Column and Foundation Design

7/30/2019 Column and Foundation Design

1/12

ProjectConcrete Design

Jobref

Part of structure

Column Design

Sheet no ref

1 / 1Drawing ref.

see RefCalc by

M Omar, A SharriffDate

01/01/2011Check by Date

Ref Calculations Output

Exp 4.1

4.4.1.2

Column is subjected to an axial load of 373kN Where axial load =

(w/2 + L/2) x Ultimate load

( 12.85/2 + 4.5/2) x 43 = 373 kN.

A horizontal load due to wind actions on the overall structure is 2.381kN

The column is 300 mm square by 3300mm long measured from top of foundation to the centre of slab.

The column is subjected to a 1 hour fire resistance on three exposed sides. Assuming the base spring and

top is unbraced.

Cover

cnom = cmin +

cdev

where cmin = max[cmin,b, cmin,dur]

where

cmin,b = diameter of bar. Assume 40 mm main bars and 10 mm links

cmin,dur= minimum cover due to environmental conditions.

Assuming primarily XC3 / XC4, secondarily XF1, cmin,dur = 35 mm

cdev = allowance in design for deviation = 10 mm try cnom = 40 + 10

=50

= 50 mm to main bars

or = 35 + 10 = 45 mm to 10 mm links (45+10 = 55mm to main bars) Therefoere use Nominal cover,

cnom = 45 mm to 10 mm links

3300mm

Column

300 x 300

Slab

Foundation


2/12


Jobref

Part of structure

Column design

Sheet no ref

1 / 1Drawing ref.

see RefCalc by




Istruct E

T5.15

BS EN

1992-1-1

6.4.5

2.4.2.4.1

5.1

Fire resistance

Fire resistance R 60 and use conservative value offi = 0.7

Therefore Minimum dimensions = 300 and a= 40mm

check on cnom: 45mm + 40mm/2 = 65 mm > 40 mm

Effective depth d

d = 350mm - 65mm =285 mm ;d =285mm

Design yield strength of reinforcement

fywd,ef = 250 + 0,25 d fywd 250 + 0.25 x 285 = 321.25 N/mm2

fywd = k x fyk / s = 1.05 x 500 / 1.15 = 457 N/mm2 therfore fywd,ef =321.25 N/mm2

Design value of concrete compressive strength

fck = 30 N/mm2 fcd = cc fck / c = 1 x 30/1.5 = 20 N/mm

2

Imperfections of columns

Imperfections as represented by an inclination , given by

i= 0hm

where 0 is the basic value: 0 = 1/200 = 0.005

h is the reduction factor for length or height = 1m is the reduction factor for number of members = 1

Therefore i = 0.005 x 1 x 1 = 0.005

Therefore flexibility of rotation restraint

k = (i / M)(E / l) where k recommended = 0.1where:

Mz = Horizontal force x l = 2.381kN x 4.5m = 10.7kNm = 10.7x 106

n/mm

cm = fck+ 8 & Ecm = 22 (fcm/10)0.3

& E = Ecm =33000N/mm2

I = 300mm x 300mm3

/12 = 75 x106

mm4

k1 = (0.005 / 12 x 106) x (33000 x 75 x 10

6/4500) = 3.30 > 0.1

therefore k1 = 3.3 & k2 = (no restrains)

Io = l x max { (1+10 x 1)1/2 = 111/2 = 3.32 ; (1 + 3.3/4.3) x (2) = 3.5}

lo = 4.5 x 3.5 = 15.74 m = 15740mm

e i = eccentricity due to imperfections & e i = i l0/2 = 0.005 x 15.74 /2 = 0.04mm

ei min = h/30 but not less than 20 mm 300/30=10mm ei min = 20mm < 30mm

ok

321.25

N/mm2

20 N/mm2


3/12


Jobref

Part of structure

Column design

Sheet no ref

1 / 1Drawing ref.

