CIVL473 Fundamentals of Steel Design

Transcription

1 CIVL473 Fundamentals of Steel Design CHAPTER 3 Design of Beams Prepared By Asst.Prof.Dr. Murude Celikag DESIGN OF STRUCTURAL ELEMENTS 3. Beams in Buildings 3.1. Laterally Restrained Beams Restrained beams should not fail by lateral instability. Their design should consider the following criteria: lateral restraint force bending capacity of section shear capacity of section combined bending and shear web bearing and buckling Dr.Murude Celikag 1

2 BS 5950: Part 1 Clause Full Lateral Restraint Full lateral restraint exists if the frictional or positive connection of a floor or other construction to the compression flange of the member is capable of resisting lateral force of not less than 2.5% of the maximum factored force in the compression flange of the member, under factored loading. This load should be considered as distributed uniformly along the flanges, provided that the dead load of the floor and the imposed load it supports together constitute the dominant loading on the beam. The floor construction should be capable of resisting this lateral force. a) Design Procedure Select section and determine the design strength of the steel Determine the section classification For slender sections, reduce the value of design strength Check the shear capacity with low shear load with high shear load Check the buckling resistance of the web Check the bearing resistance of the web Dr.Murude Celikag 2

3 b) Section Classification Due to the influence of local buckling of the flange and webs, rolled sections are classified as follows: Slender - the elastic moment capacity of the section cannot be attained Semi-compact - the elastic moment capacity, but not the plastic moment capacity, of the section can be attained Compact - the plastic moment capacity of the section can be attained, but elastic analysis of the frame is to be used Plastic - as for compact section, but there is sufficient rotation capacity for plastic design of the frame c) Bending Moment Capacity For plastic or compact sections the moment capacity M c is: M c = p y S<1.2 p y Z For slender sections: M c = p y Z where p y <p y [Clause 3.6) Dr.Murude Celikag 3

4 d) Influence of Shear According to Code the shear failure modes are as follows: the shear capacity of the web is exceeded shear buckling adjacent to supports or under point loads may occur The latter is for very slender sections and is not generally a design criterion. Therefore, the shear capacity of the web Clause P v = 0.6 p y A v For rolled I-sections, channels A v = web thickness x overall beam depth For built up sections and boxes, channels A v = web thickness x web depth e) Bending and Shear If shear load, F v > P v = 0.6 p y A v then for compact and plastic sections (Clause 4.2.6) M c = p y (S-S v 1 ) <1.2p y Z Where, 1 =2.5(F v / P v ) 1.5 For equal flange sections S v is the plastic modulus of the shear area. Dr.Murude Celikag 4

5 f) Web Buckling To prevent web buckling under point loads or reactions applied via the flange the following additional requirements are made (Clause and see figure). The buckling resistance P w is given by Pw b1 n1 tpc b 1 = stiff bearing length t = is the web thickness n 1 = length obtained by dispersion at 45 o through half the depth of the section p c = compressive strength of the web (Clause 4.7.5, Table 27(c)) Clause gives a value of of 2.5d/t on the assumption that the web is acting as a fixed-ended strut with a value of L E = 0.7d (d is the depth of the web) L r As y r L r E y E y 0.7d r y 3 I y t t A 12t d 0.7d 2.43 t t d 2.5 t L E can be changed to suit the web restraint conditions. Dr.Murude Celikag 5

6 g) Web Bearing When loads are applied directly through the flange Clause requires that the web is checked at its junction with the flange in order to ensure that the design strength is not exceeded. The local capacity (P crip ) is taken as b 1 n tpyw P crip 2 P yw n 2 is the design strength of the web is the length obtained by dispersion through the flange to the flange-t-web connection at a slope of 1: 2.5 to the plane of the flange. DESIGN OF STRUCTURAL ELEMENTS 2. Laterally Unrestrained Beams Loading a beam in its stiffer plane (the plane of the web) would induce a failure by buckling in a less stiff direction (by deflecting sideways, u, and twisting,, about a vertical axis through the web). Thus the bending strength will now be a function of the beam s slenderness analysis of lateral-torsional buckling is considerably more complex. It requires greater degree of calculations. Dr.Murude Celikag 6

