Design Pad & Chimney Foundation1

(1)

1. INTRODUCTION:

* 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.

2. TOWER GEOMETRY:

h = mm Tower Body Hieght

a = mm Tower Dimension at the Top

b = mm Tower Dimension at the Bottom

deg Leg Slope angel with Respect to Vertical

mm

deg

mm Tower First Leg Slope

mm Tower Second Leg Slope

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:

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

Note : 1600.00 1192.00 242.00 212.00 1000 11.860 Ph-x0 Ph-y0 1021.812 1016.476 Pc0 Pu0 10.330

DESIGN OF PAD & CHIMNEY FOUNDATION

* 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.

* 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.

Tower Type "A10" (Towers Include: # PI - 21) 26100 5000 15962













 





h

a

b

2 tan

1





0

h









 





h

a

b

2 tan

1









)

90 sin(

0 22

_

h







)

90 sin(

0 11

_

h

11 22

h

0

h

\\vboxsrv\conversion_tmp\scratch_3\230673500.xls.ms_office 1/25

(2)

4. FACTORED LOADS:

In Compression SFc = 1.5

In Uplift SFu = 1.5 As per SES-P-122.06

In Shear SFs = 1.5

Pc = Pc0 × SFc = KN Compression Force

Pu = Pu0 × SFu = KN Uplift Force

Ph-x = Ph-x0 × SFs = KN Resultant Shear Force in X-Direction

Ph-y = Ph-y0 × SFs = KN Resultant Shear Force in Y-Direction

KN

KN Horizontal Reaction

Moment Reaction

5. PROPOSED SIZE OF FOUNDATION:

Assumed Footing Width B = 4 m

Assumed Footing Length L = 4 m

Thickness of Pad Fd = 600mm

Assumed Footing Depth h = 3.5 m

Exposed Height, including the Eh = 500 mm 300 mm structure pad

Assumed Pedestal Size a = 600mm

Cncrete Cover cov = 85 mm

Thickness to be Ignored in T = 600 mm As per PTS (05WO307 clause 4.09 item 3) Calculating Uplift Resistance

6. MATERIAL PROPERTIES:

6.1 CONCRETE:

= KN/m³ Concrete Unit Weight

= MPa Concrete Compressive Strength

PTS (clause: 4.09 A-1)

= MPa Steel Yield Strength

PTS (clause: 5.02)

E = Mpa Steel Elastic Modulus

321.73 482.59 200000 24 28 420 2400.00 1788.00 363.00 318.00 GL B Fd L a a Ph-x Ph-y Pu OL Eh Chimney Pad h











2 0 2 0 0

(

Ph

x

)

(

Ph

y

)

Q







Q

SFs

Q

0





M



M

c



c

f



y f

(3)

6.2 SOIL PROPERTIES:

Soil Type: "S1"

= deg Angle of Internal Friction

= KN/m³ Unit Weight of Soil

C = KN/m² Soil Cohesion

f = deg Frustum Angel

7. CALCULATIONS:

7.1 DETERMINE ALLOWABLE SOIL CAPACITY:

Df = h + Fd = m Total Depth of Foundation

= KN/m³ Unit Weight of Soil

by: Joseph E. Bowles

=

= KN/m²

= KN/m² SF = 4 Safty Factor Allowable Bearing Capacity

7.2 CHECK FOUNDATION DISPLACEMENT:

= 30 SPT N value = 50 mm Tolerable Settlement B = 4 m Footing Width = 1216.034 KN/m² = Min ( , q ) = 294.112 KN/m² 35 17 Kpy = 35 4.10

Referance: Foundation Analysis and Design 2.710 12.720 25.135 9.702 25 17 0 35 82 Kpy 15 20 45 18.6 25 35 52 141 0 25 30 1176.449 294.11 40 1



S



S



1





zN BN N c q_ult 1.3. . _c  _s. . _q0.4. _s. .



a

(0.751/360)tan1



e

a

)

2 /

45 (

cos

2

1 2 2









a

N

q



Nq 1

tan

/

)

1 (







_q c

N



N

_c

















1 cos

2 tan

2 1



 py

K

N



N



qult 1.3.c.Nc 



s.z.Nq 0.4.



s.B.N



SF

q

ult allow



q

allow 60

N

c

S

75 . 0 4 . 1 60

7 .

