Design Pad & Chimney Foundation1

May 26, 2014

Date: By:

MAK

DESIGN OF PAD & CHIMNEY FOUNDATION Tower Type "A10" (Towers Include: # PI - 21)

1. INTRODUCTION: * The purpose of the foundation is to transfer the loads from the structure to the ground without causing the ground to fail in shear or to allow excessive settlement of the structure to occur. These requirements are met by ensuring the bearing pressure below the foundation does not exceed permissible bearing pressure. * The purpose of this calculation is to design Pad & Chimney foundation for "A10" type lattice steel tower structure. * The foundation is subjected to upward, downward and horizontal forces. * Foundation loads will be obtained from the tower reactions multiplied by over load capacity factors. These factored loads will be used in the sizing of the foundation. However, to design steel reinforcements reactions without the OCFs will be used.

2. TOWER GEOMETRY: h= a= b=

26100 mm 5000 mm 15962 mm

Tower Body Hieght Tower Dimension at the Top Tower Dimension at the Bottom

b  a    10.330 deg  2h 

  tan 1  h0 

1000

Leg Slope angel with Respect to Vertical

mm

b  a    tan 1   11.860 deg  2h 

h22

h11 

h0  1021.812 mm sin(90   )

Tower First Leg Slope

h22 

h0  1016.476 mm sin(90   )

Tower Second Leg Slope

h11

h0

3. LOADING IN THE VERTICAL AND HORIZONTAL DIRECTION: * The maximum vertical and horizontal reactions have been given in the PTS (clause: 4.09 A) for this kind of tower as follow: Pc0 1600.00

Pu0 1192.00

Ph-x0 242.00

Ph-y0 212.00

Note : Above Loads excludes over load capacity factors. Pc0 = Compression Force Pu0 = Uplift Force Ph-x0 = Resultant Shear Force in X-Direction Ph-y0 = Resultant Shear Force in Y-Direction

\\vboxsrv\conversion_tmp\scratch_3\230673500.xls.ms_office

1/25

May 26, 2014

Date: By:

MAK

4. FACTORED LOADS: In Compression In Uplift In Shear

SFc = SFu = SFs =

1.5 1.5 1.5

As per SES-P-122.06

Pc

= Pc0 × SFc =

2400.00 KN

Compression Force

Pu

= Pu0 × SFu =

1788.00 KN

Uplift Force

Ph-x =

Ph-x0 × SFs =

363.00 KN

Resultant Shear Force in X-Direction

Ph-y =

Ph-y0 × SFs =

318.00 KN

Resultant Shear Force in Y-Direction

Q0  ( Ph  x0 ) 2  ( Ph  y0 ) 2  321.73 KN Q  Q0  SFs 

482.59

KN

Horizontal Reaction

M0  0  M  0

Moment Reaction

5. PROPOSED SIZE OF FOUNDATION: Assumed Footing Width

B

=

4

m

Assumed Footing Length

L

=

4

m

Thickness of Pad

Fd

= 600 mm

Assumed Footing Depth

h

=

Pu

Eh

GL

Ph-y

a a

Ph-x Chimney Pad

3.5 m h

Exposed Height, including the 300 mm structure pad Assumed Pedestal Size

Eh

Fd

a

Cncrete Cover

cov

Thickness to be Ignored in Calculating Uplift Resistance

OL

= 500 mm

T

= 600 mm =

B L

85 mm

= 600 mm

As per PTS (05WO307 clause 4.09 item 3)

6. MATERIAL PROPERTIES: 6.1 CONCRETE:

c

=

24

KN/m³

Concrete Unit Weight

f c =

28

MPa

fy

=

420

MPa

E

=

Concrete Compressive Strength PTS (clause: 4.09 A-1) Steel Yield Strength PTS (clause: 5.02) Steel Elastic Modulus

200000 Mpa


2/25

May 26, 2014

Date: By:

MAK

6.2 SOIL PROPERTIES: Soil Type:

"S1"

1 =

25

deg

Angle of Internal Friction

S

=

17

KN/m³

Unit Weight of Soil

C

=

0

KN/m²

Soil Cohesion

f

=

35

deg

Frustum Angel

7. CALCULATIONS: 7.1 DETERMINE ALLOWABLE SOIL CAPACITY: Df

=

h

+

S

=

17

Kpy =

35

Fd

=

4.10

m

Total Depth of Foundation

KN/m³

Unit Weight of Soil Referance: Foundation Analysis and Design by: Joseph E. Bowles

q ult  1.3.c.N c   s .z.N q  0.4. s .B.N 

a  e  ( 0.751 / 360) tan 1   a

Nq 

=

N 

tan 1 2

12.720

 K py   1  N  =  2  cos  

qult SF

 q allow

=

Kpy 18.6 25 35 52 82 141 0

25.135

9.702

 qult  1.3.c.N c   s .z.N q  0.4. s .B.N  q allow 

15 20 25 30 35 40 45

2.710

a2  Nq = 2  cos2 (45  1 / 2)

N c  ( N q  1) / tan 1  N c =

1

=

294.11 KN/m²

1176.449

KN/m²

SF = 4 Safty Factor

Allowable Bearing Capacity

7.2 CHECK FOUNDATION DISPLACEMENT:

N 60 = 30

SPT N value

Sc

=

50 mm

Tolerable Settlement

B

=

4

Footing Width

q

m

1.4 S c  N 60 0.75 1.7  B

=

1216.034

KN/m²


 q allow

= Min ( q allow

, q)

