General Theories of Plate

4 General Theories of Plate 4.1 4.1.1 Bending Theory of Plates Derivation of the Plate Bending Equation Then, groups of equations! • Equilibrium • Geometry • Elasticity Mαβ ,αβ +q = 0 on S (145) καβ = −w,αβ (146) Mαβ = D [(1 − ν) καβ + ν κγγ δ αβ ] (147) Eliminating curvature καβ between Eq. (146) and (147), we obtain: Mαβ = −D [(1 − ν) w,αβ +ν w,γγ δ αβ ] (148) Substituting Eq. (148) into Eq. (145) reads: −D [(1 − ν) w,αβ +ν w,γγ δ αβ ] ,αβ +q = 0 (149) −D [(1 − ν) w,αβαβ +ν w,γγαβ δ αβ ] + q = 0 Note that the components of the Kronecker "δ αβ " tensor are constant and thus are not subjected to differentiation: ¯ ¯ ½ ¯1 0¯ 1 if α = β ¯ ¯ δ αβ = ¯ ¯ or δ αβ = (150) 01 0 if α 6= β Also, note that only these components: ¤αβ δ αβ = ¤αα (151) survive in the matrix multiplication for which α = β. Therefore, Eq. (149) now reads: −D [(1 − ν) w,αβαβ +ν w,γγαα ] + q = 0 (152) Because "γγ" are "dummy" indices, they can be replaced by any other indices, for example "ββ." −D [(1 − ν) w,αβαβ +ν w,ββαα ] + q = 0 (153) The order of differentiation does not matter: w,αβαβ = w,ααββ = w,ββαα Thus, two terms in Eq. (153) can now be added to give the plate bending equation: D w,ααββ = q for α, β = 1, 2 (154) Here, the index notation can be expended: w,ααββ = w,11ββ +w,22ββ (155) = w,1111 +w,2211 +w,1122 +w,2222 = w,1111 +2 w,1122 +w,2222 Now, letting "1 → x", "2 → y" leads: µ 4 ¶ ∂ w ∂4w ∂4w = q (x, y) D +2 2 2 + ∂x4 ∂x ∂x ∂y4 (156) Alternative notation can be: D ∇4 w = q (157) where Laplacian ∇2 w reads: ∇2 w = ∂2w ∂2w + ∂x2 ∂y 2 and bi-Laplican ∇4 w reads: ¡ ¢ ∇4 w = ∇2 ∇2 w ¢ ¢ ∂2 ¡ ∂2 ¡ = 2 ∇2 w + 2 ∇2 w ∂x ∂y µ 2 ¶ µ ¶ 2 ∂ ∂ w ∂2w ∂ 2 ∂ 2w ∂ 2w = 2 + + + ∂x ∂x2 ∂y2 ∂y 2 ∂x2 ∂y 2 4 4 4 ∂ w ∂ w ∂ w = +2 2 2 + 4 ∂x ∂x ∂x ∂y 4 4.1.2 (158) (159) Reduction to a System of Two Second Order Equations Denote D w,αα = −M (160) Then, from the equilibrium equation: [D w,αα ] ,ββ = q (161) we obtain a system of two linear partial differential equations of the second order: M,ββ = −q D w,αα = −M (162) or ( D ∂2M ∂2M 2 + ∂y 2 = −q ∂x ³ 2 ´ ∂ w ∂2w + = −M 2 2 ∂x ∂y (163) What is "M " ? Let us calculate Mαα : Mαα = M11 + M22 (164) = D [(1 − ν) κ11 + ν (κ11 + κ22 ) δ 11 ] +D [(1 − ν) κ22 + ν (κ11 + κ22 ) δ 22 ] = D [(1 + ν) (κ11 + κ22 )] = D (1 + ν) καα or Mαα = D καα = −D w,αα = M 1+ν (165) Therefore, Mαα = M (1 + ν) (166) = D καα (1 + ν) Now, moment sum reads: M = D καα (167) M = D [κxx + κyy ] (168) and in expanded notation it reads: 4.1.3 Exercise 1: Plate Solution Consider a simply supported plate. Square plate ( a× a) a a y x Boundary Condition General boundary condition reads: ¡ ¢ ¯ nn w,n = 0 on Γ M¡nn − M ¢ Vn − V̄n w = 0 on Γ M̄nn = 0 w=0 w=0 Mxx = 0 Myy = 0 at at at at (169) ⇒ Mnn = 0 on Γ w=0 on Γ (170) x = 0 and x = a , 0 ≤ y ≤ a y = 0 and y = a , 0 ≤ x ≤ a (171) x = 0 and x = a , 0 ≤ y ≤ a y = 0 and y = a , 0 ≤ x ≤ a (172) Loading Condition Assume for simplicity the sinusoidal load distribution: ³π x´ ³π y ´ q (x, y) = q0 sin sin (173) a a where q0 is a pressure intensity. Solution of Problem The solution of the form ³π x´ ³π y ´ w (x, y) = w0 sin sin a a (174) satisfy both the boundary conditions and the governing equations (see below). Plate Bending Equation Substituting Eq. (173) and (174) into the plate bend ing equation (156), one gets: ½ ∙³ ´ ¾ ³ π ´4 ³ π ´4 ¸ ³π x´ ³π y´ π 4 D w0 +2 + − q0 sin sin = 0 (175) a a a a a ½ ¾ ³ π ´4 ³π y´ ³π x´ − q0 sin sin =0 4 D w0 a a a In order to satisfy the above equation for all values of x and y, the coefficient in the bracket must vanish. This gives: q0 ³ a ´4 w0 = (176) 4D π ¢ £ ¡ ¢¤ ¡ where D = Eh3 / 12 1 − ν 2 . Bending Moments The various bending moments are given by: ∙ 2 ¸ ³π x´ ³π y ´ ³ π ´2 ∂ w ∂2w Mxx = −D + ν = D (1 + ν) w sin sin (177) 0 ∂x2 ∂y 2 a a a ∙ 2 ¸ ³π x´ ³π y ´ ³ π ´2 ∂ w ∂2w Myy = −D + ν = D (1 + ν) w sin sin 0 ∂y2 ∂x2 a a a ³ ´ ³ ´ ³ 2 ∂ w π 2 πx π y´ = −D (1 − ν) cos Mxy = −D (1 − ν) w0 cos ∂x∂y a a a Shear Components The shear components Qx and Qy are: ∂Mxx ∂Mxy + ∂x ∂y ∂Myy ∂Mxy Qy = + ∂y ∂x Qx = (178) Now, using the previously obtained bending moments, we get the shear components in the interior of the plate: ³ π ´3 ³π x´ ³π y´ Qx = 2 D w0 cos sin (179) a a a ³ π ´3 ³π x´ ³π y´ Qx = 2 D w0 sin cos a a a Effective Shear Components Next, let us computer the effective shear com ponents: ∂Mxy ∂y ∂Mxy Vy = Qy + ∂x Vx = Qx + (180) Using the previous results, we get: ³ π ´3 ³π x´ ³π y ´ w0 cos sin Vx = (3 − ν) D a a a ³ π ´3 ³π x´ ³π y ´ Vy = (3 − ν) D w0 sin cos a a a We now need to evaluate the effective shear on the boundaries: h h ¡ ¢3 i ¡ ¢3 i Vx / (3 − ν) D πa Vy / (3 − ν) D πa ¡ ¢ x=0 w0 sin ¡πay ¢ 0 π y −w0 sin a 0¡ ¢ x=a 0 w0 sin ¡πax ¢ y=0 −w0 sin πax 0 y=a Because our sign convention is: positive shear