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PROJECT : PAGE :

CLIENT : DESIGN BY :

JOB NO. : DATE : REVIEW BY :

Wall Pier Design Based on CBC 2001 / IBC 2003 / IBC 2006

INPUT DATA & DESIGN SUMMARY

CONCRETE STRENGTHfc' = 3 ksi

REBAR YIELD STRESS

fy = 60 ksi

WALL PIER LENGTH

L = 5 ft

WALL PIER HEIGHT

H = 12 ft

WALL PIER THICKNESS

t = 10 in

VERTICAL EDGE BARS, As2 # 9

TRANSVERSE REINFORCEMENT, Av THE DESIGN IS ADEQUATE.

2 # 4 @ 6 in, o.c. (at each face.)

FACTORED AXIAL LOAD Pu = 60 kips

FACTORED SHEAR FORCE Vu = 60 kips, (in plane)

ANALYSIS

CHECK WALL PIER DEFINITION (CBC 2001 1902A / IBC 2003 1908.1.4 / IBC 2006 1908.1.8)

L / t = 6.00 within [2.5 , 6] & H / L = 2.40 > 2

[Satisfactory]

CHECK SHEAR STRENGTH (SEC 1921.6.13.2, 1921.4.51, & 1921.3.4.2 / IBC 2003 1908.1.4 / IBC 2006 1908.1.11)

Ve = (Mpr, left, top + Mpr, right,bot) / H + Vu = 175.1 kips

< 8f(fc')

0.5bd = 216.7 kips [Satisfactory]

< f[Vc + Avfyd/s] = 197.8 kips [Satisfactory]

where d = 58.19 in

rleft = 0.003 > rmin=MIN[3(fc')0.5

/fy, 200/fy ]= 0.003 [Satisfactory]

rright = 0.003 > rmin = 0.003 [Satisfactory]

Mpr, left, top = rleft bd2fy (1.25 - 0.919rleft fy/fc') = 691 ft-kips

Mpr, right ,bot = rright bd2fy (1.25 - 0.919rright fy/fc') = 691 ft-kips

f = 0.85 (Sec 1909.3.2.3)

Av = 0.4 in2

Vc = 2(fc')0.5

bd = 0.0 kips, (Per Sec 1921.3.4.2, Vc = 0, if (Ve -Vu) 50% Ve AND Pu < Ag fc' / 20 )

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PROJECT : PAGE :

CLIENT : DESIGN BY :

JOB NO. : DATE : REVIEW BY :

INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH fc' = #REF! ksi

REBAR YIELD STRESS fy = #REF! ksi

COLUMN DIMENSUINS cx = #REF! in

cy = #REF! in

FACTORED AXIAL LOAD Pu = #REF! kFACTORED MOMENT LOAD Mu,x = #REF! ft-k

Mu,y = 0.0001 ft-k

FACTORED SHEAR LOAD Vu,x = 0 k

Vu,y = 0 k

COLUMN VERT. REINF. #### # #REF! at x dir.

#### # #REF! at y dir.

SHEAR REINF. ### legs,# #### @ 4 in o.c., x dir. #REF!

2 legs,# #### @ 4 in o.c., y dir.

ANALYSIS

f Pn (k)

f Pn (kips) f Mn (ft-kips)AT AXIAL LOAD ONLY #REF! 0

AT MAXIMUM LOAD #REF! #REF!

AT 0 % TENSION #REF! #REF!



ATe t = 0.002 #REF! #REF!

AT BALANCED CONDITION #REF! #REF!

f Mn (ft-k) ATe t = 0.005 #REF! #REF!

AT FLEXURE ONLY 0 #REF!

