Math Analysis

1. A solid circular rod of diameter d

Moment  M=1000lbf.in

Stress : S = [pic 1][pic 2]

[pic 3]

Table A-17 select a preferred fractional diameter : d = [pic 4]

The resulting factor of safety:[pic 5]

1.2

(a) For a sliding fit with 30mm basic size 

ISO symbol: 30H7/g6

Hole: 30H7 ➔ D = 30 mm

IT7 (from table A-11) ➔∆D = 0.025mm

D + ∆D = 30 +0.025 = 30.025mm.[pic 6]

 = D = 30mm[pic 7]

Shaft: 30g6 ➔ d = 30

IT6 (from talbe  A-11) ➔ ∆d = 0.016 mm

Fundamental deviation:  = - 0.009 mm  (from table A-12).[pic 8]

Sliding fit g:

= d +  = 30 + (-0.009) = 29.991 mm[pic 9][pic 10]

 = d +  - = 30 + (-0.009) – 0.016 = 29.975 mm.[pic 11][pic 12][pic 13]

Hole:                    Shaft: [pic 14][pic 15]

(b) For a clerance fit with 18mm basic size 

ISO symbol: 18H7/h6

Hole: 18H7 ➔ D = 18 mm

IT7 (from table A-11) ➔∆D = 0.021mm

DD + ∆D = 18 +0.021 = 18.021mm.[pic 16]

 = D = 18mm[pic 17]

Shaft: 318h6 ➔ d = 18 mm

IT6 (from talbe  A-11) ➔ ∆d = 0.013 mm

Fundamental deviation:  = 0 mm  (from table A-12).[pic 18]

Clearance fit h:

= d +  = 18 + 0= 18 mm[pic 19][pic 20]

 = d +  - = 30 + 0 – 0.013 = 17.987 mm.[pic 21][pic 22][pic 23]

Hole:                    Shaft: [pic 24][pic 25]

(c) For a interferece fit with 72mm basic size. 

ISO symbol: 72H7/p6

Hole: 72H7 ➔ D = 72 mm

IT7 (from table A-11) ➔∆D = 0.030mm

DD + ∆D = 72 +0.030 = 30.030mm.[pic 26]

 = D = 72mm[pic 27]

Shaft: 72p6 ➔ d = 72mm

IT6 (from talbe  A-11) ➔ ∆d = 0.019 mm

Fundamental deviation:  = +0.032 mm  (from table A-12).[pic 28]

Interference fit p:

= d + + = = 7230 + 0.032 +0.019= 72.051 mm[pic 29][pic 30][pic 31]

= d + = 72 + 0.032 = 72.032 mm.[pic 32][pic 33]

Hole:                    Shaft: [pic 34][pic 35]

(d) For a locational transition fit with 48 mm basic size. 

ISO symbol: 48H7/k6

Hole: 48H7 ➔ D = 48 mm

IT7 (from table A-11) ➔∆D = 0.025mm

DD + ∆D = 48 +0.025 = 48.025mm.[pic 36]

 = D = 48mm[pic 37]

Shaft: 48k6 ➔ d = 48 mm

IT6 (from talbe

  A-11) ➔ ∆d = 0.016 mm

Fundamental deviation:  = + 0.002 mm  (from table A-12).[pic 38]

Locational transmition fit k:

= d +  +  = 30 + (0.002) + 0.016 = 48.018 mm[pic 39][pic 40][pic 41]

 = d + = 48 + 0.002 = 48.002 mm.[pic 42][pic 43]

Hole:                    Shaft: [pic 44][pic 45]

1. We have :

a= 1.500 0.001 in [pic 46]

b = 2.000 0.003 in [pic 47]

c = 3.000 0.004 in [pic 48]

d = 6.520 0.010 in [pic 49]

1. the mean gap :   = 6.520 -1.5 -2 - 3 = 0.020 in [pic 50]

and tolerance: = 0.001 + 0.003 + 0.004 +0.010 = 0.018 [pic 51][pic 52]

• w= 0.020 0.018 in [pic 53]

1.  From part (a), = 0.002 in. Thus, must add 0.008 in to . Therefore, [pic 54][pic 55]

= 6.520 + 0.008 = 6.528 in[pic 56]

1.4

Let  be the length of  the pin , be the thickness of the clevis,  be the thickness of the snap ring , and  be the gap between the clevis and the snap ring. [pic 57][pic 58][pic 59][pic 60]

Then   [pic 61]

,[pic 62]

 , 0.004[pic 63][pic 64][pic 65]

From the geometry , need +[pic 66][pic 67]

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