1 B

N

S

q

c







q

allow

q

allow \\vboxsrv\conversion_tmp\scratch_3\230673500.xls.ms_office 3/25

(4)

7.3 DETERMINE PEDESTAL SIZE:

7.3.1 BASED ON CONCRETE BEARING :

Ap = Pc / (0.35*fc) = m² Gross Pedestal Area

a = < 600 OK Pedestal Size

7.3.2 PEDESTAL AS A SHORT COLUMN :

= Steel Ratio, Assumed

= m² Gross Pedestal Area

a = mm < 600 OK Pedestal Size

Therefore, the assumed pedestal size of 800 mm is adequate 7.4 CHECK UPLIFT RESISTANCE:

h = 3.5 m L = 4 m B = 4 m Eh = 500 mm

T = 600mm Thickness to be Ignored in As per PTS (05WO307 clause 4.09 item 3)

Calculating Uplift Resistance

= 1 / cos ( ) = 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 = Weight of Foundation + Weight of Soil Enclosed in the Frustum of an Inverted Cone of Pyramid

Volume of Concrete Pad Vf = L x B x Fd

= 4 x 4 x

= m³

Weight of Pad Wf = Vf x

= x

= KN

Volume of Concrete Chimney Vp = a x a x x

= 0.6 x 0.6 x x = m³ Weight of Chimney Wp = Vp x = x = KN mm 1.464 (h+Eh) 1.016 1.016 9.600 230.4 494.872 0.008 370.008 0.6 9.600 4 0.245 0.137 24 1.464 35.129 24





)

1 (

85 .

0 (

65 .

0

c y c P

f

P

A



















c



c



(5)

Volume of Soil Vs =

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

Ab = L x B = 4 x 4 = m² At = ( B + (2 x h3 x tan(f))² = m² Vs = m³ Weight of Soil Ws = Vs x = x = KN

Weight of Soil Replaced Wr = a x a x h3 x x

by The Pedestal = 0.6 x 0.6 x 2.9 x x

= KN

Total Resistance Force to Uplift RF = Wf + Wp + Ws - Wr

= + +

-= KN

Factor of Safety Against Uplift FS = RF / Pu0

= /

= > 1.5 OK

7.5 CHECK FOR BEARING CAPACITY OF SOIL:

7.5.1 CALCULATION ECCENTRICITY:

H = h + Fd Total Height of Chimney

= 3.5 + 0.6

= m

= H x tan ( ) Total Eccentricity

= x

= m

= x cos (45) Eccentricity on X Direction

= x

= m

= x sin (45) Eccentricity on Y Direction

= x = m 17 64.98 109.454 17 1860.711 16.00 109.454 1.016 18.040 230.4 35.129 1860.711 18.040 2108.200 2108.200 1192.00 1.769 4.100 4.100 0.182 0.747 0.747 0.707 0.528 0.747 0.707 0.528 G.L h3 B f f X Y X´ Y´



At Ab At Ab



h    3 3 S



S



_



y

e

x

e

_Total Total

e

Total

e

x e y e \\vboxsrv\conversion_tmp\scratch_3\230673500.xls.ms_office 5/25

(6)

7.5.2 CALCULATION MOMENTS:

H = h + Fd + Eh Total Height of Foundation

= 3.5 + 0.6 + 0.5 = m = Pc x = x = K N.m "About X" = Pc x = x = K N.m "About Y"

= Ph-y x H Bending Moment @ Base due

= x

= K N.m "About X"

= Ph-x x H Bending Moment @ Base due

= x = K N.m "About Y" = Wp x ( / 2 ) = x = K N.m "About X" = Wp x ( / 2 ) = x = K N.m "About Y" Total Moment: = - -= - -= K N.m = - -= - -= K N.m

7.5.3 CALCULATION RESISTING MOMENTS:

T = 600 mm Thickness to be Ignored in As per PTS (05WO307 clause 4.09 item 3)

Calculating Uplift Resistance

Fd = 600 mm

h3 = h - T " Effective Height of Soil "

= 3.5 - 0.6

= m

Kp =

Kp = Passive Earth Pressure Coefficient

= Kp x x h3 Passive Earth Pressure at

My1 2400.00 0.528 4.600 Mx1 2400.00 1268.23

Bending Moment due to Downward Force Pc

Bending Moment due to Downward Force Pc 0.528 1268.23 Mx2 318.00 4.600 1462.80 to Horizontal Force Ph - y My2 363.00 4.600 1669.80 to Horizontal Force Ph - x Mx3 35.13 0.264 9.28

Bending Moment due to Weight of Chimney

My3 Bending Moment due to

35.13 0.264 Weight of Chimney 9.28 Mx2 Mx1 Mx3 1462.80 1268.23 9.28 185.28 My2 My1 392.28 2.90 2.464 My3 1669.80 1268.23 9.28 Qtp Mxx Myy y

e

x

e

y

e

x

e

1 1 1 1



Sin Sin   S



(7)

