BRACING MEMBERS SUMMARY. OBJECTIVES. REFERENCES.

Transcription

1 BRACING MEMBERS SUMMARY. Introduce the bracing member design concepts. Identify column bracing members requirements in terms of strength and stiffness. The assumptions and limitations of lateral bracing for unrestrained beams analysis are given Diaphragms design, limitations and applications are introduced. OBJECTIVES. Understand that the design requirements for column and portal frame bracing. Understand that the design requirements for unrestrained beams lateral bracing. Understand that the design requirements for diaphragms action use and its main approaches. REFERENCES. [] ENV 99-- Eurocode Part - General rules - Supplementary rules for cold formed thin gauge members and sheeting. [] BS595-:, Structural Use of Steelwor in Building Part : code of practice for design Rolled and welded sections. [] The Steel Construction Institute, Steel Designers Manual, 5th Edt., Blacwell, 99. [4] Winter, G. Transactions of ASCE, 5, pp.87, 96. [5] Kennedy, J. B., Neville, A. M., Basic statistical Methods for engineers and Scientists, rd., Harper and Row, New Yor, pp. 4, 986. [6] Canadian Sheet Steel Building Institute, Diaphragm Action of Cellular Steel floor and Roof Dec Construction, CSSBI Suite 5, Consumers Road, Willowdale, On, MJ 4GB, 97. [7] McGuire, W., Steel Structures, Prentice-Hall, 968.

2 . INTRODUCTION. The bracing system: Must provide both force and stiffness. Is used for beams, columns and frames. Reduces effective length of columns. Reduces effective length of beams. Provides overall stability to frames. May be discrete or continuous. Consider: If is equal to zero then Q is also equal to zero, but in a system that is less than ideal,, and the situation portrayed in (c) occurs: Q = for equilibrium: P P M A P QQ or: P Note if: > P there is no sideway < P there is sideway The critical case occurs when = P EI EI [] = P cr /; The maximum load in the member is: P cr -For / >, the system is braced and can reach P cr -Any > c, gives no benefit -For < c, the system is only partially braced and P/P cr <. COUMN BRACING. Consider a two-storey column braced at mid-height. The bracing member forces the column to bucle into two half waves, i.e.:

3 Free body diagram of the top half. Q P cr Pcr [] For a three-storey column braced at each storey height: Bucling Mode Bucling Mode P cr Q Pcr [] P cr Q Not critical when compared to first mode For a four-storey column bucling in a zigzag manner the deflection at mid-height is not necessarily :

4 Free body diagram of top quarter Free body diagram of next quarter Q Q Q P and P P [A] M B Q Q P Q P P [B] Equating [A] and [B]: ; ;.44.4P [4] cr For a large number of storeys the deflections in opposite directions at interior points, remote from the ends, must be equal and the net shear on a panel height will be Q/. Therefore from the free body diagram: Q P 4P [5] l Hence: P In general Thus the bracing stiffness, corresponding to zero initial deflection at the bracing points, can be determined. The bracing force required is Q =.

5 In a real structure, however columns have initial crooedness. Forces in the bracing only occurs when the force in the column causes the brace to deform. M B Q Pcr Pcr Pcr [6] There is still the need to now both and Q [7] For compression members, normal tolerances are /5 to / of the length for plumbness Say = /5 =.; let = = (+) = Required stiffness Pcr Pcr and Q = = Q x.pcr.4pcr [8] l l Note: Assumed = /5 and = The selection of braces based on equation [8] implies: -Assume equation [8] applies to both columns and beams. -For beams deal with compressive force only. -Does not apply to plastically designed structures. -P cr can be interpreted as the elastic or inelastic load. -Does not apply to locations where brace forces are included in the analysis i.e. P- effects. Steps: - Establish brace locations. - Calculated P cr = Compressive resistance for a column and A c y for a beam. - Calculate or estimate. Pcr 4- Calculate for n equal spaces 5- Select A b i.e. the area of brace such that: 6- Chec that Q.4Pcr Ab E Pcr (units in mm b N mm mm N mm ) First Example Provide braces at / points for a WWF 6x5, 5MPa yield stress, carrying a uniformly distributed load on a span of 8 metres. Only one beam is to be braced. The beam is designed to reach its capacity for an unbraced length of 6 metres. The brace length is 4.5 metres. Evaluate the 6m unbraced length beam capacity. First the beam class has to be determined. c 5 5 -Flange i.e.: Class t x5 5 f y

