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Lecture 21 – ColumnsLecture 21 – ColumnsJuly 25, 2003CVEN 444
Lecture GoalsLecture Goals
Columns Interaction DiagramsUsing Interaction Diagrams
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagram
Consider an square column (20 in x 20 in.) with 8 #10 ( = 0.0254) and fc = 4 ksi and fy = 60 ksi. Draw the interaction diagram.
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagram
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 in 0.0254400 in
A
A
AA
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramGiven 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 kP rP
[ Point 1 ]
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramDetermine where the balance point, cb.
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramDetermine 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 0.00207
0.003 17.5 in.0.003 0.00207
10.36 in.
c
c
c
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramDetermine the strain of the steel
bs1 cu
b
bs2 cu
b
2.5 in. 10.36 in. 2.5 in. 0.00310.36 in.
0.00228
10 in. 10.36 in. 10 in. 0.00310.36 in.
0.000104
cc
cc
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagram
Determine the stress in the steel
s1 s s1
s2 s s1
29000 ksi 0.00228
66 ksi 60 ksi compression29000 ksi 0.000104
3.02 ksi compression
f E
f E
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute the forces in the column
c c 1
s1 s1 s1 c
2
2s2
0.850.85 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute the moment about the center
c s1 1 s 32 2 2 2
0.85 10.85 in.20 in.599.8 k2 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramA single point from interaction diagram, (585.6 k, 556.9 k-ft). The eccentricity of the point is defined as
6682.2 k-in 11.41 in.585.8 k
MeP
[ Point 2 ]
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramNow 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. 0.00317.5 in.
0.00257 74.5 ksi 60 ksi (compression)
10 in. 17.5 in. 10 in. 0.00317.5 in.
0.00129 37.3 ksi (compression)
cc
f
cc
f
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute the moment about the center
c s1 12 2 2
0.85 17.5 in.20 in.1012 k2 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramA single point from interaction diagram, (1314 k, 351.1 k-ft). The eccentricity of the point is defined as
4213 k-in 3.2 in.1314 k
MeP
[ Point 3 ]
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramSelect 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. 0.0036 in.
0.00175 50.75 ksi (compression)
10 in. 6 in. 10 in. 0.0036 in.
0.002 58 ksi (tension)
17.5 in. 6 in.
cc
f
cc
f
cc
s3
17.5 in. 0.0036 in.
0.00575 60 ksi (tension)f
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute 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 k0.85
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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramCompute 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. Example: Axial Load Vs. Moment Interaction DiagramMoment Interaction DiagramA single point from interaction diagram, (151 k, 471 k-ft). The eccentricity of the point is defined as
5651.2 k-in 37.35 in.151.3 k
MeP
[ Point 4 ]
Example: Axial Load vs. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagramSelect 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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagram
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. Example: Axial Load vs. Moment Moment Interaction Interaction DiagramDiagram
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. Example: Axial Load vs. Moment Moment Interaction Interaction
DiagramDiagram
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
Behavior under Combined Behavior under Combined Bending and Axial LoadsBending 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 Design for Combined Bending and Axial Load Bending and Axial Load (short column)(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 Design for Combined Bending and Axial Load Bending and Axial Load (short column)(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 Design for Combined Bending and Axial Load Bending and Axial Load (short column)(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 Design for Combined Bending and Axial Load Bending and Axial Load (short column)(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 Design for Combined Bending and Axial Load Bending and Axial Load (short column)(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 Non-dimensional Interaction DiagramsDiagrams
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 Mf A f A h
n nn n
c g c g
e versus R P PKf A f A h
or
Non-dimensional Interaction Non-dimensional Interaction DiagramsDiagrams
Design using Non-dimensional Design using Non-dimensional Interaction diagramsInteraction diagrams
Calculate factored loads (Pu , Mu ) and e for relevant load combinations
Select potentially governing case(s)
Use estimate h to calculate h, e/h for governing case(s)
1.)
2.)
3.)
Design using Non-dimensional Design using Non-dimensional Interaction diagramsInteraction diagrams
Use appropriate chart (App. A) target g
(for each governing case)
Select
4.)
5.)
n
c g
Pf A
u cg
n
c g
P fAPf A
Read Calculate required
hbAb * h & g
Design using Non-dimensional Design using Non-dimensional Interaction diagramsInteraction 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 Design using non-dimensional interaction diagramsinteraction 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 Example: Column design using Interaction Diagramsusing 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 Example: Interaction DiagramsDiagramsCompute the initial components
un
840 kips 1292 k0.65
PP
un
u
12 in.420 k-ft fte 6.0 in.
840 kMP
Example: Interaction Example: Interaction DiagramsDiagramsCompute the initial components
24 in. 5.0 in. 19.0 in.h
19.0 in. 0.7924 in.
Example: Interaction Example: Interaction DiagramsDiagramsCompute the coefficients of the column
n
ng c
1292 k16 in. 24 in. 4 ksi
0.84
PKA f
nn
g c
1292 k 6 in.e16 in. 24 in. 4 ksi 24 in.
0.21
PRA f h
Example: Interaction Example: Interaction DiagramsDiagrams
Using an interaction diagram, B-13
n n
c y
, 0.21,0.840.7 4 ksi 60 ksi
0.042
R K
f f
Example: Interaction Example: Interaction DiagramsDiagrams
Using an interaction diagram, B-14
n n
c y
, 0.21,0.840.9 4 ksi 60 ksi
0.034
R K
f f
Example: Interaction Example: Interaction DiagramsDiagramsUsing linear interpolation to find the 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 0.70.0384
Example: Interaction Example: Interaction DiagramsDiagramsDetermine 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 in 9.45 bars 10 bars1.56 in
AA
Example: Interaction Example: Interaction DiagramsDiagramsThe 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 Example: Interaction DiagramsDiagramsThe compression components are
2s1 s1 y c
c c
0.85 7.8 in 60 ksi 0.85 4 ksi
441.5 k0.85 0.85 4 ksi 16 in. 0.8546.24
C A f f
C f ba cc
Example: Interaction Example: Interaction DiagramsDiagramsThe 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 Ec c
cc
Example: Interaction Example: Interaction DiagramsDiagramsTake the moment about the tension steel
n s1 ce2aP C d d C d
e 6 in. 9.5 in. 15.5 in.
Example: Interaction Example: Interaction DiagramsDiagramsThe 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 Example: Interaction DiagramsDiagramsThe 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 Example: Interaction DiagramsDiagramsSubstitute 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 12.782c c cc
Example: Interaction Example: Interaction DiagramsDiagramsCompute Pn
Compute Mn about the center
2n 541.2 k 64.14 19.98 in. 1.27 19.98 in.
1313.7 k 1292 kP
n s1 c2 2 2 2h h a hM C d C T d
Example: Interaction Example: Interaction DiagramsDiagramsCompute 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 Example: Interaction DiagramsDiagramsCheck 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 Example: Interaction DiagramsDiagramsDetermine the tie spacing using #4 bars
b
stirrup
16spacing smallest 48
smallest dimension
16 1.41 in. 22.56 in.48 0.5 in. 24 in.
16 in.
dd
Use 16 in.