= x x Top of Pad

= KN/m²

= Kp x x Passive Earth Pressure at

= x x Bottom of Pad

= KN/m²

KN/m²

= x Qtp x h3 x a The Force due to Passive Earth

= x x x Pressure at the Chimney

= KN " X Direction"

= x Fd x B The Force due to Passive Earth

= x x Pressure at the Pad

= KN " X Direction"

= x + x Bending Moment due to

= x + x Passive Earth Pressure

= KN.m "About X"

= x Qtp x h3 x a The Force due to Passive Earth

= x x x Pressure at the Chimney

= KN " Y Direction"

= x Fd x B The Force due to Passive Earth

= x x Pressure at the Pad

= KN " Y Direction"

= x + x Bending Moment due to

= x + x Passive Earth Pressure

= KN.m "About Y"

7.5.4 CALCULATION NET MOMENTS AT THE BASE OF FOOTING:

Mx = - 0.8 x Net Bending Moment @

17 2.464 146.603 Qbp ( h3 + Fd ) 2.90 121.471 2.464 17 3.50 4.60 0.30 0.500 0.600 2.90 0.60 3.50 105.680 0.30 146.603 4.00 1.567 ( Fd / 2 ) 121.471 2.90 0.60 121.471 ( 1 / 2 ) 121.471 Ftpy 134.04 0.600 Mxp 321.688 ( 1 / 2 * ( Lb + Lt ) ) 105.680 1.567 0.60 0.5 Fbpx ( 1 / 2 ) 262.071 105.680 321.688 ( Fd + ( h3 / 3 ) ) 4.00 2.90 Fbpy ( 1 / 2 * ( Lb + Lt ) ) 134.04 0.600 4.00 105.680 ( Fd / 2 ) 105.680 1.567 321.688 0.30 Ftpy ( Fd + ( h3 / 3 ) ) Fbpy 262.071 321.688 Myp Ftpx Fbpx Ftpx 0.5 Mxx Mxp Lb = Lt = Wp Ph Wf G.L Eh = OL = Fd = T= B = Ws Pc h = h3 = S



S



(8)

= - 0.8 x Base of Footing

= KN.m "About X"

My = - 0.8 x Net Bending Moment @

= KN.m "About Y"

Mx = if ( Mx < 0 , 0 , Mx) Mx = KN.m

My = if ( My < 0 , 0 , My) My = KN.m

7.5.5 CALCULATION SOIL PRESSURE AT THE BASE OF FOOTING:

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 ) / 12

= m³

= ( L x B³ ) / 12

= m³

Calculation Vertical Loads:

Weight of Pad Wf = Vf x = x 24 = KN Weight of Chimney Wp = Vp x = x = KN Weight of Soil Ws = x = x = KN

Total Vertical Load P = Pc + Wf + Wp + Ws

= + + + = KN 35.129 930.227 3595.756 230.4 392.28 Myy Myp 185.28 262.071 -24.373 0.000 262.071 182.627

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.

35.129 182.627 21.333 21.333 930.227 ( B x L x h - a x a x h x ) 2400.000 9.600 230.4 54.719 17 1.464 24



x

I

y

I

c



c



S

















_





B

My

L

M

B

L

P

q

x

1 x

I

My

y

I

M

A

P

q

y x x











(9)

= max ( ) = KN/m² < OK = min ( ) = KN/m² > 0 OK KN/m² KN/m² KN/m² B = m L = m KN/m² KN/m² KN/m² 7.5.6 CALCULATION SOIL PRESSURE DUE TO VERTICAL LOADS:

Weight of Pad KN

Weight of Chimney KN

Weight of Soil KN

Compression Force KN

Total Downward Force = + +

+

= + + +

= KN

= Total Downward Force / ( L x B )

= / 16 = KN/m² < OK 241.856 4.00 4.00 207.613 207.613 207.613 2795.756 930.227 35.129 1600.000 Weight of Chimney 241.856 241.856 230.400 35.129 930.227 1600.000

Weight of Pad Weight of Soil

Compression Force 241.856 207.613 2 207.613 2 KN/m² 16 21.333 21.333 3595.756 0.000 2 182.627 KN/m² 16 21.333 21.333 3595.756 0.000 2 182.627 207.613 2 241.856 KN/m² 16 3595.756 0.000 2 182.627 21.333 21.333 3595.756 0.000 2 21.333 2 241.856 KN/m² 16 21.333 182.627 230.400 174.735 2795.756 , = + x + x = = - x + x = = + x - x = = - x - x = , , , , , m ax

q

m in

q

allow

q

m ax

q



m ax

q

m ax

q

1

qu

2

qu

4

qu

1

qu

1

qu

2

qu

2

qu

3

qu

3

qu

3

qu

4

qu

4

qu

(10)

