Members in Compression

Compression MembersCOLUMN STABILITY A. Flexural Buckling • Elastic Buckling • Inelastic Buckling • Yielding B. Local Buckling – Section E7 pp 16.1-39 and B4 pp 16.1-14 C. Lateral Torsional Buckling AISC Requirements CHAPTER E pp 16.1-32 Nominal Compressive Strength Pn  Fcr Ag AISC Eqtn E3-1 AISC Requirements LRFD Pu  c Pn Pu  Sum of factored loads c  resistance factor for compression  0.90 c Pn  design compressive strength . Design Strength . In Summary   Fy    0.71 r Fy or Fe  0.44 Fy Fcr       0.658 F  Fy      e  KL E if  4.877 Fe KL  200 r otherwise . 1-14 Local Stability: If elements of cross section are thin LOCAL buckling occurs The strength corresponding to any buckling mode cannot be developed .Local Stability .Section B4 pp 16. Section B4 pp 16.Local Stability .1-14 Local Stability: If elements of cross section are thin LOCAL buckling occurs The strength corresponding to any buckling mode cannot be developed . Section B4 pp 16.Local Stability .1-14 • Stiffened Elements of Cross-Section • Unstiffened Elements of Cross-Section . Local Stability .1-14 • Compact – Section Develops its full plastic stress before buckling (failure is due to yielding only) • Noncompact – Yield stress is reached in some but not all of its compression elements before buckling takes place (failure is due to partial buckling partial yielding) • Slender – Yield stress is never reached in any of the compression elements (failure is due to local buckling only) .Section B4 pp 16. 1-39 ) .Local Stability . r Reduce Axial Strength (E7 pp 16.1-14 If local buckling occurs cross section is not fully effective Avoid whenever possible Measure of susceptibility to local buckling Width-Thickness ratio of each cross sectional element:  If cross section has slender elements .Section B4 pp 16. Limiting Values AISC B5 Table B4.1 pp 16.Slenderness Parameter .1-16 . 1 pp 16.1-17 .Limiting Values AISC B5 Table B4.Slenderness Parameter . 1-18 .Limiting Values AISC B5 Table B4.1 pp 16.Slenderness Parameter . 44QFy Fcr       0.Slender Cross Sectional Element: Strength Reduction E7 pp 16.1 – B4.1-40 to 16.1-39 Reduction Factor Q:   QFy    0.71 r QFy or Fe  0.658 F  QFy     e  KL E if  4.2 pp 16.877 Fe otherwise Q: B4.1-43 . 1-40 to 16.71 r QFy or Fe  0.658 F     QFy    Fy if KL E  4.44QFy Fcr       0.Slender Cross Sectional Element: Strength Reduction E7 pp 16.2 pp 16.1 – B4.1-43 . Qa: B4.1-39 Reduction Factor Q:   e   0.877 Fe otherwise Q=QsQa Qs. Local Buckling – Section E7 pp 16. Flexural Buckling • Elastic Buckling • Inelastic Buckling • Yielding B.COLUMN STABILITY A. Lateral/Torsional Buckling .1-39 and B4 pp 16. Torsional.1-14 C. Torsional & Flexural Torsional Buckling When an axially loaded member becomes unstable overall (no local buckling) it buckles one of the three ways • Flexural Buckling • Torsional Buckling • Flexural-Torsional Buckling . Torsional Buckling Twisting about longitudinal axis of member Only with doubly symmetrical cross sections with slender crosssectional elements Cruciform shape particularly vulnerable