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CASE STUDY ON
DESIGN OF BIALONG ELEVATOR
By – Sumit Salunke
- Neha Shah
- Madhura Mali
- Malhar Sure
- Sneha Ramkrishnan
Guided by – Prof. Venugopal
CONTENTS
 INTRODUCTION
 COMPONENTS OF ELEVATOR
 KNOWN DATA’S
 STANDARD CHARTS
 CALCULATION OF DIAMETER OF THE WIRE ROPE
 CALCULATION OF STRESSES INVOLVED
 DESIGN OF CAR(ELEVATOR)
INTRODUCTION
• The Bailong Elevator (literally Hundred Dragons
Elevator) is a glass elevator built onto the side of a
huge cliff in the Wulingyuan area
of Zhangjiajie, People's Republic of China that is
326 m (1,070 ft) high.
• It is claimed to be the highest and heaviest outdoor
elevator in the world.
• On 16 July 2015, the elevator was officially
recognised by Guinness World Records as the
world’s tallest outdoor lift
• An electric lift of the kind we are examining is basically a
passenger car in a sling, to the top of which is attached a set of
wire ropes.
• The ropes pass up the lift well and over a driving pulley called a
'sheave’ on the shaft of a motor and brake unit, then down the well
again to a counterweight, which balances the weight of the car
and sling, together with some &50% of the full rated load.
• Sometimes, the ropes are not fastened directly to the sling or
counterweight, but pass around pulleys attached to them, and are
then fastened at the head of the well.
COMPONENTS OF ELEVATOR
KNOWN DATA’S
Load capacity = 4900kg x 9.18 = 48069N
Depth = 335m
Rope speed = 5m/s
Time = 120s
Designation of Wire Ropes
The wire ropes are designated by the number of strands and the number of wires in each
strand.
For example, a wire rope having six strands and seven wires in each strand is designated
× 7
rope.
STANDARD CHARTS
CALCULATION OF DIAMETER OF THE WIRE ROPE
 From table 20.7, for elevator we can take type 6 x 19 or 8 x 19
Considering 6 x 19 rope type
 FOS for elevator is generally 11
Therefore, design load for wire rope = 11 x 48069
= 528759 N
 From 20.7 table, tensile strength of 6 x 19 with 1250-1400Mpa is 435d2 (in N) where, d is
rope diameter in mm
Equating tensile strength to design load ,
435d2 = 528759 N
d = 34.86mm = 35mm
 From 20.10 for 6 x 19 ,
Wire diameter (dw) = 0.063d
Area of wire rope(A) = 0.38d2
.`. dw = 0.063 x 35 = 2.205mm
And A = 0.38 X 352 = 465.5mm2
 Various loads on ropes –
• From 20.7 , weight of rope
W = 0.0383d2
= 0.383(35)2
= 46.91 N/m
= 46.91 x 335 N …..(depth = 335m)
= 15717.36 N
• From 20.12 , sheave diameter for wire rope for elevators
Min sheave dia = 20d
Max sheave dia = 30d
But, as it is for large distance and heavy load,
we will take D = 100d
= 100 x 35
= 3500mm
Bending stress, σb = (Er x dw) / D
= ( 84000 x 2.2 ) / 3500
= 52.8 N/mm2
Er = Modulus of elasticity
Dw = Dia of wire
D = Dia of sheave drum
Equivalent bending load on rope Wb = σb x A = 52.8 x 465.5
= 24578.4 N
• Acceleration of rope and load a = v / t = 5/120
= 0.046 m/s2
Additional load due to acceleration Wa = (W+w)a / g
= (48069+15717.36)0.046 / 9.81
= 300 N
• Impact load during starting
Wst = 2(W+w) = 2(48069+15717.36)
= 127572.72 N
 Effective load on rope during normal working = W + w + Wb
= 48069+15717.36+24578.4
= 88364.76N
FoS during normal working = 528759 / 88364.76
= 6
Effective load on rope during starting = Wst + Wb
= 127572.72+24578.4
= 152151.12 N
FoS During starting = 528759 / 152151.12
= 3.7
Effective load on rope during acceleration of load
= W + w + Wb + Wa
= 48069 + 15717.36 + 24578.4 + 300
= 88664.76
FoS during acceleration = 528759 / 88664.76
= 6
Since above FoS as calculated are safe,
.`. wire rope of dia 35mm and 6 x 19 type is satisfactory.
STRESSES INVOLVED
1. Direct stress due to axial load lifted and weight of the rope
.
