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Lecture 22 - Biaxial Lecture 22 - Biaxial Columns DesignColumns Design
July 30, 2003CVEN 444
Lecture GoalsLecture Goals
Short Column Biaxial Design Slender Column Design
Biaxial Bending and Axial Biaxial Bending and Axial LoadLoadRef. PCA Notes on ACI 318-95
Unaxial bending about y-axis
Biaxial Bending and Axial Biaxial Bending and Axial LoadLoad
Ref. PCA Notes on ACI 318-95
The biaxial bending moments
Mx = P*ey
My = P*ex
Approximate Analysis Approximate Analysis MethodsMethods
Use Reciprocal Failure surface S2 (1/Pn,ex,ey)
The ordinate 1/Pn on the surface S2 is approximated by ordinate 1/Pn on the plane S’2 (1/Pn ex,ey)
Plane S2 is defined by points A,B, and C.
Approximate Analysis Approximate Analysis MethodsMethods
P0 = Axial Load Strength under pure axial compression(corresponds to point C ) Mnx = Mny = 0
P0x = Axial Load Strength under uniaxial eccentricity, ey (corresponds to point B ) Mnx = Pney
P0y = Axial Load Strength under uniaxial eccentricity, ex (corresponds to point A ) Mny = Pnex
Approximate Analysis Approximate Analysis MethodsMethodsDesign: Pu Muy, Mux Pu, Puex, Puey
Approximate Analysis Approximate Analysis MethodsMethods
Pn = Nominal axial load strength at eccentricities, ex & ey Limited to cases when
00y0x
n
00y0xnn
1111
11111
PPP
P
PPPPP
gcn 1.0 AfP
Biaxial Bending in Short Biaxial Bending in Short ColumnsColumns
1) Calculate P0
2) Calculate P0y ( Pn for e = ex, ey = 0 )
3)Calculate P0x ( Pn for ex= 0, e = ey )
4) Calculate Pn (from Bresler’s Formula )
Analysis Procedure: Reciprocal Load Method Bresler’s Formula:
Steps:00y0xn
1111
PPPP
Biaxial Bending in Short Biaxial Bending in Short ColumnsColumns
where,
nu PP
Biaxial Column ExampleBiaxial Column Example
The section of a short tied column is 16 x 24 in. and is reinforced with 8 #10 bars as shown. Determine the allowable ultimate load on the section Pn if its acts at ex = 8 in. and ey = 12 in. Use fc = 5 ksi and fy = 60 ksi.
Biaxial Column ExampleBiaxial Column Example
Compute the P0 load, compression with no moments
2 2st
0 c g st st y
2
2
n0 0
8 1.27 in 10.16 in
0.85
0.85 5 ksi 24.0 in. 24.0 in. 10.16 in
10.16 in 60 ksi
2198.4 k
0.8 2198.4 k 1758.7 k
A
P f A A A f
P rP
Biaxial Column ExampleBiaxial Column Example
Compute Pnx, by starting with ey term and assume that compression controls. Check by
Compute the nominal load, Pnx and assume second compression steel does not contribute
assume small
y
2 212 in. 21.5 in. 14.33 in.
3 3e d
n c s1 s2P C C C T
Biaxial Column ExampleBiaxial Column Example
The components of the equilibrium equation are:
Use similar triangles to find the stress in the steel, fs
c
2s1
2s
s
0.85 5 ksi 16 in. 0.8 54.4
3.81 in 60 ksi 0.85 5 ksi 212.4 kips
3.81 in
21.5 in.1 29000 ksi 0.003 1 87 ksi
C c c
C
T f
df
c c
Biaxial Column ExampleBiaxial Column Example
Compute the moment about the tension steel:
where
The resulting equation is:
1n c s12
cP e C d C d d
n
2n
9.5 in. 12 in. 21.5 in.
21.5 in. 54.4 21.5 in. 0.4
212.4 k 21.5 in. 2.5 in.
54.4 1.01 187.7
e
P c c
P c c
Biaxial Column ExampleBiaxial Column Example
Combine the two equations and solve for Pn using an iterative solution
Set the two equation equal to one another and sole for fs
and the definition:
n s
2n
54.4 212.4 3.81
54.4 1.01 187.7
P c f
P c c
2s
s
0.265 6.483
21.5 in.87 1
c
f c
f
Biaxial Column ExampleBiaxial Column Example
Combine the two equations and solve for c using an iterative technique
You are solving a cubic equation
221.5 in.87 1 0.265 6.483
cc
c (in.) fs (ksi) RHS15 37.7 66.1281910 100.05 32.9919413 56.88462 51.2831513.3 53.6391 53.3747113.315 53.48066 53.48054
Biaxial Column ExampleBiaxial Column Example
Check the assumption that Cs2 is close to zero
s2
2s2
12 in. 12 in.1 87 ksi 1 87 ksi
13.315 in.
