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218MODULE 9. STABILITY AND BUCKLINGSolution: As we saw, the non-trivial solution requiressinrPL 0, EI33rPL nπEI33The buckling loads are then given by:Pcrn n2 π 2 EI33L2which is similar to the equation (?).And the corresponding buckling deflection modes byū2 (x1 ) A sin(nπx1)L4. Determine the lowest (Euler) buckling load Pcrn 1:π 2 EI33Pcr1 L2Solution: This is obtained for(9.20)5. Compare the Euler buckling load for a simply supported beam with the one obtainedpreviously for a cantilever beam (equation (?)).Solution: In the case of acantilever beam, we found:π 2 EI22Pcr0 4L2By comparing the two equations we see that the Euler buckling load for a simplysupported beam is 4 times higher than that for a cantilever beam.Concept Question 9.4.2. Buckling of a uniform beam clamped at both endsConsider the case of a uniform beam of length L as shown in Figure 9.4.2. The beam isclamped at both ends and loaded by a uniform axial force P at (x1 L) which acts onthe beam neutral axis. The displacement ū22 satisfies the governing equation (9.4) and thesolution is given by (9.3).1. Write the boundary conditions needed to determine the constants A, B, C and D inthe solution:rrPPū2 (x1 ) A sin(x1 ) B cos(x1 ) Cx1 DEI33EI33

9.4. BUCKLING LOADS AND SHAPES FOR DIFFERENT BEAM BOUNDARY CONDITIONS21910.80.60.4ū2 /A0.20 0.2 0.4 0.6k π/Lk 2π/L 0.8 100.10.20.30.40.5x1 /L0.60.70.80.91Figure 9.7: Deformation modes of the simple supported beam.

220MODULE 9. STABILITY AND BUCKLINGe2Pe1LFigure 9.8: Uniform beam clamped at both ends.Solution:at x1 0 and x1 L: u2 (x1 0) 0u02 (x1 0) 0and u2 (x1 L) 0u02 (x1 L) 0(9.21)2. Using these boundary conditions, determine the condition on the load P for which thebeam can be in equilibrium in a deformed configuration, (i.e. we have a non-trivialsolution ū2 6 0).Solution:for ū2 (x1 0) 0:rrPPū2 (x1 0) A sin( 0) B cos( 0) C 0 DEI33EI33 B D 0for ū02 (x1 0) 0:rrrrPPPPū02 (x1 0) A cos( 0) B sin( 0) CEIEI33EI33EI33r 33P A C 0EI33for ū2 (x1 L) 0:rū2 (x1 L) A sin(rPP L) B cos( L) C L DEI33EI33for ū02 (x1 L) 0:ū02 (x1 L) rrrrPPPPA cos( L) B sin( L) CEI33EI33EI33EI33We obtain the following system:

9.4. BUCKLING LOADS AND SHAPES FOR DIFFERENT BEAM BOUNDARY CONDITIONS221 q010 1 P 01 0 EI330 0 qq 0 sin( EIP33 L)cos( EIP33 L)L 1 0 qqqqPPPPcos( L) sin( L)10EI33EI33EI33EI33ABCD For a non trivial solution: A B6 C D0000 which requires the matrix to beq singular, i.e. its determinant must vanish. Let’s callthis matrix H and define k EIP33 .k1 0k01sin(kL)L1sin(kL)cos(kL)Ldet(H) k cos(kL) 1 0k cos(kL) k sin(kL) 1 k1cos(kL)Lsin(kL)cos(kL) k k cos(kL) 1 k sin(kL) 1k cos(kL) k sin(kL) k(1 cos(kL)) k(cos(kL) Lk sin(kL)) k sin2 (kL) k cos2 (kL) k k cos(kL) k cos(kL) Lk 2 sin(kL) k 2k 2k cos(kL) Lk 2 sin(kL) kLsin(kL) 2k 1 cos(kL) 2 kLkLkL2 kL 2k 1 [1 2 sin ( )] 2 sin( ) cos( ))2222 kLkLkLkL 4k sin( ) sin( ) cos( )2222 0,which implies the three cases:(a) k 0(b) sin( kL) 0, or2k 2nπL(9.22)