x z (181) (182) in our case, shear along the boundary is: x y From the above results, we can plot the shear distribution: x y Force Balance Integrating the effective shear along the boundary, we get: Z a Z a Z ³ π ´3 ³π y ´ Vn dxt = 4 Vx |x=0 dy = 4 (3 − ν) D w0 sin dy (183) R= a a L 0 0 Then, the reduction force due to effective shear on boundaries reads: R = 2 (3 − ν) q0 ¡ a ¢2 π Now, let us complete the total load acting on the plate: Z aZ a Z ³π x´ ³π y ´ P = q (x, y) dS = q0 sin sin dx dy a a S 0 0 (184) (185) Then, the total external load acting on the plate reads: P = 4 q0 ¡ a ¢2 π (186) Notice that R and P do not balance! We did not include the corner forces. These are given by: (187) (Fcorner )x0 ,y0 = 2 (Mxy )|x=x0 ,y=y0 Because of the symmetry, all four forces are equal. So, compute the corner force at x = y = 0, (Fcorner )0,0 : ³π x´ ³ π y ´i¯ ¯ cos ¯ a a 0,0 ³ π ´2 = − 2 D (1 − ν) w0 a (Fcorner )0,0 = 2 h cos (188) Now, the vertical force balance is satisfied: R + 4 Fcorner = P ³ π ´2 ³ a ´2 − 8 D (1 − ν) w0 = 4 q0 π a π ³ π ´2 ³ a ´2 ³ a ´2 2 (3 − ν) q0 − 2 (1 − ν) q0 = 4 q0 π a π 2 (3 − ν) q0 4.1.4 ³ a ´2 (189) (190) Exercise 2: Comparison between Plate and Beam Solution Plate Solution For a square simply supported plate under loading qplate (x, y) given by: ³π x´ ³π y ´ qplate (x, y) = (q0 )plate sin sin (191) a a we found that the plate deflection is: ³π x´ ³π y ´ wplate (x, y) = (w0 )plate sin sin (192) a a with: (q0 )plate ³ a ´4 4¡ D ¢π 2 3 1 − ν (q0 )plate ³ a ´4 = E h3 π (w0 )plate = For the plate, the total load is given by: Z aZ a ³π x´ ³π y´ Pplate = (q0 )plate sin sin dy dx a a 0 0 ³ a ´2 = 4 (q0 )plate π (193) (194) Wide Beam Solution For a wide beam under line loading given by: ³π x´ qbeam (x, y) = (q0 )beam sin a we need to compute the central deflection (w0 )beam from: 0000 E I wbeam = qbeam (x) where I = ah3 /12. we get: EI Thus, (196) Assuming the deflection wbeam (x): ³π x´ wbeam (x) = (w0 )beam sin a ³ π ´4 a (w0 )beam sin ³π x´ ³π x´ = (q0 )beam sin a a (q0 )beam ³ a ´4 EI π 12 (q0 )beam ³ a ´4 = E a h3 π Now, let us compute the total forces: Z a ³π x´ Pbeam = (q0 )beam sin dx a 0 a = 2 (q0 )beam π Comparison For both total forces to be equal, we need to have: (w0 )beam = Pplate = Pbeam ³ a ´2 a 4 (q0 )plate = 2 (q0 )beam π π a (q0 )beam = 2 (q0 )plate π With a concentrated load, the beam deflection now becomes: (q0 )beam ³ a ´4 (w0 )beam = EI π 24 (q0 )plate ³ a ´4 = π E h3 π We now can compute the ratio of central deflections: 3 (1−ν 2 ) (q0 )plate ¡ a ¢4 (w0 )plate π E h3 α= = ¡ ¢ (w ) 24 (q0 )plate a 4 0 beam π E h3 (195) π (197) (198) (199) (200) (201) (202) (203) (204) ¢ π¡ = 1 − ν 2 ' 0.36 8 The above equation means that under the same total load, a plate is three times stiffer than a wide beam. The ratio α will vary slightly depending on the load distribution (sinusoidal, uniform, concentrated load, etc.). 4.1.5 Exercise 3: Finite Difference Solution of the Plate Bending Prob lem Governing Equations read: ∇2 M = −q ∇2 w = − M D (205) or in the component notation they read: ∂2M ∂2M ∂x2 + ∂y 2 = −q ∂2w ∂2w M ∂x2 + ∂y2 = − D (206) where M is the moment sum defined by: M= Mαα = D καα 1+ν (207) Case of Simply Supported Plate The boundary condition of a simple sup ported plate reads: w=0 on Γ Mnn = 0 on Γ a a (208) x n y For sides parallel to x-axis (thick lines), one gets: Mnn = Myy = 0 dw =0 → dx From the general constitutive equations, w=0 → d2 w =0 dx2 (209) → Myy = D [κyy + ν κxx ] 0 = D [κyy + ν · 0] κxx = 0 (210) (211) → κyy = 0 Therefore, M = D [καα + κββ ] = D [0 + 0] = 0 (212) Similar derivation can be performed for two edges parallel to y-axis. Then, M = 0. It can be concluded that for a simply supported plate the following boundary conditions hold: ( 2 2 ∂ M + ∂∂yM in S 2 = −q ∂x2 (213) M =0 on Γ ( 2 2 ∂ w + ∂∂yw2 = − M in S D ∂x2 (214) w=0 on Γ Therefore, the above two boundary value problems are uncoupled. The Finite Difference Technique An approximation to the first and second derivatives. z zn+1 zn h x xn xn+1 m− 1 h n−1 n, m n+1 x m+1 y ¯ dz ¯¯backward zn − zn−1 ' dx ¯n h ¯f orward dz ¯¯ zn+1 − zn ' dx ¯ h n+1 (215) ¯ ∙ ¸ d2 z ¯¯ d dz = dx2 ¯n dx dx ¡ dz ¢ ¡ dz ¢ dx n+1 − dx n = h zn+1 −zn − zn −zhn−1 h = h zn+1 − 2 zn + zn−1 = h2 ¯ d2 z ¯¯ zm+1 − 2 zm + zm−1 = ¯ 2 dy m h2 ∂2z ∂2z + 2 2 ∂x ∂y 1 = 2 (zn+1 − 2 zn + zn−1 + zm+1 − 2 zm + zm−1 ) h ∇2 z = (216) (217) (218) Top h h h Left h 0 Right Bottom 1 (zT + zB + zL + zR − 4 z0 ) (219) h2 Divide the plate into sixteen identical squares and distinguish six representative nodes: three in the interior and three at the boundary. Because of symmetry, it is enough to consider only an eighth of the plate. ∇2 z = 6 5 4 3 2 a 1 h= a 4 a Four axes of symmetry Determination of Moment For