149 Scenario: There is a world effort to rescue 33 miners trapped in a Chilean m
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149 Scenario: There is a world effort to rescue 33 miners trapped in a Chilean mine over 775 meters deep. In order to rescue these 33 miners, a new mine shaft or hole is being drilled. The diameter or width W of this shaft is 22 inches Once the hole is completed, each of the 33 miners will be retrieved one at a time using a rescue capsule. It is proposed that the trip should take 20 minutes to allow the miners to adjust to the changing oxygen levels and light (remember, these miners have not been exposed to light for 70 days). The capsule height H is 100 inches. The capsule fits inside the mine shaft and is lowered and raised to the surface via a pulley and cable system as illustrated below. As the capsule travels the mine shaft, there exists some friction between the capsule and the mine shaft. You may assume the following relation between static and kinetic friction: kinetic friction 0.95 . Also, as the capsule travels the mine shaft, the angle of wrap of the cable onto the pulley varies deg where :-5 deg Similarly, the angle ofinclination of the cable to the shaft varies ± deg where 10 deg. The length L between the bearing supports 14 inches and the angle30 UPRM graduates are recruited for this special mission due to their unique ability to speak Spanish and English along with N-m-s (Newtons-meters-seconds) and ips (inches, pounds, seconds). Your task is to select suitable bearings for the proposed system to support the rescue operation. It is required that all miners are safely hoisted (you define reliability). Select two identical ball bearings for the shaft and pulley system. List all assumptions, show "neatly" all your effort, and specify the following information: 5% Max bending stress t, in shaft (Pa): 5% Min shaft diameter d (mm): 10% Equivalent load at bearing (kN): 5% Impact factor. 10% Rated Load Co (kN): 5% Bearing Bore (mm): 5% Motor size (kW): poo pon Motor gravityExplanation / Answer
Shaft will be of 45 C8.
Inside Diameter of Shaft in mm=558.8 mm (22 Inch to Mtr).
Considering Shaft Thk =50 mm
Calculate Wt of Shaft =(0.785*((Od)^2-ID^2)*l*Density
Density -8000 Kg/me
Material :- Steel
Weight of Shaft =484843.482 Kgs.
Now calculate Wt of Capsule
Ht of Capsule=100 Inch =2540 mm.
Considering Width of capsule=300 mm.
Consider hallow capsule=2540 mm*400 mm*20 mm Thk of Steel make.
Wt of Capsule = 475.5 Kgs.(Consider above Formula and Tubular Capsule) .
Total Wt =484843.482+475.5=485318.982 kgs.
FOS AT 10 % = 48531.90 kgs
NET WEIGHT= 533850.88 Kgs.
NO OF PULLEY=2.
TOTAL HANGING WEIGHT= 533850.88
LOAD ON EACH PULLEY= 266925.44
B.M. OF THE SHAFT AT CENTER BM [SHAFT LENGTH 355 MM ]=WL/4
=266925.44*355/4=23716325.35 Kgs.
B.M. OF THE SHAFT AT CENTER BM IN N- MM ]= 237163253.53 N.
Consider Shaft for pulley
Ultimate Tensile Strength for 45 C8 =700 Mpa.(For Bearing pulley shaft).
Bending Stress=ultimate Stress/FOS.=700/6=116 Mpa =120 N/mm2.
WE KNOW = B.M./ Z HENCE Z= B.M./ =237163253.53/120=1976360.45
Z = *D3/32 HENCE D3
D=272 mm
TORQUE TO BE TRANSMITTED BY PULLEY = PULLING FORCE X PERPEND DISTANCE (PCD of Pulley=1500 mm)
=533850.888*1500=80.07*10^7 Kg-mm.
TORQUE TO BE TRANSMITTED IN N-M=80070000.00 N-m.
WE WILL REDUCE THE RATIO TO 40:1=2402100.00
NOW WE NEED 2 METER PER MINUTE SPEED=2 MPM
PM OF MAIN SPROCKET [DRIVEN] N= V/3.14/pulley dia=0.21 rpm.
Assume RATIO OF 1.5 TO 1400 RPM=1400/5.02; ratio=933.3
TORQUE RATION WILL BE ==278.88/933.33 =2573.68N-m.
EFFICIENCY OF Pulley=0.93*0.83*0.85=eff motor* Eff pulley * eff Gear Box.
HENCE POWER REQUIRED =590 Kw.
HORSE POWER OF MOTOR =590/0.75=786 Hp.
Radial Forces on Shaft
Fr=Ftp* tan @=80.07*10^8*.363(Angle = 30 Degree)
=29.07 * 10^7 Kgf.
Radial Forced on Bearing =
FrP=face width* Tangetial force /Distance=2.5*29.07*10^7/355 (Converted Inch to mtr)
= 2047 Kgf .(Equal Distance so same forces on Both bearing.)
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