Aircraft Structures for Engineering Students, 5th Edition Solution Manual

Aircraft Structures for Engineering Students Fifth Edition Solutions Manual T. H. G. Megson This page intentionally left blank 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/mm2, σ y = 0 (or vice versa) and τ xy = 45N/mm2. Thus, from Eq. (1.11) σI = 80 1 + 802 + 4 × 452 2 2 i.e. σ I = 100.2 N/mm2 From Eq. (1.12) σ II = 80 1 − 802 + 4 × 452 2 2 i.e. σ II = – 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 tan = 2θ 2 × 45 = 1.125 80 − 0 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) 100.2 − (−20.2) τ max = 60.2N / mm 2 = 2 and will act on planes at 45° to the principal planes. 4 Solutions Manual S.1.2 The principal stresses are given directly by Eqs (1.11) and (1.12) in which σ x = 50N/mm2, σ y = –35 N/mm2 and τ xy = 40 N/mm2. Thus, from Eq. (1.11) σI = 50 − 35 1 + (50 + 35)2 + 4 × 402 2 2 i.e. σ I = 65.9 N/mm2 and from Eq. (1.12) = σ II 50 − 35 1 − (50 + 35) 2 + 4 × 402 2 2 i.e. σ II = –50.9 N/mm2 From Fig. 1.8(b) and Eq. (1.10) = tan 2θ 2 × 40 = 0.941 50 + 35 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 α − 35BCsin α + 40ABsinα +40BC cosα Fig. S.1.2 (i) Solutions to Chapter 1 Problems Dividing through Eq. (i) by AB 0 = 50 cos α − 35 tan α sin α + 40sin α + 40 tan α cos α which, dividing through by cos α, simplifies to 0 = 50–35 tan2 α + 80 tan α from which 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) 5 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/mm2, 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 Solutions to Chapter 1 Problems Figs S.1.4(b) and (c) becomes the σ x , σ y , τ xy system shown in Fig. S.1.4(d) while 7 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). Solutions to Chapter 1 Problems 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/mm2 and that the x and y planes are principal planes. 9 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) 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 and CP1 = CQ12 − Q1P12 = 2 2 τ max − τ xy . Hence 2 2 σx = σ 1 − τ max + τ max − τ xy Similarly σ y = OP2 = OB − BC − CP2 in which CP2 = CP1 Thus 2 2 σy = σ I − τ max − τ max − τ xy 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 σx = My 25 × 106 × 75 = = 75 N / mm 2 I π × 1504 / 64 From the theory of the torsion of circular section shafts the shear stress at the same point is τ= xy Tr 50 × 106 × 75 = = 75N / mm 2 J π × 1504 / 32 12 Solutions Manual Substituting these values in Eqs (1.11) and (1.12) in turn and noting that σ y = 0 75 1 σI = + 752 + 4 × 752 2 2 i.e. σ I = 121.4N / mm 2 75 1 − 752 + 4 × 752 2 2 σ II = i.e. σ II = −46.4N / mm 2 The corresponding directions as defined by θ in Fig. 1.8(b) are given by Eq. (1.10) i.e. tan = 2θ 2 × 75 = 2 75 − 0 Hence θ= 31°43′(σ I ) and θ 121°43′(σ II ) = S.1.7 The direct strains are expressed in terms of the stresses using Eqs (1.42), i.e. 1 εx = [σ x − v(σ y + σ z )] E (i) εy = 1 [σ y − v(σ x + σ z )] E (ii) εz = 1 [σ z − v(σ x + σ y )] E (iii) Then 1 e =ε x + ε y + ε z = [σ x + σ y + σ z − 2v(σ x + σ y + σ z )] E i.e. = e (1 − 2v) (σ x + σ y + σ z ) E whence σ y += σz Ee −σx (1 − 2v) Solutions to Chapter 1 Problems 13 Substituting in Eq. (i) 1 E  Ee  − σ x   1 − 2v  ε x = σ x − v  so that Eε x= σ x (1 + v) − vEe 1 − 2v Thus = σx vEe E + εx (1 − 2v)(1 + v) (1 + v) or, since G = E/2(1 + ν) (see Section 1.15) σ= λ e + 2Gε x x Similarly σ= λ e + 2Gε y y and σ= λ e + 2Gε z z 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) γ= xy τ xy = εx 1 (σ x − vσ y ) E (i) = εy 1 (σ y − vσ x ) E (ii) = G 2(1 + v) τ xy E (seeSection 1.15) (iii) (see Section 1.11) (iv) The compatibility condition for plane strain is ∂ 2γ xy ∂ 2ε y ∂ 2ε x = + ∂x ∂y ∂x 2 ∂y 2 Substituting in Eq. (iv) for ε x , ε y and γ xy from Eqs (i)–(iii), respectively, gives 2 1 + v) ∂ 2τ xy ∂2 ∂2 (= ( v ) (σ x − vσ y ) σ − σ + y x ∂x ∂y ∂x 2 ∂y 2 (v) 14 Solutions Manual Also, from Eqs (1.6) and assuming that the body forces X and Y are zero ∂σ x ∂τ zy 0 + = ∂x ∂y ∂σ y ∂y + ∂τ xy ∂x (vi) 0 = (vii) Differentiating Eq. (vi) with respect to x and Eq. (vii) with respect to y and adding gives ∂ 2σ x ∂x 2 + ∂ 2τ xy ∂y ∂x + ∂ 2σ y ∂y 2 + ∂ 2τ xy ∂x ∂y 0 = or 2  ∂ 2σ ∂ 2σ y   = −  2x + 2   ∂x ∂x ∂y ∂ y   ∂ 2τ xy Substituting in Eq. (v)  ∂ 2σ ∂ 2σ y  ∂ 2 ∂2  (σ y − vσ x ) + 2 (σ x − vσ y ) −(1 + v)  2x + = 2 2  ∂x ∂y  ∂x ∂y  so that  ∂ 2σ  ∂ 2σ ∂ 2σ y  ∂ 2σ y ∂ 2σ x ∂ 2σ y  x    −(1 + v)  2x + = + − v + 2  2 2 2   ∂x  ∂x 2 ∂ y ∂ x ∂ y ∂ y     which simplifies to ∂ 2σ y ∂x 2 + ∂ 2σ x ∂y 2 + ∂ 2σ x ∂x 2 + ∂ 2σ y ∂y 2 0 = or  ∂2 ∂2  0 +  2  (σ x + σ y ) = ∂y 2   ∂x 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 (i) Pst + Pal = P From Eq. (1.40) Pst Pal = ε st = ε al Ast Est Aal Eal Solutions to Chapter 1 Problems 15 Since the bars contract by the same amount Pst P = al Ast Est Aal Eal (ii) Solving Eqs (i) and (ii) = Pst Ast Est Aal Eal = P Pal P Ast Est + Aal Eal Ast Est + Aal Eal from which the stresses are = σ st Est Eal = P σ al P Ast Est + Aal Eal Ast Est + Aal Eal (iii) The areas of cross-section are = Ast π × 752 2 = 4417.9 mm = Aal 4 π (1002 − 752 ) = 3436.1mm 2 4 Substituting in Eq. (iii) we have σ st 106 × 200 000 = 172.6N/mm 2 (compression) (4417.9 × 200 000 + 3436.1 × 80 000) σ al 106 × 80 000 = 69.1N/mm 2 (compression) (4417.9 × 200 000 + 3436.1 × 80 000) 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 alT. Therefore due to the decrease in temperature σ= Estα st= T 200 000 × 0.000012 × 150 = 360.0 N/mm 2 (tension) st σ al= Ealα alT= 80 000 × 0.000005 × 150= 60.0 N/mm 2 (tension) The final stresses in the steel and aluminium are then σ st (total) = 360.0 − 172.6 = 187.4N/mm 2 (tension) σ al (total) = 60.0 − 69.1 = −9.1N/mm 2 (compression) S.1.10 The principal strains are given directly by Eqs (1.69) and (1.70). Thus 1 2 