each interior point (1, 2, 3), we write equation ∇2 M = −q. For each boundary point (4, 5, 6), we write boundary condition M = 0 (uniform pressure). q a2 16 Point 1: 4 M2 − 4 M1 = − (220) Point 2: M1 + M4 + 2 M3 − 4 M2 = − Point 3: 2 M5 + 2 M2 − 4 M3 = − Point 4: M4 = 0 Point 5: M5 = 0 Point 6: M6 = 0 q a2 16 q a2 16 Substituting three last equations of Eq. (220) into the first three equations of Eq. (220), one ends up with the following system of linear algebraic equations: ⎧ q a2 ⎪ ⎨ 4 M2 − 4 M1 = − 16 2 (221) M1 + 2 M3 − 4 M2 = − q16a ⎪ ⎩ q a2 2 M2 − 4 M3 = − 16 whose solution is: 9 q a2 128 7 M2 = q a2 128 11 M3 = q a2 256 M1 = (222) At the plate center, Mxx = Myy so that: M= Mxx + Myy 2 Mxx = 1+ν 1+ν Mxx = 1 (1 + ν) M 2 (223) (224) At the center, M = M1 , thus, 1 (1 + ν) M1 2 1 9 q a2 = (1 + ν) 2 128 = 0.0457 q a2 Mxx = (225) This is 4.6% less than the exact solution which is (Mxx )exact = 0.0479 q a2 from the text book. Determination of Deflection For each interior point (1, 2, 3), we write equation ∇2 w = −M/D. For each boundary point (4, 5, 6), we write boundary condition w = 0. µ ¶ 2 9 qa2 a M1 a2 =− (226) Point 1: 4 w2 − 4 w1 = − D 16 128 D 16 µ ¶ 2 M2 a2 7 q a2 a =− Point 2: w1 + w4 + 2 w3 − 4 w2 = − D 16 128 D 16 µ ¶ 2 2 2 M3 a 11 q a a Point 3: 2 w5 + 2 w2 − 4 w3 = − =− D 16 256 D 16 Point 4: w4 = 0 Point 5: w5 = 0 Point 6: w6 = 0 Similarly, ⎧ 9 qa4 ⎪ ⎨ 4 w2 − 4 w1 = − 2048 D 7 w1 + 2 w3 − 4 w2 = − 2048 ⎪ 4 ⎩ q a 11 2 w2 − 4 w3 = − 4096 D q a4 D (227) Finally, the finite difference solution is: 33 q a4 q a4 = 0.00403 8196 D D 3 q a4 q a4 w2 = = 0.00293 1024 D D 35 q a4 q a4 w3 = = 0.00214 16384 D D w1 = (228) On the other hand, the exact deflection of the center point is: q a4 D (w1 )exact = 0.00416 (229) Thus, the error of the finite different solution is 3.1%. 4.2 4.2.1 Membrane Theory of Plates Plate Membrane Equation Assume that the bending rigidity is zero, D = 0. membrane. The plate becomes now a • Equilibrium of in-plane equation Nαβ ,α = 0 on S (230) Equilibrium of out-of-plane equation Nαβ w,αβ +q = 0 on S (231) • Strain-displacement relation ε◦αβ = 1 1 (uα ,β +uα ,β ) + w,α w,β 2 2 (232) • Constitutive equation £ ¤ Nαβ = C (1 − ν) ε◦αβ + ν ε◦γγ δ αβ ¢ ¡ where C = Eh/ 1 − ν 2 . This is a non-linear system of equation which is difficult to solve. corresponding system of equation for the plate bending was linear. (233) Note that 4.2.2 Plate Equation for the Circular Membrane Cylindrical coordinate system is composed of ur , uθ , uz = w. • Equilibrium of in-plane equation r ∂Nrr + Nrr − Nθθ = 0 ∂r on S (234) on S (235) Equilibrium of out-of-plane equation ¸ ∙ ∂ ∂w r +r q =0 Nrr ∂r ∂r • Strain-displacement relation ¶2 (236) Nrr = C [λrr + ν λθθ ] (237) ∂ur 1 λrr = + ∂r 2 ur λθθ = r µ ∂w ∂r • Constitutive equation Nθθ = C [λθθ + ν λrr ] 4.2.3 ¢ ¡ where C = Eh/ 1 − ν 2 . Example: Approximation Solution for the Clamped Membrane Consider a circular plate with the clamped support. q r a w0 w(r ) Membrane Solution From the symmetry and clamped boundary condition, the radial displacement ur reads: ur (r = 0) = 0 (238) ur (r = a) = 0 Thus, as a first approximation, it is appropriate to assume: ur ≡ 0 for 0 ≤ r ≤ a (239) Then, the hoop strain vanishes: εθθ = 0 Now, the radial force and the radial strain component become: Nrr = C εrr εrr 1 = 2 µ ∂w ∂r ¶2 (240) (241) With the assumption ur = 0, the in-plane equilibrium equation can not be satisfied. Consider out-of-plane equilibrium equation only. Substituting Eq. (240) and (241) into Eq. (235), one gets: " µ # ¶ ∂ C ∂w 2 ∂w r = − r q on S (242) ∂r 2 ∂r ∂r Integrating both sides once with respect to r reads: C r 2 µ ∂w ∂r ¶3 =− r2 q + c1 2 (243) At the center of the membrane, the slope should be zero. Thus, one gets: ∂w =0 at r = 0 ∂r ⇒ c1 = 0 (244) Then, Eq. (243) can be written: r ∂w qr =−3 ∂r C Integrating the above equation again reads: r 3 3 q 4/3 r + c2 w=− 4 C (245) (246) The integration constant c2 can be determined from the zero deflection condition at the clamped edge: w=0 at r = a r 3 3 q 4/3 a ⇒ c2 = 4 C (247) ¡ ¢ Recalling the definition of the axial rigidity C = Eh/ 1 − ν 2 , Eq. (246) can be put into a final form: r ∙ ³ r ´4/3 ¸ w 3 3 (1 − ν 2 ) q a = 1− (248) a 4 Eh a r ∙ ³ r ´4/3 ¸ qa 3 ' 0.73 1− Eh a In particular, the central deflection w (r = 0) = w0 is related to the load intensity by: r w0 qa = 0.73 3 (249) a Eh Bending Solution It is interesting to compare the bending and membrane re sponse of the clamped circular plate. From the page 55 of Theory of Plates and Shells (2nd Ed.) by Timoshenko and Woinowsky-Krieger, the central deflection of the plate is linearly related to the loading intensity: w0 q a3 = a 64¡D ¢ 3 1 − ν 2 q ³ a ´3 = 16 E h q ³ a ´3 ' 0.17 E h (250) Assume that a/h = 10, then Eq. (250) yields: w0 qa = 17 a Eh (251) Comparison A comparison of the bending and membrane solution is shown in the next figure. It is seen that a transition from the bending to membrane response occurs at w0 /a = 0.15 which corresponds to w0 = 1.5 h. When the plate deflection reach approximately plate thickness, the membrane action takes over the bending action in a clamped plate. If the plate is not restrained from axial motion, then the assumption ur = 0 is no longer valid, and a separate solution must be developed. 