COLUMNS: BUCKLING (DIFFERENT ENDS) Buckling of Long Straight Columns Example 4 Slide No. 1 A simple pin-connected truss is loaded and supported as shown in Fig. 1. All members of the truss are WT10 43 sections made of structural steel with a modulus of elasticity of 00 GPa and a yield strength of 50 MPa. Determine (a) the factor of safety with respect to failure by slip, and (b) the factor of safety with respect to failure by buckling.
Buckling of Long Straight Columns Slide No. Example 4 (cont d) D Figure 1 E 3 m A B C 4 m 4 m 15 kn 30 kn Buckling of Long Straight Columns Slide No. 3 Example 4 (cont d) Free-body diagram: D E R E 3 m A θ B C θ R Cy R Cx 4 m 4 m 15 kn 30 kn
Buckling of Long Straight Columns Example 4 (cont d) Find the reactions R E and R C : Slide No. 4 + + + M M F y C E 15 8 15 8 0; R ( ) + 30( 4) RE ( 3) () + 30() 4 + R () 3 Cy Note that, Cx 30 15 0; R F DE R E 80 kn (T) 1 3 0 θ tan 36.87 4 0; R 0; R Cy E Cx 45 kn 80 kn 80 kn Buckling of Long Straight Columns Example 4 (cont d) At pin C: + + C Fy 0; FDC sin 36.87 + 45 0 FDC 75 kn 75 kn (C) Fx 0; FBC 80 FDC cos 36.87 0 FBC 80 ( 75)( 0.8) 0 F 0 kn 0 kn (C) BC F DC F BC 36.87 45 80 Slide No. 5
Buckling of Long Straight Columns Example 4 (cont d) At joint B: F F DB AB 30 kn 0 kn 0 kn (C) At Joint A: + F 15 + F F AD y AD 0; sin 36.87 0 5 kn (T) F DB F B AB 0 30 A 36.87 15 F AD Slide No. 6 0 Buckling of Long Straight Columns Example 4 (cont d) Slide No. 7 Thus, the forces in the truss are as follows: D E 80 kn (T) 3 m A 5 kn (T) 30 kn (T) B 75 kn (C) 0 kn (C) 0 kn (C) 4 m 4 m C 15 kn 30 kn
Buckling of Long Straight Columns Slide No. 8 Example 4 (cont d) (a) Factor of safety with respect to slip: For WT10 43, A 5515 mm (Appendix B) For member DE, the critical member that gives the largest force: 6 6 3 P σ A 50 10 ( 5515 10 ) 1379 10 N 1379 kn max yield P FS max F DE 1379 17.4 80 Buckling of Long Straight Columns Example 4 (cont d) (b) Factor of safety with respect to failure by buckling: For WT10 43, r min 6. mm (Appendix B) For member DC, the critical member that gives the largest compressive force: L 5,000 190.84 (slender) rmin 6. 9 6 π EA π ( 00 10 )( 5515 10 ) Pcr 98.9 kn ( L / rmin ) ( 190.84) Pcr 98.9 FS 3.99 4 F 75 DC Slide No. 9
Slide No. 10 General Notes On Column Buckling 1. Boundary conditions other than simplysupported will result in different critical loads and mode shapes.. The buckling mode shape is valid only for small deflections, where the material is still within its elastic limit. 3. The critical load will cause buckling for slender, long columns. In contrast, failure will occur in short columns when the strength of material is exceeded. Between the long and short column Slide No. 11 General Notes On Column Buckling limits, there is a region where buckling occurs after the stress exceeds the proportional limit but is still below the ultimate strength. These columns are classified as intermediate and their failure is called inelastic buckling. 4. Whether a column is short, intermediate, or long depends on its geometry as well as the stiffness and strength of its material. This concept is addressed in the columns introduction page.
Slide No. 1 The Concept of Effective Length The Euler buckling formula, namely Eqs. 9 or 1 was derived for a column with pivoted ends. The Euler equation changes for columns with different end conditions, such as the four common ones found in Figs.13 and 14. While it is possible to set up the differential equation with appropriate boundary Slide No. 13 Figure 13
Figure 14 Slide No. 14 (Beer and Johnston 199) The Effective Length Concept Slide No. 15 ENES 0 Assakkaf Conditions to determine the Euler buckling formula for each case, a more common approach makes use of the concept of an effective length. The pivoted ended column, by definition, has zero bending moments at each end. The length L in the Euler equation, therefore, is the distance between
Slide No. 16 The Effective Length Concept Successive points of zero bending moment. All that is needed to modify the Euler column formula for use with other end conditions is to replace L by L. L is defined as the effective length of the column. Slide No. 17 The Effective Length Concept Definition: The effective length L (or L e ) of a column is defined as the distance between successive inflection points or points of zero moment.
The Effective Length Concept Based on the effective length concept, the Euler Buckling load formula becomes π EI Pcr (16a) Le or L e L effective length P cr π EA ( L / r) e (16b) Slide No. 18 Example 5 Slide No. 19 What is the least thickness a rectangular wood plank 4 in. wide can have, if it is used for a 0-ft column with one end fixed and one end pivoted, and must support an axial load of 1000 lb? Use a factor of safety (FS) of 5. The modulus of elasticity of wood is 1.5 10 6 psi.
