Aircraft Structures for Engineering Students 7th Edition T. H. G Megson | Solutions Manual

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Aircraft Structures for Engineering Students Seventh Edition Solutions Manual Solutions to Chapter 1 Problems S.1.1 The principal stresses are given directly by Eqs (1.11) and (1.12) in which σx... ¼80 N/mm2, σy¼0 (or vice versa), and τxy¼45 N/mm2. Thus, from Eq. (1.11), σI ¼ 80 2 + 1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p802 + 4452 i.e., σI ¼ 100:2 N=mm2 From Eq. (1.12), σII ¼ 80 2  1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p802 + 4452 i.e., σII ¼ 20:2 N=mm2 The directions of the principal stresses are defined by the angle θ in Fig. 1.8(b) in which θ is given by Eq. (1.10). Hence, tan2θ ¼ 245 800 ¼ 1:125 which gives θ ¼ 24°110 and θ ¼ 114°110 It is clear from the derivation of Eqs (1.11) and (1.12) that the first value of θ corresponds to σI while the second value corresponds to σII. Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence from Eq. (1.15), τmax ¼ 100:2  ð Þ 20:2 2 ¼ 60:2 N=mm2 and will act on planes at 45° to the principal planes. S.1.2 The principal stresses are given directly by Eqs (1.11) and (1.12) in which σx¼50 N/mm2, σ y¼–35 N/mm2, and τxy¼40 N/mm2. Thus, from Eq. (1.11), 3σI ¼ 5035 2 + 1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qð Þ 50 + 35 2 + 4402 i.e., σI ¼ 65:9 N=mm2 and from Eq. (1.12), σII ¼ 5035 2  1 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qð Þ 50 + 35 2 + 4402 i.e., σII ¼ 50:9 N=mm2 From Fig. 1.8(b) and Eq. (1.10), tan2θ ¼ 240 50 + 35 ¼ 0:941 which gives θ ¼ 21°380ð Þ σI and θ ¼ 111°380ð Þ σII The planes on which there is no direct stress may be found by considering the triangular element of unit thickness shown in Fig. S.1.2 where the plane AC represents the plane on which there is no direct stress. For equilibrium of the element in a direction perpendicular to AC, 0 ¼ 50ABcosα35BCsinα + 40ABsinα + 40BCcosα (i) Dividing through Eq. (i) by AB, 0 ¼ 50cosα35tanαsinα + 40sinα + 40tanαcosα which, dividing through by cos α, simplifies to 0 ¼ 5035tan2α + 80tanα FIG. S.1.2 4 Solutions Manualfrom which tanα ¼ 2:797 or 0:511 Hence, α ¼ 70°210 or 27°50 S.1.3 The construction of Mohr’s circle for each stress combination follows the procedure described in Section 1.8 and is shown in Figs S.1.3(a)–(d). FIG. S.1.3(a) FIG. S.1.3(b) FIG. S.1.3(c) Solutions to Chapter 1 Problems 5S.1.4 The principal stresses at the point are determined, as indicated in the question, by transforming each state of stress into a σ x, σy, τxy stress system. Clearly, in the first case, σx¼0, σy¼10 N/mm2, τxy¼0 (Fig. S.1.4(a)). The two remaining cases are transformed by considering the equilibrium of the triangular element ABC in Figs S.1.4(b), (c), (e), and (f). Thus, using the method described in Section 1.6 and the principle of superposition (see Section 5.9), the second stress system of Figs S.1.4(b) and (c) becomes the σ x, σy, τxy system shown in Fig. S.1.4(d) while the third stress system of Figs S.1.4(e) and (f) transforms into the σx, σy, τxy system of Fig. S.1.4(g). Finally, the states of stress shown in Figs S.1.4(a), (d), and (g) are superimposed to give the state of stress shown in Fig. S.1.4(h) from which it can be seen that σI¼σII ¼15 N/mm2 and that the x and y planes are principal planes. FIG. S.1.3(d) FIG. S.1.4(a) [Show More]

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