Column design

Type of columns (by design method) 1. Axially loaded columns

e M

= P =0 2. Eccentric columns

e M

= P 0

2.1. Short columns (without buckling) Pu Mu

2.2. Long (slender) columns (with buckling) Pu Mu δns 

1. Axially Loaded Columns

Safety provision

Pu ϕPn.max

where Pu = axial load on column

ϕPn.max = design axial strength

For tied columns

ϕPn.max 0.80 ϕ⁼  ^



0.85 f'c ^



Ag Ast



 fy Ast



with ϕ=0.65

For spirally reinforced columns

ϕPn.max 0.85 ϕ⁼  ^



0.85 f'c ^



Ag Ast



 fy Ast



with ϕ=0.70

where Ag = area of gross section Ast = area of steel reinforcements

Ag Ast =Ac is an area of concrete section

For tied columns Diameter of tie

Dv 10mm= for D32mm Dv 12mm= for D32mm

Spacing of tie

s48Dv s16D sb

For spirally reinforced columns

Diameter of spiral Dv 10mm Clear spacing 25mms75mm

Column steel ratio

ρg Ast

= Ag =1%8%

Determination of Concrete Section

Ag

Pu 0.80 ϕ

0.85 f'c ^



1 ρg



fy ρg

=

Determination of Steel Area

Ast Pu

0.80 ϕ  0.85 f'c Ag

0.85f'cfy

=

Example 14.1

Tributary area B 4m L 6m

Thickness of slab t 120mm

Section of beam B1 b 250mm h 500mm Section of beam B2 b 200mm h 350mm Live load for lab LL 3.00kN

m²



Materials f'c 25MPa fy 390MPa

Solution

Loads on slab

Cover 50mm 22 kN m³



Slab 120mm 25 kN m³



Ceiling 0.40kN m²



Mechanical 0.20kN m²



Partition 1.00kN m²



DL Cover Slab CeilingMechanical Partition 5.7 kN m²







LL 3 kN m²





Reduction of live load

Tributary area AT B L 24 m² For interior column KLL 4

Influence area AI KLL AT 96 m²

Live load reduction factor αLL 0.25 4.572 AI m²

  0.717



Reduced live load LL0 LL αLL 2.15 kN m²







Loads of wall

Void 30mm 30 mm 190 mm 4

Brickhollow.10 120mm Void 55 1m²

1.648 kN m²







Brickhollow.20 220mm Void 110 1m²

2.895 kN m²







Loads on column

PD.slab DL B L  136.8 kN

Pu 1.2 PD  1.6 PL 2249.063 kN

Determination of column section

Assume ρg 0.03 k b

= h k 300

 500 ϕ 0.65

Ag

Pu 0.80 ϕ

0.85 f'c ^



1 ρg



fy ρg

 Ag 1338.529 cm  ²

h Ag

k 472.322 mm

 b k h  283.393 mm

h Ceil h 50mm(  )500 mm b Ceil b 50mm(  )300 mm b

h

 



 



300 500

 



 



^mm

 Ag b h  1500 cm ²

Determination of steel area

Ast

Pu

0.80 ϕ 0.85 f'c Ag

0.85f'c fy

 Ast 30.851 cm  ²

6 π 20mm( )²

 4 6 π 16mm( )²

 4

 30.913 cm ²

Stirrups

Main bars D 20mm Stirrup dia. Dv 10mm

Spacing of tie s^ min 16 D



 48 D  vb



^{ 300 mm}

2. Short Columns

Safety provision Pu ϕPn Mu ϕMn

Equilibrium in forces



X⁼⁰ Pn C Cs=   T

Pn 0.85 f'c=   ba A's f's  As fs

Equilibrium in moments



M⁼⁰ Mn Pn e=  C h

2 a

 2

  





Cs h

2 d'

  





 T d h

 2

  







=

Mn Pn e=  0.85 f'c  ba h 2

a

 2

  





A's f's h 2  d'

  





 As fs d h

 2

  







=

Conditions of strain compatibility εs

εu

d c

= c εs εu d c

 c

=

fs Es εs=  Es εu d c

 c

=

ε's εu

c d'

= c ε's εu c d'

 c

=

f's Es ε's=  Es εu c d'

 c

=

Unknowns = 5 : a As A's f'fs s

Equations = 4 :



X⁼⁰



^M=0 2 conditions of strain compatibility

Case of symmetrical columns: As A's= Case of unsymmetrical columns: fs fy=

A. Interaction Diagram for Column Strength

Interaction diagram is a graph of parametric function, where Abscissa : Mn a( )

Ordinate: Pn a( )

B. Determination of Steel Area

Given: Mu Pu b hf'cfy

Find: As A's=

Answer: As AsN a= ( )=AsM a( )

AsN a( ) Pu

ϕ  0.85 f'c  ba f's fs

=

AsM a( ) Mu

ϕ 0.85 f'c  ba h 2

a

 2

   





f's h 2  d'

   

 fs d h

 2

   





=

f's a( ) Es εu c d'

 c fy

=

fs a( ) Es εu d c

 c fy

=

Example 14.2

Construction of interaction diagram for column strength.

