C-1Columns:Are compression members which are subjected to concentric axial compressive forces. These are to be found in trusses and as a lateral bracing members in frame building. Short columns are sometimes referred to as to as struts or stanchions.Beam-Columns:Are members subjected to combined axial compressive and bending stresses; These are found in single storey of multi-storey framed structures. These are treated independently in this course (chap. 12 in your text book). Columns Theory:Stocky columns (short) fail by yielding of the material at the cross section, but most columns fail by buckling at loads for less then yielding forces.(a) (b)

C-2For slender columns, Euler (1759) predicted the critical buckling load (Pcr) also known as Euler Buckling Load as:where: E = Young Modulus of Elasticity. I = Minor moment of Inertia.L = Unbraced length of column.Derivation of Euler Buckling Load:Solution of this differential equation:y = A cos (cx) + B sin (cx)where:, A and B are constants.

C-3From boundary conditions:y = 0 @ x = 0, andy = 0 @ x = L, we get (A = 0) and (B sin cL = 0)

if B 0, then cL = n where n = 0, 1, 2, 3

cL = ---- Euler Buckling Critical Load where: r = minor radius of gyration

C-4Example C-1Find the critical buckling load for W 12 x 50, supported in a pinned-pinned condition, and has an over-all length of 20 feet?Solution:rmin = ry = 1.96 inch (properties of section).Note: The steel grade is not a factor affecting buckling.Pcr = Fcr A = 19.1 x 14.7 = 280.8 kips

C-5For short (stocky) columns; Equation (C-2) gives high values for (Fcr), sometimes greater then proportional limit, Engessor (1889) proposed to use (Et) instead of (E) in Euler formula:where:Et = Tangent Modulus of ElasticityEt < EWhen (Fcr) exceeds (FPR), this is calledInelastic Buckling, constantly variable(Et) need to be used to predict (Fcr)in the inelastic zone.Shanley (1947), resolved this inconsistency.

C-6Depending on (L/r) value the column bucklingstrength was presented as shown by Shanley.Residual Stresses:-Due to uneven cooling of hot-rolled sections,residual stresses develop as seen here.The presence of residual stresses in almost all hot-rolled sections further complicates the issue of elastic buckling and leads towards inelastic buckling.

C-7The Previous conditions are very difficult to achieve in a realistic building condition, especially the free rotation of pinned ends. Thus an effective slenderness factor is introduced to account for various end conditions:Thus:where:K = Effective length factor.(Kl) = Effective length.(Kl/r) = Effective slenderness ratio.see commentary(C C2.2) (page 16.1-240)The Euler buckling formula (C-1) is based on: 1 Perfectly straight column. (no crookedness). 2 Load is concentric (no eccentricity). 3 Column is pinned on both ends.

C-8AISC (Chapter E) of LRFD code stipulates:Pu (factored load) c Pnwhere:Pu = Sum of factored loads on column.c = Resistance factor for compression = 0.90Pn = Nominal compressive strength = Fcr AgFcr = Critical buckling Stress. (E3 of LFRD)a) forb) forwhere:

C-9The above two equations of the LRFD code can beillustrated as below:where:* The code further stipulates that an upper value for for column should not exceed (200).* For higher slenderness ratio, Equation (E-3.3) controls and (Fy) has no effect on (Fcr).

C-10 Determine the design compressive strength (cPn) of W 14x74 with an untraced length of (20 ft), both ends are pinned, (A-36) steel is used?Example C-2Solution:Kl =1 x 20 x 12 = 240 inRmin = ry = 2.480.44 Fy = 0.44 x 36 =15.84 ksiFe 0.44 Fy Equ. E-3.2 (controls)c Pn = 0.9 x Fcr x Ag = 0.9 x (21.99) x 21.8 = 433.44 kips (Answer) Also from (table 4-22) LFRD Page 4-320 c Fcr = 19.75 ksi (by interpolation) c Pn = c Fcr Ag = 430.55 kips(much faster)

C-11For must profiles used as column, the buckling of thin elements in the section may proceed the ever-all bucking of the member as a whole, this is called local bucking. To prevent local bucking from accruing prior to total buckling. AISC provides upper limits on width to thickness ratios (known as b/t ratio) as shown here.See AISC (B4)(Page 16.1-14)

See also:Part 1 on propertiesof various sections.

