8 ELASTIC BUCKLING OF BARB AND YRAMES 49 THEORY OF ELASTIC BTABILITY The condition at the upper end of the bar requires that The critical loads for columns with some other end conditions can be obtained from the solution of the preceding case.For example,in the y-8 atz-1 case of a bar with hinged ends (Fig.2-2)it is evident from symmetry that which is satisfied if each half of the bar is in the same condition as the entire bar of Fig.2-1. 8 cos kl =0 (c) Hence the critical load for this case is obtained by substituting //2 for I Equation (c)requires that either 3=0 or coskl=0.If 8=0,there is in Eq.(2-4),which gives no deflection of the bar and hence no buckling (Fig.2-la).If cos kl =0, P.=YEI (2-5) we must have the relation =(2n1) (2-3) The case of a bar with hinged ends is probably assumed in practice more frequently than any other;it is called the fundamental case of buckling of a where n-1,2,3,....This equation determines values of k at prismatie bar. which a buckled shape can exist.The deflection 6 remains indeterminate If the bar has both ends built in (Fig.2-3),there are reactive moments and,for this ideal case,can have any value within the scope of small- that prevent the ends of the column from rotating during buckling. deflection theory. The smallest value of kl which satisfies Eq.(2-3)is obtained by taking n 1.The corresponding value of P will be the smallest critical load. and we have from which P.- 4 (2-4) This is the smallest critical load for the bar in Fig.2-1a,that is,it is the smallest axial force which can maintain the bar in a slightly bent shape. The quantity kx in Eq.(2-2)varies in this case from 0 to x/2,and the shape of the deflection curve is therefore as shown in Fig.2-16. (a) (8) Substituting n=2,3,...into Eq.(2-3),we obtain for the cor- P10.2-2 F0.23 Fra.2-4 responding values of the compressive force P。-E These end moments and the axial compressive forces are equivalent to 4 P。-25mB1. forces P applied eccentrically as shown in the figure.Inflection points are located where the line of action of P intersects the deflection curve, The quantity kx in Eq.(2-2)varies in these cases from 0 to 3x/2,from0to because at these points the bending moments are zero.The inflection 5x/2,...,and the corresponding deflection curves are shown in Fig. points and the mid-point of the span divide the bar into four equal regions, 2-1e and d.For the shape shown in Fig.2-le a force nine times larger each of which is in the same condition as the bar in Fig.2-16.Hence than the smallest critical load is necessary,and for the shape in Fig.2-1d the critical load for a column with built-in ends is found by substituting a force twenty-five times larger is required.Such forms of buckling can 1/4 for l in Eq.(2-4),which gives be produced by using a very slender bar and by applying external con- straints at the inflection points to prevent lateral deflection.Otherwise R-“ (26) these forms of buckling are unstable and have little practical mean- ing because the structure develops large deflections when the load reaches As a final example,consider the column shown in Fig.2-4a.This bar the value given by Eq.(2-4). is free to displace laterally at the upper end but is guided in such a manner 1 Note that the differential equstion (2-1)is bnsed upon an spproximate expression that the tangent to the elastic curve remains vertical.At the lower end for curvature and is valid only for amall defleetions. the column is built in.Since there is a point of inflection at the center of