CHECK FLEXURAL & AXIAL CAPACITY

Mu = ( Mu,x2

+ Mu,y2

)0.5

= ft-k, (combined bending load.) q = #REF!o

, (the direction of combined load.)

f Pmax =0.80 f [ 0.85 fc' (Ag - Ast) + fy Ast] = #REF! kips., (at max axial load, ACI 318-02, Sec. 10.3.6.2)

where f = 0.65 (ACI 318-02, Sec.9.3.2.2) #### Pu ####

Ag = #REF! in2. Ast = #REF! in

2.

a = Cbb1 = ### in (at balanced strain condition, ACI 10.3.2)

f = 0.48 + 83 et = (ACI 318-02, Fig. R9.3.2)

where Cb = d ec / (ec + es) = #REF! in et= 0.002069 ec= 0.003

d = #REF! in, (ACI 7.7.1) b1 = #REF! ( ACI 318-02, Sec. 10.2.7.3 )

fMn = 0.9 Mn = #REF! ft-kips @ Pn = 0, (ACI 318-02, Sec. 9.3.2)

fMn = ft-kips @ Pu = #REF! kips #REF! Mu ####

rmax = 0.08 (ACI 318-02, Section 10.9) rprovd = #REF!

rmin = 0.01 (ACI 318-02, Section 10.9) #REF!

CHECK SHEAR CAPACITY (ACI 318-02 Sec. 11.1.1, 11.3.1, & 11.5.6.2)

fVn = f (Vs + Vc) (ACI 318-02 Sec. 11.1.1)

#REF! Vu #REF!

where f = 0.75 (ACI 318-02 Sec. 9.3.2.3) fy = #REF! ksi

d A0 Av Vc = 2 (f c')0.5

A0 Vs = MIN (d fyAv / s , 4Vc) f Vn

x #REF! #### #REF! #REF! #REF! #REF!

y #REF! #### #REF! #REF! #REF! #REF!

smax = #REF! (ACI 318-02, Section 10.16.8.5) sprovd = 4 in

smin = 1 (ACI 318-02, Section 7.10.4.3) #REF!

Concrete Column Design Based on ACI 318-02

Daniel

#REF!

#REF!

#REF!

Tian Li

0.652

0

0

0

0

0

1

1

1

1

1

1

0 0 0 1 1 1

e

eq

( )''

2

'

'

2 0.85, 57 , 29000

0.85 2 , 0

0.85 ,

,

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C

C

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S

fksifE Ec so

Ec

c c forf c of oo

forf c o

forEss s yf

forf s yy

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===

-

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R E I N F O R C E M E N T

SIZE DIAM. AREA

# 0 #REF!

3 0.38 0.11

4 0.50 0.20

5 0.63 0.31 Mu,x = #REF! ft-k

6 0.75 0.44 Mu,y = 0.00011 ft-k

7 0.88 0.60

8 1.00 0.79

9 1.13 1.0010 1.27 1.27

11 1.41 1.56

14 1.69 2.25

18 2.26 4.00

STRESS STRAIN COMPUTATION AT MAXIMUM LOAD

1ST ESTIMATE

Axial load fPn / f = #REF! k

Distance to neutral axis c = #REF! in

Beta factor b = #REF!

Compression block a = #REF! in

E = 29000 ksi d = #REF! in

fy

= #REF! ksi D = #REF! in

fc' = #REF! ksi q = #REF!

STA A dist strain stress force arm moment

to end

1 0.00 #REF! #REF! #REF! #REF! #REF! #REF! qcritial = #REF!

2 #REF! #REF! #REF! #REF! #REF! #REF! #REF! dx = #REF!

3 #REF! #REF! #REF! #REF! #REF! #REF! #REF! dy = #REF!

4 #REF! #REF! #REF! #REF! #REF! #REF! #REF! sx = #REF!

5 #REF! #REF! #REF! #REF! #REF! #REF! #REF! sy = #REF!

6 #REF! #REF! #REF! #REF! #REF! #REF! #REF! Asx = #REF! in2

/ in

7 #REF! #REF! #REF! #REF! #REF! #REF! #REF! Asx = #REF! in2

/ in

8 #REF! #REF! #REF! #REF! #REF! #REF! #REF! cx = #REF!

