Handbook of Civil Engineering Calculations

(singke) #1

DESIGN OFA SPIRALLY


REINFORCED COLUMN


A short circular column, spirally reinforced, is to support a concentric load of 420 kips
(1868.16 kN). Design the member, using/; = 4000 lb/in
2
(27,580 kPa) and./; = 50,000
lb/in
2
(344,750 kPa).


Calculation Procedure:


  1. Assume pg = O.O25 and compute the diameter of the section
    Thus, 0.25/c' = 1000 lb/in
    2
    (6895 kPa);/ = 20,000 lb/in
    2
    (137,900 kPa). By Eq. 40, A 8 =
    420/(1 +2Ox 0.025) = 280 in
    2
    (1806.6 cm
    2
    ). Then D = (^0.785)
    05
    = 18.9 in (130.32
    mm). SetD = 19 in (131.01 mm), making^ = 283 in
    2
    (1825.9 cm
    2
    ).

  2. Select the reinforcing bars
    The load carried by the concrete = 283 kips (1258.8 kN). The load carried by the steel =
    420 - 283 = 137 kips (609.4 kN). Then the area of the steel is Ast, = 137/20 = 6.85 in^2
    (44.196 cm^2 ). Use seven no. 9 bars, each having an area of 1 in^2 (6.452 cm^2 ). Then Ast, =
    7.00 in^2 (45.164 cm^2 ). The Reinforced Concrete Handbook shows that a 19-in (482.6-
    mm) column can accommodate 11 no. 9 bars in a single row.

  3. Design the spiral reinforcement
    The portion of the column section bounded by the outer circumference of the spiral is
    termed the core of the section. Let^4c = core area, in^2 (cm^2 ); D 0 = core diameter, in (mm);
    0, = cross-sectional area of spiral wire, in^2 (cm^2 ); g = pitch of spiral, in (mm);/^ = ratio of
    volume of spiral reinforcement to volume of core.
    The ACI Code requires 1.5-in (38.1 -mm) insulation for the spiral, with g restricted to a
    maximum of DJ6. Then Dc = 19 - 3 = 16 in (406.4 mm); Ac = 201 in^2 (1296.9 cm^2 ); DJ6
    = 2.67 in (67.818 mm). Use a 2.5-in (63.5-mm) spiral pitch. Taking a 1-in (25.4-mm)
    length of column,


volume of spiral asrrDJg

(^5) volume of core irD (^2) /4
or
a sDcPs



=
—j- (
42
)
The required value of/? 5 , as given by the ACI Code is
0.45(V4C-1)/;
Ps = 7 (43)
Jy
orps = 0.45(283/201 - 1)4/50 - 0.0147; as = 2.5(16)(0.0147)/4 = 0.147 in
2
(0.9484 cm
2
).
Use H-in (12.7-mm) diameter wire with as = 0.196 in
2
(1.2646 cm
2
).




  1. Summarize the design
    Thus: column size: 19-in (482.6-mm) diameter; longitudinal reinforcement: seven no. 9
    bars; spiral reinforcement: 1/2-in (12.7-mm) diameter wire, 2.5-in (63.5-mm) pitch.

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