| 15.3 [10 pts] |
|
|
15.17 [15 pts] |
- see addendum info on page 599 (627).
- P
= 2.7; CR = 1.5
|
| 15.23
[35 pts] | - (a)
- rpmb = 400 rpm
- rpmc
= 133 rpm
- dp = 3 in
- dg
= 9 in
- p = 0.63 in
- (b)
- (i) Ta = 7.96 Nm; Tb
= 23.9 Nm; Tc = 71.6 Nm
- (ii)
Ta = 7.96 Nm; Tb = 22.7 Nm; Tc =
64.7 Nm
- (c)
- Fat
= 209 N; Far = 97.5
- Fbt
= 627 N; Fbr = 292
- Calculate
the bearing reaction force components in the horizontal and vertical planes separately,
and then find each resultant magnitude with the square root of sum of squares
of the components.
- FA = 420 N;
FB = 903 N
|
|
16.16 [20 pts] |
- Use Equation 16.3 to derive a relationship between the module (m) and
the module in the normal plane (mn).
- (a) 5.0
- (b)
- Ft = 7620 N
- Fr
= 3170 N
- Fa = 4160 N
- (c)
- Ft = 16,300 N
- Fr
= 6340 N
- Fa = 5950 N
|
| 16.21 [20 pts] |
- the cone angles (γ)
are not 45°
- Tp = 184 ft-lb
- Tg
= 460 ft-lb
- Ft = 839 lb
- Fr
= 283 lb
- Fa = 113 lb
|