8.59 ksi
2.54 in 8.59 ksi 0.85 5 ksi
11.0 kips
fc
C
This value is small relative to the others
Biaxial Column ExampleBiaxial Column Example
This Cs2 = 11 kips relatively small verses the overall load, which is
So Pnx = 733.0 kips
n s54.4 212.4 3.81
54.4 13.315 in. 212.4 k 3.81 53.48 ksi
733.0 k
P c f
Biaxial Column ExampleBiaxial Column Example
Start with ex term and assume that compression controls.
Compute the nominal load, Pny and assume second compression steel does not contribute
assume small
x
2 28.0 in. 13.5 in. 9 in.
3 3e d
n c s1 s2P C C C T
Biaxial Column ExampleBiaxial Column Example
The components of the equilibrium equation are:
c
2s1
2s
s
0.85 5 ksi 24 in. 0.8 81.6
3.81 in 60 ksi 0.85 5 ksi 212.4 kips
3.81 in
13.5 in.1 29000 ksi 0.003 1 87 ksi
C c c
C
T f
df
c c
Biaxial Column ExampleBiaxial Column Example
Compute the moment about the tension steel:
where
The resulting equation is:
1n c s12
cP e C d C d d
n
2n
5.5 in. 8 in. 13.5 in.
13.5 in. 81.6 13.5 in. 0.4
212.4 k 13.5 in. 2.5 in.
81.6 2.42 173.07
e
P c c
P c c
Biaxial Column ExampleBiaxial Column Example
Combine the two equations and solve for Pn using an iterative solution
Set the two equation equal to one another and sole for fs
and the definition:
n s
2n
81.6 212.4 3.81
81.6 2.42 173.07
P c f
P c c
2s
s
0.634 10.324
13.5 in.87 1
c
f c
f
Biaxial Column ExampleBiaxial Column Example
Combine the two equations and solve for c using an iterative technique
You are solving a cubic equation
213.5 in.87 1 0.634 10.324
cc
c (in.) fs (ksi) RHS10 30.45 73.763718 59.8125 50.925318.5 51.17647 56.159118.3 54.50602 54.02753
8.31735 54.21084 54.21043
Biaxial Column ExampleBiaxial Column Example
Check the assumption that Cs2 is close to zero
s2
2s2
8 in.1 87 ksi
8.317 in.
3.32 ksi
2.54 in 3.32 ksi 0.85 5 ksi
2.10 kips
f
C
This value is negative so it does not contribute
Biaxial Column ExampleBiaxial Column Example
This Cs2= - 2.1 kips relatively small verses the overall load, which is
So Pnx = 684.6 kips
n s81.6 212.4 3.81
81.6 8.317 in. 212.4 k 3.81 54.21 ksi
684.6 k
P c f
Biaxial Column ExampleBiaxial Column Example
Compute the nominal load
n nx ny n0
n u n
1 1 1 1
1 1 1
733.0 k 684.6 k 1758.7 k
443.2 k 0.65 443.2 k 288.1 k
P P P P
P P P
Biaxial Column ExampleBiaxial Column Example
Note: the Pnx & Pny include the corner steel bars in both calculations a more conservative solution would be to use 1/2 the steel in each direction so As= 2(1.27 in2) which would reduce Pu . (Remember fs can not be greater than 60 ksi, so that Pnx = 620.3 k and Pny= 578.4 k Pn = 360.7 k and Pu= 234.5 k )
Slender ColumnsSlender Columns
ColumnsColumns
Long with a relatively high slenderness ratio where lateral or shear walls are requiredLong with a medium slenderness ration that will cause a reduction in strengthShort where the slenderness ratio is small
Slenderness ratio = ukl
r
““Long” ColumnsLong” ColumnsSlender Columns
Column with a significant reduction in axial load capacity due to moments resulting from lateral deflections of the column (ACI Code: significant reduction 5%)
Slender = Column
““Long” ColumnsLong” ColumnsLess than 10 % of columns in “braced” or “non-sway” frames and less than half of columns in “unbraced” or “sway” frames would be classified as “slender” following ACI Code Procedure.
