BRACING REQUIREMENTS FOR LATERAL STABILITY

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1 BRACING REQUIREMENTS FOR LATERAL STABILITY By John J. Zahn, 1 M. ASCE ABSTRACT: The forces induced in braces depend on the magnitude of initial imperfections (lateral bending and twist) and magnitude of applied loads. A method for calculating these forces is presented here. Shear stiffness of attached deck is taken into account. Calculations show that, for beams which derive their lateral stability mainly from the stiffness of the attached deck, additional bracing against rotation or lateral deflection is not able to achieve an important increase in the load capacity of the systems. In those cases, bracings employed merely to straighten the initial imperfections need not be very strong. In cases where decking cannot be counted on to stabilize the roof system, bracings should be designed for stiffness (sufficient to enforce a higher buckling mode) and checked % y the methods presented here for required strength. The results are of value to designers of large flat wood roof systems. INTRODUCTION Large flat roof systems frequently employ bracing members whose primary function is to stiffen the overall system and thereby prevent lateral instability. The design of bracing members entails two considerations: (1) The amount of stiffening required to ensure stability; and (2) the required strength of bracing members. The first consideration is the subject of extensive literature (2,4,5, 6,7,8,10). It will not be treated here, although it should be pointed out that beam torsional rigidity and lateral flexural rigidity are reduced when under design load (15). This is analogous to the partial loss of lateral flexural rigidity of columns under design load and is associated with the phenomenon of buckling. The second consideration, strength of bracing, is the subject of this paper and is described after establishing the necessary theory by extending the results of Ref. 12. Bracing members theoretically contain no internal forces unless the beam they are bracing has initial imperfections. This was reviewed in Ref. 12, where the deformations under load of an initially bowed and twisted beam were analyzed. That work is extended here to include the effect of the shear stiffness of the attached deck since such stiffness is often relied upon to provide stability in wood roof systems. LITERATURE REVIEW Flint (2) looked at required bracing stiffness to achieve a desired buckling load. This philosophy has also been followed by later authors (4-8). Zuk (16), who was among the first to consider strength requirements, found that the "two percent" rule was an adequate design criterion. This 1Eng., Forest Products Lab., Forest Service, U.S. Dept. of Agriculture, Madison, Wisc. Note. Discussion open until January 1, To extend the closing date one month, a written request must be filed with the ASCE Manager of Technical and Professional Publications. The manuscript for this paper was submitted for review and possible publication on May 3, This paper is part of the Journal of Structural Engineering, Vol. 110, No. 8, August,

2 rule states that the force in the bracing is 2% of the compression force in the compression flange of the beam. Unfortunately, Zuk used the wrong amplification factor on lateral deflections by assuming it would be the same as for initially crooked columns. The correct amplification factor is given in Zahn (15). Schmidt (11) suggests that design should proceed as follows: (1) Specify permissible deflections; (2) solve for required stiffness; and (3) check required strength. In this paper, his results are verified and extended to include strength and stiffness of attached decking. Barsoum and Gallagher (1) derived a finite element method applied by Nethercot (7). It was extended to include continuous deck support by Hancock and Trahair (3). Nethercot and Rockev (8) studied the stiffness needed to force the second mode and concluded that this requirement is easily met in practical situations. They give a good historical review of earlier work. Nethercot (7) has proposed a design method for lateral and torsional restraints which does not, however, consider strength requirements. Nethercot and Trahair (9) give a design method for corrugated deck on I-beam in which strength of decking is considered. Kitipornchai and Richter (5) say that the optimum location of rigid restraints is where the area of the moment diagram is equally divided. Hancock and Trahair (4) looked at restraint location in relation to intermediate supports and concluded that the restraints should be near the intermediate supports.. Finally, O'Connor (10) studied braced columns, and found that column crookedness was more significant than lack of fit in the connection to the bracings. Theory. Consider a symmetrically loaded beam oriented with respect to x, y, z axes as shown in Fig. 1. The top edge is assumed to be hinged to a deck which deforms in shear in its own plane. Take L to be the half length and assume simple supports at x = ± L. Two kinds of bracing are assumed to act at center span: an elastic restraint against twist and an elastic restraint against lateral displacement at a distance, a, above the shear center. Of course, continuous lateral restraint is provided by the roof deck because the shear stiffness of the deck enables it to develop lateral forces at the line of attachment. A strip of deck of width S is associated with each beam where S is the beam spacing. It is assumed that all roof beams buckle congruently and consequently provide no direct lateral restraint to one another. Under load, the line joining the shear centers of the beam will undergo FIG. 1. Simple Supported Beam under Symmetric Load 1787

vertical deflection, v(x), horizontal deflection, w(x), and twist, b (x), as shown in Fig. 2. [In lateral stability, the shear center is a more natural choice than the centroid (12).

