58 THEORY OF ELASTIC BTABILITY ELASTIC BUCKLING OF BARS AND FRAMES 59 Sinee the angle of rotation of the end of the bar is sml,the diaplneement (se Fig. 2-11b)will be TABLE 2-2.CRITICAL LOADS TOR COLUMN WITH ROUNDED ENDS b▣Ra D Fro▣Fq.2-17)1 where R is the radius of the hemispherieal end of the bar. 1 Assuming a symmetrical shape of buckling (Fig.2-11c)and taking the origin of 2R 0 10 15 20 30 50 coordinates at the center of the bar,we conclude at once that the constanta A and C in the general solution (2-10)must be zero.This can be sen from the condition of symmetry which requires that the terms in Eq.(2-10)give a defeetion curve which is 制 2.7082.4592.0431.8741.7911.7431.8821.6531.6251.6111,603 symmetrical about the center of the bar.From the condition that y =0 when Pa EI/ 3.17 2.45 1.69 1,42 1.301.231.151.111.071.051.041 Equntion (2-17)can be ud to determine the eriticnl load for symmetrical buckling. If R is equnl to sero,we obtain 号-r-删 0 (218) which agrees with the usual result for a bar with pin enda.As Rinerenaes indefinitely, we appronch the condition of a bar with flat enda and Eq.(2-17)gives 艺=rP-8 which is the eritienl load for a fixed-end column.Table 2-2 givee values of H/2 and (a (e) Par from Eq.(2-17)for various values!of the ratio 1/2R. F10.2-11 2.3.The Use of Beam-column Theory in Calculating Critical Loads. -0,we get B--D,so that the equation of the deflection curve becomes Instead of the differential equation of the deflection curve being used for calculating critical loads,as was done in the two preceding articles,the y-D1-cekr〉 o problem can be solved in many cases by using resulta already obtained The bending moment at any section of the column is equal to for beam-columns.It was shown in Chap.I that at certain values of the M■-P(a-b-》 ) compressive force P the deflections and bending moments in a beam- column tend to increase indefinitely.Those values of the compressive and the bending moment at the end B is force are evidently critical values. (M).-n-P6-PRPR As an example,let us consider the case of a beam-column AB pinned at one end and fixed at the other,as shown in Fig.1-10.If the beam is Thus we have the following condition at the upper end of the bar: subjected to a uniform lateral load q,the bending moment at the built-in end [see Eq.(1-37)]is 兽--Pr是g- M=一警碧=一警 Uaing this condition with Eq.(g)givea 8 u(tan 2u-2u) (a) 1+k极an2-0 This moment increases indefinitely when the denominator of the expres- sion approaches zero,provided the numerator does not also approuch zero. 号号-一京 2-17) This condition gives tan 2u 2u 1 The following end conditions ean be uned to establish the same result:dy/dz =0 st s =0;y -8 at z -1/2 and z--1/2. 1Equation(2-17)is solved readily using tables of the function ztan;eid p.32 of Addends
60 THEORY OP ELASTIC STABILITY ELABTIC BUCKLING OF BARS AND FRAMES 61 or,substituting k=2u [see Eq.(1-13)], and Egs.(2-19)and (2-20)are replaced by the single equation tan kl kl 警-c+盼+ (2-22) This result is the same as obtained previously by integration of the differ- ential equation [see Eq.(2-11)].Thus the critical value of the compres- Solving this equation for M.and setting the denominator of the resulting sive foree is that value for which the bending moment at the built-in expression equal to zero,we obtain the equation for the critical load: end becomes infinitely large regardless of the magnitude of the lateral 1 load. +7侧+品7创=0 The same procedure can be used to determine critical loads for a bar with elastically restrained ends (Fig.1-13).When the bar is subjected Now substituting the expressions for (u)and (u)from Eqs.(1-27)and to a lateral load,the moments acting at the ends are obtained from Eqs. (1-28)and also noting that tan u=(1-cos 2u)/sin 2u,we write this equation in the form (1-43): 警-u+器+品 M (219) an“=-2EL al (2-23) 冬-a+给四+合脸7四 Msl M (2-20) Values of u found from this equation lie between the limita /2 and * The value */2 corresponds to a=0,which means that the ends of the bar are free to rotate,and the critical load is given by Eq.(2-5)for the In these equations o and 8 are coefficients of end restraint [see Eqs.(1-41)], 6 and represent the angles of rotation at the ends due to lateral load only,and the functions (and(u)are given by Egs.(1-27)and (1-28). The moments M.and M,are end moments acting on the member AB (Fig.1-13)and are positive in the directions shown.Solving Eqs.(2-19) and (2-20)for the moment Ma gives F1G.212 10.213 -[++[ 1 fundamental case.When the ends of the bar are rigidly built in,the M.= (6) coefficient a becomes infinite,the value of u is x,and the critical load is 日+]+7-6品7】 P-4E//.For intermediate values of a Eg.(2-23)can be solved readily using tables'of the function (tan z)/r. The solution for M.is obtained similarly and has the same expression in If the loading on the symmetrically supported bar is antisymmetrical the denominator. Therefore the moments M.and M become infinitely (Fig.2-13),we have large when the denominator of Eq.(b)becomes zero. Thus the equation aB bas =-6u M。=一M (④ for determining the critical condition is and the critical load is determined by the equation [日+7][a+am]-[67-o (2-21) 1 a+3远7（四-67（倒-0 For particular values of a and B,this equation can be solved for u and the or (2-24) critical load determined.This method of calculation is useful in analyz- al ing rigid frames and continuous beams,as discussed in the following Values of the critical load from this equation correspond to antisymmetric articles. buckling patterns.The equation is solved easily for any value of a by In the particular case of symmetry (Fig.2-12),we have using Table A-1,Appendix,since the expression on the left-hand side is a=B Goo=0oo Ma=Mo (d) Ibid