7.6 RECALCULATE THE FORCES WITHOUT OVER LOAD CAPACITY FACTOR:

Pc0 = KN Compression Force

Pu0 = KN Uplift Force

= KN Resultant Shear Force in X-Direction

= KN Resultant Shear Force in Y-Direction

Wf = KN Weight of Pad

Wp = KN Weight of Chimney

Ws = KN Weight of Soil

P = KN Total Vertical Load

H = h + Fd + Eh Total Height of Foundation

= 3.5 + 0.6 + 0.5 = m = Pc x = x = K N.m = Pc x = x = K N.m = Ph-y x H = x = K N.m = Ph-x x H = x = K N.m = Wp x ( / 2 ) = x = K N.m = Wp x ( / 2 ) = x = K N.m Total Moment: = - -= K N.m = - -= K N.m

Bending Moment due to

"About Y"

120.43

258.43

Weight of Chimney "About X"

Bending Moment due to Weight of Chimney

My3

Bending Moment due to Downward Force Pc "About X"

Bending Moment due to Downward Force Pc "About Y"

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

Bending Moment @ Base due to Horizontal Force Ph - x "About Y" 1113.20 Mx3 35.13 0.264 My2 242.00 4.600 4.600 Mx1 1600.00 212.00 1600.00 My1 845.49 Mx2 My2 My1 Myy Mx2 Mx1 9.28 Mxx 35.13 0.264 Mx3 4.600 975.20 9.28 My3 0.528 930.23 2795.756 845.49 0.528 Ph-x0 Ph-y0 230.40 35.13 242.00 1600.00 1192.00 212.00 y

e

x

e

y

e

x

e

(11)

Mx = - 0.8 x Net Bending Moment @

= KN.m "About X"

My = - 0.8 x Net Bending Moment @

= KN.m "About Y"

Mx = if ( Mx < 0 , 0 , Mx) Mx = KN.m

My = if ( My < 0 , 0 , My) My = KN.m

Calculation Pressure at Four Corners:

Calculate Average Pressure:

= ( + + + ) / 4

= + + +

= KN/m²

Calculate Average Pressure:

= ( + ) / 2

= KN/m²

= ( + ) / 2

= KN/m²

7.6 CHECK FOR PUNCHING SHEAR:

Assume = 20 mm Diameter for Bottom Steel in the Pad

a = mm Pedestal Size

d = Fd - cov - ( 3 / 2 ) Effective Depth for Section

= - -= mm A = L x B Foundation Area = 4 x 4 = m² b0 = 4 x ( a + d ) Perimeter Length 485 16 600 174.735 600 85 30 Mxx Mxp 120.43 262.071 -89.228 Myy Myp 258.43 262.071 48.772 0.000 0.000 2 48.772 2795.756 0.000 2 174.735 KN/m² 16 21.333 21.333 2 174.735 KN/m² 16 21.333 21.333 2795.756 0.000 2 0.000 2 174.735 KN/m² 16 21.333 21.333 2795.756 0.000 2 0.000 2 174.735 KN/m² 16 21.333 21.333 2795.756 0.000 2 0.000 174.735 174.735 174.735 174.735 174.735 174.73 174.73 174.735 174.73 174.73 = + x + x = = - x + x = = + x - x = = - x - x ₌ B =



1

q

2

q

3

q

4

q

av e

q

₁

q

2

q

3

q

4 m ax

qa

m in

qa

pb



pb



(12)

= 4 x = m = ( a + d ) x ( a + d ) Punching Area = x = m² = ( + + + ) / 4 = ( + + + ) / 4 = KN/m² = x ( A - ) = x = KN In

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

1) .

= x x x

= KN

2) . = 20 For Corner Columns

= KN

3) . = = 600 / 600 = 1

= KN

Vc = Min ( Vc1 , Vc2 , Vc3 ) Allowable Punching Shear

= KN

Check if (Punching Shear) <

If ( Qpunch <= Vc, OK , ERROR)

Qpunch = < OK

7.7 CHECK FOR ONE - WAY SHEAR:

X = - d 4

Vc (Allowable Punching Shear)

3331.193 3712.695 ( ( L - a ) / 2 ) 241.856 241.856 207.613 207.613 224.735 224.735 14.823 3712.695 3331.193 0.33 5.292 4340

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"

485 4.34 4 4 1.085 1.085 1.177 1085 3930.837 a / a 5569.042 3712.695

Critical Section for Punching Shear Fd a w w a+d B = Punching Shear Perimeter B = d Elevation q B = Punch

A

Punch

Q

A

Punch avve

qu

1

qu

2

qu

3

qu

4 avve

qu

d

b

f

V

c c 0 ' 1

3

1 

12 . . . 2 . 0 ' 0 2 d b f b d V s c c        



s c



6 .

