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Tension Structures

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Also known as tensile structures, tension

structure is a construction of elements carryingonly tension and no compression or bending .

Tensile membrane structures are most oftenused as roofs as they can economically andattractively span large distance.

Included in this group are compressive shells,tensile cable networks and air supported tensile-membranes structures.

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Tension Structure

hangers

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The cable- stayed bridge uses straight cables or rods toreduce the bending in the bridge deck

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Tensile Roof

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Millenium (O 2) Dome

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Renault Building, Swindon

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Air-supported Structures

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Surface Structures

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Folded plates

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Strength through shape

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Grandstands

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Northlight roof - avoid glare

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Hyperbolic Paraboloids

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HP Roofs

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Cooling Towers

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THTR 300

Thorium high-temperaturenuclear reactor

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Modern natural shapes

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Surface structures with stiffness

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Modern steel fabrication

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ties

Ties (members in tension) are the simplest of allmembers to design.The load required to break a tie is directlyproportional to its cross- sectional area.Effective area, A e design load/ designstrength

Ties containing holes

Effective area = (width bolt diameter) xthickness

Angle as tiesEffective area = (3a 1a 2/3a 1 + a 2 )+ a 1

a 2

a 1

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Example 1

A tie member in a pin jointed frame structure must supporta design load of 72 kN tension. Determine the requiredwidth of 6 mm thick flat mild steel bar.(Answer : 50 mm x 6 mm)

72 kN

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Example 2

Figure below shows part of a tower crane where thecounterweight is supported by inclined tie as shown. Thecharacteristic dead load of the counterweight is 150 kN.Ignore the self weight of the crane structure and otherloads that may be acting. Determine:a. The design tensile force in tieb. The required width of a 20 mm

thick mild steel tie if is to contain a

single row of 27 mm diameter boltholes(Answer : 100 mm x 20 mm)

8 m

6 m

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Example 3

Determine the dimensions of a standard mild steelequal angle to support a design load of 210 kN if itcontains 22 mm diameter bolt holes in one leg only.(Answer: 90 x 90 x 7 angle)

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Exercises

A mild steel tie member is required to support adesign tensile load of 150 kN. Compare the steel arearequired of the following two options:a) A 5 mm thick flat bar with a single row of 22 mm

diameter holes of bolts(round up width to thenearest 5 mm)b) An equal angle with only one leg of the angle

welded to form the connection.

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Cable structuresCable structure can be exciting,lightweight and highly efficient.It is usual to use cables made from avery high grade steel.

A key feature of all design involvingcables is that they are assumed tosupport only tensile loads.

Cable structures can be divided into twomain classes such asi) Lightweight cables with point loads

ii) Heavy cables with uniform loads

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Example 3(Lightweight cables with point loads) Figure shows a cable structure supporting overhead

electric railway lines. The loads are as follows:Each insulated hanger weighs an estimated 60 kgEach hanger supports a conducting line weighing

20 kg

There is the possibility of an extra imposed load oneach conducting line from snow and ice of 400 N.

All other loads, including the self-weight of thesupporting cable, may be considered to be negligible.Find:a. The design point loadsb. The unknown dimension xc. The tensile force throughout the supporting cabled. The required diameter of cable. (Assume an

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A cable structure supportingoverhead electric railway lines.

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Solutiona) The design point loads

Design dead load = 9.81 x (60 + 20) x 1.4 = 1099NDesign imposed load = 400 x 1.6 = 640 NTotal for each point load =1739 Nsay = 1.74kN

b) The unknown dimension x & tensile forces

throughout the supporting cables

A

BC D

E

FRAV

RAH

RFVRFH

x

0.5 m

1.74 kN1.74 kN 1.74 kN

1.74 kN

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SolutionPoint C

1.74 kN

FCB C FCD

l BC = (2.5 2 + 0.5 2) = 2.55 m

(V = 0) F CB x (0.5/2.55) = 1.74FCB = 8.87 kN

(H = 0) F CD = 8.87 x (2.5/ 2.55)FCD = 8. 7 kN

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SolutionPoint B

FBA

8.87 kN

Dimension ,x = 2 x 0.5 = 1.0 m

l AB = (2.5 2 + 1.0 2) = 2.69 m

(H = 0) F BA x (2.5/2.69) = 8.87 x (2.5/ 2.55)FBA = 9.36 kN

B

1.74 kN

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Solutiond) The required diameter of cable

Design stress = 1570 /5 = 314 N/mm 2

Area of cable required = Max force/ Stress= 9.36 x 10 3 / 314

= 29.8 mm 2

Cable diameter = 6.16 mmUse 7 mm diameter cable

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loadsFigure shows the Bosphorous Bridge in Turkey. The

main cables are made up from a bundle of small wire.

1074 m

Anchor block

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loadsThe following data applies :

Central span, L = 1074 mHeight of cable rise, D = 95 mConcrete anchor blocks = 42.0 m x 20.0 m x 12.0mSlope of cables at towers = same angle bothsides of towerNumber of main cables = 2Ultimate strength of cables = 1590 N/mm 2

Diameter of wires = 5 mmNumber of wires = 10412Dead load (cables + deck),G k = 147 kN/mImposed load, Q k = 20.3 kN/m

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loads

Determine:a. The design load for the suspended part of

the structureb. The total compression load, P, in one

supporting towerc. The maximum force, T, in the cabled. The value of the partial safety factor for

material strength, m, in the cable at fullload.e. The factor of safety against vertical uplift

of an anchor block.

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Solution

P

T T

T

RCH

RCV

RCH

RCVT

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Solutiona) The design load for the suspended part of the

structureDesign load = (1.4 x 147) + (1.6 x 20.3)

= 238.8 kN/mb) The total compression load, P, in one supporting

towerRCV = wl/2

= (238.8 x 1074)/2 = 128 236 kN

Load in tower = 2 x 128 236 = 256 472 kNc) The maximum force, T, in the cable

RCH= wl2 /8D = (238.8 x 1074 2)/(8x 95)= 362 434 kN

= 2 2 = 2

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SolutionMaximum cable force = 192 230 kN

d) The value of the partial safety factor for materialstrength, m, in the cable at full load.Area of steel in cable = x 2.5 2 x 10 412

= 204 466 mm 2

Stress = Force/Area = 192 230 x 10 3 /204466

= 940 N/mm 2m = 1590/ 940 = 1.69

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Solutione) The factor of safety against vertical uplift of an

anchor block.The unit weight of concrete is 24 kN/m 3Weight of anchor block = 42 x 20 x 12 x 24 = 241920 kN

RDV in one cable = 128 236/2 = 64 118 kN

Factor of safety against uplift = 241 920/64 118 =3.77