mathcad - 13-column design

8/13/2019 Mathcad - 13-Column Design

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Ag bchc= = area of gross section of column

Ac Ag Ast= = area of concrete section

Column steel ratio:

g

Ast

Ag

= 1%= 8%..

Lateral ties

According ACI 7.10.5 the diameters of lateral tie are

For longitudinal bar D 32mm : Dv 10mm

For longitudinal bar D 32mm> : Dv 12mm

The spacing of tie

s 16D s 48Dv s bc

In practice

Dv1

3

1

4..

D=

s 100mm= 150mm.. 300mm..

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Spirals

According ACI 7.10.4 the diameter of spiral: Dv 10mm

Clear spacing of spiral: 25mm s 75mm

Determination of concrete section

Ag

Pu

0.80

0.85 f'c 1 g( ) fy g+=

Determination of steel area

Ast

Pu

0.800.85 f'c Ag

0.85 f'c fy+=

Example 13.1

Tributary area B 4m:= L 8m:=

Materials f'c 25MPa:= fy 390MPa:=

Loads on slab

DL 50mm 22 kN

m3

100mm 25 kN

m3

+ 0.40kN

m2

+ 1.00kN

m2

+ 5 kN

m2

=:=

LL 2.00kN

m2

:= (for classroom)

Reduction of live load

AT B L 32 m2

=:= (tributary area)

KLL 4:= (for interior column)

AI KLLAT 128 m2

=:= (influence area)

LL 0.25 4.572

AI

m2

+ 0.654=:= LLLL 1.308 kN

m2

=

Loads of wall

Void 30mm 30 mm 190 mm 4:=

Page 78


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Brickhollow.10 120mm Void55

1m2

20 kN

m3

1.648kN

m2

=:=

Loads on column

PD.slab DL B L 160 kN=:=

PL.slab LLLL B L 41.863 kN=:=

PB1 25cm 65cm 100mm( ) 25 kN

m3

L 27.5 kN=:=

PB2 20cm 35cm 100mm( ) 25 kN

m3

B 5 kN=:=

Pwall.1 Brickhollow.10 3.5m 65cm( ) L 2m( ) 28.174 kN=:=

Pwall.2 Brickhollow.10 3.5m 35cm( ) B 20.76 kN=:=

Pcolumn 35cm 45 cm 25 kN

m3

3.5m 65cm( ) 11.222 kN=:=

PD PD.slab PB1+ PB22.5+ Pwall.1+ Pwall.22+ Pcolumn+( )5 1404.577 kN=:=

PL PL.slab5 209.316 kN=:=

Pu 1.2 PD 1.6 PL+ 2020.397 kN=:=

Verification

PL

PD PL+ 12.97 %=

PD PL+

B L 5 10.087

kN

m2

=

Pcolumn5

PD PL+ 3.477 %=

Column section

g 0.02:= 0.65:=

Ag

Pu

0.80

0.85 f'c 1 g( ) fy g+ 1357.338 cm

2=:=

k35cm

45cm:= hc

Ag

k417.75 mm=:= bc k hc 324.916 mm=:=

hc Ceil hc 50mm,( ) 450 mm=:=

Page 79


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6/34

Pn C Cs+ T=

where

C 0.85 f'c a b=

Cs

A'sf'

s=

T Asfs=

Equilibrium in moments M 0=

Mn Pne= C h

2

a

2

Csh

2d'

+ T d h

2

+=

Mn 0.85 f'c a b h

2

a

2

A'sf's h

2d'

+ Asfs d h

2

+=

Conditions of strain compatibility

s

u

d c

c=

s ud c

c= fs Ess= Esu

d c

c fy=

's

u

c d'

c=

's uc d'

c= f's Es's= Esu

c d'

c fy=

Unknowns = 5: a As, A's, fs, f's,

Equations = 4: X 0= M 0= 2 conditions of strain compatibility

Case of symmetrical column: As A's=Ast

2=

Case of unsymmetrical column: fs fy=

A. Determination of Steel Area

Given: Pu Mu, b h, f'c, fy,

Find: As A's=Ast

2=

Page 81


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Functions AsNa( )

Pu

0.85 f'c a b

f'sa( ) fsa( )=

AsMa( )

