Question 5.13: A transmission shaft of cold drawn steel 27Mn2 (Sut = 500 N/...

\text { Given } M_{t}=-100 N – m \text { to }+400 N – m .

S_{u t}=500 N / mm ^{2} S_{y t}=300 N / mm ^{2} R=90 \%(f s)=2 .

Step I Endurance limit stress for shaft

S_{e}^{\prime}=0.5 S_{u t}=0.5(500)=250 N / mm ^{2} .

From Fig. 5.24 (cold drawn steel and

\left.S_{u t}=500 N / mm ^{2}\right) ,

K_{a}=0.79 .

Assuming 7.5 < d <50 mm,

K_{b}=0.85 .

\text { For } 90 \% \text { reliability, } K_{c}=0.897 .

S_{e}=K_{a} K_{b} K_{c} S_{e}^{\prime}=0.79(0.85)(0.897)(250) .

= 150.58 N/mm².

Step II Construction of modifi ed Goodman diagram
Using the distortion energy theory,

S_{s e}=0.577 S_{e}=0.577(150.58)=86.88 N / mm ^{2} .

S_{s y}=0.577 S_{y t}=0.577(300)=173.1 N / mm ^{2} .

\left(M_{t}\right)_{m}=\frac{1}{2}\left[\left(M_{t}\right)_{\max }+\left(M_{t}\right)_{\min }\right] .

=\frac{1}{2}[400-100]=150 N – m .

\left(M_{t}\right)_{a}=\frac{1}{2}\left[\left(M_{t}\right)_{\max }-\left(M_{t}\right)_{\min }\right] .

=\frac{1}{2}[400+100]=250 N – m .

\tan \theta=\frac{\left(M_{t}\right)_{a}}{\left(M_{t}\right)_{m}}=\frac{250}{150}=1.67 .

\theta=59.04^{\circ} .

The modified Goodman diagram for this example is shown in Fig. 5.44.

Step III Permissible shear stress amplitude
Refer to Fig. 5.44. The ordinate of the point X is S_{s e} or 86.88 N/mm².

\therefore \quad S_{s a}=86.88 N / mm ^{2} .

Step IV Diameter of shaft

\text { Since } \tau_{a}=\frac{S_{s a}}{(f s)} \quad \therefore \frac{16\left(M_{t}\right)_{a}}{\pi d^{3}}=\frac{S_{s a}}{(f s)} .

\frac{16\left(250 \times 10^{3}\right)}{\pi d^{3}}=\frac{86.88}{2} .

d = 30.83 mm.