column interaction curves

Lecture 21 – Columns

July 25, 2003CVEN 444

Lecture Goals

Columns Interaction DiagramsUsing Interaction Diagrams

Example: Axial Load vs. Moment Interaction Diagram

Consider an square column (20 in x 20 in.) with 8 #10 (r = 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 in0.0254

400 in

A

A

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

[ Point 1 ]


Determine where the balance point, cb.


Determine 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 0.002070.003

17.5 in.0.003 0.00207

10.36 in.

c

c

c


Determine the strain of the steel

bs1 cu

b

bs2 cu

b

2.5 in. 10.36 in. 2.5 in.0.003

10.36 in.

0.00228

10 in. 10.36 in. 10 in.0.003

10.36 in.

0.000104

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

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

c c 1

s1 s1 s1 c

2

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 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. Moment Interaction DiagramCompute 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

6682.2 k-in11.41 in.

585.8 k

Me

P

[ Point 2 ]


Now 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.0.003

17.5 in.

0.00257 74.5 ksi 60 ksi (compression)

10 in. 17.5 in. 10 in.0.003

17.5 in.

0.00129 37.3 ksi (compression)

c

c

f

c

c

f


Compute 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



2s s s

n

3 1.27 in 0 ksi

0 k

1012 k 216 k 86 k

1314 k

T A f

P


Compute the moment about the center

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

4213 k-in3.2 in.

1314 k

Me

P

[ Point 3 ]

Example: Axial Load vs. Moment Interaction DiagramSelect 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.0.003

6 in.

0.00175 50.75 ksi (compression)

10 in. 6 in. 10 in.0.003

6 in.

0.002 58 ksi (tension)

17.5 in. 6 in.

c

c

f

c

c

f

c

c

s3

17.5 in.0.003

6 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 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


Compute 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. Moment Interaction Diagram

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

5651.2 k-in37.35 in.

151.3 k

Me

P

[ Point 4 ]


Select 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. Moment Interaction

Diagram

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 .f

Behavior under Combined Bending 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 Bending and Axial Load (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)


Spices

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

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


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


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


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 Diagrams

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 M

f A f A h

n nn n

c g c g

e versus R

P PK

f A f A h or

Non-dimensional Interaction Diagrams

Design using Non-dimensional Interaction diagrams

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

Select potentially governing case(s)

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

1.)

2.)

3.)


Use appropriate chart (App. A) target rg

(for each governing case)

Select

4.)

5.)

n

c g

P

f A u c

g

n

c g

P fA

Pf A

Read Calculate required

hbAb * h & g


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 interaction 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 using 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 DiagramsCompute the initial components

un

840 kips1292 k

0.65

PP

un

u

12 in.420 k-ft

fte 6.0 in.

840 k

M

P

Example: Interaction DiagramsCompute the initial components

24 in. 5.0 in. 19.0 in.h

19.0 in.0.79

24 in.

Example: Interaction DiagramsCompute the coefficients of the column

n

ng c

1292 k

16 in. 24 in. 4 ksi

0.84

PK

A f

nn

g c

1292 k 6 in.e

16 in. 24 in. 4 ksi 24 in.

0.21

PR

A f h

Example: Interaction Diagrams

Using an interaction diagram, B-13

n n

c y

, 0.21,0.84

0.7

4 ksi 60 ksi

0.042

R K

f f


Using an interaction diagram, B-14

n n

c y

, 0.21,0.84

0.9

4 ksi 60 ksi

0.034

R K

f f


Using linear interpolation to find the r 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 0.7

0.0384

Example: Interaction DiagramsDetermine 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 in9.45 bars 10 bars

1.56 in

A

A

Example: Interaction DiagramsThe 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 DiagramsThe compression components are

2s1 s1 y c

c c

0.85 7.8 in 60 ksi 0.85 4 ksi

441.5 k

0.85 0.85 4 ksi 16 in. 0.85

46.24

C A f f

C f ba c

c

Example: Interaction DiagramsThe 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 E

c c

c

c

Example: Interaction DiagramsTake the moment about the tension steel

n s1 ce2

aP C d d C d

e 6 in. 9.5 in.

15.5 in.

Example: Interaction DiagramsThe 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 DiagramsThe 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 DiagramsSubstitute 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 12.782

cc c

c

Example: Interaction DiagramsCompute Pn

Compute Mn about the center

2

n 541.2 k 64.14 19.98 in. 1.27 19.98 in.

1313.7 k 1292 k

P

n s1 c2 2 2 2

h h a hM C d C T d

Example: Interaction DiagramsCompute 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 DiagramsCheck 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 DiagramsDetermine the tie spacing using #4 bars

b

stirrup

16

spacing smallest 48

smallest dimension

16 1.41 in. 22.56 in.

48 0.5 in. 24 in.

16 in.

d

d

Use 16 in.

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