50 THEORY OF ELASTIC STABILITY ELASTIC BUCKLING OF BARS AND FRAMES 51 the bar (Fig.2-40),the critical load is found by substituting 1/2 for I in Referring now to the cases represented in Figs.2-1a and 2-3 and pro- Eq.(2-4),and thus it is seen that Eq.(2-5)holds for this case also. ceeding as for a bar with hinged ends,we find the following expressions In each of the preceding cases it was assumed that the column was free for the critical stresses: to buckle in any direction,and hence EI representa the smallest flexural E 常E rigidity.If a column is constrained in such a manner that buckling is (2/r) possible in one principal plane only,then EI will represent the flexural a"/2 rigidity in that plane. It is seen that in these two cases equations analogous to Eq.(2-7)for the It was assumed in the previous discussion that the bar was very slender, fundamental case can be used in caloulating the critical stress.These so that the maximum compressive stresses which occurred during buckling equations are obtained from Eq.(2-7)by substituting in place of the remained within the proportional limit of the material.( Only under actual length I of the bar a reduced length L.Thus we can write in these conditions will the preceding equations for the critical loads be general valid.To establish the limit of applicability of these formulas,let us TE consider the fundamental ease (Fig.2-2).Dividing the critical load .-L7ry (2-8) from Eq.(2-5)by the cross-sectional area A of the bar and letting In the case of a prismatic bar with one end built in and the other end free,the reduced length is twice the actual length (L=20).In the case T=4 where r represents the radius of gyration,the critical value of the com- 4-10 pressive stress is E 310 0a= (2-7) 210 This streas depends only on the modulus of elasticity E of the material 10 and on the slenderness ratio //r.The expression is valid as long as the stress remains within the proportional limit.When the proportional 100 r limit and the modulus E are known for a particular material,the limiting a.2-5 value of the slenderness ratio l/r can be found readily from Eq.(2-7). For example,for a structural steel with a proportional limit of 30,000 psi of a bar with both ends built in,the reduced length is half the actual and 30,000,000 psi,we find the minimum //r from Eq.(2-7)to be length (L=1/2).Thus the results obtained for the fundamental case about 100.Consequently,the eritical load for a bar of this material, can be used for other eases of buekling of bars by using the reduced length having hinged ends,can be calculated from Eg.(2-5)if l/r is greater than instead of the actual length of the bar. 100.If l/r is leas than 100,the compressive stress reaches the propor- 8.9.Alternate Form of the Differential Equation for Determining Critical Loads. tional limit before buckling can occur and Eq.(2-5)cannot be used.The In the preceding article it was shown that the eritical load for an ideal oolumn eould question of the buckling of bars compressed beyond the proportional be found by beginning with the diferential equstion(1-3),which expresses the curva- limit is discussed in the next chapter. ture of the bar in terms of the bending moment.An alternate method is to begin with Equation (2-7)can be represented graphically by the curve ACB in Eq.(1-5).Sinoe in determining critical loads of buckled bars the lateral load vanisbee, Fig.2-5,where the critical stress is plotted as a function of l/r.The curve the differential equation for the column is approaches the horisontal axis asymptotically,and the critical stress approaches zero as the slenderness ratio increases.The curve is also 碧+P器-0 asymptotic to the vertical axis but is applieable in this region only as long or,substituting k-P/BI, as the stress o remains below the proportional limit of the material. 兽+器-0 2-9) The curve in Fig.2-5 is plotted for the structural steel mentioned above, and point C corresponds to a proportional limit of 30,000 psi.Thus The general solution of this equntion is only the portion BC of the curve can be used. 扩=Aikz+Bcos+Cx+D (2-10)
52 THEORY OF ELASTIC STABILITY ELASTIC BUCKLING OF BARB AND FRAMES 53 The constants in this equation and the value of the eritieal lond are found from the end and the Inat two conditions give conditions of the bar.A number of particular cases will now be considered. Column with Hinged Ends.In the case of a bar with hinged enda (Fig.2-6a)the d 8in+B cos0 C =0 deflection and the bending moment are zero at the ends.and henre we have the Therefore we finally conclude that C A=0 and conditiona 0 atz -0andz-1 c0s kl =0 (2n-1) 2 Applying these conditions to the general solution [Eq.(-10)]gives which ngrees with Eq.