LOVE

2

Problem:

Design a bell crank, similar to the one shown, to carry a mild shock load. The mechanical advantage ( L1 / L2 = F2 / F1 ), the force F1, the length L1, and the material are given in the accompanying table, a) Make all significant decisions including tolerances and allowances. One approach could be to compute dimensions of the yoke connections first: t should a little less than:

An assumption for the shaft may be that, on occasion, the torque for F1 is transmitted through the shaft (ignoring bending for local convenience).

Check all dimensions for good proportion; modify as desirable.

Sketch to scale each part, showing all dimensions with tolerances necessary to manufacture.

REQ: D:

a.Determine all dimensions including tolerances and allowances.

b.Check all dimensions for good proportion

c.Sketch to scale each part.

d.All dimensions with tolerances and allowances necessary to manufacture

Given Data:

Problem No.

Load

F1

L1

AISI No.

as rolled

Mech. Advantage

115

600 lb

20 in

C1040

4

SOLUTIONS:

From given data:

Mechanical advantage = ( L1/L2 = F1/F2 ) = 4

4

2

1

=

L

L

4

1

2

=

F

F

4

20

4

1

2

=

=

L

L

)

4

(

600

)

4

(

1

2

=

=

F

F

L2 = 5 inF2 = 2,400 lb

From Table AT 7 (page 576 text)

For AISI C1040, as rolled

Sy = 50,000 psi

For Design Stresses:

Ss= 0.5 Sy / N

Ss= 0.5 ( 50,000 ) / 3

Ss= 8,333.33 psi

From table 1.1 ( page 20 text )

Design Factor ( Factor of Safety ), for repeated, one direction

N = 3, based on ultimate strength ( Su )

Solving for design stress (St):

St= Sy / N

= 50,000 psi

3

St= 16,666.67 psi

For Compression stress (Sc):

Sc= Sy / N

= 50,000 psi

3

Sc= 16,666.67 psi

For Shearing stress (Ss):

Sys= Sys / N

= 30,000 psi

3

Sys= 10,000 psi

A.) Tension across the circular section of F1 at joint A:

( @ Junction A )

F1= SdD12

4

D12= F1(4)

Sd

p

)

67

.

666

,

16

(

)

600

(

4

1

=

D

D1= 0.214 in ( 1/16 between 3/16 - 7/8 )

For Standard fraction:

1875

.

0

16

3

=

16

424

.

3

16

16

214

.

0

=

x

4

1

%

5

4

%

38

.

12

%

100

214

.

0

1875

.

0

214

.

0

-

>

=

-

x

Use D1 = 1/4 in.

The pin may fail by shearing stress in the rod (double shear):

Solving for pin diameter (d1):

Fig:

A

F

Ss

=

For area :

2

4

2

1

x

d

As

p

=

2

2

1

1

d

F

Ss

p

=

p

33

.

333

,

18

)

2

(

600

1

=

d

d1 = 0.214 in

For Standard Fraction Use d1 = 1/4 in.

The compressive stress between the pin and the rod:

Fig:

A

F

Sc

=

;

1

1

d

a

A

=

1

1

1

d

Sca

F

=

1

1

1

d

a

F

Sc

=

)

25

.

0

)(

67

.

666

,

16

(

600

1

=

a

a1 = 0.144 in

For Standard Fraction Use a1 = 5/32 in.

The compressive stress between the pin and the yoke:

Projected area on one side of a yoke is ( bd ) and for two of the yoke (bd ):

Fig:

A

F

d

b

F

Sc

1

1

1

1

2

=

=

Where:

1

1

1

1

1

1

2

d

b

F

d

a

F

=

a1 = 2 b1

Sc

d

F

b

1

1

1

2

=

(

)

(

)

67

.

666

,

16

25

.

0

2

600

1

=

b

b1 = 0.072 in

For Standard Fraction Use b1 = 5/64 in.

The rod and the yoke may fail in tension across the hole of the pin:

Fig:

For the Rod:

SdtA

F

=

1

)

)(

(

1

1

1

1

a

d

m

Sd

F

t

-

=

1

1

1

1

)

(

d

a

Sd

F

m

t

+

=

(

)

(

)

25

.

0

32

/

5

67

.

666

,

16

600

1

+

=

m

m1 = 0.48 in

For Standard Fraction Use m1 = 1/2 in.

For the Yoke:

Fig:

SdtA

F

=

1

)

2

)(

(

1

1

1

1

b

d

m

Sd

F

t

-

=

1

1

1

1

)

2

(

d

b

Sd

F

m

t

+

=

(

)

(

)

25

.

0

64

/

5

)

2

(

67

.

666

,

16

600

1

+

=

m

m1 = 0.48 in

Use: 1/2 in

Shearing stress on the lever, solving for thickness ( t1 ):

Fig:

A

F

Ss

1

=

;

t

m

A

1

=

t

m

F

Ss

1

1

=

1

1

1

Ssm

F

t

=

(

)

(

)

33

.

333

,

8

5

.

0

600

1

=

t

t1 = 0.144 in

For Standard Fraction Use t1 = 5/32 in.

For the margin of the hole, the pin may tear at the end of rod or yoke:

Fig:

F1 = Ss A ; A = 2a1 e1

)

2

(

1

1

a

Ss

F

e

=

(

)

(

)

32

/

5

2

33

.

333

,

8

600

1

=

e

e1 = 0.23 in

For Standard Fraction Use e1 = 1/4 in.

B.) FOR ANALYSIS OF JOINT B:

Tension across circular section of F2 @ joint b:

Fig:

A

F

St

2

=

For the area:

4

2

2

D

A

p

=

)

(

4

2

2

p

Sdt

F

D

=

)

)(

67

.

666

,

16

(

)

400

,

2

(

4

2

p

=

D

D2 = 0.428 in

For Standard fraction: use: D2 = 7/16 in

The pin may fail by shearing stress in the Rod (Double Shear):

Fig:

2

2

2

A

F

Ss

=

for area :

2

2

2

d

As

p

=

2

2

2

2

d

F

Ss

p

=

)

(

2

2

2

Ssd

F

d

p

=

p

)

33

.

333

,

8

(

)

2

(

400

,

2

2

=

d

d2 = 0.428 in

For Standard Fraction Use d2 = 7/16 in.

The compressive stress between the pin and the rod:

Fig:

)

(

2

2

2

d

Sc

F

a

=

(

)

(

)

16

/

7

67

.

666

,

16

400

,

2

2

=

a

a2 = 0.329 in

For Standard fraction: 1/16 between 3/16 7/8

in

313

.

0

16

5

=

16

264

.

5

16

16

329

.

0

=

x

8

3

%

5

4

%

86

.

4

%

100

329

.

0

313

.

0

329

.

0

-