Structural Evaluation of Self Supporting Tower (Calculation)__Yoppy Soleman

September 4, 2017 | Author: Yoppy Soleman | Category: Structural Analysis, Truss, Strength Of Materials, Bending, Structural Load
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Descripción: analysis of the structure of the telecommunication tower...

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Space Truss Design

SPACE TRUSS DESIGN

1. DESIGN SPECIFICATION 1.1. Design Standard 1)

The design basis of the tower applied is EIA Standard EIA-222-E “Structural Standards for Steel Antenna Tower and Antenna Supporting Structure”. The fabrication and materials of the tower will be according to the relevant Indonesian Standard.

2) The self supporting tower has square cross sections. 3) All the legs and bracings are made of equals legs angles steel. 4) All the connections in the field are made with Steel Bolts, each fitted with one spring washer and nut. 1.2. Tower Structure Design Condition 1) Tower height : 42.0 meter ( location : Limboto, North Sulawesi ) 2) Maximum wind velocity (V) : V = 120 km/hour = 33.33 m/sec. 3) Existing antennas loading ( see the drawing attachment ) : • 2 (two) Planar type antennas at 42.0 m • 1 (one) Planar type antennas at 38.0 m • 1 (one) Paraboloid grid antennas 1.20 diameter at 35.0 m • 1 (one) Paraboloid grid antennas 1.20 diameter at 42.0 m 4) Proposed antennas loading ( see the drawing attachment ) : • 1 (one) Paraboloid solid antenna 1.2 diameter at 38.0 m 1.3. Loads 1) Dead load Dead load is weight of tower, antenna, ladder, platform etc. 2) Wind load on tower structure Wind load calculation method on the tower and appurtenances are as follows F

= qz . GH . CF . AE and not to exceed 2 . qz . GH. AG

qz

= 0.613 . KZ . V2

Kz = ( z / 10 )2/7 GH = 0.65 + 0.60 / ( h / 10 )1/7 CF = 4.0 e2 – 5.9 e + 4.0 ( square cross section ) CF = 3.4 e2 – 4.7 e + 3.4 ( triangular cross section ) e

= AF / AG

AE = DF . AF Where : F

= Horizontal wind force ( N )

qz

= Velocity pressure ( Pa )

GH = Gust response factor ( 1.00 ≤ Kz ≤ 1.25 ) CF = Structure force coefficient AE = Effective projected area of structural component in one face ( m2 ) AG = Gross area of one tower face ( m2 ) Kz = Exposure coefficient ( 1.00 ≤ Kz ≤ 2.58 ) V

= Basic wind speed for the structure location ( m/s )

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

z

= Height above average ground level to midpoint of the section ( m )

h

= Ttotal height of structure ( m )

e

= Solidity ratio

AF = Projected area of flat structural component in one face of the section (m2) DF = Wind direction factor 1.00

for square cross section and normal wind direction

1.00 + 0.75e

for square cross section and 450 wind direction

3) Wind load on Antenna Wind load calculation method on the parabolic antenna is as follow : Fa = Ca x A x Kz x GH x V2 Fs = Cs x A x Kz x GH x V2 Kz = ( z /10 ) 2/7 GH = 0.65 + 0.60 / (h/10) 1/7 Where : Fa = Axial Force (lb) Fs = Side Force (lb) Ca = Wind load coefficient for axial Cs = Wind load coefficient for side Kz = Exposure coefficient ( 1.00 ≤ Kz ≤ 2.58 ) z = Height above average ground level to midpoint of the section (m) h = Total height of the structure (m) A = Normal projected area of Antenna V = Wind velocity ( m/s ) 4) Load combination Herewith the following combinations are used below : a) DL + WL at 0 degree direction (with weight of existing antenna) b) DL + WL at 45 degree direction (with weight of existing antenna) c) DL + WL at 0 degree direction (with weight of existing + proposed antenna) d) DL + WL at 45 degree direction (with weight of existing + proposed antenna) Where : DL

= Dead load weight of the structure and appurtenances.

WL

= Design wind load on antenna at above direction.

