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Aircraft Structures for Engineering Students, 5th Edition Solution Manual

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    Aircraft Structures for Engineering Students Fifth Edition Solutions Manual T. H. G. Megson

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    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 σ x = 80N/mm 2 , σ y = 0 (or vice versa) and τ xy = 45N/mm 2 . Thus, from Eq. (1.11) 2 2 I 80 1 80 4 45 2 2 σ = + + × i.e. σ I = 100.2 N/mm From Eq. (1.12) 2 2 2 II 80 1 80 4 45 2 2 σ = − + × i.e. σ II = – 20.2 N/mm 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 2 2 45 tan 2 1.125 80 0 θ × = = − which gives θ = 24°11 ′ and θ = 114°11 ′ It is clear from the derivation of Eqs (1.11) and (1.12) that the first value of θ corresponds to σ I while the second value corresponds to σ II Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence from Eq. (1.15) . 2 max 100.2 ( 20.2) 60.2N / mm 2 τ − − = = and will act on planes at 45° to the principal planes.

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    4 Solutions Manual S.1.2 The principal stresses are given directly by Eqs (1.11) and (1.12) in which σ x = 50N/mm 2 , σ y = –35 N/mm 2 and τ xy = 40 N/mm 2 . Thus, from Eq. (1.11) 2 2 I 50 35 1 (50 35) 4 40 2 2 σ − = + + + × i.e. σ I = 65.9 N/mm and from Eq. (1.12) 2 2 2 II 50 35 1 (50 35) 4 40 2 2 σ − = − + + × i.e. σ II = –50.9 N/mm From Fig. 1.8(b) and Eq. (1.10) 2 2 40 tan 2 0.941 50 35 θ × = = + which gives θ = 21°38 ′(σ I ) and θ = 111°38 ′(σ II 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 50ABcos 35BCsin 40ABsin +40BC cos α α α α = − + (i) Fig. S.1.2

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    Solutions to Chapter 1 Problems 5 Dividing through Eq. (i) by AB 0 50cos 35 tan sin 40sin 40 tan cos α α α α α α = − + + which, dividing through by cos α , simplifies to 0 = 50–35 tan 2 from which α + 80 tan α tan α = 2 . 797 or –0 . 511 Hence α = 70°21 ′ or –27°5 ′ 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)

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    6 Solutions Manual Fig. S.1.3(c) Fig. S.1.3(d) S.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 , σ y , τ xy stress system. Clearly, in the first case σ x = 0, σ y = 10 N/mm 2 , τ xy = 0 (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

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    Solutions to Chapter 1 Problems 7 Figs S.1.4(b) and (c) becomes the σ x , σ y , τ xy system shown in Fig. S.1.4(d) while

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    8 Solutions Manual Fig. S.1.4(a) Fig. S.1.4(b) Fig. S.1.4(c) Fig. S.1.4(d) the third stress system of Figs S.1.4(e) and (f) transforms into the σ x , σ y , τ xy system of Fig. S.1.4(g).

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    Solutions to Chapter 1 Problems 9 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 σ I = σ II =15N/mm 2 and that the x and y planes are principal planes.

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    10 Solutions Manual Fig. S.1.4( e ) Fig. S.1.4( f ) Fig. S.1.4(g) Fig. S.1.4(h) S.1.5 The geometry of Mohr’s circle of stress is shown in Fig. S.1.5 in which the circle is constructed using the method described in Section 1.8. From Fig. S.1.5 σ x = OP 1 = OB– BC +CP 1 (i)

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    Solutions to Chapter 1 Problems 11 Fig. S.1.5 In Eq. (i) OB = σ I , BC is the radius of the circle which is equal to τ max 2 2 2 2 1 1 1 1 max CP CQ Q P . xy τ τ = − = − and Hence 2 2 1 max max x xy σ σ τ τ τ = − + − Similarly 2 2 2 1 OP OB BC CP in which CP CP y σ = = − − = Thus 2 2 I max max y xy σ σ τ τ τ = − − − S.1.6 From bending theory the direct stress due to bending on the upper surface of the shaft at a point in the vertical plane of symmetry is given by 6 2 4 25 10 75 75 N / mm 150 / 64 x My I σ π × × = = = × From the theory of the torsion of circular section shafts the shear stress at the same point is 6 2 4 50 10 75 75N / mm 150 / 32 xy Tr J τ π × × = = = ×

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    12 Solutions Manual Substituting these values in Eqs (1.11) and (1.12) in turn and noting that σ y = 0 2 2 I 75 1 75 4 75 2 2 σ = + + × i.e. 2 I 121.4N / mm σ = 2 2 II 75 1 75 4 75 2 2 σ = − + × i.e. 2 II 46.4N / mm σ = − The corresponding directions as defined by θ in Fig. 1.8(b) are given by Eq. (1.10) i.e. 2 75 tan 2 2 75 0 θ × = = − Hence I 31 43 ( ) θ σ ′ = ° and II 121 43 ( ) θ σ ′ = ° S.1.7 The direct strains are expressed in terms of the stresses using Eqs (1.42), i.e. 1 [ ( )] x x y z v E ε σ σ σ = − + (i) 1 [ ( )] y y x z v E ε σ σ σ = − + (ii) 1 [ ( )] z z x y v E ε σ σ σ = − + (iii) Then 1 [ 2 ( )] x y z x y z x y z e v E ε ε ε σ σ σ σ σ σ = + + = + + − + + i.e. (1 2 ) ( ) x y z v e E σ σ σ − = + + whence (1 2 ) y z x Ee v σ σ σ + = − −