ε I =(−0.002 + 0.002) + 1 2 (−0.002 + 0.002) 2 + (+0.002 + 0.002) 2 16 Solutions Manual i.e. ε I = +0.00283 Similarly ε II = −0.00283 The principal directions are given by Eq. (1.71), i.e. tan 2θ = 2(−0.002) + 0.002 − 0.002 = −1 0.002 + 0.002 Hence 2θ = −45° or + 135° and θ= −22.5° or + 67.5° S.1.11 The principal strains at the point P are determined using Eqs (1.69) and (1.70). Thus 1 2 ε I =  (−222 + 45) +  (−222 + 213) 2 + (−213 − 45) 2  × 10−6 2  1 i.e. = ε I 94.0 × 10−6 Similarly ε II = −217.0 × 10−6 The principal stresses follow from Eqs (1.67) and (1.68). Hence = σI 31000 (94.0 − 0.2 × 271.0) × 10−6 1 − (0.2) 2 i.e. σ I = 1.29 N/mm 2 Similarly σ II = −814 N/mm 2 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/mm2 and σy = 0. Now subtracting Eq. (1.12) from (1.11) σ I − σ II = 2 σ x2 + 4τ xy Solutions to Chapter 2 Problems 17 i.e. 2 1.29 + 8.14 = 7 2 + 4τ xy from which τ xy = 3.17 N/mm2. The shear force at P is equal to Q so that the shear stress at P is given by 3Q = τ xy 3.17 = 2 × 150 × 300 from which = Q 95 = 100N 95.1kN. Solutions to Chapter 2 Problems S.2.1 The stress system applied to the plate is shown in Fig. S.2.1. The origin, O, of the axes may be chosen at any point in the plate; let P be the point whose coordinates are (2, 3). Fig. S.2.1 From Eqs (1.42) in which σ z = 0 3p 2p 3.5 p = − E E E 3p 3p 2.75 p εy = − −v = − E E E εx = − −v Hence, from Eqs (1.27) ∂u 3.5 p 3.5 p = − so that u = − x + f1 ( y ) ∂x E E (i) (ii) (iii) 18 Solutions Manual where f 1 (y) is a function of y. Also ∂υ 2.75 p 2.75 p =so that υ = − y + f 2 ( x) ∂y E E (iv) in which f 2 (x) is a function of x. From the last of Eqs (1.52) and Eq. (1.28) γ xy = 4 p ∂υ ∂u ∂f 2 ( x) ∂f1 ( y ) = + = + G ∂x ∂y ∂x ∂y (from Eqs (iv) and (iii)) Suppose ∂f1 ( y ) =A ∂y then f1 ( y= ) Ay + B (v) in which A and B are constants. Similarly, suppose ∂f 2 ( x) =C ∂x then f 2 ( x= ) Cx + D (vi) in which C and D are constants. Substituting for f 1 (y) and f 2 (x) in Eqs (iii) and (iv) gives 3.5 p u= − x + Ay + B E (vii) 2.75 p y + Cx + D E (viii) and = υ Since the origin of the axes is fixed in space it follows that when x = y = 0, u=v = 0. Hence, from Eqs (vii) and (viii), B = D = 0. Further, the direction of Ox is fixed in space so that, wheny = 0, ∂v/∂x = 0. Therefore, from Eq. (viii), C = 0. Thus, from Eqs (1.28) and (vii), when x = 0. ∂u 4 p = = A ∂y G Eqs (vii) and (viii) now become 3.5 p 4p u= − x+ y E G (ix) Solutions to Chapter 2 Problems 19 2.75 p y E From Eq. (1.50), G=E/2(1 +ν) =E/2.5 and Eq. (ix) becomes u= υ= (x) p (−3.5 + 10 y ) E (xi) At the point (2, 3) u= 23 p (from Eq. (xi)) E υ= 8.25 p (from Eq. (x)) E and The point P therefore moves at an angle α to the x axis given by S.2.2 An Airy stress function, φ, is defined by the equations (Eqs (2.8)): σx = ∂ 2φ ∂ 2φ ∂ 2φ σ = τ = − y xy ∂x ∂y ∂y 2 ∂x 2 and has a final form which is determined by the boundary