6 c Web i.e.: Class t 8 f y 5 It can reach: M pl = Z f y = 9 x x 5 x -6 = Nm EI M cr = z I I w z GIt EI z 6 9 x xx 9x 6 x77x xx = 67Nm x xx xx T = W y M f cr y 6 9x5x = T,5 at ( T.) T = T,5.49(.75.) T = =.69,, T 5 T T M b.rd = T w W pl.y f y / m =.69x 9 x x 5 x -6 /. =77.5Nm,5 =,5 The unfactored compressive force is mm Fc d t r f.x77.5x N For braces at / points: Pcr x475.9x 78N / mm 6 A b E x78x45 b. mm As a compression member 45 r rmin. 5mm Try 5x5x: r min = 9.7mm; A = 9mm Q.4 P cr.4xx N E 88.9 f y 5 cr 45. r 9.7x88.9 7,5 a (.) =,5.4(.7.) = =.,, N b.rd = A f y / m =. x 9 x 5 x - /. =78 N >> Q = 7.7 N Notes: P 7.7x45x..4mm ; assumed =. =.x6 = mm. AE 9x The force of 7.7N is generated in.4 mm not mm i

7 P 78 cr. Q.4 9N This is considerably less than the 7.7 N assumed and in turn generates less. - If 6mm was used rather than /5 then Q < 78 (6.4)/ =4.5 N P 4.5x45x 4.5.mm then Q < 78 (6.)/ =4.45 N AE 9x o. 4. The brace have to designed for: This is.75 Q and Q. force in the compression flange if in an elastic design or.5 force in the compression flange if in a plastic design.m r.x77.5x. Fc.4N d t 6 5 f 5- What if braces can be provided in directions so that brace can be assumed to act in tension? Then as A b = 47mm with a slenderness limit of, as a tension member: 45 r rmin 5mm Try 9x9x7: r min = 7.7mm; A = mm N pl.rd = A f y / m = x 5 x - /. =7. N O.K Q.4.4xx N P cr ' P 7.7x45x.mm << mm assumed. AE x ' 78 Q. 9. N P 9.x45x.6mm AE x P 9x45x.6mm AE x 78 ''' 78 Q.4 9 '' '' Q.6 9N ''' N This is approximately / of the required by the design standard. -Examination of Equations [6] and [7] Pcr Pcr Q Q [6] [7] [6a] [7a]

8 If Select and - As stiffness of brace, /, decreases the deflections increase. - As stiffness of brace decreases the force Q that must be developed increases. - If the brace is too soft it will need a large force therefore the design should be based on both strength and stiffness requirements.. BRACING OF PARAE MEMBERS. - Assume that bracing is of constant size. i - The extreme value of out-of-straightness for m members is m, [5]. This states that the extreme value for m members is less than for one: - Members are equally spaced. - Example m =. Q P cr M c Q Q P cr Also: Q b Q AE E Q Q Q A E P cr Q Q Qb Q Q Q Q AE E Q Q Q Q Q Q Q b Q E Q Where: =Q and Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q

9 And: Q Q Q Q Q The three simultaneous equations are; Q Q Q Q Q Q Q Q Q Q Q Q Q Q [A] [B] [C] Where: P ; Q and The previous conclusions induce a selection of: i m Second Example Will the 5 x 5 x mm selected for the first example provide adequate bracing for three parallel WWF 6x5, 5MPa yield stress, when spanning of 8 metres. The brace length is 4.5 metres. - The unfactored compressive force is equal to N. - The braces at the / points i.e.: = - Pcr x475.9x Therefore 78N / mm 6 4- i.x6 6. 9mm m Q = = 6.9 x 78 = 54. N 5- AB E 9x.N / mm B K ; K The simultaneous equations are:.57.57q.57 Q Q Q Q 54. Q [A] Q [B] Q [C] Q 576 N 5.7 N ; Q 559 N.6 N ; Q 595 N 5. N Q 5.7. Q.6 Q 5..;.. ; The capacity of 5 x 5 x mm, calculated in the first example, was: N b.rd = 78 N >> Q = 5.7 N Summary.4.. i 6. 9mm m

10 Note: (i) One brace is in tension (ii) It is tacitly assumed that the braces area installed to limit imax to. of 6mm. (iii) If braces are installed with imax = 6 mm there is a change in the forces previously calculated. (iv) If braces are installed with initial forces present, these forces should be considered. Third Example Part floor framing plan Column Cextends 7 mm above and below this level. It is prevented from bucling about its wea axis at this level only by the flexural stiffness about the wea axis of beams AC- and CE-. The diaphragm action of the floor prevents relative movement of A, B, D and E. Column loads above and below are 56 N and 78 N. Assume is / = x7/ = 7.4 mm. Question: Do beams provide adequate support so that column can be designed for = 7mm? Pc x67 - Tae P c 67N ;. 9 7 Q 7.4 Q.9( 7.4)N ;.9 N / mm Pa a ; EI P EI a a 6 6 EI EI xx.x xx5.x total N / mm. 5N a a a a no good,.9 too large, see previous graph, a stiffer system is needed, greater than double of the initially adopted. - Try a WWF x95. & WWF x5 i.e.: comparable W and Z (I z = 7x 6 mm 4, 85.5x 6 mm 4 ) total = =.55 N/mm Q = =.9 ( + 7.4) =.5mm Q = 8.96N - An alternative solution taes into account that most of the restraint comes from AC- beam. Therefore as good solution will be to increase its size and leave CE- as it is i.e.: total = =.9 N/mm.9 =.9 ( + 7.4) =.4mm Q = 9.7N 4- the bending moments developed in the members should also be checed, i.e. for AC-:

11 .69 M dy 9.7x.5. Nm.9 M b.rd = W pl.y f y / m =76x x 5x -6 /. =65Nm M M dz b, rdz M M dy b, ry..8 O. K DIAGONA BRACING FOR COUMNS. The general procedure is composed of four steps: - Select bracing, - Compute deflections; - Determine the P- effects; 4- Iterate till the desired solution convergence. Fourth Example The specified (/5) win forces acting on a braced bay and the resulting forces are show in the illustration below. Select diagonal bracing to resist the imposed loads including the P- effects. Additional information - Dead plus wind load combination case has been established to be critical. Dead loads contributing to P- effects are: evel : 648N, evel: 648N, and evel4: 84N. Bl - Storey drifts due to: bracing force B B in the storey; aae Ch Compressive force C C in the leeward column; aae TCh Tensile force T T in all the storeys above. aae - Area of column is equal to 98mm in all the levels Select bracing evel - T Q f q.5x9. 587N r 99. mm Try 5x9x long legs bac to bac with a mm gusset plate: A g = 464mm and r y = 5.8mm min

12 Adopting one bolt hole for mm diameter bolt the net area can be evaluated: A n = 464 (x6) = 4 mm Af y 464x5x Anet fu 4x4x N pl. Rd 54. 5N ; Nu. Rd N M. M.5 N 54.5N N N O.K. urd d 587 evels - and -4 Use 5x9x long legs bac to bac with a mm gusset plate, i.e.: A g = 464mm and r y = 5.8mm to provide minimum slenderness ratio. ateral deflections (specified wind) evel - Bl 9x x99 Bracing B 4. 6mm aae 9x464x Ch 9x x4 Column C. mm aae 9x98x =4.9mm evel - Bl 4x x989 Bracing B. 7mm aae 9x464x.x9 Column evel - C. 9mm 4 7x x9 evel - C. mm 9x98x 7x x4 evel - T. mm 9x98x =.4mm evel -4 Bl 86.6x x989 Bracing B. mm aae 9x464x.x9 Column evel - C. 9mm 4 evel - C. mm evel -4 C. mm evel - T. mm.x9 evel -4 T. mm 4 =.784mm

13 P- effects (D + Q) In any storey Non-iterative method ' ' V V max ' V V Q The deflected shape and factored P- effects are: 84 D Equivalent Shears, V Equivalent ateral Force, H 4.9 Q 648 D 648 D.78 Q.4 Q.9 D Q.9 D Q 7.7 D Q 6.8 D Q 7.47 D Q.9 D Q 7.47 D Q For evel -4 (84 D x.78 Q )/9 =.9 D Q For evel - (( ) D x.4 Q )/9 = 7.7 D Q For evel -4 (( ) D x 4.9 Q )/9 = 7.47 D Q For level - V ' max ' V V V The force in the bracing including the P- effects; N ' Q 7.47x.x N N NuRd 54. N O.K. a 9 d, P 5 For evels - and -4 the diagonals selected for maximum slenderness ratio will also be o.

14 5. DIAPHRAGMS. The diaphragm action: Consider: - Transfer in-plane lateral loads applied at roof and floor levels to stiffer members; - Need strength and stiffness; - Must loo at flow of forces (how forces are carried from point to point) intersheet connections and connections at boundaries are important. The shear strength of diaphragm depends on: - Seam capacity. - Edge capacity. - Purlin capacity (a controlled by fasteners). overall bucling strength. Stiffness of diaphragm depends on: -Panel warping; - Movement at seams; - Edge panel slip. The shear stiffness Wall Cladding F F a Fa G and ; G or G ' Gt a bt bt b ' F G b spring stiffness is equal to a Can cladding tae horizontal force (in-plane) from girts to reduce effective length of columns? Must establish force in girt as a function of. Cladding must provide this force and stiffness. Roof Diaphragm

15 The diaphragm acts as a deep beam to carry the shear to the ends where it must be transferred to the ground. The force in the flanges is equal to M D w 8D 4 5w vf t f V 6 84EI x Where F is the flexibility factor equal to a function of: -Dec span/average length supplied; -Dec gauge; - -Side lap connection; - -Transverse welds. And I is the moment of inertia of the flanges (which must be connected to the web).

16 Eurocode

17 BS595

18 6. CONCUDING SUMMARY. This lecture presented the approaches used for the bracing member design.