(2-3)of the preceding article. B=C-D-0 8in kl-0 Column with One End Fired and the Other Pinned.This ene is illustrated in Fig. 2-7,where the lower end of the bar is built in and the upper end is hinged.When and therefore 从■ (a) lateral buckling oceurs,s resctive force R is developed at the pinned end.The This equation determines the values of the critical load and for n =1 gives the previ- direction of this reaction is determined by noting that it must op- ous result,Eq.(2-5).The shape of the deflection curve is given by the equation pose the renetive moment at the built-in end.The end conditions for this column are .-A血红=Aim平 ) y--0tx-0 where the constant A represents the undetermined amplitude of the deflection.For the lowest critical load (n -1)the buckled shape is shown in Fig.2-6a.Forn2, =0tx=1 Using these conditions with the general aolution (2-10)gives the following equations for the constants: B+D=0 Ak+C■0 C+D■0 A sin B cos0 F1G.2-7 All four of these equations will be aatisfied by taking AB-C-D -0,in which ense the defection [ooe Eq.(2-10)]vanishes and we have the strnight form of equi- librium.In order to have the posaibility of a buckled shape of equilibrium,we needa solution of the equations other than the trivial one.Solving for A in terms of B from the firat three equation and substituting into the laat equstion gives 一B的H +B coe k0 (a) (61 (c) F0.26 and hence, tnnk缸= (2-110 3,...higher values of the critical load are obtained from Eq.(a)and the correspond- ing buckled shapes are shown in Figs.2-66 and c. Thus,in order to geta curved sbape of equilibrium aatisfying the end conditions of Columin with One End Fired and the Other Free.For the bar shown in Fig.2-1a. the bar,the transeendentnl equation (2-11)must be sstisfied. fixed at the baae and free at the upper end,the conditions at the lower end are To solve Eq.(2-11)a graphieal mothod is useful.The curves in Fig.2-8 represent tan bl as a function of H.These eurves are asymptotic to the vertical lines ki=*/2, y-是-0就z-0 3/2,...since for these values of tan becomes infinite.The roots of Eq. (2-11)are represented by the interseetion points of the above curves with the straight At the free end the bending moment and shearing force must be sero.Refer- line y=kl.The smallest root,corresponding to point A,is ring to Egs.(1-3)and (1),Art.1.2,we see that theae conditions mean that H▣4493 器0 atx▣1 and the corresponding critieal load is 碧+密-0助-1 P-20.10Br BI 0.6090于 (2-120 From the conditions at the lower end of the bar we obtain 1Also,wolutions of Eq.(2-11)are tabulated in Jahnke and Emde,"Tables of Fune- B=-DC▣-Ak tions,"4th ed.,p.30 of Addenda,Dover Publications,New York,1945
54 THEORY OF ELASTIC STABILITY ELASTIC BUCKLING OF BARS AND FRAMES 55 Thus the critical load is the same as for a bar with hinged ends having a reduced length One solution of this equntion is equal to0.691[se0E0.2-8)1. Fired-end Column.If both ends of the bar are fixed (Fig.2-9a),the end conditions 血号-0 are and therefore hl=2nx and --0 at z -0 and P.-物E7 2-13) These conditions give the following equations for determining the constanta in Eq. Noting that sin ki -0 and cos H-1 whenever sin /2=0,we find from Eqs.(e) 2-10): the following values of the constants: B+D=0 A■C■0 B■-D Ak+C■0 Ain+Bcos+Cl十D=0 (c) and the equntion for the deflection curve is Ako08从=Bki血kl+C▣0 y-B(o2平-） (2-14) Investigating the possibility of curved forms of equilibrium,we obeerve that the only If1,we obtain the lowest critical load fsee Eq.(2-6)]and the couumes the symmetrical buckled shape shown in Fig.2-96. 45 -4.493 5 (e) (b) (e) (a) (b) Fra.2-8 F1o.20 PrG.2-10 way to have a nontrivial solution of these four cquations is to have the determinant of A second aolution of Eq.(d)is obtained by setting the term in parentheses equal to the coefficients equal to zero.This determinant is zero,giving the equstion 0 0 m号-号 0 0 The lowest root of this equation is Hl/24.493,and therefore kco8-hinH1 Pr-&18251 (2-15) and setting it equal to sero gives the equstion 2(oosH-1)+ki血=0 which oorreaponds to the antisymmetrie buckling pattern shown in Fig.2-9c..How. ever,aine this eritical value ia larger than the previous value for ymmetricnl buckling, Noting that gin ki -2 sin (Hl/2)cos (kl/2)and cos h=1-2 sin(kl/2),we can it is of practical interest only in the case of a column with lateral support at midheight. write this equation in the form Coumn it Lod through a Fized Point.In the preceding examplee the direetion 血（-血）-0 of the compreasive force P waa ssumed to remain conatant during buckling of the bar. ④ Let us consider now a cnse in which a change in direction of the force P
