ANALYSIS OF STRUCTURES

Transcription

1 Mech_Eng 36 Stress Anlsis Anlsis of Structures ANAYSIS OF STRUCTURES Sridhr Krishnswm 8-1

2 Mech_Eng 36 Stress Anlsis Anlsis of Structures 8.1 ANAYSIS OF STRUCTURES: At this point, we hve developed n understnding of chrcteriztion of internl forces, geometr of deformtion, mteril response, nd how the ll fit together. We hve imposed Newton's lws nd we lso hve looked into energ principles. Our toolkit now includes quite bit of stuff: "ect" nlticl methods: using theor of elsticit equtions or the principle of minimum potentil energ "pproimte" nlticl methods: using Rleigh-Ritz pproch nd the principle of minimum potentil energ "pproimte" numericl methods: using the finite-element implementtion of the Rleigh-Ritz pproch In the rest of this course, we re going to look t severl structures nd nlze them using n convenient method from our toolkit. When confronted with the design or nlsis of structure, ou should cll upon the most pproprite method. If ou think tht problem cn be solved s simple bem, it is ridiculous to seek (nd mbe impossible to find) tedious "ect" nlticl solution using the full power of the theor of elsticit. It is equll ridiculous to rush over to the nerest computer nd strt creting finite element mesh for the problem. Mbe ll tht is required is strightforwrd Bernoulli-Euler nlsis; or if the bem is bit complicted becuse of its shpe or hs weird support conditions, mbe pencil nd pper pproimte Rleigh-Ritz technique is clled for. In other situtions, it mbe tht the finite element pproch is essentil. Some times ou m wnt quick nd dirt (ver pproimte) solution tht RR cn provide, nd then ou might wnt to follow this up with more detiled FEM nlsis of the structure. You hve to be ble to judge wht tool to bring to problem t hnd, nd ou will gin n pprecition for this we when look t severl spects of structurl nlsis, nd ou see how I hndle them. We hve lred seen how to hndle rods nd some bems. et us pick up on vrition of bem problem. Sridhr Krishnswm 8-

3 Mech_Eng 36 Stress Anlsis Elstic Supports Emple 1: Bem Simpl-Supported t its Ends nd Resting on Spring t its Midpoint Consider slender bem s shown bove. In ddition to the hinges t its ends, the bem is supported t its middle b liner elstic spring of spring constnt K (N/m). It lso crries distributed re lod of mgnitude σ o (Nm - )s shown. σ o b h z K How would ou nlze this structure for deformtion nd stresses? Right w ou might s tht we cn tr Bernoulli-Euler nlsis of the structure. To clculte the bending moment t n cross-section, the first thing ou do is to get rid of the supports, nd put in the pproprite rections. The two pins pose no problem to us, but wht bout the spring? The force the spring eerts is proportionl to its deformtion, ie R sp = K ( = /) cting upwrds on the spring, nd ou might get ques feeling tht this w lies trouble! Tr it nd see wht ou get. It turns out tht we re better off seeking n pproimte Rleigh-Ritz solution to this problem s follows. (i) Bg of Cndidte Displcements: Seek one-prmeter sinusoidl bg () = 1 sin π 0 which clerl stisfies the geometric boundr conditions: (0) = 0; () = 0, nd is smooth enough. (ii) Clculte the Potentil Energ Functionl: In this cse, energ cn be stored in two plces - the bem nd the spring (this is how the spring enters our nlsis): U=U bem + U spring U bem = EI z = 0 = π4 EI z d d d Sridhr Krishnswm 8-3

4 Mech_Eng 36 Stress Anlsis Elstic Supports U spring = 1 K { ( = / ) } = 1 K 1 The virtul work term is: W = = 0 () = b o π { } 1 o.b.d The potentil energ functionl is therefore: Π = 1 K + π 4 EI z b o π (iii) Minimizing the potentil energ functionl with respect to 1 ields: 4b o () = π K + π sin π 4 EI z 1 Remrks: (i) A simple check of dimensions is useful to ensure tht we hve not screwed up the lgebr. {Note tht K is force, nd so is EI z / (more on this strnge combintion lter); these cncel out the force dimension obtined from (σ o b) leving n ( length) s the dimension for the displcement - so, ll is well dimensionll speking!}. (ii) One cn of course improve on the solution using two or more prmeters for the RR function, or else, we cn use multi-prt RR function et ceter. For instnce, for bems tht re pinned t both =0 nd =, one-prmeter polnomil bg cn be: () = 1 1 which clerl stisfies the geometric boundr conditions tht the displcements re zero t both pins. If ou wnt prctice emples, I suggest tht ou look t ll the emples we will do in this course, nd see if ou cn come up with suitble one- or two-prmeter polnomil nd trigonometric bgs of displcements. (iii) P prticulr ttention to how we computed the virtul work term for distributed re lod in this problem. Wht if we hd concentrted force P (in N) t the mid-point; wht chnges then? Wht if we hd distributed line lod p (in N/m)? Sridhr Krishnswm 8-4

