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 22 ππEI 2 EA PIcr ==2 or using Ar, Pcr =2 L ()L / r Lr/ : Slenderness ratio Large displacements • Slenderness ratio and yield stress govern type of post-buckling response Dashed lines represent behavior without yielding Governing equation for beam column d 2 y M P = = − y dx2 EI EI d 2 y P + y = 0 dx2 EI Try : y(x) = Acos(ω x) + Bsin(ω x) d 2 y = −ω 2 (Acos(ω x) + Bsin(ω x)) dx2 P Substitute into ODE:ω = EI ⎛ P ⎞ ⎛ P ⎞ ⎜ ⎟ ⎜ ⎟ y(x) = Acos⎜ x⎟ + Bsin⎜ x⎟ ⎝ EI ⎠ ⎝ EI ⎠ Apply boundary conditions y(0) =0 and y(L) =0 ⎛ P ⎞ P ⎜ ⎟ Bsin⎜ L⎟ =0 → L = nπ n =1,2,3,.. ⎝ EI ⎠ EI n2π 2 EI P = nis the integer defining the buckling mode. L2 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 π 2 EI π 2 EI P = or using I = Ar 2 , P = cr L2 cr (L / r)2 L / r :Slenderness ratio P π 2 E Compressive (normal) stress at critical buckling load: σ = cr = cr 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 • Properties of the cross section A= 2(150)(50)=15,000mm2 25(50)(150)+(50+ 75)(50)(150) y = =75mmfrom bottom c 15,000 • Moments of inertia about the centroid of the cross section 150(50)3 50(150)3 I = +150(50)(50)2 + +150(50)(50)2 =53.13x106 mm4 x 12 12 50(150)3 150(50)3 I = + =15.625x106 mm4 y L 3000 12 12 a)Slenderness ratio = =93 • Radii of inertia r 32.3 π 2 EA I b)Critical Buckling Load Pcr = 2 = 222.75kN I x y (L / r) rx = =59.51 mmandry = =32.3mm A A 2 Pcr π E c)Critical Buckling Stress σcr = = 2 =14.85MPa ⇒r =rmin =32.3mm A (L / r) 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 2 A=3795mm xc =14.8mm 6 4 6 4 I xc = 25.3x10 mm I yc =1.01x 10 mm Solution for single channel . Radii of inertia of a single channel I I r = x =81.1 mmand r = y = 16.3 mm x A y A ⇒ r = rmin =16.3 mm L 12000 Slenderness ratio : = = 736.2 r 16.3 π 2 EA Critical Buckling load : P = = 13.82kN cr (L / r)2 Two laced channels. A = 2(3975) =7590mm3 Moments of inertia about the centroid of the cross section : 6 4 I x = 2I xc =50.6 x10 mm 2 6 4 I y = 2[]I yc + A(75 +14.8) = 63.23x10 mm Radii of inertia : I I r = x =81.7 mmand r = y = 91.3 mm → r = r =81.7 mm x A y A min L 12000 Slenderness ratio : = = 146.9 r 81.7 π 2 EA Critical Buckling load : P = = 694.3 kN cr (L / r)2 Higher buckling modes • With appropriate boundary conditions can get higher modes π 2 EI P = 4 = 4 P L2 cr P π 2 E σ = = 4 cr(2) A (L / r)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 d 2 y k 2ex P + k 2 y = − , k 2 = dx2 L EI y = 0 for x = 0 , L • General solution ex y = Asin k x + B cosk 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.