Question 6.13: A solid square rod is cantilevered at one end. The rod is 0....

L=0.6  m , F_{a}=2  kN , n=1.5, N=10^{4} \text { cycles, } S_{u t}=770  MPa , S_{y}=420  MPa  (Table A-20)

First evaluate the fatigue strength.

Since the size is not yet known, assume a typical value of k_{b} = 0.85 and check later. All other modifiers are equal to one.

Eq. (6-18):              S_{e}=0.488(0.85)(385)=160  MPa

In kpsi,                    S_{u t}=770 / 6.89=112  kpsi

Fig. 6-18:                f = 0.83

Eq. (6-14):              a=\frac{\left(f S_{u t}\right)^{2}}{S_{e}}=\frac{[0.83(770)]^{2}}{160}=2553  MPa

Eq. (6-15):             b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}}\right)=-\frac{1}{3} \log \left(\frac{0.83(770)}{160}\right)=-0.2005

Eq. (6-13):             S_{f}=a N^{b}=2553\left(10^{4}\right)^{-0.2005}=403  MPa

Now evaluate the stress.

Compare strength to stress and solve for the necessary b.

Since the size factor was guessed, go back and check it now.

Eq. (6-25):          d_{e}=0.808(h b)^{1 / 2}=0.808 b=0.808(30)=24.24  mm

Eq. (6-20):          k_{b}=\left(\frac{24.2}{7.62}\right)^{-0.107}=0.88

Our guess of 0.85 was slightly conservative, so we will accept the result of

b = 30 mm.

Checking yield,

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Eq. (6-13):  S_{f}=a N^{b}

Eq. (6-14):  a=\frac{\left(f S_{u t}\right)^{2}}{S_{e}}

Eq. (6-15):  b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}}\right)

Eq. (6-18): S_{e}=k_{a} k_{b} k_{c} k_{d} k_{e} k_{f} S_{e}^{\prime}

Eq. (6-20): k_{b}= \begin{cases}(d / 0.3)^{-0.107}=0.879 d^{-0.107} & 0.11 \leq d \leq 2 \text { in } \\ 0.91 d^{-0.157} & 2<d \leq 10 \text { in } \\ (d / 7.62)^{-0.107}=1.24 d^{-0.107} & 2.79 \leq d \leq 51 mm \\ 1.51 d^{-0.157} & 51<d \leq 254 mm \end{cases}

Eq. (6-25): d_{e}=0.808(h b)^{1 / 2}