tower design

ANALYSIS AND DESIGN OF A 30m HIGH SELF SUPPORTING FOUR LEGGED TELECOMIUNICATION TOWER

INSTRUCTED BYDr C.S. Lewangamage

NAME : NANAYAKKARA K. D.INDEX NO : 100353JDATE OF SUB : 21.01.2015

INTRODUCTIONThe self-supporting latticed tower can be considered as a cost effective structural system to mount telecommunication equipment to transmit radio waves to distance as much as possible. Also use of these self-supporting tower has following advantages. That are they dont need large area unlike for the guyed mast towers which requires to large area to fix guys. Also the self-supporting tower can further increase the high in future depending on requirement for additional fixities without changing the whole structure.Since in Yudaganawa area is rural place and has less signal for connect mobile network. Therefore the telecommunication tower is very important for this project since we are aiming to make this location as tourist attraction site. Therefore we need to make available for telecommunication facilities for this location to connect existing world. This proposed tower has four legs and mounted on foundation. Height of the tower is 30m and base area in .

(put the picture of tower)

The design requirement of tower is able to withstand design wind speed of 33.5 m/s. Because the according to wind zones of Sri Lanka, Yudaganawa is located in zone 3. Therefore the tower should be strong enough to resist this wind condition. Also the Yudaganawa is locate in southern province of Sri Lanka. Hence this area frequently experience the dramatic change of weather pattern because of the pressure drops in Bay of Bengal. Therefore this area has high possibility of subject to cyclone attack. Therefore to save expensive equipments from damaging, we need to adopt a method to dismantle form tower before the cyclone arrives to this area. Because we can detect cyclone early before it arrives. Therefore it is cost effective to adopt above method other than design tower to withstand cyclone. Hence the cyclonic resistance design is not considered. Also ice loading of tower is not considered since the Sri Lanka is tropical country and not experience any snow fall. The outline of this report included the analysis and design of the 30m tall self-supporting tower with able to withstand design wind speed of 33.5m/s. following details are also included in this report Structural details of the tower Design standards used of design Load calculation, SAP2000 modelling and design of 30m tall tower

STRUCTURAL DETAILS OF THE TOWER0

ReferenceDescriptionResults

Clause 2.6.5.1

Table 2-1

TOWER DESIGN NOTES

1. Tower designed for Exposure C2. Tower designed for basic wind speed of 33.5 m/s3. Structure Class II4. Topographic Category 1


Clause 2.6.9.1

Clause2.6.9.1.1

Table 2-6

WIND LOAD CALCULATION

Section 1

Total length of 75758 sections = 2000 2 = 4000 mm

Total length of 50505 sections = 2 (2828 + 2000) = 9656 mm

Solid Area = (4 0.075) + (9.656 0.05) = 0.7828 m2

Total Area of the Section 1 = 2 2 = 4 m2

Solidity Ratio () = 0.7828/4 = 0.1957

Design wind force (FST) = qz Gh (EPA)S

(EPA)S = Cf Df Af

whereFor Square Cross sections, Cf = 4.02 - 5.9 + 4.0 = 4 0.22 - 5.9 0.2 + 4 = 2.9985

Df = 1 (for Normal dir)

Af = 0.7828 m2

(EPA)S = 3.11 Df Af = 3.11 1 0.7828 = 2.347 m2

= 0.1957

Cf = 2.9985

Af = 0.78 m2

(EPA)S = 2.35m2


Clause 2.6.9.6

Clause 2.6.5.2

Table 2-4

Clause 2.6.6.4

Table 2-2

Table 2-3

Clause 2.67.1

Clause 2.6.9

qz = 0.613Kz Kzt Kd V2 l

Kz = 2.01(z/zg)2/

where z = 39 mzg = 274 m = 9.5Kzmin = 0.85

Hence Kz = 2.01(39/274)2/9.5 = 1.33 > 0.85 ok

Kzt = 1 (For Topography Category 1)

Kd = 0.85

V = 33.5 m/s (For Zone 3)

Importance factor (l) =1(for structure class II with wind only)

qz = 0.613 1.33 1 0.85 33.52 lqz = 779.68 N/m2

Gust effect factor (Gh) = 0.85 + 0.15[h/45.7-3] = 0.85 + 0.15 [40/45.7 -3] = 0.53

Design wind force (FST) = qz Gh (EPA)S = 779.68 0.53 2.35 = 972.33 N

Design wind force (Fw) = FST + FA + FGSince there is no Antennas and guys on this sectionFA & FG = 0

Therefore (Fw) = 972.33 N

Kz = 1.33

Kzt = 1

Kd = 0.85

l = 1

qz = 779.68 N/m2

Gh = 0.53

Fw = 0.97 kN


Clause 2.6.9.1

Clause 2.6.9.1.1

Table 2-6

Section 2 & 3

Total length of 75758 sections = 2000 4 = 8 m

Total length of 50505 sections = 4 (2828 + 2000) = 19.312 m

Solid Area = (8 0.075) + (19.312 0.05) = 1.5656 m2

Total Area of the Section 1 = 2 4 = 8 m2

Solidity Ratio () = 1.5656/8 = 0.196


(EPA)S = Cf Df Af



Af = 1.5656 m2

(EPA)S = 3.11 Df Af = 3.11 1 1.5656 = 4.69 m2

= 0.196

Cf = 2.999

Af = 1.5656 m2

(EPA)S = 4.69 m2


Clause 2.6.9.6

Clause 2.6.5.2

Table 2-4

Clause 2.6.6.4

Table 2-2

Table 2-3

Clause 2.67.1

Clause 2.6.9


Kz = 2.01(z/zg)2/

where z = 36 m (for Section 2) = 32 m (for Section 3)zg = 274 m = 9.5Kzmin = 0.85

