13. Buckling of Columns *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING

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1 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING To account for behavior of different-length columns, design codes specify several formulae that will best fit the data within the short, intermediate, and long column range. Steel columns Structural steel columns are designed on the basis of formulae proposed by the Structural Stability Research Council (SSRC). Factors of safety are applied to the formulae and adopted as specs for building construction by the American Institute of Steel Construction (AISC). 005 Pearson Education South Asia Pte Ltd 1

2 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Steel columns For long columns, the Euler formula is used. A factor of safety F.S. 3/1 1.9 is applied. Thus for design, 1π E KL KL σ allow 00 3 KL / r r r ( ) Value of slenderness ratio obtained by KL r c π E σ Y c ( 13 - ) ( 13-1) 005 Pearson Education South Asia Pte Ltd

3 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Steel columns For slenderness ratio lesser than (KL/r) the design For slenderness ratio lesser than (KL/r) c, the design eqn is ( ) / 1 Y r KL σ ( ) ( ) / 3 Y c allow KL KL r KL σ } { 3 + KL r KL r 8 8 c c r r 005 Pearson Education South Asia Pte Ltd 3

4 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Steel columns 005 Pearson Education South Asia Pte Ltd 4

5 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Aluminum columns Design equations are specified by the Aluminum Association, applicable for specific range of slenderness ratios. For a common alloy (014-T6), we have σ σ σ allow allow allow 195 MPa KL r MPa ( KL / r) MPa KL 0 r KL 1 < < 55 r 1 ( 13-4 ) 55 KL r ( 13-5 ) ( 13-6 ) 005 Pearson Education South Asia Pte Ltd 5

6 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Aluminum columns 005 Pearson Education South Asia Pte Ltd 6

7 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Timber columns Timber design formulae published by the National Forest Products Association (NFPA) or American Institute of Timber Construction (AITC). 005 Pearson Education South Asia Pte Ltd 7

8 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Timber columns NFPA s formulae for short, intermediate and long columns having a rectangular x-section of dimensions b and d (smallest dimension), σ allow σ allow σ allow KL 8.5 MPa 0 11 d 1 / KL d MPa KL / d ( ) MPa 11 < 6 < KL d KL d d 6 50 ( 13-7) ( 13-8) ( 13-9 ) 005 Pearson Education South Asia Pte Ltd 8

9 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Procedure for analysis Column analysis When using any formula to analyze a column, or to find its allowable load, it is necessary to calculate the slenderness ratio in order to determine which column formula applies. Once the average allowable stress has been computed, the allowable load in the column is determined d from P σ allow A. 005 Pearson Education South Asia Pte Ltd 9

10 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Procedure for analysis Column design If a formula is used to design a column, or to determine the column s x-sectional area for a given loading and effective length, then a trial-and-check procedure generally must be followed if the column has a composite shape, such as a wide-flange section. One way is to assume the column s x-sectional area, A, and calculate the corresponding stress σ σ P/A. 005 Pearson Education South Asia Pte Ltd 10

11 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Procedure for analysis Column design Also, with A use an appropriate design formula to determine the allowable stress allow. From this, calculate the required column area A req d P/σ allow. If A > A req d, the design is safe. When making comparison, it is practical to require A to be close to but greater than A req d, usually within -3%. A redesign is necessary if A > A req d. 005 Pearson Education South Asia Pte Ltd 11

12 *13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING Procedure for analysis Column design Whenever a trial-and-check procedure is repeated, the choice of an area is determined by the previously calculated required area. In engineering g practice, this method for design is usually shortened through the use of computer software or published tables and graphs. 005 Pearson Education South Asia Pte Ltd 1

13 EXAMPLE 13.8 An A-6 steel W member is used as a pinsupported column. Using AISC column design formulae, determine the largest load that it can safely support. E st 00(10 3 ) MPa, σ Y 50 MPa. 005 Pearson Education South Asia Pte Ltd 13

14 EXAMPLE 13.8 (SOLN) From Appendix B, A mm ; r x 117 mm; r y 674mm 67.4 mm. Since K 1 for both x and y axes buckling, slenderness ratio is largest if r y is used. Thus KL r From Eqn 13-, ( )( 1000 mm/m) ( 67.4 mm ) 1 5 m KL r c π E σ Y π ( 00)( 10 ) MPa MPa 005 Pearson Education South Asia Pte Ltd 14

15 EXAMPLE 13.8 (SOLN) Here 0 < KL/r < (KL/r) c, so Eqn 13-3 applies ( ) 1 50 MPa ( 15.66) σ allow 3 5 3( 74.18) ( ) + 3 8( 15.66) 3 8( 15.66) MPa Allowable load P on column is σ N/mm allow P P A ; P N 106 mm 005 Pearson Education South Asia Pte Ltd 15 kn

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18 EXAMPLE A bar having a length of 750 mm is used to support an axial compressive load of 60 kn. It is pin-supported at its ends and made from a 014-T6 aluminum alloy. Determine the dimensions of its x-sectional area if its width is to be twice its thickness. 005 Pearson Education South Asia Pte Ltd 18

