Question 17.7: Determine the adequacy of a rectangular-cross-section header...
Determine the adequacy of a rectangular-cross-section header with a design pressure of 150 psi and made from a seamless forging with an allowable stress S =12 500 psi. The header inner dimensions are 14 in. by 7.25 in. with a constant thickness of 1 in. All openings are reinforced. Maximum temperature is 100 °F.
1) Calculate the moment of inertia:
I_{1}=I_{2}=\frac{t^{3}}{12}=\frac{(1)^{3}}{12} =0.0833
2) Calculate bending moment ^{1} at the corner, M_{Q}, using Eq. (17.18):
M_{ Q }=\frac{P}{12}\left(\frac{h^{3} / I_{2}+H^{3} / I_{1}}{h / I_{2}+H / I_{1}}\right) (17.18)
M_{ Q }=\frac{150}{12}\left(\frac{(14)^{3} / 0.0833+(7.25)^{3} / 0.0833}{14 / 0.0833+7.25 / 0.0833}\right)= 1838.28.
3) Calculate bending moment ^{1} at midpoint of long side, M_{M}, using Eq. (17.19):
M_{ M }=M_{ Q }-\frac{P h^{2}}{8}=1838.28-\frac{(150)(14)^{2}}{8}=-1836.724) Calculate bending moment at midpoint of short side, M_{N}, using Eq. (17.20):
M_{ N }=M_{ Q }-\frac{P H^{2}}{8}=1838.28 -\frac{(150)(7.25)^{2}}{8}=852.735) Calculate bending stress at the corner of long side:
\left(S_{ b }\right)_{ QM }=\frac{(1838.28)(0.5)}{0.0833} = 11,030 psi (T in).
6) Calculate bending stress at midpoint of long side:
\left(s_{ b }\right)_{ M }=\frac{(-1836.72)(-0.5)}{0.0833} = 11,020 psi (T out).
7) Calculate bending stress at the corner of short side:
\left(S_{ b }\right)_{ QN } = 11,030 psi (T in).
8) Calculate bending moment at midpoint of short side:
\left(S_{ b }\right)_{ N }=\frac{(852.73)(0.5)}{0.0833} = 5120 psi (T in).
9) Calculate membrane stress on long side:
\left(S_{ m }\right)_{ M }=\frac{P H}{2 t}=\frac{(150)(7.25)}{2(1)} = 540 psi.
10) Calculate membrane stress on short side:
\left(S_{ m }\right)_{ N }=\frac{P h}{2 t}=\frac{(150)(14)}{2(1)} = 1050 psi.
11) Total stress at the corner of long side:
\left(S_{ t }\right)_{ QM }=\left(S_{ m }\right)_{ M }+\left(S_{ b }\right)_{ QM }= 540 + 11,030 = 11,570 psi (T in).
12) Total stress at midpoint of long side:
\left(S_{ t }\right)_{ M }=\left(S_{ m }\right)_{ M }+\left(S_{ b }\right)_{ M } =540+11,020
= 11,560 psi (T out).
13) Total stress at the corner of short side:
\left(S_{ t }\right)_{ QN }=\left(S_{ m }\right)_{ N }+\left(S_{ b }\right)_{ QN } =1050+11,030
= 12,080 psi (T in).
14) Total stress at midpoint of short side:
\left(S_{ t }\right)_{ N }=\left(S_{ m }\right)_{ N }+\left(S_{ b }\right)_{ N } =1050+5120
= 6170 psi (T in).
Since there are no butt welds, E =1.0, and SE = 12,500(1)=12,500 psi. For membrane plus bending, the allowable stress is 1.5SE =1.5(12,500)(1)= 18,750 psi. All stresses are less than allowable stress values, so 1 in. thickness is satisfactory.