Failure in Flexure. Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas

Transcription

1 Introduction to Steel Design, Tensile Steel Members Modes of Failure & Effective Areas MORGAN STATE UNIVERSITY SCHOOL OF ARCHITECTURE AND PLANNING LECTURE VIII Dr. Jason E. Charalambides Failure in Flexure! Consider an experiment where a point load is applied on a W-section at mid-point along the strong axis: 2

2 Failure in Flexure! The load was increased until P_max is reached, i.e. flexural failure is experienced.! Mn is the nominal flexural strength that the beam can sustain without failure, and it is considered based on the following failure modes Local Buckling Flange Web Lateral torsional buckling Development of fully plastic cross section failure by excessive deformation 3 Local Buckling! Buckling occurs on the upper flange and the upper portion of the web that are subjected to compression Buckling would not occur in any portion below the NA as it is subject to Tensile stresses 4

3 Local Buckling! Flange In this case the flange starts twirling, with one side bending downwards and the other upwards to compensate the compressive stress that it is subjected to, which exceeds its maximum capacity. 5 Local Buckling! Web In this case the web behaves in a manner similar to the way we saw the flange behaving in the previous slide to compensate the compressive stress that it is subjected to, which exceeds its maximum capacity. 6

4 Lateral Torsional Buckling! Lateral Torsional Buckling The center line of top flange shifts off axis, and the beam rotates, minimizing the exposure of the load on its strong xx axis, and compensating the stress taken by rotating toward the weaker yy axis. The compression flange buckles laterally, like a column. The tension flange resists the buckling and the lateral movement of the compression flange, resulting in a combined effect of lateral movement and rotation. To prevent this effect, lateral braces must be provided along the length of the beam 7 Compact Section! What is defined as a compact section? It is thick and not too long, practically the opposite of a slender element, that can develop a plastic stress distribution before buckling! What parameters define a section as compact? It has to have width-thickness rations equal to or less than the limiting values of λ p given in Table B4.1 (pp &17), and The compressive portions have to be connected to the web or webs 8

5 Bracing! Limit to moment that a beam can sustain: A beam's strength against LTB depends on the unbraced length L b : The shorter the L b, the higher the strength of the beam against LTB: Think of it in a way similar to a column's slenderness. The longer the unbraced length, the higher the slenderness factor 9 Beam Lateral Braces! In a building it is typical to have bracing by cross beams or by a concrete floor slab. A lateral brace needs to satisfy the following: Prevent lateral displacement of the compression flange, OR Prevent twist of the element 10

6 Beam Lateral Braces! In this example with the shear studs applied on the top flange. In the scenario where the bottom flange would be in compression, the concrete floor slab would not serve as a lateral brace as it would not control the lateral movement of the compression flange. 11 Beam Lateral Braces! An example where the lateral displacement of compression flange is prevented by diagonal bracing. A typical scenario is the use of angles and gussets. 12

7 Ranges Of Beam Behavior in Flexure! In all three previously mentioned scenarios of instability (FLB, WLB, LTB) occurs, it is safe to state that the failure occurred in one of the following three possible ranges of beam behavior. Failure in Plastic range Failure in Inelastic range Failure in Elastic range 13 Ranges Of Beam Behavior in Flexure! Summarizing in a tabular format the characteristics of failure and ranges: Failure mode M n Ductility Elastic Mn 0.7 M y Little or none Inelastic 0.7<Mn M p Little or none Plastic Mn = M p Large! The AISC provides different equations that govern the design of flexural members for each of these ranges of behavior 14

8 Designing for Flexure! At any point along the length of a flexural member, the following equation has to be satisfied: M u Φ M n Φ=0.9! M u is the moment generated due to design loads! M n is the nominal flexural strength of the beam M n based on Flange Local Buckling M n = Lesser of: M n based on Web Local Buckling M n based on Lateral Torsional Buckling M p 15 Nominal Moment on Local Buckling! There are two major factors that govern the behavior of a steel flexural member on local buckling: Slenderness parameter (width to thickness ratio) of flange and/or web Specified minimum Yield Stress (F y )! The symbol λ is used as a generic symbol for that purpose (pg xl) For FLB: For WLB: λ= b f 2t f λ= h t w! The values indicated above are tabulated for all rolled W shapes in the AISC user's manual in Part 1 16

