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Aircraft Structures for Engineering Students, Fourth Edition by T.H.G. Megson

By T.H.G. Megson

Plane constructions for Engineering scholars is the major self contained airplane buildings direction textual content. It covers all basic topics, together with elasticity, structural research, airworthiness and aeroelasticity. Now in its fourth version, the writer has revised and up to date the textual content all through and additional new case research and labored instance fabric to make the textual content much more obtainable. features a strategies guide to be had to all adopting lecturers. * New association aids figuring out of the basics of structural research and emphasizes purposes to aircraftstructures* New labored examples through the textual content relief realizing and relate techniques to actual international purposes* extra insurance comprises digital paintings, pressure box beams, put up buckling habit, fabric houses, composite buildings and crack propagation* an in depth plane layout venture case learn indicates the appliance of the main recommendations within the booklet* finish of bankruptcy workouts and accompanying Instructor's handbook at

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71) A similar approach may be adopted for a 60◦ rosette. 7 A bar of solid circular cross-section has a diameter of 50 mm and carries a torque, T , together with an axial tensile load, P. A rectangular strain gauge rosette attached to the surface of the bar gave the following strain readings: εa = 1000 × 10−6 , εb = −200 × 10−6 and εc = −300 × 10−6 where the gauges ‘a’ and ‘c’ are in line with, and perpendicular to, the axis of the bar, respectively. 3, calculate the values of T and P. Substituting the values of εa , εb and εc in Eq.

8 Mohr’s circle of stress Now substituting the values σx = 160 N/mm2 , 200 N/mm2 we have σy = −120 N/mm2 and σ = σ1 = τxy = ±113 N/mm2 Replacing cot θ in Eq. (ii) by 1/tan θ from Eq. (i) yields a quadratic equation in σ 2 σ 2 − σ(σx − σy ) + σx σy − τxy =0 (iii) The numerical solutions of Eq. (iii) corresponding to the given values of σx , σy and τxy are the principal stresses at the point, namely σ1 = 200 N/mm2 (given) σII = −160 N/mm2 Having obtained the principal stresses we now use Eq. 15) to find the maximum shear stress, thus τmax = 200 + 160 = 180 N/mm2 2 The solution is rapidly verified from Mohr’s circle of stress (Fig.

5 mm 1200 Nm 50 kN Fig. 2. 3 N/mm2 Fig. 2. The shear stress, τxy , at the same point due to the torque is obtained from Eq. e. 3 N/mm2 The stress system acting on a two-dimensional rectangular element at the point is shown in Fig. 11. Note that since the element is positioned at the bottom of the beam the shear stress due to the torque is in the direction shown and is negative (see Fig. 8). 9). 3 N/mm2 the negative sign arising from the fact that it is in the opposite direction to τxy in Fig. 8.

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