Buckling Column and Plate Analysis (2020-2021) PDF

Dr. Chris Harrison Dr. Mahdi Damghani Aerostructures Engineering Design and Mathematics Column Buckling (and its similarity with Plate Buckling) 2020-2021 Lecture Content The function of a wing spar Buckling in the context of aerostructures Bu...

Dr. Chris Harrison Dr. Mahdi Damghani Aerostructures Engineering Design and Mathematics Column Buckling (and its similarity with Plate Buckling) 2020-2021 Lecture Content The function of a wing spar Buckling in the context of aerostructures Buckling of columns with various boundary conditions The effect of eccentricity Similarities with buckling of plates Suggested Reading ‘Aircraft Structural Analysis’ THG Megson Chapters 8 and 9 Context Under in-plane compressive loads an initally straight or slightly curved member or panel will suddenly deform out-of-plane – this is buckling: P P Context An aircraft is a weight-optimised structure, consisting of thin skins that form a streamlined shape. As these skins are thin, they are flexible, and buckling dominates the design of the structure. The skins are therefore reinforced by internal stiffening members. Context Internal stiffening members: Wing/Tail ribs Fuselage frames Context Internal stiffening members: Spars Context Internal stiffening members: Wing stringers Stringers (wing) and longerons (fuselage) are collectively called ‘stiffeners’. Fuselage longerons Context Internal stiffening members: Stiffeners on ribs Stringers (wing) and longerons (fuselage) are collectively called ‘stiffeners’. Context: Panel Buckling The stiffners act to reduce the effective size of the panel which increases the load required to buckle it: P (Restraint on all four edges) P P  Context: Elastic or Yield? Racing Bearcat wing skin buckling – Temporary! B52 B52 Fuselage Buckling – Permanent! Column Buckling  Column Buckling A column is a straight slender rod carrying axial compressive forces. Column buckling was studied in 1744 by Euler, resulting in an ‘Euler column’ and ‘Euler buckling’. Bifurcation point EI P P PCR PCR l v v (At this point, any of these three solutions are possible.) v=0 v P PCR Critical buckling load: 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = 2 𝑙  Column Buckling Derivation nüburkulma medu When the column is buckled, it has a deformed circular shape. The c!c!m eğ"ld"ğ"nde compressive force and the displacement causes a bending moment be nonlede , burkuluyo to act on the column: 𝑑2 𝑣 nes EI PCR P 𝐸𝐼 2 = −𝑀 𝑑𝑧 𝑑2 𝑣 𝐸𝐼 2 = −𝑃𝐶𝑅 𝑣 l v v 𝑑𝑧 z M Solution gives eigenmodes: 𝑛2 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = 𝑙2 PCR 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = 2 𝑙 Note: For the full derivation of critical buckling load, PCR, refer to ‘8-1 - Derivation of Critical Buckling Load.pdf’ on blackboard.  Example 1 The steel tube shown is to be used as a pin-ended column. Determine the maximum load the column can support so that it does not buckle or yield. For steel: E = 200 GPa syield = 250 MPa Critical buckling load: 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = 2 Where second moment of area, I, for a 𝑙 tube section is: 𝜋 𝑟𝑜4 − 𝑟𝑖4 𝑟𝑖 𝐼= 4 𝑟𝑜  Example 1 The steel tube shown is to be used as a pin-ended column. Determine the maximum load the column can support so that it does not buckle or yield. For steel: E = 200 GPa syield = 250 MPa So ro = 0.075m, ri = 0.07m and I = 5.99x10-6 m4. Then: 𝜋 2 𝑥200𝑥109𝑥5.99𝑥10−6 𝑃𝐶𝑅 = = 241425 𝑁 72 At this force, the axial stress is: 𝑃𝐶𝑅 𝜎= = 106𝑥106 𝑃𝑎 = 106 𝑀𝑃𝑎 𝐴  1:20 Column Buckling 𝜋 2 𝐸𝐼 Buckling is an elastic phenomena – it depends on the 𝑃𝐶𝑅 = 2 stiffness (E and I), not the strength of the material. 