Circular Column - Structural Design Software · 2018-04-06Circular Column - Structural Design Software

PROJECT : PAGE :

CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

Circular Column Design Based on ACI 318-14

INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH = 5 ksi, (34 MPa)

REBAR YIELD STRESS = 60 ksi, (414 MPa)SECTION SIZE D = 20 in, (508 mm)

FACTORED AXIAL LOAD = 300 kips, (1334 kN)

FACTORED MAGNIFIED MOMENT = 140 ft-kips, (190 kN-m)

FACTORED SHEAR LOAD = 20 kips, (89 kN)COLUMN VERT. REINFORCEMENT 8 # 6LATERAL REINF. OPTION (0=Spirals, 1=Ties) 1 TiesLATERAL REINFORCEMENT # 4 @ 12 in. (305 mm), o.c. THE COLUMN DESIGN IS ADEQUATE.

ANALYSIS

AT AXIAL LOAD ONLY 796 0AT MAXIMUM LOAD 796 57AT 0 % TENSION 584 122AT 25 % TENSION 481 143AT 50 % TENSION 395 153

272 156AT BALANCED CONDITION 267 157

38 203AT FLEXURE ONLY 0 189

CHECK FLEXURAL & AXIAL CAPACITY

796.34 kips., (at max axial load, ACI 318-14 22.4)where 0.65 (ACI 318-14 21.2) > [Satisfactory]

F = 0.8 314 3.52

= 0.656 (ACI 318-14 21.2)

where 10.4 in 0.0021 0.003d = 17.6 in, (ACI 20.6) D = 20.0 in Cover = 1.5 in, (ACI 318 20.6.1)

155 300 kips > 140 ft-kips [Satisfactory]

= 0.08 (ACI 318-14 10.6) = 0.011

= 0.01 (ACI 318-14 10.6) [Satisfactory]

(ACI 318-11 22.5)> [Satisfactory]

where 0.75 (ACI 318-11 21.2) 60 ksi

dx 17.6 182.7 0.40 25.8 35.3 46

= 12 (ACI 318-14 25.7) = 12 in

= 1 [Satisfactory]

fc'

fy

Pu

Mu

Vu

f Pn (k)

f Pn (k) f Mn (ft-k)

AT e t = 0.002f Mn (ft-k)

AT e t = 0.005

f Pmax =F f [ 0.85 fc' (Ag - Ast) + fy Ast] =

f = Pu

Ag = in2 Ast = in2

f =0.75 + ( et - 0.002 ) (50), for Spiral

0.65 + ( et - 0.002 ) (250 / 3), for Ties

Cb = d ec / (ec + es) = et = ec =

f Mn = ft-kips @ Pu = Mu =

rmax rprovd

rmin

CHECK SHEAR CAPACITY (ACI 318-14 10 & 22.5)

f Vn = f (Vs + Vc)

Vu

f = fy =

A0 Av Vc = 2 (fc')0.5A0 Vs = MIN (d fy Av / s , 4Vc) f Vn

smax sprovd

smin

0 50 100 150 200 2500

100

200

300

400

500

600

700

800

900

e

e

' '

2'

'

2 0.85, 57 , 29000max

0.85 2 , 0

0.85 ,

,,

CC

CC

C

S

for ksifE Ec so Ec

c c forf c of oo

forf c oforEss s t

f forf s ty

e e

e ee e

ee

e e

e e ee e

= = =

=

=

PROJECT : PAGE :

CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

Magnified Moment Calculation for Circular Column Based on ACI 318-14

INPUT DATA & DESIGN SUMMARYEFFECTIVE LENGTH FACTOR k = 1 , (ACI 6.2.5)

COLUMN UNSUPPORTED LENGTH = 12 ft

LARGER FACTORED MOMENT = 200 ft-k

SMALLER FACTORED END MOMENT = 12 ft-k, (positive if single curvature.)

CONCRETE STRENGTH = 4 ksiCOLUMN DIAMETER D = 20 in

FACTORED AXIAL LOAD = 400 k

SUMMATION FOR ALL VERTICAL LOADS IN THE STORY = 1200 k

SUMMATION FOR ALL CRITICAL LOADS IN THE STORY = 13600 k, (ACI 6.6.4)

THE MAGNIFIED MOMENT: 200.0 ft-k , Nonsway

ANALYSISMAGNIFIED MOMENT - NONSWAY

r = 0.25 D = 5.0 in, ACI 6.2.5.1

28.8 < 33.28 < = = Slenderness effect may be ignored. (ACI 6.2.5)

3605.0 ksi, ACI 19.2.2

7854

7078394

3369.07

200 ft-k, ACI 6.6.4

0.624 , ACI 6.6.4.5

1.00 , ACI 6.6.4.5.2

200.0 ft-k, ACI 6.6.4.5.1 < 210.0 ft-k [Satisfactory] ,(ACI 6.6.4.3)The column is nonsway. Ignore following calculations.

MAGNIFIED MOMENT - SWAY < = = Not apply

28.8 > 22 < = = Slenderness effect must be considered. (ACI 6.2.5)

1.13 , ACI 6.6.4

314

28.80 < 62.04 [Satisfactory]

200.0 ft-k, as given

5% 10.0 ft-k, assumed conservatively

236.7 ft-k, ACI 6.6.4.6

Lu

M2

M1

fc'

Pu

S Pu

S Pc

Mu =

k Lu / r = 34 - 12(M1 / M2) =

Ec = 57000 (fc')0.5 =

Ig = p D4 / 64 = in4

k-in2 , ACI 6.6.4.4.4

k-in2 , ACI 6.6.4.4.4

M2,min = MAX[ M2 , Pu (0.6+0.03 D) ] =

Cm = MAX[ 0.6 + 0.4 (M1 / M2, min) , 0.4 ] =

Mu, ns = dns M2, min = 1.05 M2 =

k Lu / r =

Ag = p D2 / 4 = in2

Lu / r = 35 / [Pu / (fc' Ag)]0.5 =

M2s = M2 =

M2ns = M2s =

Mu, s = M2ns + ds M2s =

0.4 0.40.25

1 1 0.6E I E Ic g c gEI E Ic gd

= = = =

2

2EI

PckLu

p= =

, 1.010.75

CmMAXns PuPc

d

= =

1 , 1.0 , 2.510.75

MIN MAXs PuPc

d

= =

S S