Interaction Diagram for Concrete Columns

E702 Example Problems Interaction Diagrams for Concrete Columns

2/20/2007

D.D. Reynolds ,K.W. Kramer

2-1/2"

Problem Statement: Draw a interaction diagram for a 12" x 12" non- slender tied(non-spiral) column reinforced w/ 4 - # 8 bars bending around it's x-axis. Y

f'c = 4000 psi b = 12"

fy = 60000 psi

2-1/2"

7"

X

h = 12"

ACI-318 05 Reference

Notes Calculate ΦPnmax

Due to the fact concrete structures placed monolithically are continuous, a minimum eccentricity or minimum moment is assummed in this calculation, the code reduces the maximum axial load by 20% to account for this minimum moment.

φPn = .8φ [.85 f ' c (Ag − Ast ) + Ast f y ]

(Eq. 1-1)

Ag = b × h = 12"×12" Ag =

Section R10.3.6 & R10.3.7

Section 10.3.6.2 Eq (10-2)

144 in2

(Area of 1 - #8 bar = .79 in2) Ast = (4 * .79) = Find Φ

Φ=

2 3.16 in

0.65

φPn = .8(.65)[.85(4ksi)(144in 2 − 3.16in 2 ) + 3.16in 2 (60ksi)] ΦPnmax = Calculate points on curve

347.60 kips

It is possible to derive a group of equations to evaluate the strength of columns subjected to combined bending and axial loads. These equations are tedious to use, therefore interaction diagrams for columns are generally computed by assuming a series of strain distributions. These strain distributions correspond fo a particular point on the interaction diagram, P and M. Each steel strain is selected by multiplying an arbitrary "Z" factor and the yield strain of your steel.

ε s1 = Z × ε y

(Eq. 1-2)

The "Z" factors can range from 1 to -1000 and increments between "Z" depend on the required detail of diagram. The smaller the increment the more detailed the diagram will be.

1

Section 9.3.2.2


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Notes

With the wide range of possible "Z" factors, every designer must understand there are four mandatory points that must be calculated for each interaction diagram.These four points "Z" factors are 0, -0.5,-1.0,,-2.5, the importance of each point will now be discussed.

Z = 0 (εs1 = 0) - Strain εt = 0 in extreme layer in tension. This point marks the change from compression lap splice being allowed on all longitudinal bars to a tension lap splice. Z = -0.5 (fs1 = -0.5fy, εs1 = -0.5εy) This strain distribution affects the length of tension lap splice in a column & is customarily plotted on an interaction diagram.

Z = -1.0 (fs1 = -fy, εs1 = -εy) This is the point of balanced failure. This strain distribution marks the change from compression failures originating by crushing of the compression surface of the section to tension failures initiated by yield of the longitudinal reinforcement. - Also marks beginning of transition zone for Φ for columns in which Φ increases from 0.65 or 0.70 up to 0.90 Z = -2.5 (-2.417 if εy = .00207) This point corresponds to the tension controlled strain limit of 0.005. For this example points will be calculated in the compression controlled zone, (one with the column entirely in compression) (Z = 0.9, -0.5), the tension controlled zone (Z = -5.0) and the transition zone (Z = -1.1).

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Notes

Compression Controlled zone

ε s1 ≥ ε y

Section 10.3.3

Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis "c"

Section 10.2.2

c

εs1

εs2

εc

d2 d1 Figure 1.1 - Strain curve for a column entirely in compression d1 = 9.50 in d2 = 2.50 in

Given: Calculate ΦPn & ΦMn for point in compression contolled zone & column entirely in compression

Z = .9

εc =

0.003

f y = Esε y ∴ ε y =

εy = Calculate εs1 Strain in 1st row of steel

Es = 29000 ksi As1 = 1.58 in2 As2 = 1.58 in2

Section 10.2.3

fy Es

=

60ksi 29000ksi

(Eq. 1-3)

Section 10.2.4

0.002069

Using Equation 1-2, from previous page calculate εs1

ε s1 = .9(.002069)

ε s1 =

0.001862

As shown in Figure 1.1: "c" (distance from extreme compression fiber to neutral axis) can be calculated using similar triangles. Calculate C

⎛ ⎞ 9.5" d1 ⎞ ⎟⎟ = .003⎛⎜ c = .003⎜⎜ ⎟ ⎝ .003 − .001862 ⎠ ⎝ .003 − ε s1 ⎠

c = Calculate "a" (equivalent stress block)

