Interaction Diagram 6

Behavior under Combined Bending and Axial Loads Interaction 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 0

e

1

-

1451 k

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

0k

395 k-ft

infinity 8

8

0

-610 k

0 k-ft

0

Example: Axial Load Vs. Moment Interaction Diagram Use a series of c values to obtain the Pn verses Mn.

Column Analysis 2000

1400 1500

P (k)

1000 500 0 0

100

200

300

-500 -1000

M (k-ft)

400

500

600

Example: Axial Load Vs. Moment Interaction Diagram Max. compression Column Analysis 1200 1000

Location of the linearly varying φ.

800

φ Pn (k)

600

Cb

400 200 0 -200

0

100

200

300

-400

Max. tension

-600 -800

φ Mn (k-ft)

400

500

Design for Combined Bending and Axial Load (short column) Column Types 3) Tied Column - Bars in 2 faces (furthest from axis of bending. - Most efficient when e/h > 0.2 - rectangular shape increases efficiency

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) Spices All bars in compression

Use compression lap splice (C.12.16)

0 ≤ f s ≤ 0.5 f y on tension face →

fs > 0.5 f y

  (< 1/ 2 bars splice)   → Class B  C.12.15 ( > 1/2 bars spliced)  → Class B tension lap splice    Class A tension lap

Design for Combined Bending and Axial Load (short column) Column Shear Recall

 Nu Vc = 0.17 1 +  14 A g 

( Axial Compression )

 ' λ f bw d  c 

( C.11-4 )

If Vu > 0.5φVc ⇒ Ties must satisfy C.7.10

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

Recall 0.01 ≤ ρ ≤ 0.04 (C.10.8 )

∗⇒

( C.10.9.1)

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.

Non-dimensional Interaction Diagrams Pn f c Ag

versus

Pn or K n = f c Ag

Mn f c Ag h

versus

Pn e Rn = f c Ag h

See Figures B-12 to B-26 or ACI Common 340 Design Handbook Vol 2 Columns (ACI 340.2R-91)

Non-dimensional Interaction Diagrams

Design using non-dimensional interaction diagrams 1.) Calculate factored loads (Pu , Mu ) and e for relevant load combinations 2.) Select potentially governing case(s) 3.) Use estimate h to calculate γh, e/h for governing case(s)

Design using non-dimensional interaction diagrams 4.) Use appropriate chart (App. A) target ρg Pn f c Ag

Pf ⇒ Calculate Ag = u c required  φ Pn     f c Ag  (for each governing case)

Read

5.) Select b & h ⇒ Ag = b * h

Design using non-dimensional interaction diagrams 6.) If dimensions are significantly different from estimated (step 3), recalculate ( e / h ) and redo steps 4 & 5. Revise Ag if necessary. 7.) Select steel

⇒

Ast = ρ Ag

Design using non-dimensional interaction diagrams 8.) Using actual dimensions & bar sizes to check all load combinations ( use charts or “exact: interaction diagram).

9.) Design lateral reinforcement.

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 Diagrams Compute the initial components Pu

840 kips = = 1292 k Pn = φ 0.65  12 in.  420 k-ft   Mu ft   = 6.0 in. en = = Pu 840 k

Example: Interaction Diagrams Compute the initial components

γ h = 24 in. − 5.0 in. = 19.0 in. 19.0 in. γ= = 0.79 24 in.

Example: Interaction Diagrams Compute the coefficients of the column

Pn 1292 k Kn = = Ag f c (16 in.)( 24 in.)( 4 ksi ) = 0.84 1292 k )( 6 in.) ( Pn e Rn = = Ag f c h (16 in.)( 24 in.)( 4 ksi )( 24 in.) = 0.21

Example: Interaction Diagrams Using an interaction diagram, B-13

( Rn , K n ) = ( 0.21, 0.84 ) γ = 0.7 f c = 4 ksi f y = 60 ksi

ρ = 0.042

Example: Interaction Diagrams Using an interaction diagram, B-14

( Rn , K n ) = ( 0.21, 0.84 ) γ = 0.9 f c = 4 ksi f y = 60 ksi

ρ = 0.034

Example: Interaction Diagrams Using linear interpolation to find the ρ of the column

