supporting beams except possibly at the ends of the span.

THE SUPPORTING BEAM

Equilibrium Analysis

Figure 4 shows an element of the support beam in a deflected position,

M 129 248

Figure 3.--Free-body diagram of a support bean element.

Here p and q are distributed forces acting a distance c above the centroidal

axis. The displacements in the y and z directions are denoted by v and w,

respectively, and the angle of twist is ß, measured positive from y toward z.

The vertical displacement v is much smaller than the lateral displace-

ment w since the supporting beam is assumed to be deep; hence v is neglected.

This effectively limits the length-to-depth ratio of the supporting beams.

The equations of equilibrium are:

(5)

(6)

(7)

(8)

(9)

L o a d- D e f l e c t i o n R e l a t i o n s

The following are classical results and will not be derived here.

For details refer to (5, 7). The equation governing lateral deflection

i s

(10)

where EI2 is the flexural rigidity in the x-z plane. The equation govern-

ing twist is

FPL 43 8

(11)

where JG is the torsional rigidity.

GENERAL DIFFERENTIAL EQUATIONS

The "shear-beam" is attached to the support beam by matching dis-

placements

(12)

Using (4) and (12), equation (6) becomes

which integrates to

(13)

where sub-zero denotes evaluation at x = 0. Using (13), equation (8)

becomes

which integrates to

(14)

9

Substituting (14) into (10) we get

(15)

Differentiating (11) we get

Using (7) and (9) this becomes

Inserting q from (4) and (12) we get, finally,

(16)

where the last term can be neglected in comparison with the

left side.

At this point we nondimensionalize the main equations. Let

(17)

FPL 43 10

(17)

We will use primers to denote We repeat the main results in non-

dimensional form, retaining the original equation numbers and adding a

"hat" (^).

^ (13)

^ (14)

^ (15)

^(16)

11

It is unfortunate that the deck stiffness k occurs in two parameters τ.

and η instead of only one, but the quantity

will be zero if the point x = 0 is either:

condition 1--A free boundary

or, condition 2--A point of symmetry.

For symmetrically loaded, simply supported beams we will place the

origin at the center and argue from symmetry that and must

vanish, thus establishing condition 2. For cantilevers, we will place

the origin at the free end and appeal to condition 1. To prove condition 1 ,

recall that there can be no concentrated force transmitted between the

deck and the beam except possibly at the ends of the beam. Figure 5

shows how such a force could act at the free end of a cantilever.

M 129 241

Figure 5.--Concentrated force between deck and support beam at free end.

FPL 43 12

From statics, we have that

In nondimensional notation this means

(18)

which establishes condition 1. Thus the second term on the right in ^ equation (15) will be zero for the cases to be considered in this paper.

Then k and Fx combine into the group k + Fx and it is seen that the

stabilizing influence of the deck is exactly analogous to the effect of

an axial pull acting on the beam along the line of attachment. Thus, a

single parameter η suffices in the cases about to be considered.

STABILITY CRITERIA FOR PARTICULAR CASES

Four cases are discussed, namely:

1. Pure bending and simple support

2. Uniform load and simple support

3. End loaded cantilever

4. Uniformly loaded cantilever

For simplicity, the only cases considered are those in which c. = o; that

is, the distributed loads are assumed to act at the centroidal axis of

the beam, including that due to deck attachment. An approximate assess-

ment of the influence of c can be obtained later by using the solution

for the case c = o and adding a first-order correction term.

13

Case 1.--Pure Bending and Simple Support

M 129 243

Figure 6.--Asimply supported beam in pure bending. Attached deck is not shown.

In figure 6 the length is 2L and the ends x = +L are restrained

from rotating about the x axis. Since there are no vertical loads, we

take P to be M L . Then

and

(19)

(20)

^ ^Thus (15) and (16) become

(21)

FPL 43 14

with boundary conditions

(22)

The general solution of (21) is:

(23)

where the Ci are integration constants. Applying the first two conditions

of (22) it is noted that

C1 = C3 = 0

From the remaining two conditions of (22)

(24)

from which a nonzero solution is possible only if the determinant of the

coefficients is zero. This requires that

or

or (25)

15

which checks the classical result if η = 0, and reduces to the Euler

column formula if θ = 0; that is, if M = 0, buckling can still occur if

or (26)

which is possible since Fx can be negative (compressive). Note that L

in equation (26) is the half-length.

In the next three cases the differential equations do not have

constant coefficients, but they can be solved numerically by assuming a

power-series solution and programming a computer to set the determinant

of the boundary-condition equations to zero by trial. In this way critical

pairs (θ, η) can be found and presented graphically in lieu of a closed

expression such as (25). For ready comparison with (25) θ2 is plotted

versus η when presenting these critical relationships. Since θ2 is always

positive, this will further serve to indicate that the sign of θ is

as it is by symmetry.

Case 2.--Uniform Load and Simple Support

M 129 251

Figure 7.--A simply supported beam uniformly loaded. the attached deck (not shown) and the distributed load p act at the centtroidal axis.

FPL 43 16

Here the boundary conditions are (22), the same as in Case 1.

Again L is the half-length. P is taken to be 2 1 pL, where p is the

distributed load, and since

we have

Of course

Equation (20) still holds as in Case 1 and for the same reasons. Thus, ^ ^the differential equations (15) and (16) become:

and (27)

Details of the series solution of equations (27) subject to boundary

conditions (22) are given in the Appendix. The resulting critical con-

dition is presented in figure 8.

17

M 129 244

Figure 8.--Relation between load parameter θ and stabilizer parameter η (both dimensionless) for Case 2.