3 vertical deflection, v(x), horizontal deflection, w(x), and twist, b (x), as shown in Fig. 2. [In lateral stability, the shear center is a more natural choice than the centroid (12).] The vertical deflection will be neglected since (1) (2) by virtue of the fact that (3) (4) in which c = half depth; EI z and EI y, = flexural rigidities; and GK = torsional rigidity. By symmetry we may consider only the right half of the system. The potential energy of the beam, U B, and vertical load, U L, is given in Ref. 12 to be (5) In which subscript T denotes "total," that is (6) (7) M z is the internal bending moment about the z axis, a prime denotes d/dx, and subscript i denotes initial deflections at no load. To this must be added the strain energy of the deck and one-half the strain energy FIG. 2. Deflections of Cross Section 1788

4 of the restraints. The strain energy of the deck is (13) (8) in which S = beam spacing; and G D = in-plane shear rigidity of deck (force per length of edge). The elastic restraint against tipping exerts a torque T proportional to the rotation: (9) in which T = torque exerted by rotational bracing on beam; b o = twist at the center; and R = rotational bracing stiffness which depends upon the details of bracing method employed. The elastic restraint against lateral deflection exerts a force F proportional to the displacement at a distance a - above shear center: (10) in which F = lateral restraint force; k = lateral restraint stiffness; and w o = lateral deflection at midspan. A specific example will be given later. Note Eqs. 8, 9, and 10 assume the unloaded system has no strain energy in the deck or braces. That is, the unloaded beam is allowed to be initially bowed and twisted. Alternatively, one could assume that the deck and restraint are employed to initially straighten the beam. Then there would be strain energy in the system at no load, but no initial deflections. The present method is conservative in that it produces larger bracing forces. It is also easier to analyze. Thus, one-half the strain energy of the restraints is (11) (12) Adding Eqs. 5, 8, 11, and 12 gives the total potential energy of the system.: (13) in which terms containing only w i or b i are omitted, since their first vanation is zero. This amounts to simply shifting the arbitrary datum of U. The condition of equilibrium is that the first variation of U be zero: (14) This yields the following Euler-Lagrange differential equations: (15)..... (16)

5 and boundary conditions (17, 18) (19, 20) (21, 22) in which subzero = the value at x = 0 and L 0 denotes the preceding expression evaluated at x = L minus its value at x = 0. Since the variations at x = 0 and x = L are independent, this yields two independent conditions one at x = 0 and one at x = L. APPLICATION TO BRACING STRENGTH For simplicity, let the loading be a constant bending moment. Then p is zero and M z = M. Let the initial bow and twist be (23) (24) Then Eq. 15 becomes, after two integrations with respect to x (25) in which C 1 and C 2 are integration constants, and Eq. 16 becomes...(26) The boundary conditions are (27) (28) (29) (30) (31) (32) 1790

6 The solution of this system which satisfies every condition except Eqs. 28 and 32 is (33) (34) (35) (36) (37) and w o and b o are center deflections which can be determined by satisfying Eqs. 28 and 32. We shall consider three cases. Case 1. R = 0 and k = 0 (no bracing). Case 2. A = B = 0 (no initial bow and twist). Case 3. The general case ( R, k, A, B non-zero). Case 1. Setting R = 0 and k = 0 yields w o = Â and b o = B T (38) (39) Note that as M increases, l approaches p /2 L and both w and b approach infinity. The critical value of M is (40) in which l cr = p /2L, which agrees with the buckling load found in Ref. 13. Case 2. Here Â and B T vanish with A and B. Eqs. 28 and 32 require that either w o and b o be zero or (41) in which f is a lengthy expression that will not be shown here. It suffices - to say that f has the behavior shown in Fig. 3 where tan l L and f - - are plotted against the load parameter l L. The critical value of l L is the 1791

7 FIG. 3. Plot of Right and Left Sides of Eq. 41 intersection of the two curves. For very small k and R the critical value is only slightly greater than p/2, the value found in Case 1. As restraint stiffnesses k and R increase, the critical load increases until the second mode is enforced at l L = p. This occurs when If G D = 0 (no deck) this reduces to (42) (43) Eq. 43 can be checked against previous results. If R is zero, it agrees with Schmidt (11). If in addition a is zero, it agrees with the finite element results of Nethercot and Rockey (8) and checks Flint (2) within the error of his approximate solution. If k is zero and R is not, it agrees with Flint in requiring R to be infinite. Nethercot and Rockey get R to be 26 GK/L in that case because their finite element solution did not neglect vertical displacements as was done here and also by Flint. These results also agree with Mutton and Trahair for the special ratios of k/r that they considered (6). Case 3. In the general case the deflections approach infinity as the buckling load of Case 2 is approached. If the bracings are not stiff enough to force the second mode, then theoretically the center deflections and bracing forces approach infinity. If the bracings are stiff enough to force the second mode, then center deflections and bracing forces increase 1792