ρ = ρ0.7 +

ρ0.9 − ρ0.7

(γ − 0.7 )

( 0.9 − 0.7 ) 0.034 − 0.042 ) ( = 0.042 − ( 0.79 − 0.7 ) ( 0.9 − 0.7 )

= 0.0384

Example: Interaction Diagrams Determine the amount of steel required

Ast = ρ Ag = ( 0.0384 )(16 in.)( 24 in.) = 14.75 in 2 Select the steel for the column, using #11 bars

Ast 14.75 in 2 = = 9.45 bars ⇒ 10 bars 2 Ab 1.56 in

Example: Interaction Diagrams The areas of the steel:

Ast = 15.6 in 2 As1 = 7.8 in 2 ,

At = 7.8 in 2

The loading on the column

Example: Interaction Diagrams The compression components are

Cs1 = As1 ( f y − 0.85 f c ) = 7.8 in 2 ( 60 ksi − 0.85 ( 4 ksi ) ) = 441.5 k Cc = 0.85 f cba = 0.85 ( 4 ksi )(16 in.)( 0.85 ) c = 46.24c

Example: Interaction Diagrams The tension component is

T = As1 fs = 7.8 in 2 fs  d −c  21.5 in. − c  fs = Es   ε cu = 29000 ksi   ( 0.003) c  c     21.5 in. − c  = 87 ksi   c  

Example: Interaction Diagrams Take the moment about the tension steel

a  Pn ( e′ ) = Cs1 ( d − d ′ ) + Cc  d −  2  e′ = 6 in. + 9.5 in. = 15.5 in.

Example: Interaction Diagrams The first equation related to Pn

Pn (15.5 in.) = 441.5 k ( 21.5 in. − 2.5 in.) 0.85c   + 46.24c  21.5 in. −  2   = 8388.5 k-in. + 994.2c − 19.65c 2 Pn = 541.2 k + 64.14c − 1.27c 2

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

Pn = Cs1 + Cc − T 541.2 k + 64.14c − 1.27c = 441.5 k + 46.24c − 7.8 fs 2

7.8 f s = 1.27c − 17.9c − 99.7 2

f s = 0.1628c − 2.282c − 12.782 2

Example: Interaction Diagrams Substitute the relationship of c for the stress in the steel.  21.5 in. − c  2 = c − 2.282c − 12.782 87  0.1628  c   The problem is now a cubic solution c 15 in. 19 in. 19.5 in. 20.0 in. 19.98 in.

fs 37.7 11.45 8.92 6.52 6.62

RHS -10.38 2.64 4.63 6.70 6.62

Example: Interaction Diagrams Compute Pn

Pn = 541.2 k + 64.14 (19.98 in.) − 1.27 (19.98 in.) = 1313.7 k > 1292 k Compute Mn about the center

h a h h    M n = Cs1  − d ′  + Cc  −  + T  d −  2 2  2 2 

2

Example: Interaction Diagrams Compute Mn about the center

M n = 441.5 k (12 in. − 2.5 in.)  0.85 (19.98 in.)  +46.24 (19.98 in.) 12 in. −  2   + ( 7.8 in

2

) ( 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.

Example: Interaction Diagrams Check that Mn is greater than the required Mu

φ M n = 0.65 ( 660.5 k-ft.) = 429.33 k-ft. ≥ 420 k-ft. Check the Pn is greater than the required Pu

φ Pn = 0.65 (1313.7 k ) = 853.9 k ≥ 840 k

Example: Interaction Diagrams Determine the tie spacing using #4 bars

16d b   spacing = smallest  48dstirrup smallest dimension  16 (1.41 in.) = 22.56 in.  =  48 ( 0.5 in.) = 24 in.  16 in. Use 16 in. 