interaction diagram 6
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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.