25.05 in

a = β 1C

For f'c ≥ 4000 psi

a=

a≤h

(Eq. 1-5)

For f'c ≤ 4000 psi

β1 =

(Eq. 1-4)

0.85

"a' must be less than depth of column

β1 = .85 β1 = .85 − .05( f ' c −4000 ) β1 ≥ .65 a = .85(25.05") = 21.29" ≥ 12"

12.000 in 4

Section 10.2.7.1 Section 10.2.7.3

(Eq. 1-6)


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Notes

Calculate εs2 Strain in 2nd row of steel

As shown in Figure 1.1: εs2 can be calculated using similar triangles.

⎛ C − d2 ⎞ ⎛ 25.05"−2.5" ⎞ ε s 2 = .003⎜ ⎟ = .003⎜ ⎟ ⎝ C ⎠ ⎝ 25.05" ⎠ ε s2 =

Calculate stress in each row of steel (fs1 & fs2)

0.002701

f sx = ε sx E s

f sx ≤ f y

(Eq. 1-8)

Section 10.2.4

f s 2 = .002701(29000ksi) = 78.33 ≥ 60

f s1 = .001862(29000ksi)

f s 2 = 60 ksi

f s1 = 54 ksi Calculate force in each row of steel

(eq. 1-7)

Fx = f sx Asx Fx = Asx ( f sx − .85 f 'c )

Tension Steel:

(Eq. 1-9)

(Eq. 1-10) Compression Steel: (must subtract concrete stress when in compression since concrete will be replaced by steel)

F1 = 1.58in 2 (54ksi − .85(4ksi)) F1 =

F2 = 1.58in 2 (60ksi − .85(4ksi)) F2 = 89.43 kips

79.95 kips

.85f'c b

a a

Cc

/2

Figure 1.2 - Equivalent rectangular stress block diagram An average stress of .85 f'c is assumed uniformily distributed over an equivalent compression zone. Caclulate Cc (Concrete Compression Force)

Using Figure 1.2 - Caclulate Cc

Cc = .85 f 'c ab = .85(4ksi)(12")(12") C c = 489.60 kips

5

(Eq. 1-11)

Section 10.2.6 Section 10.2.7 Section 10.2.7.1


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Notes

b/2 = 6" Calculate Pn & Mn by applying forces to free body diagram

Pn Mn

79.95 kips

89.43 kips 489.60 kips

d2 = 2.5" a/2 = 6"

d1 = 9.5" Figure 1.3: Column free body diagram for a "Z" of .9 Using Figure 1.3, Calculate Pn by summing vertical forces:

ΣFv = 0 = 79.95kips + 489.60kips + 89.43kips − Pn Pn = 658.98 kips Using Figure 1.3 from previous page, calculate Mn by summing moments about steel in line d1: (counterclockwise being positive moment)

ΣM = 0 =

489.6kips(3.5") 89.43kips(7") 658.98kips(3.5") + − − Mn 12" 12" 12" 1' 1' 1'

Moment arms will be in inches, must convert to feet for desired Units.

M n = 2.82 kip-ft Calculate ΦPn & ΦMn

φ = .65

Section 9.3.2.2

φPn = (.65)658.98

φPn =

428.33 kips

φM n =

1.83 kip-ft

φM n = (.65)2.82 Point on curve for "Z" = .9

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Notes

εs1

c

εs2 d2

εc

d1

Figure 1.4 - Strain curve for a column in compression & tension Calculate ΦPn & ΦMn (column in compression & tension)

d1 = 9.50 in d2 = 2.50 in

Given:

Es = 29000 ksi As1 = 1.58 in2 As2 = 1.58 in2 Z = -.5

Calculate εs1 Strain in 1st row of steel

Calculate C

εc =

0.003

εy =

0.002069

ε s1 = −.5(.002069)

ε s1 =

-0.001034

⎛ ⎞ 9.5" ⎟⎟ c = .003⎜⎜ ⎝ .003 − (− .001034) ⎠

(Eq. 1-2) (Eq. 1-4)

c = 7.06 in

β1 = Calculate "a"

a = .85(7.06") = 6.00" ≤ 12"

a= Calculate εs2 Strain in 2nd row of steel

steel (fs1 & fs2)

(Eq. 1-5)

