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Chapter 10 Columns
10.1 Introduction
Column = vertical prismatic members subjected
to compressive forces
Goals of this chapter:
1. Study the stability of elastic
columns
2. Determine the critical load Pcr
3. The effective length
4. 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
allow
P
A
σ σ= <
cr
PL
AE
δ δ= <
2
2
( )sin ( )cr
LP Kθ θ∆ = ∆ (10.1)
sin θ θ∆ ≈ ∆ (10.2)
Concerns before:
New concern:
Stable?
Unstable?
2
2
( ) ( )cr
LP Kθ θ∆ = ∆
sin θ θ∆ ≈ ∆
4 /crP K L= (10.2)
The system is stable, if
The system is unstable if
Since
4 /crP K L<
4 /crP K L>
A new equilibrium state may be
established
2
2
( )sin ( )LP M Kθ θ= =
4 sin
PL
K
θ
θ
=
The new equilibrium position is:
4sin
PL
K
θ
θ
=
(10.3)
or
After the load P is applied, there are
three possibilities:
1. P < Pcr – equilibrium & θ = 0 -- stable
2. P > Pcr – equilibrium & θ = θ -- stable
3. P > Pcr – unstable – the structure
collapses, θ = 90o
10.3 Euler’s Formula for Pin-Ended Columns
Determination of Pcr for the configuration in Fig. 10.1
ceases to be stable
2
2
d y M P
y
dx EI EI
= = −
2
2
0
d y P
y
dx EI
+ =
Assume it is a beam subjected to
bending moment:
(10.5)
(10.4)
2 P
p
EI
=
2
2
2
0
d y
p y
dx
+ =
sin cosy A px B px= +
Defining:
The general solution to this harmonic function is:
(10.7)
(10.6)
(10.8)
B.C.s:
@ x = 0, y = 0 → B = 0
@ x = L, y = 0
Eq. (10.8) reduces to
0sinA pL = (10.9)
0sinA pL =
2 2
2
2
n
p
L
π
=
2 P
p
EI
=
(10.9)
Therefore,
1. A = 0 → y = 0 → the column is straight!
2. sin pL = 0 → pL = nπ → p = nπ /L →
Since (10.6)
We have
2 2
2
n EI
P
L
π
=
For n = 1
2
2cr
EI
P
L
π
=
(10.10)
-- Euler’s formula (10.11)
Substituting Eq. (10.11) into Eq. (10.6),
2
2cr
EI
P
L
π
=
2 P
p
EI
=
2 2
2
2 2
crP EI
p
EI L EI L
π π
= = =
(10.6)
(10.11)
Therefore,
Hence p
L
π
=
Equation (10.8) becomes sin
x
y A
L
π
= (10.12)
This is the elastic curve after the beam is buckled.
If P < Pcr → sin pL ≠ 0
0sinA pL =
1. A = 0 → y = 0 → the column is straight!
2. sin pL = 0 → pL = nπ →
2
2cr
EI
P
L
π
=
(10.9)
Hence, A = 0 and y = 0 → straight configuration
Critical Stress:
Introducing
2
2
( )
cr
E
L
r
π
σ =
2
2
cr
cr
P EI
A L A
π
σ = =
2
I Ar=
Where r = radius of gyration
, ,y zx
x y z
I II
r r r
A A A
= = =
Where r = radius of gyration (10.13)
L/r = Slenderness ratio
10.4 Extension of Euler’s Formula to columns with
Other End Conditions
2
2cr
e
EI
P
L
π
=
2
2
( )
cr
e
E
L
r
π
σ =
(10.11')
(10.13')
Case A: One Fixed End, One Free End
Le = 2L
Case B: Both Ends Fixed
At Point C
RCx = 0
Q = 0 → 0
VQ
It
π = =
Point D = inflection point → M = 0 → AD and DC are symmetric
Hence, Le = L/2
2
2
d y M P V
y x
dx EI EI EI
= = − −
2 P
p
EI
=
2
2
2
d y V
p y x
dx EI
+ = −
Case C: One Fixed End, One Pinned End
M = -Py - Vx
Since
Therefore,
The general solution:
The particular solution:
sin cosy A px B px= +
2
V
y x
p EI
= −
2 P
p
EI
=
V
y x
P
= −
Substituting into the particular solution, it follows
As a consequence, the complete solution is
sin cos
V
y A px B px x
P
= + − (10.16)
sin
V
A pL L
P
=
sin
V
y A px x
P
= −
(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
sin cos
V
y A px B px x
P
= + −
Taking derivative of the question,
cos
dy V
Ap px
dx P
= −
cos
V
Ap pL
P
=
sin
V
A pL L
P
=
B.C.s: @ x = L, dy/dx = θ = 0
(10.18)
(10.17)
10 17
10 18
( . )
( . )
tan pL pL= (10.19)
Solving Eq. (10.19) by trial and error,
2 2
4 4934 4 4934 20 19064. . / . /pL p L p L= ⇒ = ⇒ =
2 2
2
20 19.P EI
p P p EI
EI L
= ⇒ = =
2
2 2
20 19.