7 a) Factors Influencing Lateral Stability 1) The buckling load of the beam depends on its unbraced span, i.e. the distance between points at which lateral deflection is prevented. 2) The shape of the cross section would have some influence. 3) Beams under non-uniform moment force in the compression flange is no longer be constant 4) End restraints which inhibits development of the buckling shape is likely to increase the stability of the beam. 5) If a beam is laterally continuous, buckling involves the whole span with the more stable segments restraining the critical segment. Warning!! Beams designed as Laterally Restrained may not be adequately restrained during erection. Stability checks at this stage are also necessary. The buckling resistance (M b ) of a beam may be found by use of a number of parameters and factors: Effective Length (L E ), which allows for the effects of end restraint as well as type of beam, and the existance of destabilizing forces. Minor Axis Slenderness (λ), which includes lateral stiffness in the form of r y, and is defined by λ= L E /r y Torsional Index (x), which is a measure of the torsional stiffness of a cross section. Slenderness Factor (v), which allows for torsional stiffness and includes the ration λ/x. Slenderness Correction Factor (n), which is dependent on the moment variation along the beam. Buckling Parameter (u), which allows for section type and includes a factor for warping. Equivalent Slenderness (λ LT ), which combines the above parameters and from which the bending strength (p b ) may be derived λ LT =nuvλ In addition, an equivalent moment factor (m) is used which allows for the effect of moment variation along the beam. Dr.Murude Celikag 7

8 Dr.Murude Celikag 8

9 Response of a slender cantilever beam to vertical loading-lateral torsional buckling Similarity between strut buckling and beam buckling Effect of non-uniform moment on lateral-torsional buckling Dr.Murude Celikag 9

10 Effect of end restraint in plan or elevation on lateral-torsional buckling Buckling of beams provided with lateral bracing b) Development of Design Approach For The Basic Case Use Clause and to determine L E. Calculate the equivalent uniform moment, mm A. m = equivalent uniform moment factor cl , Table 13 = ratio of the smaller end moment to the larger end moment, Table 9 from Manual Calculate the equivalent slenderness ratio, LT =nuv = the effective length L E divided by the radius of gyration, r y. n = slenderness correction factor which is equal to 1.0 for unrestrained beam with self weight applied directly to the member between restraint points. Table 15 or 16 u = buckling parameter which may be taken as 0.9 for all rolled I-, H- or channel sections v = slenderness factor which may be obtained from Table 10 for all flanged members of uniform section. Find buckling strength, p b, and the buckling resistance moment, M b. Dr.Murude Celikag 10

11 Laterally restrained? N Y Rolled section? Determine L E taking account of support conditions using cl And p b from Table 12 for LT =0 p b from Table 11 for LT =0 Rolled section? N Y u=0.9 or use handbook value u=1.0 or calculate exact value B.2.5.1(b) v from Table 14 or B.2.5.1(d) or use safe approximation v=1.0 r y from section tables or by calculation LT = uv L E / r y cl Rolled section? Y p b from Table 11 N p b from Table 12 Basic procedure for checking lateral buckling resistance of equal flange compact section beams according to BS 5950 M b = p b S x Use of equivalent moment concept of BS 5950 Beam and loading A B C x x x x D x point of full lateral support Bending moments due to applied loads A M B M C D Beam segments whose stability must be checked A M B M C D Equivalent uniform moments for use in checking stability of each segment M AB =0.57 M B M CD =0.57 M C M BC =0.87 M C Dr.Murude Celikag 11

12 BEAM Bending only Condition I : Full lateral restraint provided (points 1-8) Condition II : Full lateral restraint not provided, loads in any position: conservative method (points 1-10) Condition III: Full lateral restraint not provided and no load other than self-weight applied directly to the member between restraint points, e.g. primary beams restrained by secondary beams. (points 1-9 and 11) Condition IV: Full lateral restraint not provided and load applied directly to the member between restraint points, e.g. primary edge beams restrained by secondary and beams Design Procedure 1) Calculate the factored load = 1.6 x LL x DL 2) Calculate the maximum factored bending moment (M x ) and the factored shear force (F v ) 3) Calculate the second moment of area (I) required to satisfy deflection limitations described in cl For simply supported beams: I = C x WL 2 where, I is the second moment of area required in cm 4 C is the deflection coefficient obtained for each loading from Fig. 3 W is the total unfactored imposed load in kn L is the span in metres Dr.Murude Celikag 12