=

294.112

KN/m²

3/25

Date: By:

May 26, 2014 MAK

7.3 DETERMINE PEDESTAL SIZE: 7.3.1 BASED ON CONCRETE BEARING : Ap

= Pc / (0.35*fc) =



a

0.245

= 494.872 mm

Gross Pedestal Area

m²

< 600 OK

Pedestal Size

7.3.2 PEDESTAL AS A SHORT COLUMN :

 =

0.008

AP 



Steel Ratio, Assumed

Pc = 0.65  (0.85  f c  (1   )    f y )

a

0.137

Gross Pedestal Area

m²

= 370.008 mm < 600 OK

Pedestal Size

Therefore, the assumed pedestal size of 800 mm is adequate

7.4 CHECK UPLIFT RESISTANCE: h

=

T

= 600 mm



3.5 m

L

=

4

m

B


= 1 / cos ( )

1.016

=

=

4

m

= 500 mm


Increasing Coefficient

f

=

35 deg

Angle of the Shearing Soil Plane with the Vertical "Frustum Angle"

h3

=

2.9 m

Soil Height Considered in Computing Uplift " Effective Height of Soil "

Resisting Force to Uplift

Eh

= Weight of Foundation + Weight of Soil Enclosed in the Frustum of an Inverted Cone of Pyramid

Volume of Concrete Pad

Vf

= = =

Weight of Pad

Wf = = =

Volume of Concrete Chimney

Vp

Weight of Chimney

Wp = = =


= = =

L x B x Fd 4 x 4 x 0.6 9.600 m³ Vf x  c 9.600 x 230.4 KN a x 0.6 x 1.464

a x 0.6 x m³

Vp x  c 1.464 x 35.129 KN

24

(h+Eh) 4

x  x 1.016

24

4/25

Date: By:

Volume of Soil

Vs =



h3 At  Ab  3

At  Ab

At

Vs

= = =

L x 4 x 16.00

G.L

B 4 m²

h3

f

f

= ( B + (2 x h3 x tan(f))² = 64.98 m²

B

= 109.454 m³ = Vs x  S 17 = 109.454 x 1860.711 KN =

Weight of Soil

Ws

Weight of Soil Replaced by The Pedestal

Wr = = =

a x a x h3 x 0.6 x 0.6 x 2.9 x 18.040 KN

Total Resistance Force to Uplift

RF

Wf + Wp + Ws - Wr 230.4 + 35.129 + 1860.711 2108.200 KN

Factor of Safety Against Uplift

MAK



At : Area at The Top Ab : Area at The Bottom Ab

May 26, 2014

= = =

FS = RF / Pu0 2108.200 = = 1.769 >

S

x 17

 x

1.016

-

18.040

/ 1192.00 OK 1.5

7.5 CHECK FOR BEARING CAPACITY OF SOIL:

Y

Y´

7.5.1 CALCULATION ECCENTRICITY: H

= = =

ey

Fd 0.6 m

Total Height of Chimney ey

ex

= = =

H x tan (  ) 4.100 x 0.182 0.747 m

= = =

eTotal

x cos (45) x 0.707 m

Eccentricity on X Direction

= = =

eTotal

x sin (45) x 0.707 m

Eccentricity on Y Direction

eTotal

ex

h + 3.5 + 4.100

0.747 0.528

0.747 0.528


X´ X

Total Eccentricity

5/25

Date: By:

May 26, 2014 MAK

7.5.2 CALCULATION MOMENTS: H

= = =

h + 3.5 + 4.600

Fd + 0.6 + m

Eh 0.5

Total Height of Foundation

Mx1 = Pc x e y = 2400.00 x 0.528 1268.23 K N.m =

Bending Moment due to Downward Force Pc "About X"

My1 = Pc x e x = 2400.00 x 0.528 1268.23 K N.m =

Bending Moment due to Downward Force Pc "About Y"

Mx2 = Ph-y x H = 318.00 x 4.600 1462.80 K N.m =

Bending Moment @ Base due to Horizontal Force Ph - y "About X"

My2 = Ph-x x H = 363.00 x 4.600 1669.80 K N.m =

Bending Moment @ Base due to Horizontal Force Ph - x "About Y"

Mx3 = Wp x ( e y / 2 ) = 35.13 x 0.264 9.28 = K N.m

Bending Moment due to Weight of Chimney "About X"

My3 = Wp x ( e x / 2 ) = 35.13 x 0.264 9.28 = K N.m Total Moment:

Bending Moment due to Weight of Chimney "About Y"

Mxx = Mx2 Mx1 = 1462.80 - 1268.23 185.28 = K N.m

-

Mx3 9.28

Myy = My2 My1 = 1669.80 - 1268.23 392.28 = K N.m

-

My3 9.28

7.5.3 CALCULATION RESISTING MOMENTS: T

= 600 mm

Fd

= 600 mm

h3

= = =

h 3.5 2.90

Kp

=

1  Sin1 1  Sin1

Kp

=

Qtp =


T 0.6 m

" Effective Height of Soil "

2.464

Kp

x S


Passive Earth Pressure Coefficient

x

h3


Passive Earth Pressure at 6/25

Date: By:

May 26, 2014 MAK

S = = Qbp = = =

2.464 x 17 121.471 KN/m²

x

2.90

Top of Pad

Kp x  S x ( h3 + Fd ) 2.464 x 17 3.50 x 146.603 KN/m²

Passive Earth Pressure at Bottom of Pad

Pc 105.680

Ph Eh = 0.500

Wp

G.L T=

0.60

h3 =

2.90

Lt = 121.471 KN/m²

Ws h = 3.50 OL =

4.60 Wf Fd = 0.600

B=

4.00

Lb = 146.603 KN/m²

0.30 1.567 Ftpx = = =

( 1 / 2 ) x Qtp x h3 0.5 x 121.471 x 105.680 KN

x a 2.90

Fbpx = = =

( 1 / 2 * ( Lb + Lt ) ) 134.04 x 0.600 321.688 KN

x x

Fd x 4.00

x

0.60

B

The Force due to Passive Earth Pressure at the Pad " X Direction"

Mxp = Ftpx Fbpx x ( Fd + ( h3 / 3 ) ) + 321.688 = 105.680 x 1.567 + x 262.071 KN.m = Ftpy = = =

( 1 / 2 ) x Qtp x h3 0.5 x 121.471 x 105.680 KN

x a 2.90

Fbpy = = =

( 1 / 2 * ( Lb + Lt ) ) 134.04 x 0.600 321.688 KN

x x

Fd x 4.00

x

The Force due to Passive Earth Pressure at the Chimney " X Direction"

x ( Fd / 2 ) 0.30

0.60

B

Myp = Ftpy Fbpy x ( Fd + ( h3 / 3 ) ) + 321.688 = 105.680 x 1.567 + x 262.071 KN.m =

Bending Moment due to Passive Earth Pressure "About X" The Force due to Passive Earth Pressure at the Chimney " Y Direction" The Force due to Passive Earth Pressure at the Pad " Y Direction"

x ( Fd / 2 ) 0.30

Bending Moment due to Passive Earth Pressure "About Y"

7.5.4 CALCULATION NET MOMENTS AT THE BASE OF FOOTING: Mx =

Mxx

-

0.8

x


Mxp

Net Bending Moment @ 7/25

Date: By:

May 26, 2014 MAK

= =

185.28 - 0.8 x -24.373 KN.m

262.071

Base of Footing "About X"

My = = =

Myy - 0.8 x 392.28 - 0.8 x 182.627 KN.m

Myp 262.071

Net Bending Moment @ Base of Footing "About Y"

Note 1 : only for safty purpose we consider in above calculation 80 % of the total resisting moment due to passive earth pressure Note 2 : if Mx < 0 then we will Mx = 0 and the same will be applicable for My. Mx = if ( Mx < 0 , 0 , Mx)

 Mx =

0.000

KN.m

My = if ( My < 0 , 0 , My)

 My =

182.627

KN.m

7.5.5 CALCULATION SOIL PRESSURE AT THE BASE OF FOOTING:

q

P Mx My  y x A Ix Iy

q

M P  My  1  x   L B  L B 

Where: Mx : Net Bending Moment @ Base of Footing "Around X". My : Net Bending Moment @ Base of Footing "Around Y". P : Total Vertical Load. A : Foundation Area. I: Second Moment of Area of The Footing About The Axis of Bending. x , y: Distance from Axis of Bending to the Position The Stress is Being Calculated. = ( B x L³ ) / = 21.333 m³

12

= ( L x B³ ) / = 21.333 m³ Calculation Vertical Loads:

12

Ix Iy

Weight of Pad

Wf = = =

Vf x  c 9.600 x 24 230.4 KN

Weight of Chimney

Wp = = =

Vp x  c 1.464 x 35.129 KN

Weight of Soil

Total Vertical Load


24

Ws = ( B x L x h - a x a x h x  ) x 17 = 54.719 x 930.227 KN = P

= = =

S

Pc + Wf + Wp + Ws 2400.000 + 230.4 + 35.129 3595.756 KN

+

930.227

8/25

May 26, 2014

Date: By:

MAK

qu1 =

3595.756 16

+

0.000 21.333

x

2

+

182.627 21.333

x

2

=

241.856

KN/m²

qu 2 =

3595.756 16

-

0.000 21.333

x

2

+

182.627 21.333

x

2

=

241.856

KN/m²

qu 3

=

3595.756 16

+

0.000 21.333

x

2

-

182.627 21.333

x

2

=

207.613

KN/m²

qu 4 =

3595.756 16

-

0.000 21.333

x

2

-

182.627 21.333

x

2

=

207.613

KN/m²

q m ax = max ( qu1 qu 2 qu 3 qu)4 , , ,

=

241.856

KN/m²

< q allow

OK

q m in = min ( qu1 , qu 2 , qu 3 , qu) 4

=

207.613

KN/m²

>

OK

0 241.856

241.856

B=

L=

207.613

KN/m²

KN/m²

4.00

4.00

241.856

KN/m²

207.613

KN/m²

m

m

KN/m² 207.613

KN/m²

7.5.6 CALCULATION SOIL PRESSURE DUE TO VERTICAL LOADS: Weight of Pad

230.400

KN

Weight of Chimney

35.129

KN

Weight of Soil

930.227

KN

Compression Force

1600.000

KN

Total Downward Force

=

=

Weight of Pad Weight of Chimney + Compression Force + 230.400 2795.756

=

 q m ax


35.129

+

+

+

930.227

Weight of Soil

+

1600.000

KN / (LxB)

= Total Downward Force

q m ax =

2795.756

/

q m ax =

174.735

KN/m²

16 < OK

9/25

Date: By:

May 26, 2014 MAK

7.6 RECALCULATE THE FORCES WITHOUT OVER LOAD CAPACITY FACTOR: Pc0 =

1600.00

KN

Compression Force

Pu0 =

1192.00

KN

Uplift Force

Ph-x0 =

242.00

KN

Resultant Shear Force in X-Direction

Ph-y0 =

212.00

KN

Resultant Shear Force in Y-Direction

Wf =

230.40

KN

Weight of Pad

Wp =

35.13

KN

Weight of Chimney

Ws =

930.23

KN

Weight of Soil

2795.756

KN

Total Vertical Load

P

=

H

= = =

h + 3.5 + 4.600

Fd + 0.6 + m

Eh 0.5

Total Height of Foundation

Mx1 = Pc x e y = 1600.00 x 0.528 845.49 = K N.m


My1 = Pc x e x = 1600.00 x 0.528 845.49 = K N.m


Mx2 = Ph-y x H = 212.00 x 4.600 975.20 = K N.m


My2 = Ph-x x H = 242.00 x 4.600 1113.20 K N.m =


Mx3 = Wp x ( e y / 2 ) = 35.13 x 0.264 9.28 = K N.m


My3 = Wp x ( e x / 2 ) = 35.13 x 0.264 9.28 = K N.m Total Moment:


Mxx = =

Mx2 Mx1 120.43 K N.m

-

Mx3

Myy = =

My2 My1 258.43 K N.m

-

My3


10/25

Date: By:

Mx = = =

Mxx - 0.8 x 120.43 - 0.8 x -89.228 KN.m

Mxp 262.071

Net Bending Moment @ Base of Footing "About X"

My = = =

Myy - 0.8 x 258.43 - 0.8 x 48.772 KN.m

Myp 262.071

Net Bending Moment @ Base of Footing "About Y"

Mx = if ( Mx < 0 , 0 , Mx)

 Mx =

0.000

KN.m

My = if ( My < 0 , 0 , My)

 My =

48.772

KN.m

May 26, 2014 MAK

Calculation Pressure at Four Corners:

q1 =

2795.756 16

+

0.000 21.333

x

2

+

0.000 21.333

x

2

=

174.735

KN/m²

q2 =

2795.756 16

-

0.000 21.333

x

2

+

0.000 21.333

x

2

=

174.735

KN/m²

q3 =

2795.756 16

+

0.000 21.333

x

2

-

0.000 21.333

x

2

=

174.735

KN/m²

q4 =

2795.756 16

-

0.000 21.333

x

2

-

0.000 21.333

x

2

=

174.735

KN/m²

Calculate Average Pressure:

q ave = ( q1 + q 2 = 174.735 + 174.735 =

+ q3 + q4 ) / 4 174.735 + 174.735 + 174.735 KN/m²

Calculate Average Pressure:

qa m ax

= =

(

174.73 + 174.73 174.735 KN/m²

)

/

2

qa m in

= =

(

174.73 + 174.73 174.735 KN/m²

)

/

2

7.6 CHECK FOR PUNCHING SHEAR:

 pb = 20 mm

Assume

a

600

Fd - cov - ( 3  pb 600 85 485 mm

d

= = =

A

= = =

L 4

=

4

b0

=

x x 16

Diameter for Bottom Steel in the Pad mm /2) 30

Pedestal Size Effective Depth for Section

B 4 m²

Foundation Area

x (a+d)

Perimeter Length B=


11/25

May 26, 2014

Date: By: = =

APunch

4

1085 x 4.34 m

4

B=

= (a+d) x (a+d) = 1.085 x 1.085 = 1.177 m²

Punching Area w a B=

qu avve

Q Punch

= ( qu1 + qu 2 + qu 3 + qu 4 ) / 4 = ( 241.856 + 241.856 + 207.613 + 224.735 KN/m² = = = =

MAK

x ( A - APunch 224.735 x 14.823 3331.193 KN

qu avve

w

4 207.613 ) / 4

a+d

) Punching Shear Perimeter

In In general, the factored shear force Q "Punch" at the critical shear section shall be less than or equal to the shear strength: Vn: ≥ Q "Punch" where the nominal shear strength Vn is: Vn = Vs + Vc Vc = nominal shear strength provided by concrete, computed if shear reinforcement is not used. Vs = nominal shear strength provided by reinforcement. " Reference ACI 318 - 02, clause 11.12.2.1"

Critical Section for Punching Shear

d

Fd q

The Nominal Shear Strength Provided by Concrete will be the Smallest of:

V c1

1) .

= =

1  3

f b0 d

0.33 x 5.292 3712.695 KN

  .d  V c 2   s  2 .  b0 

2) .

=

3930.837

= Vc = =

5569.042

x

4340

x

485

f c' .b0 .d

s

=

c

=

20

For Corner Columns

12

KN

 2  . Vc 3  1   c  

3) .

Elevation

' c

f c' .b0 .d 6

a/a

= 600

/ 600 =

1

KN

Min ( Vc1 , Vc2 , Vc3 ) 3712.695 KN

Check if (Punching Shear)

Allowable Punching Shear

Q Punch

<

Vc (Allowable Punching Shear)

If ( Qpunch <= Vc, OK , ERROR) Qpunch

=

3331.193

<

3712.695

OK

7.7 CHECK FOR ONE - WAY SHEAR: X

= ((L-a)/2) -

4

d B=


12/25

May 26, 2014

Date: By: 1.215

=

m

B=

qux  qu min  qu max  qu min . 218.015

=

Q shear

= =

MAK

X L

d

KN/m² a

1 (qux  qu max ).B. X 2 1117.486

4

Direct Shear Load B=

w a w

KN

" Reference ACI 318 - 02, clause 11.3.1.2 and clause 11.3.2.3" Allowable Direct Shear Force will be the Smallest of :

1) .