62 THEORY OF ELASTIC BTABILITY ELASTIC BUCKLING OF BARS AND FRAME8 63 the function but with w in place of 2u.As one of the limiting cases we have a=0 for pin ends;hence u=x and P.-4EI/7,which cor- If the frame in Fig.2-14 consists of four identical bars,Eq.(b)becomes responds to the antisymmetric buckling shape shown in Fig.2-60.For an4=-1 fixed ends,a becomes infinite,u4.493,and the critical load is given byEq.(2-15). and the lowest root of this equation is 2.4.Buckling of Frames.Since each member of a framework with rigid joints is in the condition of a bar with elastically restrained ends, the method deseribed in the preceding article will be used for considering u-号-2020 the buckling of frames.As a simple example,let us consider a frame and hence ABCD which is symmetrical with respect to horizontal and vertical axes P.=l6.47EI (Fig.2-14).The vertical members of the frame are compressed by axial forces P,and it is assumed that lateral movement of the joints is pre- If the horizontal bars are absolutely rigid,the right-hand side of Eq.( vented by external constraints.When the load becomes zero and therefore tan u=0,u=r,and P reaches its critical value,the vertical bars be- gin to buckle as indicated by the dotted lines. P. This buckling is accompanied by bending of the two horizontal bars AB and CD.These bars This is the case of a bar with fixed ends.Finally,if /=0 we obtain exert reactive moments at the ends of the ver- 4=第/2and tical bars and tend to resist buckling.The mo- P.=E1 E ments at the ends are proportional to the angles of rotation of the joints,and hence the vertical as for a bar with pinned ends. members are examples of bars with elastically If the horizontal members of the frame are p EIt built-in ends subjected to the action of compressive forces, FG.214 The coefficient of restraint at the ends of the as shown in Fig.2-15,the coefficients of end vertical bars is found from a consideration of the restraint a will be diminished.Instead of Eq. bending of the horizontal bars by couplea at the ends.Denoting by E/ (a)we must use the expression the flexural rigidity of the horizontal bars,the expression for the coeffi- cient a is 2EI11 a-2B1 a”btan1 (c) FG.2-15 (a) which is obtained from Eq.(1-34).The quantity ui for the horizontal Since the vertical members buckle in a symmetrical shape,the critical member is load can be obtained from Eq.(2-23)of the preceding artiele.If the bQ flexural rigidity of the vertical bars is denoted by EI,this equation =豆√Ei becomes 0”--0 In calculating the critical value of the compressive foree P,we again use (6) Eq.(2-23)of the preceding article and obtain and the critical load can be found from this equation in each particular c89e. an一 Ib tan u (2-25) 1The first discusion of problems of the atability of members ofa rectangular frame With the quantities on the right-hand side of this equation known,the was given by F.Engeaner,"Die Zusatskrafte und Nebenspannungen ciserner Fach- critical force P can be found assuming that the horisontal bars do not werkbrucken,"Berlin,1893.See also H.Zimmermann,"Knickfestigkeit der Stab- buckle first. verbindungen,"Berlin,1925. If it is desired to determine critical values of the force Q(Fig.2-15)
64 THEORY OF ELABTIC BTABILITY ELASTIC BUCKLING OF BAR8 AND FRAME8 65 the above procedure will give the equation If the flexural rigidities of the two horizontal members of the frame in Fig.2-14 are different,the end conditions for the compressed vertical 越n妙=一6越“ members are no longer the same and the critical load will be obtained 的 u from Eq.(2-21)of the preceding article. which is the same as Eq.(2-25).Thus Eq.(2-25)defines limiting values In the case shown in Fig.2-18a the vertical bar is rigidly built in at the of the two axial forces P and Q.For example,if the frame is square with base and elastically supported at the top.Then 8 is infinite,a is finite, all members having the same flexural rigidity,Eq.(2-25)becomes and Eq.(2-21)becomes tan=一an的 (2-26) 4) [+]t (2-27) This equation is plotted graphically in Fig.2-16.If the values of P and This equation can be solved for each particular case by trial and error, Q locate a point on the curved line in the figure,buckling will occur.It is using Table A-1,Appendix,for values of the functions(u)and (u). For the case shown in Fig.2-18b the coefficient 8 is zero and a is finite. When 8 approaches zero in Eq.(2-21), 16.47 the factor +应w (e) Critical must approach zero also and the equa- load curve tion for calculating the critical load is obtained by equating expression(e)to (Stabie egion zero.This gives the relation 16,47 (纠--3E1 (a) 6) 15 al (2-28) F1a.2-18 ElIE When the horizontal member is considered as a beam with one end built in and the other hinged and the flexural rigidity of this member is denoted 10.2-16 F1g.2-17 by Eli,the magnitude of a in Eq.