CHTER MECHNICS OF MTERILS 10 Ferdinand. Beer E. Russell Johnston, Jr. Columns John T. DeWolf cture Notes: J. Walt Oler Texas Tech University 006 The McGraw-Hill Companies, Inc. ll rights reserved. Columns Stability of Structures Euler s Formula for in-ended Beams Extension of Euler s Formula Sample roblem 10.1 Eccentric Loading; The Secant Formula Sample roblem 10. Design of Columns Under Centric Load Sample roblem 10.4 Design of Columns Under an Eccentric Load 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-1
Stability of Structures In the design of columns, cross-sectional area is selected such that - allowable stress is not exceeded all - deformation falls within specifications δ L E δ spec fter these design calculations, may discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles. 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10 - Stability of Structures Consider model with two rods and torsional spring. fter a small perturbation, K( θ ) restoring moment L L sin θ θ destabilizing moment Column is stable (tends to return to aligned orientation) if L θ < K( θ ) 4K < cr L 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-4
Stability of Structures ssume that a load is applied. fter a perturbation, the system settles to a new equilibrium configuration at a finite deflection angle. L sinθ K( θ ) L θ 4K cr sinθ Noting that sinθ < θ, the assumed configuration is only possible if > cr. 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-5 Euler s Formula for in-ended Beams Consider an axially loaded beam. fter a small perturbation, the system reaches an equilibrium configuration such that d y dx d y + dx M EI EI EI y 0 Solution with assumed configuration can only be obtained if π EI > cr L π E > cr L y ( r ) π E ( L r) 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-6
Euler s Formula for in-ended Beams The value of stress corresponding to the critical load, π EI > cr L π ( L r) ( r ) π E cr L > cr cr E critical stress L slenderness ratio r receding analysis is limited to centric loadings. 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-7 Extension of Euler s Formula column with one fixed and one free end, will behave as the upper-half of a pin-connected column. The critical loading is calculated from Euler s formula, π EI cr π E cr ( r) L equivalent length 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-8 4
Extension of Euler s Formula 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-9 Sample roblem 10.1 n aluminum column of length L and rectangular cross-section has a fixed end at B and supports a centric load at. Two smooth and rounded fixed plates restrain end from moving in one of the vertical planes of symmetry but allow it to move in the other plane. L 0 in. E 10.1 x 10 6 psi 5 kips FS.5 a) Determine the ratio a/b of the two sides of the cross-section corresponding to the most efficient design against buckling. b) Design the most efficient cross-section for the column. 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-10 5
Sample roblem 10.1 SOLUTION: The most efficient design occurs when the resistance to buckling is equal in both planes of symmetry. This occurs when the slenderness ratios are equal. Buckling in xy lane: 1 I ba 1 a r z z ab 1, z 0.7L rz a 1 Buckling in xz lane: 1 I y ab 1 b ry ab 1, y L ry b / 1 rz ry a 1 b 1 Most efficient design:, z, y r r z 0.7L L a 1 b / 1 a b y 0.7 a b 0.5 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-11 Sample roblem 10.1 L 0 in. E 10.1 x 10 6 psi 5 kips FS.5 a/b 0.5 Design: L ry b 1 b cr cr ( 0.5b) ( 0 in) ( FS) (.5)( 5 kips) 1500 lbs cr cr b π ( L r) ( 0.5b) E e 1500 lbs π b 18.6 1 b 6 ( 10.1 10 psi) π ( 18.6 b) 6 ( 10.1 10 psi) ( 18.6 b) b 1.60 in. a 0.5b 0.567 in. 1.5 kips 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-1 6
Eccentric Loading; The Secant Formula Eccentric loading is equivalent to a centric load and a couple. Bending occurs for any nonzero eccentricity. Question of buckling becomes whether the resulting deflection is excessive. The deflection becomes infinite when cr d y y e dx EI π ymax e sec 1 cr Maximum stress ( y + e) c 1+ max max r ec 1 1 + sec r E cr r π EI 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-1 Eccentric Loading; The Secant Formula ec 1 max Y 1 + sec r E r 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-14 7