.

2

1

0 ' 3

d

b

f

V

c c c



















Punch

Q

(13)

= m

d

= KN/m²

= Direct Shear Load

= 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) . Ag = B x Fd = x = m² = KN 2) . = KN

Vc = Min ( Vc1 , Vc2 ) Allowable Shear Force

= KN

Check if (Shear Force) < Vc (Allowable Shear Force)

If ( Qshear <= Vc, OK , ERROR)

Qshear = < 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 = = m = KN/m² 1117.486 ( ( L - a ) / 2 ) 4 1833.128 1117.486 1833.128 2093.310 0.6 4 2.400 1.215 218.015 1.7 222.167 1833.128 Fd a w w B = Critical Section for Shear B = d Elevation q a X





L

X

q

ux



um in



um ax



um in

.

X

B

q

ux u

).

.

(

2

1

m ax



d

B

f

Ag

Pc

V

_c c

.

6

14

1

' 1













_



d

B

f

Ag

Pu

V

c c

.

6 .

3 .

0

1

' 2

















shear

Q

shear

Q

L

X

q

ux u u u 1 m in m ax m in 1





(



)

(14)

= +

= KN.m

Required Rn =

= 0.9 Strength Rediction Factor

" Reference ACI 318 - 02, clause 9.3.2"

Rn = KN/m²

=

" Reference ACI 318 - 02, clause 7.12.2.1"

If ( > , , )

Required As = x B x d

= x x 485

= mm²

Use Steel with 20 mm A = mm²

= round (( Required As / A ) + 1 ) Number of Steel Reinforcement Required =

= Spacing of Steel Reinforcement

= mm

Provided As = x A

= x

= mm²

If ( Provided As > Required As , "OK" ," Revise" ) OK 7853.98 SPb ( B - 2 cov ) / ( NPb - 1 ) 160 NPb 25 314.16 7503.98 314.16 NPb 25 0.003868 4 4 1284.123 44.629 1328.752 1569.127

0.0018 Minimum As Required for Footings of Uniform Thickness, for Grade 60 Reinforcement 1.7 0.0039 0.0039 4000 Elevation Fd a ww B = Critical Section for Flexure B = d q a X1=

)

3 /

)

.

).

((

2 /

)

.

(

m ax 1 1 2 1 1 1

q

X

B

q

B

X

M

ux



ux



u



ux 2 1

.

B

d

M

ux



















\ _\

.

85 .

0

2

1

1 .

85 .

0

c y c

f

Rn

f







m in







m in



m in











 





(15)

7.8.2 UPLIFT LOAD:

= Pu - Wp - Wf Load Carried by The Pad During Uplift

= - -= KN = / ( A - AP ) A : Area of Foundation = / AP : Area of Pedestal = KN/m² X1 = = m = KN.m

Required Rn = = 0.9 Strength Rediction Factor

" Reference ACI 318 - 02, clause 9.3.2"

Rn = KN/m² = If ( > , , ) Required As = x B x d = x x = mm²

Use Steel with 20 mm A = mm²

= round (( Required As / A ) + 1 ) Number of Steel Reinforcement Required =

= ( B - 2 cov ) / ( NPt - 1 ) Spacing of Steel Reinforcement

= mm

Provided As = x A

= x

=

If ( Provided As > Required As , "OK" ," Revise" ) OK 3492.00 3769.911 0.0018 0.0018 NPt 314.16 12 0.0016 314.16 348 12

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

Punet 1522.47 15.64 4000 485 NPt SPt 0.0018 97.345 ( ( L - a ) / 2 ) 1.7 664.437 1788.00 35.129 Punet 562.652 1522.47 230.4 net

qu

2

2 1

X

B

qu

Mu

net



net



2 1

.

B

d

M

_ux



















\ _\

.

85 .

0

2

1

1 .

85 .

0

c y c

f

Rn

f



_

_

m in







m in



m in











 





(16)

7.9 PEDESTAL DESIGN:

7.9.1 CALCULATIOM MOMENT:

H = h + Eh Total Height of Pedestal

= + = m = Pc x = x = K N.m "About X" = Pc x = x = K N.m "About Y"

= x Bending Moment @ Base due

= x

= K N.m "About X"

= x Bending Moment @ Base due

= x = K N.m "About Y" = Wp x ( / 2 ) = x = K N.m "About X" = Wp x ( / 2 ) = x = K N.m "About Y" = Pu x = x = K N.m "About X" = Pu x = x = K N.m "About Y"

Total Moment with Downward Force:

= - -= - -= K N.m = - -= - -= K N.m Pd = + = + = KN Ph-y 0.528 H 1823.13 Pc Wp 1788.00 35.13 1788.00 0.528 Upwnward Force Pc

1788.00 0.528 Upward Force Pc -5.52 9.28 9.28 363.00 9.28 Mx3 1272.00 1268.23 9.28 174.48

Myc My2 My1 My3

1452.00 1268.23 0.264 Weight of Chimney Mxc Mx2 Mx1 Mx4 My4 944.83 944.83

My3 Bending Moment due to

35.13 0.264 Weight of Chimney

Mx3 Bending Moment due to

35.13

to Horizontal Force Ph - x 1452.00 4.000 to Horizontal Force Ph - y 1272.00 H Ph-x 4.000 My2 318.00

3.5 0.5 Downward Force Pc 0.528 Downward Force Pc 1268.23 My1 1268.23 Mx2 2400.00 2400.00 4.000 Mx1 Wp Ph Wf G.L Ws Pc

Loads on Foundation when Subjected to Compression y

e

x

e

y

e

x

e

y

e

x

e

(17)

Total Moment with Upward Force: = - -= - -= K N.m = - -= - -= K N.m Pf = -= -= KN

7.9.2 ESTIMATE EQUIVALENT UNIAXIAL BENDING MOMENT:

h0 = m Section Height

b = m Section Wedth

= mm Vertical Bar Diameter

= mm Ties Diameter

d = h0 - cov - Sd - Effective Depth

= - - 12

-= mm

=

= K N.m

= KN.m

7.9.3 DESIGN FOR COMPRESSION:

Try to use : NP = 32 Total Number of Steel Reinfocement in Pedestal = 25 Vertical Bar Diameter

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

= mm Minimum Bar Spacing

S =

= < CHANG NO. OF BARS

As = Area of Steel Prvided

= 15707.963 mm² Asd 40 Smin 22.625 490.50 0.65 585.98 88.44 40 Sd 12 ( Asd / 2 ) 600 85 12.5 1752.87 0.6 0.6 Asd 25 497.88 Pu Wp 1788.00 35.13 My3 1452.00 944.83 9.28 317.88

Myu My2 My4

Mx3 1272.00 944.83 9.28 Mxu Mx2 Mx4 Wp Ph Wf G.L Ws Pu

Loads on Foundation when Subjected to Uplift cov Sd Asd h0 1



1 1

1

0 





















b

h

Myu

Mxu

Mu

1 1

1

0 





















b

h

Myc

Mxc

Mc

Asd

NP

Asd

Sd

h



2 

(cov



(

/

2 )))

_

.(

4

0

NP

Asd

_

4 .

2



(18)

= 1.4 / fy Minimum Steel Ratio

= 1.4 / 420 " Reference ACI 318 - 02, clause 10.5"

=

= Maximum Steel Ratio

= As /

=

Check if the Steel Ratio is : ≤ OK

= x 0.5 = = Mc / Pc = / = m e1 = d - + = - + = mm d1 = cov + Sd + = 85 + 12 + = mm m1 = = 420 / = Pc1 = = KN > KN If ( Pc1 <= Pc , "OK" , Error" ) OK 7.9.4 CHECK FOR UPLIFT:

Pf = KN Adjusted Uplift Load Test

Mu = K N.m

= 0.5 x As Area for Tension Steel

= 0.5 x

= mm²

= 0.5 x As Area for Compression Steel

= 0.5 x

= mm²

= Maximum Concrete Strain

E = Steel Elastic Modulus

0.003 200000 36.849 15707.963 7853.98 ( h0 x b ) 0.04363 585.98 Asc Ast 15707.963 7853.98 ( h0 / 2 ) 0.0033 0.05 0.03685 0.04363 0.02182 e copm. 88.44 2400.00 Asd / 2 490.50 300 227.35 e comp. 12.5 109.5 28 17.647 10146.69 2400.000 1752.87 m in  ac tual  m ax



m in



ac tual ac tual 



_{m ax} actual





0 .

5 

_{ }

            _ _ _             d e d d m d e d e d b fc 1 1 1 2 1 1 ' ) 1 ).( 1 ( . . 2 1 1 . . . 85 . 0



'

.

85 .

