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12.1 Introduction to column buckling • Buckling: “Buckling can be defined as the sudden large deformation of structure due to a slight increase of an existing load under which the structure had exhibited little, if any, deformation before the load was increased.” No failure implied!!! Reinforced concrete steel Stability of equilibrium condition • Easy to visualize for a ball on a surface • Note that we use curvature to decide stability, but the surface can curve up with zero curvature Euler formula • For simply supported column Pcr = π 2 EI L 2 or using I = Ar , L / r : Slenderness ratio 2 Pcr = π 2 EA (L / r) 2 Large displacements • Slenderness ratio and yield stress govern type of post-buckling response Dashed lines represent behavior without yielding Governing equation for beam column d2y M P = =− y 2 dx EI EI d2y P + y=0 2 dx EI Try : y ( x) = A cos(ω x) + B sin(ω x) d2y 2 ω = − ( A cos(ω x) + B sin(ω x)) 2 dx P Substitute into ODE :ω = EI ⎛ P ⎞ ⎛ P ⎞ ⎟ ⎜ y ( x) = A cos⎜ x ⎟ + B sin ⎜⎜ x ⎟⎟ ⎝ EI ⎠ ⎝ EI ⎠ Apply boundary conditions y (0) = 0 and y ( L) = 0 ⎛ P ⎞ P ⎜ ⎟ B sin ⎜ L⎟ =0 → L = nπ n =1,2,3,.. EI EI ⎝ ⎠ n 2π 2 EI P= n is the integer defining the buckling mode. 2 L I I = Ar 2 where r = is the radius of gyration. A For a column with both ends pinned, n = 1 defines the critical buckling load Pcr = π 2 EI or using I = Ar , Pcr = L L / r :Slenderness ratio 2 2 π 2 EI (L / r)2 Compressive (normal) stress at critical buckling load : σ cr Pcr π 2E = = A (L / r)2 Example 1 A 3m column with the cross section shown is constructed from two pieces of timber, that act as a unit. If the modulus of elasticity of timber is E=13 GPa, determine a) The slenderness ratio b) Critical buckling load c) Axial stress in the column when the critical load is applied • Propertiesof the cross section A = 2(150)(50) =15,000 mm2 25(50)(150) + (50 + 75)(50)(150) yc = = 75mm from bottom 15,000 • Momentsof inertia about the centroid of the cross section 150(50)3 50(150)3 2 Ix = +150(50)(50) + +150(50)(50) 2 = 53.13 x 106 mm4 12 12 50(150)3 150(50)3 Iy = + =15.625 x 106 mm4 L 3000 12 12 a ) Slenderness ratio = = 93 r 32 . 3 • Radii of inertia Iy Ix rx = = 59.51 mm and ry = = 32.3 mm A A ⇒ r = rmin = 32.3 mm b) Critical Buckling Load Pcr = π 2 EA (L / r)2 = 222.75 kN Pcr π 2E c) Critical Buckling Stress σ cr = = =14.85 MPa A (L / r)2 Example 2 C229 x 30 structural steel channels with E=200 GPa are used for a 12 m column. Determine the total compressive load required to buckle the column if a) One channel is used b) Two channels are laced 150 mm back to back as shown Section properties for one channel A = 3795 mm 2 xc = 14.8mm I xc = 25.3 x 106 mm 4 I yc =1.01 x 106 mm 4 Solution for single channel . Radii of inertia of a single channel Ix rx = = 81.1 mm and ry = A ⇒ r = rmin = 16.3 mm Iy A = 16.3 mm L 12000 Slenderness ratio : = = 736.2 r 16.3 π 2 EA Critical Buckling load : Pcr = = 13.82 kN 2 (L / r) Two laced channels. A = 2(3975) = 7590mm 3 Moments of inertia about the centroid of the cross section : I x = 2 I x c = 50.6 x 106 mm 4 [ ] I y = 2 I yc + A(75 + 14.8) 2 = 63.23 x 106 mm 4 Radii of inertia : Iy Ix = 81.7 mm and ry = = 91.3 mm → r = rmin = 81.7 mm A A L 12000 Slenderness ratio : = = 146.9 r 81.7 π 2 EA Critical Buckling load : Pcr = = 694.3 kN 2 (L / r) rx = Higher buckling modes • With appropriate boundary conditions can get higher modes P =4 σ cr ( 2 ) π 2 EI = 4 Pcr L π 2E P = =4 2 (L / r) A 2 Buckling modes : n = 1,2,3. Imperfection approach • Put in imperfection in the form of eccentricity Eccentrically loaded pinned-end columns • Moment equation x M ( x) = − Py − Pe L Solution of eccentric load • Differential equation of equilibrium 2 d2y k ex P 2 2 , + k y = − k = dx 2 L EI y = 0 for x = 0 , L • General solution ex y = A sin k x + B cos k x − L • With boundary conditions ⎛ sin k x x ⎞ y = e ⎜⎜ − ⎟⎟ ⎝ sin k L L ⎠ Reading assignment Sections 12.3-4: Question: Why equation 12.26 does not really guarantee buckling? What does it guarantee? Source: www.library.veryhelpful.co.uk/ Page11.htm