lecture chp 9&10 column design

Lecture Chp-9&10 – Lecture Chp-9&10 – ColumnsColumns

Lecture GoalsLecture Goals

Definitions for short columnsColumns

Analysis and Design of Analysis and Design of “Short” Columns“Short” Columns

General Information

Vertical Structural members

Transmits axial compressive loads with or without moment

transmit loads from the floor & roof to the foundation

Column:


General Information

Column Types:

1. Tied

2. Spiral

3. Composite

4. Combination

5. Steel pipe


Tie spacing h (except for seismic)

tie support long bars (reduce buckling)

ties provide negligible restraint to lateral expose of core

Tied Columns - 95% of all columns in buildings are tied


Pitch = 1.375 in. to 3.375 in.

spiral restrains lateral (Poisson’s effect)

axial load delays failure (ductile)

Spiral Columns


Elastic Behavior An elastic analysis using the transformed section method would be:

stcc nAA

Pf

For concentrated load, P

uniform stress over section

n = Es / Ec

Ac = concrete area

As = steel areacs nff


Elastic Behavior The change in concrete strain with respect to time will effect the concrete and steel stresses as follows:

Concrete stress

Steel stress


Elastic Behavior An elastic analysis does not work, because creep and shrinkage affect the acting concrete compression strain as follows:


Elastic Behavior

Concrete creeps and shrinks, therefore we can not calculate the stresses in the steel and concrete due to “acting” loads using an elastic analysis.


Elastic Behavior Therefore, we are not able to calculate the real stresses in the reinforced concrete column under acting loads over time. As a result, an “allowable stress” design procedure using an elastic analysis was found to be unacceptable. Reinforced concrete columns have been designed by a “strength” method since the 1940’s.

Creep and shrinkage do not affect the strength of the member.

Note:

Behavior, Nominal Capacity and Behavior, Nominal Capacity and Design under Concentric Axial Design under Concentric Axial

loadsloads

Initial Behavior up to Nominal Load - Tied and spiral columns.

1.


loadsloads


loadsloads

stystgc0 *85.0 AfAAfP

Factor due to less than ideal consolidation and curing conditions for column as compared to a cylinder. It is not related to Whitney’s stress block.

Let

Ag = Gross Area = b*h Ast = area of long steel fc = concrete compressive strength fy = steel yield strength


loadsloads

Maximum Nominal Capacity for Design Pn (max) 2.

0maxn rPP

r = Reduction factor to account for accidents/bending

r = 0.80 ( tied )

r = 0.85 ( spiral )ACI 10.3.6.3


loadsloads

Reinforcement Requirements (Longitudinal Steel Ast)

3.

g

stg A

A

- ACI Code 10.9.1 requires

Let

08.001.0 g


loadsloads

3.

- Minimum # of Bars ACI Code 10.9.2 min. of 6 bars in circular arrangement w/min. spiral reinforcement.

min. of 4 bars in rectangular arrangement

min. of 3 bars in triangular ties

Reinforcement Requirements (Longitudinal Steel Ast)


loadsloads

3.

ACI Code 7.10.5.1

Reinforcement Requirements (Lateral Ties)

# 3 bar if longitudinal bar # 10 bar # 4 bar if longitudinal bar # 11 bar # 4 bar if longitudinal bars are bundled

size


loadsloads

3. Reinforcement Requirements (Lateral Ties) Vertical spacing: (ACI 7.10.5.2)

16 db ( db for longitudinal bars ) 48 db ( db for tie bar ) least lateral dimension of column

s s s


loadsloads

3. Reinforcement Requirements (Lateral Ties)

Arrangement Vertical spacing: (ACI 7.10.5.3)

At least every other longitudinal bar shall have lateral support from the corner of a tie with an included angle 135o.

No longitudinal bar shall be more than 6 in. clear on either side from “support” bar.

1.)