1.4. Allowable unit stress The unit stresses in the structures members do not exceed the allowable unit stresses for the materials as specified in the AISC Standard (American Institute of Steel Construction Standard) 1. Tension

: Ft = 0.60 Fy ( kg/cm2 )

2. Shear

: Fv = 0.40 Fy ( kg/cm2 )

3. Compression i) On the gross section of axially loaded compression members when kl/r is less than Cc : (kl/r)2 [ 1 - ----------] Fy 2Cc2 Fa = -----------------------------------------------

( kg/cm2 )

5/3 + [3/8(kl/r)]/8Cc - [(kl/r)3/8Cc3]

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

2π2E Where: Cc =  --------Fy ii) On the gross section of axially loaded compression members, when kl/r exceeds Cc : 12π2 E Fa = ---------------

( kg/cm2 )

23(kl/r)2 4. Bending Tension and compression on extreme fibers : Fb = 0.66 Fy ( kg/cm2 ) 5. Tension on bolts : Ft = 0.60 Fy ( kg/cm2 ) 6. Shear on bolts : Ft = 0.30 Fy ( kg/cm2 ) 7. Bearing on bolts : Ft = 1.20 Fu ( kg/cm2 ) 8. The maximum slenderness ratio (kl/r) are as follows : kl/r = 120 for compression members of legs kl/r = 150 for compression members of diagonals kl/r = 200 for tension members Notations : Ft = Allowable tensile stress ( kg /cm2 ) Fy = Minimum yield point ( kg /cm2 ) Fv = Allowable shear stress ( kg /cm2 ) Fa = Allowable compressive stress ( kg /cm2 ) k = Effective length factor l = Actual unbraced length of member ( cm ) r = Governing radius of gyration ( cm ) Cc = Column slenderness ratio E = Modulus of elascity of steel = 2,100,000 kg/cm2 Fb = Allowable bending stress ( kg /cm2 ) Fu = Minimum tensile strength ( kg /cm2 ) 1.5. Materials Steel materials to be used for the towers and appurtenances conform to the relevant Indonesian Standards and/or Japanese Industrial Standard. 1) Steel Structural Description

Tensile Strength

Minimum Yield Point Fy

( kg/cm2 )

( kg/cm2 )

Bj – 41

4100

2500

SS – 41

4100

2500

2) Bolts Description

Engineering Postgraduate Program Hasanuddin University

Ft

17

Fv

Fv

Friction Type

Bearing Type

B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

A – 325 Bolts

( kg/cm2 )

( kg/cm2 )

( kg/cm2 )

3900

1230

1476

3) Concrete Design compressive strength of concrete (f’c) at 28 days. K - 175

f’c = 175 kg/cm2

-

4) Reinforcement steel U - 24

-

Fy = 2400 kg/cm2

1.6. Structural Analysis The purpose of the structural analysis is to find the joint translations and the design axial loads in all members of the tower. Load is applied and separate load cases combined to give the most severe design conditions at various section. The structural calculation is made using SAP 90 (Structural Analysis Program 90). The program will perform the static analysis of a space truss of arbitrary geometry by the stiffness method. The truss may be subjected to loads consisting of forces acting on the joints in any directions in space. The program output consists of the joint translations, the member forces and the support reactions. The program input contains : a. Structure title b. Loading system : number of static analysis that applied to the structure. c. Group of data corresponding to the properties of the mathematical model of truss and the applied joint load : •

Group 1 : Joint coordinates



Group 2 : Support joint restraints



Group 3 : Material and member data



Group 4 : Joint loads



Group 5 : Loading combinations

The location of the joints in any structure are expressed as coordinates in a global right hand othogonal XYZ coordinate system. For the space structures the Z axis is oriented in the vertical direction positive upward, with the X and Y axes oriented in the major directions of the structure. Z+ Y+

Global Axis 0

X+

All applied joint loads, joint displacement and reactions are expressed as component in the global coordinate system. Force component and translation components are positive if they act in the positive direction of an axis. The member forces and support reactions for both conditions, tower with existing antennas and tower with existing and proposed antennas, are attached in computer output. 1.7. Design Calculation Of Foundation The calculation of foundation consists of design and control of foundation. Control of foundation includes : 1) Control of stability for uplift force :

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

Sf = W1 / T > 2.0 Where : W1= Weight of foundation and soil ( kg ) T

= Uplift force ( kg )

2.0

= Allowable safety factor

2) Control of bearing capacity of soil : Wt M F = -------- + --------------- < Q ( kg/m2) A 1/6.A . B Where :

Wt = Total vertical load includes support reaction, weight of foundation and weight of soil (kg) M = Moment load ( horisontal loads x height of foundations ) ( kgm ) A =

Area of the foundation base ( width x length of foundation ) (m2)

B =

Width of the foundation base ( m )

Q =

Allowable bearing capacity of soil.