Concrete dimension b 500mm h 200mm

Steel reinforcements As 5 π 16mm( )²

Materials f'c 25MPa

fy 390MPa

Solution

Case of axially loaded column Ag b h

Ast As A's

ϕ 0.65

ϕPn.max 0.80 ϕ ^



0.85 f'c ^



Ag Ast



fy Ast



 1490.536 kN

Case of eccentric column

β1 0.65 max 0.85 0.05 f'c 27.6MPa

ϕPn a( )^min ϕ a



( ) 0.85 f'c^



  ba A's f's a ( ) As fs a ( )



ϕPn.max



ϕMn a( ) ϕ a( ) 0.85 f'c  ba h 2

a

 2

   

 A's f's a ( ) h 2  d'

   



 As fs a ( ) d h

 2

   







  





a 0 h

100h



0 20 40 60

0 250 500 750 1000 1250 1500

Interaction diagram for column strength

ϕPn a( ) kN

ϕMn a( ) kN m

Example 14.3

Determination of steel area.

Required strength Pu 1152.27kN

Mu 42.64kN m

Concrete dimension b 500mm h 200mm

Materials f'c 25MPa

fy 390MPa

Concrete cover to main bars cc 30mm6mm 16mm

 2



Solution

Location of steel re-bars d' cc 44 mm  d hcc 156 mm

Case of axially loaded column

Ag b h ϕ 0.65

Case of eccentric column

β1 0.65 max 0.85 0.05 f'c 27.6MPa

Graphical solution

AsN a( )

0.13418 0.1342 0.13422 0.13424 0.13426 8.715 10 ^4

8.72 10 ^4 8.725 10 ^4 8.73 10 ^4 8.735 10 ^4

AsN a( ) AsM a( )

a

a 134.23mm AsN a( ) 8.722 cm ² AsM a( )8.725 cm ²

As AsN a( ) AsM a( )

2  8.724 cm ²

 5 π 16mm( )²

 4  10.053 cm ²

Analytical solution

C. Case of Distributed Reinforcements

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b

Equilibrium in forces



X⁼⁰

Pn C

Equilibrium in moments



M⁼⁰ Mn Pn e=  C h

Conditions of strain compatibility εs i

εu

di c

= c

εs i εu di c

c

= 

fs i =Es ε _{s i} Es εu d ic

c

= 

Example 14.4

Checking for column strength.

Required strength Pu 13994.6kN

Mu 57.53kN m

Materials f'c 35MPa

fy 390MPa

Solution

Determination of Concrete Section

Case of axially loaded column ϕ 0.65

Assume ρg 0.04

Ag

Pu 0.80 ϕ

0.85 f'c ^



1 ρg



fy ρg  6094.36 cm ²



Aspect ratio of column section λ b

= h λ 1

h Ag

λ 780.664 mm

 b λ h  780.664 mm

h Ceil h 50mm(  ) b Ceil b 50mm(  ) b

h

 



 



800 800

 



 



^mm

 Ag b h  6400 cm ²

Steel area

Interaction Diagram for Column Strength Distribution of reinforcements

Bars Steel area

As0 π Bars ²

Location of reinforcement rows

Concrete cover Cover 30mm 10mm 40 mm

Case of axially loaded column

ϕPn.max^ 0.80 ϕ ^



0.85 f'c ^



Ag Ast



fy Ast



ϕPn.max 14255.808 kN 

Case of eccentric column

β1 0.65 max 0.85 0.05 f'c 27.6MPa

0 1000 2000 3000 0

5000 10000 ϕPn a( )

kN Pu kN

ϕMn a( ) kN m

Mu kN m



D. Design of Circular Columns

Symbols

ns = number of re-bars Dc = column diameter Ds = diameter of re-bar circle

Location of steel re-bar

di⁼rc rs cos α^{ }

 

_{s i} ^{rc= Dc}₂ ^{rs= Ds}₂

αs i

2 π ns (i 1)

=

Depth of compression concrete

Area and centroid of compression concrete Asector 1

2  Radius Arch

Atriangle 1

2  Base Height

Equilibrium in forces



X⁼⁰

Pn C

Equilibrium in moments



M⁼⁰

Mn Pn e=  C xc

Conditions of strain compatibility εs i

Example 14.5

Required strength Pu 3437.31kN Mu 42.53kN m

Materials f'c 20MPa

fy 390MPa

Solution

Determination of concrete dimension

ϕ 0.70

Assume ρg 0.02

Ag

Pu 0.85 ϕ

0.85 f'c ^



1 ρg



fy ρg  2361.812 cm ²



Dc Ceil Ag π 4

50mm

 

 

 

 

550 mm





Ag π Dc ²

4 2375.829 cm ²



Determination of steel area

Ast

Pu

0.85 ϕ 0.85 f'c Ag

0.85f'c fy  46.597 cm ²



Ds Dc 30mm10mm 20mm

 2

   

