13. Buckling of Columns - fac.ksu.edu.safac.ksu.edu.sa/sites/default/files/5-chap13_buckling_0.pdf · 13. Buckling of Columns CHAPTER OBJECTIVES • Discuss the behavior of columns.
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13. Buckling of ColumnsCHAPTER OBJECTIVES• Discuss the behavior of
columns.• Discuss the buckling of
columns.Determine the a ial load• Determine the axial loadneeded to buckle an idealcolumn.
• Analyze the buckling withbending of a column.
• Discuss methods used to design concentric andeccentric columns.
• Spring develops restoring force F = k∆, whileapplied load P develops two horizontalcomponents, Px = P tan θ, which tends to push thepin further out of equilibrium.
13. Buckling of Columns13.2 IDEAL COLUMN WITH PIN SUPPORTS
• An ideal column is perfectly straight before loading,made of homogeneous material, and upon whichmade of homogeneous material, and upon whichthe load is applied through the centroid of the x-section.
• We also assume that the material behaves in alinear-elastic manner and the column buckles orbends in a single plane.
13. Buckling of Columns13.2 IDEAL COLUMN WITH PIN SUPPORTS
• In order to determine the critical load and buckledshape of column, we apply Eqn 12-10,
( )1-132
2M
dxdEI =υ
• Recall that this eqn assumethe slope of the elasticcurve is small andcurve is small anddeflections occur only inbending. We assume thatthe material behaves in alinear-elastic manner andthe column buckles or
13. Buckling of Columns13.2 IDEAL COLUMN WITH PIN SUPPORTS
• A column will buckle about the principal axis of thex-section having the least moment of inertia(weakest axis).
• For example, the meter stick shown willb kl b t th i d tbuckle about the a-a axis and notthe b-b axis.Th i l t b d ll t• Thus, circular tubes made excellentcolumns, and square tube or thoseshapes having I ≈ I are selectedshapes having Ix ≈ Iy are selectedfor columns.
13. Buckling of Columns13.2 IDEAL COLUMN WITH PIN SUPPORTS
• Buckling eqn for a pin-supported long slendercolumn, 2column,
( )5-132
2
LEIPcr
π=
Pcr = critical or maximum axial load on column just before it begins to buckle. This load must not causebefore it begins to buckle. This load must not cause the stress in column to exceed proportional limit.
E = modulus of elasticity of materialyI = Least modulus of inertia for column’s x-sectional
13. Buckling of Columns13.2 IDEAL COLUMN WITH PIN SUPPORTS
IMPORTANT• An ideal column is initially perfectly straight, madey p y g
of homogeneous material, and the load is appliedthrough the centroid of the x-section.
• A pin-connected column will buckle about theprincipal axis of the x-section having the least
t f i t timoment of intertia.• The slenderness ratio L/r, where r is the smallest
radius of gyration of x section Buckling will occurradius of gyration of x-section. Buckling will occurabout the axis where this ratio gives the greatestvalue.
13. Buckling of ColumnsEXAMPLE 13.1The A-36 steel W200×46 member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle
13. Buckling of ColumnsEXAMPLE 13.1 (SOLN)From table in Appendix B, column’s x-sectional area and moments of inertia are A = 5890 mm2, Ix = 45.5×106 mm4,and Iy = 15.3×106 mm4.By inspection, buckling will occur about the y-y axis.Applying Eqn 13-5, we have
13. Buckling of ColumnsEXAMPLE 13.2A W150×24 steel column is 8 m long and is fixed at its ends as gshown. Its load-carrying capacity is increased by bracing it about th i i t t th tthe y-y axis using struts that are assumed to be pin-connected to its mid-height Determine theto its mid-height. Determine the load it can support so that the column does not buckle nor material exceed the yield stress. Take Est = 200 GPa and σY = 410 MPa.
