Halfway up the Shard, London’s tallest skyscraper, you are asked to step out of the elevator at the transfer floor or “sky lobby,” a necessary inconvenience in order to reach the upper half of the building, and a symptom of the limits of elevators today. To ascend a mile-high (1.6km) tower using the same technology could necessitate changing elevators as many as 10 times because elevators traveling distances of more than 500m [1,640 ft] have not been feasible because the weight of the steel cables themselves becomes so great. Now BBC reports that after nine years of rigorous testing, Kone has released Ultrarope - a material composed of carbon-fiber covered in a friction-proof coating that weighs a seventh of the steel cables, making elevators of up to 1km (0.6 miles) in height feasible to build. Kone's creation was chosen to be installed in what's destined to become the world's tallest building, the Kingdom Tower in Jeddah, Saudi Arabia. When completed in 2020, the tower will stand a full kilometer in height, and will boast the world's tallest elevator at 660m (2,165ft). A 1km-tall tower may seem staggering, but is this the buildable limit? Most probably not, according to Dr Sang Dae Kim. “With Kingdom Tower we now have a design that reaches around 1 km in height. Later on, someone will push for 1 mile, and then 2 km,” says Kim adding that, technically speaking, a 2 km might be possible at the current time. “At this point in time we can build a tower that is 1 km, maybe 2 km. Any higher than that and we will have to do a lot of homework.”
Reply to: couldn't help myself
(Score: 2) by crutchy on Wednesday January 28 2015, @09:28AM
assuming typical mild steel grade 300 (elevator cables are no doubt higher), the allowable tensile stress would be around 0.6×300=180 MPa
take 1600 m of cable
7850×9.81×1600/1e6=123.2 N/mm^2 (MPa) tensile stress at the top due to the weight of the cable alone (irrespective of diameter)
so that leaves 180-123.2=56.8 MPa left over for payload
assume maximum acceleration in g's of about 2 (probably conservative), leaves 56.8/2=28.4 MPa left over
typical design load for a person might be something in the order of 120 kg averaged over a bunch of people. a 1.6 km express elevator gunna wanna hold a few people for a long round trip. lets say about 50-odd. that makes for a load limit of 50×120=6 metric tonnes => 58.9 kN
add maybe a tonne and a half (1.5×1×9.81=14.7 kN) for a rough guess at the weight of a large elevator cabin also, i dunno exactly what safety factors there are for elevators, but i imagine they're large lets say a factor of 2.5 for jolt and 2.5 for human occupancy makes for a design load of (58.9+14.7)×2.5×2.5=460 kN
transposing area of cable section=π×(diameter^2)/4 (assuming single round for simplicity) and pressure=force/area formulas
the diameter of round cable would need to be at least √[460×1000×4/(pi×28.4)]=144 mm
actually that doesn't seem too bad, but...
if you had 35 cables, the minimum diameter of each would still need to be √[460/35×1000×4/(π×28.4)]=24.3 mm
though for a 1.6 km elevator shaft having 35 cables might be considered reasonable merely for load path redundancy
dunno if this is all exactly right, but it was interesting to think about
(Score: 2) by crutchy on Wednesday January 28 2015, @09:28AM
density of steel is about 7850 kg/m^3
assuming typical mild steel grade 300 (elevator cables are no doubt higher), the allowable tensile stress would be around 0.6×300=180 MPa
take 1600 m of cable
7850×9.81×1600/1e6=123.2 N/mm^2 (MPa) tensile stress at the top due to the weight of the cable alone (irrespective of diameter)
so that leaves 180-123.2=56.8 MPa left over for payload
assume maximum acceleration in g's of about 2 (probably conservative), leaves 56.8/2=28.4 MPa left over
typical design load for a person might be something in the order of 120 kg averaged over a bunch of people. a 1.6 km express elevator gunna wanna hold a few people for a long round trip. lets say about 50-odd. that makes for a load limit of 50×120=6 metric tonnes => 58.9 kN
add maybe a tonne and a half (1.5×1×9.81=14.7 kN) for a rough guess at the weight of a large elevator cabin
also, i dunno exactly what safety factors there are for elevators, but i imagine they're large
lets say a factor of 2.5 for jolt and 2.5 for human occupancy makes for a design load of (58.9+14.7)×2.5×2.5=460 kN
transposing area of cable section=π×(diameter^2)/4 (assuming single round for simplicity) and pressure=force/area formulas
the diameter of round cable would need to be at least √[460×1000×4/(pi×28.4)]=144 mm
actually that doesn't seem too bad, but...
if you had 35 cables, the minimum diameter of each would still need to be √[460/35×1000×4/(π×28.4)]=24.3 mm
though for a 1.6 km elevator shaft having 35 cables might be considered reasonable merely for load path redundancy
dunno if this is all exactly right, but it was interesting to think about