$F = \sigma A$

$F = 3.5 [ \, \frac{1}{4} \pi(430^2) \, ]$

$F = 508\,270.42 \, \text{N}$

$P = F$

$(\sigma_{\text{bolt}} A) \, n = 508\,270.42 \, \text{N}$

$(80 - 55)[ \, \frac{1}{4} \pi (40^2) \, ]n = 508\,270.42$

$n = 16.19$ say 17 bolts *answer*

**Circumferential stress (consider 1-m strip):**

$F = pA = 3.5 [ \, 430(1000) \, ]$

$F = 1\,505\,000 \, \text{N}$

$2T = F$

$2[ \, \sigma_t (1000)(10)] = 1\,505\,000$

$\sigma_t = 75.25 \, \text{MPa}$ *answer*

*Discussion:*

To avoid steam leakage, it is necessary to tighten the bolts initially in order to press the gasket to the flange. If the internal pressure will cause 110 MPa of stress to each bolt causing it to fail, leakage will occur. If the failure is sudden, the cap may blow.