8 until the second mode buckling load is reached (l L = p), at which load the member snaps into the second mode shape for which center deflections are zero and the bracing forces are relieved. There is a very real danger, however, that the bracings will fail before the second mode critical load is reached, even though theoretically they were stiff enough to force the second mode. That is, the second mode will not really be enforced unless the bracings are sufficiently strong. To solve for the forces in bracings at loads below the buckling load, one proceeds as follows: 1. From the load, M, calculate the load parameter, l. Here one might increase M by a load factor if using a limit-state design. 2. Assume initial imperfections A and B. These, too, might be increased by a judgment factor to account for uncertainty. 3. From A, B, and l get Â and B Ù using Eqs. 36 and 37. These are the values that the center deflections would attain (in addition to initial values) if there were no center restraints. 4. Use Eqs. 28 and 32 to solve for w o and b o the center deflections actually attained in the presence of center restraints. 5. Finally, calculate bracing force F, in which (44) and bracing torque, T, in which (45) All of this could be done once and for all algebraically but the resulting equations are cumbersome and cost the user as much computational effort as following the previous sequence. 6. The strength of the deck should also be checked. The shear force per inch of edge in the deck is given by (13): (46) From Eqs. 33 and 34, w' and b ' are a maximum at x = L: (47) (48) These expressions are not singular at l L = p/2 even though Â and B T go to infinity there. For purpose of computation near l L = p/2, it is better to rewrite Eqs. 47 and 48 as..... (49) 1793

9 (50) EXAMPLE APPLICATION We consider the following case: Beam. Half-length L = 360 in. (9,100 mm); width, b = in. (130 mm); depth, h = 35 in. (890 mm); spacing, S = 240 in. (6,100 mm); and modulus, E = psi (12.4 GPa). For many wood species, G is approximately E /16 and for narrow rectangular cross sections, K is approximately 4 I y. Thus, we may assume (51) for narrow rectangular wood beams. Deck. Stiffness G D = 450 lb/in. (79 kn/m). This is typical of nominal 2-in. (exactly 38-mm) tongue-and-groove wood decking (14). Its strength is about 6.2 lb/in. (1.1 kn/m). Most other deck constructions are considerably stiffer and stronger than this. Load. Assume the beam is rated at f = 2,400 psi (16.5 MPa) in bending. Then the bending moment at design load is (52) Restraints. Before specifying any restraints, let us see how well this beam does without them. The buckling moment is, from Eq. 40 so that the factor of safety against buckling is (53) (54) This would seem to be adequate until one takes a closer look. If there were initial deflections as large as (55) (56) then, under design load, these would increase to (57) (58) 1794

10 FIG 4.-Center Deflections and Deck Shear Force versus Load from Eqs. 36 and 37, and from Eqs the deck shear force would be (59) These values may not seem large but, unfortunately, they grow rapidly with an increase in load as Fig. 4 shows. Note that the deck fails at only in.-lb (331 kn m). This means that the factor of safety is only (60) which hardly seems adequate. Thus, it is worth inquiring whether center constraints stiff enough to force the second mode would significantly increase the failure load. The second mode buckling load is in.-lb, (870 kn m), from Eq. 40 with l = p /L. To obtain this modest increase in buckling load, the bracings must not only be stiff enough but also strong enough to force the second mode. Therefore look next at the forces induced in the bracings and in the deck when the beam is braced at midspan. Suppose that there are 4- by 12-in. nominal (exactly 89-mm by 286- mm) purlins crossing at midspan. Rotational restraint can be obtained by placing 45 braces from bottom of beam to purlins as shown in Fig. 5( a ). The stiffness, R, is derived from the bending stiffness of the purlins. Fig. 5( b ) shows purlin bending when all main beams roof buckle congruently (as they are constrained to do by the purlins). Fig. 5( c ) extracts one period from the system. On this free body diagram (61) (62) (63) 1795

11 FIG. 5. Analysis of Torque Exerted by Bracing Therefore, the rotational restraint stiffness is Assume further that the end walls of the structure transmit a small elastic lateral restraint, k, along the axis of the purlins. Say that (65) Eq. 42 shows that these restraints are more than stiff enough to force the second mode. The forces developed in these bracings are (66) (67) is the distance from shear center of main beams to centroid of purlins, and (68) where Eq. 66 is obtained from the free body diagram in Fig. 6. The effect of adding these restraints is shown in Fig. 7 where bracing forces and deck shear force are plotted against load. Note that the deck (64)

