Question 9.4: Factor of Safety for a Stepped Shaft under Torsional Shock L...
Factor of Safety for a Stepped Shaft under Torsional Shock Loading
A stepped shaft of diameters D and d with a shoulder fillet radius r has been machined from AISI 1095 annealed steel and fixed at end A (Figure 9.6). Determine the factor of safety n, using the maximum shear stress theory incorporated with the Goodman fatigue relation.
Given: Free end C of the shaft is made to rotate back and forth between 1.0° and 1.5° under torsional minor shock loading. The shaft is at room temperature.
Data:
L=300 mm , \quad d=30 mm , \quad D=60 mm , \quad r=2 mm , \\ K_{s t}=1.5 \quad \text { (by Table } 9.1 \text { ), } \quad G=79 GPa \quad \text { (from Table B.1) } \\ S_u=658 \text { and } H_B=192 \quad \text { (by Table B.4) }
Design Assumption: A reliability of 95% is used.
From the geometry of the shaft, D = 2d and L_{AB} = L_{BC} = L. The polar moment of inertia of the shaft segments are
J_{B C}=\frac{\pi d^4}{32} \quad J_{A B}=\frac{\pi D^4}{32}=16 J_{B C}
in which
J_{B C}=\frac{\pi}{32}(0.030)^4=79.52\left(10^{-9}\right) m ^4
The total angle of twist is
\phi=\frac{T L}{G}\left\lgroup \frac{1}{16 J_{B C}}+\frac{1}{J_{B C}} \right\rgroup
or
T=\frac{16 G J_{B C} \phi}{17 L}
Substituting the numerical data, this becomes T = 19,708.5 \phi. Accordingly, for \phi_{\max } = 0.0262 rad and \phi _{min} = 0.0175 rad, it follows that T_{max} = 516.4 N·m and T_{min} = 344.9 N·m.
Hence,
T_{m} =430.7 N ⋅ m T_{a} = 85 8. N ⋅m
The modified endurance limit, using Equation 7.6, is
S_e=C_f C_r C_s C_t\left(1 / K_f\right) S_e^{\prime}
where
C_f=A S_u^b = 4.51(658^{–0.265}) = 0.808 (by Equation 7.7 and Table 7.2)
C_{r} = 0.87 (from Table 7.3)
C_{s} = 0.85 (by Equation 7.9)
C_s=\left\{\begin{array}{ll} 0.85 & (13 mm <D \leq 50 mm ) \quad\left(\frac{1}{2}<D \leq 2 in .\right) \\ 0.70 & (D>50 mm ) \quad(D>2 in .) \end{array}\right. (7.9)
C_{t} = 1 (for normal temperature)
S_{e} ^{\prime} = 0.29S_{u} = 190.8 MPa (applying Equation 7.4)
and
K_{t} = 1.6 (from Figure C.8, for D/d = 2 and r/d = 0.067)
q = 0.92 (from Figure 7.9, for r = 2 mm and H_{B} = 192 annealed steel)
K_{f} = 1 + 0.92(1.6 – 1) = 1.55 (using Equation 7.13b)
Therefore,
S_{e} = (0.808)(0.87)(0.85)(1)(1/1.55)(190.8) = 73.55 MPa
We now use Equation 9.13 with M_{m} = M_{a} = 0 to estimate the factor of safety:
\frac{S_u}{n}=\frac{32}{\pi D^3}\left[K_{s b}\left\lgroup M_m+\frac{S_u}{S_e} M_a \right\rgroup ^2+K_{s t}\left\lgroup T_m+\frac{S_u}{S_e} T_a \right\rgroup ^2\right]^{1 / 2} (9.13)
\frac{S_u}{n}=\frac{32}{\pi d_{B C}^3}\left[K_{s t}\left\lgroup T_m+\frac{S_u}{S_e} T_a \right\rgroup ^2\right]^{1 / 2}
Introducing the numerical values,
\frac{658\left(10^6\right)}{n}=\frac{32}{\pi(0.03)^3}\left[1.5\left\lgroup 430.2+\frac{658}{73.55} 86.2 \right\rgroup ^2\right]^{1 / 2}
This gives n = 1.19
| Table 7.2 Surface Finish Factors C_{f} |
|||
| A | |||
| Surface Finish | MPa | ksi | b |
| Ground | 1.58 | 1.34 | −0.085 |
| Machined or cold drawn | 4.51 | 2.7 | −0.265 |
| Hot rolled | 57.7 | 14.4 | −0.718 |
| Forged | 272.0 | 39.9 | −0.995 |
| Table 7.3 Reliability Factors |
|
| Survival Rate (%) | C_{r} |
| 50 | 1.00 |
| 90 | 0.89 |
| 95 | 0.87 |
| 98 | 0.84 |
| 99 | 0.81 |
| 99.9 | 0.75 |
| 99.99 | 0.70 |
| Table 9.1 Shock Factors in Bending and Torsion |
|
| Nature of Loading | K_{sb}, K_{st} |
| Gradually applied or steady | l.0 |
| Minor shocks | l.5 |
| Heavy shocks | 2 |