see RefCalc by




5.8.3.2

5.8.8.2

First order moments

M02 = Mz + ei x Ned2.381kN x 4.5m + 0.04 x 373 = 25.63 kNm

Slenderness

Slenderness = lo/i

where i = ( I/A)

i = Radius of gyration

= 15740mm / (75x 106/(300x300)) =545

Slenderness lim

lim = 20 ABC / n0.5

here A = 1 / (1+ 0.2ef)

(if ef is not known A may be taken as 0.7)A = 0.7

B = (1 + 2)^0.5 = 1.1

C = 1.7 rm C = 1.7 0 = 1.7

where:

rm = M01 / M02 = 0 / M02 rm = 0

n = relative normal force = NEd / Acfcd

n = 373 x 103 /

(3002

x 20) = 0.15

Therefore lim =(20 0.7 1.1 1.7 / 0.150.5

) = 67.6 < 545

> lim Therefore column is slender about z axis.

Second order moment

M2 = nominal 2nd order moment = Ned x e2

e = 0.1x l2

x [ Kr x Kx ( fyd / (Es 0.45d))]

Kr = (nu n) / (nu nbal) 1.0

where : nu = 1 + = & 1 + 0.39 = 1.39n = NEd/Ac fcd n = 0.15 & nbal = 0.4

Kr= (1.39 0.15 ) / ( 1.39 0.5) = 1.41 > 1 Therefore Kr = 1.0

K= 1 + ef 1

where = 0.35 + (fck/200) (/150)

= 0.35 + (30/200) (545/150) =-3.13

K=1 +(-3.13) x 1.6 = -4.28 < 1.0 therefore use = K= 1

e2 =0.1 x (15.74 x 103)2

[1 x 1 x (321.25 / (205000 0.45 x 285))] = 303mm

M2 =303 x 10-3

x 373 =113 kNm

545

0.7

1.1

1.7

303mm


4/12


Jobref

Part of structure

Column design

Sheet no ref

1 / 1Drawing ref.

see RefCalc by




Design Moment

Med = M02 + M2 = 37.2+ 113=150.2 kNm

Ned = 373 kN

Therefore Calculating As required by using Column charts

d/h = 285 mm / 300 mm =0.95 use chart for 0.95

M/bh2

fck = 150.2 x 106

/ ( 300 x 3002

x 30) = 0.19N/bhfck = 373 x 10

3/ ( 300 x 300 x 30)m= 0.14

Therefore: chart value = 0.3

As = 0.3 x b x h x fck / fyk = 0.3 x 300 x 300 x 30 / 500 = 1620 mm2

Check minimum reinforcement!

Rules for detailing

Longitudinal reinforcement

As,min maximum of (0.1NEd/fyd ; 0.002Ac)

As,min = 0.1 x 373x 103

/ 321.25 = 116mm2

0.002 x 3002

= 180mm2

As,req = 1620mm2

As, prov = 1625mm2

As, max < 0.04 Ac < 0.04 x 3002

= 3600mm2

> 1625mm2

Transversal reinforcement

Smax = 20 x 30= 600mm

Smax = < 300mm

Smax = 400

Therefore Smax = 300mm


5/12


Jobref

Part of structure

Foundation designSheet no ref

1 / 1Drawing ref.

see RefCalc by




Concise

EC2

Design BasisThe limit states of stability strength and serviceability needs to be considered.

Design Action

Permanent load

Slab = 25 x 3.5 x 2.25 x 0.22 = 43.31 kN

Roof = 2.35 x 3.5 x 2.25 = 18.51 kN

Beam = 25 x 0.5 x 0.3 x ( 3.5 + 2.25) = 21.56 kN

Column = 25 x 0.3 x 0.3 x 2.38 = 5.36 kN

Sum of permanent load = 43.31 + 18.51 + 21.56 + 5..36 + 0.2 ( services) = 88.94 kN

Therefore Gk= 88.94 kN

Variable load

Imposed load = 0.6 kN/m2

x 3.5 x 2.25 = 4.725kN

Horizontal load ( wind action )