9 Nominal Moment on Local Buckling! To determine the Mn based on FLB and WLB, the value of the slenderness ratio is to be compared to the slenderness ratio for a compact and a non-compact element (λr and λp respectively.)! Table Β4.1b in section B4 provides all the necessary formulae. See examples for hot rolled W shape: For FLB: λ= b f 2t f λ p =0.38 λ r =1.0 For WLB: λ= h t w λ p =3.76 λ r =5.70 Where λ p is the limiting slenderness parameter for compact element and λ r for non-compact 17 Nominal Moment w.r.t. Unbraced Length 18

10 Determining Compactness! For an element to be considered compact the following conditions must be satisfied: λ λ p For FLB: For WLB: b f 2t f 0.38 h t w 3.76 Defined as Compact Section! If the above conditions are met, the section is compact, and failure by local buckling will occur in the plastic range. 19 Determining Compactness! An element that lies in the area of the graph where Inelastic LTB is anticipated is Non-Compact: λ p <λ λ r For FLB: For WLB: 0.38 < b f 2t f < h t w 5.70 Defined as Non-Compact Section! If the above conditions are met, failure by local buckling occurs in the inelastic range, at a moment between 0.7 M y and M p. 20

11 Determining Compactness! An element that lies in the area of the graph that Elastic LTB is anticipated is considered Slender: λ>λ r For FLB: For WLB: b f 2t f >1.0 h t w >5.70 Defined as Slender Section! If the above conditions are met, failure by local buckling occurs in the elastic range, at a moment below 0.7 My. 21 Effect of Web Local Buckling on Mn! For sections with a compact web, WLB has no effect on M n, which is controlled by FLB, LTB, or M p.! For sections with a non-compact or slender web: WLB is considered to reduce the amount of web area available to resist the applied moment Buckled portion considered ineffective in resisting moment 22

12 Effect of Web Local Buckling on Mn! For sections with a non-compact or slender web: (cont) Buckled portion is ineffective, Remaining portions of the cross section can continue to resist moment until FLB, LTB, or yielding occurs, Equations to account for the effects of WLB on M n are given in chapter F, Sections F4 and F5 23 Practical Considerations! Flange Local Buckling: Most rolled shapes are compact for FLB at Fy=50 ksi.! Web Local Buckling: All rolled W-shapes are compact for WLB for Fy=50 ksi.! Local Buckling may control Mn for the following cases: Rolled W-shapes with high strength steels (F y 50 ksi ) Welded W-shapes made of thin plates Angles, WT's, and shapes other than standard rolled W-shapes 24

13 Fundamental Nomenclature! C b - Lateral Torsional Buckling modification factor (F1-1)! Beam Section Properties that LTB depends upon: J St. Venant's torsional constant Cw Warping torsion constant Iy Moment of inertia about y axis Sx Elastic section modulus h0 Distance between flange centroids r_ts -Effective radius of gyration for LTB r ts = I y C w S x! It can be approximated to the radius of gyration of the compression flange + 1/6 of the web! The above values are available except of the Cb that can be calculated 25 The Lateral Torsional Buckling Modification Factor! The lateral buckling can be substantially restrained by the end conditions and affected by loading patterns.! Therefore a factor that helps better define the anticipated behavior of an element, based on the factors affect this behavior has been developed and refined to today's Cb: 12.5 M C b = max F M max +3M A +4M B +3M C Where: M max is the absolute value of max. moment, MA, MB, and MC is the value at quarter, mid, and three quarter points of the unbraced segment. 26

14 Some Standard Values of Cb Diagrams copied for educational purposes only 27 Some Standard Values of Cb Also see Table 3-1 on pg of AISC Manual Diagrams copied for educational purposes only 28

15 Determining The Limit State of the Unbraced Length L p =.76 r y E F F2-5 y 2 E L r =1.95 r ts h F F y S 0 x + ( J c ) +6.79( 0.7 F 2 y S x h 0 E ) J c The coefficient c (in Jc) is equal to 1 for symmetric shapes. For channels refer to AISC manual F2-8b, pg Determining The Limit State of the Unbraced Length For L b L p ΦM n =ΦM p =F y Z x F2-1 For L p <L b L r ΦM n =Φ C b[ M p (Mp 0.75 F y S x )( L b L p L F2-2 r L p )] ΦM p For L b >L r ΦM n =Φ F r S x ΦM p F2-3 30

16 The critical Stress! The Critical Stress Fcr, is used in the formulae above mentioned and under these conditions it is given by: F cr = C bπ 2 E ( L 2 b r ts ) J c S x h 0 ( L 2 b r ts ) F In Class Exercise Determine the largest value of uniformly distributed load that can be applied on a W-shape (W18x50) spanning 20 ft. Assume shear is OK. Use A992 steel: 32

