columna jai breslertarea

8/14/2019 Columna JAI BreslerTarea

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PROJECT : PAGE :CLIENT : DESIGN BY :

JOB NO. : DATE : REVIEW BY :

INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH f c ' = 6 ksi

REBAR YIELD STRESS f y = 60 ksi

SECTION SIZE C x = 40 inC y = 36

FACTORED AXIAL LOAD P u = 1700 k

FACTORED MAGNIFIED MOMENT Mu,x = 1800 ft-k

Mu,y = 100 ft-k

FACTORED SHEAR LOAD Vu,x = 130 k THE COLUMN DESIGN IS ADEQUATE.

Vu,y = 150 kCOLUMN VERT. REINFORCEMENT 8 # 8 at x dir.

3 # 8 at y dir.LATERAL REINF. OPTION (0=Spirals, 1=Ties) 1 TiesLATERAL REINFORCEMENT 4 4 @ 12 in o.c., at x dir.

3 4 @ 12 in o.c., at y dir.

ANALYSIS

f Pn (k)

f Mn (ft-k)

Pn (k) Mn (ft-k) AT AXIAL LOAD ONLY 4315 0 AT MAXIMUM LOAD 4315 1196 AT 0 % TENSION 3724 1744 AT 25 % TENSION 3149 2121 AT 50 % TENSION 2700 2305 AT e t = 0.002 2064 2433 AT BALANCED CONDITION 2043 2457

AT e t = 0.005 1517 2917 AT FLEXURE ONLY 0 1273

CHECK FLEXURAL & AXIAL CAPACITYf P max =F f [ 0.85 f c ' (Ag - Ast ) + f y Ast ] = 4315 kips., (at max axial load, ACI 318-11, Sec. 10.3.6.2)

where f = 0.65 (ACI 318-11, Sec.9.3.2.2) > P u [Satisfactory]

F = 0.8 Ag = 1440 in2 Ast = 17.38 in2

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

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

where C b = d ec / (ec + es ) = 22.2 in et = 0.0021 ec =

d = 37.5 in, (ACI 7.7.1) D = 40.0 in Cover = 1.5 in, (ACI 318 7.7.1)

f Mn = 2757 ft-kips @ P u = 1700 kips > Mu = 1803 ft-kips [Satisfactory]

r max = 0.08 (ACI 318-11, Section 10.9) r provd = 0.012

r min = 0.01 (ACI 318-11, Section 10.9) [Satisfactory]

CHECK SHEAR CAPACITY (ACI 318-11 Sec. 11.1.1, 11.2.1, & 11.4.6.2)

f Vn = f (Vs + Vc) (ACI 318-11 Sec. 11.1.1)

> Vu [Satisfactory]

where f = 0.75 (ACI 318-11 Sec. 9.3.2.3) f y = 60 ksi

d A0 Av Vc = 2 (f c ')0.5 A0 Vs = MIN (d f y Av / s , 4V c) f Vn

x 37.5 1240 0.80 192.1 150.0 257y 33.5 1240 0.60 192.1 100.5 219

0.656

0.003

(Total 22 # 8)

in

(ACI 318-11, Fig. R9.3.2)f =

Concrete Column Design Based on ACI 318-11

=

legs,#legs,#

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 500 1000 1500 2000 2500 3000 3500

'

'

2

'

'

2 0.85, 57 , 2

0.85 2 , 0

0.85 ,

,

,

C

C

C C

C

S

f f E E c so

E c

c c for f f oo

for f c o for E s s s t

f for f s t y

e

e e

e e

e e

e e e

e e

= = =

=

=

e

e




JOB NO. : DATE : REVIEW BY :

INPUT DATA & DESIGN SUMMARY

EFFECTIVE LENGTH FACTOR k = 1.6 , (ACI 10.10.6.3 or 10.10.7.1)COLUMN UNSUPPORTED LENGTH Lu = 12 ft

LARGER FACTORED MOMENT M2 = 200 ft-k

SMALLER FACTORED END MOMENT M1 = 100 ft-k, (positive if single curvature.)