19 EXAMPLE (SOLN) Since KL 750 mm is the same for x-x and y-y axes buckling, largest slenderness ratio is determined using smallest radius of gyration, using I min I y : KL KL 1 ( 750 ) r y I y / A 3 b ( 1/1) b( b )/ b( b) [ ] Since we do not know the slenderness ratio, we apply Eqn 13-4 first, ( 3 P ) N 195 N/mm ; 195 N/mm A b b ( ) b 1.40 mm 005 Pearson Education South Asia Pte Ltd 19 ( 1)

20 EXAMPLE (SOLN) Checking slenderness ratio, KL > 1 r 1.40 Try Eqn 13-6, which is valid for KL/r 55; P A MPa ( KL / r) ( 3 ) b b ( ) ( 598.1/ b) b 7.05 mm 005 Pearson Education South Asia Pte Ltd 0

21 EXAMPLE (SOLN) From Eqn (1), KL > 55 OK! r 7.05 Note: It would be satisfactory to choose the x-section with dimensions 7 mm by 54 mm. 005 Pearson Education South Asia Pte Ltd 1

22 EXAMPLE A board having x-sectional dimensions of 150 mm by 40 mm is used to support an axial load of 0 kn. If the board is assumed to be pin-supported at its top and base, determine its greatest allowable able length L as specified by the NFPA. 005 Pearson Education South Asia Pte Ltd

23 EXAMPLE (SOLN) By inspection, board will buckle about the y axis. In the NFPA eqns, d 40 mm. Assuming that Eqn 13-9 applies, we have ( 3) 0 10 N P A 3718 MPa ( KL / d ) 3718 L ( 150 mm)( 40 mm) ( 1 / 40 mm ) L 1336 N/mm mm 005 Pearson Education South Asia Pte Ltd 3

24 EXAMPLE (SOLN) Here KL ( ) d mm 40 mm 33.4 Since 6 < KL/d 50, the solution is valid. 005 Pearson Education South Asia Pte Ltd 4

25 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING A column may be required to support a load acting at its edge or on an angle bracket attached to its side. The bending moment MPe Pe, caused by eccentric loading, must be accounted for when column is designed. Use of available column formulae Stress distribution ib ti acting over x-sectional area of column shown is determined from both axial force P and bending moment M Pe. 005 Pearson Education South Asia Pte Ltd 5

26 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Use of available column formulae Maximum compressive stress is P Mc σ max + A I ( 13-30) A typical stress profile is also shown here. If we assume entire x-section is subjected to uniform stress σ max, then we can compare it with σ allow, which is determined from formulae given in chapter If σ max σ allow, then column can carry the specified load. 005 Pearson Education South Asia Pte Ltd 6

27 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Use of available column formulae Otherwise, the column s areaa A is increased and a new σ max and σ allow are calculated. This method of design is rather simple to apply ppy and works well for columns that are short or intermediate length. Calculations of σ allow is usually done using the largest slenderness ratio for the column regardless of the axis about the column experiences bending. 005 Pearson Education South Asia Pte Ltd 7

28 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Interaction formula It is sometimes desirable to see how the bending and axial loads interact when designing an eccentrically loaded column. We will consider the separate contributions made to the total column area from the axial force and the moment. If allowable stress for axial load is (σ a ) allow, then required area for the column needed d to support the load P is P A a σ ( a ) allow 005 Pearson Education South Asia Pte Ltd 8

29 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Interaction formula Similarly, if allowable bending stress is (σ b ) allow, then since I Ar, required area of column needed to resist eccentric moment is determined from flexure formula, Mc Ab σ r ( ) b allow 005 Pearson Education South Asia Pte Ltd 9

30 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Interaction formula Thus, total area A for the column needed to resist both axial force and bending moment requires that A a or P Mc + A ( ) + b σ a ( σ b ) allow allow P / A Mc / Ar + ( σ ) ( σ ) a allow σ a σ b allow + r 1 σ b σ ( ) ( ) a allow b allow A 1 ( 13-31) 005 Pearson Education South Asia Pte Ltd 30

31 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Interaction formula σ a axial stress caused by force P and determined from σ a P/A, where A is the x-sectional area of the column. σ b bending stress caused by an eccentric load or applied moment M; σ b is found from σ b Mc/I, where I is the moment of inertia of x-sectional area computed about the bending or neutral axis. σ ( σ ) ( ) a a P σ allow σ b + σ b allow Mc + σ ( ) ( ) a allow b allow 005 Pearson Education South Asia Pte Ltd 31 1 r A ( 13-31)

32 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Interaction formula (σ a ) allow allowable axial stress as defined by formulae given in chapter 13.6 or by design code specs. Use the largest slenderness ratio for the column, regardless of which axis it experiences bending. (σ b ) allow allowable bending stress as defined by code specifications. σ a + σb σ 1 a σ ( ) ( ) allow allow r P + Mc σ r 1 a σ b ( ) ( ) allow b allow ( ) 005 Pearson Education South Asia Pte Ltd 3

