Bia Axial Column Design

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.


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


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.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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)


Slenderness Ratio for columns

Fixed-Pinned Connection

Partial restrained Connection

(c)

(d)


Slenderness Ratio for columns in frames


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


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


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.


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


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


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


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


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


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


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


The magnification factor

mns

u

c

0.89

800 k1 10.75 0.75 7354.3 k

1.04 1.0

C

P

P


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


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


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


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


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


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


Combined force

n c s1

1405.7 k 378.35 k 406.0 k

1378.05 k

P C C T


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


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


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


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


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


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


Combined force

n c s1

1546.26 k 378.35 k 313.74 k

1610.9 k

P C C T


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


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


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 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

Bia Axial Column Design

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