BIAXIAL BENDING
Consideration of Columns with Axial Load and Biaxial Bending
1.
Occurrence. Will be presented in any situations where beams frame into column at
right angles, or bridge pier, etc.
2.
Criteria for determining nominal strength are the same as for columns in uniaxial
bending (from material standpoint). Problem is complicated by the fact that the
neutral axis of the bending is no longer parallel to a major axis. Optimum column
dimensions are not likely to be equal if projected eccentricities vary.
3.
Graphical presentation of the problem
Figure 1. Interaction surface for compression plus biaxial bending: (a) uniaxial bending about
Y- axis; (b) uniaxial bending about X-axis; (c) biaxial bending about diagonal axis; and (d)
interaction surface.
1
2
Current Methods of Analysis
Bresler, B. “Design Criteria for Reinforced Concrete Columns Under Axial Load and Biaxial
Loading.” Journal of the American Concrete Institute, Vol. 57, No. 5, November 1960, pp.481490.
Bresler based his analysis on an assumption of a number of possible “Failure Surfcace” in three
dimensions.
Failure Surface 1 -
Failure point defined as a function of axial load and eccentricities.
Failure Surface 2 -
Similar basis with 1 failure point defined as function of 1/ pn , ex , ey
Bresler reasoned:
1.
The failure surface is too complicated to exactly define.
2.
An acceptable approximation could be defined by a plane which passes through three
points which could be found by conventional (uniaxial bending) analysis.
•
3
Reciprocal Load Mehtod
Bresler’s reciprocal load equation derived from the geometry of the approximate plane:
1
1
1
1
=
+
−
pn pnx 0 pny 0 p0
where
pn = Approximate value of ultimate load in biaxial bending with eccentricity of ex , ey .
pnx 0 = Ultimate load when only eccentricity e y is present ( ex = 0 )
pny 0 = Ultimate load when only eccentricity ex is present ( ey = 0 )
p0 = Ultimate load for concentrically loaded column.
This procedure is acceptably accurate for design purposes provided pn ≥ 0.1P0 . If
Pn < 0.1P0 , it would be more accurate to neglect the axial force entirely and to calculate
the section for biaxial bending only.
ACI strength reduction factors do not change the development in any fundamental way as long
as the φ factor is constant for all columns.
1
1
1
1
=
+
−
φ pn φ pnx 0 φ pny 0 φ p0
Note that:
• It is necessary to use the uniaxial curves without the horizontal cutoff in obtaining values
for the above equation.
Tied Columns
0.8φ p0
• φ pn ≤ 
0.85φ p0 Spiral Columns
4
Load Contour Method
The load contour method is based on representing the failure surface of the Figure 1 give above
by a family of curves corresponding to constant value of pn . The general form of these curves
can be approximated by a non-dimensional interaction equation:
α
α2
1
 M ny 
 M nx 
 = 1.0

 + 
M
0
ny
 M nx 0 


where:
M nx = Pn ey
M nx 0 = M nx when M ny = 0
M ny = Pn ex
M ny 0 = M ny when M nx = 0
This equation gives the surface of design strength. The parameter alpha is
1.15 < α < 1.55 for square and rectangular columns where β is tabulated for specific
•
•
•
Strength
Geometry
Material strength
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Design Example for Biaxial Bending
ex = 6"
15”
ey = 3"
f c′ = 4 ksi
Ast = 8 in 2
f y = 60 ksi
2.5”
Pu = 275 kips
12”
2.5”
Check the adequacy of the trial design
(a)
using the reciprocal load method
(b)
using load contour method
2.5”
2.5”
(a) Using the reciprocal load method - Solution:
About Y-axis
15

= 0.75

16

As
8

ρt =
=
= 0.033 Use Graph A.7
bh 240

e 6

=
= 0.30

h 20

γ=
Read from graph and simplify
φ Pn
 A = 1.75 → φ Pn = 1.75 × 240 = 420 kips
 g

φ Pn = 3.65 → φ P = 3.65 × 240 = 876 kips
n
 Ag
About X-axis
7
γ = = 0.58 say 0.6
12
6
7

= 0.58

12

As
8

=
= 0.033 Use Graph A.6
ρt =
bh 240

e 3

=
= 0.25

h 12

γ=
Read from graph and simplify
φ Pn
 A = 1.8 → φ Pn = 1.8 × 240 = 432 kips
 g

φ Pn = 3.65 → φ P = 3.65 × 240 = 876 kips
n
 Ag
The reciprocal method gives:
1
1
1
1
=
+
−
φ pn φ pnx 0 φ pny 0 φ p0
1
1
1
1
=
+
−
= 0.00356
φ pn 432 420 876
φ pn = 281 kips > Pu = 275 kips
Therefore the design is adequate.
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Using load contour method Solution
About Y-axis
Pu = φ Pn = 275 kips 