Mu

0.85 f'c a b h

2

a

2

f'sh

2d'

fs dh

2

+

=

Answer: As AsNa( )= AsMa( )=

Example 13.2

Required strength Pu

1200kN:=

Mu 30kN m:=

Concrete dimension b 300mm:=

h 300mm:=

Materials f'c 25MPa:=

fy 390MPa:=

Solution

d h 30mm 10mm+ 20mm

2+

250 mm=:=

d' 30mm 10mm+ 20mm

2+ 50 mm=:=

1 0.65 max 0.85 0.05f'c 27.6MPa

6.9MPa

min 0.85

0.85=:=

Es 2 10

5

MPa:= u 0.003:=

c a( )a

1

:=

fsa( ) min Esu d c a( )

c a( ) fy,

:= fs150mm( ) 250 MPa=

f'sa( ) min Esu c a( ) d'

c a( ) fy,

:= f's100mm( ) 345 MPa=

dt d:=

Page 82


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a( ) t u

dt c a( )

c a( )

0.65 max1.45 250 t+

3

min 0.9

:= 90mm( ) 0.824=

AsNa( )

Pu

a( )0.85 f'c a b

f'sa( ) fsa( ):=

AsMa( )

Mu

a( )0.85 f'c a b

h

2

a

2

f'sa( )h

2d'

fsa( ) dh

2

+

:=

a1 257.9mm:= a2 258mm:=

0.2579 0.257954.05 10

4

4.06 10 4

4.07 10 4

4.08 10 4

4.09 10 4

AsNa( )

AsMa( )

a

a 257.94mm:= a

h0.86=

AsNa( ) 4.071 cm2= AsMa( ) 4.071 cm2=

Ast AsNa( ) AsMa( )+ 8.141 cm2

=:=

6 14mm( )

2

4 9.236 cm

2=

Ag b h:= Ag 900 cm2

=

Page 83


9/34

Asteel N( ) a h

N

i ORIGIN

f f'sa( ) f

sa( )

continue( ) f 0=if

AsN

Pu

a( )0.85 f'c a b

f

continue( ) AsN 0if

fd f'sa( )h

2d'

fsa( ) dh

2

+

continue( ) fd 0=if

AsM

Mu

a( )0.85 f'c a b

h

2

a

2

fd

continue( ) AsM 0if

AAsM AsN

AsN

Zi

a

h

AsN

Ag

AsM

Ag

A

i i 1+

a a a a+, h..for

csort ZT

ORIGIN 3+,( )

:=

Z Asteel 10000( ):=

a Z0 0,

h 257.94 mm=:=

AsN Z0 1,Ag 4.071 cm

2=:=

AsM Z0 2,Ag 4.071 cm

2=:=

Ast AsN AsM+ 8.141 cm2

=:=

Page 84


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Z

0 1 2 3

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

0.86 -34.52310 -34.52310 -53.16910

0.86 -34.51810 -34.53210 -33.02210

0.86 -34.52810 -34.51410 -33.07810

0.86 -34.51310 -34.54110 -36.08310

0.86 -34.53310 -34.50510 -36.11610

0.859 -34.53710 -34.49610 -39.14710

0.86 -34.50910 -34.5510 -39.15110

0.859 -34.54210 -34.48710 0.012

0.86 -34.50410 -34.55910 0.012

0.859 -34.54710 -34.47810 0.015

0.86 -34.49910 -34.56810 0.015

0.859 -34.55210 -34.46910 0.018

0.86 -34.49410 -34.57710 0.018

0.859 -34.55710 -34.4610 0.021

0.86 -34.48910 -34.58610 0.022

0.859 -34.56210 -34.45110 ...

=

B. Interaction Diagram for Column Strength

Interaction diagram is a graph of parametric funct ion Mn Pn,( )

Pna( ) a( ) 0.85 f'c a b A'sf's+ Asfs( ) Pn.max=

Mna( ) a( ) 0.85 f'c a b h

2

a

2

A'sf's h

2d'

+ Asfs dh

2

+

=

where fsa( ) Esu d c

c fy=

f'sa( ) Esu c d'

c fy=

Example 13.3

Concrete dimension b 300mm:= h 300mm:=

d' 30mm 8mm+ 16mm

2+ 46 mm=:=

d h 30mm 8mm+ 16mm

2+

254 mm=:=

Page 85


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Steel reinforcements As 3 16mm( )

2

4 6.032 cm

2=:=

A's 3 16mm( )