6.004 in

⎛ 7.06"−2.5" ⎞ ⎟ ⎝ 7.06" ⎠

ε s 2 = .003⎜ ε s2 =

Calculate stress in each row of

0.85

(eq. 1-7)

0.001938

f s1 = −.001034(29000ksi)

f s 2 = .001938(29000ksi)

f s 2 = 56.21 kips

f s1 = -30.00 ksi

F1 = 1.58in 2 (− 30ksi) F2 = 1.58in 2 (56.21ksi − .85(4ksi))

Calculate force in each row of steel

F1 =

(Eq. 1-8)

F2 = 83.44 kips

-47.40 kips 7

(Eq. 1-9) (Eq. 1-10)


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Notes

Caclulate Cc

Cc = .85(4ksi)(6.004")(12")

(Eq. 1-11)

Section 10.2.6

C c = 244.98 kips b/2 = 6" Calculate Pn & Mn by applying forces to free body diagram

Pn Mn

83.44 kips

-47.4 kips

244.98 kips d2 = 2.5" a/2 = 3.002" d1 = 9.5"

Figure 1.5: Column free body diagram for a "Z" of -.5 Tensile strength of concrete shall be neglected in axial and flexural calculations of reinforced concrete. Using Figure 1.5 , calculate Pn by summing vertical forces:

ΣFv = 0 = −47.4kips + 244.98kips + 83.44kips − Pn

Pn =

281.02 kips

Using Figure 1.5, calculate Mn by summing moments about steel in line d1: (counterclockwise being positive moment)

ΣM = 0 =

244.98kips(6.5") 83.44kips(7") 281.02kips(3.5") + − − Mn 12" 12" 12" 1' 1' 1'


φ = .65 φPn = (.65)281.02

φPn =

182.67 kips

φM n =

64.62 kip-ft

φM n = (.65)99.41 Point on curve for "Z" = -.5

8

Section 10.2.5


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Notes

.002 ≤ ε s1 ≤ .005

Transition Zone Calculate ΦPn & ΦMn (column in compression & tension)

Calculate εs1 Strain in 1st row of steel Calculate C

d1 = 9.50 in d2 = 2.50 in

Given:

εc =

0.003

εy =

0.002069

ε s1 = −1.1(.002069 )

β1 =

Calculate stress in each row of steel (fs1 & fs2)

0.85 (Eq. 1-5)

4.590 in

⎛ 5.40"−2.5" ⎞ ⎟ ⎝ 5.40" ⎠

ε s 2 = .003⎜ ε s2 =

(eq. 1-7)

0.001612

f s1 = −.002276(29000ksi) = −66 ≥ 60ksi f s 2 = .001612(29000ksi)

(Eq. 1-8) (Eq. 1-8)

f s 2 = 46.75 kips

F1 = 1.58in 2 (− 60ksi)

F2 = 1.58in (46.75ksi − .85(4ksi)) 2

F1 = Caclulate Cc

(Eq. 1-4)

5.40 in

f s1 = -60.00 ksi Calculate force in each row of steel

5.401961

a = .85(5.4") = 4.59" ≤ 12"


ε s1 =

⎞ ⎛ 9.5" ⎟⎟ c = .003⎜⎜ ⎝ .003 − (− .002276) ⎠

c= Calculate "a" (equivalent stress block)

Es = 29000 ksi As1 = 1.58 in2 As2 = 1.58 in2

(Eq. 1-9) (Eq. 1-10)

F2 = 68.49 kips

-94.80 kips

Cc = .85(4ksi)(4.59")(12")

C c = 187.27 kips

9

(Eq. 1-11)


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Notes


Pn Mn

68.49 kips

-94.8 kips

187.27 kips

a/2 = 2.295" d2 = 2.5" d1 = 9.5" Figure 1.6: Column free body diagram for a "Z" of -1.1

Using Figure 1.6, calculate Pn by summing vertical forces:

ΣFv = 0 = −94.8kips + 68.49kips + 187.27 kips − Pn Pn =

160.96 kips

Using Figure 1.6 calculate Mn by summing moments about steel in line d1: (counterclockwise being positive moment)

ΣM = 0 =

187.27kips(7.205") 68.49kips(7") 160.96kips(3.5") + − − Mn 12" 12" 12" 1' 1' 1'


