10 Deflections due to Bending

Transcription

1 1 Deflections due to Bending 1.1 The Moment/Curvture Reltion Just s we took the pure bending construction to be ccurte enough to produce useful estimtes of the norml stress due to bending for lodings tht included sher, so too we will use the sme moment/curvture reltionship to produce differentil eqution for the trnsverse displcement, v(x) of the bem t every point long the neutrl xis when the bending moment vries long the bem. M b EI dφ ds The moment/curvture reltionship itself is this differentil eqution. All we need do is express the curvture of the deformed neutrl xis in terms of the trnsverse displcement. This is stright forwrd ppliction of the clssicl clculus s you hve seen perhps but my lso hve forgotten. Tht s ok. For it indeed cn be shown tht 1 : v(x) ρ v(x) φ ~ dv/ x s dv dφ ds 3 -- dv Note in this exct reltionship, the independent vrible is s, the distnce long the curved, deformed neutrl, x xis.

2 64 Chpter 1 There now we hve it once given the bending moment s function of x ll we need do is solve this non-liner, second order, ordinry differentil eqution for the trnsverse displcement v(x). But hold on. When ws the lst time you solved second order, non liner differentil eqution? eonhrd Euler ttcked nd resolved this one for some quite sophisticted end-loding conditions bck in the eighteenth century but we cn get wy more cheply by mking our usul ssumption of smll displcement nd rottions. Tht is we tke dv < 1 which sys tht the slope of the deflection is smll with respect to 1.. Or equivlently tht the rottion of the cross section s mesured by φ (dv/) is less thn 1., one rdin. In this we note the dimensionless chrcter of the slope. Our moment curvture eqution cn then be written more simply s d dv x M b ( x) EI Exercise 1.1 Show tht, for the end loded bem, of length, simply supported t the left end nd t point /4 out from there, the tip deflection under the lod is given by 3 ( 3 16) EI A B C /4 The first thing we must do is determine the bending moment distribution s function of x. No problem. The system is stticlly determinte. We first determine the rections t A nd B from n isoltion of the whole. We find R A 3, directed down, nd R B 4 directed up.

3 Deflections due to Bending 65 An isoltion of portion to the right of the support t B looks very much like Glileo s cntilever. In this region we find constnt sher force equl in mgnitude to the end lod nd linerly vrying bending moment which, t x/4 is equl to -(3/4). We note tht the sher between x nd x < /4 equls R A, the rection t the left end, nd tht the bending moment must return to zero. The discontinuity in the sher force t B llows the discontinuity in slope of M b t tht point. Our liner, ordinry, second order differentil eqution for the deflection of the neutrl xis becomes, for x < /4 R A ( - )/ B R B / V(x) ( - )/ - M b (x) x x - (-) x > dv M b ( x) 3 x < x M b - (-x) EI EI M b -x (/ -1) Integrting this is stright forwrd. I must be creful, though, not to neglect to introduce constnt of integrtion. I obtin dv ( x ) + C EI 1 This is the slope of the deflected neutrl xis s function of x, t lest within the domin <x</4. Integrting once more produces n expression for the displcement of the neutrl xis nd, gin, constnt of integrtion. 3 vx ( ) ( x 3 6) + C EI 1 x + C Here then is n expression for the deflected shpe of the bem in the domin left of the support t B. But wht re the constnts of integrtion? We determine the constnts of integrtion by evluting our expression for displcement v(x) nd/or our expression for the slope dv/ t points where we re sure of their vlues. One such boundry condition is tht, t x the displcement is zero, i.e., vx ( ) x Another is tht, t the support point B, the displcement must vnish, i.e., vx ( ) x 4

4 66 Chpter 1 These yield C nd C 1 nd we cn write, for x < /4 ( 3EI) vx ( ) ( x 3 x 16) EI So fr so good. We hve pinned down the displcement field for the region left of the support point t B. Now for the domin /4 < x <. The liner, ordinry, second order differentil eqution for the deflection, gin obtined from the moment/curvture reltion for smll deflections nd rottions, becomes dv ( x) EI Integrting this twice we obtin, first n expression for the slope, then nother for the displcement of the neutrl xis. To wit: vx ( ) dv ( x x ) + D EI 1 nd [ x x 3 6] + D EI 1 x + D Now for some boundry conditions: It ppers t first look tht we hve but one condition, nmely, t the support point B, the displcement must vnish. Yet we hve two constnts of integrtion to evlute! The key to resolving our predicment is reveled by the form of the eqution for the slope; we need to fix the slope t some point in order to evlute D 1.Wedo this by insisting tht the slope of the bem is continuous s we pss over the support point B. Tht is, the two slopes, tht of v(x) evluted t the left of B must equl tht of v(x) evluted just to the right of B. Our boundry conditions re then, for x > /4: vx ( ) x 4 nd dv x where the right hnd side of this lst eqution hs been obtined by evluting the slope to the left of B t tht support point. Spring you the detils, which you re encourged to plough through t your leisure, I - nd I hope you - obtin 3 D 1 ( 5 3) nd D EI ( 1 96) EI

5 Deflections due to Bending 67 So, for /4 < x < we cn write: 3 vx ( ) --- [ 1 15( x ) + 48( x ) 16( x ) 3 ] 9 Setting x we obtin for the tip deflection: 3 v ( ) ( 3 16) EI where the negtive sign indictes tht the tip deflects downwrd with the lod directed downwrd s shown. The process nd the results obtined bove prompt the following observtions: The results re dimensionlly correct. The fctor 3 /(EI) hs the dimensions of length, tht is F 3 /[(F/ ) 4 ]. We cn spek of n equivlent stiffness under the lod nd write EI K where K ( 16 3) E.g., n luminum br with circulr cross section of rdius 1.in, nd length 3. ft. would hve, with I π r 4 /4.785 in 4, nd E 1x1 6 psi, n equivlent stiffness of K898 lb/inch. If it were but one foot in length, this vlue would be incresed by fctor of nine. This lst speks to the sensitivity of stiffness to length: We sy the stiffness goes s the inverse of the length cubed. But then, the stiffness is even more sensitive to the rdius of the shft: it goes s the rdius to the fourth power. Finlly note tht chnging mterils from luminum to steel will increse the stiffness by fctor of three - the rtio of the E s. The process ws lengthy. One hs to crefully estblish n pproprite set of boundry conditions nd be meticulous in lgebric mnipultions. It s not the differentil eqution tht mkes finding the displcement function so tedious; it s, s you cn see, the discontinuity in the loding, reflected in the necessity of writing out different expression for the bending moment over different domins, nd the mtching of solutions t the boundries of these regions tht mkes life difficult. Fortuntely, others hve lbored for century or two crnking out solutions to this quite ordinry differentil eqution. There re reference books tht provide full coverge of these nd other useful formule for bem deflections nd mny other things. One of the clssicl works in this regrd is Rork nd Young, FOR- MUAS FOR STRESS AND STRAIN, 5th Edition, McGrw-Hill, We summrize selected results s follow.. It took me three psses through the problem to get it right.