Assuming half of the wind directly hitting the wall acting on the foundation

Aref = (4.5 x 2.38 )/ 2 = 5.355

Fwe = 1 x (-0.323 x 5.4) = -1.74 KN

Fwi = 1 x ( -0.04 x 5.44) = -0.216 kN

Therefore the total force acting on foundation = -1.74(- 0.216) = - 1.52 kN

Axial load

ULS

NED = 1.35 Gk + 1.5 Qk = 1.35 x 88.94 + 1.5 x 4.725 = 131.63 kN

HORIZONTAL LOAD (ULS)

NED = 1.35 x 0 + 1.5 x -1.52 = -2.28 kN

Design Moment

From column calculation Design Moment (Med = M02 + M2 = 37.2+ 113=150.2 kNm)

Therefore determining ULS ( Geo) Assuming 50% permanent action & 50 % variable action

MED = 29.95 KNm

Member size of foundation

To satisfy eccentricity

e = Med/ Ned = 29.95 / 131.63 = 0.228

To achieve e d/6 try d = b = 1.36m

Try 1.5m x 1.5m x 0.8m deep e < d/6

Ground bearing pressure:

88.94 kN

4.725kN

- 1.52 kN

131.63 kN

-2.28 kN


6/12


Jobref

Part of structure

Wind load calculation

Sheet no ref

1 / 1Drawing ref.

see RefCalc by




Exp 4.1

Istruct E

Table B2

Annex C

Table C1

Size of foundation = 131.63 / 120 = 1.097 m2

( Area of foundation)Therefore the size of foundation is> 1.097 Thus a moment will add up to the ground pressureHence try 2m x 2 m x 0.8m

Checking ground bearing pressure

Nfound = 25 x 2 x 2 x 0.8 = 80 kN

Structural resistance = NED, found = 1.35 x 80 = 108kN

P = NED+ NED, found = 373 kN + 108 = 481 kN

2 = 481 / 2 x 2 29.95 x 6 /( 2 x 22) = 120.25 22.46

= 142.85 < 168 & = 97.65

Flexibility of foundation

d = 2 & d column = 0.3 , h = 0.8 ,Vmax = 0.8

Therefore V = 2/20.3/2 = 0.85> 0.8 Therefore foundation is flexible.

Cover

Cover

cnom = cmin +cdev

where cmin = max[cmin,b, cmin,dur]

where

cmin,b = diameter of bar. Assume 20 mm main bars

cmin,dur= minimum cover due to environmental conditions = 40mm

Assuming primarily XC3 / XC4, secondarily XF1, cmin,dur = 35 mm

cdev = allowance in design for deviation= 10 mm try cnom= 40 + 10 =50

= 50 mm to main bars

d = 800- 50mm = 75020/2 = 740mm

Yield Strength of reinforcement

Fywd, eff = 250 + 0.25 Fywd = 250 + 0.25 x 740 = 435 N/mm2

Fywd = k x fyk/ = 1.05 x 500/1.15 = 457 N/mm2

K=(fe/fy) k1.05 for steel grade A

Concrete compressive strength

Fck = 30 N/mm2

, fcd = = 1 x 30 / 1.5 = 20 N/mm2

Flexure design

Design moment at face of column

Mean GBP = (120.25 + 142.25)/ 2 = 131.25 kN/ m2

M = w x v2/2 = 131.25 x 0.85

2/2 = 47.41 knm

K = MED / bd2fck = 47.41x10

6/ ( 1000 x 740

2x 30)= 0.003

Limit z/d to 0.95

ok

OK

740mm

435 N/mm2

20 N/mm2

z/d = 0.97


7/12


Jobref

Part of structure


Sheet no ref

1 / 1Drawing ref.

see RefCalc by




Table 6.1

Z = 0.95 x 740 = 703mm

Klim > Kf ; Therefore NO compression reinforcement needed

Reinforcement required

As = M / ( 0.87z Fyk) = 47.71 x 106

/ 0.87 x 703 x 500 = 156 mm2

As, min = 0.26 fcm, btd/fyk

Fctm = 3.08 N/mm2

Bt = 1

Asmin = 0.26 x 3.08 x 1000 x 740/500 = 1185 > 156 & 0.0013 x 1000 x 740 = 962 > 156