CONCRETE STRENGTH f c ' = 4 ksi

COLUMN DIMENSIONS h = 20 inb = 20 in

FACTORED AXIAL LOAD P u = 400 k

SUMMATION FOR ALL VERTICAL LOADS IN THE STORY S P u = 1200 k

SUMMATION FOR ALL CRITICAL LOADS IN THE STORY S P c = 13600 k, (ACI Eq. 10-21)

THE MAGNIFIED MOMENT: Mu = 236.7 ft-k , Sway

ANALYSISMAGNIFIED MOMENT - NONSWAY

r = 0.3 h = 6.0 in, ACI 10.10.1.2

k Lu / r = 38.4 > 34 - 12(M 1 / M2) = 28 < = = Slenderness effect must be considered. (ACI Eq 10-7)

Ec = 57000 (f c ')0.5 = 3605.0 ksi, ACI 8.5.1

Ig = b h 3 / 12 = 13333 in4

1.2E+07 k-in2 , ACI 10.10.6.1

2234.18 k, ACI Eq (10-13)

M2,min = MAX[ M2 , P u (0.6+0.03 h) ] = 200 ft-k, ACI 10.10.6.5

Cm = MAX[ 0.6 + 0.4 (M 1 / M2, min ) , 0.4 ] = 0.8 , ACI Eq(10-16)

1.05 , ACI Eq (10-12)

Mu, ns = d M2, min = 210.2 ft-k, ACI Eq (10-11) > 1.05 M2 = 210.0 ft-k [Unsatisfactory] ,(ACI

The column is sway. See calculation as follows.

MAGNIFIED MOMENT - SWAY

k Lu / r = 38.4 > 22 < = = Slenderness effect must be considered. ACI Eq (10-6)

1.13 , ACI Eq (10-21)

Ag = b h = 400 in2

Lu / r = 24.00 < 35 / [P u / (f c ' Ag)]0.5 = 70.00 [Satisfactory]

M2s = M2 = 200.0 ft-k, as given

M2ns = 5% M2s = 10.0 ft-k, assumed conservatively

Mu, s = M2ns + ds M2s = 236.7 ft-k, ACI Eq (10-19)

Note: For column subject to bending about both principal axis, the moment about each axis shall be magnified separately basedon the conditions corresponding to that axis.

Magnified Moment Calculation for Concrete Column Based on ACI 318-11

0.4 0.40.25

1 1 0.6

E I E I c g c g EI E I c g d

= = = =

2

2

EI P c

k L u

= =

, 1 .01

0.75

C m MAX P u

P c

d

= =

1, 1.0 , 2 .5

10.75

MIN MAX s P u P c

d

= =

S S




JOB NO. : DATE : REVIEW BY :Rectangular Concrete Column Design

INPUT DATAf 'c = 4 ksi

f y = 60 ksix = 48 iny = 32 in

Bar Size ==> # 11 1.375No. of A sx = 8 8 # 11 1.48489 11.88

No. of A sy = 9 9 # 11 1.48489 13.36Total Bars ==> # 30 # 11

r = 3.0%P u = 1700 k

Mux = 2900 ft-k e x = 20.5 in

Muy = 1200 ft-k e y = 8.5 in

CHECK COLUMN CAPACITY BY THE BRESLER METHOD

1700.0 < 4407.9 ok 2428.6 < 3145.4 ok 1.199 > 1.0 NG

ANALYSIS

= 0.003 d = 45.295 in 29.295 ind' = 2.705 in 2.705 in

Es = 3E+07 3E+07 As = 12.48 in214.04 in2

= 0.70 b = 32 in 48 in

Mn = 4142.9 ft-k 1714.3 ft-k

= 0.85 Mno = 5485.1 ft-k 3864.7 ft-k

P o = 3820.9 k 5457.1 k

MOMENT STRENGTH (Pn=0)

-11175204.10078 in 3.70173 in

29612.2 psi 23425.7 psi

4034.01 ft-k 3052.89 ft-k

AXIAL LOAD STRENGTH (Mn=0)

7871.28 k = 7871.28 k

BALANCED CONDITION

22.7862 in 14.7372 in

60000 psi 60000 psi

2479.13 k 2405.11 k

5515.32 ft-k 3881.41 ft-k

x DIRECTION y DIRECTION

ue

1

f

2

' '' '1

'1

' 3.4

1.7

s s s s su u y c

c

b f f f d A A E A A E A E s s su yc

b f

eee

==

''

s u s

c d f E c

e ==

' ' ''110.85

2n sc s

ccb d d f f d M A

= =

'

0.85 ( )o g st st c y f f P A A A= =

1 sub

su y

d E a

f E

e e

==

'' 1

su y su sb

su

f E d f E d E

e e

e

= =

' ''0.85b s sbc sb yb f f f a P A A= =

' ' ''0.852 2 2 2

bb s sbc sb y

h h d ab h f f f a d M A A = =

11 1 1n

ox oy o

P

P P P

1.0nynx

nox noy

M M M M

0.80 0.70u o P P

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