222MODULE 9. STABILITY AND BUCKLING(c) sin( kL) 2kL2cos( kL) 0, or2tan(kLkL) .22(9.23)In Case (?), we can replace k by its original expression and obtainrP2nπ EI33Lhence:Pcrn 4n2 π 2 EI33L2In this case, the displacement ū2 reads: 2nπx1 ) 1ū2 B (cos(kx1 ) 1) B cos(L In Case (?), the solution is also a series of numbers due to the periodicity of tangentialfunction. The solutions can be obtained numerically, and the first two are 8.97( 2.85π),LL15.454.92π( ),whichleadtoLLPcr k 2 EI33 80.76EI338.18π 2 EI33 238.72EI3324.19π 2 EI33( ),( ), .L2L2L2L2Set B 1 and then coefficients A, C, D can be determined by solving the reducedlinear system.The first two deformation modes from Case (?) and Case (?) are plotted in Figure9.4.2.3. Determine the Euler critical load Pcr0 and compare the expression obtained with thosefound for the simply-supported and the cantilever beam.Solution:Pcr0 4π 2 EI33L2we obtain:(Pcr0 )clamped 4 (Pcr0 )simply supported 4 (Pcr0 )cantilever

9.4. BUCKLING LOADS AND SHAPES FOR DIFFERENT BEAM BOUNDARY CONDITIONS2232k 2π/L1.5k 2.85π/Lk 4π/L1k 4.92π/Lū2 /B0.50 0.5 1 1.5 200.10.20.30.40.5x1 /L0.60.70.80.9Figure 9.9: Deformation modes of the clamped beam.1

224MODULE 9. STABILITY AND BUCKLINGe2aPbLe1Figure 9.10: Clamped beam at both ends with an intermediate support at x a.Concept Question 9.4.3. Buckling of a uniform beam clamped at both ends with an intermediate supportConsider the uniform beam of length L, clamped at both ends (Figure 9.8) loaded by a forceP at the right end (x1 L) along the beam neutral axis. An additional support is placedat the cross-section x1 a, as shown in the figure.1. The analysis is done considering the left and right regions as separate solutions andthen enforcing compatibility at the support. The transverse displacement is denotedū2 and ũ2 in the first and second region, respectively.Determine the general form of the transverse displacement ū2 and ũ2 in both regions.Solution: The differential equationsFor convenience, we introduce k 2 P/EI33 .governing the transverse displacement in both regions, 1 and 2 are the following:P 00ū 0 for 0 x1 aEI33 2P 00(IV )ũ2 ũ 0 for a x1 LEI33 2(IV )ū2 Thus, both, ū2 and ũ2 have the same following form:ū2 (x1 ) A1 sin(kx1 ) B1 cos(kx1 ) C1 x1 D1ũ2 (x1 ) A2 sin(kx1 ) B2 cos(kx1 ) C2 x1 D22. Determine the boundary conditions on the beam:conditions read for ū2 :Solution: The boundaryū02 (x1 0) 0 kA1 C1 0ū2 (x1 0) 0 B1 D1 0ū2 (x1 a) 0 A1 sin(ka) B1 cos(ka) C1 a D1 0and for ũ2 :ũ02 (x1 L) 0 kA2 cos(kL) kB2 sin(kL) C2 0ũ2 (x1 L) 0 A2 sin(kL) B2 cos(kL) C2 L D2 0ũ2 (x1 a) 0 A2 sin(ka) B2 cos(ka) C2 a D2 0