5 Mech_Eng 36 Stress Anlsis Elstic Supports 8. EASTIC FOUNDATIONS: Consider long bem tht is supported b liner elstic springs of spring constnt K (N/m) t severl points seprted b distnce '' prt. An emple of this is rilw trck with the springs being the ties. If the bem deflection is (), then the springs provide rection force: F i = K ( = i ) t the i'th spring where i is the loction of th i'th spring. If the spring spcing '' is smll compred to other relevnt length scles, it is simpler to think of the bed of discrete springs s being mde of continuum of springs clled n elstic foundtion. The rection force from the springs re now distributed s line lod (N/m): q() = { K / } () k () where k (in N/m ) is clled the foundtion modulus. Elstic foundtions ctull re quite common - the cn be the ground or floor on which bem rests tht resists b ppling line lod t point (force/length) tht is proportionl to the displcement t the point. Given set of discrete springs, we cn smer them s described bove to get n equivlent elstic foundtion. We cn lso conceptull reverse the process bove to go from n elstic foundtion to n equivlent set of discrete springs. Emple : Bem Simpl-Supported t its Ends nd Resting on n Elstic Foundtion σ o b elstic foundtion of modulus k (Nm - ) h z Now consider the bem we sw in emple 1, but let us replce the discrete spring t its midpoint b n elstic foundtion. As before, we cn use the Rleigh-Ritz technique with the sme bg of cndidte displcements. Sridhr Krishnswm 8-5

6 Mech_Eng 36 Stress Anlsis Elstic Supports (i) Bg of Cndidte Displcements: Seek one-prmeter sinusoidl bg () = 1 sin π 0 which clerl stisfies the geometric boundr conditions: (0) = 0; () = 0, nd is smooth enough. (ii) Clculte the Potentil Energ Functionl: In this cse, energ cn gin be stored in two plces - the bem nd the elstic foundtion : U=U bem + U foundtion We lred know how to clculte the energ stored in the bem, so let us concentrte on the foundtion. Here is where we should conceptull run the "smering" process bckwrds into "lumping" process. A continuous foundtion of foundtion modulus k (N/m ) cn be lumped over length 'd' into discrete liner spring of spring constnt K={k.d} (in N/m). The energ stored in this equivlent discrete spring is: (1/) (kd){ ()}. Summing over ll such equivlent springs tht mke up the entire foundtion, we hve: U foundtion = = = 0 = 0 1 u () { } { k.d} k () { }.d Evluting the bove (everthing else we just borrow from the previous emple, becuse we hve not chnged nthing else), we get the potentil energ functionl for the given structure to be: Π = 1 4 k + π 4 EI z b o π 1 (iii) Minimizing the potentil energ functionl, we find: 4b o () = π k + π sin π 4 EI z Vritions: (i) Consider set of discrete lods P i (N) cting t the qurter nd hlf points of the bem; nd redo the bove emple. (ii) Replce one of the end pins, nd use fied end insted. (iii) Tr different - mbe polnomil - bg of displcements. Sridhr Krishnswm 8-6

7 Mech_Eng 36 Stress Anlsis Bending of Pltes 8.3 BENDING OF PATES: A plte is thin structure tht is trnsversel loded, nd is therefore in bending just like bem, but now in two-dimensions. h z b Assumptions: Figure 1 -We will restrict ttention to onl rectngulr pltes of sides '' nd 'b' tht re of constnt thickness 'h' nd which re thin (h<< &b). - octe n z coordinte sstem such tht the z-plne coincides with the mid-surfce of the plte. - Bending of pltes dels with the deformtion nd stresses in the plte cused b trnsverse -directed lods. As we observed for Bernoulli-Euler bems, we find tht the stress nd strin sttes for such pltes re considerbl simplified, t lest to fir pproimtion. In prticulr, we find tht: (i) The onl importnt strin components re the in-plne (z) components: = u ; = u z zz z ; = 1 z nd the other components re negligible. Tht is: u z + u z ; (1) = = 0 () = 1 z = 1 u + = 0 (3) u z + z = 0 (4) (ii) The stress components ; ; z re ll negligible. Sridhr Krishnswm 8-7

