Page 391 - Aircraft Stuctures for Engineering Student
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372 Stress analysis of aircraft components
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
M, = -100 x 1.9 = -190kNm
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,, = 4 x 900 x 295’ + 2 x 1200 x 295’ = 5.22 x lo8 1ll111~
and
-190 x lo6
0z:r = = -0.364~~
5.22 x lo8 Yr
Hence
Pi = P3 = -P4 = -P6 = -0.364 X 295 X 900 = -96642N
and
P2 -Ps = -0.364 x 295 x 1200 = -128 856N
At a section 1.9 m from the built-in end
M, = -100 x 2.1 = -2lOkNm
and the section dimensions are width = 1.22m, depth = 0.61 m so that
I,, = 4 x 900 x 305’ + 2 x 1200 x 305’ = 5.58 x lo8 mm4
and
-210 x lo6
0z>r = = -0.376~~
5.58 x lo8 Yr
Hence
Pi = P3 = -P4 = -P6 = -0.376 x 305 x 900 = -103212N
and
P2 = -Ps = -0.376 x 305 x 1200 = -137616N
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 = 8760N while
the tensile load in boom 5 increases by 8760N. Therefore, the change in boom load
per unit length is given by
6570
AP1 = AP3 = -AP4 = -AP6 = -
32.85 N
200
and