Question

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:

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Answer 1

(a) 1.90 kpsi

(b) 0.40 kpsi

(c) 0.61 in.

(d) 0.009

(a) 8 MPa

(b) 1.30 cm⁴

(c) 2.04 cm⁴

(d) 62.2 MPa

Step-by-step explanation:

(a) σ = M/Z, where M = 1770 lbf·in and Z = 0.943 in³.

1770/0.943 = 1876.988 lbf/in² = 1.90 kpsi

(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

9440 /23.8 = 396.639 lbf/in² = 0.4 kpsi

(c) y = Fl³/(3EI)

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740 × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

∴ σ = 1000/(0.025 × 0.005) = 8 MPa

(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

(c) I = π·d⁴/64 where d = 25.4 mm.

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.