Funicular of forces• Divide the uniformly distributed load into a sufficient number of segments
• Assign a force to each segment
• The maximum force in the cable can be obtained by measuring the segment O‐I:
iAB FF 5,8
AR
BR
sinA
ABRF Or:
iF
O
I
For a given cable cross section and strength, the position of “O” can be determined graphically
Move the origin from “O” to “O’”
Note that doubling the distance between O and the vertical line the sag f is reduced by one half
However, the force in the cable has significantly increased!
Deflection: larger self weight induce smaller deflection due to concentrated load!
∙lRatio between point loadand self‐weight (unif. distributed)
Influence of the bridge self‐weight on the deflection caused by a
concentrated load – a graphic‐staticsapproach
Cable subjected only to concentratedload P• Consider a cable with sag f=L/5 subjected to a concentrated load P
Cable
• Consider a cable with sag f=L/5 subjected to a concentrated load P
Cable afterloading
Cable subjected only to concentratedload P
Deformation of the cable for different positions of ”Q” (analitically derived curves)
∙l
Ratio betweenpoint load and self‐weight (unif. distributed)
Position of ”Q” for maximum deflection (1/5 ofthe span)
How can we increase the stiffness of a cable?
in other words
how can we reduce de verticaldeflection of a cable caused by a pointload?
Methods to reduce the deformation of cable structure
Pre‐stress the hangers
Cable with a certain bending stiffness (beam/cable)
Increase self‐weight
Use a stiffer stiffening girder which can redistribute the concentrated load
What is the effect of the self weight?
Larger self‐weight induce larger tension “T” in the cable. “T” can be regarded as a pre‐stressing force.
The higher the pre‐ stress force, the stiffer is the cable (compare to a guitar string)
Light deck (small pre‐stress in the cable)
Large deformations to concentrated load
Heavy deck (large pre‐stress in the cable)
Small deformations to concentrated load
Increase the stiffness of the deck (in otherwords: use a stiff stiffneing girder)
”soft” suspension bridge
Increase the stiffness of the deck (in otherwords: use a stiff stiffneing girder)
suspension bridge with stiffeninggirder
If the deck is stiff enough, the shape of the cable remains a parabola under loading. This means that the cable must be subjected to uniformlydistributed load!
Suspension bridges with stiffening girder• The stiffening girder transforms the concentratedload into a distibuted set of equal vertical pullsthat are compatible with the shape of the cable
• All the differences between the actual loading and the loading that corresponds to the shape of the cable are absorbed by the beam
Melan’s theory• Hooke’s law applies for all the components
• Hangers closely placed (the suspending force can be considered as uniformly distributed)
• Stretching of the cable negligible
• Elongation of the hangers and misalignment negligible
Melan’sdifferential equation
"".1 beamQg MyHH Cable force equilibrium:
Derivate two times:
syHH Qg ''''.2
qxM
2
2
Remember that:
Melan’sdifferential equation
sqgIE IV .3Differential equation for the girder:
Put 2. in 3.: syHH Qg ''''.2
''''.4 yHHqgIE QgIV
Melan’sdifferential equation
''''''''.5 qqggIV HyHHyHqgIE
Equation 4. can be rewritten in the following way:
Melan’sdifferential equation
It can be assumed that the cable carries the self‐weight alone (i.e. the girder is not contributing to carry the self weight)
gyHMyH ggg ''.6
Put 6. in 5.:
''''
''''''.7
yHHHqIE
HyHHgqgIE
qqgIV
qqgIV
yHHHMM qqgq .8
Or integrating Eq. 7 two times:
M: Bending moment in the girder
Mq: Bending moment due to “q” acting on a simply supported beam
Note that for slender girders (i.e. EI/Lsmall) the equation of the cable alone leads to similar results as the Melan’sequation
Example: Severn bridge (UK)
‐ the bridge becomes more rigid (» 25%), due to the truss behaviour
‐ reduced tendency to oscillate (flutter).
‐ However, the constantly changing forces in the hangers can create fatigue problems
Towers/Pylons
• Rigid towers for multispan suspension bridges to provide enough stiffness to the bridge
• Flexible towers are commonly used in long‐span suspension bridges, ‐
• Rocker towers occasionally for relatively short‐span suspension bridges.
Great belt Bridge, Pylon
• Height above sea level: 254 m
• Legs at top: 6.5 m x 7.5 m
• Legs at base: 14 m x 15 m
• Caisson: 78 m x 35 m
• Wall thickness of legs: 1.7 m
• Concrete per pylon: 51,250 m3
Cables
• A cable is a highly flexible member
• A cable transmits primarily axial forces
• A cable can be made of : • a bundle of steel wires• A bundle of strands• A boundle of several cables
Wires• Are produced from high‐strength steel bars by rolling or cold drawing (the initial area is reduced)
• The cold‐forming process results in‐ an increase in the tensile and yield stress and‐ a decrease in the ductility of the
Diametre 1‐7 mm
Wires
• The optimum wire diameter is 5,0‐5,5mm
• A larger diameter makes the wire too stiff
• A smaller diameter requires more wires and more labour.
• The wire material has an ultimate strength up to 1600 ‐ 1800N/mm2
Strand• is produced from a series of wires that are wound together in a helical, parallel or Z‐lock fashion
Cables/ropesParallel wire cables are composed of a series of parallel wires
Strand cables are composed of parallel or helically combined strands
Locked‐coil cables (which were invented for better corrosion protection). These cables are less flexible than the othertypes.
Mechanical properties of cables• the tensile strength of the wires is high (it is also normally inversely proportional to the wire diameter). Normally fu= 1500‐1800 N/mm2.
• The diagram for this steel has no yield plateau and so the yield strength is conventionally defined as the stress at which the plastic deformation is 0,2%. fy= 80 to 90% of fu
• The modulus of elasticity of cables is generally smaller than that of the steel material (wire) of which they are composed
Stiffness of cables
• For parallel wire cables E = 200 N/mm2
• For locked‐coil cables E = 160 N/mm2
• For strand cables E = 150 N/mm2
Design values for cables (check producers web‐page)
Fu = ks Am . fuThe tensile resistance of a cable is:
Am is the “metallic area”of a cable:
ks = 0,76 – 1,0
f = 0,55 – 0,86
The design tensile resistance is
FRd = Fu /M (M = 2,0 for cables)
d=cable diameter
Attaching the cables –socketed fitting
The broomed end of the cable is then inserted into a conical steel basket
Molten zinc is poured into the basket to embed the wires and make the attachment