Aircraft Structures for Engineering Students, 5th Edition Solution Manual

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Aircraft Structuresfor Engineering StudentsFifth EditionSolutions ManualT. H. G. Megson

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Solutions ManualSolutions to Chapter 1 ProblemsS.1.1The principal stresses are given directly by Eqs (1.11) and (1.12) in whichσx= 80N/mm2,σy= 0 (or vice versa) andτxy= 45N/mm2. Thus, from Eq. (1.11)22I8018044522σ=++×i.e.σI= 100.2 N/mmFrom Eq. (1.12)222II8018044522σ=−+×i.e.σII= – 20.2 N/mmThe directions of the principal stresses are defined by the angleθin Fig. 1.8(b) inwhichθis given by Eq. (1.10). Hence2245tan 21.125800θ×==−which givesθ= 24°11′andθ= 114°11′It is clear from the derivation of Eqs (1.11) and (1.12) that the first value ofθcorresponds toσIwhile the second value corresponds toσIIFinally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15).Hence from Eq. (1.15).2max100.2( 20.2)60.2N / mm2τ− −==and will act on planes at 45° to the principal planes.

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4Solutions ManualS.1.2The principal stresses are given directly by Eqs (1.11) and (1.12) in whichσx=50N/mm2,σy= –35 N/mm2andτxy= 40 N/mm2. Thus, from Eq. (1.11)22I50351(5035)44022σ−=+++×i.e.σI= 65.9 N/mmand from Eq. (1.12)222II50351(5035)44022σ−=−++×i.e.σII= –50.9 N/mmFrom Fig. 1.8(b) and Eq. (1.10)2240tan 20.9415035θ×==+which givesθ= 21°38′(σI)andθ= 111°38′(σIIThe planes on which there is no direct stress may be found by considering thetriangular element of unit thickness shown in Fig. S.1.2 where the plane AC representsthe plane on which there is no direct stress. For equilibrium of the element in a directionperpendicular to AC)050ABcos35BCsin40ABsin+40BC cosαααα=−+(i)Fig. S.1.2

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Solutions to Chapter 1 Problems5Dividing through Eq. (i) by AB050cos35 tansin40sin40 tancosαααααα=−++which, dividing through by cosα,simplifies to0 = 50–35 tan2from whichα+80 tanαtanα= 2.797or–0.511Henceα=70°21′or–27°5′S.1.3The construction of Mohr’s circle for each stress combination follows the proceduredescribed 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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6Solutions ManualFig. S.1.3(c)Fig. S.1.3(d)S.1.4The principal stresses at the point are determined, as indicated in the question, bytransforming each state of stress into aσx,σy,τxystress system. Clearly, in thefirst caseσx= 0,σy= 10 N/mm2,τxy= 0 (Fig. S.1.4(a)). The two remaining casesare transformed by considering the equilibrium of the triangular element ABC inFigs S.1.4(b), (c), (e) and (f). Thus, using the method described in Section 1.6and the principle of superposition (see Section 5.9), the second stress system of

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Solutions to Chapter 1 Problems7Figs S.1.4(b) and (c) becomes theσx,σy,τxysystem shown in Fig. S.1.4(d)while

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8Solutions ManualFig. 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 theσx,σy,τxysystem ofFig. S.1.4(g).

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Solutions to Chapter 1 Problems9Finally, the states of stress shown in Figs S.1.4(a), (d) and (g) are superimposedto give the state of stress shown in Fig. S.1.4(h) from which it can be seen thatσI=σII=15N/mm2and that thexandyplanes are principal planes.

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10Solutions ManualFig. S.1.4(e)Fig. S.1.4(f)Fig. S.1.4(g)Fig. S.1.4(h)S.1.5The geometry of Mohr’s circle of stress is shown in Fig. S.1.5 in which the circle isconstructed using the method described in Section 1.8.From Fig. S.1.5σx= OP1= OB– BC +CP1(i)

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Solutions to Chapter 1 Problems11Fig. S.1.5In Eq. (i) OB =σI, BC is the radius of the circle which is equal toτmax22221111maxCPCQQ P.xyττ=−=−andHence221maxmaxxxyσστττ=−+−Similarly2221OPOBBCCP in which CPCPyσ==−−=Thus22Imaxmaxyxyσστττ=−−−S.1.6From bending theory the direct stress due to bending on the upper surface of the shaftat a point in the vertical plane of symmetry is given by62425107575 N / mm150/ 64xMyIσπ××===×From the theory of the torsion of circular section shafts the shear stress at the samepoint is62450107575N / mm150/ 32xyTrJτπ××===×

