Question 17.8: A rectangular-cross-section header is made from two C-sectio...
A rectangular-cross-section header is made from two C-sections and butt welded along the center of the short sides.The weld joints are spot examined with the backing strip left in place (from Table UW-12 of the ASME Code, VIII-1, E =0.8). The design pressure is 115 psi, and the material is SA-515 Grade 70 with an allowable stress S =17,500 psi. The long side is 13.5 by 1 in. thick, and the short side is 6 by 0.625 in. thick. One long side contains a row of 1.5 in. diameter holes on 3.75 in. centers. Is the design acceptable?
1) Calculate the ligament efficiency of the long side with the tube holes:
e_{ m }=e_{ b }=\frac{3.75-1.5}{3.75} =0.60
2) Calculate the moments of inertia of both the long side and the short side:
I_{1}=\frac{t_{1}^{3}}{12}=\frac{(0.625)^{3}}{12} =0.0203
I_{2}=\frac{t_{2}^{3}}{12}=\frac{(1)^{3}}{12} =0.0833
3) Calculate the bending moment 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{115}{12}\left[\frac{(13.5)^{3} / 0.0833+(6)^{3} / 0.0203}{13.5 / 0.0833+6 / 0.0203}\right]= 841.35.
4) Calculate the bending moment at the midpoint of the long side, M_{M}, using Eq. (17.19):
M_{ M }=M_{ Q }-\frac{P h^{2}}{8}=841.35 -\frac{(115)(13.5)^{2}}{8}=-1778.505) Calculate the bending moment at the midpoint of the short side, MN, using Eq. (17.20):
M_{ N }=M_{ Q }-\frac{P H^{2}}{8}=841.35-\frac{(115)(6)^{2}}{8} =323.85
6) Calculate the bending stress at the corner of the long side with the holes:
\left(S_{ b }\right)_{ QM }=\frac{M_{ Q } c_{2}}{I_{2}}=\frac{(841.35)(0.5)}{(0.0833)} =5050 psi
7) Calculate the bending stress at the midpoint of the long side with the holes ( e_{ b } =0.60; E =1.0):
\left(S_{ b }\right)_{ M }=\frac{M_{ M } c_{2}}{I_{2} e_{ b }}=\frac{(1778.50)(0.5)}{(0.0833)(0.6)} =17,790 psi
8) Calculate the bending stress at the corner of the short side (E =1.0):
\left(S_{ b }\right)_{ QN }=\frac{M_{ Q } c_{1}}{I_{1}}=\frac{(841.35)(0.3125)}{(0.0203)} =12,950 psi
9) Calculate the bending stress at the midpoint of the short side (E =0.80):
\left(S_{ b }\right)_{ N }=\frac{M_{ N } c_{1}}{I_{1}}=\frac{(323.85)(0.3125)}{(0.0203)} =4990 psi
10) Calculate the membrane stress on the long side ( e_{ m } =0.60; E =1.0):
\left(S_{ m }\right)_{ M }=\frac{P H}{2 t e_{ m }}=\frac{(115)(6)}{2(1)(0.6)} =580 psi
Allowable stress = SE = (17,500)(1.0)
= 17,500 psi.
11) Calculate the membrane stress on the short side (E =0.80):
\left(S_{ m }\right)_{ N }=\frac{P h}{2 t_{1}}=\frac{(115)(13.5)}{2(0.625)} =1240 psi
Allowable stress = SE = (17,500)(0.8)
= 14,000 psi.
12) Total stress at the corner of long side:
\left(S_{ t }\right)_{ QM }=\left(S_{ m }\right)_{ M }+\left(S_{ b }\right)_{ QM } =580+5050
= 5630 psi
Allowable stress = 1.5SE = 1.5(17,500)(1.0)
= 26,250 psi.
13) Total stress at midpoint of long side:
\left(S_{ t }\right)_{ M }=\left(S_{ m }\right)_{ M }+\left(S_{ b }\right)_{ M } =580+17,790
= 18,370 psi
Allowable stress = 1.5SE = 1.5(17,500)(1.0)
= 26,250 psi.
14) Total stress at the corner of short side:
\left(S_{ t }\right)_{ QN }=\left(S_{ m }\right)_{ N }+\left(S_{ b }\right)_{ QN }= 1240 + 12,950 = 14,190 psi
Allowable stress = 1.5SE = (1.5)(17,500)(0.80)
= 21,000 psi.
15) Total stress at midpoint of short side:
\left(S_{ t }\right)_{ N }=\left(S_{ m }\right)_{ N }+\left(S_{ b }\right)_{ N }= 1240 + 4990 = 6230 psi
Allowable stress = 1.5SE = (1.5)(17,500)(0.8)
= 21,000 psi.
All calculated stresses are less than the allowable stresses and are acceptable.