(2-28)is 4EI/6 and the equation for seen that the critical value of P decreases as Q increases,and vice versa, the eritical load becomes as would be expected.If the values of P and Q locate a point below the curved line,no buckling will occur;hence the portion of the graph below 烟-皱 (2-29) the curve represents a stable region. Other forms of buckling for the frame in Fig.2-14 are possible.For Assuming,for instance,that b =I and I=I1,we find from Eq.(2-29) that example,in Fig.2-17 is shown a buckled configuration in which the mem- bers have inflection points at their mid-points.This case corresponds 0=-是 2u-h-3.83P.-47B☑ to the antisymmetric buckling case of the preceding article,and the criti- cal load is found from Eq.(2-24)by substituting In the limiting case where a becomes infinite,Eq.(2-28)coincides with a=6EI Eq.(2-11)and we have the solution for a column fixed at one end and b (d④ pinned at the other (Fig.2-7). which represents the coefficient of end restraint.Critical loads for the In order for this discussion to be exst,it must be asaumed that the compressive foree P is applied to the vertical bar before the rigid conpection is made.Then there buckling mode of Fig.2-17 are larger than those for the symmetrical will be no bending in the horisontal bar before buckling occurs.The amall changes in mode of Fig.2-14 and hence are not usually so important. the lengths of the bars at buckling are neglected alao
66 THEORY OF ELASTIC STABILITY ELASTIC BUCKLING OF BARS AND FRAMES 67 In the previous discussion it was assumed that the ends of the com- column formulas.Previously,formulas were derived for the bending of pressed members do not displace laterally.Let us consider now the case continuous beams subjected to longitudinal compression in addition to shown in Fig.2-19 in which a frame with compressed vertical members is transverse bending (see Art.1.10).The critical values of the compressive free to move laterally at the top.If the frame has a vertical axis of forces will now be obtained as the values at which even a slight lateral symmetry,each vertical member can be considered separately as a com- load will produce infinite defection.As in Art.1.10,the compressive pressed bar free at the lower end and elastically built in at the upper end. forces in the continuous beam are assumed to be constant within each Taking the coordinate axes as shown in the figure,the differential equa- span but may vary from one span to the next. tion of the deflection curve of the bar AB is Considering two consecutive spans of a bar on several supports (Fig. 1-14)the relation between the three consecutive bending moments at the 骆=-P supports is given by Eq.(1-44): The solution of this equation,satisfying condi- Mad+2.[e0+2监ze tions at the lower end,is +n2芸a)=-62a+回 a-1 y A sin k () There will be as many equations of this type as there are statically indeter- At the upper end the angles and 0 must be minate moments,provided the ends of the continuous beam are simply FG.219 equal and since the horizontal bar BC is bent by supported.If the ends are fixed,then two additional equations,expressing the P I 2 two couples,each equal to P(y),the condition at the upper end is 3P condition of fixity,must be used in addi- (儡-Po -12 tion to Egs.(a).The coefficients in these equations contain the funetions Fra.2-20 or,by using expression (f), ()and (u)and depend on the magnitudes of the compressive force P. 无co8L- Pb sin kl The critical values of these forces are those values for which the bending 6EI1 (g) moments,as solved from Eq.(a),become infinitely large.This requires that the determinant of the left-hand side of Eqs.(a)be made equal to If the horizontal bar is absolutely rigid,EI1=,and we obtain zero.In this way an equation for calculating the critical values of the compressive forces is obtained.1 c0g4=0 从-五 R.- Let us consider a bar on three supports,with hinged ends,and com- pressed by forces P applied at the ends(Fig.2-20).In this case there is In the general case,Eq.(g)can be represented in the following form: only one unknown moment M:and Eq.(a)becomes hana-职 () 2w[+2]=-oa+ and the critical value of the load P can be found for any numerical value The critical value of the compressive force is now obtained from the of the ratio Il/Tb.Assuming that all three bars of the frame are identi- condition that Mf:becomes infinite,which means that cal,we obtain kl tan kl=6 from which w+0经ew-0 (6) 4=1,35and R.=182E7 There is an exeeptional case in which buckling may oceur with all bending moments at the supporta equal to zero.This takes place when all spans of the bar have such proportion8and8 ach compre8 aive forees that41■w,-···.In this case 2.5.Buckling of Continuous Beams.In calculating critical com- buekling of each span is not influenced by the adjaeent spana and the critical values of pressive forces for continuous beams,we can again make use of beam- the forees are caleulated for each apan as for s bar with hinged enda