6 68 Chpter 1 End-loded Cntilever v(x) For <x< v(x) [ 3 /()][3(x/) -(x/) 3 ] v(x) v mx 3 /(3EI) t x Couple, End-loded Cntilever v(x) For < x < v(x) [ M /(EI)] (x/) v(x) M b v mx M /(EI) t x Uniformly oded Cntilever v(x) w o (x) For <x < v(x) [ w o 4 /(4EI)] (x/) [(x/) - 4(x/) + 6] v(x) v mx w o 4 /(8EI) t x Uniformly oded Simply-Supported Bem v(x) w o (x) For < x < v(x) [w 4 /(4EI)] (x/) [1 - (x/) + (x/) 3 ] v mx [5w o 4 /(384EI)] t x/

7 Deflections due to Bending 69 Couple, End-loded Simply-Supported Bem v(x) For < x < v(x) - [M /()] (x/) [1 - x / ] (x) M b oint od, Simply-Supported Bem b For <x<(-b) v mx [ M /(9 3 EI)] t x/ 3 v(x) [ 3 /()] (b/) [ -(x/) 3 + (1-b / )(x/) ] v 3 /[9 3 EI)] (b/)[1- b / ] 3/ t x(/ 3) (1-b / ) mx For (-b) <x < v(x) [ 3 /()] (b/) { (/b) [(x/)- (1-b/)] 3 - (x/) 3 +(1-b / )(x/)} With these few reltionships we cn construct the deflected shpes of bems subjected to more complex lodings nd different boundry conditions. We do this by superimposing the solutions to more simple loding cses, s represented, for exmple by the cses cited bove. Exercise 1. Show tht the expression obtined for the tip deflection s function of end lod in the previous exercise cn be obtined by superimposing the displcement fields of two of the cses presented bove. We will consider the bem deflection t the tip to be the sum of two prts: One prt will be the deflection due to the bem cting s if it were cntilevered to wll t the support point B, the middle figure below, nd second prt due to the rottion of the bem t this imgined root of the cntilever t B the figure left below. A B + A B A B rigid We first determine the rottion of the bem t this point, t the support B. To do this we must imgine the effect of the lod pplied t the tip upon the deflected shpe bck within the region < x </4. This effect cn be represented s n equivlent force system t B cting internlly to the bem. Tht is, we cut

8 7 Chpter 1 wy the portion of the bem x>/4 nd show n equivlent verticl force cting downwrd of mgnitude nd clockwise couple of mgnitude (3/4) cting t B. Now the force produces no deflection. The couple M produces rottion which we will find by evluting the slope of the displcement distribution for couple, end-loded, simply supported bem. From bove we hve, letting lower cse l stnd in for the spn from A to B, A l/4 B M (3/4) φ B v(x) -[ Ml /()] (x/l) [1-x /l ] so dv/ - [Ml/()] (1-3x /l ) This yields, for the slope, or rottion t the support point B, dv/ B - Ml/(3EI). The couple is M 3/4 so the rottion t B is φ B The deflection t the tip of the bem where the lod is pplied due to this rottion is, for smll rottions, ssuming this portion rottes s rigid body, by rigid body ( 3 4) φ B EI where the negtive sign indictes tht the tip displcement due to this effect is downwrd. 3 We now superimpose upon this displcement field, the displcement of bem of length 3/4, imgined cntilevered t B, tht is displcement field whose slope is zero t B. We hve, for the end loded cntilever, tht the tip displcement reltive to the root is -l 3 /(3EI) where now the lower cse l stnds in for the length 3/4 nd we hve noted tht the lod cts downwrd. With this, the tip deflection due to this cntilever displcement field is A B cntilever /(64EI) (3/4) So the finl result, the totl deflection t the tip is, s before, rigid body + cntilever /(1). 3. In this problem M is tken positive in opposition to our usul convention for bending moment. I hve left off the subscript b to void confusion.

9 Deflections due to Bending 71 Exercise 1.3 Estimte the mgnitude of the mximum bending moment due to the uni- o, force/length w form loding of the cntilever bem which is lso supported t its end wy from the wll. Dej vu! We posed this chllenge bck in n erlier chpter. There we mde n estimte bsed upon the mximum bending moment within uniformly loded simply supported bem. We took w /8 s our estimte. We cn do better now. We use superpositioning. We will consider the tip deflection of uniformly loded cntilever. We then consider the tip deflection of n end loded cntilever where the end lod is just the rection force (the unknown rection force becuse the problem is stticlly indeterminte) t the end. Finlly, we then sum the two nd figure out wht the unknown rection force must be in order for the sum to be zero. This will be reltively quick nd pinless, to wit: R + R R 3 /(3EI) w o (x) w o -w 4 /(8EI) For the end-loded cntilever we obtin R R 3 /(3EI) where the deflection is positive up. For the uniformly loded cntilever we hve w o -w 4 /(8EI). The two sum to zero if nd only if R ( 3 8) w o We hve resolved stticlly indeterminte problem through the considertion of displcements nd insisting on deformtion pttern comptible with the constrint t the end tht the displcement there be zero. With this, we cn determine the rections t the root nd sketch the sher force nd bending moment distribution. The results re shown below.