⎛ 250 ⎞ ⎛ 250 ⎞ ⎟ = .65 + (.002276 − .002 )⎜ ⎟ = .673 ⎝ 3 ⎠ ⎝ 3 ⎠

φ = .65 + (ε s1 − .002 )⎜

φM n = (.673)105.45

φPn = (.673)160.96

φPn =

108.33 kips

φM n =

70.97 kip-ft

Point on curve for "Z" = -1.1

10

Figure R9.3.2


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Notes

Tension Controlled Zone

ε s1 ≥ .005

Calculate ΦPn & Given: ΦMn for point in Tension controlled zone & column in compression & tension

Calculate εs1 Strain in 1st row of steel

Calculate C

d1 = 9.50 in d2 = 2.50 in

As2 = 1.58 in2 Z = -5

εc =

0.003

εy =

0.002069

ε s1 = −5(.002069)

β1 =

in each row of steel (fs1 & fs2)

2.14 in 0.85

(Eq. 1-7)

-0.000505

f s1 = −.010345(29000ksi) = −300 ≥ 60ksi f s 2 = −.000505(29000ksi)

F1 = 1.58in 2 (− 60ksi) F1 =

Caclulate Cc

(Eq. 1-5)

1.819 in

f s1 = -60.00 ksi Calculate force in each row of steel

(Eq. 1-8) (Eq. 1-8)

f s 2 = -14.65 kips

F2 = 1.58in 2 (− 14.65ksi)

(Eq. 1-9)

F2 = -23.15 kips

-94.80 kips

Cc = .85(4ksi)(1.819")(12")

Cc =

(Eq. 1-2)

(Eq. 1-4)

⎛ 2.14"−2.5" ⎞ ε s 2 = .003⎜ ⎟ ⎝ 2.14" ⎠ ε s2 =

Calculate stress

-0.010345

a = .85(2.14") = 1.819" ≤ 12"


ε s1 =

⎞ ⎛ 9.5" ⎟⎟ c = .003⎜⎜ ⎝ .003 − (− .010345) ⎠ c=

Calculate "a"

Es = 29000 ksi As1 = 1.58 in2

74.22 kips

11

(Eq. 1-11)


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Notes


Pn Mn

-23.15 kips

-94.8 kips

74.22 kips

a/2 = .91" d2 = 2.5" d1 = 9.5" Figure 1.7: Column free body diagram for a "Z" of -5

Using Figure 1.7, calculate Pn by summing vertical forces:

ΣFv = 0 = −94.8kips − 23.15kips + 74.22kips − Pn

Pn =

-43.73 kips

Using Figure 1.7 calculate Mn by summing moments about steel in line d1: (counterclockwise being positive moment)

ΣM = 0 =

74.22kips(8.59") 23.15kips(7") − 43.73kips(3.5") − − − Mn 12" 12" 12" 1' 1' 1'


φ = .9

Section 9.3.2.2

φM n = (.9)52.38

φPn = (.9 )(− 43.73)

φPn =

-39.36 kips

φM n =

47.14 kip-ft

Point on curve for "Z" = -5

12


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Notes

Calculate ΦPnt

When column is entirely in tension the designer shall assume the concrete in the column will not contribute to tension strength, only reinforcement shall resist tension.

Section 10.2.5

Pnt = − Ast f y (Area of 1 - #8 bar = .79 in2) 2 3.16 in

Ast = (4 * .79) =

Pnt = 3.16in 2 (60ksi)

Pnt = -189.60 kips Find Φ

0.9

Φ=

Section 9.3.2.2

φPnt = .9(− 189.6kips) ΦPnt = -170.64 kips

εs1

ΦPn

ΦMn

0.002069 0.0018621 -0.001034 -0.002276 -0.010345

347.60 347.60 182.67 108.33 -39.36 -170.64

0.00 1.83 64.62 70.97 47.14 0.00

Column Interaction Diagram

400.00

300.00

200.00

ΦPn (Kips)

Draw interaction diagram using points calculated.

100.00

0.00 0.00

10.00

20.00

30.00

40.00

-100.00

-200.00 ΦMn (k-ft)

13

50.00

60.00

70.00

80.00


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Notes

Finally an actual iteraction diagram will be shown with using more values of "Z", to get a more detailed curve.

Column Interaction Diagram

400.00

300.00

ΦPn (Kips)

200.00

100.00

0.00 0.00

10.00

20.00

30.00

40.00

-100.00

-200.00 ΦMn (k-ft)

14

50.00

60.00

70.00

80.00

Interaction Diagram for Concrete Columns

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