W = Load lifted,
w = Weight of the rope, and
A = Net cross-sectional area of the rope.
 direct stress = (W+w) / A
 = (48069 + 15717.36 ) / 465.5
 = 137.02 N/mm2
2. Bending stress when the rope winds round the sheave or drum.
σb = (Er x dw) / D
= ( 84000 x 2.2 ) / 3500
= 52.8 N/mm2
The bending stress induced depends upon many factors
such as construction of rope, size of wire, type of centre and the amount of restraint in the
grooves.
The approximate value of the bending stress in the wire as proposed by Reuleaux, is
Er = Modulus of elasticity of the wire rope,
dw = Diameter of the wire,
D = Diameter of the sheave or drum
3. Stresses during starting and stopping.
σst = 2(W+w) / A= 2(48069+15717.36) / 465.5
= 274.055 N / mm2
During starting and stopping, the rope and the
supported load are to be accelerated. This induces additional load in the rope
4. Stress due to change in speed.
σa = (W+w)a / gA
= (48069+15717.36)0.046 / (9.81 x 465.5 )
= 0.642 N / mm2
5. Effective stress
• Effective stress in the rope during normal working
= σd + σb = 137.02 + 52.8 = 189.82 N/mm2
• Effective stress in the rope during starting
= σst + σb = 274.055 + 52.8 = 326.855 N/mm2
• and effective stress in the rope during acceleration of the load
= σd + σb + σa =137.02 + 52.8 + 0.642 = 190.462 N/mm2
DESIGN OF CAR OF THE ELEVATOR
• The sizes of human occupants fix the car size, in that the car interior must be
at least 2 m high to provide headroom.
• The maximum number of passengers (i.e. the safe number) is sixteen, so the car
floor is not permitted to exceed 2.8 m' and we envisage the car as a box, whose
inside dimensions are 2.2 m high, 2.0 m wide and 1.4 m from front to back.
CONCLUSION
• Could successfully calculate the rope diameter required
• Successfully found the type of stresses involved and their magnitude
• Could predict the size of the car required in the elevator
REFERENCES
• A Textbook of Machine Design by R.S.Khurmi & J.K. Gupta
• https://en.wikipedia.org/wiki/Bailong_Elevator
• https://www.topchinatravel.com/china-attractions/bailong-elevator.htm
• http://www.industrytap.com/highest-outdoor-elevator-in-the-world-hundred-dragons-elevator/328
Design of bailong elevator   case study

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Design of bailong elevator case study

  • 1. CASE STUDY ON DESIGN OF BIALONG ELEVATOR By – Sumit Salunke - Neha Shah - Madhura Mali - Malhar Sure - Sneha Ramkrishnan Guided by – Prof. Venugopal
  • 2. CONTENTS  INTRODUCTION  COMPONENTS OF ELEVATOR  KNOWN DATA’S  STANDARD CHARTS  CALCULATION OF DIAMETER OF THE WIRE ROPE  CALCULATION OF STRESSES INVOLVED  DESIGN OF CAR(ELEVATOR)
  • 3. INTRODUCTION • The Bailong Elevator (literally Hundred Dragons Elevator) is a glass elevator built onto the side of a huge cliff in the Wulingyuan area of Zhangjiajie, People's Republic of China that is 326 m (1,070 ft) high. • It is claimed to be the highest and heaviest outdoor elevator in the world. • On 16 July 2015, the elevator was officially recognised by Guinness World Records as the world’s tallest outdoor lift
  • 4. • An electric lift of the kind we are examining is basically a passenger car in a sling, to the top of which is attached a set of wire ropes. • The ropes pass up the lift well and over a driving pulley called a 'sheave’ on the shaft of a motor and brake unit, then down the well again to a counterweight, which balances the weight of the car and sling, together with some &50% of the full rated load. • Sometimes, the ropes are not fastened directly to the sling or counterweight, but pass around pulleys attached to them, and are then fastened at the head of the well. COMPONENTS OF ELEVATOR
  • 5. KNOWN DATA’S Load capacity = 4900kg x 9.18 = 48069N Depth = 335m Rope speed = 5m/s Time = 120s
  • 6. Designation of Wire Ropes The wire ropes are designated by the number of strands and the number of wires in each strand. For example, a wire rope having six strands and seven wires in each strand is designated × 7 rope.
