Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 115 views

    ACI 318-08 Ex002

    Uploaded by

    Antonius Ajalah
    This document summarizes an example problem that tests the demand/capacity ratio calculation for a reinforced concrete column subjected to axial load and moment. The example compares the ratio calculated using SAP2000 software to an independent hand calculation. The geometry, material properties, loading and reinforcement details of the rectangular column are provided. The technical feature tested is the tied reinforced concrete column demand/capacity ratio. The results from SAP2000 and the independent calculation match exactly, with no percentage difference.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    ACI 318-08 Ex002

    Uploaded by

    Antonius Ajalah
    115 views4 pages
    This document summarizes an example problem that tests the demand/capacity ratio calculation for a reinforced concrete column subjected to axial load and moment. The example compares the ratio calculated using SAP2000 software to an independent hand calculation. The geometry, material properties, loading and reinforcement details of the rectangular column are provided. The technical feature tested is the tied reinforced concrete column demand/capacity ratio. The results from SAP2000 and the independent calculation match exactly, with no percentage difference.

    Original Description:

    ACI 318-08 Ex002

    Copyright

    © © All Rights Reserved

    Available Formats

    PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    This document summarizes an example problem that tests the demand/capacity ratio calculation for a reinforced concrete column subjected to axial load and moment. The example compares the ratio calculated using SAP2000 software to an independent hand calculation. The geometry, material properties, loading and reinforcement details of the rectangular column are provided. The technical feature tested is the tied reinforced concrete column demand/capacity ratio. The results from SAP2000 and the independent calculation match exactly, with no percentage difference.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    115 views4 pages

    ACI 318-08 Ex002

    Uploaded by

    Antonius Ajalah
    This document summarizes an example problem that tests the demand/capacity ratio calculation for a reinforced concrete column subjected to axial load and moment. The example compares the ratio calculated using SAP2000 software to an independent hand calculation. The geometry, material properties, loading and reinforcement details of the rectangular column are provided. The technical feature tested is the tied reinforced concrete column demand/capacity ratio. The results from SAP2000 and the independent calculation match exactly, with no percentage difference.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4
    At a glance
    Powered by AI
    The example tests checking the demand to capacity ratio of a reinforced concrete column subjected to axial load and bending moment using SAP2000 software by comparing the output to independent hand calculations.

    The example is testing the demand to capacity ratio of a reinforced concrete column subjected to a factored axial load of 398.4 kips and bending moment of 332 k-ft using SAP2000 software.

    The hand calculation involved determining the neutral axis depth, calculating the axial and moment capacities, checking if the assumptions of compression-controlled failure and yielding of the compression steel were valid, and calculating the strength reduction factor.

    Software Verification

    PROGRAM NAME: SAP2000


    REVISION NO.: 0

    ACI 318-08 Example 002

    P-M INTERACTION 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 area
    of 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

    Muy=332k-ft 22"

    A A
    2.5" 14"

    10

    Section A-A

    Material Properties Section Properties Design Properties

    E= 3600 k/in2 b = 14 in fc = 4 k/in2


    = 0.2 d = 19.5 in fy = 60 k/in2
    G= 1500 k/in2

    ACI 318-08 Example 002 - 1


    Software Verification
    PROGRAM NAME: SAP2000
    REVISION NO.: 0

    TECHNICAL FEATURES TESTED


    Tied Reinforced Concrete Column Demand/Capacity Ratio

    RESULTS COMPARISON
    Independent results are hand calculated and compared.

    Percent
    Output Parameter SAP2000 Independent 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.

    ACI 318-08 Example 002 - 2


    Software Verification
    PROGRAM NAME: SAP2000
    REVISION NO.: 0

    HAND CALCULATION

    COLUMN STRENGTH UNDER COMPRESSION CONTROL

    fc = 4 ksi fy = 60 ksi
    b = 14 inch d = 19.5 inch
    Pu = 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
    cb = dt = (19.5) = 11.54 inch
    87 + f y 87 + 60

    2) From the equation of equilibrium:

    Pn = Cc + Cs T
    where
    Cc = 0.85 f c' ab = 0.85 4 14a = 47.6a
    ( )
    Cs = As' f y - 0.85 f c' = 4 ( 60 - 0.85 4 ) = 226.4 kips
    Assume compression steels yields, (this assumption will be checked later).
    (
    T = As f s = 4 f s f s < f y )
    Pn = 47.6a + 226.4 - 4 f s (Eqn. 1)

    3) Taking moments about As:

    a '
    Pn =
    1
    (
    Cc d - 2 + Cs d - d )

    e'
    The plastic centroid is at the center of the section and d " = 8.5 inch
    e' = e + d " = 10 + 8.5 = 18.5 inch.
    1 a
    Pn = 47.6a 19.5 - + 226.4 (19.5 - 2.5 )
    18.5 2

    ACI 318-08 Example 002 - 3


    Software Verification
    PROGRAM NAME: SAP2000
    REVISION NO.: 0

    Pn = 50.17 a - 1.29a 2 + 208 (Eqn. 2)

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

    a = 0.85 13.45 = 11.43 inch

    Substitute in Eqn. 2:
    Pn = 50.17 11.43 - 1.29 (11.43) + 208 = 612.9 kips
    2

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

    19.5 -13.45
    fs = 87 = 39.13 ksi
    13.45
    s = t = f s Es = 0.00135

    6) Substitute a = 13.45 inch and fs = 39.13 ksi in Eqn. 1 to calculate Pn2:


    Pn2 = 47.6 (11.43) + 226.4 - 4 ( 39.13) = 613.9 kips
    Which is very close to the calculated Pn2 of 612.9 kips (less than 1% difference)
    10
    M n = Pn e = 612.9 = 510.8 kips-ft
    12

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


    13.45 - 2.5
    s' = ( 0.003) = 0.00244 > y = 0.00207 ksi
    13.45
    Compression steels yields, as assumed.

    8) Calculate ,
    dt = d = 19.5 inch, c = 13.45 inch
    19.45 -13.45
    t (at the tension reinforcement level) = 0.003 = 0.00135
    13.45

    Since t < 0.002 , then = 0.65

    Pn = 0.65 ( 612.9 ) = 398.4 kips


    M n = 0.65 ( 510.8 ) = 332 k-ft.

    ACI 318-08 Example 002 - 4

    You might also like