'  Pc  f c  Vc1  1  .B.d  14 Ag  6

=

2) .

1833.128

Ag

= = =

B

x

4 2.400

2093.310

Critical Section for Shear

0.6

KN X

'  0.3.Pu  f c  Vc 2  1  .B.d Ag  6 

=

Fd x m²

d

Fd

KN q

Vc = =

Min ( Vc1 , Vc2 ) 1833.128 KN

Check if (Shear Force)

Allowable Shear Force

Q shear

Elevation

Vc (Allowable Shear Force)

<

If ( Qshear <= Vc, OK , ERROR) Qshear

=

1117.486

<

1833.128

OK

7.8 FLEXURE DESIGN: 7.8.1 COMPRESSION LOAD: The Critical Section for The Design for Flexural Reinforcement Should be Taken at The Face of The Column. X1

= ((L-a)/2) 1.7 = m

qux1  qu m in  (qu m ax  qu m in ) =

222.167

X1 L

KN/m²


13/25

May 26, 2014

Date: By:

MAK

M ux1  (qux1 . X 12 .B ) / 2  (( q u m ax  q ux1 ). B. X 1 ) / 3) = =

1284.123 1328.752

44.629 + KN.m

Required Rn

=

M ux1  .B.d 2



=

0.9



 m in

KN/m²

B=

 2 Rn 1  1  0.85. f c\ 

0.85. f c\ fy

4

Critical Section for Flexure

Minimum As Required for Footings of Uniform Thickness, for Grade 60 Reinforcement " Reference ACI 318 - 02, clause 7.12.2.1"

0.0018

  If ( 

 m in , 

>

  Required As

 , m in

= = =

d

Fd



x B x d 0.0039 x 4000 7503.98 mm²

x 485

q Elevation

 20 mm  A = 314.16 mm²

NPb = round (( Required As / A

) + 1 )

Number of Steel Reinforcement Required

25

=

=

X1= 1.7

)

0.0039

Use Steel with

SPb =

a a w w

  

0.003868 =

4

Strength Rediction Factor " Reference ACI 318 - 02, clause 9.3.2"

1569.127

Rn =



B=

( B - 2 cov ) / ( NPb - 1 ) 160

Provided As

Spacing of Steel Reinforcement

mm

=

NPb

=

25

=

7853.98

x A x

314.16 mm²

If ( Provided As > Required As , "OK" ," Revise" )


 OK

14/25

Date: By:

May 26, 2014 MAK

7.8.2 UPLIFT LOAD: Punet = Pu - Wp - Wf = 1788.00 - 35.129 1522.47 KN =

Load Carried by The Pad During Uplift -

230.4

qu net =

Punet / ( A - AP ) 15.64 = 1522.47 / 97.345 = KN/m²

X1

A : Area of Foundation AP : Area of Pedestal

= ((L-a)/2) 1.7 = m

Mu net  qu net  B  =

562.652

Required Rn

=

 m in

KN.m

M ux1  .B.d 2



664.437

Rn =



X 12 2

=

0.9

Strength Rediction Factor " Reference ACI 318 - 02, clause 9.3.2"

KN/m²

0.85. f c\  2 Rn  1  1   fy 0.85. f c\   =

0.0018

  If (    Required As

 , m in

)

0.0018 = = =



x B x d 0.0018 x 4000 3492.00 mm²

x

485

 20 mm  A = 314.16 mm²

Use Steel with

NPt = round (( Required As / A

) + 1 )

Number of Steel Reinforcement Required

12

=

SPt = ( B - 2 cov ) / ( NPt - 1 ) =

0.0016

Minimum As Required for Footings of Uniform Thickness, for Grade 60 Reinforcement " Reference ACI 318 - 02, clause 7.12.2.1"

 m in , 

>

 

348

Provided As

= = =

mm NPt x A 12 x 314.16 3769.911

If ( Provided As > Required As , "OK" ," Revise" )


Spacing of Steel Reinforcement

 OK

15/25

Date: By:

May 26, 2014 MAK

7.9 PEDESTAL DESIGN: 7.9.1 CALCULATIOM MOMENT: H

= = =

h

+ 3.5 4.000

Eh + m

Total Height of Pedestal 0.5

Mx1 = Pc x e y = 2400.00 x 0.528 1268.23 K N.m =


My1 = Pc x e x = 2400.00 x 0.528 1268.23 K N.m =


Mx2 = = =

Ph-y H x 318.00 x 4.000 1272.00 K N.m


My2 = = =

Ph-x H x 363.00 x 4.000 1452.00 K N.m


Mx3 = Wp x ( e y / 2 ) = 35.13 x 0.264 9.28 = K N.m


My3 = Wp x ( e x / 2 ) = 35.13 x 0.264 9.28 = K N.m


Mx4 = Pu x e y = 1788.00 x 0.528 944.83 = K N.m

Bending Moment due to Upward Force Pc "About X"