4.3 4.3.1 Buckling Theory of Plates General Equation of Plate Buckling • Equilibrium of in-plane equation Nαβ ,α = 0 on S (252) Equilibrium of out-of-plane equation Mαβ ,αβ +Nαβ w,αβ +q = 0 on S (253) • Strain-displacement relation 1 1 (uα ,β +uα ,β ) + w,α w,β 2 2 καβ = − w,αβ ε◦αβ = • Constitutive equation for axial force and axial strain £ ¤ Nαβ = C (1 − ν) ε◦αβ + ν ε◦γγ δ αβ ¡ ¢ where C = Eh/ 1 − ν 2 , and another one for moment and curvature Mαβ = D [(1 − ν) καβ + ν κγγ δ αβ ] £ ¡ ¢¤ where D = Eh3 / 12 1 − ν 2 . (254) (255) (256) By combining Eq. (254) and (256), one gets: Mαβ ,αβ = −D w,ααββ (257) Substituting Eq. (257) into Eq. (253) leads: −D w,ααββ +Nαβ w,αβ +q = 0 on S (258) The buckling problem is specified by: q≡0 (259) Now, changing signs leads the general out-of-plane equation for the buckling of the plates: D w,ααββ −Nαβ w,αβ = 0 (260) where the second term in the left hand is non-linear due to Nαβ which should be obtained from: Nαβ ,α = 0 4.3.2 (261) Linearized Buckling Equation of Rectangular Plates The nonlinear buckling equation can be separated into two linear equations: one for in-plane equation for Nαβ and another one for w. a x Px Px b ux y Px k 2 Pc Post-buckling (δw ≠ 0) Pre-buckling (δw = 0) k ux Pre-Buckling Problem Recall that: µ ¶ 1 ∂uα ∂uβ 1 ∂w ∂w ◦ εαβ = + + 2 ∂xβ ∂xα 2 ∂xα ∂xβ (262) ¤ Eh £ ◦ ◦ (1 − ν) ε + ν ε δ (263) αβ γγ αβ 1 − ν2 In the pre-buckling problem, the linear equilibrium equations are obtained by omitting the nonlinear terms in the governing equations Eq. (260) and (261). The resulting equations are now: D w,ααββ = 0 Nαβ = Nαβ ,β = 0 For the pre-buckling trajectory, δw = 0, one gets the equilibrium equation: Nαβ ,β = 0 (264) where Nαβ = ¤ Eh £ (1 − ν) ε◦αβ + ν ε◦γγ δ αβ 2 1−ν µ ¶ 1 ∂uα ∂uβ ◦ εαβ = + 2 ∂xβ ∂xα (265) (266) and boundary condition: ¢ ¡ Nnn − N̄nn δun = 0 on Γ (267) Here, it is assumed that the unknown membrane force tensor Nαβ is equal to ◦ : the similar quantity known from the pre-buckling solution Nαβ ◦ Nαβ = −Nαβ where the compressive pre-buckling membrane force are defined as positive. Post-Buckling Problem Now, the governing equation for buckling of plates reads: ◦ w, D ∇4 w + Nαβ αβ = q + Boundary Conditions (268) ◦ is defined as: where membrane force tensor in the pre-buckling solution Nαβ ◦ Nαβ = λ Ñαβ ¯ ¯ ¯ Ñxx Ñxy ¯ ¯ = λ ¯¯ Ñyx Ñyy ¯ (269) where N̄αβ is the known direction from the pre-buckling analysis, and η is unknown load amplitude. Now, the nonlinear buckling equation becomes a linear eigenvalue problem: D ∇4 w + λ Ñαβ w,αβ = 0 (270) where λ ≥ 0 is eigenvalues, and w is eigenfunctions. 4.3.3 Analysis of Rectangular Plates Buckling Simply Supported Plate under In-Plane Compressive Loading Consider a plate simply supported on four edges. The plate is subjected to an in-plane compressive load Px uniformly distributed along the edges x = [0, a]. a x Px Px b y h From equilibrium equations, one gets: ¯ ◦ ¯ ◦ ¯ ¯ Nxx Nxy ◦ ¯ ¯ Nαβ = ¯ ◦ ◦ ¯ Nyx Nyy ¯ ¯ ¯ Ñxx Ñxy ¯ ¯ ¯ =λ ¯ Ñyx Ñyy ¯ ¯ ¯ Px ¯¯ 1 0 ¯¯ = b ¯0 0¯ (271) Introducing Eq. (271) into Eq. (268) leads: D ∇4 w + Px w,xx = 0 b (272) Boundary condition for this simply supported plate are written as: w=0 on Γ Mnn = 0 on Γ (273) where the moment components read: Mxx = −D (w,xx +ν w,yy ) = 0 (274) Myy = −D (w,yy +ν w,xx ) = 0 Thus, one gets: w = w,xx = 0 on x = [0, a] w = w,yy = 0 on x = [0, b] (275) Equation (272) is a constant-coefficient equation, and a solution of the following form: ³m π x´ ³n π y ´ sin for m, n = 1, 2 (276) w = c1 sin a b satisfies both the differential equation and the boundary conditions. Introduction into Eq. (272) gives: ∙³ ³ m π ´2 ³ n π ´2 ³ n π ´4 ¸ P ³ m π ´2 m π ´4 x D +2 + − =0 (277) a a b b b a ⇒ ³ π a ´2 ∙³ m ´2 ³ n ´2 ¸2 Px =D + b m a b (278) where for the discrete values of Px Eq. (272) has nontrivial solutions. The critical load can be determined by the smallest eigenvalue, i.e. n = 1 for all values of a: " # ³ π a ´2 ³ m ´2 µ 1 ¶2 Px =D + (279) b m a b µ ¶ " µ ¶2 # 1 π 2 D a b 2 ³ m ´2 + = 2 b m a b µ ¶ a 2 π2 D m b = 2 + b a mb Now, the critical load (Px )cr can be written as: (Px )cr = kc where kc = µ π2 D b mb a + a mb ¶2 (280) (281) where coefficient kc is a function of aspect ratio a/b and wavelength parameter m. 6 