^2

 450 mm



ns ceil π Ds 100mm

 



 



^{ 15}

 As0 π 20mm( )²

4  3.142 cm ²



Ast ns As0  47.124 cm ² s π Ds

ns 94.248 mm



Interaction diagram for column strength

ϕPn.max 0.85 ϕ ^



0.85 f'c ^



Ag Ast



fy Ast



 3448.996 kN

β1 0.65 max 0.85 0.05 f'c 27.6MPa

0 100 200 300 0

1000 2000 3000

Interaction diagram for column strength

ϕPn a( ) kN Pu kN

ϕMn a( ) kN m

Mu kN m



3. Long (Slender) Columns

Stability index

Q ΣPu Δ0 Vu Lc

=

where

ΣPu Vu = total vertical force and story shear Δ0 = relative deflection between column ends Lc = center-to-center length of column

Q0.05 : Frame is nonsway (braced) Q0.05 : Frame is sway (unbraced)

Unbraced Frame Braced Frame

Shear Wall Braced Frame

Brick Wall

Ties

Slenderness of column

The column is short, if

In nonsway frame: k Lu

r min 34 12 M1 M2



 40

 



 





In sway frame: k Lu

r 22 where

M1 min MA MB⁼







M2 max MA MB⁼







= minimum and maximum moments at the ends of column

Lu = unsuppported length of column r = radius of gyration

r I

= A

I A = moment of inertia and area of column section k = effective length factor

k ⁼k ψA ψB







ψA ψB = degree of end restraint (release)

ψ

EIc Lc

 



 

 

EIb Lb

 



 

 

=

ψ=0 : column is fixed ψ=∞ : column is pinned

Moments of inertia

For column Ic 0.70Ig= For beam Ib 0.35Ig=

Ig = moment of inertia of gross section

Determination of effective length factor Way 1. Using graph

Way 2. Using equations For braced frames:

ψA ψB 4

π



k

  

2

 ψA ψB

2 1

π k tan π



k

 





 

 



 

 







2 tan π 2 k

  





π k

 =1

For unbraced frames:

ψA ψB π



k

  

2

 36

6 ψA ψB^







π k tan π



k

  

=

Way 3. Using approximate relations

In nonsway frames:

k ⁼0.7^0.05 ψA ψB^







^1.0

k =0.85 0.05 ψmin 1.0 ψmin min ψA ψB⁼







In sway frames:

Case ψm 2

k

20ψm

20  1 ψm

=

Case ψm 2

k =0.9 1 ψm

ψm ψA ψB

= 2

Case of column is hinged at one end k =2.00.3 ψ

ψ is the value in the restrained end.

Moment on column

Mc M2 δns=  M2.min δns

where

M2.min Pu 15mm 0.03h= (  )

Moment magnification factor

δns Cm

Euler's critical load Pc π²EI

Example 14.6

Required strength Pu 6402.35kN PD

PL

Length of column Lc 7.8m

Upper and lower columns

ba

Upper and lower beams

ba1

Materials f'c 30MPa

fy 390MPa

Solution

Determination of concrete dimension

ϕ 0.65

Proportion of column section k b

= h k 60

Determination of steel area Ag b h  3.6 10³cm² Cover 40mm 10mm 25mm

 2 62.5 mm



d11 Cover Δs h Cover 2

Slenderness of column

Stability index Q 0 Radius of gyration r h

12

0.173 m





Modulus of elasticity wc 24kN m³



Ec 44MPa wc

kN

Degree of end restraint

Ia1 0.35 ba1 ha1 ³

Effective length factor

Checking for long column

M1 MA if MA  MB

MB otherwise

 M2 MB if MA  MB

MA otherwise



The_column "is short" k Lu

r min 34 12 M1

"is long" otherwise



The_column "is long"

Case of long column

βd 1.2 PD

δns max Cm

Interaction diagram for column strength c a( ) a

0 200 400 600 800 1000 1200 0

1000 2000 3000 4000 5000 6000 7000

Interaction diagram for column strength

ϕPn a( ) kN Pu kN

ϕMn a( ) kN m

Mc kN m



15. Footing Design

A. Determination of Footing Dimension

Required area of footing

Areq PD PL

= qe where

PD PL = dead and live loads on footing qe = effective bearing capacity of soil

qe qa 20kN m³

H

= 

qa = allowable bearing capacity of soil with FS=2.53

20kN m³

= average density of soil and concrete

H = depth of foundation

Checking for maximum stress of soil under footing qmax qu

qmax P

B L 1 6 e L





 





e L

6

 if 4P

3 B (L2 e ) e L

 6 if

=

where

qu = design bearing capacity of soil

qu qa 1.2PD 1.6 PL  PD PL

= 

P = axial load on footing

P=1.2 PD P0^







 1.6 PL P0 20kN

m³

 BH L

=

e = eccentricity of load

e M

= P

L B = long and width of footing

B. Determination of Depth of Footing

Checking for Punching

Vu ϕVc where

Vu = punching shear

Vc = punching shear strength

ϕ=0.75 is a strength reduction factor for shear

Punching shear

Vu qu A A0^{= }







A=B L

A0 ⁼



bc d



^



hc d



Punching shear strength

Vc 4 f'c=  b0d (in psi)

