LECTURE Third Edition COLUMNS: BUCKLNG (PNNED ENDS) A. J. Clark School of Engineering Department of Civil and Environmental Engineering 6 Chapter 10.1 10.3 b Dr. brahim A. Assakkaf SPRNG 003 ENES 0 Mechanics of Materials Department of Civil and Environmental Engineering Universit of Marland, College Park LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 1 ntroduction is a mode of failure generall resulting from structural instabilit due to compressive action on the structural member or element involved. Eamples Overloaded metal building columns. Compressive members in bridges. Roof trusses. Hull of submarine.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. ntroduction Eamples (cont d) Metal skin on aircraft fuselages or wings with ecessive torsional and/or compressive loading. An thin-walled torque tube. The thin web of an -beam with ecessive shear load A thin flange of an -beam subjected to ecessive compressive bending effects. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 3 ntroduction n view of the above-mentioned eamples, it is clear that buckling is a result of compressive action. Overall torsion or shear, as was discussed earlier, ma cause a localized compressive action that could lead to buckling. Eamples of buckling for commonl seen and used tools (components) are provided in the net few viewgraphs.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 4 ntroduction igure 1 LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 5 ntroduction igure 1 (cont d)
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 6 ntroduction igure 1 (cont d) LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 7 ntroduction n ig. 1, (a) to (d) are eamples of temporar or elastic buckling. While (e) to (h) of the same figure are eamples of plastic buckling The distinctive feature of buckling is the catastrophic and often spectacular nature of failure.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 8 ntroduction igure. Reinforced Concrete LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 9 ntroduction igure 3. Steel Beam
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 10 ntroduction The collapse of a column supporting stands in a stadium or the roof of a building usuall draws large headlines and cries of engineering negligence. On a lesser scale, the reader can witness and get a better understanding of buckling b tring to understand a few of the tests shown in ig. 1. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 11 The Nature of n the previous chapters, we related load to stress and load to deformation. or these non-buckling cases of aial, torsional, bending, and combined loading, the stress or deformation was the significant quantit in failure. of a member is uniquel different in that the quantit significant in failure is
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 1 The Nature of the buckling load itself. The failure (buckling) load bears no unique relationship to the stress and deformation at failure. Our usual approach of deriving a loadstress and load-deformation relations cannot be used here, instead, the approach to find an epression for the buckling load P cr. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 13 The Nature of is unique from our other structural-element considerations in that it results from a state of unstable equilibrium. or eample, buckling of a long column is not caused b failure of the material of which the column is composed, but b determination of what was a stable state of equilibrium to an unstable one.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 14 The Nature of Mechanism of Let s consider ig. 4, 5, and 6, and stud them ver carefull. n ig. 4, some aial load P is applied to the column. The column is then given a small deflection b appling the small lateral force. f the load P is sufficientl small, when the force is removed, the column will go back to its original straight condition. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 15 The Nature of Mechanism of P < P cr igure 4 P < Pcr P < Pcr Before During After Before During After Stable Equilibrium
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 16 The Nature of Mechanism of The column will go back to its original straight condition just as the ball returns to the bottom of the curved container. n ig. 4 of the ball and the curved container, gravit tends to restore the ball to its original position, while for the column the elasticit of the column itself acts as restoring force. This action constitutes stable equilibrium. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 17 The Nature of Mechanism of The same procedure can be repeated for increased value of the load P until some critical value P cr is reached, as shown in ig. 5. When the column carries this load, and a lateral force is applied and removed, the column will remain in the slightl deflected position. The elastic restoring force of the column is not sufficient to return the column to its original
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 18 The Nature of Mechanism of P P cr igure 5 P Pcr P Pcr Before During Before During After After Precarious Equilibrium LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 19 The Nature of Mechanism of straight position but is sufficient to prevent ecessive deflection of the column. n ig. 5 of the ball and the flat surface, the amount of deflection will depend on the magnitude of the lateral force. Hence, the column can be in equilibrium in an infinite number of slightl bent positions. This action constitutes neutral or precarious equilibrium.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 0 The Nature of Mechanism of f the column is subjected to an aial compressive load P that eceeds P cr, as shown in ig. 6, and a lateral force is applied and removed, the column will bend considerabl. That is, the elastic restoring force of the column is not sufficient to prevent a small disturbance from growing into an ecessivel large deflection. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 1 The Nature of Mechanism of P > P cr igure 6 P > Pcr P > Pcr Before During After Before Small disturbance ν During Possible buckle or collapse After Unstable Equilibrium