Standard Hot-Rolled Shapes are NOT susceptible Built-Up Members should be investigated . double-angle.Flexural Torsional Buckling Combination of Flexural and Torsional Buckling Only with unsymmetrical cross sections 1 Axis of Symmetry: channels. equal length single angles No Axis of Symmetry: unequal length single angles . structural tees. Torsional Buckling   ECw  1 Fe    2   K z L   GJ  I x  I y 2 Eq.200 ksi for structural steel) J = Torsional Constant . E4-4 Cw = Warping Constant (in6) Kz = Effective Length Factor for Torsional Buckling (based on end restraints against twisting) G = Shear Modulus (11. Lateral Torsional Buckling 1-Axis of Symmetry  Fe    2  Fey  Fez  4 Fey Fez H    1  1  2H   Fey  Fez     2E Fey   K y L ry  2 H  1 xo .t centroid of section . E4-5   2 ECw  1 Fez    GJ  2 2   K L z   Ag r o 2 o r x y  2 o 2 o Ix  Iy Ag Coordinates of shear center w. yo xo2  yo2 r 2 o AISC Eq.r. E4-6 .Lateral Torsional Buckling No Axis of Symmetry  Fe  Fex   Fe  Fey   Fe  Fez   2  xo  F  Fe  Fey      ro  2 e  yo  F  Fe  Fex     ro  2 e Fe is the lowest root of the Cubic equation 2 0 AISC Eq. torsional or flexural torsional) Theory of Elastic Stability (Timoshenko & Gere 1961) Flexural Buckling  2E Fe   KL / r  2 Torsional Buckling Flexural Torsional 2-axis of symmetry Buckling 1 axis of symmetry AISC Eqtn E4-4 AISC Eqtn E4-5 Flexural Torsional Buckling No axis of symmetry AISC Eqtn E4-6 .Definition of Fe Fe: Elastic Buckling Stress corresponding to the controlling mode of failure (flexural.In Summary . 44 Fy    0.877 Fe otherwise Pn  Fcr Ag .658    Fy  Fy Fe Fcr     if Fe  0.Column Strength     0. 43 Fy 50 OK Inelastic Buckling  2E  2 29.658     50    37.3 kips .44   Fy  0.71  113  87.44 ksi 2 2  KL r   87.9 in2 rx=3.000 Fe    37. Assume (KxL) = 25.50 4.59( 23.658 (50)  28.43  200 r rx 3.5 ft.05 in KL K x L 25.9)  683. (KyL) = 20 ft. and (Kz L) = 20 ft FLEXURAL Buckling – X axis WT 12X81 Ag=23.5  12    87.43  Fy   Fe Fcr  0.71 E 29.000  4.50 in ry=3.EXAMPLE Compute the compressive strength of a WT12x81 of A992 steel.59 ksi Pn  Fcr Ag  28. 05 in KyL ry  20  12  78.9 2 tf 2 .22 in4 Cw=43.000   46.22 ksi 2  78.05  2E Fey  KyL r y   2 OK  2 29.EXAMPLE FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry) WT 12X81 Ag=23.69  y=2.87 23.22 in Shear Center Ix=293 in4 Iy=221 in4 x0  0 J=9.9 in2 rx=3.70 in tf=1.8 in6 y0  y  Ix  Iy r x y  Ag 2 o 2 o 2 o 293  221  0  ( 2.50 in ry=3.69  200 3.09)   25. 87    167.22 in4 Cw=43.50 in ry=3.8 in6   2 ECw  1 Fez    GJ  2 2   K L z   Ag r o   2 ( 29. 9 25 .8)  1   11.70 in tf=1.22) 2 2        20 12 23 .EXAMPLE FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry) WT 12X81 Ag=23.05 in y=2.000)( 43.200(9.22 in Ix=293 in4 Iy=221 in4 J=9.4ksi .9 in2 rx=3. 9 in 2 H  1 0   2.22 in4 Cw=43.4  0.8312    46.22  167.8312 25.22 in 4 Fey Fez H   Fey  Fez   1 1  46.05 in  Fey  Fez  Fe    2H   y=2.70 in tf=1.22 167.63ksi .4   Ix=293 in4 Iy=221 in4 J=9.4  4 46.50 in ry=3.EXAMPLE FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry) WT 12X81 Ag=23.8312     1 1  2 2 0.8 in  6  53.090   1  0.87 2 xo2  yo2 r 2 o rx=3.22  167. 