10 7 Chpter 1 w o /8 w o 5w o /8 3w o /8 We see tht there re two positions where the bending moment must be inspected to determine whether it ttins mximum. At the root we hve M b (1/8)w o while t x(5/8) its mgnitude is (9/18)w o. The moment thus hs mximum mgnitude t the root. It is there where the stresses due to bending will be mximum, there where filure is most likely to occur 4. V(x) (5/8) +3w o /8 x 5w o /8 M b (x) M b mx (9/18)w o x w o /8 1. Buckling of Bems Buckling of bems is n exmple of filure mode in which reltively lrge deflections occur while no member or prt of the structure my hve experienced frcture or plstic flow. We spek then of elstic buckling. Bems re not the only structurl elements tht my experience elstic buckling. Indeed ll structures in theory might buckle if the loding nd boundry conditions re of the right sort. The tsk in the nlysis of the possibility of elstic instbility is to try to determine wht lod levels will bring on buckling. Not ll elstic instbility qulifies s filure. In some designs we wnt buckling to occur. The Tin Mn s oil-cn in The Wizrd of Oz ws designed, s were ll oil cns of this form, so tht the bottom, when pushed hrd enough, dished in with snp, snpped through, nd displced just the right mount of oil out the long conicl nozzle. Other ltch mechnisms rely upon snp-through to lock fstener closed. But generlly, buckling mens filure nd tht filure is often ctstrophic. To see why, we first consider simple mechnism of little worth in itself s mechnicl device but vluble to us s n id to illustrting the fundmentl phe- 4. If we compre this result with our previous estimte mde bck in exercise 3.9, we find the ltter ws the sme mgnitude! But this is by chnce; In the simply supported bem the mximum moment occurs t midspn. Here it occurs t the left end.

11 Deflections due to Bending 73 nomenon. The link shown below is to be tken s rigid. It is fstened to ground through frictionless pin but liner, torsionl spring of stiffness K T resists rottion, ny devition from the verticl. A weight is suspended from the free end. We seek ll possible sttic equilibrium configurtions of the system. On the left is the one we postulte working from the perspective of smll deflections nd rottions; tht is up until now we hve lwys considered equilibrium with respect to the φ undeformed configurtion. We cn cll this trivil solution; the br does not rotte, the rection moment of the torsionl spring is zero, the rigid link is in compression. K T The one on the right is more interesting. But note; we consider equilibrium with M T - respect to deformed configurtion. Moment equilibrium requires tht M T sinφ or, ssuming liner, torsionl spring ie., M T K T φ: φ sinφ Our chllenge is to find vlues of φ which stisfy this eqution for given lod. We wish to plot how the lod vries with displcement, the ltter mesured by either sinφ or φ itself. Tht is the trditionl problem we hve posed to dte. But note, this eqution does not solve so esily; it is nonliner in the displcement vrible φ. Now when we enter the lnd of nonliner lgebric equtions, we enter world where strnge things cn hppen. We might, for exmple, encounter multi-vlued solutions. Tht is the cse here. One brnch of our solution is the trivil solution φ for ll vlues of. This K T 1 /K T non-trivil y(k T /) φ (KT/)>1 (KT/) (KT/)<1 1 trivil φ / ysinφ is the verticl xis of the plot below. (We plot the nondimensionl lod (/K T ) versus the nondimensionl horizontl displcement ( /). But there exists nother

12 74 Chpter 1 brnch, one reveled by the geometric construction t the right. Here we hve plotted the stright line y (K T /)φ, for vrious vlues of the lod prmeter, nd the sine function y sinφ on the sme grph. This shows tht, for (K T /) lrge, (or smll), there is no nontrivil solution. But when (K T /) gets smll, (or grows lrge), intersections of the stright line with the sine curve exist nd the nonliner equilibrium eqution hs nontrivil solutions for the ngulr rottion. The trnsition vlue from no solutions to some solutions occurs when the slope of the stright line equls the slope of the sine function t φ, tht is when (K T /)1, when K T Here is criticl, very specil, vlue if there ever ws one. If the lod is less thn K T /, less thn the criticl vlue then we sy the system is stble. The link will not deflect. But beyond tht ll bets re off. Wht is the locus of equilibrium sttes beyond the criticl lod? We cn, for this simple problem, construct this brnch redily. We find nontrivil solutions, pirs of (/K T ) nd φ vlues tht stisfy equilibrium most redily by choosing vlues for φ nd using the eqution to compute the required vlue for the lod prmeter. I hve plotted the brnch tht results bove on the left. You cn imgine wht will hppen to our structure s we pproch nd exceed the criticl vlue; up until the criticl lod ny deflections will be insensible, in fct zero. This ssumes the system hs no imperfections sy in the initil lignment of the link reltive to verticl. If the ltter existed, the link would show some smll ngulr rottions even below the criticl lod. Once pst the criticl lod we see very lrge deflections for reltively smll increments in the lod, nd note the rigid br could swing either to the right or to the left. Another wy to visulize the effect of exceeding the criticl lod is to imgine holding the mechnism stright while you tke the lod up, then, once pst the criticl vlue, sy by %, let go... nd stnd bck. The system will jump towrd either equilibrium stte possible t tht lod to the left or to the right 5. The criticl vlue is clled the buckling lod, often the Euler buckling lod. Before turning to the buckling of bems, we mke one finl nd importnt observtion: If in the equilibrium eqution tken with respect to the deformed configurtion, we sy tht φ, the devition from the verticl is smll, we cn pproximte sinφ φnd our equilibrium eqution tkes the liner, homogeneous form φ K T 5. In relity the system would no doubt bounce round good bit before returning to sttic equilibrium.

13 Deflections due to Bending 75 Now how do the solutions of this compre to wht we hve previously obtined? At first glnce, not very well. It ppers tht the only solution is the trivil one, φ. But look! I cn lso clim solution if the brcket out front is zero. This will hppen for specil vlue, n eigenvlue, nmely when /K T 1.. Now tht is significnt result for, even though we cnnot sy very much bout the ngulr displcement, other thn it cn be nything t ll, we hve determined the criticl buckling lod without hving to solve nonliner eqution. We will pply this sme procedure in our nlysis of the buckling of bems. To do so we need to develop n eqution of equilibrium for bem subject to compressive lod tht includes the possibility of smll but finite trnsverse displcements. We do this by considering gin differentil element of bem but now llow it to deform before writing equilibrium. The figure below shows the deformed element cted upon by compressive lod s well s sher force nd bending moment. w o (x) x V+ V V M b + M b w o (x) M b x x Force equilibrium in the horizontl direction is stisfied identiclly if we llow no vrition of the xil lod 6. Force equilibrium of the differentil element in the verticl direction requires V w o x + ( V + V ) while moment equilibrium bout the sttion x yields ( x) M b w o ( V + V ) x + ( M b + M b ) + v 6. The buckling of verticl column under its own distributed weight would men tht would vry long the xis of the bem.