cr
e
EI EI
P
L L
π
= =
Since
Therefore,
Case C
Solving for Le
Le = 0.699L ≈ 0.7 L
Summary
10.5*
Eccentric Loading; the Scant Formula
Secant Formula:
(10.36)
2
1
1
2
max
sec( )e
P
A ec P L
r EA r
σ
=
+
1
1
2
sec( )eP L
EA r
≈If Le/r << 1,
Eq. (10.36) reduces to
2
1
maxP
ecA
r
σ
=
+
(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
-- σ < σy
Reality: may violate these assumptions
-- use empirical equations and rely lab data
Facts:
1. Long Columns: obey Euler’s Equation
2. Short Columns: dominated by σy
3. Intermediate Columns: mixed behavior
Test Data:
Empirical Formulas:
Real Case Design using Empirical Equations:
Two Approaches:
1. Allowable Stress Design
2. Load & Resistance Factor Design
1. For L/r ≥ Cc [long columns]:
[Euler’s eq.]
Structural Steel – Allowable Stress Design
Approach I -- w/o Considering F.S.
2
2
( / )cr
E
L r
π
σ =
2
2
1
2
( / )
[ ]cr Y
c
L r
C
σ σ= −
2. For L/r ≤ Cc [short & interm. columns]:
where
2
2 2
c
Y
E
C
π
σ
=
1. L/r ≥ Cc :
(10.43)
Approach II -- Considering F.S.
2
2
1 92. . ( / )
cr
all
E
F S L r
σ π
σ = =
2. L/r ≤ Cc :
21
1
2
/
[ ( ) ]
. . .
cr Y
all
c
L c
F S F S C
σ σ
σ = = − (10.45)
10.7 Design of Columns under an Eccentric Load
centric bendingσ σ σ= +
max
P Mc
A I
σ = +
(10.56)
(10.57)
(I) Allowable Stress Method
(II) Interaction Method
Two Approaches:
1. The section is far from the ends
2. σ < σy
I. Allowable-Stress Method
all
P Mc
A I
σ+ ≤ (10.58)
-- σall is obtained from Section 10.6.
-- The results may be too conservative.
II. Interaction Method
1
/ /
all all
P A Mc I
σ σ
+ ≤
1
/ /
( ) ( )all centric all bending
P A Mc I
σ σ
+ ≤
(10.59)
(10.60)(Interaction Formula)
( )all centric
σ -- determined using the largest Le
Case A: If P is applied in a plane of
symmetry:
1max max/ //
( ) ( ) ( )
x x x x
all centric all bending all bending
M z I M x IP A
σ σ σ
+ + ≤ (10.61)
Case B: If P is NOT Applied in a Plane of Symmetry:

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Column lessons

  • 2. 10.1 Introduction Column = vertical prismatic members subjected to compressive forces Goals of this chapter: 1. Study the stability of elastic columns 2. Determine the critical load Pcr 3. The effective length 4. 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 allow P A σ σ= < cr PL AE δ δ= < 2 2 ( )sin ( )cr LP Kθ θ∆ = ∆ (10.1) sin θ θ∆ ≈ ∆ (10.2) Concerns before: New concern: Stable? Unstable?
  • 5. 2 2 ( ) ( )cr LP Kθ θ∆ = ∆ sin θ θ∆ ≈ ∆ 4 /crP K L= (10.2) The system is stable, if The system is unstable if Since 4 /crP K L< 4 /crP K L> A new equilibrium state may be established
  • 6. 2 2 ( )sin ( )LP M Kθ θ= = 4 sin PL K θ θ = The new equilibrium position is: 4sin PL K θ θ = (10.3) or
  • 7. After the load P is applied, there are three possibilities: 1. P < Pcr – equilibrium & θ = 0 -- stable 2. P > Pcr – equilibrium & θ = θ -- stable 3. P > Pcr – unstable – the structure collapses, θ = 90o
  • 8. 10.3 Euler’s Formula for Pin-Ended Columns Determination of Pcr for the configuration in Fig. 10.1 ceases to be stable 2 2 d y M P y dx EI EI = = − 2 2 0 d y P y dx EI + = Assume it is a beam subjected to bending moment: (10.5) (10.4)
  • 9. 2 P p EI = 2 2 2 0 d y p y dx + = sin cosy A px B px= + Defining: The general solution to this harmonic function is: (10.7) (10.6) (10.8) B.C.s: @ x = 0, y = 0 → B = 0 @ x = L, y = 0 Eq. (10.8) reduces to 0sinA pL = (10.9)
  • 10. 0sinA pL = 2 2 2 2 n p L π = 2 P p EI = (10.9) Therefore, 1. A = 0 → y = 0 → the column is straight! 2. sin pL = 0 → pL = nπ → p = nπ /L → Since (10.6) We have 2 2 2 n EI P L π = For n = 1 2 2cr EI P L π = (10.10) -- Euler’s formula (10.11)
  • 11. Substituting Eq. (10.11) into Eq. (10.6), 2 2cr EI P L π = 2 P p EI = 2 2 2 2 2 crP EI p EI L EI L π π = = = (10.6) (10.11) Therefore, Hence p L π = Equation (10.8) becomes sin x y A L π = (10.12) This is the elastic curve after the beam is buckled.
