Question 9.4: Factor of Safety for a Stepped Shaft under Torsional Shock L......

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(\frac{1}{16 J_{B C}}+\frac{1}{J_{B C}}\right)

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 \cdot m \text { and } T_{\min }=344.9  N \cdot m . Hence,

T_m=430.7  N \cdot m \quad T_a=85.8  N \cdot 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}     (7.6)

S_e=C_f C_r C_s C_t\left(\frac{1}{K_f}\right) S_e^{\prime}

where

C _f=A_u^b=4.51\left(658^{-0265}\right)=0.808 (by Equation (7.7) and Table 7.2)

C_f=A S_u^b    (7.7)

C_r=0.87 (from Table 7.3)

C_{ s }=0.85 (by Equation (7.9))

C_s= \begin{cases}0.85 & (13  mm <D \leq 50  mm ) \quad\left(\frac{1}{2}<D \leq 2 \text { in. }\right) \\ 0.70 & (D>50  mm ) \quad(D>2 in .)\end{cases}    (7.9)

C_t=1 (for normal temperature)

S_e^{\prime}=0.29 S_u=190.8  MPa (applying Equation (7.4))

\text { Iron} \left(S_{e s}^{\prime}\right)=0.32 S_u       (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))

K_f=1+q\left(K_t-1\right)     (7.13b)

Therefore,

S_e=(0.808)(8.87)(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(M_m+\frac{S_u}{S_e} M_a\right)^2+K_{s t}\left(T_m+\frac{S_u}{S_e} T_a\right)^2\right]^{1 / 2}      (9.13)

\frac{S_u}{n}=\frac{32}{\pi d_{B C}^3}\left[K_{s t}\left(T_m+\frac{S_u}{S_e} T_a\right)^2\right]^{1 / 2}     (9.17)

Introducing the numerical values,

\frac{658\left(10^6\right)}{n}=\frac{32}{\pi(0.03)^3}\left[1.5\left(430.2+\frac{658}{73.55} 86.2\right)^2\right]^{1 / 2}

This results in n=1.19.