My4 = Pu x e x = 1788.00 x 0.528 944.83 = K N.m

Bending Moment due to Upwnward Force Pc "About Y" Pc

Total Moment with Downward Force: Ph

Mxc = Mx2 Mx1 = 1272.00 - 1268.23 -5.52 = K N.m

-

Myc = My2 My1 = 1452.00 - 1268.23 174.48 = K N.m

-

Pd

Pc Wp = + = 1788.00 + 35.13 = 1823.13 KN


Mx3 9.28

G.L

Wp Ws

My3 9.28 Wf

Loads on Foundation when Subjected to Compression

16/25

Date: By:

Total Moment with Upward Force:

May 26, 2014 MAK

Pu

Mxu = Mx2 Mx4 = 1272.00 - 944.83 317.88 = K N.m

-

Mx3 9.28

Myu = My2 My4 = 1452.00 - 944.83 497.88 = K N.m

-

My3 9.28

Ph

G.L

Wp Ws

Wf

Pf

Pu = = 1788.00 = 1752.87

Wp 35.13 KN Loads on Foundation when Subjected to Uplift

7.9.2 ESTIMATE EQUIVALENT UNIAXIAL BENDING MOMENT: h0 b Asd Sd

= = = =

d

= = =

0.6 0.6 25 12

m m mm mm

h0 - cov 600 490.50 mm

1 =

Section Height Section Wedth Vertical Bar Diameter Ties Diameter -

Sd 85

-

( Asd / 2 ) 12.5 12 -

Effective Depth

0.65

 h0  1   1 Mu  Mxu  Myu   =  b  1

585.98

K N.m

 h0  1   1 Mc  Mxc  Myc  =  b  1

88.44

KN.m

7.9.3 DESIGN FOR COMPRESSION: Try to use :

NP = Asd =

32 25

Total Number of Steel Reinfocement in Pedestal Vertical Bar Diameter

Smin = ( If 1.5Asd > 40 mm, 1.5 Asd , 40 mm ) cov Smin

=

S

= =

As

=

40

Minimum Bar Spacing

4.(h0  2  (cov Sd  ( Asd / 2)))  Asd NP 22.625

 . Asd 4

=

mm

< 2

40

 NP

h0

CHANG NO. OF BARS Area of Steel Prvided Asd

Sd

15707.963 mm²


17/25

Date: By:

 m in = 1.4 / fy = =

 m ax =

May 26, 2014 MAK

Minimum Steel Ratio " Reference ACI 318 - 02, clause 10.5"

1.4 / 420 0.0033 0.05

Maximum Steel Ratio

 actual = As =

/ ( h0 x b ) 0.04363

 m in ≤  actual  actual ≤  m ax

Check if the Steel Ratio is : Check if the Steel Ratio is :

  0.5   actual





= 0.04363 x

OK OK 0.5 = 0.02182

e copm. = Mc / Pc = 88.44 / 2400.00 = 0.03685 m d - ( h0 / 2 ) + 490.50 300 227.35 mm

e1

= = =

d1

= cov + Sd + = 85 + 12 + = 109.5 mm

m1 =

fy 0.85. f c'

e comp. + 36.849

Asd / 2 12.5

= 420



/

' Pc1 = 0.85. f c .b.d    1 

 

=

10146.69

KN >

If ( Pc1 <= Pc , "OK" , Error" )

28

e1  d

2400.000

=

17.647

2 e  d e    1  1   2. .(m1  1).(1  1 )  1   d  d d    

KN OK

7.9.4 CHECK FOR UPLIFT: Pf

=

Mu =

1752.87

KN

585.98

K N.m

Adjusted Uplift Load Test

Ast = = =

0.5 x As 0.5 x 15707.963 7853.98 mm²

Area for Tension Steel

Asc = = =

0.5 x As 0.5 x 15707.963 7853.98 mm²

Area for Compression Steel

c

=

0.003

E

=

200000


Maximum Concrete Strain Steel Elastic Modulus

18/25

Date: By:

May 26, 2014 MAK

In order to calculate the iteraction curve due to the uplift force, it is necessary to calcuate the bending capacity of the section Mu0 when the axial load is zero, Also we need to determine Pu00 the maximum axial capacity of the section in the absence of the moment. we will assume the x value " distance of the center of rotation from the center line of the section" and then it will be adjusted to be as foolow x

y

=

144.64 mm

= fy = 420

/ /

Distance of the Neutral Axis from the Compression Face

E 200000

=

0.0021

s

= (( x - d1 ) / x ) x

fs

= ( s /  y )  f y

Cs

= Asc . ( fs - 0.85 fc )

Ts

= Ast x fy

Cc

=

0.85 x fc x ß2 x x x b

=

Eq

=

Ts

= 585.201

-

= 0.00073 145.759

=

=

Cc

c

Strain in the Compression Steel Mpa

=

957.867

7853.98

x 420 =

-

Cs

1755.604

KN

Compression Force in Steel

3298.672 KN

KN

Tension Force in Steel

Compression Force in Concrete

Check Equation if ( Ts - Cc - Cs = 0 , OK , Change x ) =

0.5 * ( d -d1 )

=

190.5 mm

Tension Steel Distance from the center

Xcs =

0.5 * ( d -d1 )

=

190.5 mm

Compression Steel Distance from the center

Xts

Xc

=

d

-

Xst

-

ß2 * 0.5*x =

238.529 mm

Center of Concrete Compression Block from the Center

Muo = 0.9 * ( Ts.Xst + Cs . Xcs + Cc. Xc ) = 1106.670 KN.m

Section Moment Capacity when P = 0

Pu00 = =

Section Axial Load when M = 0

0.9 . ( fy . As ) 5937.610 KN

 Pf IR    Pu00 = =

  