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. The Nature of Mechanism of Depending on the magnitude of P, the column either will remain in the bent position or will completel collapse and fracture, just as the ball will roll off the curved surface in ig. 6. This tpe of behavior indicates that for aial loads greater than P cr, the straight position of a column is one of unstable equilibrium in that a small disturbance will tend to grow into an ecessive deformation. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 3 The Nature of Definition can be defined as the sudden large deformation of structure due to a slight increase of an eisting load under which the structure had ehibited little, if an, deformation before the load was increased.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 4 Critical Load The purpose of this analsis is to determine the minimum aial compressive load for which a column will eperience lateral deflection. Governing Differential Equation: Consider a buckled simpl-supported column of length L under an eternal aial compression force P, as shown in the left schematic of ig. 7. The transverse displacement of the buckled column is represented b δ. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 5 Critical Load P igure 7 P P (a) (b)
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 6 Critical Load Governing Differential Equation: The right schematic of ig. 7 shows the forces and moments acting on a cross-section in the buckled column. Moment equilibrium on the lower free bod ields a solution for the internal bending moment M, P + M 0 (1) LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 7 Critical Load Governing Differential Equation (cont d): Recall the relationship between the moment M and the transverse displacement for the elastic curve, d E M () d Eliminating M from Eqs. 1 and results in the governing equation for the buckled slender column,
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 8 Critical Load Governing Differential Equation (cont d): d P + 0 (3) d E Solution: The governing equation is a second order homogeneous ordinar differential equation with constant coefficients and can be solved b the method of characteristic equations. The solution is found to be, LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 9 Critical Load Solution (cont d): ( ) Asin p + B cos p (4) Where p P/E. The coefficients A and B can be determined b the two boundar conditions, (0) 0 and (L) 0, which ields, B 0 Asin pl 0 (5)
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 30 Critical Load Solution (cont d): The coefficient B is alwas zero, and for most values of m L the coefficient A is required to be zero. However, for special cases of m L, A can be nonzero and the column can be buckled. The restriction on m L is also a restriction on the values for the loading ; these special values are mathematicall called eigenvalues. All other values of lead to trivial solutions (i.e. zero deformation). LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 31 Critical Load Solution (cont d): sin pl 0 pl 0, π, π, 3π, L, nπ or π π 3π nπ p 0,,,, L, L L L L Since p P/E, therefore, P (6) π E ( ) π E ( 3) π E n π E 0,,,, L, (7) L L L L
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 3 Critical Load Solution (cont d): Or nπ P E for n 0,1,,3L L (8) The lowest load that causes buckling is called critical load (n 1). P cr π E (9) L LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 33 Critical Load, P cr The critical buckling load (Euler ) for a long column is given b π E P cr (9) L where E modulus of elasticit of the material moment of inertia of the cross section L length of column
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 34 Critical Load Equation 9 is usuall called Euler's formula. Although Leonard Euler did publish the governing equation in 1744, J. L. Lagrange is considered the first to show that a non-trivial solution eists onl when n is an integer. Thomas Young then suggested the critical load (n 1) and pointed out the solution was valid when the column is slender in his 1807 book. The "slender" column idea was not quantitativel developed until A. Considère performed a series of 3 tests in 1889. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 35 Critical Load Shape function: Substituting the epression of P in Eq. 9, into Eq. 4, and noting that B 0, the shape function for the buckled shape () is mathematicall called an eigenfunction, and is given b, nπ sin (10) L ( ) A
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 36 Critical Stress The critical buckling normal stress σ n is found as follows: When the moment of inertia in Eq. 9 is replaced b Ar, the result is Pcr π E σ cr A ( L / r) where A cross-sectional area of column r radius of gration A (11) LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 37 Critical Stress The critical buckling normal stress is given b π E σ cr (1) L / r ( ) Where r radius of gration A (L/r) slenderness ratio of column