2 12 For a given a/b, m may be chosen to yield the smallest eigenvalue. In order to minimize kc in Eq. (281), treating m as a continuous variable produces: µ ¶µ ¶ mb a b a ∂kc − =2 + =0 (282) a mb a b m2 ∂m where the first bracket can not be zero, so the second bracket should be zero: ⇒ b a 1 − =0 a b m2 (283) a b kc = 4 (284) Now, one gets: m= Here, this is valid when a/b is integer and when considering a very long plates. Transition from m to m+1 half-waves occurs when the two corresponding curves have equal ordinates, i.e. from Eq. (281): kc |m = kc |m+1 ⇒ ⇒ (m + 1) b a mb a + = + a mb a (m + 1) b a p = m (m + 1) b (285) (286) p a = m (m + 1) b Example 1 For m = 1, a/b = (287) √ 2 Example 2 For a very large m, i.e. a very long plate, a/b ' m. Thus, kc = 4 is now independent of m. A very long plate buckles in half-waves, whose lengths approach the width of the plate: ³π x´ ³n π y ´ w = c1 sin sin b b Thus, the buckled plate subdivides approximately into squares. Various Boundary Conditions of Plate under In-Plane Compressive Load ing The critical buckling load reads: (Px )cr = kc π2 D b 16 14 12 10 kc 8 6 c A ss C c ss free E ss B free D c c ss 4 Loaded edges clamped. Loaded edges simply supported. 2 0 0 1 2 a b 3 4 5 Figure by MIT OCW. Influence of boundary conditions on the buckling coefficients of plates subjected to in-plane compressive loading Various Boundary Conditions of Plate under In-Plane Shear Loading The critical buckling load per unit length reads: (Nxy )cr = kc π2 D b2 where the dimension of Nxy is [N/m]. 15 2 rcr = � D kc b2h 13 Clamped Edges 11 kc 9 Simply Supported Edges 7 5 3 0 1 2 a b 3 4 5 Critical values of shear stress for plates subjected to in-plane loading. Figure by MIT OCW. Limiting Case: Wide Plates Consider a wide plate for which a/b ¿ 1. From the diagram, we see that if a/b < 1, then m is set to be equal to unity, i.e. just one wavelength in the x-direction. a x b y The buckling formula thus becomes: π2 D (Nx )cr = kc |m=1 b2 µ ¶ π2 D b a 2 = 2 + b a b µ ¶ ³ π 2 D a ´2 b a 2 = 2 + a b a b ∙ ¸ ³ π2 D a ´2 2 = 2 1+ a b (288) If a/b ¿ 1, then the second term in the bracket can be neglected so that the buckling load per unit length becomes: (Nx )cr = π2 D a2 which is called Sezawa’s formula for wide plates. Example 3 Here, relative merits of stiffening a large panel are investigated in the longitudinal or in the transverse direction. It is assumed that the stiffeners provide for a simply supported boundary conditions. Consider the case of longitudinal stiffeners. L σcr s From the von Karman formula, the buckling load per unit length for each divided part reads: 4π 2 D (Nx )cr = s2 Now, the buckling stress can be calculated: (Nx )cr h 2 4π D = 2 s h (σ cr )longitudinal = Consider the case of transverse stiffeners. L s σcr From the Sezawa’s formula, the buckling load per unit length along the loaded edges reads: π2 D (Nx )cr = 2 s Now, the buckling stress can be calculated: (Nx )cr h π2D = 2 s h (σ cr )transverse = Thus, it is concluded that (σ cr )longitudinal = 4 (σ cr )transverse This shows advantages of longitudinal stiffeners over transverse stiffeners. 4.3.4 Derivation of Raleigh-Ritz Quotient Recall the total potential energy of system and other corresponding definitions: Π = (Ub − Vb ) + (Um − Vm ) (289) where each term for buckling problems will be discussed in the following. Term Relating to Plate Bending Response In the buckling problem, the work done by external load causing bending response considered as zero: Vb = 0 (290) The bending energy can be expressed: Z 1 Ub = Mαβ καβ dS (291) 2 S Z D = [(1 − ν) καβ + ν κγγ δ αβ ] καβ dS 2 S Z h i D = (1 − ν) καβ καβ + ν (κγγ )2 dS 2 S Z h i D = (1 − ν) (κ11 κ11 + κ12 κ12 + κ21 κ21 + κ22 κ22 ) + ν (κ11 + κ22 )2 dS 2 S Z n h io D = (κ11 + κ22 )2 − 2 (1 − ν) κ11 κ22 − (κ12 )2 dS 2 S Here, the term in the square bracket is called Gaussian curvature: κ11 κ22 − (κ12 )2 = κI κII (292) where κI and κII are the principal curvatures. For plates with straight edges, Gaussian curvature vanishes, so one gets: Z D (κ11 + κ22 )2 dS (293) Ub = 2 S The integrand of the above equation can be written in terms of the transverse displacement: ¡ ¢2 (κ11 + κ22 )2 = −∇2 w 2 (294) 2 =∇ w ∇ w Now, the bending energy reads: D Ub = 2 Z S ∇2 w ∇2 w dS (295) Term Relating to Plate Membrane Response The work done by external load causing membrane response reads: Z Z N̄nn un dl + N̄tn ut dl (296) Vm = Γ Γ ◦ = λN ˜αβ is determined from the In the buckling problem, the axial force Nαβ pre-buckling solution and is considered as constant, so the membrane energy reads: Z Um = −λ Ñαβ ε◦αβ dS (297) ∙ ¸ ZS 1 1 = −λ Ñαβ (uα ,β +uα ,β ) + w,α w,β dS 2 2 ZS Z λ = −λ Ñαβ uα ,β dS − Ñαβ w,α w,β dS 2 S S Here, the first term can be extended in