14 76 Chpter 1 The lst term in moment equilibrium is only there becuse we hve tken equilibrium with respect to the slightly deformed configurtion. In the limit s x we obtin the two differentil equtions: dv nd w o ( x) dm b dv + V + It is importnt to distinguish between the lower cse nd upper cse vee; the former is the deflection of the neutrl xis, the ltter the sher force. We obtin single equilibrium eqution in terms of displcement by first differentiting the second eqution with respect to x, then eliminte the term dv/ using the first eqution. I obtin dm b dv + w x o ( x) d We now phrse the bending moment in terms of displcement using the linerized form of the moment-curvture reltion, M b /(EI) d v/ nd obtin 4 dv dv ( EI) + 4 Along the wy I hve ssumed tht the pplied distributed lod w (x) is zero. This is no gret loss; it will hve little influence on the behvior if the end lod pproches the buckling lod. It is stright forwrd mtter to tke its effect into ccount. In order to focus on the buckling mechnism, we leve it side. This is fourth order, ordinry, liner differentil, homogeneous eqution for the trnsverse displcement of the neutrl xis of bem subject to n end lod,. It must be supplemented by some boundry conditions on the displcement nd its derivtives. Their expression depends upon the prticulr problem t hnd. They cn tke the form of zero displcement or slope t point, e.g., v or dv

15 Deflections due to Bending 77 They cn lso involve higher derivtives of v(x) through conditions on the bending moment t point long the bem, e.g., condition on dv M b ( EI) or on the sher force dv dv V ( EI) 3 This lst is resttement of the differentil eqution for force equilibrium found bove, which, since we incresed the order of the system by differentiting the moment equilibrium eqution, now ppers s boundry condition 3 Exercise 1.4 A bem of ength, moment of inerti in bending I, nd mde of mteril with Young's modulus E, is pinned t its left end but tied down t its other end by liner spring of stiffness K. The bem is subjected to compressive end lod. v(x) x EI Show tht the Euler buckling lod(s) re determined from the eqution sin ( EI) K Show lso tht it is possible for the system to go unstble without ny elstic deformtion of the bem. Tht is, it deflects upwrd (or downwrd), rotting bout the left end s rigid br. Construct reltionship n expression for the stiffness of the liner spring reltive to the stiffness of the bem when this will be the cse, the most likely mode of instbility. We strt with the generl solution to the differentil eqution for the deflection of the neutrl xis, tht described by the function v(x). vx ( ) c 1 + c x + c 3 sin -----x + c EI 4 cos -----x EI etting λ (/EI) this cn be written more simply s vx ( ) c 1 + c x + c 3 sinλx + c 4 cosλx In this, c 1, c, c 3, nd c 4 re constnts which will be determined from the boundry conditions. The ltter re s follows:

16 78 Chpter 1 At the left end, x, the displcement vnishes so v() nd since it is pinned, free to rotte there, the bending moment must lso vnish M b () or d v/ while t the right end, x, the end is free to rotte so the bending moment must be zero there s well: M b () or d v/ The lst condition (we need four since there re four constnts of integrtion) requires drwing n isoltion of the end. We see from force equilibrium of the tip of the bem tht V V+F or V +Kv() FK v() v positive up where F is the force in the spring, positive if the end moves upwrd, nd V, the sher force, is consistent with our convention set out prior to deriving the differentil eqution for v(x). Expressing V in terms of the displcement v(x) nd its derivtives we cn write our fourth boundry condition s: d 3 v/ 3 + (/EI)dv/ - (K/EI) v() or d 3 v/ 3 + λdv/ - β v() where I hve set β K/EI. To pply these to determine the c s, we need expressions for the derivtives of v(x), up to third order. We find dv/ c + c 3 λ cosλx - c 4 λ sinλx d v/ - c 3 λ sinλx - c 4 λ cosλx d 3 v/ 3 - c 3 λ 3 cosλx + c 4 λ 3 sinλx With these, the boundry conditions become t x. v(): c 1 + c 4 d v/ : + c 4 t x. d v/ : (λ sin λ) c 3 ( λ - β )c - (sin λ) c 3 Now these re four, liner homogeneous equtions for the four constnts, c 1 - c 4. One solution is tht they ll be zero. This, if your were to report to your boss would ern your very erly retirement. The trivil solution is not the only one. In fct there re mny more solutions but only for specil vlues for the end lod,

17 Deflections due to Bending 79 (λ). We know from our prior studies of systems of liner lgebric equtions tht the only hope we hve for finding non zero c's is to hve the determinnt of the coefficients of the liner system be zero. The eigenvlues re obtined from this condition. Rther thn evlute the determinnt, we will proceed by n lternte pth, no less decisive. From the second eqution we must hve c 4. Then, from the first we must hve c 1. Turning to the third eqution, we might conclude tht c 3 is zero s well. Tht would be mistke. For we might hve sinλ. Now consider the lst two s two equtions for c nd c 3. The determinnt of the coefficients is ( λ β ) sin( λ) which, when set to zero, cn be written K sin ( EI) This cn be mde zero in vrious wys. We cn hve (/K) 1 or we cn set sin [ /(EI)] 1/ which hs roots /(EI) π, 4π, 9π,... The criticl eigenvlue will be the lowest one, the one which gives the lowest vlue for the end lod. We see tht this depends upon the stiffness of the liner spring reltive to the stiffness of the bem s expressed by EI/ 3. For the mode of instbility implied by the eqution (/K) 1, we must hve K < π EI/ or K/ (EI/ 3 ) < π If this be the cse, then the coefficient of c in the lst of our four equtions will be zero. At the sme time, sinλ will not be zero in generl so c 3 must be zero. The only non-zero coefficient is c nd, our generl solution to the differentil eqution is simply v(x) c x This prticulr buckling mode is to be red s rigid body rottion bout the left end. If, on the other hnd, the inequlity goes the other wy, then nother mode of instbility will be encountered when π EI/ Now c must vnish but c 3 cn be rbitrry. Our deflected shpe is in ccord with v(x) c 3 sin λx