Effective LengthEffective Length
The effective length - klu
lu - It measures the clear distance between floors.
k - a factor, which represents the ratio of the distance between points of zero moments in the columns
K FactorK Factor
beams of /
columns of /
u
u
lEI
lEI
A andB are the top and bottom factors of the column. For a hinged end is infinite or 10 and for a fixed end is zero or 1
K FactorK Factor
A andB are the top and bottom factors of the column.
For a Braced Frame:(Non-sway)
A B
min
0.70 0.05 1.0 smaller of
0.85 0.05 1.0
kk
k
K FactorK FactorFor a Sway Frame:
a) Restrained @both ends
b) One hinged or free end
Non-sway frames:
Sway frames:
mm avg m
m m
20if 2.0 : 1
20
if 2.0 : 0.9 1
k
k
2.0 0.3k
0 1.0
1.0 typically k=1.5
k
k
K FactorK Factor
The general assumptions are
- Structure consists of symmetric rectangular frames
- The girder moment at a joint is distributed to columns according to their relative stiffness
- All columns reach their critical loads at the same time
General FormulationGeneral Formulation
Modulus of Elasticity
Reinforced Moment (ACI 10.11.1) c
c5.1
c
57000
33
f
fwE
column afor 70.0
beam afor 35.0
g
g
II
II
General FormulationGeneral Formulation
Area
Moment of inertia shall be divided by (1 + d) with sustain lateral loads
gAA
d =Max. factored sustain lateral load
Max. factored axial load
K FactorK FactorUse the values to obtain the K factors for the columns.
““Long” ColumnLong” Column
Lateral deflection - increases moment
Eccentrically loaded pin-ended column.
M = P*( e + )
““Long” ColumnLong” ColumnEccentrically loaded pin-ended column.
o = first-order deflection due to Mo
a = second-order deflection due to Po
““Long” ColumnLong” Column
OA - curve for end moment
OB - curve for maximum column moment @ mid-height)
Eccentrically loaded pin-ended column.
Axial capacity is reduced from A to B due to increase in maximum moment due to ’s (slenderness effects)
““Long” Columns Long” Columns
From ACI Sec. 12.10.2 , the slenderness effects may be neglected if
k = effective length factor (function of end restraints)
Non-sway frames
Sway frames
2
1
ratio sslendernes
u 1234M
M
r
kL
k
k
0.1
0.15.0
““Long” Column - Long” Column - Slenderness RatioSlenderness Ratio
Slenderness Ratio for columns
Pinned-Pinned Connection
Fixed-Fixed Connection
(a)
(b)
““Long” Column - Long” Column - Slenderness RatioSlenderness Ratio
Slenderness Ratio for columns
Fixed-Pinned Connection
Partial restrained Connection
(c)
(d)
““Long” Column - Long” Column - Slenderness RatioSlenderness Ratio
Slenderness Ratio for columns in frames
““Long” Column - Long” Column - Slenderness RatioSlenderness Ratio
Slenderness Ratio for columns in frames
““Long” ColumnLong” Column
Unsupported height of column from top of floor to bottom of beams or slab in floor
Radius of gyration
= 0.3* overall depth of rectangular columns = 0.25* overall depth of circular columns
lu =
r =A
I
““Long” ColumnLong” Column
singular curvature double curvature
Ratio of moments at two column ends, where M2 > M1 (-1 to 1 range)
M1/M2 =
0
2
1 M
M0
2
1 M
M
““Long” ColumnsLong” ColumnsM1/M2 = Ratio of moments at two column ends
where M2 > M1 (-1.0 to 1.0 range)
- single curvature
- double curvature
1.0k and
5.0
2
1
M
M
is typically conservative (non-sway frames)
Note Code (10.12.2) M1/M2 -0.5 non-sway frames
““Long” ColumnLong” Column
framessway -non veconservati typicallyis
0.1 and
5.0
2
1
kM
M
(non-sway frames)Note: Code 10.12.2 5.0
2
1 M
M
Possible range of = 22 to 40
r
ukl
Moment Magnification Moment Magnification in Non-sway Framesin Non-sway Frames
If the slenderness effects need to be considered. The non-sway magnification factor, ns, will cause an increase in the magnitude of the design moment.
where
mns
u
c
1.0
10.75
C
P
P
c ns 2M M
Moment Magnification Moment Magnification in Non-sway Frames in Non-sway Frames
The components of the equation for an Euler bucking load for pin-end column
and the stiffness, EI is taken as
2
c 2
u
EIP
kl
c g s se c g
conservativelyd d
0.2 0.4
1 1
E I E I E IEI EI
Moment Magnification in Moment Magnification in Non-sway Frames Non-sway Frames
A coefficient factor relating the actual moment diagram to the equivalent uniform moment diagram. For members without transverse loads
For other conditions, such as members with transverse loads between supports, Cm = 1.0
1m
2
0.6 0.4 0.4M
CM
Moment Magnification Moment Magnification in Non-sway Frames in Non-sway Frames
The minimum allowable value of M2 is
The sway frame uses a similar technique, see the text on the components.