Vc 0.332 f'c=  b0d (in MPa)

b0⁼

_ 

bc d



^



hc d

 _

^2

Checking for Beam Shear

Vu1 ϕVc1 Vu2 ϕVc2 where

Vu1 Vu2 = beam shears

Vc1 Vc2 = beam shear strength

Beam shears

Vu1 qu B L 2

hc

 2 d

 



 





=

Vu2 qu L B 2

bc

 2 d

 



 





=

Beam shear strength Vc1 0.166 f'c=   dB

Vc2 0.166 f'c=   dL

C. Determination of Steel Area

Steel re-bars in long direction

Required strength

q1 qu B=  L1 L

2 hc

 2

=

Mu1 q1 L1 ²

= 2

Design section: rectangular singly reinforced beam of Bd

Steel re-bars in short direction

Required strength

q2 qu L=  L2 B

2 bc

 2

=

Mu2 q2 L2 ²

= 2

Design section: rectangular singly reinforced beam of Ld

Example 15.1

Required strength PD 484.71kN

PL 228.56kN PL

PD PL  0.32 Mu 5.03kN m

Dimension of column stub bc 350mm hc 350mm

Depth of foundation H 2.0m

Allowable bearing capacity of soil qa 178.33kN m²

3

2.5 213.996 kN m²







Materials f'c 25MPa

fy 390MPa

Solution

Determination of Dimension of Footing

Effective bearing capacity of soil

qe qa 20kN

Required area of footing

Areq PD PL

qe 4.099 m²



Footing proportion k B

= L k 2

Design bearing capacity of soil

qu qa 1.2 PD 1.6 PL

Checking for maximum stress of soil Pu 1.2 PD B L H 20 kN

qmax Pu

B L 1 6 e

 qmax 276.981 kN

m²





qmax

qu 0.975 Soil "is safe" if qmax qu

"is not safe" otherwise

 Soil "is safe"

Determination of depth of footing

Punching shear

A0 d( )^



bc d



^



hc d



^{A B L}

Vu d( )^ qu A A0 d^



 ( )



Vu 320mm( ) 1066.154 kN

Punching shear strength

b0 d( )^

_ 

bc d



^



hc d

 _

^2

ϕ 0.75

ϕVc d( ) ϕ 0.332 MPa f'c MPa

 b0 d( )d

 ϕVc 320mm( )1067.712 kN

Beam shears

Vu1 d( ) qu B L 2

hc 2

  d

 



 





 Vu1 300mm( )326.858 kN

Vu2 d( ) qu L B 2

bc 2

  d

 



 





 Vu2 300mm( )313.357 kN

Beam shear strength

ϕVc1 d( ) ϕ 0.166 MPa f'c MPa

  dB

 ϕVc1 300mm( )373.5 kN

ϕVc2 d( ) ϕ 0.166 MPa f'c MPa

  dL

 ϕVc2 300mm( )392.175 kN

c 50mm 20mm 20mm 2

 80 mm



dmin 150mm c 70 mm

d ddmin

dd 50mm Vu d( )ϕVc d( )

 

^



Vu1 d( )ϕVc1 d( )



^



Vu2 d( )ϕVc2 d( )



while

d



d 320 mm h dc 400 mm

Steel reinforcements

ρshrinkage (return 0.0020) if fy 50ksi 0.0018

return

( ) if fy 60ksi max 0.0018 60ksi

 fy 0.0014

 



 

return



 



 



^otherwise



ρshrinkage 0.0018

Re-bars in long direction

b B Ln L

Re-bars in short direction

b L Ln B

16. Design of Pile Caps

1. Determination of Pile Cap

Number of required piles

n PD PL

= Qe where

PD PL = dead and live loads on pile cap Qe = effective bearing capacity of pile

Qe Qa 20kN m³

3 D ( )²

 H

= 

20kN m³

= average density of soil and concrete

D = pile size

H = depth of foundation

Distance between piles = 2 D 4 D

Distance from face of pile to face of pile cap = D

2 200mm

Checking for pile reaction

Ri P n

Myx

 i

1 n

k xk

 

²





 M

xy

 i

1 n

k yk

 

²





 Qu

=

where

P = load on pile cap

Mx My = moments on pile cap

Qu = design bearing capacity of pile

Qu Qa 1.2 PD 1.6 PL PD PL

= 

2. Depth of Pile Cap

Case of punching

Vu ϕ Vc 

where

Vu = punching shear

Vu ⁼



Routside^=Qu



^noutside

Vc = punching shear strength

Vc 0.332 f'c=  b0d

b0⁼

_ 

bc d



^



hc d

 _

^2

Case of beam shear

Vu1 ϕVc1 Vu2 ϕVc2

where

Vu1 Vu2 = beam shears

Vc1 Vc2 = beam shear strength

Vu1 max⁼

^ _ 

Rleft^



^Rright

^ _

Vu2 max⁼

^ _ 

Rbottom^



^Rtop

^ _

Vc1 0.166 f'c=   dB

Vc2 0.166 f'c=   dL

3. Determination of Steel Reinforcements

In long direction

Required moment: Mu1 max



Rleft



^xleft ^hc2

 