18 8 Chpter 1 nd is sketched below. Note in this cse the liner spring t the end neither extends nor contrcts. v(x) A sin λx A Observe There re still other specil, or eigenvlues, which ccompny other, higher, mode shpes. The next one, corresponding to /EI 4π, would pper s full sine wve But since the lower criticl mode is the most probble, you would rrely see this wht remins s but mthemticl possibility. If we let K be very lrge reltive to (EI/ 3 ), we pproch bem pinned t both ends. The buckled bem would pper s in the figure bove. Conversely, If we let K be very smll reltive to the bem s stiffness, we hve sitution much like the system we previously studied nmely rigid br pinned t n end but restrined by torsionl spring. In fct, we get the sme buckling lod if we set the K T of the torsionl spring equl to K. We only see the possibility of buckling if we consider equilibrium with respect to the deformed configurtion. 1.3 Mtrix Anlysis of Frme Structures We return to the use of the computer s n essentil tool for predicting the behvior of structures nd develop method for the nlysis of internl stresses, the deformtions, displcements nd rottions of structures mde up of bems, bem elements rigidly fixed one to nother in some pttern designed to, s is our hbit, to support some externlly pplied lods. We cll structures built up of bem elements, frmes. Frmes support modern skyscrpers; your bicycle is frme structure; cntilevered blcony might be girded by frme. If the structure s members re intended to support the externlly pplied lods vi bending, the structure is frme.

19 Deflections due to Bending 81 As we did with truss structures, structures tht re designed to support the externlly pplied lods vi tension nd compression of its members, we use displcement method. Our finl system of equtions to be solved by the mchine will be the equilibrium equtions expressed in terms of displcements. The figure shows frme, building frme, subject to side loding, sy due to wind. These structurl members re not pinned t their joints. If they were, the frme would collpse; there would be no resistnce to the shering of one floor reltive to nother. The structurl members re rigidly fixed to one nother t their joints. So the joints cn trnsmit bending moment from one element to nother. The figure t the left shows how we might model the frme s collection of discrete, bem elements. The number of elements is 7 11 quite rbitrry. Just how mny elements is 3 sufficient will depend upon severl fctors, Y e.g., the sptil vribility of the externlly 6 X 1 pplied lods, the homogeniety of the mterils out of which the elements re mde, the M Y 1 desired "ccurcy" of the results. 5 9 X 1 In this model, we represent the structure s n M 1 ssemblge of 16 elements, 8 horizontl, two for ech floor, nd 8 verticl. At ech node there re three degrees of freedom: horizontl displcement, verticl displcement, nd the rottion t the node. This is one more degree of freedom thn ppered t ech node of our (two-dimensionl) truss structure. This is becuse the geometric boundry conditions t the ends, or junction, of bem element include the slope s well s the displcement. With 1 nodes nd 3 degrees of freedom per node, our structure hs totl of 36 degrees of freedom; there re 36 displcements nd rottions to be determined. We indicte the pplied externl force components cting t but two of the nodes: X 1,Y 1 ct t node 1 for exmple; M 1 stnds for n pplied couple t the sme node, but we do not show the corresponding components of displcement nd rottion. Equilibium in terms of displcement will require 36 liner, simultneous equtions to be solved. This presents no problem for our mchinery. We will construct the system of 36 equilibrium equtions in terms of displcement by directly evluting the entries of the whole structure s stiffness mtrix. To do this, we need to construct the stiffness mtrix for ech individul bem element, then ssemble the stiffness mtrix for the entire structure by superimpositioning. Wht this mens will become cler, I hope, in wht follows.

20 8 Chpter 1 Our pproch differs from the wy we treted truss structures. There we constructed the equilibrium equtions by isolting ech node of the truss; then wrote down set of force-deformtion reltions for ech truss member; then nother mtrix eqution relting the member deformtions to the displcement components of the nodes. We then eliminted the member forces in terms of these nodl displcements in the equilibrium equtions to obtin the overll or globl stiffness mtrix for the entire truss structure. Only t this point, t the end of our construction, did we point out tht tht ech column of the stiffness mtrix cn be interpreted s the forces required to mintin equilibrium for unit displcement corresponding to tht column, ll other displcements being held to zero. This is the wy we will proceed from the strt, now, constructing first the stiffness mtrix for horizontl bem element. The figure t the right shows such n element. At ech of its end nodes, we S 1, v 1 S, v llow for n xil, trnsverse force nd couple. These re ssumed to be F 1, u 1 F u positive in the directions shown. The displcement components t ech of the Q 1, φ 1 Q φ two ends - v 1,v, in the trnsverse direction, u 1, u, in the xil direction - nd the slopes t the ends, φ 1 nd φ, re lso φ indicted; ll of these re positive in the φ 1 directions shown. v The bottom figure shows possible u 1 v deformed stte where the displcements 1 n rottions re shown more clerly (sve u ). Our first tsk is to construct the entries in the stiffness mtrix for this bem element. It will hve the form: F 1 u 1 A bsic feture of mtrix multipliction, S of this expression, is the following: ech 1 v 1 column my be interpreted s the force Q 1 φ 1 nd moment components, the left side of F u the eqution, tht re required to mintin deformed configurtion of unit S v displcement corresponding to tht column nd zero displcements nd rot- Q φ tions otherwise. For exmple, the entries in the first column my be interpreted s F 1, S 1...Q for the displcement u 1 1 nd v 1 φ 1 u v φ

21 Deflections due to Bending 83 This turns out to be prticulrly esy column to fill in, for this deformtion stte, shown in the figure, only F 1, u 1 F requires xil forces F1 nd F to induce this displcement nd mintin equilibrium. In fct, if u 1 1, then the force required to produce this uint displcement is just (AE/) 1. This mend tht F, in this cse, must, in order for equilibrium to hold, be equl to - (AE/). No other forces or moments re required or engendered. Hence ll other elements of the first column of the stiffness mtrix, the column corresponding to u 1, must be zero. The elements of the fourth column, the column corresponding to AE AE F the displcement u, re found just ?? ?? u 1 s esily. In this cse F (AE/) S 1???? v 1 nd F 1 - (AE/) for equilibrium. Our stiffness mtrix now hs the form shown t the right. We continue considering the second column nd envision the displce configurtion v 1 1 nd u 1 φ 1 u v φ. Q 1 F S Q???? AE AE ?? ?????????? φ 1 u v φ S The figure t the left shows the 1 Q deformed stte. In this it ppers we 1 S will require S1 nd Q1 but no xil V 1 1 Q force. Remember, we ssume smll displcements nd rottions so the xil force does not effect the bending of the bem element (nd, similrly, bending of the bem does not induce ny xil deformtion). Only if we llow greter thn smll displcements nd rottions will n xil lod effect bending; this ws the cse in buckling but we re not llowing for buckling. To determine wht end force S 1, nd wht end couple, Q 1, re required to mke the verticl displcement t the end of cntilever equl 1 nd the slope there zero - it s cntilever becuse the verticl displcement nd rottion t the right end must be zero -we mke use of the known expressions for the tip displcement nd rottion due to n end lod nd n end couple, then superimpose the two to ensure verticl displcement of 1 nd rottion of zero. For verticl, end lod S 1, we hve from the reltionships given on pge 18,