3) Control of sliding force : SF = Wt . φ / H > 1.5 Where : SF = Wt =

Safety factor Total vertical load includes support reaction, weight of foundation

and weight of soil (kg) φ

= Coefficient of soil friction

H = Horisontal loads ( kg ) 1.5 = Allowable safety factor 2. STRUCTURAL CALCULATION The structural analysis is made using SAP 90. Input and output program is shown as attachment.

Deflection, sway and twist are calculated as follows : a. Deflection

: Dxn : Joint displacement at a point n Dxn’ : Joint displacement at a point n’ Dxn – Dxn’

b. Sway angle

= arc tan ( --------------------------------------------------------- ) Distance between point n and point n’ Dxn – Dxn’

c. Twist angle

= arc tan ( ---------------------------------------------------------- ) Distance between point n and point n’

1) Tower without proposed antenna a. Deflection

= 6.4177 cm

b. Dxn

= 5.2096 cm

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

Dxn’

= 5.8865 cm

d

= 250 cm

Sway angle

= arc tan (( 5.8865 – 5.2096 ) / 250 ) = 0.1551 degree

c. Dxn

= 5.8865 cm

Dxn’

= 6.4173 cm

d

= 300 cm

Twist angle

= arc tan (( 6.4173 – 5.8865 ) / 300 ) = 0.1014 degree

2) Tower with proposed antenna a. Deflection

= 6.5947 cm

b. Dxn

= 5.2534 cm

Dxn’

= 5.9388 cm

d

= 250 cm

Sway angle

= arc tan (( 5.9388 – 5.2534 ) / 250 ) = 0.1570 degree

c. Dxn

= 6.4763 cm

Dxn’

= 5.9388 cm

d

= 300 cm

Twist angle

= arc tan (( 6.4763 – 5.9388 ) / 300 ) = 0.1027 degree

Sway and twist at 120 km/hour wind velocity without proposed antennas as follows : Deflection (cm)

Actual 6.4177

Allowable 42

Sway angle (degree) Twist angle (degree)

0.1551 0.1014

0.5 0.5

Sway and twist at 120 km/hour wind velocity with proposed antennas as follows : Deflection (cm)

Actual 6.5947

Allowable 42

Sway angle (degree) Twist angle (degree)

0.1570 0.1027

0.5 0.5

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

3. FOUNDATION ANALYSIS 3.1. Column Anchorage Bolt Calculation 1 1) Steel Bar Bj 37

--------- Fy = 2400 kg / cm2

2 2 ) Notation : Fy

= Yield strength of steel

Fv

= Allowable shear strength of anchor bolt

Ft

= Allowable tensile stress of anchor bolt

Fts = Allowable tensile stress for bolt subject to combine tension and stress Fcv = Allowable bond stress of concrete fv

= Actual shear stress of anchor bolt

ft

= Actual tensile stress of anchor bolt

f’c

= Compressive strength of concrete

A

= Total area of anchor bolt

P

= Total compression of tower base per one leg

T

= Total uplift force at tower base per one leg

S

= Total shear force at tower base per one leg

Le

= Required embeded length of anchor bolt in concrete

3 3 ) Maximum forces at tower base a.

Tower with existing antenna : T = 21110 – 488.09 = 20621.91 kg S = 2377 kg

b.

Tower with existing and proposed antenna : T = 21110 – 488.09 = 20621.09 kg S = 2387 kg

1 4 ) Allowable tensile stress of anchor bolts Fv

= 0.3 Fy = 0.3 x 2400 = 720 kg/cm2

Ft

= 0.6 Fy = 0.6 x 2400 = 1440 kg/cm2

a.

Tower with existing antenna : Number of anchor bolt = 6 φ ¾ “ A

= 6 x ( 0.25π x 1.9052 ) = 6 x 2.85 = 17.1 kg/cm2

fv

= S / A = 2377 / 17.1 = 139.0 kg/cm2 < Fv ………….Ok !

Fts = 1.4 Ft – 1.6 fv

= ( 1.4 x 1440 ) – ( 1.6 x 139.0 ) = 2016 – 222.40 = 1793.60 kg

Fts > Ft ---------------ft b.

use Ft = Fts = 1440 kg/cm2

= T / A = 20621.09 / 17.1 = 1205.91 kg/cm2 < Ft ……… Ok !

Tower with existing and proposed antenna : Number of anchor bolt = 6 φ ¾ “ ----- A = 17.1 cm2 fv

= S / A = 2387 / 17.1 = 139.59 kg/cm2 < Fv ……………. Ok !

Fts = ( 1.4 x 1440 ) – ( 1.6 x 139.59 ) = 2016 – 223.344 = 1792.656 kg/cm2 Fts > Ft ---------------ft

use Fts = Ft = 1440 kg/cm2

= T / A = 20621.09 / 17.1 = 1205.91 kg/cm2 < Ft ……… Ok !