Hence For Section 2, Kz = 2.01(36/274)2/9.5 = 1.31 > 0.85 ok Hence For Section 3, Kz = 2.01(32/274)2/9.5 = 1.28 > 0.85 ok


Kd = 0.85



For Section 2, qz = 0.613 1.31 1 0.85 33.52 l qz = 766.65 N/m2For Section 3, qz = 0.613 1.28 1 0.85 33.52 l qz = 747.87 N/m2


Design wind force (FST) = qz Gh (EPA)S For Section 2 (FST) = 766.65 0.53 4.69 = 1912.16 NFor Section 3 (FST) = 747.87 0.53 4.69 = 1865.32 N


Therefore for Section 2 (Fw) = 1912.16 N for Section 2 (Fw) = 1865.32 N

Kz = 1.31

Kz = 1.28

Kzt = 1

Kd = 0.85

L = 1

qz = 766.65 N/m2

qz = 747.87 N/m2

Gh = 0.53

Fw = 1.91 kNFw = 1.87 kN


Clause 2.6.9.1

Clause 2.6.9.1.1

Table 2-6

Section 4

Total length of 10010010 sections = 6018 2 = 12.036 m


Total length of 50505 sections=2(1075+1238)+2000+2325 = 8.951 m

Solid Area = (12.0360.1) + (15.2060.075) + (8.9510.05) = 2.7916 m2

Total Area of the Section 4 = 6 (2 + 2.65)/2 = 13.95 m2

Solidity Ratio () = 2.7916/13.95 = 0.2


(EPA)S = Cf Df Af



Af = 2.7916 m2

(EPA)S = 3.07 Df Af = 3.07 1 2.7916 = 8.32 m2

= 0.2

Cf = 2.98

Af = 2.7916 m2

(EPA)S = 8.3 m2


Clause 2.6.9.6

Clause 2.6.5.2

Table 2-4

Clause 2.6.6.4

Table 2-2

Table 2-3

Clause 2.67.1

Clause 2.6.9


Kz = 2.01(z/zg)2/


Hence For Section 2, Kz = 2.01(27/274)2/9.5 = 1.23 > 0.85 ok


Kd = 0.85



For Section 2, qz = 0.613 1.23 1 0.85 33.52 l qz = 721.6 N/m2




Therefore for Section 2 (Fw) = 3188.78 N

Kz = 1.23

Kzt = 1

Kd = 0.85

l = 1

qz = 721.6 N/m2

Gh = 0.53

Fw = 3.19 kN


Clause 2.6.9.1

Clause 2.6.9.1.1

Section 5



Total length of 50505 sections=2*(735 + 1563 + 1470 + 1787 + 735 + 899 + 1678 + 1798 +1827 + 899) +3300 + 2650 = 32.732 m

Solid Area = (24.070.1) + (27.4460.075) + (32.7320.05) = 6.1 m2

Total Area of the Section 5 = 12 (3.951 + 2.65)/2 = 39.606 m2

Solidity Ratio () = 6.1/39.606 = 0.15


(EPA)S = Cf Df Af


= 0.15

Cf = 3.19


Table 2-6

Clause 2.6.9.6

Clause 2.6.5.2

Table 2-4

Clause 2.6.6.4

Table 2-2

Table 2-3

Clause 2.67.1

Clause 2.6.9


Af = 6.1 m2

(EPA)S = Cf Df Af = 3.19 1 6.1 = 19.44 m2


Kz = 2.01(z/zg)2/




Kd = 0.85








Df = 1

Af = 6.1 m2

(EPA)S = 19.4m2

Kz = 1.13

Kzt = 1

Kd = 0.85

l = 1

qz = 662.56 N/m2

Gh = 0.53

Fw = 6.84 kN


Clause 2.6.9.1

Clause 2.6.9.1.1

Section 6



Total length of 50505 sections=2(1226+1962+2451+1912+1226+1062+1886+2125+1793+1063)= 33.412 m

Solid Area = (24.020.12) + (30.250.075) + (33.4120.05) = 56.82 m2

Total Area of the Section 6 = 12 (3.951 + 5.25)/2 = 55.2 m2

Solidity Ratio () = 6.82/55.2 = 0.12


(EPA)S = Cf Df Af


= 0.12

Cf = 3.35


Table 2-6

Clause 2.6.9.6

Clause 2.6.5.2

Table 2-4

Clause 2.6.6.4

Table 2-2

Table 2-3

Clause 2.67.1

Clause 2.6.9


Af = 6.82 m2

(EPA)S = Cf Df Af = 3.35 1 6.82 = 22.85 m2


Kz = 2.01(z/zg)2/




Kd = 0.85




Gust effect factor (Gh) = 0.85 + 0.15[h/45.7-3] = 0.85 + 0.15 [40/45.7 -3] = 0.53 (0.85 0.53 1.00) hence ok




Df = 1

Af =6.82 m2

(EPA)S = 22.85m2

Kz = 0.9

Kzt = 1

Kd = 0.85

l = 1

qz = 660.77 N/m2

Gh = 0.53

Fw = 0.54 kN

tower design

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