2.)


loadsloads

Examples of lateral ties.


loadsloads

ACI Code 7.10.4

Reinforcement Requirements (Spirals )

3/8 “ dia.(3/8” smooth bar, #3 bar dll or wll wire)

size

clear spacing between spirals 3 in. ACI 7.10.4.31 in.


loadsloads

Reinforcement Requirements (Spiral)

sDA

c

sps

4Core of VolumeSpiral of Volume

Spiral Reinforcement Ratio, s

sDDA

41

:from 2c

csps


loadsloadsReinforcement Requirements (Spiral)

y

c

c

gs *1*45.0

f

f

A

A ACI Eqn. 10-5

psi 60,000 steel spiral ofstrength yield

center) (center to steel spiral ofpitch spacing

spiral of edge outside toedge outside :diameter core 4

area core

entreinforcem spiral of area sectional-cross

y

c

2c

c

sp

f

s

D

DA

A

where


loadsloads4. Design for Concentric Axial Loads

(a) Load Combination

u DL LL

u DL LL w

u DL w

1.2 1.61.2 1.0 1.60.9 1.3

P P PP P P PP P P

Gravity:

Gravity + Wind:

and

etc. Check for tension


loadsloads4. Design for Concentric Axial Loads

(b) General Strength Requirement

un PP = 0.65 for tied columns

= 0.7 for spiral columns

where,


loadsloads

4. Design for Concentric Axial Loads

(c) Expression for Design

08.00.01 Code ACI gg

stg

AA

defined:


loadsloads

ucystcgn

steel

85.0

concrete

85.0 PffAfArP

or

ucygcgn 85.085.0 PfffArP


loadsloads

85.085.0 cygc

ug

fffr

PA

* when g is known or assumed:

cg

u

cy

st 85.085.0

1 fAr

P

ffA

* when Ag is known or assumed:

Example: Design Tied Column for Example: Design Tied Column for Concentric Axial Load Concentric Axial Load

Example: Design Tied Column Example: Design Tied Column for for Concentric Axial Load Concentric Axial Load

Design tied column for concentric axial load

Pdl = 150 k; Pll = 300 k; Pw = 50 k

fc = 4500 psi fy = 60 ksi

Design a square column aim for g = 0.03. Select longitudinal transverse reinforcement.

Example: Design Tied Column Example: Design Tied Column for for Concentric Axial Concentric Axial LoadLoad

Determine the loading

u dl ll

u dl ll w

1.2 1.6

1.2 150 k 1.6 300 k 660 k

1.2 1.0 1.6

1.2 150 k 1.0 300 k 1.6 50 k 560 k

P P P

P P P P

u dl w0.9 1.3

0.9 150 k 1.3 50 k 70 k

P P P

Check the compression or tension in the column


For a square column r = 0.80 and = 0.65 and = 0.03

ug

c g y c

2

2g

r 0.85 0.85

660 k0.85 4.5 ksi

0.65 0.8 0.03 60 ksi 0.85 4.5 ksi

230.4 in

15.2 in. 16 in.

PAf f f

A d d d


For a square column, As=Ag= 0.03(15.2 in.)2 =6.93 in2

ust c g

y c

2

2

1 0.85r0.85

160 ksi 0.85 4.5 ksi

660 k * 0.85 4.5 ksi 16 in0.65 0.8

5.16 in

PA f A

f f

Use 8 #8 bars Ast = 8(0.79 in2) = 6.32 in2

Example: Design Tied Column Example: Design Tied Column for for Concentric Axial Concentric Axial LoadLoadCheck P0

0 c g st y st

2 2 2

n 0

0.85

0.85 4.5 ksi 256 in 6.32 in 60 ksi 6.32 in

1334 k0.65 0.8 1334 k 694 k > 660 k OK

P f A A f A

P rP

Example: Design Tied Column Example: Design Tied Column for for Concentric Axial Concentric Axial LoadLoadUse #3 ties compute the spacing

b stirrup# 2 cover

# bars 1

16 in. 3 1.0 in. 2 1.5 in. 0.375 in.2

4.625 in.

b d ds

< 6 in. No cross-ties needed


Stirrup design

b

stirrup

16 16 1.0 in. 16 in. governs 48 48 0.375 in. 18 in.

smaller or 16 in. governs

ds d

b d

Use #3 stirrups with 16 in. spacing in the column

Behavior under Combined Behavior under Combined Bending and Axial LoadsBending and Axial Loads

Usually moment is represented by axial load times eccentricity, i.e.


Interaction Diagram Between Axial Load and Moment ( Failure Envelope )

Concrete crushes before steel yields

Steel yields before concrete crushes

Any combination of P and M outside the envelope will cause failure.