22 84 Chpter 1 3 v 1 S 3EI 1 nd dv 1 φ 1 S EI 1 For n end couple, Q 1, we hve v 1 Q EI 1 nd φ Q EI 1 If both verticl force nd couple ct, then, superimposing, we obtin the following two equtions for determining the verticl displcement nd the rottion t the end of the bem: v 3 1 S 3EI 1 Q EI 1 S 1 -φ 1 v 1 End lod. End couple φ 1 S -v 1 EI Q EI 1 φ 1 Q 1 Now we set the verticl displcement equl to unity, v 1 1, nd the slope zero, φ 1, nd solve these equtions for the required end lod S 1 nd the required end couple Q 1. We obtin: S 1 (1EI/ 3 ) nd Q 1 (/ ) The force nd couple t the right end of the element re obtined from equilibrium. Without even drwing free body digrm (living dngerously) we hve: S - (1EI/ 3 ) nd Q (/ ) Thus the elements of the second column of the mtrix re found nd our Q 1 S S 1 stiffness mtrix now hs the form shown. v 1 1 Q

23 Deflections due to Bending 85 F 1 S 1 Q 1 F S Q AE AE ? ?? 1EI --- 3?????? AE AE ? ?? 1EI ---??? 3??? u 1 v 1 φ 1 u v φ The elements of the third column, corresponding to unit rottion φ 1 1 nd ll other displcements nd rottions zero, re found gin from two simultneous equtions but now we find, with v 1 nd φ 1 1: S 1 (/ ) nd Q 1 (4EI/) Agin, the force nd S Q couple t the right end of the element re obtined from equilibrium. Now we hve: S - (/ ) nd Q (EI/) With this, our stiffness mtrix becomes s shown. F 1 S 1 Q 1 F AE EI AE ???? 4EI?? AE AE ?? 1EI --- 3?? EI?? u 1 v 1 φ 1 u v φ

24 86 Chpter 1 roceeding in similr wy t the right end of the bem element, we cn construct the elements of the finl two columns corresponding to unit displcement, v 1 (the fifth colunm) nd unit rottion t the end, φ 1, (the sixth column). Our finl result is: F 1 S 1 Q 1 F S Q AE EI EI AE EI EI AE - 1EI EI AE EI EI u 1 v 1 φ 1 u v φ Exercise 1.5 Construct the stiffness mtrix for the Y simple frme structure shown. 1,V 1 Y,V X 1,U 1 X,U We employ the sme pproch s used in Μ 1 Φ 1 Μ Φ constructing the stiffness mtrix for the single, horizontl bem element. We consider b unit displcement of ech degree of freedom U 1,V 1, Φ 1,U,V, nd Φ constructing the corresponding column of the stiffness mtrix of the whole structure in turn. Note how we use cpitl letters to specify the displcements nd rottions of ech node reltive to globl coordinte reference frme. V 1 Φ 1 b X 1 AE/ U 1 1 1EI/b 3 /b X -AE/ We strt, tking U 1 1, nd consider wht force nd moment components re required to both produce this displcement nd ensure equilibrium t the nodes. Since ll other displcement nd rottion components re zero, we drw the deformed configurtion t the left. Refering to the previous figure, we see tht we must hve horizontl force of mgnitude AE/ + 1EI/b 3 pplied t node #1, nd moment of mgnitude /b to mintin this deformed configurtion. There is no verticl force required t node #1. At node #, we see we must pply, for equilibrium of the horizontl bem element, n equl nd opposite force to X 1 AE/. No other externlly pplied forces re required.

25 Deflections due to Bending 87 Thus, the first column of our stiffness mtrix ppers s shown t the right: Entries in the second column re obtined by setting V1 1 nd ll other displcement components zero. The deformed stte looks s below; X 1 Y 1 M 1 X Y M AE EI b 3?????????? b????? AE ??????????????? U 1 V 1 Φ 1 U V Φ Y 1 Φ 1 AE/b b V 1 1 1EI/ 3 / / -1EI/ 3 The forces nd moments required to engender this stte nd mintin equilibrium re obtined from the locl stiffness mtrix for single bem element on the previous pge. Thus, the second column of our stiffness mtrix cn be filled in: X 1 Y 1 M 1 X Y M AE EI b 3???? b AE AE 1EI b 3???????????? 1EI --- 3???????? U 1 V 1 Φ 1 U V Φ Continuing in this wy, next setting F1 1, ll other displcements zero, sketching the deformed stte, reding off the required force nd moment components to mintin this deformed stte nd superimposing corresponding compo-

26 88 Chpter 1 nents t ech of the two nodes, produces the stiffness mtrix for the whole structure. We obtin: X 1 Y 1 M 1 X Y M AE EI b 3 b AE AE b 1EI EI b b 4EI + AE - 1EI AE 1EI b 3 1EI AE 1EI b 3 EI b EI b 4EI 4EI + b U 1 V 1 Φ 1 U V Φ 1.4 Energy Methods Just s we did for Truss Stuctures, so the sme perspective cn be entertined for bems. A Virtul Force Method for Bems The intent here is to develop wy of computing the displcements of (stticlly determinte) bem under rbitrry loding nd just s rbitrry boundry conditions without mking explicit reference to comptibility condiditions, i.e. without hving to integrte the differentil eqution for trnsverse displcements. The pproch mimics tht tken in the section on Force Method #1 s pplied to stticlly determinte truss structures. We strt with the comptibility condition which reltes the curvture, κ 1/ρ, to the trnsverse displcement, v(x), ρ v(x) v(x) x d κ vx ( )

