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2012

Joseph Caesar B. Torres

5MEB

Machine Design II

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I.C.E

Motor

Water

Power

Wind

Power Pump

Compressor

Mixer

Crusher

Elevator

DriverMechanical

System

Shaft

Belt

Chain

RopesDriven

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F1 - tight side tensionF2 - slack side tension

Fc - centrifugal tension

f - coefficient of friction (text) - arc of contactb - belt widtht - belt thickness (text)

S - allowable stress on belt = 400 - efficiency of joint (text) ex: - belt density (text)V - belt speed N -

g -

(Open Belt)

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For angle of contact,:

For velocity:

Tension Ratio

For the cross-sectional area:

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Where:

* + * +

For medium ply leather

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DriverDriven

Driver

DrivenDriven

DrivenDriven

Driver

idler

Direct Drive

Dual Drive:

USEFUL EQUATIONS:

, -

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Length of chain:

For

:

, -

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For transmitted power:

Solving for Torque:

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A

BC

A dual chain drive consists of a double strand RC 120 has a driving sprocket A

has 11 teeth rotating at 100 rpm. While the driven sprocket B has 17 teeth andrequires 5KW power. The driven sprocket C has 21 teeth and requires 7 KW

power. For the system shown, find:

a) Torque on shaft B and shaft Cb) Service factor of the drivec) Tension on various sections

6 x 7

6 x 19

6 x 37

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D, ropeStrands x Wire

Application:

1. Elevators

8 to 12

2. Hoist 3 to 53. Haulage4. Derricks5. Tramways6. Cable Car7. Guard Roil etc

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Where:

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Nomenclature:

Fu - ultimate load or breaking load

F.S. - factor of safety

Ft - total load

Fb - equivalent bending loadk - 12x106 constant for steel ropeA - Cross-sectional area

dw - wire diameter

Ds - sheave or drum diameter

Wp - weight of passenger

Ws - weight of skip or cage

Wr - weight of rope (text)

Wacc - load due to acceleration=

Values for C for wire rope

Ropediameter

C Ropediameter

C

1.09

1.046

1.083 1.04 1.076 1.07

1.064 1.054

Recommended minimum factor of safety

Track 3.2

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CableGuy wires 3.5Mine Hoist 5 to 8Haulage 6Cranes 6

Sling 8Elevators 8 to 12

Where:

Therefore:

T1T2T3T4T5

P

Q

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Band Brake Block Brake Disk Brake

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= 270Tension Ratio

F1 = 4.11 F2

Tq = 95.5 Nm > torque to be absorbed

Tq = (F1 + F2) x F1F2 =

= = 191 N4.11F2F2 = 191

F2 = = 61 N

F1 = 4.11(61) = 252 N

m @ 0 = 0

0 = F1 (280) + F2 (280)Q (1750)

For the band break shown, find Q need to absorb 18kW at 500 rpm, f=0.3

Calculate the tensile stress on band if band width and thickness is 120 mm and 12mm, respectively.

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= cos-1 = 41.41

= 36041.4190

= 228.59

m @ 0 = 0

Tq = 343.77 Nm 0 = F2 (150mm)Q (760mm)

F1F2 = Q = 195.81 Nm

3.31F2F2 = F2 = 992.12 N St =

F1 = 3283.92 N St =

x St = 2.28 MPa

The 300 mm diameter brake drum is used in a band brake that absorbs

18kW at 520 rpm, f= 0.4. Estimate the force Q to operate the system. What is themax tensile stress on the band of width and thickness are 120mm and 12mm,respectively. What is the band pressure?

Calculate the power loss at 1500 rpm for the double block brake drum shown.

f= 0.3

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T = Torque

P = Axial Force

f = Coefficient of Friction

D = Outside Diameter

d = Inside Diameter

P = pressuren = No. of Friction Surface

N = No. of Plate (Driver and

Driven)

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Where:

T - torqueDp - diametral pitch

n - number of teeth

y - lewis tooth form factor

Cp - circular pitch

K - 3 to 4

Barth Equation For Allowable Stress Where:

So - basic static sale stress

Cv - velocity factor

Common Standard Diametral Pitch1 2 5 9 16

1 2 6 10 18

1 3 7 12 20

1 4 8 14 24

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f = 1.35 ; pulsating load

f = 1.5 ; shock load

For gear tooth to be satisfactory on standpoint of dynamic loading Sd< Sef

flexural endurance limit.For gear tooth to be satisfactory on standpoint of wear Wd < Ww

For Common Material Static Gate Stress

SD, psi

Cast iron ordinary 8,000

Cast iron medium grade 10,000

Cast iron highest grade 15,000

Cast iron 0.20% C untreated 20,000

Cast iron 0.20% C heat treated 28,000

Forged Carbon Steel

SAE1020 Cage hardened 18,000

SAE1030 Untreated 20,000

SAE1035 Untreated 23,000

SAE1040 Untreated 25,000

SAE1045 Heat Treated 25,000

GearPinion

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VVelocity

Da Db - Pitch Diameter

na nbnumber of teeth

Dpdiametral pitch

Na Nbrpm

CL - centerline distance

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