Keep using anchor bolt 6 φ ¾ “ Required embedded length of anchor bolt :

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

Fcv

= 0.53  f’c = 0.53  175 = 7.0 kg/cm2

Le

= T / ( Fcv x 6 x π x d ) = 20621.09 / ( 7 x 6 x 3.14 x 1.905 ) = 82.1 cm

Use

Le = 85 cm

3.2. Column Base Plate 1) Steel Concrete

: Bj – 37

Fy = 2500 kg/cm2

: K – 175

Fp = 0,35 f’c = 0,35 x 175 = 61.25 kg/cm2

2) The formula to calculate column base plate is shown as follows : Ar = P / Fp

( m2 )

Ab ≥ Ar then check fp ≤ Fp Ab = B x B t = ( 6M / Fb )½ Fb = 0.75 Fy = 0.75 x 2500 = 1875 kg/cm2 Where :

P

= Total compression at tower base per one leg ( kg )

Ar

= Required area of column base plate ( m2 )

Ab = Designed area of column base plate ( m2 ) B

= Length of base plate ( cm )

fp

= Actual bearing pressure ( kg/cm2 )

Fp

= Allowable bearing strength stress ( kg/m2 )

tp

= Required thickness of base plate ( cm, mm)

M

= Moment at the edge of base plate ( kgm, kgcm)

Fb

= Allowable bending stress of base plate ( kg/cm2 )

Fy

= Yield strength of steel ( kg/cm2 )

f’c

= Compressive strength of concrete ( kg/cm2 )

m

= Distance from steel structural to the edges of base plate ( cm )

fb

= Bending stress ( kg/cm2 )

The calculation is shown as follows below : a. Tower without proposed antennas Column base plate area The existing column base plate : 600 mm x 600 mm x 25 mm Maximum compression force ( P ) = 26980 kg Applied load at support join = 488.09 kg P Total = 26980 + 488.09 = 27468.09 kg A = 60 x 60 = 3600 cm2 fp = P / A = 26980 / 3600 = 7.494 kg/cm2 < Fp …………. Ok ! Column base plate thickness Use m = (60 – 15) / 2 cm = 22.5 cm M = ½ q m2 = ½ x 7.494 x 22.52 = 1896.92 kgcm check the stress : fb = ( 6M / tp2 )= (6 x 1896.92 / 2.52 ) = 1821.042 kg/cm2 < Fb (1875 kg/cm2 )………….Ok ! b. Tower with proposed antennas Maximum compression force ( P ) = 27200 kg Applied load at support join = 488.09 kg P Total = 27200 + 488.09 = 27688.09 kg fp = P / A = 27688.09 / 3600 = 7.691 kg/cm2 < Fp …………. Ok ! Column base plate thickness

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

Use m = (60 – 15) / 2 cm = 22.5 cm M = ½ q m2 = ½ x 7.691 x 22.52 = 1946.78 kgcm check the stress : fb = ( 6M / tp2 )= (6 x 1946.78 / 2.52 ) = 1868.91 kg/cm2 < Fb (1875 kg/cm2 )………….Ok ! Keep using column base plate : 600 mm x 600 mm x 25 mm

3.3. Design and Control Of Foundation 3.3.1. Tower with existing antennas 1) Design load :

H

= 2377

kg

( max horizontal reaction )

V

= 26980 kg

( max vertical reaction )

T

= 21110 kg

( max uplift reaction )

V1

= 488.09 kg

( dead load at support join )

P

= V + V1 = 26980 + 488.09 = 27468.09 kg

Tt

= T – V1 = 21110 – 488.09 = 20621.91 kg

From data above the design foundation will be checked for uplift force, bearing capacity of soil and horizontal loads (sliding). Design of foundation : 800

200

GroungLevel 1950

Soil

2850

700 3000

2) Check stability for uplift force Concrete volume ( Vc ) : Pedestal column : 0.80 x 0.80 x 2.15

= 1.376

m3

Footing : 3.0 x 3.0 x 0.70

= 6.300

m3

= 7.676

m3

Soil volume for anti uplifting ( Vs ) : Vs

= (( 3.0 x 3.0 ) – ( 0.80 x 0.80 )) x 1.95= 16.30

m3

Weight of concrete and soil : W1 = W+ Ws = 7.676 x 2.4 + 16.30 x 1.6

= 44.5024 t

S.F = W1 / T = 44.502 / 21.110 = 2.11 > 2.0 ………………..Ok ! 3 ) Bearing capacity of soil The allowable bearing capacity of soil is 0.267 kg/cm2 = 2.67 t/m2 ( Bearing capacity data was gathered from Tower Name / Date Plate ) 4 ) Check of compressive force Wt = 44.502 + 27.468

