Local Buckling. Local Buckling in Columns. Buckling is not to be viewed only as failure of the entire member

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1 Local Buckling MORGAN STATE UNIVERSITY SCHOOL OF ARCHITECTURE AND PLANNING LECTURE V Dr. Jason E. Charalamides Local Buckling in Columns Buckling is not to e viewed only as failure of the entire memer Especially in elements that carry such specific geometric forms that the parts can fail, e.g. W shapes, leading to failure of the ensemle, it is necessary to view the parts as well. 2

2 Local Buckling in Columns AISC Section B4 addresses the issue of local uckling That is the failure of the plate parts of the form: Local uckling of We Local uckling of Flange The strength equations in Chapter E are ased on the assumption that the element will fail as a whole. This should e examined after it has een determined that local uckling did not occur. 3 Local Buckling due to Axial Load The AISC manual addresses this issue in tale B4.1a, pg : It should e noted that plate elements (that have open surfaces that do not race among themselves, e.g. W shapes, angles, etc.) are more susceptile to local uckling. Compare factors used for plate elements vs closed elements such as rectangular or circular sections. 4

3 Local Buckling due to Flexural Load B4.1 is used in case there is flexural load applied: The suject shall e addressed at a later stage when flexural ehavior is studied when comined with axial loading 5 Local Buckling Strength Let's address it from the aspect of Critical Stress: Let F cr LB = P cr LB A The controlling width-thickness ratio of plate element is: t This is not easy to visualize immediately. In the tale the ratio is mentioned as /t ut how does that translate to each type of section? 6

4 Local Buckling Strength For a W shape: Flange: t = f 2t f i.e. the ase we consider is that of the portion that is not raced. We: t = h t w Note that as /t increases, the Fcr value decreases 7 Design Considerations Relation of /t and λr If t λ r F cr LB F y For λr see tale B4.1a Flange: λ r =0.56 E F y We: λ r =1.49 E F y 8

5 Design Considerations Therefore, Local uckling for a rolled shape will not control if: Flange: f 2t f 0.56 E F y We: h t 1.49 E F y 9 Design Considerations Slender and Unstiffened elements: If then t >λ r F cr <F y Local uckling may control column strength ased on local uckling per section E7 (AISC ) for memers with slender elements. This shall e considered eyond the scope of this course for Architecture majors. 10

6 Oservations Local Buckling rarely controls for rolled W-Shapes Local Buckling may control for High Fy Other shapes such as angles, HSS etc. Note: HSS sections should, in concept, not e so prone to local uckling, since all four walls are stiffened elements. However, many of the HSS sections have very thin walls, where local uckling may control the strength. The /t limits for walls of HSS memers in compression is: t >1.40 E Fy For Fy = 46ksi, this works out to e /t=35. Many HSS sections exceed this limit. So, local uckling is an issue for many of the HSS sections. 11 In Class Exercise Check if local uckling would control for a W12x79 with A992 steel (Fy=50 ksi) h We: 1.49 OK t w E 20.7<1.49 F y =35.9 f Flange: 0.56 OK 2t f E 8.22<0.56 F y =

7 Built up compression memer Prolem Statement: A uilt up compression memer consists of two C shapes and two plates as indicated. The memer is used as a concentrically axially loaded element, with effective unraced lengths as indicated elow. Determine the design strength of this element: h_plate h pl := 11in _plate pl := 0.5in Area_Channel C10x25 A c 7.35in 2 := Area_plate A pl := h pl pl = 5.5 in 2 Mom of Inertia y I yc := 3.34in 4 3 h pl pl Mom of Inertia y I ypl := = in 4 Mom of Inertia x I xc 91.1in 4 := We thickness t w := 0.526in Width of we T:= 8in h := T 3 pl h pl Mom of Inertia x I xpl := = in 4 Depth of Chanel d := 10in 12 Centroid of channel x ar :=.617in K factor K:= 1 Young's Modulus of Elasticity E := 29000ksi Yield Stress: F y := 50ksi Unraced length on x axis: L ux := 22ft Ultimate Strength: F u := 65ksi Unraced length on y axis: L uy := 25ft Factor of Safety phi ϕ := 0.9 A g := 2A ( pl + A c ) A g = 25.7 in 2 Solution: 1) Determining the cumulative Moment of Inertia the governing radius of gyration, and check for local uckling of we 2 2 d pl I y := 2I xc + I ypl + A pl I y = in 4 h pl I x := 2I yc + I xpl + A c x 2 ar I x = in 4 I x I y r x := A g in 2 in r x = in r y := A g in 2 in r y = in r := min( r x, r y ) r = 4.27 in governing radius of gyration is not determinant KL ux KL uy λ x := λ r x = λ y := λ x r y = λ := max( λ x, λ y ) λ = governing unraced length y is determinant E λ w := 1.49 λ F w = y 2) Calculating Euler's Buckling Stress F E := π 2 E ( λ) 2 h t w ksi ( 41.19) 2 = ksi F E = ksi 3) Determining if the uckling will e elastic or inelastic. = We_Local_Buckling := if h t w < λ w "OK",, "Fail" We_Local_Buckling = "OK" Buckling := E if ( λ) 4.71, "Inelastic", "Elastic" F OR F y Buckling := if 2.25, "Inelastic", "Elastic" Buckling = "Inelastic" y F E 4) Calculating the Buckling Stress (Fcr) and the load capacity of the section: F y F E F cr if Buckling = "Inelastic" :=, F y,.877 F E F cr = ksi ΦP n := ϕ A g F cr ΦP n = kip

8 Solution Method 2: Using Tale 4-22: 1) Determining the governing slenderness ratio We already did that right? λ = ) Using tale we locate the KL/r value corresponding to the Fy used for factorized critical stress: ΦF cr := 0.9 F cr = ksi Note: From our previous calculations: F cr = ksi Therefore: ϕ F cr = ksi 3) Calculating the capacity of the element: ΦP n := ϕ A g F cr ΦP n = kip