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  • 12. CALCULATION OF DIAMETER OF THE WIRE ROPE  From table 20.7, for elevator we can take type 6 x 19 or 8 x 19 Considering 6 x 19 rope type  FOS for elevator is generally 11 Therefore, design load for wire rope = 11 x 48069 = 528759 N  From 20.7 table, tensile strength of 6 x 19 with 1250-1400Mpa is 435d2 (in N) where, d is rope diameter in mm Equating tensile strength to design load , 435d2 = 528759 N d = 34.86mm = 35mm  From 20.10 for 6 x 19 , Wire diameter (dw) = 0.063d Area of wire rope(A) = 0.38d2 .`. dw = 0.063 x 35 = 2.205mm And A = 0.38 X 352 = 465.5mm2
  • 13.  Various loads on ropes – • From 20.7 , weight of rope W = 0.0383d2 = 0.383(35)2 = 46.91 N/m = 46.91 x 335 N …..(depth = 335m) = 15717.36 N • From 20.12 , sheave diameter for wire rope for elevators Min sheave dia = 20d Max sheave dia = 30d But, as it is for large distance and heavy load, we will take D = 100d = 100 x 35 = 3500mm Bending stress, σb = (Er x dw) / D = ( 84000 x 2.2 ) / 3500 = 52.8 N/mm2 Er = Modulus of elasticity Dw = Dia of wire D = Dia of sheave drum Equivalent bending load on rope Wb = σb x A = 52.8 x 465.5 = 24578.4 N
  • 14. • Acceleration of rope and load a = v / t = 5/120 = 0.046 m/s2 Additional load due to acceleration Wa = (W+w)a / g = (48069+15717.36)0.046 / 9.81 = 300 N • Impact load during starting Wst = 2(W+w) = 2(48069+15717.36) = 127572.72 N
  • 15.  Effective load on rope during normal working = W + w + Wb = 48069+15717.36+24578.4 = 88364.76N FoS during normal working = 528759 / 88364.76 = 6 Effective load on rope during starting = Wst + Wb = 127572.72+24578.4 = 152151.12 N FoS During starting = 528759 / 152151.12 = 3.7 Effective load on rope during acceleration of load = W + w + Wb + Wa = 48069 + 15717.36 + 24578.4 + 300 = 88664.76 FoS during acceleration = 528759 / 88664.76 = 6 Since above FoS as calculated are safe, .`. wire rope of dia 35mm and 6 x 19 type is satisfactory.
  • 16. STRESSES INVOLVED 1. Direct stress due to axial load lifted and weight of the rope . W = Load lifted, w = Weight of the rope, and A = Net cross-sectional area of the rope.  direct stress = (W+w) / A  = (48069 + 15717.36 ) / 465.5  = 137.02 N/mm2
  • 17. 2. Bending stress when the rope winds round the sheave or drum. σb = (Er x dw) / D = ( 84000 x 2.2 ) / 3500 = 52.8 N/mm2 The bending stress induced depends upon many factors such as construction of rope, size of wire, type of centre and the amount of restraint in the grooves. The approximate value of the bending stress in the wire as proposed by Reuleaux, is Er = Modulus of elasticity of the wire rope, dw = Diameter of the wire, D = Diameter of the sheave or drum
  • 18. 3. Stresses during starting and stopping. σst = 2(W+w) / A= 2(48069+15717.36) / 465.5 = 274.055 N / mm2 During starting and stopping, the rope and the supported load are to be accelerated. This induces additional load in the rope
  • 19. 4. Stress due to change in speed. σa = (W+w)a / gA = (48069+15717.36)0.046 / (9.81 x 465.5 ) = 0.642 N / mm2
  • 20. 5. Effective stress • Effective stress in the rope during normal working = σd + σb = 137.02 + 52.8 = 189.82 N/mm2 • Effective stress in the rope during starting = σst + σb = 274.055 + 52.8 = 326.855 N/mm2 • and effective stress in the rope during acceleration of the load = σd + σb + σa =137.02 + 52.8 + 0.642 = 190.462 N/mm2
  • 21. DESIGN OF CAR OF THE ELEVATOR • The sizes of human occupants fix the car size, in that the car interior must be at least 2 m high to provide headroom. • The maximum number of passengers (i.e. the safe number) is sixteen, so the car floor is not permitted to exceed 2.8 m' and we envisage the car as a box, whose inside dimensions are 2.2 m high, 2.0 m wide and 1.4 m from front to back.
  • 22. CONCLUSION • Could successfully calculate the rope diameter required • Successfully found the type of stresses involved and their magnitude • Could predict the size of the car required in the elevator
  • 23. REFERENCES • A Textbook of Machine Design by R.S.Khurmi & J.K. Gupta • https://en.wikipedia.org/wiki/Bailong_Elevator • https://www.topchinatravel.com/china-attractions/bailong-elevator.htm • http://www.industrytap.com/highest-outdoor-elevator-in-the-world-hundred-dragons-elevator/328