Case 3.--End Loaded Canti lever

M 129 240

Figure 9.--A cantilever beam under a concentrated load at centroid of the end cross section. Attached deck is not shown.

FPL 43 18

In figure 9, L is the length of the beam and P is the end load, Then

The basic torque-twist relation (11) is

(11)

Since we are considering only the case where c = 0, and no torque Mx is applied at the free end, it is concluded that

^ ^ ^ Obviously, M yo = 0 so the differential equations (15) and (16) become

(28)

with the boundary conditions

(29)

19

The eigenvalues of system ( 2 8 , 29) were obtained by assuming a s e r i e s

so lu t ion i n a manner e n t i r e l y analogous t o Case 2 . The c r i t i c a l con-

d i t i o n i s presented i n f igure 10.

129 250

Figure 10.--Relation between load parameter θ and stabiIizer parameter η (both dimensionless) for Case-3.

FPL 43 20

Case 4 Loaded Cantilever

M 129 247

Figure 11.--A cantilever beam uniformly loaded. Attached deck (not shown) and distributed load p act at the centroidal axis.

In figure 11, L is the length of the beam and P is taken to be 1 2 pL.

^Again Mzo = 0. Thus,

^ ^ and the differential equations (15) and (16) are

(30)

subject to the same boundary conditions (29) as in Case 3. The eigen-

values of system (29, 30), as obtained by a series solution, are pre-

sented in figure 12.

21

129 249

Figure 12.--Relation between load parameter θ and stabiIizer parameter η (both dimensionless) for Case-4.

DISCUSSION AND RECOMMENDATIONS

Particular stability criteria are presented in equation (25) and in

figures 8, 10, and 12. These can be used to predict the lateral-buckling

load of flat roof and floor systems employing widely spaced deep beams

with an attached deck. The "diaphragm action" of the deck, that is, the

stabilizing influence obtained from its shear rigidity, is here quanti-

tatively accounted �or in the parameter ? for the four cases considered.

FPL 43 22 GPO 821-778-3

The assumptions made in the derivation are that the beam is thin

and deep--so that vertical deflections were neglected--and the deck is

taken to be attached at the centroidal axis (c = 0). It is believed

the influence of c will be small for relatively long beams, but this

question deserves further study. The theoretical results contained here

should be checked experimentally.

23

LITERATURE CITED

(1) Flint, A. R. 1951. The influence of restraints on the stability of beams.

Structural Engineer, Vol. 29(9):235-246.

(2) 1950. The stability and strength of slender beams. Engineering,

Vol. 170:545-549.

(3) Hooley, R. F., and Madsen, B. 1964. Lateral stability of glued laminated beams. Jour. Amer.

Soc. Engrs., Part I, Vol. 90(ST3) :201-218.

(4) Mitchell, A. G. M. 1899. Elastic stability of long beams under transverse forces.

Phil. Mag., Vol. 48: 298-309.

(5) Prescott, J. 1918. The buckling of deep beams. Phil. Mag., Series 6, 36:

297-314; Vol. 39:194-223, 1920.

(6) Temple, G., and Bickley, W. G. 1933. Rayleigh's principle and its applications to engineering.

Dover, NewYork.

(7) Timoshenko, S. 1936. Theory of elastic stability. McGraw-Hill, New York.

FPL 43 24

APPENDIX

D e t a i l s o f S e r i e s So lu t ion f o r Case 2

Given t h e d i f f e r e n t i a l equat ions

and t h e boundary cond i t i ons

(22)

we assume a solution in the form

with

(A2)

so that ao and bo are integration constants. Note (A1) is even in ξ

so that only the last two boundary conditions (22) remain to be

satisfied. They require that

25

(27.1)

(27.2)

(A1)

(A3)

System (A3) has a nontrivial solution only if the determinant of co-

efficients vanishes. Let H be this determinant. Then,

(A4)

is the critical condition.

In (A4) the subscripted variables are all functions of η and θ.

We shall want to specify η and solve for θ by trial. Hence it is

convenient to write

(A5)

where the double-subscripted variables are functions of η. Then (A1)

becomes

FPL 43 26

(A6)

(A7)

where, by (A2)

Substituting (A6) into the differential equations (27), we

from (27.1)

obtain,

(A8)

27

and from (27.2)

(A9)

From (A8) and (A9) four recursion relations are obtained, as follows:

Equating coefficients of in (A8), we obtain

(A10)

FPL 43 28

Equating coefficients of in (A8), we obtain

(A11)

Equating coefficients of in (A9), we obtain

(A12)

Equating coefficients of in (A9) , we obtain

(A13)

These relations can be made to hold for all integers n, including

n = 0, if we define

that is ,

(A14)

and agree that if any subscript is negative, the subscripted variable is

zero. Critical condition (A4) now reads:

29

(A15)

Now, given any value for θ, the recursion relations (A10) through

(A13) and the initial values (A7) and (A14) will generate the arrays

Amn, Bmn, Cmn, and Dmn.Summing these over n will yield one-dimensional arrays of coefficients of powers of θ in (A15). It is found that (A15)

is a polynominal in powers of θ 2 . (This could have been expected from

the fact that the system is symmetrical about the x-z plane and there-

fore the sense of the critical load in the y direction--that is, the

sign of θ--should be immaterial.) The critical value of θ 2 correspond-

ing to the given η is then the lowest root of the polynominal (A15).

An IBM 1620 computer was programmed to follow this procedure automatically

and figure 8 was generated. A similar method was used in Cases 3 and 4

to produce figures 10 and 12.

FPL 43 30 1.5-33 GPO 821-778-2