𝑙 But we can also re-cast this equation in terms of a critical stress, as s = P/A : 𝐼 𝜋 2𝐸 r is the radius of gyration 𝑟 = 𝐴 𝜎𝐶𝑅 = 𝑙 2 𝑟 1:5 𝑙 is the slenderness ratio 𝑟 Rule of thumb: Euler buckling formula appropriate for slenderness ratio > 70 for pin-jointed column, and > 17.6 for fixed-fixed column, for Aluminium. Slenderness ratio 17.3 69.3 (square cross-section) Column Buckling 𝜋 2 𝐸𝐼 Buckling is an elastic phenomina – it depends on the 𝑃𝐶𝑅 = 2 stiffness (E and I), not the strength of the material. 𝑙 For stable equibrium, the load can be removed, and the column returns to its undeformed position. Force increases with deflection = Stable = Safe EI P P PCR Neutral l v Force reduces with deflection = Unstable = Collapse! v=0 v P Effect of Half-Wavelength P n=3 𝑛2 𝜋 2 𝐸𝐼 n=Number of half- 𝑃𝐶𝑅 = waves along the length. 𝑙2 Column restrained at these points. P P P n=2 v v l x v x x n=1 v=0 v P P P n=1 n=2 n=3  Effect of Boundary Conditions Fixed-Free Pinned-Pinned Fixed-Fixed P P P le=l l l le=0.5l le=2l P P 𝜋 2 𝐸𝐼 le=kl is the effective length 𝑃𝐶𝑅 = 2 and is the length of a pin- 𝑙𝑒 ended column that would have the same critical load. Note: For the derivation of the critical load for fixed-fixed boundary conditions, refer to ‘8-2 - Derivation of Fixed-Fixed Critical Buckling Load.pdf’ on blackboard. Effect of Boundary Conditions 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = k =1 (𝑘𝑙)2 k =0.5 k =0.7 k =2 k =1 k =2  Example 2 The Scottish Aviation Twin Pioneer aircraft employs an aluminium truss to support the wing. The geometry of the main and jury strut are shown below. B 0.4m B-B: Jury Strut B 1.1m A 20mm 100mm A A-A: Main Strut 1.1m 45mm 25mm 175mm  Example 2 When landing, the truss supports the weight of the outer part of the wing under a vertical decceleration of 1.5g, and the axial compressive force in the main strut and the jury strut are 750 kN and 525 kN, respectively. The connection between the struts can be considered to be pin-jointed. The ends of the struts are built in. Check that the truss and jury strut do not buckle, using a safety factor of 1.5.  Example 2 𝑛2 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = 𝑘𝑙 2 For aluminium: E = 70 GPa 1.1m Main Strut 45mm 25mm 1.1m 175mm 0.2𝑥0.0453 𝜋𝑥0.01254 𝐼= −2 = 1.29𝑥10−6 𝑚4 12 4 𝑙 = 2.2𝑚 𝑃𝐶𝑅 = 2947𝑘𝑁 𝑘 = 0.5 (Fixed – Fixed) 𝑃𝐶𝑅 𝑛=2 = 1965𝑘𝑁 Safe! 1.5  Example 2 𝑛2 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = 0.4m 𝑘𝑙 2 For aluminium: E = 70 GPa Jury Strut: Jury Strut 0.1𝑥0.023 𝐼= = 6.67𝑥10−8 𝑚4 12 20mm 100mm 𝑙 = 0.4𝑚 𝑘 = 0.7 (Fixed – Pinned) 𝑛=1 𝑃𝐶𝑅 𝑃𝐶𝑅 = 2349𝑘𝑁 = 1567𝑘𝑁 Safe! 1.5 Limitations of Euler Theory 𝜋 2 𝐸𝐼 𝜎𝐶𝑅 = 𝑙𝑒 2 𝑟 𝑙𝑒 is the slenderness ratio 𝑟 Euler theory valid Euler theory invalid 𝑙𝑒 𝑟 Below this point the material is likely to Yield at a lower load than PCR. This depends upon the Youngs modulus and Yield strength of the material.  