By T.H.G. Megson
Based at the author's best-selling textual content "Aircraft buildings for Engineering Students", this short publication covers the fundamentals of structural research as utilized to airplane constructions. insurance of elasticity, strength equipment and digital paintings set the degree for discussions of airworthiness/airframe lots and tension research of plane elements. quite a few labored examples, illustrations, and pattern difficulties exhibit how one can practice the options to reasonable occasions. Self-contained, this value-priced ebook is a superb source for someone studying the topic. It covers the middle recommendations in approximately two hundred fewer pages via elimination a few not obligatory subject matters like structural vibrations and aeroelasticity. It deals systematic step-by-step systems within the labored examples. it really is self-contained, with whole derivations for key equations.
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Extra resources for Introduction to Aircraft Structural Analysis
9) gives ∂ 4φ ∂ 4φ ∂ 4φ + 2 2 2 + 4 = 2 × 60Dy − 120Dy = 0 4 ∂x ∂x ∂y ∂y Therefore, the biharmonic equation is satisﬁed, and the stress function is valid. From Fig. 3, σy = 0 at y = h so that, from Eq. (i) 2A + 2BH + 10Dh3 = 0 (iv) Also, from Fig. 3, σy = −q at y = −h so that, from Eq. (i) 2A − 2BH − 10Dh3 = −q (v) Again, from Fig. 3, τxy = 0 at y = ±h giving, from Eq. (iii) 2Bx + 30Dxh2 = 0 so that 2B + 30Dh2 = 0 (vi) At x = 0, there is no resultant moment applied to the beam; that is, h Mx=0 = −h h σx y dy = (6Cy2 − 20Dy4 ) dy = 0 −h that is, Mx=0 = [2Cy3 − 4Dy5 ]h−h = 0 or C − 2Dh2 = 0 (vii) Subtracting Eq.
33) becomes (CE)2 (1 + 2εy ) = (CF)2 (1 + 2εn ) + (FE)2 (1 + 2εn+π/2 ) − 2(CF)(FE)γ From Fig. 16(a), (CE)2 = (CF)2 + (FE)2 and the preceding equation simpliﬁes to 2(CE)2 εy = 2(CF)2 εn + 2(FE)2 εn+π/2 − 2(CF)(FE)γ Dividing by 2(CE)2 and transposing, γ= εn sin2 θ + εn+π/2 cos2 θ − εy sin θ cos θ Substitution of εn+π/2 and εn from Eqs. 13 PRINCIPAL STRAINS If we compare Eqs. 34) with Eqs. 9), we observe that they may be obtained from Eqs. 9) by replacing σn by εn , σx by εx , σy by εy , τxy by γxy /2, and τ by γ /2.
18 Strain gauge rosette. 38 CHAPTER 1 Basic Elasticity The principal stresses are now obtained by substitution of εI and εII in Eqs. 52). 68) and Solving Eqs. 66) gives and A typical rosette would have α = β = 45◦ , in which case the principal strains are most conveniently found using the geometry of Mohr’s circle of strain. Suppose that the arm a of the rosette is inclined at some unknown angle θ to the maximum principal strain as in Fig. 18. Then, Mohr’s circle of strain is as shown in Fig. 19; the shear strains γa , γb , and γc do not feature in the analysis and are therefore ignored.