9 #REF! #REF! #REF! #REF! #REF! #REF! #REF! cy = #REF!

10 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!

11 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!

12 #REF! #REF! #REF! #REF! #REF! #REF! #REF!

CONC #REF! #REF!

S #REF! #REF! #REF!

Ec = #REF! ksi eo = #REF!

conc sec thick to end b control D ec fc force arm moment to end b to end b

1 #REF! #REF! #REF! 0.00 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!

2 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!



5 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!






S #REF! #REF! #REF! #REF!

3RD ITERATION


Distance to neytral axis c = #REF! in



E = 29000 ksi d = #REF! in

fy = #REF! ksi D = #REF! in


STA A dist strain stress force arm momentto end

1 0.00 #REF! #REF! #REF! #REF! #REF! #REF!








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CONC #REF! #REF!

S #REF! #REF!




2 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!3 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!









5TH ITERATION



Beta factor b = #REF!Compression block a = #REF! in

E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!


conc sec thick to end b control ec fc force arm moment to end b to end b












7TH ITERATION





E = 29000 ksi d = #REF! in




to end






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CONC #REF! #REF!

S #REF! #REF!

Ec

=#REF! ksi

eo

=#REF!













STRESS-STRAIN COMPUTATIONS AT POINT OF 0% TENSION

Stress at last row of bars fs = 0 ksi




E = 29000 ksi d = #REF! in




to end



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12 #REF! #REF! #REF! #REF! #REF! #REF! #REF!CONC #REF! #REF!

S #REF! #REF!











9 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!10 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!



Stress at last row of bars fs = #REF! ksi




E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














STRESS-STRAIN COMPUTATIONS AT BALANCED CONDITION





E = 29000 ksi d = #REF! in




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to end









9 #REF! #REF! #REF! #REF! #REF! #REF! #REF!10 #REF! #REF! #REF! #REF! #REF! #REF! #REF!



CONC #REF! #REF!

S #REF! #REF!








6 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!7 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!





STRESS-STRAIN COMPUTATIONS AT Pu = 0.1 * f'c * Ag < = = Not used

1ST ESTIMATE

Axial load Pn = #REF! k




E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














3RD ITERATION

Axial load Pu = #REF! k

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E = 29000 ksi d = #REF! in




to end












CONC #REF! #REF!

S #REF! #REF!

Ec

= #REF! ksi eo

= #REF!













5TH ITERATION





E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!













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7TH ITERATION





E = 29000 ksi d = #REF! in


fc'

=#REF! ksi q = #REF!


to end












CONC #REF! #REF!

S #REF! #REF!














STRESS STRAIN COMPUTATION AT PURE MOMENT CONDITION

1ST ESTIMATE

Axial load Pu = 0 k




E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!







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3RD ITERATION

Axial load Pu = 0 k

Distance to neytral axis c = #REF! inBeta factor b = #REF!


E = 29000 ksi d = #REF! in




to end












CONC #REF! #REF!

S #REF! #REF!














5TH ITERATION

Axial load Pu = 0 k




E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!




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S #REF! #REF! #REF! #REF!7TH ITERATION

Axial load Pu = 0 k




E = 29000 ksi d = #REF! in




to end












CONC #REF! #REF!

S #REF! #REF!














STRESS STRAIN COMPUTATION AT MAXIMUM LOAD1ST ESTIMATE





E = 29000 ksi d = #REF! in




to end










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CONC #REF! #REF!

S #REF! #REF!


conc sec thick to end b control DA ec fc force arm moment to end b to end b



3 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!4 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!








3RD ITERATION





E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














5TH ITERATION





E = 29000 ksi d = #REF! in




to end



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CONC #REF! #REF!

S #REF! #REF!














7TH ITERATION





E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














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f Pn (kips) f Mn (ft-kips)0 #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! 0

f Pn (kips) f Mn (ft-kips)#REF! #REF!