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12Solutions ManualSubstituting these values in Eqs (1.11) and (1.12) in turn and noting thatσy= 022I7517547522σ=++×i.e.2I121.4N / mmσ=22II7517547522σ=−+×i.e.2II46.4N / mmσ= −The corresponding directions as defined byθin Fig. 1.8(b) are given by Eq. (1.10)i.e.275tan 22750θ×==−HenceI31 43 ()θσ′=°andII121 43 ()θσ′=°S.1.7The direct strains are expressed in terms of the stresses using Eqs (1.42), i.e.1 [()]xxyzvEεσσσ=−+(i)1 [()]yyxzvEεσσσ=−+(ii)1 [()]zzxyvEεσσσ=−+(iii)Then1 [2 ()]xyzxyzxyzevEεεεσσσσσσ=++=++−++i.e.(12 ) ()xyzveEσσσ−=++whence(12 )yzxEevσσσ+=−−

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Solutions to Chapter 1 Problems13Substituting in Eq. (i)112xxxEevEvεσσ=−−−so that(1)12xxvEeEvvεσ=+−−Thus(12 )(1)(1)xxvEeEvvvσε=+−++or, sinceG=E/2(1 +ν) (see Section 1.15)2xxeGσλε=+Similarly2yyeGσλε=+and2zzeGσλε=+S.1.8The implication in this problem is that the condition of plane strain also describesthe condition of plane stress. Hence, from Eqs (1.52)1 ()xxyvEεσσ=−(i)1 ()yyxvEεσσ=−(ii)2(1)(seeSection 1.15)xyxyxyvGEτγτ+==(iii)The compatibility condition for plane strain is22222(see Section 1.11)xyyxxyxyγεε∂∂∂=+∂ ∂∂∂(iv)Substituting in Eq. (iv) forεx,εyandγxyfrom Eqs (i)–(iii), respectively, gives222222(1)()()xyyxxyvvvxyxyτσσσσ∂∂∂+=−+−∂ ∂∂∂(v)

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14Solutions ManualAlso, from Eqs (1.6) and assuming that the body forcesXandYare zero0zyxxyτσ∂∂+=∂∂(vi)0yxyyxστ∂∂+=∂∂(vii)Differentiating Eq. (vi) with respect toxand Eq. (vii) with respect toyand adding gives2222220xyyxyxyxxyxyτστσ∂∂∂∂+++=∂ ∂∂ ∂∂∂or222222xyyxxyxyτσσ∂∂∂= −+∂ ∂∂∂Substituting in Eq. (v)22222222(1)()()yxyxxyvvvxyxyσσσσσσ∂∂∂∂−++=−+−∂∂∂∂so that222222222222(1)yyyxxxvvxyxyxyσσσσσσ∂∂∂∂∂∂−++=+−+∂∂∂∂∂∂which simplifies to222222220yyxxxyxyσσσσ∂∂∂∂+++=∂∂∂∂or2222()0xyxyσσ∂∂++=∂∂S.1.9Suppose that the load in the steel bar isPstand that in the aluminium bar isPal. Then,from equilibriumstalPPP+=(i)From Eq. (1.40)stalstalststalalPPA EA Eεε==

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Solutions to Chapter 1 Problems15Since the bars contract by the same amountstalststalalPPA EA E=(ii)Solving Eqs (i) and (ii)ststalalstststalalststalalalA EA EPPPPA EA EA EA E==++from which the stresses arestalstalststalalststalalEEPPA EA EA EA Eσσ==++(iii)The areas of cross-section are22222stal75(10075 )4417.9 mm3436.1mm44AAππ×−====Substituting in Eq. (iii) we have62st10200000172.6N/mm(compression)(4417.92000003436.180000)σ×==×+×62al108000069.1N/mm(compression)(4417.92000003436.180000)σ×==×+×Due to the decrease in temperature in which no change in length is allowed the strainin the steel isαstTand that in the aluminium isαalT. Therefore due to the decrease intemperature2ststst2000000.000012150360.0 N/mm(tension)ETσα==××=2alalal800000.00000515060.0 N/mm(tension)ETσα==××=The final stresses in the steel and aluminium are then2st(total)360.0172.6187.4N/mm(tension)σ=−=2al(total)60.069.19.1N/mm(compression)σ=−= −S.1.10The principal strains are given directly by Eqs (1.69) and (1.70). Thus22I11( 0.0020.002)( 0.0020.002)( 0.0020.002)22ε=−++−++ ++

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16Solutions Manuali.e.I0.00283ε= +SimilarlyII0.00283ε= −The principal directions are given by Eq. (1.71), i.e.2( 0.002)0.0020.002tan 210.0020.002θ−+−== −+Hence245 or135θ = −°+°and22.5 or67.5θ = −°+°S.1.11The principal strains at the pointPare determined using Eqs (1.69) and (1.70). Thus226I11( 22245)( 222213)( 21345)1022ε−=−++−++ −−×i.e.6I94.010ε−=×Similarly6II217.010ε−= −×The principal stresses follow from Eqs (1.67) and (1.68). Hence6I231000(94.00.2271.0)101(0.2)σ−=−××−i.e.2I1.29 N/mmσ=Similarly2II814 N/mmσ= −SincePlies on the neutral axis of the beam the direct stress due to bending is zero.Therefore, atP,σx=7 N/mm2andσy= 0. Now subtracting Eq. (1.12) from (1.11)22III4xxyσσστ−=+

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