a similar way shown in Eq. (132): Z Z ³ ´ −λ Ñαβ uα ,β dS = −λ Ñnn un + Ñtn ut dl S ZΓ −λ Ñαβ ,β uα dS ZS ³ ´ = −λ Ñnn un + Ñtn ut dl (298) Γ where the in-plane equilibrium is applied: −λ Ñαβ ,β = 0 Now, the membrane energy can be expressed: Z ³ Z ´ λ Um = −λ Ñnn un + Ñtn ut dl − Ñαβ w,α w,β dS 2 S Γ Thus, the term relating to membrane response can be summarized: Z ³ Z ´ λ Um − Vm = −λ Ñαβ w,α w,β dS Ñnn un + Ñtn ut dl − 2 S Z Γ Z ¯nn un dl − N ¯tn ut dl − N Γ Γ Z ³ Z ³ ´ ´ = −λ Ñnn − N̄nn un dl − −λ Ñtn − N̄tn ut dl Γ Γ Z λ − Ñαβ w,α w,β dS 2 S Z λ =− Ñαβ w,α w,β 2 S where the boundary conditions on Γ are applied. (299) (300) (301) Total Potential Energy and Its Variations Now, one gets the total potential energy: Π = (Ub − Vb ) + (Um − Vm ) Z Z λ D 2 2 = ∇ w ∇ w dS − Ñαβ w,α w,β dS 2 S 2 S The first variation of the potential energy can be obtained: Z £ ¡ 2 ¢ 2 ¡ ¢¤ D δΠ = δ ∇ w ∇ w + ∇2 w δ ∇2 w dS 2 S Z λ − Ñαβ (δw,α w,β +w,α δw,β ) dS 2 S Z Z ¡ 2 ¢ 2 = D ∇ w δ ∇ w dS − λ Ñαβ w,α δw,β dS ZS Z S = D ∇2 w ∇2 δw dS − λ Ñαβ w,α δw,β dS S (302) (303) S where Ñαβ is considered as constant under the variation. Similarly, the second variation of the potential energy reads: Z Z ¡ ¢ δ 2 Π = D δ ∇2 w ∇2 δw dS − λ Ñαβ δw,α δw,β dS (304) S S Z Z 2 2 D ∇ δw ∇ δw dS − λ Ñαβ δw,α δw,β dS S S Raleigh-Ritz Quotient Application of the Trefftz condition for invertability, δ 2 Π = 0, determines the load intensity: R D S ∇2 δw ∇2 δw dS λ= R (305) S Ñαβ δw,α δw,β dS Here, choose a trial function for w: w=Aφ (306) where A is the undetermined magnitude, and φ = φ̂ (x, y) is a normalized shape function. Then, the variation of the trial function reads: δw = δA φ Now, the load intensity reads: D λ= R R S ∇2 (δA φ) ∇2 (δA φ) dS Ñαβ (δA φ) ,α (δA φ) ,β dS R D δA ∇2 φ δA ∇2 φ dS =R S S Ñαβ δA φ,α δA φ,β dS R D S ∇2 φ ∇2 φ dS =R S Ñαβ φ,α φ,β dS S (307) (308) The Raleigh-Ritz quotient is defined as: D λ= R S R S ∇2 φ ∇2 φ dS Ñ αβ φ,α φ,β dS (309) Example 4 As a special case, consider 1-D case: ¯ ¯ ¯1 0¯ Ñαβ = ¯¯ ¯¯ 00 then, the Raleigh-Ritz quotient becomes: R D S ∇2 φ ∇2 φ dS λ= R 2 S (φ,x ) dS Example 5 Similarly, consider 2-D compression case: ¯ ¯ ¯1 0¯ Ñαβ = δ αβ = ¯¯ ¯¯ 01 then, the Raleigh-Ritz quotient becomes: R D S ∇2 φ ∇2 φ dS λ= R S φ,α φ,α dS R D S ∇2 φ ∇2 φ dS i =R h 2 2 (φ, ) + (φ, ) dS x y S 4.3.5 Ultimate Strength of Plates The onset of buckling stress σ cr does not necessarily means the total collapse of the plate. Usually, there is redistribution of stresses, and the plate takes additional load until the ultimate strength σ u is reached. Von Karman Analysis of the Effective Width For a simply supported plate, the buckling load is: 4π 2 D b 4π 2 E h3 = 12 (1 − ν 2 ) b Pcr = and the corresponding buckling stress is: Pcr σ cr = bh µ ¶2 4π 2 E h = 2 12 (1 − ν ) b µ ¶2 π2 E h = 2 3 (1 − ν ) b (310) (311) The normalization of buckling stress by the yield stress reads: E σ cr π2 = 2 σy 3 (1 − ν ) σ y = (1.9)2 σ y ¡ b ¢2 E µ ¶2 h b (312) ¶2 (313) h σ cr = σy µ 1.9 β where β is a non-dimensional parameter defined by: r σy b β= E h (314) The relation between the normalized buckling stress versus β is plotted in the next figure. 1.9 On further loading the plate beyond σ cr , a greater proportion of the load is taken by the regions of the plate near the edge. Von Karman assumed that these edge regions, each of the width bef f /2, carry the stress up to the yield while the center is stress free. σ (y) σ x b y Before Buckling After Buckling beff 2 σy beff 2 σy σ (y) b Actual Stress von Karman Model The edge zones are at yield, i.e. σ cr /σ y = 1, but the width of the effective portion of the plate is unknown: σ cr = σy from which one obtains: σy E (1.9)2 ³ ´2 = 1 bef f h s bef f = 1.9 h E σy Taking, for example, E/σ y = 900 for mild steel, one gets: √ bef f = 1.9 900 = 57h (315) (316) (317) This is somehow high, but there is not much difference from the empirically deter mined values of bef f = 40h ∼ 50h. The total force at the point of ultimate load is: P = σ y bef f h (318) Now, the average ultimate stress can be calculated: σu = P bh (319) bef f b hp = 1.9 Eσ y b = σy The average ultimate stress can be normalized by the yield stress: p h Eσ y σu = 1.9 b σy σy σu 1.9 = σy β (320) (321) The average ultimate stress is plotted with respect to β in the next figure. 1.9 Comparison of the ultimate and buckling load solution is shown in the next figure. 