Lecture 21 – ColumnsLecture 21 – ColumnsJuly 25, 2003CVEN 444

Lecture GoalsLecture Goals

Columns Interaction DiagramsUsing Interaction Diagrams

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.

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

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

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

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

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

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

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

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

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

2s s s

n

3 1.27 in 0 ksi

0 k 1012 k 216 k 86 k 1314 k

T A f

P

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

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

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

0.85

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

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

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

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 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 ]

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

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

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

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

Behavior under Combined Behavior under Combined Bending and Axial LoadsBending and Axial LoadsInteraction Diagram Between Axial Load and Moment ( Failure Envelope )

Concrete crushes before steel yields

Steel yields before concrete crushes

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

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

Column Types

Tied Column - Bars in 2 faces (furthest from axis of bending.

- Most efficient when e/h > 0.2

- rectangular shape increases efficiency

3)

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

Typically longitudinal bars spliced just above each floor. (non-seismic)

Type of lap splice depends on state of stress (ACI 12.17)

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

SpicesAll bars in compression Use compression lap splice

(ACI 12.16)

15.12 ACI

splice lap tension B Classspliced) bars 1/2 (

B Class)splice bars 2/1(

lapA tension Class

5.0

face on tension 5.00

ys

ys

ff

ff

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

Column Shear

4-11 ACI 2000

12 wc

g

uc dbf

A

NV

Recall

( Axial Compression )

5.0 If cu VV Ties must satisfy ACI 11 and ACI Sec. 7.10.5

Design for Combined Design for Combined Bending and Axial Load Bending and Axial Load (short column)(short column)Additional Note on Reinforcement Ratio

10.9.1 ACI 0.08 0.01 Recall For cross-section larger than required for loading:

Min. reinforcement may be computed for reduced effective area, Ag, ( 1/2 Ag (total) )

Provided strength from reduced area and resulting Ast must be adequate for loading.

(ACI 10.8.4 )

Non-dimensional Interaction Non-dimensional Interaction DiagramsDiagrams

See Figures B-12 to B-26

or ACI Common 340 Design Handbook Vol 2 Columns (ACI 340.2R-91)

n n

c g c g

versus P Mf A f A h

n nn n

c g c g

e versus R P PKf A f A h

or

Non-dimensional Interaction Non-dimensional Interaction DiagramsDiagrams

Design using Non-dimensional Design using Non-dimensional Interaction diagramsInteraction diagrams

Calculate factored loads (Pu , Mu ) and e for relevant load combinations

Select potentially governing case(s)

Use estimate h to calculate h, e/h for governing case(s)

1.)

2.)

3.)

Design using Non-dimensional Design using Non-dimensional Interaction diagramsInteraction diagrams

Use appropriate chart (App. A) target g

(for each governing case)

Select

4.)

5.)

n

c g

Pf A

u cg

n

c g

P fAPf A

Read Calculate required

hbAb * h & g

Design using Non-dimensional Design using Non-dimensional Interaction diagramsInteraction diagrams

If dimensions are significantly different from estimated (step 3), recalculate ( e / h ) and redo steps 4 & 5.

Revise Ag if necessary.

Select steel

6.)

7.) gst AA

Design using non-dimensional Design using non-dimensional interaction diagramsinteraction diagrams

Using actual dimensions & bar sizes to check all load combinations ( use charts or “exact: interaction diagram).

Design lateral reinforcement.

8.)

9.)

Example: Column design Example: Column design using Interaction Diagramsusing Interaction Diagrams

Determine the tension and compression reinforcement for a 16 in x 24 in. rectangular tied column to support Pu= 840 k and Mu = 420 k-ft. Use fc = 4 ksi and fy = 60 ksi. Using the interaction diagram.

Example: Interaction Example: Interaction DiagramsDiagramsCompute the initial components

un

840 kips 1292 k0.65

PP

un

u

12 in.420 k-ft fte 6.0 in.

840 kMP

Example: Interaction Example: Interaction DiagramsDiagramsCompute the initial components

24 in. 5.0 in. 19.0 in.h

19.0 in. 0.7924 in.

Example: Interaction Example: Interaction DiagramsDiagramsCompute the coefficients of the column

n

ng c

1292 k16 in. 24 in. 4 ksi

0.84

PKA f

nn

g c

1292 k 6 in.e16 in. 24 in. 4 ksi 24 in.

