Math Analysis

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Therefore, the dimension a=0.004=1.546 in[pic 68]

tolerance [pic 69]

1.5

Convert the following to appropriate ips units:

- A stress, [pic 70]

= 150/6.89 = 21.8 kpsi[pic 71]

- A force, F = 2kN.

F= 2 /4.45 = 0.449 kip = 449 lbf

- A moment M = 150 N.m

M= 150/0.113 = 1330 lbf .in = 1.33 kip .in

- An area A= 1500 [pic 72]

A= 1500/ = 2.33 [pic 73][pic 74]

- A second moment of area, I = 750 [pic 75]

I= 750/= 18.0 [pic 76][pic 77]

- A modulus of elasticity E = 145 GPa

E= 145/6.89 = 21.0 Mpsi

- A speed, v = 75[pic 78]

v = 75/1.61 = 46.6 mi/h

- A volume, V = 1 liter

V= 1000/946 = 1.06 qt

1.6

Convert the following to appropriate SI units:

- A length, l = 5 ft

l = 5(0.305) = 1.53 m

- A stress,[pic 79]

= 90(6.89) = 620 MPa[pic 80]

- A pressure, p = 25 psi

P= 25(6.89) = 172 kPa

- A section modulus, Z = 12 [pic 81]

Z =12(16.4) = 197 [pic 82]

- A unit weight, w = 0.208 [pic 83]

w = 0.208(175) = 36.4 N/m

- A deflection,[pic 84]

= 0.001 89(25.4) = 0.0480 mm[pic 85]

- A velocity, v = 1200 [pic 86]

v = 1200(0.0051) = 6.12 m/s

- A unit strain,[pic 87]

= 0.002 15(1) = 0.002 15 mm/mm[pic 88]

- A volume, V = 1830 [pic 89]

V= 1830[pic 90]

1.7

Generally, final design results are rounded to or fixed to three digits because the given data cannot justify a greater display. In addition, prefixes should be selected so as to limit number strings to no more than four digits to the left of the decimal point. Using these rules, as well as those for the choice of prefixes, solve the following relations:

- [pic 91]

[pic 92]

- [pic 93]

[pic 94]

- [pic 95]

[pic 96]

- [pic 97]

[pic 98]