 



 

Rright



^xright ^hc2

 

 









 

 

=



Design section: Rectangular singly reinforced of B d

In short direction

Required moment: Mu2 max



Rbottom



^xbottom ^bc2

 

 



 

Rtop



^xtop ^bc2

 

 









 

 

=



Design section: Rectangular singly reinforced of Ld

Example 16.1

Pile size D 300mm Lp 9m

Allowable bearing capacity of pile Qa 351.5kN

Loads on pile cap PD 1769.88kN PL 417.11kN

My 33.92kN m Mx 56.82kN m

Depth of foundation H 1.5m

Column stub bc 350mm hc 500mm

Materials f'c 25MPa

fy 390MPa

Diameters of main bar D1

D2

  



16mm 16mm

 



 





Concrete cover c 75mm

Depth of concrete crack hshrinkage 200mm Diameter of shrinkage rebar Dshrinkage 12mm

Solution

Design of pile

Required strength of pile concrete

Ag D D

f'c.pile Qa 1 4Ag

15.622 MPa



 Use f'c.pile 20MPa

Steel re-bars

Ast 0.005 Ag 4.5 cm ² 4 π 16mm( )²

 4  8.042 cm ²

Dimension of pile cap

Effective bearing capacity of pile Qe Qa 20kN

Number of piles

n PD PL

Qe  6.684



Required number of piles ceil n( ) 7

Location of pile

X

Dimension of pile cap

B max Y( ) min Y( ) min D

Checking for pile reactions

Qu Qa 1.2 PD  1.6 PL

Determination of Depth of Pile Cap Punching shear

Outside d( ) X hc

Punching shear strength ϕ 0.75

b0 d( )^

_ 

hc d



^



bc d

 _

^2

ϕVc d( ) ϕ 0.332 MPa f'c

 MPab0 d( )d

 ϕVc 700mm( )3921.75 kN

Beam shears

Left d( ) X hc

Beam shear strength

ϕVc1 d( ) ϕ 0.166 MPa f'c

Depth of pile cap

Cover cD1 D2

Steel Reinforcements

ρshrinkage (return 0.0020) if fy 50ksi 0.0018

return

( ) if fy 60ksi max 0.0018 60ksi

 fy 0.0014

 



 

return



 



 



^otherwise



In long direction b B

s1 Floor

b c D1

In short direction b L

R Mn

s2 Floor

b c D2

Shrinkage reinforcement

b 1m hshrinkage h if hshrinkage 0= hshrinkage otherwise



As ρshrinkage b h shrinkage 3.6 cm ²

As0 π Dshrinkage ²

 4 n As

 As0

sshrinkage Floor b n5mm

Dimension of pile cap B= 2.20 m

L= 2.60 m

Depth of pile cap h= 750 mm

Direction Length (mm) Dia. (mm) NOS Spacing (mm)

Long 2.43 16 15 145

Short 2.03 16 18 140

Top N/A 12 N/A 310

17. Slab Design

A. Design of One-Way Slabs

La = length of short side Lb = length of long side

La

Lb 0.5 : the slab in one-way La

Lb 0.5 : the slab is two-way

Thickness of one-way slab

Simply supported Ln

20

One end continuous Ln

24 Both ends continuous Ln 28

Cantilever Ln

10

Analysis of one-way slab

Design scheme: continuous beam

Determination of bending moments: using ACI moment coefficients

Design of one-way slab

Design section: rectangular section of 1m x h Type section: singly reinforced beam

Example 17.1

Span of slab Ln 2m 20cm1.8 m

Live load LL 12kN

m²



Materials f'c 20MPa fy 390MPa

Solution

Thickness of one-way slab

tmin Ln

28  64.286 mm



Use t 100mm

Loads on slab

Cover 50mm 22 kN m³

1.1 kN m²







Slab t 25 kN m³

2.5 kN m²







Ceiling 0.40kN m²



Mechanical 0.20kN m²



Partition 1.00kN m²



DL Cover Slab CeilingMechanical Partition 5.2 kN m²







wu 1.2 DL 1.6 LL 25.44 kN m²







Bending moments Msupport 1

11wuLn² 7.493 kN m

 1m





Mmidspan 1

16wuLn² 5.152 kN m

 1m





Steel reinforcements

β1 0.65 max 0.85 0.05 f'c 27.6MPa

ρmin max

0.249MPa f'c

 MPa

ρshrinkage (return 0.0020) if fy 50ksi 0.0018

return

( ) if fy 60ksi max 0.0018 60ksi

 fy 0.0014

 



 

return



 



 