70 in tf=1.8 in6  Fy  Fy Fe Fcr     if Fe  0.44 Fy  0.22 in4 Cw=43.22 in Ix=293 in4 Iy=221 in4 J=9.658   y=2.877 Fe otherwise .9 in2 rx=3.05 in   0.63   ry=3.44(50)  22.50 in Elastic or Inelastic LTB? 0.0ksi  Fe  43.44 Fy    0.EXAMPLE FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry) WT 12X81 Ag=23. 7kips Compare to FLEXURAL Buckling – X axis Pn  Fcr Ag  21.9 in2 rx=3.9)  683.70 in tf=1.3 kips .95( 2.70)  739.50 in ry=3.658   Fy  Fy Fe    50 43.22 in Ix=293 in4 Iy=221 in4 J=9.63  50  28.82( 23.05 in y=2.22 in4 Cw=43.658      0.8 in6  Fcr   0.EXAMPLE FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry) WT 12X81 Ag=23.59ksi  Pn  Fcr Ag  30. 22 pp (4-318 to 4-322) .Column Design Tables Assumption : Strength Governed by Flexural Buckling Check Local Buckling Column Design Tables Design strength of selected shapes for effective length KL Table 4-1 to 4-2. (pp 4-10 to 4-316) Critical Stress for Slenderness KL/r table 4. Use (a) Table 422 and (b) column load tables (a) LRFD .77  200 r ry 2.48 Fy=50 ksi Table has integer values of (KL/r) Round up or interpolate Pcr  22.67ksi Pn  Pcr Ag  22.8)  494ksi . Assume pinned ends and L=20 ft.67( 21.Table 4-22 – pp 4-318 Maximum KL KL (1)( 20)(12)    96.EXAMPLE Compute the available compressive strength of a W14x74 A992 steel compression member. EXAMPLE Compute the available compressive strength of a W14x74 A992 steel compression member. Assume pinned ends and L=20 ft. Use (a) Table 422 and (b) column load tables (b) LRFD Column Load Tables Tabular values based on minimum radius of gyration Maximum KL  (1)( 20)  20 ft c Pn  494kips Fy=50 ksi . Determine available compressive strength Ag  17 rx  5.55  200 rx 5.51 Enter table 4.24(17)  616kips .55 (LRFD) Pcr  36.22 with KL/r=54.Example II A W12x58.24ksi Pn  Pcr Ag  36.28 K y L 1(8)(12)   38.51 K x L 1(24)(12)   54.25  200 ry 2. A992 steel is used.28 ry  2. 24 feet long in pinned at both ends and braced in the weak direction at the third points. 28 K y L 1(8)(12)   38. 24 feet long in pinned at both ends and braced in the weak direction at the third points.51 K x L 1(24)(12)   54.09ksi c Pn Fcr  Ag  410kips c c .22 with KL/r=54.28 ry  2.55 (ASD) Fcr  24. A992 steel is used.55  200 rx 5.Example II A W12x58.25  200 ry 2. Determine available compressive strength Ag  17 rx  5.51 Enter table 4. A992 steel is used.28 CAN I USE Column Load Tables? K y L 1(8)(12)   38.25  200 ry 2.55  200 rx 5. Determine available compressive strength K x L 1( 24)(12)   54.Example II A W12x58. 24 feet long in pinned at both ends and braced in the weak direction at the third points.51 Not Directly because they are based on min r (y axis buckling) If x-axis buckling enter table with KxL KL  rx ry . 43 ft rx ry 2. 24 feet long in pinned at both ends and braced in the weak direction at the third points.28 K y L 1(8)(12)   38. Determine available compressive strength K x L 1( 24)(12)   54.55  200 rx 5.25  200 ry 2.Example II A W12x58.51 X-axis buckling enter table with K x L (1)( 24) KL    11. A992 steel is used.1 Pn  616kips .