12 FIG. 6. Free Body Diagram of Beam Section at Center Restraint Showing How Restraint Torque T is Produced by Forces P in Diagonal Braces now fails at 3.74 million in.-lb (423 kn m) so that the factor of safety has become (69) At this load the forces in the bracings are very small. Nominal 2- by 4- in. (exactly 38-mm by 89-mm) diagonals would be more than sufficient. A factor of safety of only 1.49 seems rather small. If a factor of safety of 2.0 is imposed, then the maximum initial imperfections that can be tolerated are easily solved for. With M = 5.02 million in.-lb (567 kn m), one finds that (70) This is plotted in Fig. 8 where the safe and unsafe regions are labeled. FIG. 7. Effect of Adding Center Restraints to Case Shown in Fig

13 FlG. 8. Safe Region for a Factor of Safety of 2 A better way to assure safety would be to use a stronger deck. If the 2-in. (exact 38-mm) deck used above were overlaid with sheets of 3/8 in. (10 mm) plywood, its strength and stiffness would increase to (14) overlaid deck (71) (72) Fig. 9 shows that this deck fails at 9.26 million in.-lb (105 MN m) with no center restraints and initial imperfections given by Eqs. 55 and 56. The factor of safety is now (73) Fig. 10 shows the tolerable initial imperfections if a factor of safety of 2.0 is required. The safe region is FIG. 9. Effect of Adding Plywood Overlay to 2-in. Deck 1798

14 FIG. 10. Safe Region for Factor of Safety (74) (This has a different slope than Fig. 8 because here the center rotational restraint is not acting.) One sees that bracings are not needed for safety or stability when a strong stiff deck is employed. If they were used merely to straighten an initially crooked member, the required strength would be merely whatever is required to straighten the member. With assumed initial deflections of 2 in. and 5 the bracing forces to straighten the member would be lateral restraint: (75) restraint torque:..... (76) 45 brace force: (77) Again, 2 by 4 bracings are sufficient. This shows that deck strength and stiffness are the most important factors in this design. This is likely to be the case whenever a deck is essential to beam stability. In this example the beam alone would buckle at only (78) indicating that the deck is indeed all-important to system stability. To be effective in achieving stability the deck must be attached to the compression edge of the beam. In this example, if the deck were attached to the tension edge (c negative), the buckling load would be reduced to 4.7 x 10 6 in.lb (530 kn m) for the 2-in. deck. The design moment would then be more than half the critical load. Given the initial imperfections we have been assuming, the deck would fail at only 1.68 x 10 6 in.-lb 1799

15 (190 kn m) or 67% of design load! Even the reinforced deck would fail at only 78% of design load. CONCLUSION In roof designs where the stability of the main beams is derived mainly from the stiffness of attached decking, additional bracing against rotation or lateral deflection is not very significant. The first evidence for this is the fact that midspan bracing only increased the buckling load by 11% in this example that is, from 6.94 x 10 6 in.-lb (784 kn m) to 7.70 x 10 6 in.-lb (870 kn m). Since the deck is such a vital element, its strength governs system failure. Under these conditions, center restraints cannot do much to increase the ultimate load capacity of the roof system. Their most important function would be to straighten initially imperfect beams and hold them in line, a function that does not demand much strength. In cases where the attached deck is very light and not the main source of structural stability for the roof system, bracings must not only be stiff enough to enforce the second mode but also strong enough to do so, the required strength depending upon the magnitude of initial imperfections of the beam. A method for calculating the forces induced in the braces is presented in Case 3 of this paper. APPENDIX I.-REFERENCES 1800

16 APPENDIX II.-NOTATION The following symbols are used in this paper: amplitudes of cosine deflections; distance from shear center to lateral load, or to lateral restraint; width of beam; integration constants; distance from shear center to location of deck attachment; flexural and torsional rigidities of beam; lateral restraint force; deck shear force per length of edge; stress; deck shear stiffness, force per unit length of edge; purlin force developed by rotational restraint. See Fig. 5; beam depth; initial state; lateral stiffness, force/distance; half length; bending moment; axial force in 45 braces, or column load; distributed vertical load, force/distance; shear in purlins; torsional stiffness, torque/angle; beam spacing, distance; torque; potential energy; vertical deflection; lateral deflection; coordinate axes; L/2 twist deflection; total lateral midspan deflection; 1801

17 lateral midspan deflection before primary load (P or M ) applied; total midspan twist; midspan twist before primary load (P or M) applied; and load parameter defined by Eq

FEM-Design Useful Examples

FEM-Design Useful Examples 1 Example 1: A simple example... 5 1.1 Buckling load... 5 1.2 Design... 6 1.2.1 1 th order theory... 6 1.2.2 2 nd order theory... 10 2 Example 2: A frame type example... 15 2.1