0

_c y

f

c



(19)

x = mm Distance of the Neutral Axis from the Compression Face

= fy / E

= 420 / =

= (( x - d1 ) / x ) x = Strain in the Compression Steel

fs = = Mpa

Cs = Asc . ( fs - 0.85 fc ) = KN Compression Force in Steel

Ts = Ast x fy = x 420 = KN Tension Force in Steel

Cc = = KN Compression Force in Concrete

Eq = Ts - Cc - Cs =

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

Xts = = mm Tension Steel Distance from the center

Xcs = = mm Compression Steel Distance from the center

Xc = d - Xst - = mm

= 0.9 * ( Ts.Xst + Cs . Xcs + Cc. Xc ) Section Moment Capacity when P = 0

= KN.m

= Section Axial Load when M = 0

= KN

= +

= <

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

Final Vertical Reinforcement for Pedestal:

NP = 32 Number of Steel Reinforcement Bars

= 25 Diameter of Steel Reinforcement Bars

0.5 * ( d -d1 )

5937.610

0.261 0.497

Asd

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 144.64 200000 0.0021 3298.672 190.5 ß2 * 0.5*x 145.759 957.867 7853.98 0.5 * ( d -d1 ) 190.5 0.85 x fc x ß2 x x x b 1755.604 585.201 0.00073 0.758 1

238.529 Center of Concrete Compression Block from the Center Muo 1106.670 Pu00 0.9 . ( fy . As ) y



s



c



y y s/ ) f (



1 . 1 0 1 . 1 00





























Mu

Pu

Pf

IR

(20)

7.9.5 TIES:

Use Steel Bars of Diameter 12 mm @ 200 mm Spacing

ds = 12 mm Diameter of Ties

Ass = Area of Ties

= 0

Ns = Number of Ties

=

7.10 CHECK FORCE TRANSFER AT INTERFACE OF COLUMN AND FOOTING:

1- Bearing Strength of Column:

fc = 28 MPa Compreesive Strength of the pedestal Concrete

=

A1 = Ap = 0

Bearing Strength of Column

" Reference ACI 318 - 02, clause 10.17"

= KN > OK

2- Bearing Strength of Footing:

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.

4 4 0.6 0.6 1.2 0.6 1.2 1.2 0.6 1.2 360000 5569.2 3595.756

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.

21 0.65 (( h + Eh - cov ) / 200 ) + 1 113.10 45 45 a w B = B = Fd= a Fd =

4

2 s

d





) . . 85 . 0 .( _f' _A₁ P_nb



_c







(21)

A1 = Ap = 0

A2 = ( + + ) x ( + + )

= 0

= > =

= KN > OK

7.11 REQUIRED DOWEL BARS BETWEEN COLUMN AND FOOTING:

" Reference ACI 318 - 02, clause 15.8.2.1"

= x Ap Ap Area of Pedestal

= x

= 0

Provide 5 Bars of 25 dia ( As = 0 )

7.12 DEVELOPMENT OF DOWEL REINFORCEMENT IN COMPREESION:

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

This length may be reduced to account for excess reinforcement.

As ( required ) / As ( provided ) = / = Reqiured dc = x = mm 1800 2290.9 496.08 1800 Asmin 0.005 0.005 360000

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

11138.4 3595.756

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.

9000000 5 2 2 360000 0.6 1.2 1.2 0.6 1.2 1.2 2290.9 0.786 420 200

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.

496.08 496.08 0.786 389.78 1 2

/ A

A

_

1 2

/ A

A

1 2 1 '

).

.

85 .

0 .(

A

f

P

nb



c





b c

d

f

f y

_



'

4

db fy  04 . 0



(22)

Available length for development in footing :

= Footing Thickness - Cover - 2.( Footing Bar Diameter ) - Dowel Bar Diameter

= - -

-= > OK Therefore, the dowels can be fully developed in the footing

7.13 CALCULATION MATERIAL QUANTITIES:

7.13.1 EXCAVATION: L = m B = m h = m Fd = mm H = h + Fd = m X = m = deg Y = H x tan ( s ) = x = m

Vs = Volume of Soil to be Excavated

Ab = Bottom Area = m² At = ( L + 2 X + 2 Y ) * ( B + 2 X + 2 Y ) Top Area = m² Vs = m³ 7.13.2 CONCRETE QUANTITY: L = m Footing Length B = m Footing Width h = m Footing Depth Fd = mm Thickness of Pad 4 4 3.5 600 0.5 20 450 389.78 600 85 40 25

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.

