ME 323 – Mechanics of Materials Lecture 42: Failure analysis – Buckling of columns Joshua Pribe Fall 2019 Lecture Book: Ch. 18 Stability and equilibrium What happens if we are in a state of unstable equilibrium? Stable Neutral Unstable 2 Buckling experiment There is a critical stress at which buckling occurs depending on the material and the geometry How do the material properties and geometric parameters influence the buckling stress? 3 Euler buckling equation Consider static equilibrium of the buckled pinned-pinned column 4 Euler buckling equation We have a differential equation for the deflection with BCs at the pins: d 2v EI+= Pv( x ) 0 v(0)== 0and v ( L ) 0 dx2 The solution is: P P A = 0 v(s x) = Aco x+ Bsin x with EI EI PP Bsin L= 0 L = n , n = 1, 2, 3, ... EI EI 5 Effect of boundary conditions Critical load and critical stress for buckling: EI EA P = 22= cr L2 2 e (Legr ) 2 E cr = 2 (Lreg) I r = Pinned- Pinned- Fixed- where g Fixed- A pinned fixed fixed is the “radius of gyration” free LLe = LLe = 0.7 LLe = 0.5 LLe = 2 6 Modifications to Euler buckling theory Euler buckling equation: works well for slender rods Needs to be modified for smaller “slenderness ratios” (where the critical stress for Euler buckling is at least half the yield strength) 7 Summary L 2 E Critical slenderness ratio: e = r 0.5 gYc Euler buckling (high slenderness ratio): LL 2 E EI If ee : = or P = 2 rr cr 2 cr L2 gg c (Lreg) e Johnson bucklingI (low slenderness ratio): r = 2 g Lr LLeeA ( eg) If : =−1 rr cr2 Y gg c 2 Lr ( eg)c with radius of gyration 8 Summary Effective length from the boundary conditions: Pinned- Pinned- Fixed- LL= LL= 0.7 FixedLL=- 0.5 pinned fixed e fixede e LL= 2 free e 9 Example 18.1 Determine the critical buckling load Pcr of a steel pipe column that has a length of L with a tubular cross section of inner radius ri and thickness t. The material has Young’s modulus E and yield strength σY. Use pinned-fixed boundary conditions. 10 Example 18.2 The steel compression strut BC of the frame ABC is a tube with an outer diameter of d = 48 mm and a wall thickness of t = 5 mm. Determine the factor of safety against elastic buckling if a distributed load of 10 kN/m is applied to AB. Let E = 210 GPa and σY = 340 MPa. 11 Example 18.8 (additional examples) Members (1), (2), and (3) have Young’s modulus E = 107 psi 3 and σY = 60 × 10 psi. Each member has a solid circular cross section with diameter d = 1 in. A force P = 10 kips is applied to joint C. Determine the maximum length L that can be used without buckling based on Euler’s theory of buckling. 12 Real-world example: Railroad track buckling What would cause the railroad track to buckle as shown in the picture? G. Yang and M.A. Bradford, Engineering Failure Analysis 92 (2018), pp. 107-120. 13.