The Design of Modern Steel Bridges - TEDI
The Design of Modern Steel Bridges - TEDI
The Design of Modern Steel Bridges - TEDI
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196 <strong>The</strong> <strong>Design</strong> <strong>of</strong> <strong>Modern</strong> <strong>Steel</strong> <strong>Bridges</strong><br />
ends. If the weight <strong>of</strong> the cable is uniform along its length, the deflected shape<br />
<strong>of</strong> the cable will be a catenary. <strong>The</strong> analysis <strong>of</strong> the catenary shape is much<br />
more complicated than that <strong>of</strong> a parabola, which would have been the deflected<br />
shape <strong>of</strong> the cable if its weight were uniformly distributed on the horizontal<br />
projected length <strong>of</strong> the cable. <strong>The</strong> error caused by this approximation on the<br />
length, sag or tension <strong>of</strong> the cable is found to be insignificant for practical<br />
implications in even the extreme real-life cases. <strong>The</strong> relationship between the<br />
sag and the tension in the cable can be obtained by equating the bending<br />
moment at mid-length <strong>of</strong> the cable to zero and is given by<br />
where<br />
f ¼ wL2<br />
8H<br />
¼ gC2<br />
8s<br />
f ¼ vertical sag<br />
w ¼ weight <strong>of</strong> cable per unit horizontal length<br />
L ¼ horizontal span between cable anchorages<br />
H ¼ horizontal component <strong>of</strong> the cable tension<br />
g ¼ weight per unit volume <strong>of</strong> the cable material<br />
C ¼ inclined distance between cable anchorages<br />
s ¼ tensile stress in the cable.<br />
ð7:1Þ<br />
It may be noted that the vertical sag <strong>of</strong> the cable is inversely proportional to<br />
the tension in it.<br />
Due to the sag the curved length <strong>of</strong> the cable is larger than the chord length by<br />
8f 2<br />
3L sec3 y<br />
where y is the angle <strong>of</strong> the cable with the horizontal.<br />
Due to the tension s the cable will have stretched elastically by an amount<br />
Cs/E; hence the length <strong>of</strong> the unstressed cable (i.e. lying on the ground)<br />
should be<br />
C þ<br />
3C sec2 Cs<br />
ð7:2Þ<br />
y E<br />
<strong>The</strong> self-weight <strong>of</strong> the cable reduces the cable tension in the bottom half <strong>of</strong><br />
the cable and increases the cable tension in the top half; the tension in the cable<br />
will be maximum at its top end and minimum at the bottom; the maximum<br />
tension is given by<br />
where<br />
8f 2<br />
H 1 þ h<br />
( )<br />
2<br />
1=2<br />
4f<br />
þ<br />
L L<br />
H ¼ horizontal component <strong>of</strong> the cable tension at the ends and<br />
h ¼ vertical distance between the cable anchorages.