Note:


Axial Load and Moment Interaction Diagram – General


Resultant Forces action at Centroid

( h/2 in this case )s2

positive is ncompressio

cs1n TCCP

Moment about geometric center

2*

22*

2* 2s2c1s1n

hdTahCdhCM

Columns in Pure Columns in Pure TensionTension

Section is completely cracked (no concrete axial capacity)

Uniform Strain y

N

1iisytensionn AfP

ColumnsColumnsStrength Reduction Factor, (ACI Code 9.3.2)

Axial tension, and axial tension with flexure. = 0.9

Axial compression and axial compression with flexure.

Members with spiral reinforcement confirming to 10.9.3

Other reinforced members

(a)

(b)

ColumnsColumnsExcept for low values of axial compression, may be increased as follows:

when and reinforcement is symmetric

and

ds = distance from extreme tension fiber to centroid of tension reinforcement.

Then may be increased linearly to 0.9 as Pn decreases from 0.10fc Ag to zero.

psi 000,60y f

70.0s

h

ddh

ColumnColumn

ColumnsColumnsCommentary:

Other sections:

may be increased linearly to 0.9 as the strain s increase in the tension steel. Pb

Design for Combined Design for Combined Bending and Axial Load Bending and Axial Load

(Short Column)(Short Column)

Design - select cross-section and reinforcement to resist axial load and moment.

Design for Combined Design for Combined Bending and Axial Load Bending and Axial Load

(Short Column)(Short Column)Column Types

Spiral Column - more efficient for e/h < 0.1, but forming and spiral expensive

Tied Column - Bars in four faces used when e/h < 0.2 and for biaxial bending

1)

2)

General ProcedureGeneral Procedure

The interaction diagram for a column is constructed using a series of values for Pn and Mn. The plot shows the outside envelope of the problem.

General Procedure for General Procedure for Construction of IDConstruction of ID

Compute P0 and determine maximum Pn in compression

Select a “c” value (multiple values)Calculate the stress in the steel components.Calculate the forces in the steel and concrete,Cc,

Cs1 and Ts.Determine Pn value.Compute the Mn about the center.Compute moment arm,e = Mn / Pn.

General Procedure for General Procedure for Construction of IDConstruction of ID

Repeat with series of c values (10) to obtain a series of values.

Obtain the maximum tension value. Plot Pn verse Mn. Determine Pn and Mn.

Find the maximum compression level.Find the will vary linearly from 0.65 to 0.9

for the strain values The tension component will be = 0.9

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagram

Consider an square column (20 in x 20 in.) with 8 #10 ( = 0.0254) and fc = 4 ksi and fy = 60 ksi. Draw the interaction diagram.


Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi

2 2st

2 2g

2st

2g

8 1.27 in 10.16 in

20 in. 400 in

10.16 in 0.0254400 in

A

A

AA

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramGiven 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi

0 c g st y st

2 2

2

0.85

0.85 4 ksi 400 in 10.16 in

60 ksi 10.16 in

1935 k

P f A A f A

n 0

0.8 1935 k 1548 kP rP

[ Point 1 ]

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramDetermine where the balance point, cb.

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramDetermine where the balance point, cb. Using similar triangles, where d = 20 in. – 2.5 in. = 17.5 in., one can find cb

b

b

b

17.5 in.0.003 0.003 0.00207

0.003 17.5 in.0.003 0.00207

10.36 in.

c

c

c

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramDetermine the strain of the steel

bs1 cu

b

bs2 cu

b

2.5 in. 10.36 in. 2.5 in. 0.00310.36 in.

0.00228

10 in. 10.36 in. 10 in. 0.00310.36 in.

0.000104

cc

cc


Determine the stress in the steel

s1 s s1

s2 s s1

29000 ksi 0.00228

66 ksi 60 ksi compression29000 ksi 0.000104

3.02 ksi compression

f E

f E

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute the forces in the column

c c 1

s1 s1 s1 c

2

2s2

0.850.85 4 ksi 20 in. 0.85 10.36 in.598.8 k

0.85

3 1.27 in 60 ksi 0.85 4 ksi

215.6 k

2 1.27 in 3.02 ksi 0.85 4 ksi

0.97 k neglect

C f b c

C A f f

C


2s s s

n c s1 s2 s

3 1.27 in 60 ksi

228.6 k

599.8 k 215.6 k 228.6 k 585.8 k

T A f

P C C C T

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute the moment about the center

c s1 1 s 32 2 2 2

0.85 10.85 in.20 in.599.8 k2 2

20 in. 215.6 k 2.5 in.2

20 in. 228.6 k 17.5 in.2

6682.2 k-in 556.9 k-ft

h a h hM C C d T d

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramA single point from interaction diagram, (585.6 k, 556.9 k-ft). The eccentricity of the point is defined as