1.1

0.261 0.758

 Mu    Mu0 + <

  

1.1

0.497 1

If ( IR <=1 , "OK" , " ERROR" )

OK

Final Vertical Reinforcement for Pedestal: NP =

32

Number of Steel Reinforcement Bars

Asd =

25

Diameter of Steel Reinforcement Bars


19/25

Date: By:

May 26, 2014 MAK

7.9.5 TIES: Use Steel Bars of Diameter 12 mm @ 200 mm Spacing ds

=

Ass =

12 mm

Diameter of Ties

 d

Area of Ties

2 s

4 113.10

= Ns

0

= (( h + Eh - cov ) / 200 ) + 1 21 =

Number of Ties

7.10 CHECK FORCE TRANSFER AT INTERFACE OF COLUMN AND FOOTING: 1- Bearing Strength of Column: fc



A1

= = =

28 MPa 0.65 Ap = 360000

Compreesive Strength of the pedestal Concrete 0

Pnb   .( 0.85 . f c' . A1 ) =

5569.2

Bearing Strength of Column " Reference ACI 318 - 02, clause 10.17" 3595.756

KN >

OK

2- Bearing Strength of Footing: The bearing strength of the footing is increased by a factor Sqrt ( A2 / A1 ) <= 2 due to the large footing area permitting a greater distribution of the column load. 4

B=

" Reference ACI 318 - 02, clause 15.8"

1.2

a B=

4

0.6

a w 45 45

1.2

Fd =

Fd=

0.6

0.6 1.2

0.6

1.2

A1 is the column (loaded) area and A2 is the plan area of the lower base of the largest frustum of the pyramide , cone or tapered wedge cotained wholly within the support and for having its upper base the loaded area , and having side slopes of 1 vertical to 2 horizontal.


20/25

Date: By:

Ap

=

A2

1.2 = ( 9000000 =

A2 / A1

=

360000

A1

MAK

0 0.6

+

) x (

1.2

+

1.2

+

0.6

+

1.2

)

0 5

=

May 26, 2014



2

>

2

A2 / A1 =

Note that bearing on the column concrete will allways govern until the strength of the column concrete exceeds twice that of the footing concrete

Pnb   .(0.85. f c' . A1 ). 11138.4

=

A2 A1

KN >

3595.756

OK

7.11 REQUIRED DOWEL BARS BETWEEN COLUMN AND FOOTING: Even though bearing on the column and footing concrete is adequate to transfer the factored loads, a minimum area of reinforcement is required across the interface. " Reference ACI 318 - 02, clause 15.8.2.1" Asmin = 0.005 x Ap Ap Area of Pedestal = 0.005 x 360000 1800 = 0 Provide

5 Bars of 25 dia ( As =

2290.9

0)

7.12 DEVELOPMENT OF DOWEL REINFORCEMENT IN COMPREESION: Shorter development lengths are required for bars in compression than in tension since the weakening effect of flexural tension cracks in the concrete is not present. The development length for deformed bars or deformed wire in compression is dc = (db.fy) / (4.Sqrt (f c)), but not less than 0.04.db.fy or 200mm. " Reference ACI 318 - 02, clause 12.3" dc1 =

dc1 =

fy 4

f c'

 db

496.08 mm

dc2 = 0.04  f y  db =

420

mm

dc3 =

200

mm



dc

= max ( dc1 , dc2 , dc3 ) = 496.08 mm

This length may be reduced to account for excess reinforcement. As ( required ) / As ( provided ) = Reqiured dc =

496.08

x


1800 0.786

=

/

2290.9

=

0.786

389.78 mm

21/25

May 26, 2014

Date: By:

MAK

Available length for development in footing : = Footing Thickness - Cover - 2.( Footing Bar Diameter ) - Dowel Bar Diameter =

600

-

85

=

450

>

389.78

40

-

25

-

Therefore, the dowels can be fully developed in the footing

OK

Note: If case the avilable development length is less than the required length, either increase footing depth or use larger number of smaller size dowels. Also note that if the footing dowels are bent for placement on top of the footing reinforcement , the bent portion cannot be considered effective for development the bars in compression.

7.13 CALCULATION MATERIAL QUANTITIES: 7.13.1 EXCAVATION: L

=

4

m

B

=

4

m

h

=

3.5

m

Fd

=

600

mm

H

=

X

=

Y

Eh =

0.5

G.L

3.5

h=

20

h

+

Fd

0.5

m

20

deg

=

4.1

m

Fd =

=

0.6 X

Y

= = =

B=

4

X

H x tan ( s ) 4.1 x 0.364 1.492 m





Vs =

H At  Ab  At  Ab 3

Ab

= =

(L+2X)*(B+2X) 25 m²

Bottom Area

At

= =

(L+2X+2Y)*(B+2X+2Y) 63.75 m²

Top Area

Vs

=

175.9

Volume of Soil to be Excavated

m³

7.13.2 CONCRETE QUANTITY: L

=

4

m

Footing Length

B

=

4

m

Footing Width

h

=

3.5

m

Footing Depth

Fd

=

600

mm

Thickness of Pad


22/25

May 26, 2014

Date:

Eh

=

500

mm

a

=

600

mm

By:

MAK

x

1.016

A=

4.46

Exposed Height, including the 300 mm structure pad Pedestal Size

Volume of Concrete Pad

Vf

Volume of Concrete Chimney

Vp

Total Volume of Concrete

Vc

x B 4 x 9.600 m³

x

x 0.6 1.464

x

= = =

L

= = =

a

= = =

Vf Vp + 9.600 + 1.464 11.064 m³

a x m³

Fd 4

x

0.6

(h+Eh) 0.6 x

x

 4

7.13.3 BACKFILL: Vbf = = = =

Vs Vc 175.9 - 11.064 175.9 - 11.064 164.72 m³

- Volume of Exposed Height of Chimney - ( a x a x ( Eh - 0.3 )) 0.072 -

7.13.4 STEEL QUANTITY: 1- Chimney Rebars (Vertical Bars ): OL = = A1

h

+

Eh m

+

Fd

= OL 4.6 = = 4.390

Fd m

+

dc 0.6

4.6

A1 x 4.390 4.46



B

= = =

12 x 12 x 0.30

Asd 25 m

LP

= = =

A + 4.46 4.76

B + m

x m

0.3898

B=

= = =

A

+

m

0.30

m

1.016

Total Pedestal Vertical Bar Length 0.30

2- Chimney Rebars (Ties ): a1

LS

= = = = =

- 2. cov 0.6 0.17 0.43 m

200

a

( 4 x a1 ) + 0.3 2.02 m

a1

0.43

Total Tie Length a1

0.43

3- Footing Rebars: \\vboxsrv\conversion_tmp\scratch_3\230673500.xls.ms_office

23/25

May 26, 2014

Date: By:

C1

= (( Fd - 2. cov ) / 2 ) + 0.1 = 0.315 m

C2

= = =

LPB = = = LPT =

MAK

( 2 x cov ) 4 0.17 3.83 m

L

( 2 x C1 ) C2 + 3.83 0.63 + 4.46 m LPB

=

4.46

C1

m

C2

=

3.83

C1= 0.315

Total Pad Bar Length

7.13.5 STEEL WEIGHT: 1- Chimney Rebars (Vertical Bars ): Asd = NP =

25 mm 32

LP ( Total )

LPW = = =

= = =

Diameter of Steel Reinforcement Bars Number of Steel Reinforcement Bars LP x NP 4.76 x 32 152.39 m

Total Pedestal Vertical Bars Length

LP ( Total ) x W 152.39 x 3.856 587.644 Kg

W

: Weight of Chimney Rebar (Per Meter)

2- Chimney Rebars (Ties ): ds = NS =

12 mm 21 mm

LS ( Total )

LSW = = =

= = =

Diameter of Ties Number of Ties LS x NS 2.02 x 21 41.56 m

LS ( Total ) x W 41.56 x 0.888 36.927 Kg

W

: Weight of Ties (Per Meter)

3- Footing Rebars:

 pb

=

NPP = = =

20 mm ( NPb + 50 + 74

LPP ( Total ) = = = LPPW = = =

Diameter for Bottom Steel in the Pad NPt ) 24

2

Number of Pad Bars

NPP x ( LPB , LPT ) 74.00 x 4.46 330.04 m

LS ( Total ) x W 330.04 x 2.468 814.539 Kg

S.S.E.M Saudi Services For Electro Mechanic Works Ltd. \\vboxsrv\conversion_tmp\scratch_3\230673500.xls.ms_office

x

W

: Weight of Bars (Per Meter)

Date: By:

May 26, 2014 MAK 24/25

May 26, 2014

Date: By:

MAK

DESIGN OF PAD & CHIMNEY FOUNDATION 0 Tower Type "A10" (Towers Include: # PI - 21)

SUMMURY FOR DESIGN Pc

=

2400.00

KN

B

=

4.00

Pu

=

1788.00

KN

L

=

4.00

m

Ph-x

363.00

=

Ph-y

318.00

=

Fd

KN

Geometry

Factored Loads

Pc

KN

Q

=

482.59

KN

M

=

0.00

KN

600.00

=

h

mm

3.50

=

Ph-y

m Eh

GL

a

OL

=

500.00

mm

a

=

600.00

mm

Ph-x Chimney Pad

m

Eh

a

h

Fd

Material Properties

B

c

=

24.00

KN/m³

cov

=

85.00

mm

f c

=

28.00

MPa

T

=

600.00

mm

fy

=

420.00

MPa

QUANTITY SURVEY " One Footing" EXCAVATION

1

=

25.00

deg

S

=

17.00

KN/m³

C

=

0.00

KN/m²

f

=

35.00

deg

L

Vs

=

175.9

CONCRETE Volume of Pad 10 Vf = m³ Volume of Chimney 1 Vp = m³ Total Volume of Concrete 11 Vc = m³

m³

BACKFILLING Vbf

=

164.7

m³

STEEL Steel of Pad 815 Kg LPPW = Steel of Chimney 625 Kg LPSW = Total Weight of Concrete 1439 Kg Total =

STEEL REINFORCEMENT PAD C2 C1

=

3.83

0.315

=

Top reinforcement for Pad C1

NPt

=

12

CHIMNEY Chimney Vertical Bars

Ties ( Stirrups ) 200

a1=

0.43

A= C1

0.315

=

C1

NPb

=

a1= C2

=

3.83

Bottom reinforcement for Pad


4.46

25 0.43

B=

0.30

25/25

Design Pad & Chimney Foundation1

Recommend Documents