2

4 6.032 cm

2=:=


fy 390MPa:=

Solution

Ag b h 900 cm2

=:=

Ast As A's+ 12.064 cm2

=:=

Case of axially loaded column

0.65:=

Pn.max 0.80 0.85 f'c Ag Ast( ) fyAst+:=

Pn.max 1029.588 kN=

Case of eccentric column

1 0.65 max 0.85 0.05

f'c 27.6MPa

6.9MPa

min 0.85

0.85=:=

c a( ) a

1

:=

u 0.003:= Es 2 105MPa:=

fsa( ) min Esu d c a( )

c a( ) fy,

:= fs200mm( ) 47.7 MPa=

f'sa( ) min Esu c a( ) d'

c a( ) fy,

:= f's100mm( ) 365.4 MPa=

dt d:=

a( ) t u

dt c a( )

c a( )

0.65 max1.45 250 t+

3

min 0.9

:= 100mm( ) 0.773=

Pna( ) min a( ) 0.85 f'c a b A'sf'sa( )+ Asfsa( )( ) Pn.max,:=

Mna( ) a( ) 0.85 f'c a b

h

2

a

2

A'sf'sa( )

h

2 d'

+ Asfsa( ) d

h

2

+

:=

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a 0h

100, h..:=

0 10 20 30 40 50 60 70 80

0

100

200

300

400

500

600

700

800

900

1000

1100

1200Interaction diagram for column strength

Pn

a( )

kN

Mna( )

kN m

Page 87


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C. Case of Distributed Reinforcements

Equilibrium in forces

Pn C

1

n

i

Ti

=

= 0.85 f'c a b

1

n

i

As i,

fs i,

( )=

=

Equilibrium in moments

Mn Pne= Ch

2

a

2

1

n

i

Ti

di

h

2

=

+=

Mn 0.85 f'c a b h

2

a

2

1

n

i

As i,

fs i,

di

h

2

=

+=

Condition of strain compatibility

s i,

u

di

c

c=

s i, u

di

c

c=

fs i, Ess i,

= Esud

i c

c= with f

s i, fy

Unknows = 2 n 1+ a As i,

, fs i,

,

Equations = n 2+ X 0= M 0=n Conditions of strain compatibility

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Additional conditions

As i,

niA

s0=

Interaction Diagram for Column Strength

Abscissa Mna( ) 0.85 f'c a b h

2

a

2

1

n

i

As i,

fs i,

di

h

2

=

+

=

Ordinate Pna( ) 0.85 f'c a b

1

n

i

As i,

fs i,

( )=

Pn.max=

Example 13.4


Steel reinforcements

As

7

7

4

4

4

4

4

7

7

32mm( )

2

4

:= d

70

177.5

285

392.5

500

607.5

715

822.5

930

mm:=


Solution

ORIGIN 1:=

Case of axially load column

Ag b h:= ns rows As( ):= ns 9=

Ast As:= Ast 386.039 cm2= gAst

Ag

0.055=:=

0.65:=

Pn.max 0.80 0.85 f'c Ag Ast( ) fyAst+:=

Pn.max 20984.038 kN=

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1 0.65 max 0.85 0.05f'c 27.6MPa

6.9MPa

min 0.85

0.724=:=

c a( )a

1

:=

u 0.003:= Es 2 105MPa:=

fsi a,( ) s u

di

c a( )

c a( )

sign s( )min Es s fy,( )

:= fs2 200mm,( ) 214.516 MPa=

dt max d( ):= dt 930 mm=

a( ) t u

d

t

c a( )

c a( )

0.65 max1.45 250 t+

3

min 0.9

:= 300mm( ) 0.794=

Pna( ) min a( ) 0.85 f'c a b

1

ns

i

Asi

fsi a,( )=

Pn.max,

:=

M

n

a( ) a( ) 0.85 f'

c

a b h

2

a

2

1

ns

i

A

si

f

s

i a,( ) di

h

2

=+

:=

a 0h

100, h..:=

Page 90


16/34

0 1000 2000 3000 4000 5000 60000

5000

10000

15000

20000

25000

Interaction diagram for column strength

Pna( )

kN

Mna( )

kN m

Determination of Steel Area

X 0= Pn 0.85 f'c a b1

n

i

As i,

fs i,

( )=

= 0.85 f'c a b As01

n

i

ns i,

fs i,

( )=

=

AsNa( )