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 38 Critical Load and Stress The Euler buckling load and stress as given b Eq. 9 or Eq. 1 agrees well with eperiment if the slenderness ratio is large (L/r > 140 for steel columns). Short compression members (L/r < 140 for steel columns) can be treated as compression blocks where ielding occurs before buckling. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 39 Critical Load and Stress Man columns lie between these etremes in which neither solution is applicable. These intermediate-length columns are analzed b using empirical formulas to be described later. When calculating the critical buckling for columns, (or r) should be obtained about the weak ais.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 40 Review of Parallel-Ais Theorem for Radius of Gration n dealing with columns that consist of several rolled standard sections, it is sometimes necessar to compute the radius of gration for the entire section for the purpose of analzing the buckling load. t was shown that the parallel-ais theorem is a useful tool to calculate the second LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 41 Review of Parallel-Ais Theorem for Radius of Gration Moment of area (moment of inertia) about other aes not passing through the centroid of the overall section. n a similar fashion, the parallel-ais theorem can be used to find radii of gration of a section about different ais not passing through the centroid.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 4 Review of Parallel-Ais Theorem for Radius of Gration Consider the two channels, which are laced a distance of a back to back. d a Lacing bars C C C r C Aoverall Asec Asec ( + Asecd ) ( Asecr Asecd ) Asec( r d ) C + C + C Asec( r + d ) r A overall C A sec r C + d r C (13) (14) LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 43 Parallel-Ais Theorem for Radius of Gration Eqs. 13 and 14 indicate that the radius of gration for the two channels is the same as that for one channel, and d a Lacing bars C r r + + C ( a ) (15) where a + C d C
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 44 Eample 1 A 3-m column with the cross section shown in ig. 8 is constructed from two pieces of timber. The timbers are nailed together so that the act as a unit. Determine (a) the slenderness ratio, (b) the Euler buckling load (E 13 GPa for timber), and (c) the aial stress in the column when Euler load is applied. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 45 Eample 1 (cont d) 50 mm igure 8 50 mm 50 mm 150 mm 50 mm 150 mm
N.A. 75 mm LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 46 50 mm 50 mm 50 mm 150 mm Eample 1 (cont d) Properties of the cross section: A ( 150)( 50) 15,000 mm 5( 50 150) + ( 50 + 75)( 50 150) 150 mm 50 mm C 1 3 1 1 r A 15,000 1 3 1 1 3 3 ( 50)( 15) + ( 150)( 75) ( 100)( 5) 3 ( 150)( 50) + ( 50)( 150) min A 3 15.65 10 15,000 1 3 6 15.65 10 3.3 53.13 10 mm 75.0 mm from bottom 3 6 4 6 mm 4 LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 47 Eample 1 (cont d) (a) Slenderness Ratio: L 3000 Slendernes s ratio 93 r 3.7 (b) Euler Load: P cr π E L π (c) Aial Stress: σ 9 6 ( 13 10 )( 15.65 10 ) () 3 P cr 3.75 A 15 10 14.85 MPa (C).75 kn
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 48 Eample A WT6 36 structural steel section is used for an 18-ft column. Determine (a) The slenderness ratio. (b) The Euler buckling load. Use E 9 10 3 ksi. (c) The aial stress in the column when Euler load is applied. LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 49 Eample (cont d) or a WT6 36 section (see ig 9, or Appendi B of Tetbook: A.6 in r 1.48 in Pcr (c) σ A ( L / r) 10 min L 18 1 (a) 145.9 146 (slender) r 1.48 π EA π ( 9,000)( 10.6) (b) Pcr 14.4 kips 145.9 14.4 13.43 ksi (C) 10.6
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 50 igure 9 Eample (cont d) LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 51 Eample 3 Two C9 30 structural steel channels are used for a column that is 1 m long. Determine the total compressive load required to buckle the two members if (a) The act independentl of each other. Use E 00 GPa. (b) The are laced 150 mm back to back as shown in ig. 10.
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 5 Eample 3 (cont d) igure 10 Lacing bars 150 mm LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 53 Eample 3 (cont d) (a) Two channels act independentl: f the two channels are not connected and each acts independentl, the slenderness ratio is determined b using the minimum radius of gration r min of the individual section or a C9 30 section (see ig 11, or Appendi B of Tetbook): r min r 16.3 mm A 3795 mm
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 54 of igure Long 11 Straight Eample 3 (cont d) LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 55 150 mm Eample 3 (cont d) 3 L 1 10 736. (slender) r 16.3 π EA π P cr ( L / r) Lacing bars 9 6 ( 00 10 )[( )( 3795 10 )] ( 736.) (b) or a C9 30 section (see ig 11, or Appendi B of Tetbook): r 81.8 mm 14.8 mm min 5.3 10 C 14.8 mm 6 mm 4 C 3 7.64 10 N 7.6 kn 1.01 10 6 mm
LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 56 150 mm Eample 3 (cont d) 6 6 ( 5.3 10 ) 50.6 10 mm r C 6 ( + Ad ) 1.01 [ 10 + 3795( 75 + 14.8) ] C Lacing bars r A 63.3 10 ( 3795) 6 91.3 mm A 50.6 10 63.3 10 ( 3795) 6 6 mm 81.7 r r P cr min C 14.8 mm 81.7, therefore, π EA ( L / r ) min π L r min 1 10 81.7 9 6 ( 00 10 )[ ( 3795 10 )] ( 146.9) 3 146.9 694.3 kn LECTURE 6. : (pinned ends) (10.1 10.3) Slide No. 57 Eample 3 (cont d) An alternate solution for finding r and r : Using Eqs. 13 and 15, r r r C r 81.8 mm C + 91.3 mm ( a + ) ( 16.3) + ( 75 + 14.8) C Therefore, rmin r 81.8 mm The slight difference in the result is due to round-off errors.