^otherwise



ρshrinkage 0.0018

Top rebars

b 1m d t 20mm 10mm

Bottom rebars

Mu Mmidspan b 5.152 kN m 

Mn Mu

0.9  5.724 kN m 



R Mn b d ²

1.018 MPa





ρ 0.85 f'c

fy 1 1 2 R

0.85 f'c









 

 





 0.003

 ρρmax 1

As^ max ρ b



 d ρ shrinkage b t



^{2.019 cm 2}

As0 π 10mm( )²

 4 n As

 As0 smax min 3 t(  450mm ) s min Floor b

n10mm

   

^smax

   

^{300 mm}



Link rebars

As ρshrinkage b t1.8 cm ²

As0 π 10mm( )²

 4 n As

 As0 smax min 5 t(  450mm ) s min Floor b

n10mm

   

^smax

   

^{430 mm}



B. Design of Two-Way Slabs

Design methods:

- Load distribution method - Moment coefficient method - Direct design method (DDM) - Equivalent frame method - Strip method

- Yield line method

(1) Load Distribution Method

Principle: Equality of deflection in short and long directions

fa fb=

αa wa La ⁴

 EI αb wb Lb ⁴

 EI

=

Case αa=αb wa wb

Lb⁴ La⁴

= 1

λ⁴

= λ La

= Lb

wa wb =wu

From which, wa wu 1 1 λ⁴



=

wb wu λ⁴ 1 λ⁴



=

For λ 1 1

1λ⁴

0.5 λ⁴

1λ⁴

0.5

For λ 0.8 1

1λ⁴

0.709

 λ⁴

1λ⁴

0.291



For λ 0.6 1

1λ⁴

0.885

 λ⁴

1λ⁴

0.115



For λ 0.5 1

1λ⁴

0.941

 λ⁴

1λ⁴

0.059



For λ 0.4 1

1λ⁴

0.975

 λ⁴

1λ⁴

0.025



Example 17.2

Slab dimension La 4.3m Lb 5.5m

Live load LL 2.00kN

m²



Materials f'c 20MPa fy 390MPa

Solution

Thickness of two-way slab

Perimeter^



La Lb



^2

tmin Perimeter

180 108.889 mm

Loads on slab

SDL 50mm 22 kN

Load distribution

λ La

Bending moments

Ma.neg 1

11waLa² 11.992 kN m

 1m





Ma.pos 1

16waLa² 8.245 kN m

 1m





Mb.neg 1

11wbLb² 7.33 kN m

 1m





Mb.pos 1

16wbLb² 5.04 kN m

 1m





Steel reinforcements

β1 0.65 max 0.85 0.05 f'c 27.6MPa

ρmin max

0.249MPa f'c

 MPa

ρshrinkage (return 0.0020) if fy 50ksi 0.0018

return

( ) if fy 60ksi max 0.0018 60ksi

 fy 0.0014

 



 

return



 



 



^otherwise



ρshrinkage 0.0018

Top rebar in short direction

b 1m d t 20mm10mm 10mm

ρ 0.85 f'c

Bottom rebar in short direction

Mu Ma.pos b 8.245 kN m 

Top rebar in long direction

Mu Mb.neg b 7.33 kN m 

As0 π 10mm( )²

 4 n As

 As0 smax min 2 t(  450mm ) s min Floor b

n10mm

   

^smax

   

^{240 mm}



Bottom rebar in long direction

Mu Mb.pos b 5.04 kN m 

Mn Mu

0.9  5.599 kN m 



R Mn

b d ²

0.775 MPa





ρ 0.85 f'c

fy 1 1 2 R

0.85 f'c









 

 





 0.002

 ρρmax 1

As^ max ρ b



 d ρ shrinkage b t



^{2.16 cm 2}

As0 π 10mm( )²

 4 n As

 As0 smax min 2 t(  450mm ) s min Floor b

n10mm

   

^smax

   

^{240 mm}



Shrinkage rebars

b 1m

As ρshrinkage b t 2.16 cm ²

As0 π 10mm( )²

 4 n As

 As0 smax min 5 t(  450mm ) s min Floor b

n10mm

   

^smax

   

^{360 mm}



(2) Moment Coefficient Method

Negative moments

Ma.neg Ca.neg wu=  La²

Mb.neg Cb.neg wu=  Lb²

Positive moments

Ma.pos Ca.pos.DL wD=  La²Ca.pos.LL wL La²

Mb.pos Cb.pos.DL wD=  Lb²Cb.pos.LL wL Lb²

where Ca.neg Cb.neg Ca.pos.DLCa.pos.LLCb.pos.DLCb.pos.LL are tabulated moment coefficients

wD 1.2 DL=  wL 1.6 LL=  wu 1.2 1.6 LL=  

Example 17.3

Slab dimension La 5.0m 25cm4.75 m Lb 5.5m 20cm5.3 m

Live load for office LL 2.40kN m²



Materials f'c 20MPa fy 390MPa

Boundary conditions in short and long directions Simple

Continuous

 



 



0 1

 



 





Short Continuous

Continuous

 



 