0.21

PRA f h

Example: Interaction Example: Interaction DiagramsDiagrams

Using an interaction diagram, B-13

n n

c y

, 0.21,0.840.7 4 ksi 60 ksi

0.042

R K

f f

Example: Interaction Example: Interaction DiagramsDiagrams

Using an interaction diagram, B-14

n n

c y

, 0.21,0.840.9 4 ksi 60 ksi

0.034

R K

f f

Example: Interaction Example: Interaction DiagramsDiagramsUsing linear interpolation to find the of the column

0.9 0.70.7 0.7

0.9 0.7

0.034 0.0420.042 0.79 0.7

0.9 0.70.0384

Example: Interaction Example: Interaction DiagramsDiagramsDetermine the amount of steel required

Select the steel for the column, using #11 bars

st g

2

0.0384 16 in. 24 in.

14.75 in

A A

2st

2b

14.75 in 9.45 bars 10 bars1.56 in

AA

Example: Interaction Example: Interaction DiagramsDiagramsThe areas of the steel:

The loading on the column

2st

2 2s1 t

15.6 in

7.8 in , 7.8 in

A

A A

Example: Interaction Example: Interaction DiagramsDiagramsThe compression components are

2s1 s1 y c

c c

0.85 7.8 in 60 ksi 0.85 4 ksi

441.5 k0.85 0.85 4 ksi 16 in. 0.8546.24

C A f f

C f ba cc

Example: Interaction Example: Interaction DiagramsDiagramsThe tension component is

2s1 s s

s s cu

7.8 in

21.5 in.29000 ksi 0.003

21.5 in.87 ksi

T A f f

d c cf Ec c

cc

Example: Interaction Example: Interaction DiagramsDiagramsTake the moment about the tension steel

n s1 ce2aP C d d C d

e 6 in. 9.5 in. 15.5 in.

Example: Interaction Example: Interaction DiagramsDiagramsThe first equation related to Pn

n

2

2n

15.5 in. 441.5 k 21.5 in. 2.5 in.

0.85 46.24 21.5 in.2

8388.5 k-in. 994.2 19.65

541.2 k 64.14 1.27

P

cc

c c

P c c

Example: Interaction Example: Interaction DiagramsDiagramsThe second equation comes from the equilibrium equation and substitute in for Pn

n s1 c

2s

2s

2s

541.2 k 64.14 1.27 441.5 k 46.24 7.8

7.8 1.27 17.9 99.7

0.1628 2.282 12.782

P C C T

c c c f

f c c

f c c

Example: Interaction Example: Interaction DiagramsDiagramsSubstitute the relationship of c for the stress in the steel.

The problem is now a cubic solution

c fs RHS

15 in. 37.7 -10.3819 in. 11.45 2.6419.5 in. 8.92 4.63 20.0 in. 6.52 6.70 19.98 in. 6.62 6.62

221.5 in.87 0.1628 2.282 12.782c c cc

Example: Interaction Example: Interaction DiagramsDiagramsCompute Pn

Compute Mn about the center

2n 541.2 k 64.14 19.98 in. 1.27 19.98 in.

1313.7 k 1292 kP

n s1 c2 2 2 2h h a hM C d C T d

Example: Interaction Example: Interaction DiagramsDiagramsCompute Mn about the center

n

2

441.5 k 12 in. 2.5 in.

0.85 19.98 in.46.24 19.98 in. 12 in.

2

7.8 in 6.62 ksi 21.5 in. 12 in.

4194.25 k-in. 3241.4 k-in. 490.54 k-in.7926.2 k-in. 660.5 k-ft.

M

Example: Interaction Example: Interaction DiagramsDiagramsCheck that Mn is greater than the required Mu

Check the Pn is greater than the required Pu

n 0.65 660.5 k-ft.

429.33 k-ft. 420 k-ft.

M

n 0.65 1313.7 k

853.9 k 840 k

P

Example: Interaction Example: Interaction DiagramsDiagramsDetermine the tie spacing using #4 bars

b

stirrup

16spacing smallest 48

smallest dimension

16 1.41 in. 22.56 in.48 0.5 in. 24 in.

16 in.

dd

Use 16 in.