If the vertical plane of symmetry in the beam of P.21.3 is swept back by 25° as shown in plan in Fig. P.21.4, the span of 3 m is retained and the 40 kN load is applied as in P.21.3, calculate the flange and stringer loads and the shear flow distribution at a section 1.5 m from the built-in end.
As in S.21.3 and referring to Figs. P.21.4, S.21.4(a), and S.21.4(b),
\begin{aligned}& P_{z, 1}=-P_{z, 2}=-133300 N \\&P_{z, 3}=P_{z, 5}=-P_{z, 4}=-P_{z, 6}=-66650N\end{aligned}The solution now continues in Table S.21.4.
From Table S.21.4,
Then S_{x,w}=0,\,S_{y,w}=40-13.4=26.6\,\mathrm{kN}
The q_{\mathrm{b}} shear flows are the same as in S.21.3, so taking moments about boom 2,
26.6\times10^{3}\times550=59.2\times225\times500+29.6\times250\times225+2\times500\times225q_{s,0}from which q_{s,0}=27.6\ \mathrm{N/mm.}
The final shear flows are then as shown in Fig.S.21.4(c).
Table S.21.4 | |||||||||||||||||||||
Boom | \begin{aligned}& P z, r \\& (k N)\end{aligned} | \delta x_r / \delta z | \delta y_r / \delta z | \begin{gathered}P _{ x , r } \\( k N )\end{gathered} | \begin{gathered}P _{ y , r } \\( k N )\end{gathered} | \underset{( k N )}{ P _{ r }} | \begin{gathered}\xi_r \\( mm )\end{gathered} | \begin{gathered}\eta _{ r } \\( m m )\end{gathered} | \begin{aligned}& P _{x, r} \eta_r \\& ( kNm )\end{aligned} | \begin{aligned}& P_{y, r} \xi_r \\& ( kNm )\end{aligned} | |||||||||||
1 | -133.3 | 0.47 | –0.025 | –62.7 | 3.3 | –147.3 | 250 | 112.5 | 7.05 | –0.83 | |||||||||||
2 | 133.3 | 0.47 | 0.025 | 62.7 | 3.3 | 147.3 | 250 | 112.5 | 7.05 | 0.83 | |||||||||||
3 | -66.7 | 0.47 | –0.025 | –31.3 | 1.7 | –73.7 | 0 | 112.5 | 3.52 | 0 | |||||||||||
4 | 66.7 | 0.47 | 0.025 | 31.3 | 1.7 | 73.7 | 0 | 112.5 | 3.52 | 0 | |||||||||||
5 | -66.7 | 0.47 | –0.025 | –31.3 | 1.7 | –73.7 | 250 | 112.5 | 3.52 | –0.42 | |||||||||||
6 | 66.7 | 0.47 | 0.025 | 31.3 | 1.7 | 73.7 | 250 | 112.5 | 3.52 | 0.42 |