10.04.2013 Views

The Design of Modern Steel Bridges - TEDI

The Design of Modern Steel Bridges - TEDI

The Design of Modern Steel Bridges - TEDI

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

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.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!