27 Deflections due to Bending 89 nd tke totlly unmotivted step, multiplying both side of this eqution by function of x which cn be nything whtsoever, we integrte over the length of the bem: This rbitrry function bers n sterisk to distinguish it from the ctul bending moment distribution in the structure. At this point, the function M* could be ny function we wish, but now we mnipulte this reltionship, integrting the right hnd side by prts nd so obtin Integrting by prts once gin, we hve then consider the function M*(x) to be bending moment distribution, ny bending moment distribution tht stisfies the equilibrium requirement for the bem, i.e., κ M * ( x) M * d ( x) vx ( ) d κ M * ( x) M * dv κ M * ( x) M * dv d M * ( x) p * ( x) v(x) x So p*(x) is rbitrry, becuse M*(x) is quite rbitrry - we cn envision mny different pplied lods functions. With this our comptibility reltionship pre-multiplied by our rbitrry function, now red s bending moment distribution, becomes κ M * ( x) M * dv x dv dm * d * dm * ( M ) v + v v(x) p(x) d * ( M ) v + p * ( x) vx ( ) y p(x) V Mb x

28 9 Chpter 1 (Note: the dimensions of the quntity on the left hnd side of this lst eqution re force times length or work. The dimensions of the integrl nd boundry terms on the right hnd side must be the sme). Now we choose p*(x) in specil wy; we tke it to be unit lod t single point long the bem, ll other loding zero. For exmple, we tke p * ( x) 1 δ( x ) unit lod in the verticl direction t distnce long the x xis. Crrying out the integrtion in the eqution bove, we obtin just the displcement t the point of ppliction, t x, i.e., v ( ) κ M * ( x ; ) M * dv + We cn put this lst eqution in terms of bending moments moment/curvture reltionship, nd obtin: v ( ) M( x) ---- M * ( x ; ) M * dv + EI lone using the And tht is our specil method for determining displcements of stticlly determinte bem. It requires, first, solving equilbrium for the "ctul" bending moment given the "ctul" pplied lods. We then solve nother equilibrium problem - one in which we pply unit lod t the point where we seek to determine the displcement nd in the direction of the sought fter displcement. With this bending moment distribution determined from equilibrium, we crry out the integrtion in this lst eqution nd there we hve it. Note: We cn lwys choose the "strred" loding such tht the "boundry terms" in this lst eqution ll vnish. Some of the four terms will vnish becuse of vnishing of the displcement or the slope t boundry. (The unstrred quntities must stisfy the boundry conditions on the ctul problem).grnting this, we hve more simply v ( ) M( x) ---- EI M * x ; ( ) We emphsze the difference between the two moment distributions ppering in this eqution; M(x), in plin font, is the ctul bending moment distribution in the bem, given the ctul pplied lods. M*(x), with the sterisk, on the other hnd, is some originlly rbitrry, bending moment distribution which stisfies equilibrium - n equilibrium solution for the bending moment corresponding to unit lod in the verticl direction t the point x. d * ( M ) v d * ( M ) v

29 v(x) 1 p(x) M*(x;) 1(-x) x Deflections due to Bending 91 x p*(x) 1δ(x-) p*(x) 1δ(x-) As n exmple, we consider cntilever bem subject to distributed lod p(x), which for now, we llow to be ny function of distnce long the spn. We seek the verticl displcement t x. We determine the bending moment distribution corresponding to the unit lod t. This is shown in the figure t the left, t the bottom of the frme. With this, our expression for the displcement t becomes: v ( ) M*(x;) Note tht: i) M* t x nd S* ( dm*/) t x re zero so the two boundry terms in the more generl expression for v() vnish. ii) At the root, v nd f ( dv/) so the two boundry terms in the more generl expression for v() vnish. M( x) ---- ( x) EI iii) Finlly, note tht since the bending moment M*(x;) is zero for x>, the limit of integrtion in the bove eqution cn be set to. v(x) p(x) p x We now simplify the exmple by tking our distributed lod to be constnt, p(x) p. From the free body digrm t the right, we find M( x) p ( x) nd so the integrl left to evlute is: M(x) p (-x) v ( ) p ( x) p --- ( x) x EI EI ( ) ( x)

30 9 Chpter 1 which, upon evlution yields v ( ) p --- ( ) 4EI Agin, the significnt thing to note is tht we hve produced n expression for the trnsverse displcement of the bem without confronttion with the differentil eqution for displcement! Our method is force method requiring only the solution of (moment) equilibrium twice over. A Virtul Displcement Method for Bems The gme now is to construct the stiffness mtrix for bem element using displcement nd deformtion/displcement considertions lone. We consider bem element, uniform in cross-section nd of length, whose end displcements nd rottions re presecribed nd we re sked to determine the end forces nd moments required to produce this system of displcements. In the figure t the right, we show the bem element, deformed with prescribed y V end displcements v 1,v, nd prescribed M S end rottions φ 1, φ. The tsk is to find x the end forces, S 1,S nd end moments S Q Q 1,Q, tht will produce this deformed 1 φ stte nd be in equilibrium nd we wnt Q 1 φ 1 to do this without hving to consider v equilibrium explicitly. We strt with equilibrium: 7 v 1 v(x) x dv dm + V( x) nd tke totlly unmotivted step, multiplying the first of these equtions, the one ensuring force equilbirium in the verticl direction, by some function v*(x) nd the second, the one ensuring moment equilbrium t ny point long the element, by nother function φ (x), then integrte the sum of these products over the length of the element. dv v * dm ( x) + + V( x) ϕ * ( x) The functions v*(x) nd φ*(x) re quite rbitrry; t this stge in our gme they could be nything whtsoever nd still the bove would hold true, s long s the sher force nd bending moment vry in ccord with the equilibrium requirements. They ber n sterisk 7. Note: our convention for positive sher force nd bending moment is given in the figure.

