| 7.8
[15 pts] |
- total maximum elongation (including static deflection) = 2.44 in
- maximum
tensile stress = 34.8 ksi
|
| 8.23
[35 pts] | - Assume
CT = CR = 1 (i.e., neglect temperature and reliability effects).
- CS
= 0.74
- assume a nominal CG(axial) of 0.8 for small diameters
(< 2 in)
- be sure to use the appropriate 103-cycle strengths
from Table 8.1
- be sure to comply with note c in Table 8.1
- Sn(bending)
= 36.6 ksi; Sn(axial) = 32.6 ksi Sn(torsion)
= 21.2 ksi
- draw a log-log S-N curve for each loading case
- use
logarithmic interpolation to calculate S(6x104)
- S(6x104,
bending) = 55 ksi
- S(6x104, axial) = 48 ksi
- S(6x104,
torsion) = 36 ksi
|
| 8.32
[30 pts] | - Change
"Figure 4.39" to "Figure 4.36" in the problem statement.
- Assume
CT = CR = 1 (i.e., neglect temperature and reliability effects).
- Kt
= 1.63; q = 0.79; CS = 0.86 (explain where these
numbers come from)
- assume the average surface finish at the point of highest
stress is between "machined" and "commercial polish," but
closer to "commercial polish"
- assume: S'n=
0.5Su; Su (in ksi) = 0.5 * BHN;
Sus = 0.8Su; Sys = 0.58Sy
(see pp. 88, 247, 295, 300) - apply the safety factor to the allowable torque
and use a load line, or use the Goodman relation directly to find the allowable
torques
- construct a constant (infinite) life τa
vs. τm diagram
- reversed torsion:
Tmax = 1320 in*lb
- for steady torque T with alternating torque
2T (note: τa/τm=
2), Tmax = 580 in*lb
|
| 8.33
[20 pts] | - Assume
CT = CR = 1 (i.e., neglect temperature and reliability effects).
- assume
linear elastic response (F = kΔ)
- assume S'n
= 100 ksi (approximation from Fig. 8.6)
- use CS = 0.54
- use
q = 1
- see p. 302 concerning the gradient factor
- justify the values
used (including those listed above)
- plot an infinite life Σa-Σm
diagram to determine if the load state will result in failure
- ans:
the load state is close to (just above) the failure threshold
|