9.4. BUCKLING LOADS AND SHAPES FOR DIFFERENT BEAM BOUNDARY CONDITIONS2253. Are the previously found boundary conditions enough to compute the solution on bothsides of the additional support? If not, what other conditions must be satisfied by ū2Solution: At thisand ũ2 on both sides of the additional support?stage we are two equations short to fully compute the transverse displacement on bothside of the additional support. We are still missing continuity conditions of the slopeand the bending moments – We have already enforced continuity of displacements atthe intermediate support by applying the null displacement condition at x1 a in theprevious question. These continuity conditions read:ū02 (x1 a) ũ02 (x1 a)ū002 (x1 a) ũ002 (x1 a)(continuity of the slopes)(continuity of the bending moments)4. Apply the boundary conditions only and show that the displacements: ū2 and ũ2 canrespectively be written as:ū2 A ((cos(ka) 1)(sin(kx1 ) kx1 ) (sin(ka) ka)(cos(kx1 ) 1))ũ2 C ((cos(k(L a)) 1)(sin(k(L x1 )) k(L x1 )) (sin(k(L a)) k(L a))(cos(k(L x1 )) 1))Solution:Replacing Bi , Ci , Di by Ai , we can obtain: sin(ka) ka(cos(kx1 ) 1)ū2 A1 sin(kx1 ) kx1 cos(ka) 1 sin(k(L a)) k(L a)ũ2 A2 sin(k(L x1 )) k(L x1 ) (cos(k(L x1 )) 1)cos(k(L a)) 1Then we can define A A1 /(cos(ka) 1) and C A2 /(cos(k(L a)) 1) to simplifythe expression.5. Apply the additional conditions to both ū2 and ũ2 and derive a system of two equationsdepending on: A, C. What condition should satisfy the system of equation so thatnon-trivial solutions are found?Solution:Apply the two continuity conditions and use trigonometric idenities, we can obtain:A[2 2 cos(ka) ka sin(ka)] C[2 2 cos(kâ) kâ sin(kâ)] 0A[ sin(ka) ka cos(ka)] C[ sin(kâ) kâ cos kâ] 0where â L a. The non-trivial solution can be found when the determinant of the2-by-2 linear system is zero, i.e. when0 [2 2 cos(ka) ka sin(ka)] [sin(kâ) kâ cos kâ] [2 2 cos(kâ) kâ sin(kâ)] [sin(ka) ka cos(ka)]

226MODULE 9. STABILITY AND BUCKLING6. Let us introduce the following non-dimensional quantities u a/L and α kL, andrewrite the previously found condition.Solution: With definition u a/L,α kL, and the â L a we just defined above, we can get identities: ka αu andkâ α(1 u). The previous condition now can be written as0 [2 2 cos(αu) αu sin(αu)] [sin(α(1 u)) α(1 u) cos(α(1 u))] [2 2 cos(α(1 u)) α(1 u) sin(α(1 u))] [sin(αu) αu cos(αu)]7. Determine the location of the intermediate support (a a ) for which the bucklingload will attain a maximum, hence the best location of the intermediate support toavoid buckling.Solution:For a a , the buckling load will attain a maximum value and corresponds herein tou a /L. The previously found condition is an equation of α and u. In other word,α may be viewed as an implicit function of u. The previous equation can be writtenas follows using trigonometric identities:f (α(u), u) [2 2 cos(αu) αu sin(αu)] [sin(α(1 u)) α(1 u) cos(α(1 u))] [2 2 cos(α(1 u)) α(1 u) sin(α(1 u))] [sin(αu) αu cos(αu)] 2 α u(1 u) 2 sin(α) α cos(α) α cos(αu) cos(α(1 u)) 2 sin(α(1 u)) 2 sin(αu) 2α(1 u) cos(α(1 u)) 2αu cos(αu) 0 .Let’s derive the function f with respect to u: f dα f 0. α du uThus,dα du f α 1 f u ,assuming that f / α 6 0. The condition:dα 0,duis then equivalent to: f 0, uwhich leads to:(1 2u)α2 sin(α) α2 sin(α(1 2u)) 2α2 (1 u) sin(α(1 u)) 2α2 u sin(αu) 0 .