6682.2 k-in 11.41 in.585.8 k

MeP

[ Point 2 ]

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramNow select a series of additional points by selecting values of c. Select c = 17.5 in. Determine the strain of the steel. (c is at the location of the tension steel)

s1 cu

s1

s2 cu

s2

2.5 in. 17.5 in. 2.5 in. 0.00317.5 in.

0.00257 74.5 ksi 60 ksi (compression)

10 in. 17.5 in. 10 in. 0.00317.5 in.

0.00129 37.3 ksi (compression)

cc

f

cc

f


c c 1

2s1 s1 s1 c

2s2

0.85 0.85 4 ksi 20 in. 0.85 17.5 in.1012 k

0.85 3 1.27 in 60 ksi 0.85 4 ksi

216 k

2 1.27 in 37.3 ksi 0.85 4 ksi

86 k

C f b c

C A f f

C


2s s s

n

3 1.27 in 0 ksi

0 k 1012 k 216 k 86 k 1314 k

T A f

P


c s1 12 2 2

0.85 17.5 in.20 in.1012 k2 2

20 in. 216 k 2.5 in.2

4213 k-in 351.1 k-ft

h a hM C C d

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramA single point from interaction diagram, (1314 k, 351.1 k-ft). The eccentricity of the point is defined as

4213 k-in 3.2 in.1314 k

MeP

[ Point 3 ]

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramSelect c = 6 in. Determine the strain of the steel, c =6 in.

s1 cu

s1

s2 cu

s2

s3 cu

2.5 in. 6 in. 2.5 in. 0.0036 in.

0.00175 50.75 ksi (compression)

10 in. 6 in. 10 in. 0.0036 in.

0.002 58 ksi (tension)

17.5 in. 6 in.

cc

f

cc

f

cc

s3

17.5 in. 0.0036 in.

0.00575 60 ksi (tension)f


c c 1

s1 s1 s1 c

2

2s2

0.85

0.85 4 ksi 20 in. 0.85 6 in.

346.8 k0.85

3 1.27 in 50.75 ksi 0.85 4 ksi

180.4 k C

2 1.27 in 58 ksi

147.3 k T

C f b c

C A f f

C


2s s s

n

3 1.27 in 60 ksi

228.6 k 346.8 k 180.4 k 147.3 k 228.6 k 151.3 k

T A f

P


c s1 1 s 32 2 2 2

0.85 6 in.346.8 k 10 in.

2

180.4 k 10 in. 2.5 in.

228.6 k 17.5 in. 10 in.

5651 k-in 470.9 k-ft

h a h hM C C d T d

Example: Axial Load Vs. Example: Axial Load Vs. Moment Interaction DiagramMoment Interaction DiagramA single point from interaction diagram, (151 k, 471 k-ft). The eccentricity of the point is defined as

5651.2 k-in 37.35 in.151.3 k

MeP

[ Point 4 ]

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramSelect point of straight tension. The maximum tension in the column is

2n s y 8 1.27 in 60 ksi

610 k

P A f

[ Point 5 ]


Point c (in) Pn Mn e

1 - 1548 k 0 0

2 20 1515 k 253 k-ft 2 in

3 17.5 1314 k 351 k-ft 3.2 in

4 12.5 841 k 500 k-ft 7.13 in

5 10.36 585 k 556 k-ft 11.42 in

6 8.0 393 k 531 k-ft 16.20 in

7 6.0 151 k 471 k-ft 37.35 in

8 ~4.5 0 k 395 k-ft infinity

9 0 -610 k 0 k-ft


Column Analysis

-1000

-500

0

500

1000

1500

2000

0 100 200 300 400 500 600

M (k-ft)

P (k

)

Use a series of c values to obtain the Pn verses Mn.

Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction

DiagramDiagram

Column Analysis

-800

-600

-400

-200

0

200

400

600

800

1000

1200

0 100 200 300 400 500

Mn (k-ft)

Pn

(k)

Max. compression

Max. tension

Cb

Location of the linearly varying

lecture chp 9&10 column design

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