... 3.75in, T∗= 7173lbf·in; (c) (d/D)∗= 1/√3 = 0.57716–19 (a) Uniform wear: pa= 82.2kPa, F = 949N;(b) Uniform pressure: pa= 79.1kPa, F = 948N16–23Cs= 0.08, t = 5.30in16–26 ... Answers to Selected Problems Appendix B1013B–1 Chapter 11–810 ≤ F ≤ 10.5 lbf, 13.5 ≤ N ≤ 14.2 lbf,K =0.967,19.3 ≤ T...