Question 21.3: Solve Example 21.2 by considering the differences in boom lo...
Solve Example 21.2 by considering the differences in boom load at sections of the beam either side of the specified section.
In this example the stringer areas do not vary along the length of the beam but the method of solution is identical.
We are required to find the shear flow distribution at a section 2 m from the built-in end of the beam. We therefore calculate the boom loads at sections, say 0.1 m either side of this section. Thus, at a distance 2.1 m from the built-in end
The dimensions of this section are easily found by proportion and are width = 1.18 m, depth = 0.59 m. Thus the second moment of area is
I_{x x}=4 \times 900 \times 295^2+2 \times 1200 \times 295^2=5.22 \times 10^8 \mathrm{~mm}^4and
\sigma_{z, r}=\frac{-190 \times 10^6}{5.22 \times 10^8} y_r=-0.364 y_rHence
P_1=P_3=-P_4=-P_6=-0.364 \times 295 \times 900=-96642 \mathrm{~N}
and
P_2=-P_5=-0.364 \times 295 \times 1200=-128856 \mathrm{~N}
At a section 1.9 m from the built-in end
M_x = −100 × 2.1 = −210 kN m
and the section dimensions are width = 1.22 m, depth = 0.61 m so that
and
\sigma_{z, r}=\frac{-210 \times 10^6}{5.58 \times 10^8} y_r=-0.376 y_rHence
P_1=P_3=-P_4=-P_6=-0.376 \times 305 \times 900=-103212 \mathrm{~N}
and
P_2=-P_5=-0.376 \times 305 \times 1200=-137616 \mathrm{~N}
Thus, there is an increase in compressive load of 103 212 − 96 642 = 6570 N in booms 1 and 3 and an increase in tensile load of 6570 N in booms 4 and 6 between the two sections. Also, the compressive load in boom 2 increases by 137 616 − 128 856 = 8760 N while the tensile load in boom 5 increases by 8760 N. Therefore, the change in boom load per unit length is given by
\Delta P_1=\Delta P_3=-\Delta P_4=-\Delta P_6=\frac{6570}{200}=32.85 \mathrm{~N}and
\Delta P_2=-\Delta P_5=\frac{8760}{200}=43.8 \mathrm{~N}The situation is illustrated in Fig. 21.9. Suppose now that the shear flows in the panels 12, 23, 34, etc. are q_{12}, q_{23}, q_{34}, etc. and consider the equilibrium of boom 2, as shown in Fig. 21.10, with adjacent portions of the panels 12 and 23. Thus
q_{23}+43.8-q_{12}=0
or
q_{23}=q_{12}-43.8
Similarly
The moment resultant of the internal shear flows, together with the moments of the components P_{y,r} of the boom loads about any point in the cross-section, is equivalent to the moment of the externally applied load about the same point. We note from Example 21.2 that for moments about the centre of symmetry
\sum_{r=1}^6 P_{x, r} \eta_r=0 \quad \sum_{r=1}^6 P_{y, r} \xi_r=0Therefore, taking moments about the centre of symmetry
\begin{aligned}100 \times 10^3 \times 600=& 2 q_{12} \times 600 \times 300+2\left(q_{12}-43.8\right) 600 \times 300 \\&+\left(q_{12}-76.65\right) 600 \times 600+\left(q_{12}+32.85\right) 600 \times 600\end{aligned}from which
q_{12} = 62.5 N/mm
whence
\begin{array}{ll}q_{23}=19.7 \mathrm{~N} / \mathrm{mm} & q_{34}=-13.2 \mathrm{~N} / \mathrm{mm} \quad q_{45}=19.7 \mathrm{~N} / \mathrm{mm} \\q_{56}=63.5 \mathrm{~N} / \mathrm{mm} & q_{61}=96.4 \mathrm{~N} / \mathrm{mm}\end{array}so that the solution is almost identical to the longer exact solution of Example 21.2.
The shear flows q_{12}, q_{23}, etc. induce complementary shear flows q_{12}, q_{23}, etc. in the panels in the longitudinal direction of the beam; these are, in fact, the average shear flows between the two sections considered. For a complete beam analysis the above procedure is applied to a series of sections along the span. The distance between adjacent sections may be taken to be any convenient value; for actual wings distances of the order of 350–700 mm are usually chosen. However, for very small values small percentage errors in P_{z,1} and P_{z,2} result in large percentage errors in ΔP. On the other hand, if the distance is too large the average shear flow between two adjacent sections may not be quite equal to the shear flow midway between the sections.