9.5. BUCKLING OF BEAMS WITH IMPERFECTIONS227We immediately identify that u 1/2 satisfies this equation, and therefore dα(u ) 0.du By substituting u u 1/2 into the previous equation f (α(u), u) 0, we can obtain:α 4π .In sum, the location of the intermediate support that maximizes the buckling load isin the middle of the beam. The buckling load is equal to:Pcr 9.516π 2 EI.L2Buckling of beams with imperfectionsSo far, we have assumed idealized beams with mathematically exact geometries, made ofperfectly homogeneous materials and loads perfectly aligned with the centroid axis. In reality, beams have imperfections due to the fabrication process and cannot be considered ashomogeneous or geometrically exact. In this section, we study the effect of these imperfections.Concept Question 9.5.1. Buckling of a simply supported beam with an imperfectionWe will account for any geometric imperfection in the material as an eccentricity in theapplication of the load. Consider a simply-supported beam of length L as shown in Figure9.5.1. The uniform compressive load applied at the free end has an eccentricity e.e2bbe1ePLFigure 9.11: Simply supported beam with eccentric end load1. what do you think is the main difference with the idealized buckling problem? Howdoes the influence of the eccentricity affect the analysis?Solution: The maindifference is that the load P now produces a bending moment even in the undeformedconfiguration. We will call this the primary bending moment. This is the moment thatwe would need to consider in the absence of structural instability considerations, i.e.in linear beam theory. The analysis changes in a fundamental way, as now the problemhas non-homogeneous boundary conditions.

228MODULE 9. STABILITY AND BUCKLING2. how do you think the governing equation changes with respect to the idealized bucklingSolution: It doesn’t change at all, the differential equation is still theproblem?homogeneous equation used in the idealized case, as there is no distributed transverseforce, i.e. p2 03. Write the boundary conditions needed to determine the constants A, B, C and D inthe solution:rrPPx1 ) B cos(x1 ) Cx1 Dū2 (x1 ) A sin(EI33EI33Solution:at x1 0 and x1 L: u2 (x1 0) 0M3 (x1 ) EI33 u002 (x1 0) P eand u2 (x1 L) 0M3 (x1 ) EI33 u002 (x1 L) P e4. Apply the boundary conditions and find the solution ū2 .Solution: for ū2 (x1 0) 0:rPP 0) B cos( 0) C 0 Dū2 (x1 0) A sin(EI33EI33 B D 0rfor M3 (x1 0) P e:ū002 (x1rP 0) P A sin( 0) P B cos(EI33 P B P e B erP 0)EI33hence D e.for ū2 (x1 L) 0:rū2 (x1 L) A sin(for M3 (x1 L) P e:rPP L) e cos( L) C L e 0EI33EI33rPM3 (x1 L) AP sin( L) eP cos(EI33q(1 cos( EIP33 L))q A esin( EIP33 L)rP L) P eEI33

9.5. BUCKLING OF BEAMS WITH IMPERFECTIONS229hence, C 0. we finally obtain the displacement ū2 as a function of e:q rr (1 cos( EIP L)) PP33qsin(ū2 (x1 ) ex1 ) cos(x1 ) 1 sin( EI33EI33P L)EI335. Notice that we did obtain a fully defined solution in this case. No bifurcation ofequilibrium in this case? How come? What happens to the solution as P approachesSolution: Wethe critical load?found a solution because this is a non-homogeneous problem, i.e. solving the governingequation responds to the question: how does the beam deform under the applicationof the load p0 (and how does the influece of P modify that response. This is a verydifferent question from: under what conditions (what values of P ) would this beam beable to be in equilibrium in the deformed configuration (in addition to the undeformedconfiguration)? So there is no bifurcation of equilibrium.It can be seen in the solution that when P approaches the critical load the displacementsgrow unboundedly, i.e., the beam would fail.6. Determine the relation P f (ū2 (L/2)) at the middle of the beam and plot this expression for different ratios e/LqSolution: Let’s use k EIP33 (1 cos(k L))LLsin(k ) cos(k ) 1sin(k L)22()L(1 cos(2k 2 ))LLesin(k ) cos(k ) 122sin(2k L2 )()(1 cos(2k L2 ))LLesin(k ) cos(k ) 1222 sin(k L2 ) cos(k L2 )()(1 cos(2k L2 ))Le cos(k ) 1L22 cos(k 2 )()2 sin2 (k L2 ))Le cos(k ) 122 cos(k L2 )()1e 1cos(k L2 ))Lū2 (x1 ) e2 and we have:4P EI33 2 arccos2LeLū2 ( 2 ) e!