31 Deflections due to Bending 93 to distinguish them from the ctul displcement nd rottion t ny point long the bem element. Now we mnipulte this reltionship, integrting by prts, noting tht: nd write Now from the figure, we identify the internl sher force nd bending moments cting t the ends with the pplied end forces nd moments, tht is S 1 - V(), Q 1 - M(), S + V(), nd Q +M(). so we cn write v * V ϕ * + M dv v * ( x) v * V nd dm ϕ * ( x) ϕ * M V( x) dv * V( x) dϕ * M( x) dv * ϕ * ( x) + M( x) dϕ * v 1 * * * * S 1 + ϕ 1 Q 1 + v S + ϕ Q V( x) dv * ϕ * ( x) + M( x) dϕ * We now restrict our choice of the rbitrry functions v*(x) nd φ*(x) 8.We ssocite the first with trnsverse displcement nd the second with rottion while requiring tht there be no trnsverse sher deformtion, i.e, plne cross sections remin plne nd perpendiculr to the neutrl xis. In this cse dv * ϕ * ( x) so the first term in the integrl on the right hnd side vnishes, leving us with the following; v 1 * * * * dv * S 1 + φ 1 Q 1 + v S + φ Q M( x) d x 8. Actully we hve lredy done so, insisting tht they re continuous to the extent tht their first derivtives exist nd re integrble, in order to crry through the integrtion by prts.

32 94 Chpter 1 We cn cst the integrnd into terms of member deformtions lone (nd member stiffness, EI), by use of the moment curvture reltionship nd write dv M( x) EI v 1 * * * * dv dv * S 1 + φ 1 Q 1 + v S + φ Q EI And tht will serve s our specil method for determining the externl forces nd moments, cting t the ends, given the prescribed displcement field v(x). The ltter must be in ccord with equilibrium, our strting point; tht is dm dv so from the moment/curvture reltion 4 4 Thus v(x), our prescribed displcement long the bem element, hs the form: d dv x x vx ( ) + 1 x + x + 3 x 3 nd With this, our reltionship mong end forces, end moments, nd prescribed displcements becomes v 1 * * * * dv * S 1 + φ 1 Q 1 + v S + φ Q EI ( x) The coeficients nd 3, s well s nd 1, cn be relted to the end displcements nd rottions, v 1, φ 1, v nd φ but we defer tht tsk for the moment. Insted, we choose v*(x) to S 1 give unit displcement v 1 *1nd S the other end displcement nd rottions to be. There re mny func- Q Q 1 tions tht will do the job; we tke v* 1 1 φ* 1 v* φ*

33 95 Chpter 1 v * ( x) πx + cos----- differentite twice, nd crry out the integrtion to obtin S 1 EI 6 3 Now we relte the s to the end displcements nd rottions. With v 1 v( ) v φ 1 dv 1 nd v ( ) φ dv we solve for the s nd obtin v 1 1 φ v φ v -- φ v φ v φ 3 which then yields the following result for the end force required, S 1, for prescribed end displcements nd rottions v 1, φ 1, v, nd φ. 3 S 1 1EI --- v 1 φ 1EI v + φ 3 3 The sme ploy cn be used to S 1 obtin the end moment Q 1 required for S prescribed end displcements nd Q 1 1 rottions v 1, φ 1, v, nd φ. We need Q but choose our rbitrry function v*(x) to give unit rottion φ 1 * v* 1 v* dv*/ 1 t the left end, x, nd φ* φ* 1 1 the other end displcements nd rottionsto be. There re mny functions tht will do the job; we tke v * ( x) x ( x ) 3 + x ( ) 5

34 96 Chpter 1 differentite twice, nd crry out the integrtion to obtin Q 1 EI which then yields the following result for the end moment required, Q 1, for prescribed end displcements nd rottions v 1, φ 1, v, nd φ. Q 1 v 4EI 1 φ 1 EI + v + φ In similr wy, expressions for the end force nd moment t the right end, x re obtined. utting this ll together produces the stiffness mtrix: S 1 Q 1 S Q 1EI EI EI EI 1EI EI EI 4EI v 1 φ 1 v φ Once gin we obtin symmetric mtrix; why this should be is not mde cler in tking the pth we did. Tht this will lwys be so my be deduced from the boxed eqution on previous pge, nmely v 1 * * * * dv dv * S 1 + φ 1 Q 1 + v S + φ Q EI by choosing our rbitrry function v*(x) to be identicl to the prescribed displcement field, v(x). We obtin: Now nd 3 my be expressed in terms v 1 S 1 + φ 1 Q 1 + v S + φ Q EI ( x) ( x)

35 97 Chpter 1 of the prescribed end displcements nd rottions vi the reltionships derived erlier, which, in mtrix form, is 3 v φ v φ G v 1 φ 1 v φ Our bsic eqution then becomes: v 1 S 1 + φ 1 Q 1 + v S + φ Q EI v 1 φ 1 v φ G T 4 1x 1x 36x G v 1 φ 1 v φ so the symmetry is pprent. If you crry through the mtrix multipliction nd the integrtion with respect to x you will recover the stiffness mtrix for the bem element.

36 98 Chpter 1 Design Exercise 1.1 Your tsk is to design clssroom demonstrtion which shows how the torsionl stiffness of structure depends upon mteril properties nd the geometry of its stucturl elements. Your professor hs proposed the following s wy to illustrte torsionl stiffness nd, t the sme time, the effects of combined loding on shft subjected to bending nd torsion. A smll number, N, circulr rods or tubes re uniformly distributed round nd fstened to two reltively rigid, end pltes. The rods re rigidly fixed to one of the end pltes, sy the plte t the left shown bove. The other ends of the rods re designed so tht they cn be either free to rotte bout their xis or not; tht is, the right ends cn be fixed rigidly to the plte or they cn be left free to rotte bout their xis. If the rods re free to rotte bout their xis, then ll resistnce to rottion of the entire structure is due to the resistnce to bending of the N rods. (Note not ll N rods re shown in the figure). If, on the other hnd, they re rigidly fixed t both ends to the pltes, then the torsionl stiffness of the overll structure is due to resistnce to torsion of the N rods s well s bending. A preliminry design of this pprtus is needed. In prticulr: The torsionl stiffness of the entire structure due to bending of the rods is to be of the sme order of mgnitude s the torsionl stiffness of the entire structure due to torsion of the rods. The overll torsionl stiffness should be such tht the rottion cn be mde visible to the nked eye for torques whose ppliction does not require excess mchinery. The pprtus should not fil, yield or brek during demonstrtion. It should work with rods of two different mterils. It should work with hollow tubes s well s solid shfts. Attention should be pid to how the ends of the rods re to be fstened to the pltes.