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Stability of columns. - PowerPoint PPT Presentation

Stability of columns

Columns and sturts: Structural members subjected to compression and which are relatively long compared to their lateral dimensions are called columns or Struts. Generally, the term column is used to denote vertical members and the term strut denotes inclined members Examples: strut in a truss, Piston rods, side links in forging machines, connecting rods etc.

Stable, Neutral and unstable EquilibriumStable equilibrium: A stable equilibrium is one in which a body in static equilibrium on being displaced slightly, returns to its original position and continues to remain in equilibrium.

Neutral equilibrium: A neutral equilibrium is one in which a body in equilibrium, on being displaced does not returns to its original position, but its motion stops and resumes its equilibrium state in its new position.

Unstable equilibrium: An unstable equilibrium is one in which a body in equilibrium on being slightly disturbed, moves away from its equilibrium position and loses its state of equilibrium.

Buckling Load: The maximum load which a column can support before becoming unstable is known as buckling load or crippling load or critical load The buckling takes place about the axis having minimum radius of gyration or least moment of inertia. At this stage, the maximum stress in the column will be less than the yield stress (crushing stress) of the material.

Safe load: It is the load to which a column is subjected to and is well below the buckling load. It is obtained by dividing the buckling load by a suitable factor of safety.

safe load = buckling load / factor of safety Stability factor: The ratio of critical load to the allowable load on a column is called stability Factor

MODES OF FAILURE OF THE COLUMNS The load carrying capacity of a short column depends only on its cross sectional area(A) and the crushing stress of the material(cu). The crushing load Pu for axially loaded short column is given by Pcu =cu A . The safe load on the column is obtained by dividing the crushing load by suitable factor of safety. i.e., Psafe =Pcu/ FS The mode of failure of columns depends upon their lengths and depending on the mode of failure columns are classified asa. Short columns b. Long columns Short Columns: A short column buckles under compression as shown in figure and fails by crushing. The load causing failure is called crushing load.

Long columns: Long columns, which are also called slender columns, when subjected to compression, deflects or bends in a lateral direction as shown in the figure. The lateral deflection of the long column is called buckling. The load carrying capacity of long column depends upon several factors like the length of the column, M.I of its crosssection, Modulus of elasticity of the material, nature of its support, in addition to area of cross section and the crushing strength of the material. Critical load denotes the maximum load carrying capacity of the long column. The long column fails when there is excessive buckling .ie when the load on the column exceeds critical load.

SC 10Short columns fails by crushing or yielding of the material under the load P1Long column fails by buckling at a substantially smaller load P2 (less than P1). The buckling load is less than the crushing load for a long columnThe value of buckling load for long column is low whereas for short column the value of buckling load is relatively high.

sc-11Failure of long columns(contd)Stress due to buckling b = ( M.ymax)/ I = {(P.e). ymax}/ I = ( P.e) / ZWhere e = maximum bending of the column at the centreConsider a long column of uniform cross sectional area A throughout its length L subjected to an axial compressive load P. The load at which the column just buckles is known as buckling load or crippling load. Stress due to axial load c = P/AP

Le

Sc-12Failure of long columns(contd)max = c + b Extreme stress at centre of column will be the sum of direct compressive stress and buckling stressIn case of long columns, the direct compressive stresses are negligible when compared to buckling stress. So always long columns fail due to buckling.

Modes of failures (contd.) Intermediate Columns: These are columns which have moderate length, length lesser than that of long columns and greater than that of short columns. sc-13In these columns both bulging and buckling effects are predominant. They show the behavior of both long columns and short columns when loaded.

