The header in Example 17.9 is to be built with a shortened length where L1 = 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?

Question

The header in Example 17.9 is to be built with a shortened length where [latex]L_1[/latex] = 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?

Accepted Answer

Calculate the header geometry properties as follows:

[latex]\frac{L_{1}}{H}=\frac{12}{7.25}=1.66[/latex]

[latex]\frac{L_{1}}{h}=\frac{12}{14}=0.86[/latex]

Since [latex]L_{1} / h<1.0[/latex], the axes of the header must be reoriented for analysis. In the new terms, [latex]L_{1}[/latex] =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.

[latex]\alpha=\frac{H}{h}=\frac{7.25}{12}=0.6042[/latex]

K = 0.6042

Check if the strengthening factors apply:

[latex]\frac{L_{1}}{H}=\frac{14}{7.25}=1.93[/latex] [latex]C_{1}=0.99[/latex] and [latex]C_{2}=0.99[/latex]

[latex]\frac{L_{1}}{h}=\frac{14}{12}=1.17[/latex] [latex]C_{1}=0.62[/latex] and [latex]C_{2}=0.68[/latex]

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

[latex]\left(S_{b} ight)_{N}= Eq .17 .25 imes C_{1}[/latex] (17.41)

[latex]\left(S_{b} ight)_{N}=\frac{M_{N} t_{1}}{2 I_{1}} imes \frac{1}{E}[/latex] (17.25)

[latex]\left(S_{b} ight)_{N}=\frac{(150)(0.5)}{12(0.0833)(1)}\left[1.5(7.25)^{2}-(12)^{2} \frac{1+(0.6042)^{3}}{1+0.6042} ight] (0.99)[/latex]

[latex]\left(S_{b} ight)_{N}=2280 \ psi[/latex]

Calculate the bending stress at the midpoint of the long-side using Table 17.1, first equation:

Table 17.1
Figure	Location of Weld Between	Bending Stress at Joint [latex]\pm\left(S_{b} ight)_{j}( psi )[/latex]
17.3 a	M and Q	[latex]\frac{P c}{12 I_{2} E}\left\{h^{2}\left[1.5-\left(\frac{1+\alpha^{2} K}{1+K} ight) ight]-6 d_{j}^{2} ight\}[/latex]
17.3 a	N and Q	[latex]\frac{P c}{12 I_{1} E}\left[1.5 H^{2}-h^{2} \frac{\left(1+\alpha^{2} K ight)}{1+K}-6 d_{j}^{2} ight][/latex]
17.3 b	M and Q	[latex]\frac{Pc}{2 I_{22} E}\left\{\frac{h^{2}}{2 N}\left[\left(K_{2}-k_{1} k_{2} ight)+\alpha^{2} k_{2}\left(K_{2}-k_{2} ight) ight]-\frac{h^{2}}{4}+d_{j}^{2} ight\}[/latex]
17.3 b	[latex]M_{1}[/latex] and [latex]Q_{1}[/latex]	[latex]\frac{P c}{2 I_{2} E}\left\{\frac{h^{2}}{2 N}\left[\left(K_{1} k_{1}-k_{2} ight)+\alpha^{2} k_{2}\left(K_{1}-k_{2} ight) ight]-\frac{h^{2}}{4}+d_{j}^{2} ight\}[/latex]
17.3 c	A and B	[latex]\frac{c}{I_{1} E}\left(M_{A}+\frac{P d_{j}^{2}}{2} ight)[/latex]
17.3 c	D and C	[latex]\frac{c}{I_{1} E}\left[M_{A}+\frac{P}{2}\left(L^{2}+2 a L-2 a l_{1}-l_{1}^{2}+d_{j}^{2} ight) ight][/latex]
17.3 d	M and Q	[latex]\frac{P p h^{2}}{24 Z_{21} E}\left[3-2\left(\frac{1+\alpha_{1}^{2} k}{1+k} ight)-\frac{12 d_{j}^{2}}{h^{2}} ight][/latex]
17.3 d	N and Q	[latex]\frac{P p}{24 Z_{11} E}\left[3 H^{2}-2 h^{2}\left(\frac{1+\alpha_{1}^{2} k}{1+k} ight)-12 d_{j}^{2} ight][/latex]
17.3 e	A and B	[latex]\frac{1}{Z_{21} E}\left(M_{A}+P \frac{p d_{j}^{2}}{2} ight)[/latex]
17.3 e	B and C	[latex]\frac{c}{I_{2} E}\left(M_{A}+P \frac{p d_{j}^{2}}{2} ight)[/latex]
17.3 e	F and E	[latex]\frac{1}{Z_{11} E}\left\{M_{A}+P \frac{p}{2}\left[\left(L+L_{11} ight)^{2}+2 a\left(L+L_{11}-l_{1}-l_{11} ight)-\left(l_{1}+l_{11} ight)^{2}+d_{j}^{2} ight] ight\}[/latex]
17.3 e	E and D	[latex]\frac{c}{I_{1} E}\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} ight)+d_{j}^{2} ight] ight\}[/latex]
17.5 a	A and B	[latex]\frac{P c}{I_{2} E}\left(\frac{-L C_{1}}{6 A}+\frac{d_{j}^{2}}{2} ight)[/latex]
17.5 b	A and B	[latex]\frac{P p}{Z_{11} E}\left(\frac{-L C_{2}}{6 A_{3}}+\frac{d_{j}^{2}}{2} ight)[/latex]
Courtesy American Society of Mechanical Engineers, from Table 13-18.1 of ASME Code VIII-1.

[latex]\left(S_{b} ight)_{M}=\frac{(150)(0.5)(12)^{2}}{12(0.0833)(1)}\left[1.5-\frac{1+(0.6042)^{3}}{1+0.6042} ight](0.62)=4950 \ psi[/latex]

Question 17.11: The header in Example 17.9 is to be built with a shortened l...

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