φ Pn 275
 Use Graph A.7
=
= 1.15

Ag 240

φ M nx 0
Ag h
= 0.62
φ M nx 0 = 0.62 × 240 × 20 = 2980 in − kips
About X-axis
Pu = φ Pn = 275 kips 

φ Pn 275
 Use Graph A.6
=
= 1.15

Ag 240

φ M nx 0
Ag h
= 0.53
φ M nx 0 = 0.53 × 240 × 12 = 1530 in − kips
M uy = Pu ex = 275 × 6 = 1650 in − kips
M ux = Pu ey = 275 × 3 = 825 in − kips
The reciprocal method:
α
α2
1
 M ny 
 M nx 
 = 1.0

 + 
M
M
 nx 0 
 ny 0 
1.15
 825 


 1530 
1.15
 1650 
+

 2980 
= 0.491 + 0.507 = 0.998 < 1.000 This column is adequate.
log 0.5
log 0.5
β = 0.56 → α =
= 1.19
log β
log 0.56
Note: Consider consider biaxial bending when estimated eccentricity ratio approaches or exceed
0.2.
From Bresler α =
8
Design Example
Problem: Select a tied column cross-section to resist factored loads and moments of Pu = 420
kips, Mux = 70 ft-kips, and Muy = 80 ft-kips. Use 8#8 bars in each face and No. 3 ties.
f c′ = 4 ksi
f y = 60 ksi
Solution:
8 No. 8 bars → As = 8 × 0.79 = 6.32 in 2
Pu = 0.8φ 0.85 f c′Ac + As f y 
Pu = 0.8 × 0.65  0.85(4)( Ag − 6.32) + (6.32)(60)  = 420
Solve for gross cross sectional area
Ag = 132 in 2
Select:
b = h = 11.45 → use b = h = 12 in
Try a 12 inch by 12 inch column, use 1.5 inch cover, No. 3 ties, and assume No. 8 bars:
12 − 2(1.5 + 0.375 + 0.5)

= 0.60 

12

As
6.32

=
= 0.044
ρt =

bh 12 × 12
γ=
M ux 70 × 12

=
= 0.486 
 Pux
Ag h 144 × 12
 A = 2.4 ⇒ Pux = 2.4 × 144 = 345.6 kips


 g
ρt = 0.044
 Graph A.6 → 
φ P0 = 4.15 ⇒ P = 4.15 × 144 = 517.6 kips

γ = 0.60
0

 Ag

9

80 ×12
= 0.555
 Puy
Ag h 144 ×12
= 1.75 ⇒ Pux = 1.75 ×144 = 252 kips


A

 g
ρt = 0.044
 Graph A.5 → 

φ P0 = 4.15 ⇒ P = 4.15 × 144 = 517.6 kips
γ = 0.60
0
 Ag



M uy
=
The reciprocal method gives:
1
1
1
1
=
+
−
φ pn φ pnx 0 φ pny 0 φ p0
1
1
1
1
=
+
−
φ pn 345.6 252 597.6
φ pn = 195 kips < Pu = 420 kips ---- This column is not adequate.
Try a 14 inch by 14 inch column, use 1.5 inch cover, No. 3 ties, and assume No. 8 bars:
14 − 2(1.5 + 0.375 + 0.5)

= 0.66 

14

As
6.32

ρt =
=
= 0.032
bh 14 × 14

γ=
M ux

70 × 12

=
= 0.3
 Pux

Ag h 14 × 14 ×14
 A = 2.9 

 g

ρt = 0.032
 → Fig A.7 → 

φ
P
0


= 3.6  γ = 0.66
γ = 0.75


 Ag


  Pux
= 2.84 → Pux = 2.84 × 14 ×14 = 557 kips

Ag
M ux
70 × 12



=
= 0.3
 Pux

Ag h 14 × 14 ×14
 A = 2.8   φ P0 = 3.6 → P = 3.6 × 14 × 14 = 713.4 kips

ux

 g
ρt = 0.032
  Ag
 → Fig A.6 → 

φ P0 = 3.6 
γ = 0.60


 Ag


10


80 × 12
= 0.35
 Pux

Ag h 14 × 14 × 14
 A = 2.75


 g

ρt = 0.032
 → Fig A.7 → 

 φ P0 = 3.6  
γ = 0.75
γ = 0.66

 Ag


  Pux
= 2.69 → Pux = 2.69 × 14 × 14 = 528 kips

Ag
M uy

80 × 12


=
= 0.35
 Pux

Ag h 14 × 14 × 14
 A = 2.65  φ P0 = 3.6 → P = 3.6 × 14 × 14 = 713.4 kips

ux

 g
  Ag
ρt = 0.032
 → Fig A.6 → 

 φ P0 = 3.6 
γ = 0.60


 Ag



M uy
=
The reciprocal method gives:
1
1
1
1
=
+
−
φ pn φ pnx 0 φ pny 0 φ p0
1
1
1
1
=
+
−
φ pn 557 528 717.4
φ pn = 435 kips > Pu = 420 kips ---- This column is adequate.
Need to finish detailing
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