1.9 Thin plates buckle before reaching ultimate load. Thick plates yield before buckle. Under the uniaxial loading, the relation between an applied load and the cor responding displacement is schematically shown all the way to collapse in the next figure. Px x Pu Pcr post-buckling Px Px collapse or crash ux pre-buckling y ux Empirical Formulas • Foulkner correction σu 2 1 = − 2 σy β β (322) • Gerard (Handbook of elastic stability) σu = 0.56 σy à gh2 A s E σy !0.85 (323) where g is the sum of the number of cuts and the number of flanges after the cuts, A is the cross sectional area A = bh, and the coefficients 0.56 and 0.85 are empirical constants. Example 6 Consider a plate which has one cut and two flanges. b h Then, g =1+2 =3 Now, s !0.85 3h2 E bh σ y à s !0.85 h E = 1.42 b σy σu = 0.56 σy = à 1.42 β 0.85 Modifications in Codes In the original von Karman formula, the effective width ratio reads: r bef f σ cr = (324) b σy In the ANSI specification, imperfection is considered: µ ¶ r r bef f σ cr σ cr = 1 − 0.218 b σy σy (325) In UK specification, the effective width ratio is defined as: " µr ¶4 #−0.2 bef f σ cr = 1 + 14 − 0.35 b σy (326) eff y A comparison of theoretical (von Karman) and experimental predictions for the effective width in compressed steel plates 4.3.6 Plastic Buckling of Plates Stocky plates with low b/h ratio will yield before buckling at the point B. After additional load, the plate will deform plastically on the path BC until conditions are met for the plate to buckle in the plastic range. C B A 1.9 Stowell’s Theory for the Buckling Strain Stowell developed the theory of plastic buckling for simply-supported square plates loaded in one direction. b x b thickness h y The critical buckling strain εcr was derived by him in the form: µ ¶ i p π2 h 2 h εcr = 2 + 1 + 3 (Et /Es ) 9 b (327) where the tangent modulus Et and the secant modulus Es are defined by: Et = dσ dε ; Es = σ ε (328) σ Et σy Es ε For the materials obeying the power hardening low: µ ¶n ε σ = σr εr (329) where σ r and εr are the reference stress and strain. Now, the tangent and secant modulus are: µ ¶ dσ σ r ε n−1 Et = =n (330) dε εr εr µ ¶ σ σ r ε n−1 Es = = ε εr εr Substituting these expression back into the buckling equation (327), one gets: εcr π2 = 9 µ ¶2 √ ¢ h ¡ 2 + 1 + 3n b (331) The exponent n varies usually between n = 0 (perfectly ¡ √plastic ¢material) and n = 1 (elastic material). This makes the coefficient 2 + 1 + 3n vary in the range 3 ∼ 4. For a realistic value of n = 0.3, the buckling strain becomes: εcr µ ¶2 h = 3.7 b (332) Having determined εcr , the corresponding buckling stress is calculated from the power law. Approximate Solution for the Buckling Strain stress relation: Consider an elastic plane E (εxx + νεyy ) 1 − ν2 E σ yy = (εyy + νεxx ) 1 − ν2 σ xx = (333) Using the solution for the pre-buckling state, i.e. σ xx = σ cr and σ yy = 0 leads: E (εxx + νεyy ) 1 − ν2 E 0= (εyy + νεxx ) 1 − ν2 σ cr = (334) from which one gets: εyy = −νεxx (335) σ cr = Eεxx = Eεcr The critical elastic buckling stress is: σ cr π2 E = kc 12 (1 − ν 2 ) µ ¶2 h b (336) So, the corresponding critical elastic buckling strain read: εcr π2 = kc 12 (1 − ν 2 ) µ ¶2 h b (337) Equation (337) is more general than a similar expression Eq. (327) given by the Stowell theory because it applies to all type of boundary conditions. At the same time, Stowell’s equation was derived only for the simply supported boundary conditions. In particular, for kc = 4, Eq. (337) predicts: εcr µ ¶2 h = 3.6 b (338) which should be compared with the coefficient 3.7 of Eq. (332) in the Stowell’s theory. For a plastic material or very high hardening exponent, the prediction of both method are much closer. 4.3.7 Exercise 1: Effect of In-Plane Boundary Conditions, δw = 0 No Constraint in In-Plane Displacement Consider no constraint in in-plane ◦ = 0. displacement in y-direction, Nyy expansion due to Poison's effect uy a x ND ND a ux y The membrane force tensor reads: ◦ Nαβ ¯ ◦ ¯ ¯N 0¯ ¯ = ¯¯ 0 0¯ (339) From the constitutive equation, one gets: ¤ Eh £ ◦ ◦ εyy + ν ε◦xx = 0 = −Nyy 2 1−ν =⇒ ε◦yy = −ν ε◦xx Nyy = (340) By applying the geometric equation between strain and the displacement and con sidering δw = 0, here, one gets the relation between ux and uy : ∂uy ∂ux = −ν ∂y ∂x (341) Integrating both sides over the plate length leads: Z a Z a ∂uy ∂ux dy = −ν dx ∂y 0 0 ∂x uy = −ν ux (342) Constraint in In-Plane Displacement Consider a plate fully constrained in the y-direction, uy = 0. D N yy = νN D a x D N xx = ND ND ND a ux y Consequently, one also gets: ε◦yy = ∂uy =0 ∂y (343) Under the uniform compression, ε◦xx reads: ε◦xx = ∂ux ux = ∂x a (344) From the constitutive relation, the membrane forces reads: ¤ Eh £ ◦ Eh ux ◦ εxx + ν ε◦yy = = −Nxx 2 1−ν 1 − ν2 a ¤ Eh £ ◦ ux Eh ◦ Nyy = εyy + ν ε◦xx = ν = −Nyy 2 2 1−ν 1−ν a Nxx = Finally, the membrane force tensor reads: ¯ ◦ ¯ ¯ Nxx 0 ¯ ◦ ¯ Nαβ = ¯¯ ◦ ¯=λ 0 Nyy where λ= Eh ux 1 − ν2 a ¯ ¯ ¯1 0 ¯ ¯ ¯ ¯0 ν ¯ (345) (346) (347) 4.3.8 Exercise 2: Raleigh-Ritz Quotient for Simply Supported Square Plate under Uniaxial Loading Consider a simply supported square plate