#REF! #REF!

#REF! #REF!

f Pn (kips) f Mn (ft-kips)0 #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! #REF! #REF!

#REF! 0

f Pn (kips) f Mn (ft-kips)POINT 1 #REF! #REF!

#REF! #REF!

#REF! #REF!


#REF! #REF!

#REF! #REF!


#REF! #REF!

#REF! #REF!


#REF! #REF!

#REF! #REF!


#REF! #REF!

#REF! #REF!


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#REF! #REF!

#REF! #REF!

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2ND ESTIMATE





E = 29000 ksi d = #REF! in

fy

= #REF! ksi D = #REF! in



to end













CONC #REF! #REF!

S #REF! #REF!


to end b conc sec thick to end b control ec fc force arm moment to end b

#REF! #REF! 1 #REF! #REF! #REF! 0.00 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!

#REF! #REF! 2 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!



#REF! #REF! 5 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!


#REF! #REF! 7 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!#REF! #REF! 8 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!




4TH ITERATION





E = 29000 ksi d = #REF! in



STA A dist strain stress force arm momentto end









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CONC #REF! #REF!

S #REF! #REF!


to end b conc sec thick to end b control D ec fc force arm moment to end b


#REF! #REF! 2 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!#REF! #REF! 3 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!









6TH ITERATION



Beta factor b = #REF!Compression block a = #REF! in

E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














8TH ITERATION





E = 29000 ksi d = #REF! in




to end






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CONC #REF! #REF!

S #REF! #REF!

Ec

=#REF! ksi

eo

=#REF!


















E = 29000 ksi d = #REF! in




to end



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12 #REF! #REF! #REF! #REF! #REF! #REF! #REF!CONC #REF! #REF!

S #REF! #REF!













STRESS-STRAIN COMPUTATIONS AT POINT OF et = 0.002





E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














STRESS-STRAIN COMPUTATIONS AT POINT OF et = 0.005





E = 29000 ksi d = #REF! in




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to end












CONC #REF! #REF!

S #REF! #REF!













2ND ESTIMATE

Axial load Pn = #REF! k




E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














4TH ITERATION


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8TH ITERATION





E = 29000 ksi d = #REF! in


fc'

=#REF! ksi q = #REF!


to end












CONC #REF! #REF!

S #REF! #REF!














2ND ESTIMATE

Axial load Pu = 0 k




E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!







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4TH ITERATION

Axial load Pu = 0 k

Distance to neutral axis c = #REF! inBeta factor b = #REF!


E = 29000 ksi d = #REF! in




to end












CONC #REF! #REF!

S #REF! #REF!














6TH ITERATION

Axial load Pu = 0 k




E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!




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S #REF! #REF! #REF! #REF!8TH ITERATION

Axial load Pu = 0 k




E = 29000 ksi d = #REF! in




to end












CONC #REF! #REF!

S #REF! #REF!














2ND ESTIMATE





E = 29000 ksi d = #REF! in




to end










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CONC #REF! #REF!

S #REF! #REF!


to end b conc sec thick to end b control DA ec fc force arm moment to end b



#REF! #REF! 3 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!#REF! #REF! 4 #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF! #REF!








4TH ITERATION





E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














6TH ITERATION





E = 29000 ksi d = #REF! in




to end



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CONC #REF! #REF!

S #REF! #REF!














8TH ITERATION





E = 29000 ksi d = #REF! in




to end













CONC #REF! #REF!

S #REF! #REF!














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to end b to end b

#REF! #REF! #REF! #REF!






#REF! #REF! #REF! #REF!#REF! #REF! #REF! #REF!



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to end b to end b


#REF! #REF! #REF! #REF!#REF! #REF! #REF! #REF!








to end b to end b











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to end b to end b











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to end b to end b











to end b to end b











case 2 case 3

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to end b to end b











to end b to end b











case 2 case 3

case 2 case 3

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Date post:	14-Apr-2018
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