0.5 0.6 4 3.5 4 4 3.5 600 175.9 4.1 4.1 0.364 1.492 63.75 ( L + 2 X ) * ( B + 2 X ) 25 X 20 G.L Eh = Fd = B = h = X Y



At Ab At Ab



H _ _ _ 3

(23)

Eh = mm

300 mm structure pad

a = mm Pedestal Size

Volume of Concrete Pad Vf = L x B x Fd

= x x

= m³

Volume of Concrete Chimney Vp = a x a x x

= x x x

= m³

Total Volume of Concrete Vc = +

= +

= m³

7.13.3 BACKFILL:

Vbf = - - Volume of Exposed Height of Chimney

= -

-= -

-= m³

7.13.4 STEEL QUANTITY:

1- Chimney Rebars (Vertical Bars ):

OL = h + Eh + Fd = m m A1 = OL - Fd + dc = - + = m m A = A1 x = x = m B = 12 x = 12 x = m

LP = A + B Total Pedestal Vertical Bar Length

= +

= m

2- Chimney Rebars (Ties ):

a1 = a

-= - a1

= m

LS = Total Tie Length

= m a1 3- Footing Rebars: 2.02 0.43 2. cov 0.6 0.17 0.43 ( 4 x a1 ) + 0.3 0.43 25 0.30 4.46 0.30 4.76 4.390 1.016 4.46 Asd 0.30 4.6 0.6 0.3898 4.390 164.72 4.6 4.46 1.016 Vf Vp 9.600 1.464 1.464 11.064 11.064 ( a x a x ( Eh - 0.3 )) 4 4 0.6 0.6 0.6 4 (h+Eh) 500 600

Exposed Height, including the

175.9 11.064 0.072 Vs Vc 9.600 175.9

A =

B =

200



(24)

C1 = (( Fd - 2. cov ) / 2 ) + 0.1 = m C2 = L -= -= m = C2 + C1 C2 = C1= = + = m

= = m Total Pad Bar Length

7.13.5 STEEL WEIGHT:

1- Chimney Rebars (Vertical Bars ):

= 25 mm Diameter of Steel Reinforcement Bars

NP = 32 Number of Steel Reinforcement Bars

LP ( Total ) = LP x NP Total Pedestal Vertical Bars Length

= x 32

= m

= W : Weight of Chimney Rebar (Per Meter)

= x

= Kg

2- Chimney Rebars (Ties ):

ds = 12 mm Diameter of Ties

NS = 21 mm Number of Ties

LS ( Total ) = LS x NS

= x 21

= m

= W : Weight of Ties (Per Meter)

= x

= Kg

3- Footing Rebars:

= 20 mm Diameter for Bottom Steel in the Pad

= + x 2 Number of Pad Bars

= +

=

LPP ( Total ) = x

= x

= m

= W : Weight of Bars (Per Meter)

= x

= Kg

Saudi Services For Electro Mechanic Works Ltd.

S.S.E.M

Date:

May 26, 2014

By:

MAK 152.39 41.56 0.888 36.927 2.02 41.56 LPPW LS ( Total ) x W 74 74.00 330.04 330.04 2.468 814.539 ( LPB , LPT ) 4.46 NPP NPP ( NPb NPt ) 50 24 LSW LS ( Total ) x W 4.76 152.39 LPW LP ( Total ) x W 587.644 3.856 LPT LPB 4.46 LPB 0.63 3.83 4.46 ( 2 x C1 ) Asd 3.83 0.315 4 0.17 3.83 ( 2 x cov ) 0.315 pb



(25)

= KN = m = KN = m = KN = mm = KN = m = KN = mm = KN = mm = = mm = = mm = = = m³ = m³ = Kg = = m³ = Kg = = m³ = m³ = Kg = = C1 = C1 = a1= C1 = C1 = a1= =

DESIGN OF PAD & CHIMNEY FOUNDATION

0

Tower Type "A10" (Towers Include: # PI - 21)

SUMMURY FOR DESIGN

Fac tor ed L oad s Pc Pu Ph-x Ph-y Q M 363.00 318.00 482.59 0.00 2400.00 1788.00 B L Fd h Eh a Geom et ry 4.00 4.00 600.00 3.50 500.00 600.00 cov M at er ial Pr op er ties C f 420.00 25.00 deg 85.00 T 600.00 KN/m² KN/m³ Vs 175.9 Steel of Chimney NPt NPb 12 25 Top reinforcement for Pad 4.46 Total 0.30 C2 C2 3.83 3.83 0.315 0.315 0.43 0.43

QUANTITY SURVEY " One Footing"

Chimney Vertical Bars 1439 STEEL REINFORCEMENT CHIMNEY BACKFILLING Vbf 164.7 Vp Ties ( Stirrups ) 35.00 Bottom reinforcement for Pad PAD KN/m³ MPa MPa deg 24.00 17.00 0.00 28.00 EXCAVATION Steel of Pad

Total Weight of Concrete

LPPW 815 LPSW 625 STEEL CONCRETE Vf 10 Volume of Pad Vc 1 11 Volume of Chimney Total Volume of Concrete

GL B Fd L a a Ph-x Ph-y Pc OL Eh Chimney Pad h

A =

B =

200 c



c

f



y

f

1



S

