Solution Design

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Given Data: MATERIAL: AISI C3140 Mechanical Advantage = 4 Force 1= 600lbs Length 1= 20 in SOLUTION: For solving F2: F 1 = 600 lbs F 2 F 1 = 4 F 2 4F 1 F 2 2400 lb For solving L2: L 1 = 20in 2

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Solution Design

Transcript of Solution Design

Given Data:MATERIAL: AISI C3140 Mechanical Advantage = 4Force 1= 600lbsLength 1= 20 inSOLUTION:For solving F2:F1 = 600 lbs =

For solving L2: L1 = 20in

MATERIAL: AISI C3140From Page 571 Table AF1Sy= 58000 lbs

From Page 20Assuming mild shock, repeated, one direction:N=3 (Yield Strength)

Design Stress for Shear:

Design Stress for Tension and Compression:

A.TENSION ACROSS THE CIRCULAR SECTION JOINT A:

= =

For Standard Fraction: X =

Use

B. SHEAR BETWEEN ROD AND PIN

= =

For Standard Fraction: X =

Use

C. COMPRESSIVE STRESS BETWEEN PIN AND ROD

= =

For Standard Fraction: X =

Use D. COMPRESSIVE STRESS BETWEEN PIN AND YOKE = =

For Standard Fraction: x =

Use

E. ROD AND YOKE TENSION ACROSS THE PIN:

ROD: = =

YOKE: = =

For Standard Fraction: X =

Use: F. SHEARING STRESS ON THE LEVER:

= =

For Standard Fraction: X =

Use

G. THE PIN MAY TEAR AT THE END OF THE ROD OR YOKE:

= =

For Standard Fraction: x =

Use

ANALYZING JOINT B: A.TENSION ACROSS THE CIRCULAR SECTION AT JOINT B:

= =

For Standard Fraction: x =

B. SHEAR BETWEEN ROD AND PIN

= =

For Standard Fraction:

x =

Use

C. COMPRESSIVE STRESS BETWEEN PIN AND ROD

= =

For Standard Fraction: x =

Use

D. COMPRESSIVE STRESS BETWEEN PIN AND YOKE

= =

For Standard Fraction: x =

Use E. ROD AND YOKE TENSION ACROSS THE PIN:

= =

For Standard Fraction: x =

Use

= =

For Standard Fraction: X =

Use F. SHEARING STRESS ON THE LEVER:

= =

For Standard Fraction: x =

Use

G. THE PIN MAY TEAR AT THE END OF THE ROD OR YOKE:

= =

For Standard Fraction: x =

Use

ANALYZING JOINT C :

Shaft is subjected to torsional stress:

For Shaft Diameter:

=

For Standard Fraction: x =

Use

For Hub Diameter:For steel diameter hub, page 388

For Standard Fraction: x =

FOR THE KEY DIMENSION:From table AT 19: Page 594 Shaft Diameter:

Tolerance on b, (in): - 0.0025in

FOR THE SIZE OF THE KEY: (L)For Shear: =

For Compression: =

Therefore; Use L.s = 2.74in. with 1/2 x 3/8 cross sectional area

FOR THE LENGTH OF THE HUB: From Chapter 10, page 283, typical hub length falls between 1.25 to 2.4 shaft diameter.

TOLERANCES AND ALLOWANCESFor the yoke at Section A and B, use loose running fit (Rc-9)Page 83: Tolerances and Allowances:FOR THE HOLE:+ Hole Tolerance

Nominal Size - 0.000

FOR THE SHAFT:+ 0.000

Nominal Size Allowance - Shaft Tolerance

At Yoke A:

From Table 3.1: RUNNING AND SLIDING FITS:NOMINAL SIZE RANGE(in)RC 9: (x10-3)

HoleShaft

0.20 0.40+3.5-5.0

0-7.2

Hole Tolerance = 0.0035 - 0.0000 = 0.0035 in.Shaft Tolerance = - 0.005 - (- 0.0072) = 0.0022in.Allowance = 0.000 - (- 0.005) = 0.005 in.

FOR THE HOLE:+ 0.0035 Tolerance

0.25 in- 0.000

+ 0.000

FOR THE SHAFT:

- 0.0022 Tolerance

At Yoke B:

From Table 3.1: RUNNING AND SLIDING FITS:NOMINAL SIZE RANGE(in)RC 9: (x10-3)

HoleShaft

0.40 0.71+4.0-6.0

0-8.8

Hole Tolerance = 0.004 - 0.000 = 0.004 in.Shaft Tolerance = - 0.006 - (- 0.0088) = 0.0028in.Allowance = 0.000 - (- 0.006) = 0.006 in.FOR THE HOLE:- 0.000

+ 0.004 Tolerance

0.438 in

FOR THE SHAFT:+ 0.000

- 0.0028 Tolerance

At Yoke C:

From Table 3.1: RUNNING AND SLIDING FITS:NOMINAL SIZE RANGE(in)RC 9: (x10-3)

HoleShaft

1.19 1.97+6.0-8.0

0-12.0

Hole Tolerance = 0.006 - 0.000 = 0.006 in.Shaft Tolerance = - 0.008 - (- 0.012) = 0.004in.Allowance = 0.000 - (- 0.008) = 0.008 in.

FOR THE HOLE:- 0.000

+ 0.006 Tolerance

1.812 in

FOR THE SHAFT:+ 0.000

- 0.004 Tolerance

SUMMARY OF COMPUTED AND ADJUSTED VALUES

PartsComputed values( inches )Computed values(inches)

L12020

D10.25

d10.25

a10.125

b10.0625

m10.5

t10.125

e10.25

L255

D20.4375

d20.4375

a20.3125

b20.15625

m20.8125

t20.3125

e20.4375

Ds1.8125

Dh3.25

b0.5

t0.375

Ls2.742.74

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