1. Chapter Columns 

2. 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

3. 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)

4. 10.2 Stability of Structures
Concerns before:
Stable?
Unstable?
New concern:
(10.1)
(10.2)

5. 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

6. The new equilibrium position is:
(10.3) or

7. 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

8. 10.3 Euler’s Formula for Pin-Ended Columns
Determination of Pcr for the configuration in Fig ceases to be stable
Assume it is a beam subjected to bending moment:
(10.4)
(10.5)

9. 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 =  B = 0
@ x = L, y = 0
Eq. (10.8) reduces to
(10.9)

10. 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)

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.

12. 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

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

14. 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

15. 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

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

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

18. (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

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

20. 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

21. Summary

22. 10.5* Eccentric Loading; the Scant Formula

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

25. 10.6 Design of Columns under a Centric Load

26. 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

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

28. Empirical Formulas:

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

30. 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

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

32. 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

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

34. 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

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