When the Eastern Span of the San Francisco Bay Bridge is replaced in mid-2013, it will be the world's longest self-anchored suspension bridge. To hold the cables that hold the road, crews have just installed a gargantuan 450-ton cable saddle.
The cable saddle is what holds the main suspension cables in place: basically a 900,000 pound, 13.7 foot tall shock absorber that sits on top of the tower, shifting the load from the cable to the tower. Without the saddles, these cables would take the brunt of the strain and likely
pinch (thanks Ben!) — not what you want from the mile-long 30-inch-wide steel rope (comprised of 17,399 steel wire strands) that holds up the side of one of the nation's busiest bridges.
Take away the saddles and a suspension bridge like this wouldn't work. Although the new Eastern span will run 1,542 feet, it will do so with just a single tower. The main cables attach at the either end of the span and run through the cable saddle at the tops of the tower with the rest of the structure hanging off of them. This massive saddle will hold the cable needed to support the deck. Crews hoisted it more than 500 feet (over water, mind you) to the top of the tower during its 12-hour installation last Thursday. The cables themselves will go up later this year. [via ABC News]
UPDATE: UPDATE: Eagle-eyed reader Ben Weaver, S.E. (California-licensed structural engineer) points out that the support cables would not actually snap straight away without the cable saddle, instead they would just pinch against the edges of the tower (which is so much better):
Basically the curve in the saddle enables a nice even pressure to develop along the length of the bearing area of the cable, and thus creating a larger bearing area. If there were no saddle (or no curved transition) the cable would bear on two discreet points, a very small bearing area, creating stress concentrations in the cable in those areas. A good analogy is pushing with 50 lbs each with the palm of your hand or with the tip of a pencil. The area of your palm is, say, 12 sq in, while the area of the tip of a pencil is say 1/4000 sq in. Thus the resulting stresses are 50/12=2.5 psi and 50/(1/4,000)=200,000 psi. You get the picture.
Top image courtesy of Bay Bridge Public Information Office