Sample roblem 10. The uniform column consists of an 8-ft section of structural tubing having the cross-section shown. 6 E 9 10 psi. a) Using Euler s formula and a factor of safety of two, determine the allowable centric load for the column and the corresponding normal stress. b) ssuming that the allowable load, found in part a, is applied at a point 0.75 in. from the geometric axis of the column, determine the horizontal deflection of the top of the column and the maximum normal stress in the column. 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-15 Sample roblem 10. SOLUTION: Maximum allowable centric load: - Effective length, ( 8 ft) 16 ft 19 in. L e - Critical load, π EI π cr 6.1 kips 6 4 ( 9 10 psi)( 8.0 in ) ( 19 in) 006 The McGraw-Hill Companies, Inc. ll rights reserved. - llowable load, cr all FS 6.1kips 1.1 kips all.54 in all 1.1kips 8.79 ksi 10-16 8
Sample roblem 10. Eccentric load: - End deflection, π ym e sec π ( 0.075 in) sec 1 y m 0.99 in. 1 cr 1.1kips 1 +.54 in - Maximum normal stress, ec π m 1 + sec r cr m.0 ksi ( 0.75 in)( in) ( 1.50 in) π sec 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-17 Design of Columns Under Centric Load revious analyses assumed stresses below the proportional limit and initially straight, homogeneous columns Experimental data demonstrate - for large L e /r, cr follows Euler s formula and depends upon E but not Y. - for small L e /r, cr is determined by the yield strength Y and not E. - for intermediate L e /r, cr depends on both Y and E. 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-18 9
Design of Columns Under Centric Load Structural Steel merican Inst. of Steel Construction For L e /r > C c π E cr ( L / r) e FS 1.9 cr all FS For L e /r > C c ( / r) cr Y 1 Cc 5 L / 1 / + e r L e r FS 8 Cc 8 Cc cr all FS t L e /r C c π E 1 cr Y C c Y 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-19 Design of Columns Under Centric Load luminum luminum ssociation, Inc. lloy 6061-T6 L e /r < 66: all [ 0. 0.16( / r) ] ksi [ 19 0.868( L / r) ]Ma L e /r > 66: 51000 ksi 51 10 Ma all ( / r) ( / r) lloy 014-T6 L e /r < 55: all [ 0.7 0.( / r) ] ksi [ 1 1.585( L / r) ]Ma L e /r > 66: 54000 ksi 7 10 Ma all ( / r) ( / r) e e 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-0 10
Sample roblem 10.4 SOLUTION: With the diameter unknown, the slenderness ratio can not be evaluated. Must make an assumption on which slenderness ratio regime to utilize. Calculate required diameter for assumed slenderness ratio regime. Evaluate slenderness ratio and verify initial assumption. Repeat if necessary. Using the aluminum alloy 014-T6, determine the smallest diameter rod which can be used to support the centric load 60 kn if a) L 750 mm, b) L 00 mm 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-1 Sample roblem 10.4 For L 750 mm, assume L/r > 55 Determine cylinder radius: all 7 10 Ma ( L r) 60 10 N 7 10 Ma πc 0.750 m c/ c 18.44mm c cylinder radius r radius of gyration I 4 πc 4 c πc Check slenderness ratio assumption: L L 750mm 81. > 55 r c / 18.44 mm ( ) assumption was correct d c 6.9 mm 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-11
Sample roblem 10.4 For L 00 mm, assume L/r < 55 Determine cylinder radius: L all 1 1.585 Ma r 60 10 N 0. m 6 1 1.585 10 a / πc c c 1.00mm Check slenderness ratio assumption: L L 00 mm 50 < 55 r c / 1.00 mm ( ) assumption was correct d c 4.0 mm 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10 - Design of Columns Under an Eccentric Load llowable stress method: Mc + all I Interaction method: Mc I + 1 ( all ) centric ( all ) bending n eccentric load can be replaced by a centric load and a couple M e. Normal stresses can be found from superposing the stresses due to the centric load and couple, + centric max + Mc I bending 006 The McGraw-Hill Companies, Inc. ll rights reserved. 10-4 1