56 THEORY OF ELASTIC STABILITY occurs.Assume,for example,that the force P is produced by the tension of a eable ELASTIC BUCKLING OF BARS AND FRAMES 57 which always paeses through the fixed point C on the x axis,as ghown in Fig.2-10. The lower end of the column is built in,and the upper end is free to move laterally. This problem differs from the usunl Euler ease (see Fig.2-1)because during buck- TABLX 2-1.CRITICAL LOADB FOR COLUMN WITH LOAD THROUOH A FIXED POIMT [From Eq.(2-16)] ling there is a shearing foree at the upper end of the bar.This foree is equal to the horizontal component of the tensile foree P in the cable (Fig.2-106),and since for small deflections the vertical component of the force can be taken equal to P,we 0 0.2 0.4 0.6 0.8 1.0 1.2 1.6 obtain V--P 长 4.493 4.438 4.348 4,173 3.790 2.654 2.289 Pu Substituting this expression for V into the general equation for the sbearing foree EIA 2.05 2.00 1.91 1,76 1.46 0.714 0.531 [Eq.(1-4)]gives the following condition at the upper end of the bar (-) 1 器+P鼎- 2.0 3,0 4.0 5.0 8.0 10 20 or 器+毫-智 的 2.029 1.837 1.758 1.716 1.657 1.638 1,602 A second condition at the upper end of the bar is that the bending moment is zero,or Pu 器-0“红-1 EI/ 0.417 0.342 0.313 0.298 0.278 0.272 0.260 0.256 At the lower end of the bar the conditions are If e is greater than as assumed in Fig.2-106,the right-hand side of Eq.(2-16)is y-皇-02-0 negntive and the smallest value of ki which satisfies the equation'is between /2 and (nee Fig.2-8).This means that the critical load is grenter than g//40,which was found previously for the case shown in Fig.2-1.This can be explained by noting that Again evsluating the constants in the general solution (2-10),we obtnin from the the tranaverse foree P8/c counteracts the tendeney for lateral buckling and henee a conditions at the lower end larger critical load is required.If c increases,the value of ti approaches the value /2,and when c finslly beoomes infinitely large,we have B D =0 Ak +C-0 and from the conditions at the upper end -员P- c-t which is the same result as for the previous csse (Fig.2-1)when the load remains As血kH+B con kt=O always vertical. When e-,the fixed point C coineides with the lower end of the bar (Fig.2-10c), Solving these cqustions for the constanta and then substituting into Eq.(2-10), the right-hand side of Eq.(2-16)vanishes,and we obtain we obtain y-是Ian(eos:-)+:-i恤 网 -r P 挂. As a final condition,the deflection st the upper end of the bar is &and hence,from Eq. which is the same as for the fundnmental case of buckling.This can be explained by (e),we conclude that noting that when the line of action of P passes through the base of the column,the bending moment at that point vanishes and the bar is in the same condition asbar mH-(-) 2-16) with hinged ends. If the distance e is less than l,the right hand aide of Eq.(2-16)is poaitive and the Equstion (2-16)gives the value of the eritieal lond for any particular value of the amallest value of ki which astisfies the equation is somewhere between r and 3/2. ratio c/.The solution of this equation is facilitated by the use of tablest of the The deflection curve then has an inflection point D as shown in Fig.2-10d.Finally, funetion (tan r)/Table 2-1 gives values of kl and P for various values of e/!a when c -0,Eq.(2-16)becomes the same as Eq.(2-11)and we have the case of a bar determined from Eq.(2-16). pinned at the top and fixed at the base (Fig.2-7). Bar with Rounded Enda.When a bar with rounded ends buekles laterally(Fig. See fbid.,p.32 of Addends. 2-11),there will be a displacement b of the line of action of the compressive forees P. The trivial solution,k=0,ia excluded as usual