33 *13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING Interaction formula Eqn is sometimes referred to as the interaction formula. This approach requires a trial-and-check procedure. Designer needs to choose an available column and check to see if the inequality is satisfied. If not, a larger section is picked and the process repeated. American Institute of Steel Construction specifies the use of Eqn only when the axial-stress ratio σ a /(σ a ) allow Pearson Education South Asia Pte Ltd 33

34 EXAMPLE 13.1 Column is made of 014-T6 aluminum alloy and is used to support an eccentric load P. Determine the magnitude of P that can be supported if column is fixed at its base and free at its top. Use Eqn Pearson Education South Asia Pte Ltd 34

35 EXAMPLE 13.1 (SOLN) K. Largest slenderness ratio for column is KL ( 1600 mm ) r 3 ( 1/1)( 80 mm)( 40 mm) /[ 40 mm 80 mm ] [ ] ( )( ) 77.1 By inspection, Eqn 13-6 must be used (77.1 > 55). σ MPa ( KL r) ( 77.1) MPa allow 4.9 MPa 005 Pearson Education South Asia Pte Ltd 35

36 EXAMPLE 13.1 (SOLN) Actual maximum compressive stress in the column is determined from the combination of axial load and bending. P ( Pe) c σ max + A I P P( 0 mm)( 40 mm) + 40 mm( 80 mm) 3 1/1 40 mm 80 mm ( )( )( ) P Assuming that this stress is uniform over the x- section, instead of just at the outer boundary, σ σ P allow max ; P N 6.30 kn 005 Pearson Education South Asia Pte Ltd 36

37 EXAMPLE The A-36 steel W column is pin-connected at its ends and subjected to eccentric load P. Determine the maximum allowable value of P using the interaction method if allowable bending stress is (σ b ) allow 160 MPa, E 00 GPa, and σ Y 50 MPa. 005 Pearson Education South Asia Pte Ltd 37

38 EXAMPLE (SOLN) K 1. The geometric properties for the W are taken from the table in Appendix B. A r y 3790 mm I x mm d 157 mm 6 mm 4 We consider r y as it lead to largest value of the slenderness ratio. I x is needed since bending occurs about the x axis (c( 157 mm/ 78.5 mm). To determine the allowable bending compressive stress, we have KL 1( 4 m)( 1000 mm/m) r 38. mm 005 Pearson Education South Asia Pte Ltd 38

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40 EXAMPLE (SOLN) Then KL/r < (KL/r) c and so Eqn 13-3 must be used. ( ) 1 50 MPa ( 15.66) σ allow 3 5 3( ) ( ) { + } 3 8( 15.66) 3 8( 15.66) MPa Assuming that t this stress is uniform over the x- section, instead of just at the outer boundary, 005 Pearson Education South Asia Pte Ltd 40

41 EXAMPLE (SOLN) Applying the interaction formula Eqn yields σ a σ + b 1 σ σ P / 3790 mm N/mm ( ) ( ) + a P allow b allow 6 ( 750 mm)( 157 mm / ) / ( ) mm 160 N/mm P kn Checking application of interaction method for steel section, we require 4 1 σ a σ allow ( 3 ) ( N / 3790 mm ) 0.15 < 0.15 OK! N/mm 005 Pearson Education South Asia Pte Ltd 41

42 EXAMPLE Timber column is made from two boards nailed together so the x-section has the dimensions shown. If column is fixed at its base and free at its top, use Eqn to determine the eccentric load P that can be supported. 005 Pearson Education South Asia Pte Ltd 4

43 EXAMPLE (SOLN) K. Here, we calculate KL/d to determine which eqn to use. Since σ allow is determined using the largest slenderness ratio, we choose d 60 mm. This is done to make the ratio as large as possible, and thus yield the lowest possible allowable axial stress. This is done even though bending due to P is about the x axis. KL ( 100 mm ) 40 d 60 mm 005 Pearson Education South Asia Pte Ltd 43

44 EXAMPLE (SOLN) Allowable axial stress is determined using Eqn 13-9 since 6 < KL/d < 50. Thus 3718 MPa 3718 MPa σ allow.34 MPa KL / d 40 ( ) ( ) Applying Eqn with σ allow σ max, we have.34 σ allow + N/mm P Mc A I P 60 mm 10 P 3.35 kn ( mm) + P ( 80 mm )( 60 mm ) ( 1/1)( 600 mm)( 10 mm) Pearson Education South Asia Pte Ltd 44

45 CHAPTER REVIEW Buckling is the sudden instability that occurs in columns or members that support an axial load. The maximum axial load that a member can support just before buckling occurs is called the critical load P cr. The critical load for an ideal column is determined from the Euler eqn, P cr π EI/(KL), where K 1 for pin supports, K 0.5 for fixed supports, K 0.7 for a pin and a fixed support, and K for a fixed support and a free end. 005 Pearson Education South Asia Pte Ltd 45