Chapter 13: Buckling of Columns Mechanics of Solids : ENG 2333 Professor: Elham Sahraei Based on: Mechanics of Materials, Hibbeler, Pearson, 10th Edition Fall 2021 1 ENG2333-Professor: Elham Sahraei Critical Load Not only must a member satisfy specific strength and deflection requirements but it must also be stable. Stability is particularly important if the member: • is long and slender • supports a compressive loading that becomes large enough to cause the member to suddenly deflect laterally or sidesway These members are called columns and the lateral deflection is called buckling. The maximum axial load that a column can support when it is on the verge of buckling is called the critical load, π·ππ . Any additional loading will cause the column to buckle. 2 ENG2333-Professor: Elham Sahraei Critical Load Stable equilibrium: πΉ > 2ππ₯ → ππ πΏΤ2 > 2ππ → π < ππΏ 4 Unstable equilibrium: πΉ < 2ππ₯ → ππ πΏΤ2 < 2ππ → π > ππΏ 4 Neutral equilibrium (critical load): πΉ = 2ππ₯ → ππ πΏΤ2 = 2ππ → πππ = πππ is independent of π → the transition point when π = πππ is called bifurcation point. The bar will be in neutral equilibrium for any small value of π. If the load exceeds its critical loading, this loading will require the column to undergo a large deflection. = πβ Two-bar mechanism consisting of weightless rigid bars that are pin connected ππΏ 4 For small π: β≈ π πΏΤ2 tan π ≈ π Restoring spring force: πΉ = ππ πΏΤ2 Disturbing force: 2ππ₯ = 2ππ 3 ENG2333-Professor: Elham Sahraei Ideal Column with Pin Supports π2π£ π2π£ π πΈπΌ = π = −ππ£ → + π£=0 ππ₯ 2 ππ₯ 2 πΈπΌ Ideal Column: homogeneous linear elastic material, perfectly straight before loading Homogenous linear differential equation. π£ = πΆ1 sin π π₯ + πΆ2 cos πΈπΌ π π₯ πΈπΌ B.C.: @π₯ = 0: π£ = 0 → πΆ2 = 0 π = −ππ£ @π₯ = πΏ: π£ = 0 → πΆ1 sin Tendency of a column to remain stable or become unstable when subjected to an axial load depends on its ability to resist bending. → sin π πΏ =0 → πΈπΌ π2 π 2 πΈπΌ π= πΏ2 4 π πΏ =0 πΈπΌ π πΏ = ππ πΈπΌ π = 1,2,3, … ENG2333-Professor: Elham Sahraei Ideal Column with Pin Supports π2 π 2 πΈπΌ π= πΏ2 A column will buckle about the principal axis of the cross section having the least moment of inertia (the weakest axis), provided it is supported the same way about each axis. Engineers try to keep the moment of inertia the same in all directions (Circular tubes, square tubes, and shapes having πΌπ₯ ≈ πΌπ¦ ) π = 1,2,3, … π2 πΈ πΌ π = 1 → πππ = πΈπ’πππ πΏπππ πΏ2 ππ₯ π£ = πΆ1 sin πΏ The above equation can be written in terms of stress and using πΌ = π΄π 2 as: π 2 πΈ (π΄π 2 ) π πππ = → πΏ2 π΄ = πππ ππ π2πΈ = πΏΤπ 2 Critical stress πππ ≤ ππ πΏ/π: slenderness ratio Measure of column’s flexibility 5 Smallest radius of gyration of the column, π = πΌ Τπ΄ ENG2333-Professor: Elham Sahraei Example 1 6 ENG2333-Professor: Elham Sahraei Columns Having Various Types of Supports Fixed at one end and free at the other π2π£ π2π£ π π πΈπΌ = π = π(πΏ − π£) → + π£ = πΏ ππ₯ 2 ππ₯ 2 πΈπΌ πΈπΌ Nonhomogeneous linear differential equation. π π π£ = πΆ1 sin π₯ + πΆ2 cos π₯ +πΏ πΈπΌ πΈπΌ B.C.: @π₯ = 0: π£ = 0 → πΆ2 = −πΏ ππ£ @π₯ = 0: = 0 → πΆ1 = 0 ππ₯ π = π(πΏ − π£) @π₯ = πΏ: π£ = πΏ → πΏ cos π πΏ = 0 → cos πΈπΌ π ππ πΏ= , π = 1,3,5, … πΈπΌ 2 7 → π£ = πΏ 1 − cos π π₯ πΈπΌ π πΏ =0 πΈπΌ π 2 πΈπΌ πππ = 4πΏ2 ENG2333-Professor: Elham Sahraei Columns Having Various Types of Supports Euler Formula: π2 πΈ πΌ πΏ2 π2 πΈπΌ πππ = 4πΏ2 For pinned columns: πππ = For fixed/free columns: To use Euler formula for columns having different types of support, modify the column length πΏ to represent the distance between points of zero moment on the column. This distance is called effective length, πΏπ . Effective length: πΏπ = πΎπΏ πΎ: effective-length factor π2 πΈ πΌ πππ = πΎπΏ 2 π2 πΈ πππ = πΎπΏ/π 2 πΏπ = 2πΏ πΎπΏ/π: effective-slenderness ratio 8 ENG2333-Professor: Elham Sahraei Example 2 9 ENG2333-Professor: Elham Sahraei Example 3 10 ENG2333-Professor: Elham Sahraei Example 3-Continue 11 ENG2333-Professor: Elham Sahraei