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    Solutions to Chapter 1 Problems 13 Substituting in Eq. (i) 1 1 2 x x x Ee v E v ε σ σ     = − −     −     so that (1 ) 1 2 x x vEe E v v ε σ = + − − Thus (1 2 )(1 ) (1 ) x x vEe E v v v σ ε = + − + + or, since G = E/ 2(1 + ν ) (see Section 1.15) 2 x x e G σ λ ε = + Similarly 2 y y e G σ λ ε = + and 2 z z e G σ λ ε = + S.1.8 The implication in this problem is that the condition of plane strain also describes the condition of plane stress. Hence, from Eqs (1.52) 1 ( ) x x y v E ε σ σ = − (i) 1 ( ) y y x v E ε σ σ = − (ii) 2(1 ) (seeSection 1.15) xy xy xy v G E τ γ τ + = = (iii) The compatibility condition for plane strain is 2 2 2 2 2 (see Section 1.11) xy y x x y x y γ ε ε ∂ ∂ ∂ = + ∂ ∂ ∂ ∂ (iv) Substituting in Eq. (iv) for ε x , ε y and γ xy from Eqs (i)–(iii), respectively, gives 2 2 2 2 2 2 ( 1 ) ( ) ( ) xy y x x y v v v x y x y τ σ σ σ σ ∂ ∂ ∂ + = − + − ∂ ∂ ∂ ∂ (v)

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    14 Solutions Manual Also, from Eqs (1.6) and assuming that the body forces X and Y are zero 0 zy x x y τ σ ∂ ∂ + = ∂ ∂ (vi) 0 y xy y x σ τ ∂ ∂ + = ∂ ∂ (vii) Differentiating Eq. (vi) with respect to x and Eq. (vii) with respect to y and adding gives 2 2 2 2 2 2 0 xy y xy x y x x y x y τ σ τ σ ∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂ ∂ ∂ or 2 2 2 2 2 2 xy y x x y x y τ σ σ   ∂ ∂ ∂   = − +   ∂ ∂ ∂ ∂   Substituting in Eq. (v) 2 2 2 2 2 2 2 2 (1 ) ( ) ( ) y x y x x y v v v x y x y σ σ σ σ σ σ   ∂ ∂ ∂ ∂   − + + = − + −   ∂ ∂ ∂ ∂   so that 2 2 2 2 2 2 2 2 2 2 2 2 (1 ) y y y x x x v v x y x y x y σ σ σ σ σ σ     ∂ ∂ ∂ ∂ ∂ ∂     − + + = + − +     ∂ ∂ ∂ ∂ ∂ ∂     which simplifies to 2 2 2 2 2 2 2 2 0 y y x x x y x y σ σ σ σ ∂ ∂ ∂ ∂ + + + = ∂ ∂ ∂ ∂ or 2 2 2 2 ( ) 0 x y x y σ σ   ∂ ∂ + + =     ∂ ∂   S.1.9 Suppose that the load in the steel bar is P st and that in the aluminium bar is P al . Then, from equilibrium st al P P P + = (i) From Eq. (1.40) st al st al st st al al P P A E A E ε ε = =

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    Solutions to Chapter 1 Problems 15 Since the bars contract by the same amount st al st st al al P P A E A E = (ii) Solving Eqs (i) and (ii) st st al al st st st al al st st al al al A E A E P P P P A E A E A E A E = = + + from which the stresses are st al st al st st al al st st al al E E P P A E A E A E A E σ σ = = + + (iii) The areas of cross-section are 2 2 2 2 2 st al 75 (100 75 ) 4417.9 mm 3436.1mm 4 4 A A π π × − = = = = Substituting in Eq. (iii) we have 6 2 st 10 200000 172.6N/mm (compression) (4417.9 200000 3436.1 80000) σ × = = × + × 6 2 al 10 80000 69.1N/mm (compression) (4417.9 200000 3436.1 80000) σ × = = × + × Due to the decrease in temperature in which no change in length is allowed the strain in the steel is α st T and that in the aluminium is α al T . Therefore due to the decrease in temperature 2 st st st 200000 0.000012 150 360.0 N/mm (tension) E T σ α = = × × = 2 al al al 80000 0.000005 150 60.0 N/mm (tension) E T σ α = = × × = The final stresses in the steel and aluminium are then 2 st (total) 360.0 172.6 187.4N/mm (tension) σ = − = 2 al (total) 60.0 69.1 9.1N/mm (compression) σ = − = − S.1.10 The principal strains are given directly by Eqs (1.69) and (1.70). Thus 2 2 I 1 1 ( 0.002 0.002) ( 0.002 0.002) ( 0.002 0.002) 2 2 ε = − + + − + + + +

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    16 Solutions Manual i.e. I 0.00283 ε = + Similarly II 0.00283 ε = − The principal directions are given by Eq. (1.71), i.e. 2( 0.002) 0.002 0.002 tan 2 1 0.002 0.002 θ − + − = = − + Hence 2 45 or 135 θ = − ° + ° and 22.5 or 67.5 θ = − ° + ° S.1.11 The principal strains at the point P are determined using Eqs (1.69) and (1.70). Thus 2 2 6 I 1 1 ( 222 45) ( 222 213) ( 213 45) 10 2 2 ε −   = − + + − + + − − ×     i.e. 6 I 94.0 10 ε − = × Similarly 6 II 217.0 10 ε − = − × The principal stresses follow from Eqs (1.67) and (1.68). Hence 6 I 2 31000 (94.0 0.2 271.0) 10 1 (0.2) σ − = − × × − i.e. 2 I 1.29 N/mm σ = Similarly 2 II 814 N/mm σ = − Since P lies on the neutral axis of the beam the direct stress due to bending is zero. Therefore, at P , σ x =7 N/mm 2 and σ y = 0. Now subtracting Eq. (1.12) from (1.11) 2 2 I II 4 x xy σ σ σ τ − = +

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