Question 16.10: The vessel given in Example 16.3 is to be constructed from S...
The vessel given in Example 16.3 is to be constructed from SA-516 Grade 60 material and designed for 475 psi at 650 °F. The weld joint efficiency is E =1.0. What are the total longitudinal stresses on both the windward and leeward sides at the support line? What are the allowable tensile and compressive stresses?
From 1A of the ASME Code, II-D, for SA-516 Grade 60 at 650 °F, the allowable tensile stress is S_{t} =15 800 psi. The weight of the shell and upper head above the support line is determined from Example 16.6 as
W = 65,210 + 0.5(3320) = 66,870 lb.
The overturning moment due to the wind loading at the support line is determined from Example 16.5 as
M_{0} = (2,216,950)(12) = 26,603,340 in.-lb.
Determine the total longitudinal stress \sigma_{ L } using Eq. (16.21) as follows:
\sigma_{ L }=+P\left(\frac{R}{2 t}-0.2\right) \pm \frac{W}{\pi D_{ m } t} \pm \frac{4 M_{ e }}{\pi D_{ m }^{2} t} (16.21)
\sigma_{L}=+475\left(\frac{30}{2 \times 1}-0.2\right) \pm \frac{66,870}{\pi(61)(1)} \pm \frac{4(26,603,340)}{\pi(61)^{2}(1)}\sigma_{L}=+7030 \pm 350 \pm 9100
Windward-Side Stresses
1) Pressure + dead load+ wind load = +7030 −350 + 9100 = 15 780 psi tension.
2) No pressure + dead load+ wind load = 0 −350 + 9100 = 8750 psi tension.
3) Dead load only = 350 psi compression.
Leeward-Side Stresses
1) Internal pressure + dead load + wind load = + 7030 – 350 – 9100 = 2420 psi compression.
2) No internal pressure + dead load+ wind load = 0 – 350 – 9100 = 9450 psi compression.
Allowable Stress – Tension
S_{t} = 15,800 psi for SA-516 Grade 60 at 650°F
Allowable Stress – Compression
A=\frac{0.125}{R_{ o } / t}=\frac{0.125}{31 / 1}=0.0040and from Figure 8.11 and also Figure CS-2 of the ASME Code, II-D, the value of B is 11,500. This gives an allowable stress of S_{c} =11,500 psi. All calculated stresses are less than the allowable stresses.