8 Mech_Eng 36 Stress Anlsis Bending of Pltes (iii) The mid-surfce of the plte (the =0 plne) is unstretched nd unstressed nd is clled the neutrl surfce of the plte. These ssumptions re obviousl just two-dimensionl nlog of the Bernoulli-Euler bem theor ssumptions, nd s in tht theor, ou might guess tht the re not entirel consistent. The resulting (Kirchoff) plte bending theor is gin onl pproimtel correct; but it ields prett decent results for the in-plne components. As in the Bernoulli-Euler bem theor, we will single out the out-of-plne displcement s our fvorite quntit, nd cll it the plte deflection. We will cst ll other quntities in terms of this plte deflection. -From (): = = 0 = (, z) (5) -From (3) nd (5): = 1 u + = 0 u = (, z) u = (,z) + f 1 (, z) But noting tht the neutrl surfce (=0) is unstretched, we cn set f 1 (,z) = 0, nd so u = (,z) (6) -Resoning ectl s bove, but strting with (4) nd (5), we find u z = z (,z) (7) Remrk: The phsicl mening of (5), (6) nd (7) tken together is tht n -directed stright line (verticll through the plte thickness) sts stright line, suffering verticl displcement nd rottion bout both - nd z-directions. We hve now cst ll the displcements in terms of our fvored cndidte, the plte deflection. Computing the strins, using the strin-displcement reltions (1), we hve: = u = (6) } (,z) (8) Sridhr Krishnswm 8-8

9 Mech_Eng 36 Stress Anlsis Bending of Pltes zz = u (7) } z z = u z (,z) (9) z = 1 u z + u z = z (, z) (10) Net, we cn use Hooke's lw for n isotropic mteril nd impose our plte ssumptions, to get the reduced stress-strin reltions: = 1 E zz = 1 E zz 0 } + zz 0 } + zz 1 E 1 E { } zz zz { } z = 1 + E z Inverting the bove to epress stresses in terms of strins, nd then using (8-10), we find: = zz = z = E ( 1 + zz ) = E E ( 1 + zz ) = E E 1+ z z z (11) (1) (13) Note tht the stresses nd strins re zero on the mid-plne, s we hd demnded. At this point, we cn continue s we did for bems, nd consider smll plte element nd derive the differentil equtions for plte bending, but we will choose nother route insted. Now tht we hve the stresses nd strins, it is strightforwrd mtter to clculte the strin-energ stored in plte bending. In prticulr, the strin-energ densit for plte bending becomes: { } U o,plte = zz zz z z nd using (8-13), the totl strin-energ stored in plte bending becomes: Sridhr Krishnswm 8-9

10 Mech_Eng 36 Stress Anlsis Bending of Pltes where D = U plte = V = D U o dv A + z (1 ) z. d.dz z Eh 3 is clled the fleurl rigidit of rectngulr plte. Note the similrit 1(1 ) (t lest for the first term in the integrl) between plte nd bem bending. For rectngulr plte ll of whose edges re constrined such tht the re not llowed to displce (ie =0 on ll the edges, s using simple-supports nd fied supports), the bove somewht mess eqution simplifies to: U plte = D u u +. d.dz (onl for the cse: u z =0 on ll the edges) A Tht is, the second term drops out. One cn show this using the -d counterprt of integrtion b prts: A. z z.d.dz = S z..d A. 3 z.d.dz = z..d S = 0 since =0 on S S. z.dz = 0 since =0 on S A. z.d.dz where we repetedl swp n re integrl over A to boundr integrl over S. We cn now use energ methods, in prticulr Rleigh-Ritz, to solve plte bending problems. Sridhr Krishnswm 8-10