0.85 f'c a bPu

1

n

i

ns i,

fs i,

( )=

=

M 0= Mn 0.85 f'c a b h

2

a

2

1

n

i

As i,

fs i,

di

h

2

=

+=

Mn 0.85 f'c a b h

2

a

2

As01

n

i

ns i, fs i,

di

h

2

=

+=

Page 91


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AsMa( )

Mu

0.85 f'c a b

h

2

a

2

1

n

i

ns i,

fs i,

di

h

2

=

=

Example 13.5

Required strength Pu 19000kN:=

Mu 300kN m:=


Concrete cover Cover 40mm:=

Diameter of stirrup Dv 10mm:=

Diameter of coner bar D0 32mm:=

Distribution of reinforcements

n

7

7

4

4

4

4

4

7

7

:=


Solution

Case of axially loaded column

Ag b h 0.7m2

=:=

0.65:=

Ast

Pu

0.80 0.85 f'c Ag

0.85 f'c fy+ 277.568 cm

2=:= 48

32mm( )2

4 386.039 cm

2=


Page 92


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d1

Cover Dv+D0

2+ 66 mm=:=

ns rows n( ) 9=:= h d

12

ns 1 108.5 mm=:=

i 2 ns..:= di di 1 +:=

dt max d( ):= dt 934 mm=

1 0.65 max 0.85 0.05f'c 27.6MPa

6.9MPa

min 0.9

0.724=:=

c a( ) a

1

:=

a( ) t u

dt c a( )

c a( )

0.65 max1.45 250 t+

3

min 0.9

:= 300mm( ) 0.797=

fsi a,( ) s u

di

c a( )

c a( )

sign s( )min Ess fy,( )

:= fs2 300mm,( ) 347.354 MPa=

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Asteel No( )

h d1

No

k 1

nf

1

ns

i

nifsi a,( )( )

=

continue( ) nf 0=if

AsN

0.85 f'c a bPu

a( )

nf

continue( ) AsN 0if

nfd

1

ns

i

nifsi a,( ) di

h

2

=

continue( ) nfd 0=if

AsM

Mu

a( )0.85 f'c a b

h

2

a

2

nfd

continue( ) AsM 0if

A AsM AsN

Zk

a

h

AsN

Ag

AsM

Ag

A

Ag

k k 1+

a d1

d1 +, h..for

csort ZT

4,( )

:=

Z Asteel 20000( ):= rows Z( ) 766=

a Z1 1,

h 980.806 mm=:=

AsN Z1 2, Ag 1.917 cm

2=:=

Page 94


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AsM Z1 3, Ag 1.918 cm

2=:=

As

AsN AsM+

2n:= As 92.044 cm

2=

Ast 277.568 cm2

=

As max AsAst, 0.01 Ag,( ):= As 277.568 cm2

=

D. Design of Circu lar Column

Area and centro id of segment

acosrc a

rc

= rc

Dc

2=

Asector

Radius Arch

2=

rcrc 2

2= r

c

2=

Page 95


21/34

AtriangleBase Height

2=

rcsin ( ) 2 rc cos ( )

2= rc

2sin ( ) cos ( )=

Asegment Asector Atriangle= rc2 sin ( ) cos ( )( )=

dA1

2rc rc d=

rc2

2d=

xc

2 rc

3cos ( )=

Axc

d

rc

3

3cos ( )

d=2 rc

3 sin ( )

3=

xsector

Axc

d

A=

2 rc3

sin ( )

3

1

rc2

=2 rc

3

sin ( )

=

xtriangle2

3rc cos ( )=

xsegment

xsectorAsector xtriangleAtriangle

Asegment=

xsegment

2 rc

3

sin ( )

rc

2

2 rc

3cos ( ) rc

2 sin ( ) cos ( )

rc2 sin ( ) cos ( )( )

=

xsegment

2 rc

3

sin ( )3

sin ( ) cos ( )=

Location of steel re-bars

di

rc rscos i( )= rs rc Cover=

i

2

ns

i 1( )=

ns= number of steel re-bars

Equilibrium in forces

Page 96


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Pn C

1

ns

i

Ti

=

= 0.85 f'c Ac

1

ns

i

As i,

fs i,

( )=

=

Equilibrium in moments

Mn Pne= C xc

1

ns

i

Ti d

i rc( )

