Chapter 10 Columns 

10.1 Introduction Column = vertical prismatic members subjected to compressive forces Goals of this chapter: Study the stability of elastic columns Determine the critical load Pcr The effective length Secant formula

Previous chapters: -- concerning about (1) the strength and (2) excessive deformation (e.g. yielding) This chapter: -- concerning about (1) stability of the structure (e.g. bucking)

10.2 Stability of Structures Concerns before: Stable? Unstable? New concern: (10.1) (10.2)

The system is unstable if Since (10.2) The system is stable, if The system is unstable if A new equilibrium state may be established

The new equilibrium position is: (10.3) or

After the load P is applied, there are three possibilities: 1. P < Pcr – equilibrium &  = 0 -- stable 2. P > Pcr – equilibrium &  =  -- stable 3. P > Pcr – unstable – the structure collapses,  = 90o

10.3 Euler’s Formula for Pin-Ended Columns Determination of Pcr for the configuration in Fig. 10.1 ceases to be stable Assume it is a beam subjected to bending moment: (10.4) (10.5)

Defining: @ x = 0, y = 0  B = 0 @ x = L, y = 0 (10.6) (10.7) The general solution to this harmonic function is: (10.8) B.C.s: @ x = 0, y = 0  B = 0 @ x = L, y = 0 Eq. (10.8) reduces to (10.9)

1. A = 0  y = 0  the column is straight! (10.9) Therefore, 1. A = 0  y = 0  the column is straight! 2. sin pL = 0  pL = n  p = n /L  (10.6) Since We have (10.10) For n = 1 -- Euler’s formula (10.11)

Substituting Eq. (10.11) into Eq. (10.6), Therefore, Hence Equation (10.8) becomes (10.12) This is the elastic curve after the beam is buckled.

1. A = 0  y = 0  the column is straight! (10.9) 1. A = 0  y = 0  the column is straight! 2. sin pL = 0  pL = n  If P < Pcr  sin pL  0 Hence, A = 0 and y = 0  straight configuration

L/r = Slenderness ratio Critical Stress: Introducing Where r = radius of gyration Where r = radius of gyration (10.13) L/r = Slenderness ratio

10.4 Extension of Euler’s Formula to columns with Other End Conditions Case A: One Fixed End, One Free End (10.11') (10.13') Le = 2L

Hence, Le = L/2 Case B: Both Ends Fixed At Point C RCx = 0 Q = 0  Point D = inflection point  M = 0  AD and DC are symmetric Hence, Le = L/2

M = -Py - Vx Case C: One Fixed End, One Pinned End Since Therefore, The general solution: The particular solution:

into the particular solution, it follows Substituting into the particular solution, it follows As a consequence, the complete solution is (10.16)

(10.16) B.C.s: @ x = 0, y = 0  B = 0 @ x = L, y = 0  (10.17) Eq. (10.16) now takes the new form

B.C.s: @ x = L, dy/dx =  = 0 Taking derivative of the question, (10.18) (10.17) (10.19)

Le = 0.699L  0.7 L Solving Eq. (10.19) by trial and error, Since Therefore, Solving for Le Case C Le = 0.699L  0.7 L

Summary

10.5* Eccentric Loading; the Scant Formula

Secant Formula: If Le/r << 1, Eq. (10.36) reduces to (10.36) (10.37)

10.6 Design of Columns under a Centric Load

10.6 Design of Columns under a Centric Load Assumptions in the preceding sections: -- A column is straight -- Load is applied at the center of the column --  < y Reality: may violate these assumptions -- use empirical equations and rely lab data

Test Data: Facts: 1. Long Columns: obey Euler’s Equation 2. Short Columns: dominated by y 3. Intermediate Columns: mixed behavior

Empirical Formulas:

Real Case Design using Empirical Equations: 1. Allowable Stress Design Two Approaches: 2. Load & Resistance Factor Design

Structural Steel – Allowable Stress Design Approach I -- w/o Considering F.S. 1. For L/r  Cc [long columns]: [Euler’s eq.] 2. For L/r  Cc [short & interm. columns]: where

Approach II -- Considering F.S. 1. L/r  Cc : (10.43) 2. L/r  Cc : (10.45)

10.7 Design of Columns under an Eccentric Load (10.56) 1. The section is far from the ends 2.  < y (10.57) Two Approaches: (I) Allowable Stress Method (II) Interaction Method

I. Allowable-Stress Method (10.58) -- all is obtained from Section 10.6. -- The results may be too conservative.

II. Interaction Method Case A: If P is applied in a plane of symmetry: (10.59) (Interaction Formula) (10.60) -- determined using the largest Le

Case B: If P is NOT Applied in a Plane of Symmetry: (10.61)