13. Buckling of ColumnsEXAMPLE 13.2 (SOLN)Buckling behavior is different about the x and yyaxes due to bracing. Buckled shape for each case is shown. The effective length for buckling about the x-x axis is (KL)x = 0.5(8 m) = 4 m.F b kli b t thFor buckling about the y-yaxis, (KL)y = 0.7(8 m/2) = 2.8 m.W t I 13 4 106 4 d I 1 83 106 4
NOTE: From Eqn 13-11, we see that buckling always occur about the column axis having the largest slenderness ratio. Thus using data for the radius of gyration from table in Appendix B,
( ) 4.60mm2.66mm/m1000m4 ==⎟
⎠⎞
⎜⎝⎛
xrKL
( ) 3.114mm5.24
mm/m1000m8.2==⎟
⎠⎞
⎜⎝⎛
yrKL
Hence, y-y axis buckling will occur, which is the same conclusion reached by comparing Eqns 13-11 for
13. Buckling of Columns*13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING
• To account for behavior of different-length columns,design codes specify several formulae that will bestg yfit the data within the short, intermediate, and longcolumn range.
Steel columns• Structural steel columns are designed on the basis
of formulae proposed by the Structural StabilityResearch Council (SSRC).F t f f t li d t th f l d• Factors of safety are applied to the formulae andadopted as specs for building construction by theAmerican Institute of Steel Construction (AISC)
13. Buckling of Columns*13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING
Procedure for analysisColumn designColumn design• If a formula is used to design a column, or to
determine the column’s x-sectional area for a givengloading and effective length, then a trial-and-checkprocedure generally must be followed if the columnhas a composite shape, such as a wide-flangesection.O i t th l ’ ti l• One way is to assume the column’s x-sectional
area A', and calculate the corresponding stressσ'= P/A'.
13. Buckling of Columns*13.6 DESIGN OF COLUMNS FOR CONCENTRIC LOADING
Procedure for analysisColumn designColumn design• Also, with A' use an appropriate design formula to
determine the allowable stress allow.• From this, calculate the required column area
Areq’d = P/σallow.req d allow
• If A' > Areq’d, the design is safe. When makingcomparison, it is practical to require A' to be close tobut greater than Areq’d, usually within 2-3%. Aredesign is necessary if A' < Areq’d.
13. Buckling of ColumnsEXAMPLE 13.6An A-26 steel W250×149 member is used as a pin-supported column. Using AISC column gdesign formulae, determine the largest load that it can safely support. Est = 200(103) MPa, σY = 250 MPa.
From Appendix B,A = 19000 mm2; r = 117 mm; r = 67 4 mmA = 19000 mm ; rx = 117 mm; ry= 67.4 mm.Since K = 1 for both x and y axes buckling, slenderness ratio is largest if r is used Thusslenderness ratio is largest if ry is used. Thus
13. Buckling of ColumnsEXAMPLE 13.8A 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 f 2014 T6 l i llfrom a 2014-T6 aluminum alloy.Determine the dimensions of its x-sectional area if its width is to bex-sectional area if its width is to be twice its thickness.
Since KL = 750 mm is the same for x-x and y-y axes buckling, largest slenderness ratio is determinedbuckling, largest slenderness ratio is determined using smallest radius of gyration, using Imin = Iy:
( ) 125987501KLKL ( )( ) ( ) ( )[ ]
( )11.25982/212/1
7501/ 3 bbbbbAI
KLrKL
yy===
Since we do not know the slenderness ratio, we apply Eqn 13-24 first,q ,
13. Buckling of ColumnsEXAMPLE 13.9A board having x-sectional dimensions of 150 mm by 40 mm yis used to support an axial load of 20 kN. If the board is assumed to be pin-supported at its top and base, determine its greatest allo abledetermine its greatest allowable length L as specified by the NFPA.
13. Buckling of ColumnsEXAMPLE 13.9 (SOLN)By inspection, board will buckle about the y axis. In the NFPA eqns, d = 40 mm. Assuming that Eqn 13-29 applies, we have
13. Buckling of Columns*13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING
• A column may be required tosupport a load acting at its
d l b k tedge or on an angle bracketattached to its side.