=

+=

Mn 0.85 f'c Ac xc

1

ns

i

As i,

fs i,

di

rc( )=

+=

where Ac rc2 sin ( ) cos ( )( )=

xc

2 rc

3

sin ( )3

sin ( ) cos( )=

Condition of strain compatibility

s i,

u

di

c

c=

s i,u

di

c

c=

fs i,

Ess i,= Esu

di

c

c fy=

Example 13.6

Required strength Pu 3437.31kN:=

Mu 42.53kN m:=


fy 390MPa:=

Solution

Case of axially l oaded co lumn

0.70:=

Assume g 0.025:=

Ag

Pu

0.85

0.85 f'c 1 g( ) fy g+:= Ag 1.896 10

3 cm

2=

Page 97


23/34

Diameter of column Dc CeilAg

4

50mm,

:= Dc 500 mm=

Ag

Dc2

4:= Ag 1.963 10

3 cm

2=

Ast

Pu

0.85 0.85 f'c Ag

0.85 f'c fy+:= Ast 43.514 cm

2=

g

Ast

Ag

:= g 0.022=

Ds Dc 2 30mm 8mm+ 20mm

2+

:= Ds 404 mm=

n

s

14:= As0

20mm( )2

4

:= Ast

n

s

A

s0

43.982 cm2

=:=

sDs

ns

90.657 mm=:=

Pn.max 0.85 0.85 f'c Ag Ast( ) fyAst+:= Pn.max 3447.594 kN=

Case of Eccentric Column

rc

Dc

2

:= rsDs

2

:=

i 1 ns..:= si

360deg

ns

i 1( ):=

di

rc rscos si

:=

1 0.65 max 0.85 0.05f'c 27.6MPa

6.9MPa

min 0.85

0.85=:=

c a( )a

1

:= dt max d( ):= dt 452 mm=

a( ) t u

dt c a( )c a( )

0.70 max1.7 200 t+

3

min 0.9

:= 150mm( ) 0.879=

a( ) acosrc a

rc

:=

xca( )2 rc

3

sin a( )( )3

a( ) sin a( )( ) cos a( )( ):=

Aca( ) rc2a( ) sin a( )( ) cos a( )( )( ):=

Page 98


24/34

fsi a,( ) s u

di

c a( )

c a( )

sign s( )min Es s fy,( )

:=

Pna( ) min a( ) 0.85 f'c Aca( )

1

ns

i

As0fsi a,( )( )=

Pn.max,

:=

Mna( ) a( ) 0.85 f'c Aca( ) xca( )

1

ns

i

As0fsi a,( ) di rc( )=

+

:=

a 0Dc

100, Dc..:=

0 100 200 3000

1000

2000

3000

Interaction diagram for column strength

Pna( )

kN

Pu

kN

Mna( )

kN m

Mu

kN m,

Page 99


25/34

13.3. Long (Slender) Columns

Stability index

QPu 0VuLc

=

where PuVu, = total vertical load and story shear

0= relative deflection between the top and bottom of story

Lc= center-to-center length of column

Q 0.05 : the column is nonsway (braced)

Q 0.05> : the column is sway (unbraced)

Unbraced Frame Braced Frame

Shea

rWall

Braced Frame

Brick Wall

Ties

Slenderness of Column

Column is short, if

For nonsway columns:k Lu

r34 12

M1

M2

40

For sway columns:k Lu

r22

where

Lu= unsupported length of column

r = radius of gyration of column section rI

A

=

Page 100


26/34

For rectangular column r h

12= 0.289 h=

I A, = moment of inertia and area of column section

M1 min Ma Mb,( )= M2 max MaMb,( )=

MaMb, = moments at the ends of column

k ka b,( )= is an effective length factor

EIc

EIb

= is a degree of end restraint

(degree of end release)

Page 101


27/34

For nonsway column

ab

4

k

2

a b+

21

k

tan

k

+

2 tan

2 k

k

+ 1=

For sway column

ab

k

2

36

6 a b+( )

k

tan

k

=

Approximate values of k

In nonsway frames:

k 0.7 0.05 A B+( )+ 1.0=

k 0.85 0.05 min+ 1.0=

min min A B,( )=

In sway frames:

Case m 2

mathcad - 13-column design

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