Effect of Imperfections An initial imperfection (or end moment) can be represented as an eccentric load. The eccentricity is e: e e EI P P 𝑑2 𝑣 𝐸𝐼 2 = −𝑀 𝑑𝑧 𝑑2 𝑣 l v z v 𝐸𝐼 2 = −𝑃 𝑣 + 𝑒 𝑑𝑧 M The solution is: P 𝑃 𝑙 𝑃 𝑃 𝑣 = 𝑒 𝑡𝑎𝑛 𝑠𝑖𝑛 𝑧 + 𝑐𝑜𝑠 𝑧 −1 𝐸𝐼 2 𝐸𝐼 𝐸𝐼 Effect of Imperfections e EI 𝑃 𝑙 𝑃 𝑃 𝑣 = 𝑒 𝑡𝑎𝑛 𝑠𝑖𝑛 𝑧 + 𝑐𝑜𝑠 𝑧 −1 P 𝐸𝐼 2 𝐸𝐼 𝐸𝐼 Some observations from this equation: l v The displacement can be calculated explicitly (unlike Euler buckling). The displacement is linear with e (but non-linear with P). 𝜋 Because 𝑡𝑎𝑛 → ∞, this means the deflection is 2 P 𝑃 𝑙 𝜋 infinite when =.Or: 𝐸𝐼 2 2 𝜋 2 𝐸𝐼 𝑃𝐶𝑅 = 2 𝑙 Effect of Imperfections The maximum stress in the column can be calculated from the previous equation. This gives the secant formula: 1 𝑒𝑡 𝑃 𝑙 𝜎𝑚𝑎𝑥 = 𝑃 + 𝑠𝑒𝑐 PCR 𝐴 2𝐼 𝐸𝐼 2 A = Cross sectional area t = Thickness P Increasing eccentricity smax Buckling of Plates a The plate is simply supported along all four edges. Nx is a uniformly distributed load along N b N the edge of the plate. Units N/m. The bending stiffness of the plate is given the parameter D: 𝐸𝑡 3 𝐷= 12 1 − 𝜈 Critical buckling load: 𝑘𝜋 2 𝐷 𝑁𝐶𝑅 = 𝑏2 By inspection it is clear that there are similarities with Euler column buckling. D is similar to EI. b is similar to l. What about k? Buckling of Plates a 𝑘𝜋 2 𝐷 𝑁𝐶𝑅 = N b N 𝑏2 k is known as the plate buckling coefficient. 2 𝑚𝑏 𝑎 k 𝑘 = 𝑚𝑖𝑛 𝑎 + 𝑚𝑏 a/b  a Buckling of Plates N b N 𝑎 𝑚 = 1, = 1 𝑏 𝑘𝜋 2 𝐷 𝑁𝐶𝑅 = 𝑎 𝑏2 𝑚 = 2, =2 𝑏 k 𝑚 = 3, 𝑎 =3 𝑏 a/b  Key Learning Points Buckling is a key design driver for aircraft structures. Buckling is an elastic phenomina. Stringers and longerons are aircraft structural components that can be approximated as an Euler column. An Euler column has a critical compressive load at which buckling occurs. The boundary conditions affect the buckling load, calculated using effective length. Imperfections can mean failure before critical buckling load. Many structural relationships in Euler columns are the same for plates. Thanks for your attention! [email protected]  Tutorial 1 The structural member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields. Section properties are as; Assume modulus of elasticity of 200 GPa and material stress allowable of 250MPa.  Tutorial 2 A 3m column with the following cross section is 150 mm constructed of material with E=13GPa and is simply supported at its two ends. – Determine Euler buckling 50 mm load – Determine stress associated 50 mm 50 mm 50 mm with the buckling load  Tutorial 3 Vertical stabiliser 1 The vertical stabiliser of an aircraft 5 shown is made from 3mm thick aluminium with material properties: E=70GPa, v=0.3 The stabiliser is swept aft with a ratio of 1:5 2m Rudder Fairing Approximating the stabiliser as a panel, and assuming edges are simply supported, 1. Calculate the critical buckling force from bending of the stabiliser panel 2. Decide which is better for buckling: a. Two equally spaced stringers running parallel to the leading edge of the stabaliser, or b. A rib at the centre of the stabiliser, running normal to the leading edge: or ?  Tutorial 4 For the loading given, which panels are more susceptible to buckling phenomenon. Ignoring the effects of membrane shear stresses and assuming that only membrane direct stresses due to wing bending act on panels, i.e. direct stresses in the wing spanwise direction, determine the conservative buckling load of panels in Rib Bay 1. To enable your calculations, consider conservative boundary conditions for the panels of rib bay 1 to represent spars and ribs. Specify whether the panels buckle or not. If they do buckle, put forward a solution to delay buckling of panels.