13 When more then one load is imposed on the beam the principle of superposition may be used. or S x M p y x where axis M x S x p y is the applied moment is the plastic modulus of section about major is the design strength of steel 4) Choose a section such that its I x and S x are greater than the required values, find out about the section classification 5) Determine the value of the moment capacity (M cx ) M cx = p y S x <1.2 p y Z for plastic or compact sections M cx = p y Z x for semi-compact or slender section Where, S x is the plastic modulus of section about major axis Z x is the elastic modulus of section about major axis p y is the design strength of steel Dr.Murude Celikag 13

14 6) Calculate the shear capacity of the web P v = 0.6 p y A v For load parallel to web for I,H, channel and RHS A v =td If shear force F v < P v no further checks are required If F v > P v then either increase the section size or reduce the value of moment capacity, M cx M cx M cx 2.5 F P v v 1.5 p y D 2 t 4 7) Check for web buckling: The buckling resistance P w is given by P w b 1 n1 tpc b 1 = stiff bearing length t = is the web thickness n 1 = length obtained by dispersion at 45 o through half the depth of the section p c = compressive strength of the web (Clause 4.7.5, Table 27(c)) Clause gives a value of of 2.5d/t on the assumption that the web is acting as a fixed-ended strut with a value of L E =0.7d (d is the depth of the web) Dr.Murude Celikag 14

15 8) Check for web bearing: The local capacity (P crip ) is given by P yw is the design strength of the web n 2 of b ntpyw Pcrip 2 1 is the length obtained by dispersion through the flange to the flange-t-web connection at a slope 1:2.5 to the plane of the flange. 9) Determine the effective length L E from the two cases a) Beams with lateral restraints at ends only: L E should be taken from Table 7 and 8 of Manual. If the end conditions differ then take the mean value. b) Beams with effective intermediate lateral restraints as well as at their ends. L E for parts of the beam may be obtained from the following: i. Parts of beam between restraints: actual distance between the restraints. ii. Part of the beam between the end of the beam and the first internal lateral restraint: Mean value of (i) and the value given by Table 7 or 8 of the Manual. Dr.Murude Celikag 15

16 10)Choose a trial section and grade of steel and check that the maximum M x on any portion of the beam between lateral restraints does not exceed the buckling resistance moment M b of the section obtained from M b =p b S x where p b is the bending strength of the member (tables in Appendix B for the design strength p y, the slenderness and the torsional index x) S x is the plastic modulus of the section L E n r y a) for beams without intermediate lateral restraints may be taken as: n=0.86 n=0.94 for central point loads for all other loads b) for beams with intermediate lateral restraints, cantilevers and beams subject to destabilizing loads: n =1.0 Less conservative values of n may be obtained from Tables 12 and 13 (Manual) x is the torsional index which may be taken as D/T Dr.Murude Celikag 16

17 11)Choose a trial section and grade of steel and check that the equivalent uniform factored moment M on any portion of the beam between adjacent lateral restraints, does not exceed the buckling resistance moment M b of the section chosen. M = mm A m is the equivalent uniform moment factor, Table 9 is the maximum M x on the portion of the member M A considered. M b = p b S x S x is the plastic modulus of the section p b is the bending strength of the member, Table 11 for the design strength p y, the equivalent slenderness LT Equivalent slenderness ratio, LT = nuv = the effective length L E divided by the radius of gyration, r y. n = slenderness correction factor which is equal to 1.0 for condition III u = buckling parameter which may be taken as 0.9 for all rolled I-, H- or channel sections v = slenderness factor which may be obtained from Table 10 for 12) Same all flanged as point members 11) except of uniform that section. Take N =0.5, x =D/T m = 1.0 for condition IV n = slenderness correction factor Tables 12, 13 and 14. For cantilevers and destabilizing loads n=1.0 Dr.Murude Celikag 17