Question 17.11: The header mentioned in Example 17.9 is to be built with a s...
The header mentioned in Example 17.9 is to be built with a shortened length L_{1} =12 in. What is done to the header axes for analysis, and what are the stresses at the midpoints of the short side and the long side?
Calculate the header-geometry properties as follows:
\frac{L_{1}}{H}=\frac{12}{7.25} =1.66
\frac{L_{1}}{h}=\frac{12}{14} =0.86
Since L_{1}/h < 1.0, the axes of the header must be reoriented for analysis. In the new terms, L_{1} =14 in., h=12 in., and H =7.25 in.
The first assumption is that the original flat end closures, which have now become sides, are also 1 in. thick. On that basis, I =0.0833 remains. We have
\alpha=\frac{H}{h}=\frac{7.25}{12} =0.6042
K = 0.6042.
Check if the strengthening factors apply:
\frac{L_{1}}{H}=\frac{14}{7.25} =1.93 C_{1} = 0.99 and C_{2} = 0.99
\frac{L_{1}}{h}=\frac{14}{12} =1.17 C_{1} = 0.62 and C_{2} = 0.68.
Strengthening factors apply to both short side and long side.
Calculate the bending stress at the midpoint of the short side using Eq. (17.41):
\left(S_{ b }\right)_{ N }=\text { Eq. }(17.25) \times C_{1} (17.41)
\left(S_{ b }\right)_{ N }=\frac{(150)(0.5)}{12(0.0833)} \times\left[1.5(7.25)^{2}-(12)^{2} \frac{1+(0.6042)^{3}}{1+0.6042}\right] (0.99)
\left(S_{ b }\right)_{ N } = 2280 psi.
Calculate the bending stress at the midpoint of the long side using Table 17.1, first equation:
\left(S_{ b }\right)_{ M }=\frac{(150)(0.5)(12)^{2}}{12(0.0833)} \times\left[1.5-\frac{1+(0.6042)^{3}}{1+0.6042}\right] (0.62)= 4950 psi.
Table 17.1 Bending stress values in rectangular headers.
| Figure | Location of weld between | Bending stress at joint, \pm\left(S_{b}\right)_{j} (psi) |
| 17.3a | M and Q | \frac{P c}{12 I_{2}}\left\{h^{2}\left[1.5-\left(\frac{1+\alpha^{2} K}{1+K}\right)\right]-6 d_{ j }^{2}\right\} |
| 17.3a | N and Q | \frac{P_{c}}{12 I_{1}}\left[1.5 H^{2}-h^{2} \frac{\left(1+\alpha^{2} K\right)}{1+K}-6 d_{ j }^{2}\right] |
| 17.3b | M and Q | \frac{P c}{2 I_{22}}\left\{\frac{h^{2}}{2 N}\left[\left(K_{2}-k_{1} k_{2}\right)+\alpha^{2} k_{2}\left(K_{2}-k_{2}\right)\right]-\frac{h^{2}}{4}+d_{ j }^{2}\right\} |
| 17.3b | M_{1} and Q_{1} | \frac{P c}{2 I_{2}}\left\{\frac{h^{2}}{2 N}\left[\left(K_{1} k_{1}-k_{2}\right)+\alpha^{2} k_{2}\left(K_{1}-k_{2}\right)\right]-\frac{h^{2}}{4}+d_{ j }^{2}\right\} |
| 17.3c | A and B | \frac{c}{I_{1}}\left(M_{ A }+\frac{P d_{ j }^{2}}{2}\right) |
| 17.3c | D and C | \frac{c}{I_{1}}\left[M_{ A }+\frac{P}{2}\left(L^{2}+2 a L-2 a l_{1}-l_{1}^{2}+d_{ j }^{2}\right)\right] |
| 17.3d | M and Q | \frac{P p h^{2}}{24 Z_{21}}\left[3-2\left(\frac{1+\alpha_{1}^{2} k}{1+k}\right)-\frac{12 d_{ j }^{2}}{h^{2}}\right] |
| 17.3d | N and Q | \frac{P p}{24 Z_{1}}\left[3 H^{2}-2 h^{2}\left(\frac{1+\alpha_{1}^{2} k}{1+k}\right)-12 d_{j}^{2}\right] |
| 17.3e | A and B | \frac{1}{Z_{21}}\left(M_{ A }+P \frac{p d_{ j }^{2}}{2}\right) |
| 17.3e | B and C | \frac{c}{I_{2}}\left(M_{ A }+P \frac{p d_{ j }^{2}}{2}\right) |
| 17.3e | F and E | \frac{1}{Z_{11}}\left\{M_{ A }+P \frac{p}{2}\left[\left(L+L_{11}\right)^{2}+2 a\left(L+L_{11}-l_{1}-l_{11}\right) -\left(l_{1}+l_{11}\right)^{2}+d_{ j }^{2}\right]\right\} |
| 17.3e | E and D | \frac{c}{I_{1}}\left\{M_{ A }+P \frac{p}{2}\left[L^{2}+2 L L_{11}+L_{11}^{2}-2 l_{1} l_{11}-l_{11}^{2}-l_{1}^{2} +2 a\left(L+L_{11}-l_{1}-l_{11}\right)+d_{ j }^{2}\right]\right\} |
| 17.5a | A and B | \frac{P c}{I_{2}}\left(\frac{-L C_{1}}{6 A}+\frac{d_{ j }^{2}}{2}\right) |
| 17.5b | A and B | \frac{P p}{Z_{11}}\left(\frac{-L C_{2}}{6 A_{3}}+\frac{d_{ j }^{2}}{2}\right) |
| Source: Courtesy of American Society of Mechanical Engineers, from Table 13-18.1 of ASME Code VIII-1. | ||