Long Continuous Continuous

 



 





Solution

Thickness of two-way slab Perimeter^



La Lb



^2

tmin Perimeter

180 111.667 mm



t 1

30 1



50

  

^{La (158.333 95) mm}



Use t 120mm

Loads on slab

Cover 50mm 22 kN m³

1.1 kN m²







Slab t 25 kN m³

3 kN m²







Ceiling 0.40kN m²



Partition 1.00kN m²



SDL Cover CeilingPartition 2.5 kN m²







DL SDL Slab 5.5 kN m²





 wD 1.2 DL

LL 2.4 kN m²



 wL 1.6 LL

wu 1.2 DL 1.6LL 10.44 kN m²







Moment coefficients

Table 12.3a

Coefficients for negative moments in short direction of slab

m Case 1 Case 2 Case 3 Case 4 Case 5 Case 6 Case 7 Case 8 Case 9 1.00 0.000 0.045 0.000 0.050 0.075 0.071 0.000 0.033 0.061 0.95 0.000 0.050 0.000 0.055 0.079 0.075 0.000 0.038 0.065 0.90 0.000 0.055 0.000 0.060 0.080 0.079 0.000 0.043 0.068 0.85 0.000 0.060 0.000 0.066 0.082 0.083 0.000 0.049 0.072 0.80 0.000 0.065 0.000 0.071 0.083 0.086 0.000 0.055 0.075 0.75 0.000 0.069 0.000 0.076 0.085 0.088 0.000 0.061 0.078 0.70 0.000 0.074 0.000 0.081 0.086 0.091 0.000 0.068 0.081 0.65 0.000 0.077 0.000 0.085 0.087 0.093 0.000 0.074 0.083 0.60 0.000 0.081 0.000 0.089 0.088 0.095 0.000 0.080 0.085 0.55 0.000 0.084 0.000 0.092 0.089 0.096 0.000 0.085 0.086 0.50 0.000 0.086 0.000 0.094 0.090 0.097 0.000 0.089 0.088 Table 12.3b

Coefficients for negative moments in long direction of slab

m Case 1 Case 2 Case 3 Case 4 Case 5 Case 6 Case 7 Case 8 Case 9 1.00 0.000 0.045 0.076 0.050 0.000 0.000 0.071 0.061 0.033 0.95 0.000 0.041 0.072 0.045 0.000 0.000 0.067 0.056 0.029 0.90 0.000 0.037 0.070 0.040 0.000 0.000 0.062 0.052 0.025 0.85 0.000 0.031 0.065 0.034 0.000 0.000 0.057 0.046 0.021 0.80 0.000 0.027 0.061 0.029 0.000 0.000 0.051 0.041 0.017 0.75 0.000 0.022 0.056 0.024 0.000 0.000 0.044 0.036 0.014 0.70 0.000 0.017 0.050 0.019 0.000 0.000 0.038 0.029 0.011 0.65 0.000 0.014 0.043 0.015 0.000 0.000 0.031 0.024 0.008 0.60 0.000 0.010 0.035 0.011 0.000 0.000 0.024 0.018 0.006 0.55 0.000 0.007 0.028 0.008 0.000 0.000 0.019 0.014 0.005 0.50 0.000 0.006 0.022 0.006 0.000 0.000 0.014 0.010 0.003

ORIGIN 1

Index 1 2

J Index

Long₁ Long1 ₂1

 Case Table

I J 2

vs2 pspline Vλ Vbneg(  ) Cb.neg interp vs2 Vλ(  Vbnegλ)0.037 vs3 pspline Vλ VaposDL(  ) Ca.pos.DL interp vs3 Vλ(  VaposDLλ) 0.022 vs4 pspline Vλ VbposDL(  ) Cb.pos.DL interp vs4 Vλ(  VbposDLλ) 0.014 vs5 pspline Vλ VaposLL(  ) Ca.pos.LL interp vs5 Vλ(  VaposLLλ)0.034 vs6 pspline Vλ VbposLL(  ) Cb.pos.LL interp vs6 Vλ(  VbposLLλ)0.022

Bending moments

Ma.neg Ca.neg wu La² 13.043 kN m

Steel reinforcements

β1 0.65 max 0.85 0.05 f'c 27.6MPa

ρmin max

0.249MPa f'c

 MPa

ρshrinkage (return 0.0020) if fy 50ksi 0.0018

return

( ) if fy 60ksi max 0.0018 60ksi

 fy 0.0014

 



 

return



 



 



^otherwise



ρshrinkage 0.0018

Top rebars in short direction

b 1m d t 20mm10mm 10mm

 2

   

  85 mm



Mu Ma.neg b 13.043 kN m 

Bottom rebars in short direction

Mu Ma.pos b 6.266 kN m 

Top rebars in long direction

Mu Mb.neg b  10.729 kN m 

Mn Mu

0.9  11.921 kN m 



R Mn

Bottom rebars in long direction

Mu Mb.pos b 4.911 kN m 

Shrinkage rebars b 1m

(3) Direct Design Method (DDM)

