6 Reinforced Columns

Reinforced ColumnsReinforced Columns

Concrete Construction

Lecture GoalsLecture Goals

• Columns– Short Column Design– Long Column Design

Behavior under Combined Bending and Behavior under Combined Bending and Axial LoadsAxial 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

hdT

ahCd

hCM

Columns in Pure TensionColumns in Pure Tension

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

ColumnsColumns

Commentary:

Other sections:

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

Design for Combined Bending and Design for Combined Bending and Axial Load (short column)Axial Load (short column)

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

Design for Combined Bending and Design for Combined Bending and Axial Load (short column)Axial Load (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 construction General Procedure for construction of an interaction diagramof an interaction diagram

– Compute P0 and determine maximum Pn in compression

– Select a c value.• 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 construction General Procedure for construction of an interaction diagramof an interaction diagram

– 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. Moment Example: Axial Load Vs. Moment Interaction DiagramInteraction Diagram

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

8 1.27 in 10.16 in

20 in. 400 in

A

A


Given 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 k

P rP


Determine where the balance point, cb.

n 0

0.8 1935 k 1548 k

P rP


Determine where the balance point, cb. Using similar triangles you can find cb

bb

b

17.5 in. 0.00317.5 in.

0.003 0.003 0.00207 0.003 0.00207

10.36 in.

cc

c


Determine the strain of the steel

bs1 cu

b

bs2 cu

b

2.5 in. 10.36 in. 2.5 in.0.003 0.00228

10.36 in.

10 in. 10.36 in. 10 in.0.003 0.000104

10.36 in.

c

c

c

c


Determine the stress in the steel

s1 s s1

s2 s s1

29000 ksi 0.00228

66 ksi 60 ksi compression

29000 ksi 0.000104

3.02 ksi compression

f E

f E


Compute the forces in the column

c c 1

2s1 s1 s1 c

2s2

0.85 0.85 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

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


Compute the moment about the center

c s1 1 s 32 2 2 2

0.85 10.85 in.20 in.599.8 k

2 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


A single point from interaction diagram, (585.6 k, 556.9 k-ft). The eccentricity of the point is defined as

Now select a series of additional points by selecting values of c. Select c = 17.5 in.

6682.2 k-in11.41 in.

585.8 k

Me

P


Determine the strain of the steel, c =17.5 in.

s1 cu

s1

s2 cu

s2

2.5 in. 17.5 in. 2.5 in.0.003 0.00257

17.5 in.

74.5 ksi 60 ksi (compression)

10 in. 17.5 in. 10 in.0.003 0.00129

17.5 in.

37.3 ksi (compression)

c

c

f

c

c

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 k

2 2

20 in. 216 k 2.5 in.

2

4213 k-in 351.1 k-ft

h a hM C C d


A single point from interaction diagram, (1314 k, 351.1 k-ft). The eccentricity of the point is defined as

Now select a series of additional points by selecting values of c. Select c = 6 in.

4213 k-in3.2 in.

1314 k

Me

P


Determine the strain of the steel, c =6 in.

s1 cu

s1

s2 cu

s2

s2 cu

2.5 in. 6 in. 2.5 in.0.003 0.00175

6 in.

50.75 ksi (compression)

10 in. 6 in. 10 in.0.003 0.002

6 in.

58 ksi (tension)

17.5 in. 6 in.

c

c

f

c

c

f

c

c

s2

17.5 in.0.003 0.00575

6 in.

60 ksi (tension)f



c c 1

2s1 s1 s1 c

2s2

0.85 0.85 4 ksi 20 in. 0.85 6 in.

346.8 k

0.85 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


A single point from interaction diagram, (151 k, 471 k-ft). The eccentricity of the point is defined as

Select point of straight tension

5651.2 k-in37.35 in.

151.3 k

Me

P


The maximum tension in the column is

2n s y 8 1.27 in 60 ksi

610 k

P A f

ExampleExample

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

ExampleExampleColumn 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.

ExampleExample

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

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6 Reinforced Columns