11 Mech_Eng 36 Stress Anlsis Bending of Pltes Emple 3: Rectngulr plte simpl-supported on ll sides nd crring distributed re lod p 0 N/m over its entire surfce. z (i) Bg of Displcements: The geometric boundr conditions re: ( = 0,z) = 0; ( =,z) = 0; (,z = 0) = 0; (,z = b) = 0; nd therefore the following one-prmeter bg is legl: (,z) = A 1 sin π πz sin b where A 1 is the Rleigh-Ritz prmeter we need to determine. (ii) Clculte the potentil energ functionl: The energ stored is onl in the plte, nd since ll sides of the plte re constrined not to displce, we cn use the simplified formul we developed erlier to get: U plte = D = D A b z= 0 = D A 1 = 4 + = z A b + b. d.dz b z= 0 8 D b + b b.a 1 The virtul work for this cse is: W = b z =0 =0 b ( p o.d.dz) = p o A 1 sin πz b.dz z = 0 =0 sin π z.dz sin π sin πz b.dz = 4p o b π A 1 The potentil energ functionl is therefore: Π = 4 8 D b + b b h b.a 1 4p o b π A 1 sin πz b Sridhr Krishnswm 8-11 = 0.d.dz sin π.d.dz (iii) Minimizing the potentil energ functionl with respect to A 1, we get:

12 Mech_Eng 36 Stress Anlsis Bending of Pltes (,z) = 16 1 b π 6 D b + p o.sin π πz.sin b Remrks: (i) Better ("ect") solutions cn be obtined b considering: (,z) = n=1 m =1 sin mπ nπz.sin b (ii) A one-prmeter polnomil bg tht is suitble for this problem is: (,z) = A 1 1 z b 1 z b (iii) Vritions on the bove theme: - wht if there is concentrted force P o (N) t the mid-point? - or line lod q (N/m) long the center-line prllel to the -is? - or wht if there is liner elstic spring supporting the plte t its mid-point? Tr pling round with such vritions to get better hndle on these things. Sridhr Krishnswm 8-1

13 Mech_Eng 36 Stress Anlsis Buckling of Columns 8.4 BUCKING OF COUMNS: We hve thus fr looked into equilibrium solutions for vrious structures. The net question I wnt to ddress is this: re these solutions stble? First, wht do we men b stbilit? Consider rigid bll on frictionless surfce, where the onl significnt force is grvittionl cting downwrds s shown grvit At which position do ou think is the bll in stble sttic equilibrium? - Position 3 is clerl not in sttic equilibrium; ou epect the bll to roll down. - Position cn be in sttic equilibrium, nd I bet ou will s tht it is stble. - Position 4 cn lso be in sttic equilibrium if ou do blncing ct, but is unstble. - Position 1 is in sttic equilibrium, nd is sid to be neutrll stble. Wht went into our decisions bout the stbilit of the sttic equilibrium solutions? If ou think bout it bit, ou will find tht: to check for stbilit of n equilibrium stte, we slightl perturb the bod bout its equilibrium stte, nd see whether it returns to tht stte. Pushing the bll bout point, we find tht it would like to come bck to due to the ction of grvit s shown; hence is stble equilibrium stte. At 4, even though we m be ble to blnce the bll just so tht it is in sttic equilibrium there, n slight perturbtion will cuse the bll to roll downhill w from point 4; hence it is n unstble equilibrium stte. At 1, slight (s long s it is not too fr tht it goes down the slope) perturbtion will simpl move it to new equilibrium point; hence 1 is neutrll stble point. Now tht we understnd tht investigting the stbilit of equilibrium sttes requires us to (conceptull) slightl perturb the bod bout the equilibrium stte, let us look t couple of more interesting emples. Tke the cse of rigid uniform br hinged t one end; once gin, onl grvit cting downwrds is importnt. It is possible, of course, to blnce the rod in either position s shown (the second one, if ou re ver skilled). But clerl, onl stble unstble one of these is stble. grvit Sridhr Krishnswm 8-13