  • 12. If P < Pcr → sin pL ≠ 0 0sinA pL = 1. A = 0 → y = 0 → the column is straight! 2. sin pL = 0 → pL = nπ → 2 2cr EI P L π = (10.9) Hence, A = 0 and y = 0 → straight configuration
  • 13. Critical Stress: Introducing 2 2 ( ) cr E L r π σ = 2 2 cr cr P EI A L A π σ = = 2 I Ar= Where r = radius of gyration , ,y zx x y z I II r r r A A A = = = 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 2 2cr e EI P L π = 2 2 ( ) cr e E L r π σ = (10.11') (10.13') Case A: One Fixed End, One Free End Le = 2L
  • 15. Case B: Both Ends Fixed At Point C RCx = 0 Q = 0 → 0 VQ It π = = Point D = inflection point → M = 0 → AD and DC are symmetric Hence, Le = L/2
  • 16. 2 2 d y M P V y x dx EI EI EI = = − − 2 P p EI = 2 2 2 d y V p y x dx EI + = − Case C: One Fixed End, One Pinned End M = -Py - Vx Since Therefore, The general solution: The particular solution: sin cosy A px B px= + 2 V y x p EI = −
  • 17. 2 P p EI = V y x P = − Substituting into the particular solution, it follows As a consequence, the complete solution is sin cos V y A px B px x P = + − (10.16)
  • 18. sin V A pL L P = sin V y A px x P = − (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 sin cos V y A px B px x P = + −
  • 19. Taking derivative of the question, cos dy V Ap px dx P = − cos V Ap pL P = sin V A pL L P = B.C.s: @ x = L, dy/dx = θ = 0 (10.18) (10.17) 10 17 10 18 ( . ) ( . ) tan pL pL= (10.19)
  • 20. Solving Eq. (10.19) by trial and error, 2 2 4 4934 4 4934 20 19064. . / . /pL p L p L= ⇒ = ⇒ = 2 2 2 20 19.P EI p P p EI EI L = ⇒ = = 2 2 2 20 19. cr e EI EI P L L π = = Since Therefore, Case C Solving for Le Le = 0.699L ≈ 0.7 L
  • 23. Secant Formula: (10.36) 2 1 1 2 max sec( )e P A ec P L r EA r σ = + 1 1 2 sec( )eP L EA r ≈If Le/r << 1, Eq. (10.36) reduces to 2 1 maxP ecA r σ = + (10.37)
  • 24.
  • 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 -- σ < σy Reality: may violate these assumptions -- use empirical equations and rely lab data
  • 27. Facts: 1. Long Columns: obey Euler’s Equation 2. Short Columns: dominated by σy 3. Intermediate Columns: mixed behavior Test Data:
  • 29. Real Case Design using Empirical Equations: Two Approaches: 1. Allowable Stress Design 2. Load & Resistance Factor Design
  • 30. 1. For L/r ≥ Cc [long columns]: [Euler’s eq.] Structural Steel – Allowable Stress Design Approach I -- w/o Considering F.S. 2 2 ( / )cr E L r π σ = 2 2 1 2 ( / ) [ ]cr Y c L r C σ σ= − 2. For L/r ≤ Cc [short & interm. columns]: where 2 2 2 c Y E C π σ =
  • 31. 1. L/r ≥ Cc : (10.43) Approach II -- Considering F.S. 2 2 1 92. . ( / ) cr all E F S L r σ π σ = = 2. L/r ≤ Cc : 21 1 2 / [ ( ) ] . . . cr Y all c L c F S F S C σ σ σ = = − (10.45)
  • 32. 10.7 Design of Columns under an Eccentric Load centric bendingσ σ σ= + max P Mc A I σ = + (10.56) (10.57) (I) Allowable Stress Method (II) Interaction Method Two Approaches: 1. The section is far from the ends 2. σ < σy
  • 33. I. Allowable-Stress Method all P Mc A I σ+ ≤ (10.58) -- σall is obtained from Section 10.6. -- The results may be too conservative.
  • 34. II. Interaction Method 1 / / all all P A Mc I σ σ + ≤ 1 / / ( ) ( )all centric all bending P A Mc I σ σ + ≤ (10.59) (10.60)(Interaction Formula) ( )all centric σ -- determined using the largest Le Case A: If P is applied in a plane of symmetry:
  • 35. 1max max/ // ( ) ( ) ( ) x x x x all centric all bending all bending M z I M x IP A σ σ σ + + ≤ (10.61) Case B: If P is NOT Applied in a Plane of Symmetry:
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