2,min u 0.6 0.03M P h
Design of Long Columns- Design of Long Columns- ExampleExample
A rectangular braced column of a multistory frame building has floor height lu =25 ft. It is subjected to service dead-load moments M2= 3500 k-in. on top and M1=2500 k-in. at the bottom. The service live load moments are 80% of the dead-load moments. The column carries a service axial dead-load PD = 200 k and a service axial live-load PL = 350 k. Design the cross section size and reinforcement for this column. Given A = 1.3 and B = 0.9. Use a d’=2.5 in. cover with an sustain load = 50 % and fc = 7 ksi and fy = 60 ksi.
Design of Long Columns- Design of Long Columns- ExampleExample
Compute the factored loads and moments are 80% of the dead loads
u D L
1u D L
2u D L
1.2 1.6 1.2 200 k 1.6 350 k
800 k
1.2 1.6 1.2 2500 k-in 1.6 0.8 2500 k-in
6200 k-in.
1.2 1.6 1.2 3500 k-in 1.6 0.8 3500 k-in
8680 k-in.
P P P
M M M
M M M
Design of Long Columns- Design of Long Columns- ExampleExample
Compute the k value for the braced compression members
Therefore, use k = 0.81
A B
min
0.7 0.05 0.7 0.05 1.3 0.9
0.81 1.0
0.85 0.05 0.85 0.05 0.9
0.895 1.0
k
k
Design of Long Columns- Design of Long Columns- Example Example
Check to see if slenderness is going to matter. An initial estimate of the size of the column will be an inch for every foot of height. So h = 25 in.
n0.81 25 ft 12 in./ft
32.4r 0.3 25 in.
6200 k-in.32.4 34 12 25.43
8680 k-in.
kl
Design of Long Columns- Design of Long Columns- Example Example
So slenderness must be considered. Since frame has no side sway, M2 = M2ns, s =0 Minimum M2
2,min u
2
0.6 0.03 800 k 0.6 0.03 25 in.
1080 k-in. 8680 k-in.
M P h
M
Design of Long Columns- Design of Long Columns- Example Example
Compute components of concrete
The moment of inertia is
1.51.5c c
6 3
33 33 150 7000
5.07x10 psi 5.07x10 ksi
E w f
33
g
4
25 in. 25 in.
12 12
32552 in
bhI
Design of Long Columns- Design of Long Columns- ExampleExample
Compute the stiffness
3 4c g
d
7 2
0.4 5.07x10 ksi 32552 in0.4
1 1 0.5
4.4x10 k-in
E IEI
Design of Long Columns- Design of Long Columns- ExampleExample
The critical load is
2 7 22
c 2 2
u
4.4x10 k-in
12 in.0.81 25 ft
ft
7354.3 k
EIP
kl
Design of Long Columns- Design of Long Columns- ExampleExample
Compute the coefficient
1m
2
0.6 0.4
6200 k-in.0.6 0.4 0.89 0.4
8680 k-in.
MC
M
Design of Long Columns- Design of Long Columns- ExampleExample
The magnification factor
mns
u
c
0.89
800 k1 10.75 0.75 7354.3 k
1.04 1.0
C
P
P
Design of Long Columns- Design of Long Columns- ExampleExample
The design moment is
Therefore the design conditions are
c ns 2 1.04 8680 k-in. 9027.2 k-in.M M
c c800 k & M 9027.2 k-in.
9027.2 k-in.e 11.28 in.
800 k
P
Design of Long Columns- Design of Long Columns- ExampleExample
Assume that the = 2.0 % or 0.020
Use 14 # 9 bars or 14 in2
2 2s 0.02 25 in. 12.5 inA
2s
2cs
7.0 in
7.0 in
A
A
Design of Long Columns- Design of Long Columns- ExampleExample
The column is compression controlled so c/d > 0.6. Check the values for c/d = 0.6
1
0.6 0.6 22.5 in. 13.5 in.