14 Mech_Eng 36 Stress Anlsis Buckling of Columns et us now dd liner elstic spring of spring constnt K (N/m) to the bove uniform rigid br of length, nd put it in the following equilibrium stte (we will turn grvit off for simplicit): P (N) P (N) δ K(N/m) perturb to K(N/m) (grossl eggerted) The upright position cn be in sttic equilibrium, with the rod compressing bit due to the ill-pplied lod. To check for stbilit, perturb the rod bout its compressed equilibrium stte, b smll displcement δ t the top s shown. Tking moments bout the pin t the bottom (nd recognizing tht the perturbtion is smll), we find: Moment due to lod P tht tkes the sstem w from the verticl equilibrium stte is Pδ (cting nticlockwise) The restoring moment (due to the force from the spring) tht wnts to tke the sstem bck into its equilibrium position is (k δ) (cting clockwise) where the moment rm is tken s in view of the smll perturbtions ssumed. Therefore: if Pδ < (k δ) : br wnts to return to equilibrium stte; if Pδ > (k δ) : spring cnnot restore the equilibrium stte; if Pδ = (k δ) : sstem cn st in perturbed stte. Therefore: P cr = k is the lod which demrctes the cse of stble equilibrium nd unstble equilibrium - i.e, the sstem (given rod nd spring) is stble for ll lods P<P cr ; nd is not stble for higher lods. We cn imgine tht this issue of stbilit of equilibrium sttes might rise for deformble structures s well, nd indeed it does. We will see onl one cse in this course: tht of deformble rod under n ill compressive lod - lso known s column. Sridhr Krishnswm 8-14

15 Mech_Eng 36 Stress Anlsis Buckling of Columns FORCE-BAANCE APPROACH TO BUCKING OF COUMNS: Consider stright clindricl column whose ends re pinned, nd which crries compressive il lod of P (N) s shown. The sttic equilibrium solution to this P P problem is well know to ou. Solution: Column sts stright, but compresses such tht: stress σ = - P/A strin ε = -P/AE where A is the cross-sectionl re, nd E is the Young's modulus of the mteril of the rod. () Now, is this sttic equilibrium stte stble? To check this, we need to perturb the rod bout its stright compressed stte. Tht is, let us conceptull slightl perturb the rod lterll b n mount () which cn be rbitrr ecept tht it must stisf the end conditions tht (0)=0; nd ()=0 (becuse of the pins). et us see if the deformed stte (shown dotted in the skeletl figure bove) is possible equilibrium stte: if it is, then ou must see tht the stright compressed position is not stble equilibrium stte if ou hve been following wht we hve been sing so fr. Note tht the deformed stte is just bem, nd so we cn use the Bernoulli-Euler mchiner to see if it is possible to obtin perturbed stte. The il lod 'P' produces moment -P. () t n cross-section distnce '' from the top pin. In ddition, if there re n rection forces produced b the P pin t the top, this contributes moment F R. t n cross-section. So, the net cross-sectionl moment is: F R M() = -P. ()+ F R.. (We might lso hve rection moment M R if the top hd been fied end rther thn hinge). M() Using this in the Bernoulli-Euler bem eqution, we find: EI z d d = M() = P. + F R. (*) where I z is the cross-sectionl moment of inerti. It is convenient to get rid of the unknown rection force (nd rection moment if we hd n) b tking two derivtives of (*) with respect to ''. Then: EI z d 4 d 4 + P. d d = 0. (**) Sridhr Krishnswm 8-15

16 Mech_Eng 36 Stress Anlsis Buckling of Columns This is the governing differentil eqution for column buckling, nd in fct is vlid for uniform column with n end condition (hinges or clmps). If we find non-zero solution to (**), then the perturbed stte is possible, nd so the column is not stble in its verticl compressed stte. The generl solution to (**) is: () = c 1 sin. + c cos. + c 3 + c 4 where = P EI z. Cse 1: Rod pinned t both ends. Here we require tht: displcements t the pins be zero: (0) = 0; ()= 0; nd moment t the pins be zero: u '' '' (0) = 0; () = 0; where double-primes indicte twice-differentition with respect to ''. Thus: (0) = 0 c + c 4 = 0 () = 0 c 1 sin. + c cos. + c 3 + c 4 = 0 '' (0) = 0 c = 0 '' () = 0 c 1 sin. c cos. = 0 which cn be written in mtri form s: sin. cos sin. cos. 0 0 c 1 c c 3 c 4 0 = One obvious solution is tht ll the c's re zero, which mens tht () = 0, nd so the perturbed stte is not possible; which mens tht the compressed verticl stte is stble equilibrium. However, non-zero solutions re possible if the determinnt of the coefficient mtri bove were to vnish: ie., sin. = 0. This cn hppen if:. = nπ n = 1,,3... Tht is, t certin discrete levels of the compressive lod 'P' (note tht = P EI z ), it is possible for perturbed stte to be obtined, which mens tht the column is not stble in its verticl position t these lods: P n,cr = n π EI z n = 1,,3... nd the corresponding perturbed stte tht is possible is: () = c 1 sin nπ The lowest of these lods is when n=1: Sridhr Krishnswm 8-16

