lutz1985Q4

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Fig. 1. Finite bracing at spacing S

rom such calculations can grossly overestimate the bracing

tiffness required. One must remember that the length Le in

Eq. 6 defines the required buckling length of the member.

Using a S which is less than Le in Eq. 6 presumes a shorter

mode shape, which requires more stiffness than is necessary.

For a small S , relative to Le, Eq. 5 provides an accurate

olution for bracing stiffness. However, as S increases

elative to Le , using Eq. 5 results in a substantial error.

Solutions are presented in Ref. 3 for critical buckling

oads ( P cr ) for cases when brace stiffness is less than the

tiffness required to force the column to buckle in its highestmode, i.e., the number of one-half waves equal to the number

f braces plus one. Solutions are presented for one, two, three

nd four intermediate brace points, as well as for columns

ontinuously braced.

Shown in Fig. 2 are the solutions from Ref. 3 for cases

with three and four intermediate braces replotted usin

absissa of K iS/P E . It can be seen from Fig. 2 that the bra

expression, Eq. 5, provides a solution for P cr with

accuracy to approximately three-quarters of their maxim

P cr /P E value. Based on bracing Eqs. 5 and 6, bra

stiffness over the entire range of S less than or equal t

can be conservatively calculated from the expression

K 1 = [2.5 + 1.5(S/Le)4][ P cr SLe

2]

This equation provides a transition between

continuous bracing equation, which has a numecoefficient of approximately 2.5, and the solution for f

full bracing, which has a maximum numerical coefficien

4.

The accuracy of Eq. 7 is illustrated in Fig. 2, whe

has been plotted for the cases of four and three intermed

brace points. Also shown in Fig. 2 is the expression 4 P cr /

When the number of intermediate braced points is

than three, it has been found that Eq. 7 becomes ov

conservative as S approaches Le.

BRACING STIFFNESS FOR LESS THAN THREE

INTERMEDIATE SUPPORTS

Provided in Ref. 2 for a single intermediate brace, Fig.

the relationship:

P cr = P E + (3/16)( K i L) for P E < P cr ≤π2 E t I/S

2

Fig. 2. Solution for three or more intermediate supports

64 ENGINEERING JOURNAL / AMERICAN INSTITUTE OF STEEL CONSTRUCT

3 by American Institute of Steel Construction, Inc. All rights reserved. This publication or any part thereof must not be reproduced in any form without the written permission of the

8/14/2019 lutz1985Q4


Fig. 3. One intermediate brace

f 2S is substituted for L, one obtains

K i = (8/3)( P cr – P E )/S (8)

where,

P E = the buckling load with no intermediate support

= π2 E t I/L

2

This relationship is plotted in Fig. 4.

Also plotted in Fig. 4 for the one intermediate support

ondition is the commonly used expression K i = 2 P cr /S (see

Ref. 1). It is recommended the designer requiring more

ccuracy use Eq. 8.For the situation where two intermediate supports exist

Fig. 5) the "exact" solution from Ref. 2 is again replotted in

Fig. 4 in terms of K iS/P E along with the commonly used

quation K i = 3 P cr /S (see Ref. 2). One can see from Fig. 4 the

discrepancy between the exact solution and the commonly

used expression for K i when full P cr is not necessary.

The empirical equation:

K i = 3 P cr S/Le2

is suggested for more accurate solutions, as shown in Fig

EVALUATION OF THE TANGENT MODULUS OF

ELASTICITY

The tangent modulus E t equals E as long as the mem

remains elastic. Using the CRC expression to represen

inelastic region,

P cr = P y[1 – .5( Le /rC c)2]

for P cr > .5 P y.

Solving for Le in Eq. 10,

L rC P P e c cr y= −2 1( / )

with rC I A E F c y= / /2 2π

= π 2 EI P y/ by definition,

the equivalent length L P P P e y cr y= −π ( / )( / )4 1 EI

= π E I P t cr / from Eq. 3From this

E t = (4 EP cr /P y)(1 – P cr /P y)

when P cr /P y > .5 or Le /r < C c.

Using E t in Eq. 3, the procedure for determining the

stiffness K i can be used for all P cr < P y.

Fig. 4. Solution for one and two intermediate supports

FOURTH QUARTER / 1985

3 by American Institute of Steel Construction, Inc. All rights reserved. This publication or any part thereof must not be reproduced in any form without the written permission of the

8/14/2019 lutz1985Q4


8/14/2019 lutz1985Q4


Solution:

. Determine the stiffness provided ( K act ) by the strong axis

of the W12×26 bracing the weak axis of the W12×40.

Assume the two intermediate supports on the W12×40

exert forces in the same direction. The minimum stiffness

of the supports can be found from the deflection equation

for a simple beam loaded with two equal symmetric

loads.

∆ = ( Px/6 EI )(31a – 3a2 – x

2)

∆ = ( Pa/6 EI )(31a – 4a2)

∆/ P = [(8)/6 EI ][3(24)(8) – 4(8)2] 1,728

= 4,423,680/6 EI

= 4,423,680/[6(29,000)(204)]

= .125 kip/in.

Therefore, the stiffness K act = 1/.125 = 8.0 = kips/in. The

axial stiffness of the W8×31 ties can be neglected. The

connections of the ties to the columns are assumed to be

rigid.

Fig. 6. Example 3

If fully braced, Le /r = 8(12)/1.93 = 49.7 and F a = 18.38 k

P allow = 18.38(11.8) = 216.9kips

and P cr ~ . ( . )− =216 9 17 369kips

and K req = 6 P cr /S = 6(369)/(8)(12) = 23 kips/in.

∴ the W12×40 is only partially braced.

2. From Eq. 17, for two intermediate braces,

K req = 6 P cr S/Le2

For P E < P cr < π2 E t I/S

2.

From Eq. 11, L rC P P e c cr y= −2 1( / )

P cr can be found by substituting Le from Eq. 11 into

17, and substituting K act for K req .

The result is,

P cr = P y/[3SP y/( K act r 2C c

2)+1]

Thus, P cr = P y/(3)(96)(424.8)/[(8.0)(1.93)2(126.1)

2] +

= .795 P y

= 337.6kips

and Le = (1.93)(126.1) 2(1 .795)−

= 155.8 in. = 12.98 ft

Since P cr > .5 P y, an acceptable solution has been fo

Had P cr been less than half of P y,

Le2 = π

2 EI/P cr (Euler buckling) and substitution into

17 would yield

P EIK S cr act = π2 6/ .

3. Determine the allowable buckling load of the W12×40

l/r x = 24(12)/5.13 = 56.14

Le /r y = 155.8/1.93 = 80.75 controls

F a = 15.27, thus the allowable load is

15.27(11.8) = 180.2 kips

The brace force required from Eq. 13 is K act d. If d

with d o considered to be .375 in., F req = 8.0(.375) =

kips

The stress in the W12×26 would be

3.0 kips (8 ft)(12)/33.4 = 8.62 ksi

REFERENCES

1. Timoshenko and Gere Theory of Elastic Stability

Ed., McGraw-Hill Book Company, New York.

2. Winter, George Lateral Bracing of Columns and Be

Trans. ASCE, Vol. 125, 1960, pp. 807-845.

3. Green, Giles G., George Winter and T. R. Cuyke

Light Gage Steel Columns in Wall-braced Pa

Bulletin No. 35, Part 2, Engineering Experim

Station, Cornell University, October 1947.

FOURTH QUARTER / 1985

3 by American Institute of Steel Construction Inc All rights reserved This publication or any part thereof must not be reproduced in any form without the written permission of the

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