Lecture 22 Biaxial Columns Design

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Lecture 22 - Biaxial Columns Design

• July 30, 2003
• CVEN 444

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Lecture Goals

• Short Column Biaxial Design
• Slender Column Design

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Biaxial Bending and Axial Load
Ref. PCA Notes on ACI 318-95
Unaxial bending about y-axis
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Biaxial Bending and Axial Load
Ref. PCA Notes on ACI 318-95
The biaxial bending moments Mx Pey My Pex
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Approximate Analysis Methods
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 S2
(1/Pn ex,ey) Plane S2 is defined by points A,B,
and C.
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Approximate Analysis Methods
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
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Approximate Analysis Methods
Design Pu Muy, Mux Pu, Puex, Puey
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Approximate Analysis Methods
Pn Nominal axial load strength at
eccentricities, ex ey Limited to cases when
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Biaxial Bending in Short Columns
Analysis Procedure Reciprocal Load Method
Breslers Formula
Steps
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 Breslers Formula )
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Biaxial Bending in Short Columns
where, f 0.65
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Biaxial 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 f Pn if its acts at ex 8 in. and ey
12 in. Use fc 5 ksi and fy 60 ksi.
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Biaxial Column Example
Compute the P0 load, compression with no moments
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Biaxial 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
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Biaxial Column Example
The components of the equilibrium equation
are Use similar triangles to find the
stress in the steel, fs
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Biaxial Column Example
Compute the moment about the tension
steel where The resulting equation is
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Biaxial 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
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Biaxial Column Example
Combine the two equations and solve for c using
an iterative technique You are solving a cubic
equation
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Biaxial Column Example
Check the assumption that Cs2 is close to zero
This value is small relative to the others
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Biaxial Column Example
This Cs2 11 kips relatively small verses the
overall load, which is So Pnx 733.0 kips
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Biaxial 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
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Biaxial Column Example
The components of the equilibrium equation are
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Biaxial Column Example
Compute the moment about the tension
steel where The resulting equation is
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Biaxial 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
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Biaxial Column Example
Combine the two equations and solve for c using
an iterative technique You are solving a cubic
equation
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Biaxial Column Example
Check the assumption that Cs2 is close to zero
This value is negative so it does not contribute
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Biaxial Column Example
This Cs2 - 2.1 kips relatively small verses the
overall load, which is So Pnx 684.6 kips
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Biaxial Column Example
Compute the nominal load
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Biaxial 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 )
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Slender Columns
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Columns
Slenderness ratio

• Long with a relatively high slenderness ratio
  where lateral or shear walls are required
• Long with a medium slenderness ration that will
  cause a reduction in strength
• Short where the slenderness ratio is small

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Long Columns
Slender Columns
Slender Column
Column with a significant reduction in axial load
capacity due to moments resulting from lateral
deflections of the column (ACI Code significant
reduction 5)
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Long Columns
Less 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.
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Effective 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
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K Factor
YA and YB are the top and bottom factors of the
column. For a hinged end Y is infinite or 10 and
for a fixed end Y is zero or 1
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K Factor
For a Braced Frame(Non-sway)
YA and YB are the top and bottom factors of the
column.
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K Factor
For a Sway Frame a) Restrained _at_both
ends b) One hinged or free end Non-sway
frames Sway frames
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K 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
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General Formulation
Modulus of Elasticity Reinforced Moment (ACI
10.11.1)
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General Formulation
Area Moment of inertia shall be divided by (1
bd) with sustain lateral loads
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K Factor
Use the Y values to obtain the K factors for the
columns.
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Long Column
Eccentrically loaded pin-ended column.
Lateral deflection - increases moment
M P( e D )
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Long Column
Eccentrically loaded pin-ended column.
Do first-order deflection due to Mo Da
second-order deflection due to Po
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Long Column
Eccentrically loaded pin-ended column.
OA - curve for end moment OB - curve for maximum
column moment _at_ mid-height)
Axial capacity is reduced from A to B due to
increase in maximum moment due to Ds
(slenderness effects)
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Long Columns
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Long Column - Slenderness Ratio
Slenderness Ratio for columns
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Long Column - Slenderness Ratio
Slenderness Ratio for columns
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Long Column - Slenderness Ratio
Slenderness Ratio for columns in frames
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Long Column - Slenderness Ratio
Slenderness Ratio for columns in frames
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Long 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
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Long Column
double curvature
singular curvature
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Long Columns
M1/M2 Ratio of moments at two column ends
where M2 gt M1 (-1.0 to 1.0 range) - single
curvature - double curvature
is typically conservative (non-sway frames)
Note Code (10.12.2) M1/M2 -0.5 non-sway frames
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Long Column
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Moment Magnification in Non-sway Frames
If the slenderness effects need to be considered.
The non-sway magnification factor, dns, will
cause an increase in the magnitude of the design
moment. where
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Moment Magnification 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
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Moment Magnification in 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
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Moment Magnification in Non-sway Frames
The minimum allowable value of M2 is The sway
frame uses a similar technique, see the text on
the components.
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Design of Long Columns- Example
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 M12500 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 YA
1.3 and YB 0.9. Use a d2.5 in. cover with
an sustain load 50 and fc 7 ksi and fy
60 ksi.
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Design of Long Columns- Example
Compute the factored loads and moments are 80 of
the dead loads
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Design of Long Columns- Example
Compute the k value for the braced compression
members Therefore, use k 0.81
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Design of Long Columns- 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.
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Design of Long Columns- Example
So slenderness must be considered. Since frame
has no side sway, M2 M2ns, ds 0 Minimum M2
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Design of Long Columns- Example
Compute components of concrete The moment of
inertia is
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Design of Long Columns- Example
Compute the stiffness
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Design of Long Columns- Example
The critical load is
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Design of Long Columns- Example
Compute the coefficient
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Design of Long Columns- Example
The magnification factor
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Design of Long Columns- Example
The design moment is Therefore the design
conditions are
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Design of Long Columns- Example
Assume that the r 2.0 or 0.020 Use 14 9
bars or 14 in2
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Design of Long Columns- Example
The column is compression controlled so c/d gt
0.6. Check the values for c/d 0.6
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Design of Long Columns- Example
Check the strain in the tension steel and
compression steel.
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Design of Long Columns- Example
The tension steel
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Design of Long Columns- Example
Combined forces
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Design of Long Columns- Example
Combined force
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Design of Long Columns- Example
Moment is
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Design of Long Columns- Example
The eccentricity is Since the e 11.28 in. lt
13.62 in. The section is in the compression
controlled region f 0.65. You will want to
match up the eccentricity with the design.
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Design of Long Columns- Example
Check the values for c/d 0.66
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Design of Long Columns- Example
Check the strain in the tension steel and
compression steel.
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Design of Long Columns- Example
The tension steel
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Design of Long Columns- Example
Combined forces
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Design of Long Columns- Example
Combined force
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Design of Long Columns- Example
Moment is
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Design of Long Columns- Example
The eccentricity is Since the e 11.28 in.
The reduction factor is equal to f 0.65.
Compute the design load and moment.
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Design of Long Columns- Example
The design conditions are
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Design of Long Columns- Example
Design the ties Provide 3 ties, spacing will be
the minimum of Therefore, provide 3 ties _at_
18 in. spacing.
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Using Interaction Diagrams

• Determine eccentricity.
• Estimate column size required base on axial load.
• Determine e/h and required fPn/Ag
• Determine which chart to use.

• Select steel sizes.
• Design ties by ACI code
• Design sketch