ACI 318-08 Ex002

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Software VerificationPROGRAM NAME: SAP2000

REVISION NO.: 0

ACI 318-08 Example 002 - 1

ACI 318-08 Example 002

P-MINTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN

EXAMPLE DESCRIPTION

The Demand/Capacity ratio for a given axial loading and moment is tested in this

example.

A reinforced concrete column is subjected factored axial load Pu = 398.4 k and

moments Muy = 332 k-ft. This column is reinforced with 4 #9 bars. The total areaof reinforcement is 8.00 in2. The design capacity ratio is checked by hand

calculations and result is compared.

GEOMETRY,PROPERTIES AND LOADING

Pu=398.4 kips

Material Properties

E = 3600 k/in2

= 0.2G = 1500 k/in2

Section Properties Design Properties

fc= 4 k/in2

fy = 60 k/in2

b = 14 ind = 19.5 in

A

10

Section A-A

14"2.5"

22"

A

Muy=332k-ft

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REVISION NO.: 0

ACI 318-08 Example 002 - 2

TECHNICAL FEATURES TESTED

Tied Reinforced Concrete Column Demand/Capacity Ratio

RESULTS COMPARISON

Independent results are hand calculated and compared.

Output Parameter SAP2000 IndependentPercent

Difference

Column Demand/Capacity Ratio 1.000

1.00 0.00%

COMPUTER FILE: ACI 318-08 Ex002

CONCLUSION

The computed results show an exact match with the independent results.

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REVISION NO.: 0

ACI 318-08 Example 002 - 3

HAND CALCULATION

COLUMN STRENGTH UNDER COMPRESSION CONTROL

fc= 4 ksi fy= 60 ksi

b = 14 inch d = 19.5 inchPu= 398.4 kips Mu= 332 k-ft

1) Because e= 10 inch < (2/3)d = 13 inch., assume compression failure. This assumption will be

checked later. Calculate the distance to the neutral axis for a balanced condition, cb:

Position of neutral axis at balance condition:

87 87

= = 19.5 = 11.5487 + 87 + 60

b t

y

c df

inch

2) From the equation of equilibrium:

= n c s

P C C T

where= 0.85 = 0.85 4 14 = 47.6

'

c cC f ab a a

= -0.85 = 4 60 -0.85 4 = 226.4' 's s y cC A f f kipsAssume compression steels yields, (this assumption will be checked later).

4s s s s yT = A f = f f < f = 47.6 + 226.4 - 4n sP a f (Eqn. 1)

3) Taking moments about As:

1= 2 '

n c s'aP C d - C d - d

e

The plastic centroid is at the center of the section and "d = 8.5 inch

=10+8.5 =18.5' "

e = e + d inch.

1

= 47.6 19.5 226.4 19.5 - 2.52

n

aP a -

18.5

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REVISION NO.: 0

ACI 318-08 Example 002 - 4

2= 50.17 1.29 + 208

nP a - a (Eqn. 2)

4) Assume c= 13.45 inch, which exceed cb(11.54 inch).

= 0.85 13.45 = 11.43a inch

Substitute in Eqn. 2:

2

= 50.17 11.43 1.29 11.43 + 208 = 612.9 n

P - kips

5) Calculatefs from the strain diagram when c = 13.45 inch.

19.5 -13.45= 87 = 39.13

13.45

sf ksi

s t s s

= f E = 0.00135

6) Substitute a = 13.45 inchandfs = 39.13 ksi in Eqn. 1 to calculatePn2:

= 47.6 11.43 + 226.4- 4 39.13 613.9n2P kipsWhich is very close to the calculatedPn2of 612.9 kips (less than 1% difference)

10= 612.9 510.8

12

n nM = P e kips-ft

7) Check if compression steels yield. From strain diagram,

13.45 - 2.5

= 0.003 = 0.00244 > 0.0020713.45

'

s y ksi

Compression steels yields, as assumed.

8) Calculate ,

= = 19.5t

d d inch, c= 13.45 inch

t(at the tension reinforcement level) =

19.45 -13.450.003 = 0.00135

13.45

Since 0.002 t

, then = 0.65

= 0.65 612.9 398.4nP kips

= 0.65 510.8 332 nM k-ft.

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ACI 318-08 Ex002