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.

Example: Axial Load Vs. Moment Interaction Diagram

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

Example: Axial Load Vs. Moment Interaction Diagram

Point c (in) Pn Mn e

1 - 1451 k 0 0

2 17.5 1314 k 351 k-ft 3.2 in

3 12.5 841 k 500 k-ft 7.13 in

4 10.36 585 k 556 k-ft 11.42 in

5 8.0 393 k 531 k-ft 16.20 in

6 6.0 151 k 471 k-ft 37.35 in

7 4.5 0 k 395 k-ft infinity 8

8 0 -610 k 0 k-ft

Example: Axial Load Vs. Moment Interaction Diagram

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.

1400

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

φφφφMn (k-ft)

φφ φφPn

(k)

Max. compression

Max. tension

Cb

Location of the linearly varying φ.

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)

Design for Combined Bending and Axial Load (short column)

Spices

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

Type of lap splice depends on state of stress (C.12.17)

Design for Combined Bending and Axial Load (short column)

SpicesAll bars in compression Use compression lap splice

(C.12.16)

s y

s y

0 0.5 on tension face Class A tension lap

( 1/ 2 bars splice)

Class BC.12.15

( 1/2 bars spliced)

0.5 Class B tension lap splice

f f

f f

≤ ≤ → < → > > →

Design for Combined Bending and Axial Load (short column)

Column Shear

( )c

'uc w

g

0.17 1 C.11-414

NV f b d

Aλ

= +

Recall

( Axial Compression )

5.0 If cu ⇒> VV φ Ties must satisfy C.7.10

Design for Combined Bending and Axial Load (short column)

Additional Note on Reinforcement Ratio

( )Recall 0.01 0.04 C.10.9.1ρ≤ ≤

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.

≥

(C.10.8 )

⇒∗

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 γh, e/h for governing case(s)

1.)

2.)

3.)

Design using non-dimensional interaction diagrams

Use appropriate chart (App. A) target ρg

(for each governing case)

Select

4.)

5.)

n

c g

P

f A⇒ u c

g

n

c g

P fA

P

f A

φ=

Read Calculate required

hbAb * h & g =⇒

Design using non-dimensional interaction 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 ρ=⇒

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

φ= = =

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.hγ = − =

19.0 in.0.79

24 in.γ = =

Example: Interaction Diagrams

Compute 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

γ

ρ

=== =

=

Example: Interaction Diagrams

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

γ

ρ

=== =

=

Example: Interaction Diagrams

Using linear interpolation to find the ρ 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

ρ ρρ ρ γ−= + −−

−= − −

−=

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 Aρ= =

=

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

ε

= =− − = =

− =

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.

Mφ == ≥

( )n 0.65 1313.7 k

853.9 k 840 k

Pφ == ≥

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.