= 71.970 t

M

= 6.774

= 2.377 x 2.85

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tm

B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

Z

= Section modulus of footing base

Z

= 3.0 x 3.0 x 3.0 / 6

fe

= Compressive stress of footing base

A

= Area of foundation base = 3.00 x 3.00

fe

= 71.970/ 9.0 + 6.774 / 4.5 = 7.999 t/m2 > 2.67 t/m2………….Fail

= 4.500

m3

= 9.00

m2

The dimension of foundation is designed based on the nomogram. As shown in calculation above, the bearing capacity of soil is unable to support the existing tower. In fact, the soil is bearable. Possibly this is due to the difference in type and dimension between the existing tower foundation and the designed foundation above. 5) Factor of safety against sliding Wt

= 71.970 t

H

= 2.377 t

θ

= Coefficient of friction = 0.45

SF

= Wt x θ / H = 71.970 x 0.45 / 2.377 = 13.62 > 1.50 …………. Ok !

3.3.2. Tower with existing and proposed antennas 1) Design load :

H

= 2387 kg

( max horizontal reaction )

V

= 27200 kg

( max vertical reaction )

T

= 21160 kg

( max uplift reaction )

V1

= 488.09 kg

( dead load at support join )

P

= V + V1 = 27200+ 488.09 = 27688.09 kg

Tt

= T – V1 = 21160 – 488.09 = 20671.09 kg

2) Check stability for uplift force Concrete volume ( Vc ) : Pedestal column : 0.80 x 0.80 x 2.15

= 1.376

m3

Footing : 3.0 x 3.0 x 0.70

= 6.300

m3

= 7.676

m3

Soil volume for anti uplifting ( Vs ) : Vs

= (( 3.0 x 3.0 ) – ( 0.80 x 0.80 )) x 1.95= 16.30

m3

Weight of concrete and soil : W1 = W+ Ws = 7.676 x 2.4 + 16.30 x 1.6

= 44.502 t

S.F = W1 / T = 44.502 / 21.160 = 2.103 > 2.0 ………………..Ok ! 3 ) Bearing capacity of soil The allowable bearing capacity of soil is 0.267 kg/cm2 = 2.67 t/m2 4 ) Check of compressive force Wt = 44.502 + 27.688

= 72.190 t

M

= 2.387 x 2.85

= 6.803

tm

Z

= Section modulus of footing base

Z

= 3.0 x 3.0 x 3.0 / 6

= 4.500

m3

fe

= Compressive stress of footing base

A

= Area of foundation base = 3.00 x 3.00

= 9.00

m2

fe

= 72.190/ 9.00 + 6.803 / 4.500= 8.022 t/m2 > 2.67 t/m2………….Fail

5) Factor of safety against sliding Wt

= 72.190 t

Engineering Postgraduate Program Hasanuddin University

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

H

= 2.387 t

θ

= Coefficient of friction = 0.45

SF

= Wt x θ / H = 72.190 x 0.45 / 2.387 = 13,61 > 1.50 …………. Ok !

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

4. CONCLUSION AND RECOMMENDATION We have carefully analysed the existing tower of Limboto structure for the proposed additional antenna at 120 km/hr wind velocity. The following major conclusions have been drawn from this analysis :

-

The existing tower has strength enough to support the existing configuration and the proposed antennas at 120 km/hr maximum wind velocity.

-

The anchor bolt and the base plate has strength enough to resist the forces at support joint. Additional force at the tower base (maximum) due to the proposed antennas is less than 2.10 % of support reaction at tower without proposed antennas.

-

The designed foundation has strength enough to resist the uplift and shear forces. The designed foundation has not strength enough to resist the compressive force. It means that the bearing capacity of soil is unable to support the structure. This is, possibly, due to the difference in type and dimension between the existing tower foundation and the designed foundation.

-

The minimum required of bearing capacity to support the tower with existing and proposed antennas is about 8.0 t/m2.

Luwuk, Januari 2001

Yoppy Soleman

Engineering Postgraduate Program Hasanuddin University

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

Space Truss Design

800 200

GL

1950

Soil

2850

700 3000

Engineering Postgraduate Program Hasanuddin University

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B. OF DUCTILE STEEL STRUCTURES.  Yoppy Soleman, 2005

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