0.7 13.5 in. 9.45 in.
c d
a c
Design of Long Columns- Design of Long Columns- ExampleExample
Check the strain in the tension steel and compression steel.
s1 cu
cs1 s s1
cs1
13.5 in. 2.5 in.0.003
13.5 in.
0.00244
29000 ksi 0.00244
70.76 ksi 60 ksi
c d
c
f E
f
Design of Long Columns- Design of Long Columns- ExampleExample
The tension steel
s cu
s s s
22.5 in. 13.5 in.0.003 0.002
13.5 in.
29000 ksi 0.002
58 ksi
d c
c
f E
Design of Long Columns- Design of Long Columns- ExampleExample
Combined forces
c c
2s1 cs cs c
2s s
0.85 0.85 7 ksi 25 in. 9.45 in.
1405.7 k
0.85 7 in 60 ksi 0.85 7 ksi
378.35 k
7 in 58 ksi
406.0 k
C f ba
C A f f
T A f
Design of Long Columns- Design of Long Columns- ExampleExample
Combined force
n c s1
1405.7 k 378.35 k 406.0 k
1378.05 k
P C C T
Design of Long Columns- Design of Long Columns- ExampleExample
Moment is
n c s12 2 2 2
9.45 in.1405.7 k 12.5 in.
2
378.35 k 12.5 in. 2.5 in.
406.0 k 22.5 in. 12.5 in.
18773 k-in
h a h hM C C d T d
Design of Long Columns- Design of Long Columns- ExampleExample
The eccentricity is
Since the e = 11.28 in. < 13.62 in. The section is in the compression controlled region = 0.65. You will want to match up the eccentricity with the design.
n
n
18773 k-in
1378.05 k
13.62 in.
Me
P
Design of Long Columns- Design of Long Columns- ExampleExample
Check the values for c/d = 0.66
1
0.66 0.66 22.5 in. 14.85 in.
0.7 14.85 in. 10.395 in.
c d
a c
Design of Long Columns- Design of Long Columns- ExampleExample
Check the strain in the tension steel and compression steel.
s1 cu
cs1 s s1
cs1
14.85 in. 2.5 in.0.003
14.85 in.
0.00249
29000 ksi 0.00249
72.35 ksi 60 ksi
c d
c
f E
f
Design of Long Columns- Design of Long Columns- ExampleExample
The tension steel
s cu
s s s
22.5 in. 14.85 in.0.003
14.85 in.
0.00155
29000 ksi 0.00155
44.82 ksi
d c
c
f E
Design of Long Columns- Design of Long Columns- ExampleExample
Combined forces
c c
2s1 cs cs c
2s s
0.85 0.85 7 ksi 25 in. 10.395 in.
1545.26 k
0.85 7 in 60 ksi 0.85 7 ksi
378.35 k
7 in 44.82 ksi
313.74 k
C f ba
C A f f
T A f
Design of Long Columns- Design of Long Columns- ExampleExample
Combined force
n c s1
1546.26 k 378.35 k 313.74 k
1610.9 k
P C C T
Design of Long Columns- Design of Long Columns- ExampleExample
Moment is
n c s12 2 2 2
10.395 in.1545.26 k 12.5 in.
2
378.35 k 12.5 in. 2.5 in.
313.74 k 22.5 in. 12.5 in.
18205.2 k-in
h a h hM C C d T d
Design of Long Columns- Design of Long Columns- ExampleExample
The eccentricity is
Since the e 11.28 in. The reduction factor is equal to = 0.65. Compute the design load and moment.
n
n
18205.2 k-in
1610.9 k
11.30 in.
Me
P
Design of Long Columns- Design of Long Columns- ExampleExample
The design conditions are
u n
u n
0.65 1610.9 k
1047.1 k 800 k OK!
0.65 18205.2 k-in
11833.4 k-in. 9027.2 k-in. OK!
P P
M M
Design of Long Columns- Design of Long Columns- ExampleExample
Design the ties
Provide #3 ties, spacing will be the minimum of:
Therefore, provide #3 ties @ 18 in. spacing.
stirrup
bar
48 48 0.375 in. 18 in.
smallest 16 16 1.128 in. 18 in. controls
h 25 in.
d
s d
Using Interaction Using Interaction Diagrams Diagrams
Determine eccentricity.Estimate column size required base on axial load.Determine e/h and required Pn/Ag
Determine which chart to use.
Select steel sizes.Design ties by ACI codeDesign sketch