230MODULE 9. STABILITY AND BUCKLINGso4P 2 arccos2Pcrπwith Pcr eLū2 ( 2 ) e!π 2 EI33L27. Draw the function f for several values of the ratio e/L and interpret the result.Solution: For low values of the load P : 1) an increase of the load leads to anessentially linear growth of the deflection at the center, 2) increasing the eccentricitye increases the primary moment (for a fixed P ) and thus the deflection.As the load increases and becomes a significant fraction of the critical load, the behaviordeviates from the solution of linear elasticity and the secondary moment produced bythe load P with the extra moment arm corresponding to the deflection further increasesthe deflection. When the load gets close to the critical value, the deflections growunbounded.The plots also bears the interpretation that as the imperfections disappear there is asmooth transition to the solution of the bifurcation problem.1.21P/Pcr0.80.60.4e/L 0.01e/L 0.05e/L 0.1e/L 0.2e/L 0.5e/L 1.00.200246ū2 (L/2)/L810Figure 9.12: simply supported beam with eccentric end load8. Find the distribution of the bending momentSolution: It follows directly that:qrrh 1 cos EIP LiPP33qM3 EI33 u002 P esinx1 cosx1EI33EI33sin EIP33 L

9.6. OTHER ISSUES IN BUCKLING INSTABILITY9. Interpret the result in the limits P 0, Pcr231Solution:P 0, M3 P ei.e. the primary bending moment obtained when equilibrium in the undeformed configuration is consideredP Pcr ,PL π M3 EI33i.e. as the load reaches the critical load, the moment grows unbounded.9.6Other issues in buckling instabilityConcept Question 9.6.1. We saw that beams and columns under states of strong compression buckle.1. Is this always true? If not, what other considerations come into play and when wouldSolution: Buckling occurs if the compressive force approaches thethat happen?. If the beam is long, the critical load iscritical value which in all cases scales as π 2 EIL2low and the beam buckles under fairly low loads which implies that the stresses in thematerial are low as well (poor structural efficiency).For short beams, the critical load increases quadratically with the reduction in length,which means that the likelihood of buckling decreases, whereas the stress σ11 PA canincrease with P to high values and reach material limits.2. In order to start looking at this problem, let’s write the critical load for general boundary conditions as:EIEIEIPcr cπ 2 2 π 2 π 2 02LLL( )2c {z}L0where we define c as the coefficient of fixity which depends on the boundary condition(e.g. c 1 for simply supported, c 4 for clamped-clamped, c 1/4 for cantilever,etc). L0 Lc as the equivalent length for buckling.3. In order to compare the competition between buckling and material failure by compression, evaluate give an expression for the stress in the material when the load approachesthe critical valueSolution: This is simply: ρ 2PcrI2 Eσ11 π 02 π2EALAL0 {z }ρ2

232MODULE 9. STABILITY AND BUCKLINGwhere we have defined:ρ rIAas the radius of gyration. (Interpretation?)04. Define the beam slenderness ratio as λ Lρ and plot the “buckling stress” as afunction of λ. Superimpose in your plot the material limiting stress (yielding, crushing)and define regions of beam response as a function of the slenderness ratio (buckling,crushing or squashing and transition between the two.Solution:

212 MODULE 9. STABILITY AND BUCKLING 3.Show that enforcing equilibrium of internal and external moments leads to an ODE of the type: u 00 3 (x 1) k 2 u 3 (x 1) k 2 and nd k Solution: Setting M int M ext, we obtain: EI 22 u 00