17 Mech_Eng 36 Stress Anlsis Buckling of Columns P Euler = π EI z which is clled the Euler buckling lod of column tht is pinned t both ends. When the compressive lod hppens to be the Euler buckling lod, the column cn buckle into sinusoidl shpe (of mplitude c 1 tht cnnot be determined b our simple theor). Remrks: (i) In the design of structures tht hve compressive il lods (for instnce trusses), it is importnt tht not onl do we tke cre tht ll components re strong enough to sustin the epected stresses, it is lso essentil tht we ensure tht buckling of n element does not occur. (ii) Note tht slender column (one whose length is lrge compred to its cross-sectionl dimensions represented b I z ) buckles more redil thn ft one. Actull, nondimensionl mesure of this slenderness is: S = A /I z where A is the cross-sectionl re. (iii) Show tht if the rod were in tension, the verticl stretched equilibrium stte is stble. (All ou need to show is tht the governing differentil eqution for column buckling (**) with 'P' replced b '-P' hs no non-zero solution.) Cse : A pinned-clmped column Net, let us consider column tht is fied t one end nd is clmped t the other. Here, the displcements t both ends re zero; the moment t the pin is zero, nd the slope t the fied end is zero. (0) = 0 c + c 4 = 0 P () = 0 c 1 sin. + c cos. + c 3 + c 4 = 0 '' (0) = 0 c = 0 ' () = 0 c 1 cos. c sin. + c 3 = 0 The onl non-trivil solution is if: sin. cos cos. sin. 1 0 = 0 which ields:. = tn.. The smllest root when this occurs is. = Thus, the criticl buckling lod for pinned-clmped column is: P cr = 0.16 EI Z higher thn for pinned-pinned column.. Note tht this is Sridhr Krishnswm 8-17

18 Mech_Eng 36 Stress Anlsis Buckling of Columns Remrks: Even though our nlsis indictes tht these columns re unstble (cn buckle) onl t discrete criticl lods, in relit n slight imperfections in the columns will cuse the columns to undergo rther lrge lterl deflections s we pproch the criticl lods. It is thus wise to design compressive structures with decent fctor of sfet to void buckling. RAYEIGH-RITZ APPROACH TO BUCKING OF COUMNS: For columns tht re not uniform, we cn use the Rleigh-Ritz technique to compute the buckling lods. The onl thing here tht mbe somewht not redil pprent to ou is the clcultion of the virtul work term. P P We need to clculte the distnce 'e' moved b the pplied lod (in the e de=ds-d d direction of the lod) when we lterll perturb the rod b (). This is shown gretl eggerted in the figures. Noting tht the neutrl is of the bem is unstretched (ber in mind tht we d ds re now tlking bout this from the compressed equilibrium stte), we find tht n elementl length ds of the bem deforms s shown, nd so de = ds d = ( d) + ( d ) d. Integrting over the entire bem: e = e = 0 =0 1 ds d 1 du d 0 d d d 1/ 1 d where the lst pproimtion is ssuming smll deformtions. The virtul work is therefore P.e, nd we cn evlute the potentil energ functionl. Emple: A free-fied column, of uniform fleurl rigidit EI z nd length. Sridhr Krishnswm 8-18

19 Mech_Eng 36 Stress Anlsis Buckling of Columns (i) Bg of displcements: () = 1 is smooth nd stisfies the gbc's tht displcement nd slope be zero t the fied end (=0). (ii) Clculte the potentil energ functionl: Π = 0 EI z d d d = 1 EI z P 3 P 0 d d (iii) Minimize the potentil energ functionl with respect to 1 : d either 1=0 (trivil solution which mens tht buckled stte not possible) P or P cr = 3EI z Remrks: which is the criticl buckling lod. (i) The ect solution using the force-blnce pproch is: P cr,ect =.4674 EI z (ii) Using better two prmeter bg: () = 1 which is better. + 3 ields P cr =.49 EI z (iii) Unfortuntel, pproimte RR solutions for the buckling lods lws over-predict, nd this is not good becuse it is not conservtive! So when, using RR for buckling lods in n rel design sitution, give ourself plent of sfet fctor; or else crr lrge libilit insurnce! Sridhr Krishnswm 8-19