subjected to uniform compressive load in the x-direction. a ND x ND a y Then, the membrane force tensor reads: ¯ ¯ ¯ ¯ ◦ ◦ ¯1 0¯ Nαβ = N ¯ ¯ = λ 00 ¯ ¯ ¯1 0¯ ¯ ¯ ¯0 0¯ (348) The plate will deform into a dish, so for the trial function take the following: ³π x´ ³π y ´ φ = φ̂ (x, y) = sin sin (349) a a Now, in order to obtain the load intensity, we first calculate φ,x and ∇2 φ : ³π x´ ³π y´ π φ,x = cos sin a ³ π ax ´ ³ πay ´ π φ,y = sin cos a a a ³π x´ ³π y ´ sin a a a ³ π ´2 ³π x´ ³π y ´ φ,yy = − sin sin a a a φ,xx = − ³ π ´2 sin ∇2 φ = φ,xx +φ,yy ³ π ´2 ³π x´ ³π y ´ = −2 sin sin a a a ³ π ´4 ³π x´ ³π y ´ ∇2 φ ∇2 φ = 4 sin2 sin2 a a a (350) (351) (352) (353) Now, the Raleigh-Ritz quotient is calculated: R D S ∇2 φ ∇2 φ dS ◦ N =λ = R 2 S (φ,x ) dS ¡ ¢4 R ¡ ¢ ¡ ¢ 4D πa sin2 πax sin2 πay dS S ¡ ¢ ¡ ¢ = ¡ ¢2 R 2 π x sin2 π y dS π a a S cos a ³ π ´2 = 4D a ◦ Ncr = λ = 4D µ π a ¶2 (354) (355) This is the classical buckling solution, and it is exact because of the right guess of the displacement field. 4.4 4.4.1 Buckling of Sections Transition from Global and Local Buckling Euler buckling load of a simply-supported column reads: (Pcr )column = π 2 EI l2 (356) where I is the bending rigidity of the column. Consider a section column which is composed of several thin plates, then the Euler buckling load can be considered as a global buckling load of the column. ( Pcr )column l EI On the other hands, local buckling force of a simply-supported plate reads: Ncr = kc π 2 D b2 (357) £ ¡ ¢¤ where D = Eh3 / 12 1 − ν 2 and kc = [(mb) /a + a/ (mb)]2 . Thus, the total local buckling load can be obtained: (Pcr )plate = kc π 2 D b (358) N cr h b a Transition from global to local buckling can be calculated by (Pcr )column = (Pcr )plate : π2 EI π2 D = k c l2 b kc π 2 E h3 = 12 (1 − ν 2 ) b I kc = 3 bh 12 (1 − ν 2 ) Example 7 Consider a square box column. µ ¶2 l b (359) (360) h b b l Then, one obtains: kc = 4 2 I = hb3 3 Now, the global buckling load reads: (Pcr )column = π 2 E I 2π 2 E hb3 = l2 3 l2 and the local buckling load from four plates can be calculated: (Pcr )f our plates = 4kc π 2 E h3 12 (1 − ν 2 ) b Then, applying (Pcr )column = (Pcr )f our plates , one gets: µ b2 hl ¶2 ' 2.20 Thus, the local and global buckling loads become same when b2 ' 1.5hl For example, if b = 40h, then l = 60b. Transition from the local to global buckling for an open channel section with lips is shown in the figure below. Buckling coefficients and modes for a hat section 4.4.2 Local Buckling The remainder of this section deals only with the local buckling. Dividing both sides of Eq. (357) by the plate thickness b gives the expression of the buckling stress σ cr : µ ¶2 Ncr kc π 2 E h σ cr = = (361) 2 h 12 (1 − ν ) b Consider two adjacent plates of a section of the prismatic column. P late 1 h1 b1 h2 b2 P late 2 Compatibility and equilibrium conditions at junction of adjoining walls of a section In general, there will be a restraining moment acting at the corner line between P late 1 and P late 2. The buckling stresses for those two plates are: µ ¶2 k1 π 2 E h1 (σ cr )1 = 2 12 (1 − ν ) b1 µ ¶2 k2 π 2 E h2 (σ cr )2 = 2 12 (1 − ν ) b2 (362) Before buckling, stresses in the entire cross-section are the same. So, at the point of buckling, one gets: (363) (σ cr )1 = (σ cr )2 from which the buckling coefficient k2 is relating to k1 : µ ¶ h1 b2 2 k2 = k1 h2 b1 (364) The total buckling load on the angle element is: Pcr = (σ c )1 h1 b1 + (σ c )2 h2 b2 " µ ¶ # µ ¶2 π2 E h1 2 h2 = k1 h1 b1 + k2 h2 b2 12 (1 − ν 2 ) b1 b2 " # µ ¶ h1 b2 2 (h2 )3 π2 E (h1 )3 = + k1 k1 12 (1 − ν 2 ) b1 h2 b1 b2 µ ¶ π 2 Ek1 h1 2 = A 12 (1 − ν 2 ) b1 (365) where A is the sectional area of two plate A = b1 h1 +b2 h2 . From this derivation, the conclusion is that only one buckling coefficient is needed to calculate the buckling load of the section consisting of several plates. Determination of the buckling coefficient is a bit more complicated because of the existence of the edge bending moment. This can be illustrated in an example of a box column with a rectangular cross section with the same thickness h. b1 b2 uniform thickness h The wider flange will be ready to buckling first while the narrow plate is not ready to buckle. When the second plate buckles, the first plate would have buckled long before. Thus, there is an interaction between left plates, and a compromise must be established because left plates must buckle at the same time. The buckling coefficient as a function of the ratio b2 /b1 is plotted in the figure next. h h Buckling coefficients for box sections In the limiting case of a square box (b1 = b2 = b), k1 = 4 and the edge interactive moment between adjacent plates is zero. Some useful graphs and formulas for typical sections are given next.