Eulers Theory (For long columns)Assumptions:The column is initially straight and of uniform lateral dimensionThe material of the column is homogeneous, isotropic, obeys Hookes lawThe stresses are within elastic limitThe compressive load is axial and passes through the centroid of the sectionThe self weight of the column itself is neglected. The column fails by buckling alone

Eulers Theory (For long columns)A Bending moment which bends the column as to present convexity towards the initial centre line of the member will be regarded as positiveBending moment which bends the column as to present concavity towards the initial centre line of the member will be regarded as negativeSign convention for Bending Moments

Eulers Formula for Pin-Ended Beams (both ends hinged) Consider an axially loaded long column AB of length L. Its both ends A and B are hinged. Due to axial compressive load P, let the deflection at distance x from A be y.d2ydx2 L y x P P A B The bending moment at the section is given by EI= - P y -ve sign on right hand side, since as x increases curvature decreases

At x =0, y =0,we get c1=0 (from eq.1)Also at x=L, y =0 we get c2 .sin [LP/(EI)] =0If c2 = 0, then y at any section is zero, which means there is no lateral deflection which is not trueTherefore sin [LP/(EI)] =0 This is the linear differential equation, whose solution is Y = c1.cos [xP/(EI)] + c2.sin[x P/(EI)] (1) Where c1 and c2 are the constants of integration. They can be found using the boundary conditions.

sin [LP/(EI)] =0=> [LP/(EI)] = 0, , 2 ,n Taking least non zero value we get [LP/(EI)] = Squaring both sides and simplifying This load is called critical or buckling load or crippling load

Note: L is the actual length of respective column and Le is to be considered in calculating Euler's buckling load

caseEnd conditionEquivalent length(Le) Eulers Buckling load1Both ends hingedLe=L PE= ( 2E I) / Le2

2One end fixed, other end freeLe=2LPE= ( 2EI) / 4L2

3One end fixed, other end pin jointedLe=L / 2PE= 2( 2EI) / L2

4Both ends fixedLe=L/2PE= 4( 2 EI) / L2

Extension of Eulers formula

Slenderness ratio: It is the Ratio of the effective length of the column to the least radius of gyration of the cross sectional ends of the column. The Effective length: of a column with given end conditions is the length of an equivalent column with both ends hinged, made up of same material having same cross section, subjected to same crippling load (buckling load) as that of given column.

Based on slenderness ratio ,columns are classified as short ,long and intermediate columns. Generally the slenderness ratio of short column is less than 32 ,and that of long column is greater than 120, Intermediate columns have slenderness ratio greater than 32 and less than 120.

Limitation of Euler's theory Pcr= ( 2 EI) / Le 2 But I =Ak2 Pcr/A= 2E/(Le/K)2

cr = 2E/(Le/K)2 Where cr is crippling stress or critical stress or stress at failure The validity of Eulers theory is subjected to condition that failure is due to buckling. The Eulers formula for crippling is The term Le/K is called slenderness ratio. As slenderness ratio increases critical load/stress reduces. The variation of critical stress with respect to slenderness ratio is shown in figure 1. As Le/K approaches to zero the critical stress tends to infinity. But this cannot happen. Before this stage the material will get crushed.

c = 2E/(Le/K)2 Le/K= (2E / c) For steel c = 320N/mm2 and E =2 x 105 N/mm2 Limiting value (Le/K) is given by (Le/K)lim = (2E / c) = 2 2 105/320) = 78.54 Hence, the limiting value of crippling stress is the crushing stress. The corresponding slenderness ratio may be found by the relation cr = cHence if Le /k < (Le /k)lim Euler's formula will not be valid.

Empirical formula or Rankine - Gordon formula PR = crippling load by Rankines formula Pc = crushing load = c .A PE = buckling load= PE= ( 2 EI) / Le 2 We know that, Eulers formula for calculating crippling load is valid only for long columns. But the real problem arises for intermediate columns which fails due to the combination of buckling and direct stress. The Rankine suggested an empirical formula which is valid for all types of columns. The Rankines formula is given by,

For short columns: The effective length will be small and hence the value of PE =( 2 EI) / Le2 will be very large. Hence 1/ PE is very small and can be neglected.therefore 1/ PR= 1/ Pc or PR =PcFor long column: we neglect the effect direct compression or crushing and hence the term 1/ Pc can be neglected.therefore 1/ PR= 1/ PE or PR =PEHence Rankines formula, 1/ PR= 1/ Pc + 1/ PE is satisfactory for all types of columns

(I=AK2)