• The bending moment M = Pe• The bending moment M = Pe,caused by eccentric loading,must be accounted for whencolumn is designed.
Use of available column formulaeSt di t ib ti ti ti l f• Stress distribution acting over x-sectional area ofcolumn shown is determined from both axial force Pand bending moment M = Pe.
13. Buckling of Columns*13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING
Use of available column formulae• Maximum compressive stress isMaximum compressive stress is
( )30-13max IMc
AP+=σ
• A typical stress profile is also shown here.• If we assume entire x-section is subjected to uniform• If we assume entire x-section is subjected to uniform
stress σmax, then we can compare it with σallow, whichis determined from formulae given in chapter 13.6.g p
• If σmax ≤ σallow, then column can carry the specifiedload.
13. Buckling of Columns*13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING
Interaction formulaσa = axial stress caused by force P and determinedσa axial stress caused by force P and determined
from σa = P/A, where A is the x-sectional area of the column.
σb = bending stress caused by an eccentric load or applied moment M; σb is found from σb = Mc/I, where I is the moment of inertia of x-sectional area computed about the bending or neutral axis.
13. Buckling of ColumnsEXAMPLE 13.10Column is made of 2014-T6 aluminum alloy and is used to ysupport an eccentric load P. Determine the magnitude of P that
b t d if l i fi dcan be supported if column is fixed at its base and free at its top. Use Eqn 13-30Eqn 13-30.
Actual maximum compressive stress in the column is determined from the combination of axial load anddetermined from the combination of axial load and bending. ( )
IcPe
AP
max +=σ
( )( )( )
( )( )( )PP
mm80mm4012/1mm40mm20
mm80mm40 3+=
Assuming that this stress is uniform over the x-
( ) ( )( )( )P00078125.0=
Assuming that this stress is uniform over the xsection, instead of just at the outer boundary,
13. Buckling of ColumnsEXAMPLE 13.11The A-36 steel W150×30 column is pin-connected at its ends and subjected to eccentric load P. Determine the maximum
ll bl l f P i thallowable value of P using the interaction method if allowable bending stress isbending stress is (σb)allow = 160 MPa, E = 200 GPa, and σY = 250 MPa.Y
K = 1. The geometric properties for the W150×30 are taken from the table in Appendix B.taken from the table in Appendix B.
mm101.17mm3790 462 ×== IA x
We consider r as it lead to largest value of the
mm157mm2.38 == dry
We consider ry as it lead to largest value of the slenderness ratio. Ix is needed since bending occurs about the x axis (c = 157 mm/2 = 78.5 mm). To ( )determine the allowable bending compressive stress, we have ( )( ) 71104mm/m1000m41KL
13. Buckling of ColumnsEXAMPLE 13.12Timber column is made from two boards nailed together so gthe x-section has the dimensions shown. If column is fixed at its base and free at its top, use Eqn 13-30 to determine the eccentric load Pdetermine the eccentric load Pthat can be supported.
K = 2. Here, we calculate KL/d to determine which eqn to use. Since σallow is determined using theeqn to use. Since σallow is determined using the largest slenderness ratio, we choose d = 60 mm.This is done to make the ratio as large as possible, g p ,and thus yield the lowest possible allowable axial stress.This is done even though bending due to P is about the x axis. ( )mm12002KL ( ) 40
• Buckling is the sudden instability that occurs incolumns or members that support an axial load.
• The maximum axial load that a member cansupport just before buckling occurs is called thecritical load Pcr.
• The critical load for an ideal column is determinedfrom the Euler eqn, Pcr = π2EI/(KL)2, whereK = 1 for pin supports, K = 0.5 for fixed supports,K = 0 7 for a pin and a fixed support and K = 2 forK = 0.7 for a pin and a fixed support, and K = 2 fora fixed support and a free end.