Total static moment

M0 wu L2 Ln²

= 8

Longitudinal distribution of moments Mneg Cneg M0= 

Mpos Cpos M0= 

Lateral distribution of moments

Mneg.col Cneg.col Mneg=  Mneg.mid Cneg.mid Mneg=  Mpos.col Cpos.col Mpos=  Mpos.mid Cpos.mid Mpos= 

Example 17.4

Slab dimension La 4m Lb6m Live load for hospital LL 3.00kN

m²



Materials f'c 25MPa fy 390MPa

Solution

Section of beam in long direction

L Lb 6 m

Section of beam in short direction

L La 4 m

Determination of slab thickness

Perimeter^



La Lb



^2

tmin Perimeter

180 111.111 mm



Assume t 120mm

In long direction

bw bb h hb hf t

Ib I1 I2  4.61710⁵cm⁴

Is La hf ³

 12 wc 24kN

m³

 Ec 44MPa wc

kN m³

 



 



1.5

 f'c

 MPa  2.58710⁴MPa



α Ec Ib

Ec Is  8.016

 αb α

In short direction

bw ba h ha hf t

hw h hf

b^ min bw 2 hw



  bw 8 hf 



^{0.56 m}

A1 bw h x1 h

 2 A2



b bw



hf x2 hf

 2 xc x1 A1  x2 A2

A1 A2 112.326 mm



I1 bw h ³

12 ^ A1 x1 xc^







²



I2

b bw

 

hf³

12 ^ A2 x2 xc^







²



Ib I1 I2  7.05310⁴cm⁴ Iba Ib

Is Lb hf ³

12  8.64 10⁴cm⁴



α Ec Ib

Ec Is  0.816

 αa α

Required thickness of slab

αm αa 2 αb 2

4  4.416



β Lb

La 1.5



LnLb 20cm 5.8 m

hf max

"DDM is not applied" otherwise



hf  126.876 mm

Loads on slab

DL 50mm 22 kN

In long direction

L1 Lb 6 m Ln L1 ba 5.8 m L2 La 4 m α1 αb 8.016 Total static moment

M0 wu L2 Ln²

8 191.748 kN m 



Longitudinal distribution of moments Mneg 0.65 M0 124.636 kN m  Mpos 0.35 M0 67.112 kN m 

Lateral distribution of moments

k1 L2

Cneg.col linterp2

Cpos.col linterp2 0

Ccol.beam linterp 0

Mneg.col.beam Ccol.beam Mneg.col  90.05 kN m  Mneg.col.slab Ccol.slab Mneg.col  15.891 kN m 

Mpos.col.beam Ccol.beam Mpos.col 48.488 kN m  Mpos.col.slab Ccol.slab Mpos.col  8.557 kN m 

bcol min L1 L2







4 2 2 m



bmid L2 bcol  2 m

Top rebars in column strip

b bcol d t 20mm10mm 10mm

R Mn

Bottom rebars in column strip

Mu Mpos.col.slab 8.557 kN m 

Top rebars in middle strip b bmid

As^ max ρ b



 d ρ shrinkage b t



^{6.494 cm 2}

As0 π 10mm( )²

 4 n As

 As0 smax min 2 t(  450mm ) s min Floor b

n10mm

  



^smax

  



^{240 mm}



Bottom rebars in middle strip

Mu Mpos.mid 10.067 kN m  

Mn Mu

0.9  11.185 kN m 



R Mn

b d ²

0.774 MPa





ρ 0.85 f'c

fy 1 1 2 R

0.85 f'c









 

 





 0.002

 ρρmax 1

As^ max ρ b



 d ρ shrinkage b t



^{4.32 cm 2}

As0 π 10mm( )²

 4 n As

 As0 smax min 2 t(  450mm ) s min Floor b

n10mm

  



^smax

  



^{240 mm}



In short direction

L1 La 4 m Ln L1 bb 3.75 m L2 Lb 6 m α1 αa 0.816

Total static moment

M0 wu L2 Ln²

8 120.234 kN m 



Longitudinal distribution of moments Mneg 0.65 M0 78.152 kN m  Mpos 0.35 M0 42.082 kN m 

Lateral distribution of moments

k1 L2 L1  1.5

 k2 α1 L2

L1 1.224



Cneg.col linterp2 0

Cpos.col linterp2 0

Ccol.beam linterp 0

Mneg.col.beam Ccol.beam Mneg.col  39.858 kN m  Mneg.col.slab Ccol.slab Mneg.col  7.034 kN m 

Mpos.col.beam Ccol.beam Mpos.col 21.462 kN m  Mpos.col.slab Ccol.slab Mpos.col  3.787 kN m 

bcol min L1 L2







4 2 2 m



bmid L2 bcol  4 m

Top rebars in column strip b bcol

Mu Mneg.col.slab 7.034 kN m  

Mn Mu

0.9  7.815 kN m 

0.9  7.815 kN m 

In document Mathcad in Concrete Structures ACI 318-05 6th Edition (Page 69-163)