Try H20 B : As, prov = 7 x 320 = 2240

Asreq/As,prov = 1185/2240 = 0.53 : Therefore aggregate needs to be limited to 0 20mm

Punching shear design

Ved =Ned = 373KN

A1 = 3002

= 0.09m2

Ved, red = Ved -

= A1 x ( GBPSelf weight of foundation)

=0.09 x ( 120.25( 25 x 0.8) = 9kN

Ved,red= 3739 = 364 kN

Shear stress at column face

Ved,max = Ved,red/ud

D = mean effective depth of slab (dy + dz)/2

Dy = hCnom0.5 = 800500.5 x 20 = 740

Dz = hCnom1.5 = 800501.5 x 20 = 720

D = ( 740 + 720 ) / 2 = 730mm

Vo = 300 x 4 = 1200

K = 0.7

W1 = c12/2 + c1c2 + 4c2d + 16 d

2+2dc1

W1 = 3002/2 + 300 x 300 + 4 x 300 x 730 + 16 x 730

2+ 2 x 730 x 300 = 10913419mm2

U2 = 2 x 730 = 1460

U1 = 1460 x 2 = 9174mm

= 1 + k Med/ved x u1/w1

156 mm2

Use

Asmin

364 kN


8/12


Jobref

Part of structure


Sheet no ref

1 / 1Drawing ref.

see RefCalc by




Bs EN 1992(6.50)

= 1 + 0.6 x 47.41/373 x 103 x 9173 / 10913418 = 1 N/mm2Ved,max = 1 x 373 x 10

3/ 1200 x 730 = 0.43 N/mm2

Shear resistance of strutVrd,max = 0.5v, fcdFcd = 30/1.5 = 20

V = 0.6 ( 1FCK/250) = 0.6 ( 130/250) = 0.530.5 x 0.53 x 20= 5.3 N/mm2

Vrd, max = 5.3 N/mm2 > Ved, max = 0.43 N/mm2

Maximum shear strength Vrd

Vrd = Crdc k ( 100, fck)1/3

Crdc = 0,12

K = 1 + 200/d = 1 + 200/730 = 1.52

Vrdc = 0.12 x 1.52 ( 100 x 0.003 x 30 )1/3

= 0.36N/mm2

As.bwd = 2240/ 1000 x740 = 0.003

Vrdc= Vmin x d/a

Vrdc = 0.035 fck x k3/2 x 2d/a

Vrdc = 0.035 x 303/2

x 1.523/2

x 2 = 0.36 N/mm2

Uout, eff = x Ved / ( Vrdc x d)

= 1.52 x 373 x 103/ 0.36 x 730 = 2157mm

Distance from column face = Uout eef4h / 2

= 2157mm4 x 800 / 2 = 0.36N/mm2

Shear stress at Uo2

Ved,red (Uo2) = 0.5d

A0.5 = ( 0.5d + dcol/2)

2

x = (730 x 0.5 + 300/2)2

x = 0.83Ved,red = Ved - Ved

Ved = 0.83 x ( 120.2525 x 0.8 ) = 83.2kN

Ved,red = 37383.2 = 289.6

Perimeter at 0.5 d = 0.5 x 730 = 365mm

Uo5 = 365 x 2 x + 300 x 4 = 3493.4mm

Ved at U2 = x Ved /( Uo5 x d )

1.52 x 113 x 103 / ( 3493 x 730 ) = 0.06 N/mm2 < 1.52N/mm2 = Ved

Therefore No shear reinforcement required

ok

0.36N/m

m2

2157mm

0.36N/mm2


9/12


Jobref

Part of structure


Sheet no ref

1 / 1Drawing ref.

see RefCalc by




Therefore N


10/12


Jobref

Part of structure


Sheet no ref

1 / 1Drawing ref.

see RefCalc by





11/12


Jobref

Part of structure


Sheet no ref

1 / 1Drawing ref.

see RefCalc by





12/12


Jobref

Part of structure


Sheet no ref

1 / 1Drawing ref.

see RefCalc by




Documents

Column and Foundation Design