Example 5 (cont d) The rectangular cross section is Slide No. 0 b 4 in t? 3 bt I 1 For the end conditions specified in the problem, Fig. 13d (or Fig. 14c) gives L e 0. 7L Slide No. 1 Example 5 (cont d) Figure 13
Example 5 (cont d) Since FS 5 is required, but P cr π EI Pcr Le From which ( FS ) (1000) 5,000 lb P 5 3 6 4t π (1.5 10 ) 1 5,000 [ 0.7( 0 1) ] t 3.06 in Slide No. Example 6 Slide No. 3 An L10 76 6.4-mm aluminum alloy (E 70 GPa) angle is used for fixed-end, pivoted-end column having an actual length of.5 m. Determine the maximum safe load for the column if a factor of safety 1.75 with respect to failure by buckling is specified.
Example 6 (cont d) For fixed-end, pivoted-end column, Fig. 13d (or Fig. 14c) gives the equivalent length as Slide No. 4 L e (.5) 1.75 m 1,750 mm 0.7L 0.7 For L10 76 6.4 section (see Fig. 15 or Appendix B of textbook): A 1090 mm r 16.5 mm min Figure 15 Slide No. 5 Example 6 (cont d)
Example 6 (cont d) Therefore, Slide No. 6 L r min 1750 106.06 (slender) 16.5 P max Pcr FS π π EA ( L / r ) FS min 9 6 ( 70 10 )( 1090 10 ) ( ) 1.75 106.06 π EA FS ( L / r ) min 38,54.54 N 38.3 kn Example 7 Slide No. 7 Determine the maximum load that a 50- mm 75-mm.5-m long aluminum alloy bar (E 73 GPa) can support with a factor of safety of 3 with respect to failure by buckling if it is used as a fixed-end, pivoted column. For fixed-end, pivoted column, Fig. 13d (or Fig. 14c) gives (.5) 1.75 m effective length, L 0.7L 0.7
Slide No. 8 Example 7 (cont d) Computation of the section properties: 50 mm 75 mm A 75 I r min min ( 50) ba 1 I A min 3 3,750 mm ( 50) 75 1 3 0.7813 10 3750 0.7813 10 6 6 mm 14.43 mm 4 Example 7 (cont d) Therefore, the slenderness ratio of the column can be obtained: and L r min π EA 1.75 10 14.43 π 3 11.8 (slender) 9 6 ( 73 10 )( 3750 10 ) cr max FS FS( L / rmin ) 3( 11.8) P P 61. kn Slide No. 9
Example 8 Slide No. 30 A structural steel (E 9,000 ksi) column 0 ft long must support an axial compressive load of 00 kip. The column can be considered pivoted at one end and fixed at the other end for bending about one axis and fixed at both ends for bending about the other axis. Select the lightest wide-flange or American standard section that can be used for the column. Slide No. 31 Example 8 (cont d) First case: fixed-end, pivoted column L P I ex cr x L 0.7 x π EI L x x P crl π E x ( 0) 00 14 π 14 ft ( 1) ( 9,000) 19.7 in
Slide No. 3 Example 8 (cont d) Second case: fixed ends column L I ey y L 0.5 y P L cr y π E ( 0) 00 10 π 10 ft ( 1) ( 9,000) 10.06 in Use a W54 33 section Example 9 Slide No. 33 A 5 mm-diameter tie rod AB and a pipe strut AC with an inside diameter of 100 mm and a wall thickness of 5 mm are used to support a 100-kN load as shown in Fig. 16. Both the tie rod and the pipe strut are made of structural steel with modulus of elasticity of 00 GPa and a yield strength of 50 MPa. Determine
Slide No. 34 Example 9 (cont d) (a) the factor of safety with respect to failure by slip. (b) the factor of safety with respect to failure by buckling Slide No. 35 Example 9 (cont d) B Figure 16 A.5 m 100 kn C 4.5 m 6 m
Example 9 (cont d) At joint A: A 1.5 0 1 4.5 0 θ tan.6, α tan 36.87 6 6 + Fx 0; FAB cosθ + FAC cosα 0 FAB cos.6 + FAC cos 36.87 0 F 0.8667F AB AC y (17) + Fx 0; FAB sinθ FAC sinα 100 0 FAB sin.6 FAC sin 36.87 100 0 F 1.56F + 60 (18) AB AC θ α Slide No. 36 F AB F AC x Example 9 (cont d) Substituting for F AB in Eq. 17 into Eq. 18, gives 0.8667FAC 1.56FAC + 60 Thus, F 107.14 kn 107.14 kn (C) AC From Eq. 18, with F AC known, we have F AB 1.56FAC + 60 1.56( 107.14) + 60 9.86 kn (T) Slide No. 37
Example 9 (cont d) Calculate the cross-sectional areas for AB and AC: π π ( 5) 490.9 mm, ( 150) ( 100) 9817 mm A AB A AC 4 (a) FS due to Failure by Slip: σ FS AB FAB 9.86 10 AAB 490.9 10 50 1.3 189.16 AB 6 3 4 189.16 10 [ ] 6 N/m 189. MPa Slide No. 38 Slide No. 39 Example 9 (cont d) σ FS AB FAC 107.14 10 6 AAC 9817 10 50.9 10.914 AC 3 10.914 10 N/m 10.914 MPa (C) (b) FS due to Failure by Buckling: Member AC is the compression member with a compressive force of 107.14 kn: 6 π ( 150) ( 100) 19.94 10 mm I AC 64 [ ] 4 6
Example 9 (cont d) (b) FS for Buckling (cont d): Slide No. 40 6 I 19.94 10 r 45.07 A 9817 ( 6) + ( 4.5) 1000 L 166.41(slender) r 45.07 9 6 π EA π ( 00 10 )( 9817 10 ) 3 P 699.8 10 N 700 kn cr FS AC ( L / r) P F cr AC 700 107.14 ( 166.41) 6.53