column interaction curves
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Transcript of 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.
Example: Axial Load vs. Moment 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
Example: Axial Load vs. Moment Interaction Diagram
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 ]
Example: Axial Load vs. Moment Interaction Diagram
Determine where the balance point, cb.
Example: Axial Load vs. Moment Interaction Diagram
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
Example: Axial Load vs. Moment Interaction Diagram
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
Example: Axial Load vs. Moment Interaction Diagram
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
Example: Axial Load vs. Moment Interaction DiagramCompute 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. 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
Example: Axial Load vs. Moment Interaction Diagram
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 ]
Example: Axial Load vs. Moment Interaction Diagram
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
Example: Axial Load vs. Moment Interaction Diagram
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
Example: Axial Load vs. Moment Interaction Diagram
Compute 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. Moment Interaction Diagram
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
Example: Axial Load vs. Moment Interaction Diagram
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
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 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
Example: Axial Load vs. Moment Interaction Diagram
Compute 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. Moment Interaction Diagram
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 ]
Example: Axial Load vs. Moment Interaction Diagram
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 ]
Example: Axial Load vs. Moment Interaction Diagram
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. 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.
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)
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 (ACI 12.17)
Design for Combined Bending and Axial Load (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 Bending and Axial Load (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 Bending and Axial Load (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 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.)
Design using Non-dimensional Interaction diagrams
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
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
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
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
Example: Interaction Diagrams
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.