Question 5.14: A spherical pressure vessel, with a 500 mm inner diameter, i...

\text { Given For vessel } D_{i}=500 mm .

p_{i}=\text { zero to } 6 N / mm ^{2} \quad S_{u t}=440 N / mm ^{2} .

S_{y t}=242 N / mm ^{2} \quad R=50 \%(f s)=3.5 \quad K_{b}=0.85 .

Step I Endurance limit stress for vessel

S_{o}^{\prime}=0.5 S_{u t}=0.5(440)=220 N / mm ^{2} .

From Fig. 5.24 (cold drawn steel and

\left.S_{u t}=440 N / mm ^{2}\right) .

K_{a}=0.82 .

K_{b}=0.85 .

\text { For } 50 \% \text { reliability, } K_{c}=1.0 .

S_{e}=K_{a} K_{b} K_{c} S_{e}^{\prime}=0.82(0.85)(1.0)(220) .

= 153.34 N/mm².

Step II Construction of modifi ed Goodman diagram
For a spherical pressure vessel

\sigma_{t}=\frac{p_{i} D_{i}}{4 t} .

\sigma_{\max }=\frac{p_{\max } D_{i}}{4 t}=\frac{6(500)}{4 t}=\left(\frac{750}{t}\right) N / mm ^{2} .

\sigma_{\min .}=0 .

\sigma_{a}=\sigma_{m}=\frac{1}{2} \sigma_{\max }=\frac{1}{2}\left(\frac{750}{t}\right)=\left(\frac{375}{t}\right) N / mm ^{2} .

\tan \theta=\frac{\sigma_{a}}{\sigma_{m}}=1 \quad \text { or } \quad \theta=45^{\circ} .

The modified Goodman diagram for this example is shown in Fig. 5.45

Step III Permissible stress amplitude
Refer to Fig. 5.45. The coordinates of the point X are determined by solving the following two equations simultaneously

(i) Equation of line AB

\frac{S_{a}}{153.34}+\frac{S_{m}}{440}=1           (a).

(ii) Equation of line OX

\frac{S_{a}}{S_{m}}=\tan \theta=1               (b).

Solving the two equations,

S_{a}=S_{m}=113.71 N / mm ^{2} .

Step IV Thickness of plate

\text { Since } \quad \sigma_{a}=\frac{S_{a}}{(f s)} \quad \therefore \quad\left(\frac{375}{t}\right)=\frac{113.71}{3.5} .

t = 11.54 mm.