17 Beam subjected to LTB Problem Statement: Determine the largest value of uniformly distributed load that can be applied on the W-shape (W18x50). Assume shear is OK.Use A992 steel Unbraced Length: L b := 20ft Section modulus: S x := 88.9in 3 Plastic modulus: Z x := 101in 3 Moment of Inertia on yy I yy := 40.1in 4 radius of gyration y r y := 1.65in effective radius of gyration r ts := 1.98in Dist flange centroids h 0 := 17.4in Warping constant C w 3040in 6 := Torsional constant J := 1.24in 4 Young's Modulus of Elasticity E := 29000ksi Yield Stress: F y := 50ksi Ultimate Strength: F u := 65ksi LTB factor for non-uniform moment diagrams Solution: C b := ) Determining the plastic moment and the elastic moment By M. Engelhardt Factor of Safety phi ϕ := 0.9 ΦM p := ϕ Z x F y ΦM p = k' ΦM r := ϕ 0.7S x F y ΦM r = k' Check Local buckling, OK per table 3-2, (pg 325) 2) Calculating the Limit State lengths 2 E E J.7 F y S x h 0 L p := 1.76 r y L F p = 5.83 ft L r := 1.95 r ts L y.7 F y S x h 0 E J r = ft OR the above values of Lp and Lr can be obtained from table 3-2 (pg 325) 3) Calculating the Nominal Moment Since Lb>Lr we need to apply the formulae F2-3 and F2-4 C b π 2 2 E J L b F cr := S L b x h 0 r ts F cr = ksi ΦM n := ϕ F cr S x ΦM n = k' r ts 4) Calculating the maximum Uniformly Distributed load possible based on Moment capacity: 8 ΦM n w max := w 2 max = 4.1 kip ft L b OR we could go to charts, p at Lb=20ft, read up to W18x50, find ΦMn at Cb=1 that gives us 180k', multiply 180 by 1.14 which gives us 205k', and then use the formula in step 4 to determine that the maximum happens to be 4.1 kips per foot! Note that W18x50 is given in a dotted line that indicates that the section is not compact.

18 Beam subjected to LTB Problem Statement: By M. Engelhardt Determine the largest value of uniformly distributed load that can be applied on the W-shape (W18x50). Assume shear is OK.Use A992 steel. Length is 20 ft with lateral supports at ends and midpoint Unbraced Length: L b := 10ft Section modulus: S x := 88.9in 3 Plastic modulus: Z x := 101in 3 Moment of Inertia on yy I yy := 40.1in 4 radius of gyration y r y := 1.65in effective radius of gyration r ts := 1.98in Dist flange centroids h 0 := 17.4in Warping constant C w 3040in 6 := Torsional constant J := 1.24in 4 Young's Modulus of Elasticity E := 29000ksi Yield Stress: F y := 50ksi Ultimate Strength: F u := 65ksi LTB factor for non-uniform moment diagrams Solution: C b := 1.3 1) Determining the plastic and elastic moments Factor of Safety Φ ϕ := 0.9 M p := Z x F y M p = k' M y := S x F y M y = k' Elast_fail_Mom := M y Elast_fail_Mom = k' 2) Calculating the Limit State lengths 2 E E J.7 F y S x h 0 L p := 1.76 r y L F p = 5.83 ft L r := 1.95 r ts L y.7 F y S x h 0 E J r = ft OR the above values of Lp and Lr can be obtained from table 3-2 (p 325) 3) Calculating the Nominal Moment Since Lb<Lr we need to apply the formula F2-2 L b L p ΦM n = ϕ C b M p ( M p 0.7M y )... ϕ C L r L b M p M p 0.7M y p ΦM n := if ϕ C b M p M p 0.7M y ( ) Check Local buckling, OK per table 3-2 L b L p L r L p ( ) ( ) L b L p L r L p L b L p L r L p = k' BUT.9 M p, ϕ C b M p M p 0.7M y, ϕ M p ΦM n = k' 4) Calculating the maximum Uniformly Distributed load possible based on Moment capacity: 8 ΦM n w max := ( 2L b ) 2 w max = 7.58 kip Note that we used 2Lb because the total length of the beam is ft actually twice the length of the unbraced sections! ΦM p :=.9 M p ΦM p = k' OR we could go to charts, p at Lb=10ft, read up to W18x50, find PhiMn at Cb=1 that gives us 324k', multiply 324 by 1.3 which gives us 421k'. That is already larger thatn 0.9 Mp, so 0.9Mp governs at 379 k'... and then use the formula in step 4 to determine that the maximum happens to be 7.58 kips per foot!