circular column - structural design software · 2018-04-06circular column - structural design...
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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