C-12Depending on their ( b/t ) ratios (referred to as ) ,sections are classified as:Compact sections are those with flanges fully welded (connected) to their web and their: p (AISC B4)b) Non compact Sections:p r (B4)c) Slender Section: > r (B4)

Certain strength reduction factors (Q) are introduced for slendermembers. (AISC E7). This part is not required as most sectionselected are compact.

C-13Example C-3Determine the design compressive strength(c Pn) for W 12 x 65 column shown below,(Fy = 50 ksi)?From properties: Ag =19.1 in2rx = 5.28 inry = 3.02 inSolution:c Pn = 0.9 x Fcr Ag = 0.9 x 40.225 x 19.1 = 691.5 kipsA) By direct LRFD

C-14B) From Table (4.22) LRFD

Evaluate = = 54.55

Enter table 4.22 (page 4 318 LRFD)cFc = 36.235 ksi (by interpolation)Pn = Fc x Ag = 692.0 kips

C) From (Table 4.1 LRFD)Enter table (4.1 ) page 4.17 LFRD with (KL)y = 13.7Pn = 691.3 kips (by interpolation).

C-15Design with Columns Load Table (4) LFRD:-Design with Column Load Table (4) LFRD: The selection of an economical rolled shape to resist a given compressive load is simple with the aid of the column load tables. Enter the table with the effective length and move horizontally until you find the desired design strength (or something slightly larger). In some cases, Usually the category of shape (W, WT, etc.) will have been decided upon in advance. Often the overall nominal dimensions will also be known because of architectural or other requirements. As pointed out earlier, all tabulated values correspond to a slenderness ratio of 200 or less. The tabulated unsymmetrical shapes the structural tees and the single and double-angles require special consideration and are covered later.

C-16EXAMPLE C - 4A compression member is subjected to service loads of 165 kips dead load and 535 kips live load. The member is 26 feet long and pinned in each end. Use (A572 Gr 50) steel and select a W14 shape.SOLUTIONCalculate the factored load:Pu = 1.2D + 1.6L = 1.2(165) + 1.6(535) = 1054 kips Required design strength cPn = 1054 kipsFrom the column load table for KL = 26 ft, a W14 176has design strength of 1150 kips.ANSWERUse a W14 145, But practically W14 132 is OK.

C-17EXAMPLE C - 5Select the lightest W-shape that can resists a factored compressive load Pu of 190 kips. The effective length is 24 feet. Use ASTM A572 Grade 50 steel.SOLUTIONThe appropriate strategy here is to fined the lightest shape for each nominal size and then choose the lightest overall. The choices are as follows.W4, W5 and W6: None of the tabulated shape will work.W8:W8 58, cPn = 194 kipsW10:W10 49, cPn = 239 kipsW12:W12 53, cPn = 247 kipsW14:W14 61, cPn = 276 kipsNote that the load capacity is not proportional to the weight (or cross-sectional area). Although the W8 58 has the smallest design strength of the four choices, it is the second heaviest.ANSWERUse a W10 49.

C-18B) Design for sections not from Column Load Tables:For shapes not in the column load tables, a trial-and-error approach must be used. The general procedure is to assume a shape and then compute its design strength. If the strength is too small (unsafe) or too large (uneconomical), another trial must be made. A systematic approach to making the trial selection is as follows.Assume a value for the critical buckling stress Fcr. Examination of AISC Equations E3-2 and E3-3 shows that the theoretically maximum value of Fcr is the yield stress Fy.

2) From the requirement that cPn Pu, letcAgFcr Pu and

Select a shape that satisfies this area requirement.Compute Fcr and cPn for the trial shape.Revise if necessary. If the design strength is very close to the required value,the next tabulated size can be tried. Otherwise, repeat the entire procedure,using the value of Fcr found for the current trial shape as a value for Step 16) Check local stability (check width-thickness ratios). Revise if necessary.

C-19Example C-6Select a W18 shape of A36 steel that can resist a factored load of 1054 kips.The effective length KL is 26 feet.Solution:Try Fcr = 24 ksi (two-thirds of Fy):RequiredTry W18 x 192:Ag = 56.4 in2 > 48.8in2

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