conditions relating to a particular problem. Since φ =Ay 3 + By 3 x + Cyx (i) ∂ 4φ ∂ 4φ ∂ 4φ 0 0 0 = = = ∂x 4 ∂y 4 ∂x 2 ∂y 2 and the biharmonic equation (2.9) is satisfied. Further = σx ∂ 2φ = 6 Ay + 6 Byx ∂y 2 (ii) ∂ 2φ = 0 ∂x 2 (iii) ∂ 2φ = −3By 2 − C ∂x ∂y (iv) = σy τ xy = − The distribution of shear stress in a rectangular section beam is parabolic and is zero at the upper and lower surfaces. Hence, when y = ±d/2, τ xy = 0. Thus, from Eq. (iv) −4C (v) B= 3d 2 20 Solutions Manual The resultant shear force at any section of the beam is –P. Therefore ∫ d /2 − d /2 τ xy t dy = − P Substituting for τ xy from Eq. (iv) ∫ d /2 − d /2 −P (−3By 2 − C )t dy = which gives  Bd 3 Cd  2t  + P = 2   8 Substituting for B from Eq. (v) gives C= 3P 2td (vi) It now follows from Eqs (v) and (vi) that −2P (vii) td 3 At the free end of the beam where x =l the bending moment is zero and thus σ x = 0 for any value of y. Therefore, from Eq. (ii) 6A + 6Bl = 0 whence B= A= 2Pl td 3 (viii) Then, from Eq. (ii) = σx 12 Pl 12 P y − 3 xy 3 td td or 12 P(l − x) (ix) y td 3 Equation (ix) is the direct stress distribution at any section of the beam given by simple bending theory, i.e. σx = σx = My I where M = P (l –x) and I = td3/12. The shear stress distribution given by Eq. (iv) is = τ xy 6 P 2 3P y − 2td td 3 Solutions to Chapter 2 Problems 21 or τ xy = 6P  2 d 2  y −  4  td 3  (x) Equation (x) is identical to that derived from simple bending theory and may be found in standard texts on stress analysis, strength of materials, etc. S.2.3 The stress function is = φ w (15 h 2 x 2 y − 5 x 2 y 3 − 2h 2 y 3 + y 5 ) 3 20h Then ∂ 2φ w = = (30h 2 y − 10 y3 ) σ y 2 ∂x 20h3 ∂ 2φ w = (−30 x 2 y − 12h 2 y + 20 y 3 ) = σ x 2 3 ∂y 20h ∂ 2φ w = −τ xy (30h 2 x − 30 xy 2 ) = ∂x ∂y 20h3 ∂ 4φ =0 ∂x 4 ∂ 4φ w = (120 y ) 4 ∂y 20h3 ∂ 4φ w = (−60 y ) 2 2 ∂x ∂y 20h3 Substituting in Eq. (2.9) ∇ 4φ = 0 so that the stress function satisfies the biharmonic equation. The boundary conditions are as follows: ● At y= h, σ y = w and τ xy = 0 which are satisfied. ● At y = –h, σ y = –w and τ xy = 0 which are satisfied. ● At x =0, σ x = w/20h3 (–12h2y + 20y3) ≠ 0. Also h w h σ x d= (−12h 2 y + 20 y 3 )dy y −h 20h3 − h w = [−6h 2 y 2 + 5 y 4 ]h− h 3 20h =0 i.e. no resultant force. ∫ ∫ 22 Solutions Manual Finally ∫ h −h w h (−12h 2 y 2 + 20 y 4 )dy 3 −h 20h w = [−4h 2 y 3 + 4 y 5 ]h− h 3 20h =0 ∫ σ x y dy= i.e. no resultant moment. S.2.4 The Airy stress function is = φ p [5( x3 − l 2 x)( y + d ) 2 ( y − 2d ) − 3 yx( y 2 − d 2 ) 2 ] 120d 3 Then ∂ 4φ ∂ 4φ 3 pxy ∂ 4φ 3 pxy = 0 = − = 4 4 3 2 2 ∂x ∂y d ∂x ∂y 2d 3 Substituting these values in Eq. (2.9) gives 0 + 2× 3 pxy 3 pxy − 3 = 0 2d 3 d Therefore, the biharmonic equation (2.9) is satisfied. The direct stress, σ x , is given by (see Eqs (2.8)) σx = ∂ 2φ px [5 y ( x 2 − l 2 ) − 10 y 3 + 6d 2 y ] = 2 3 20d ∂y When x = 0, σ x = 0 for all values of y. When x = l pl (−10 y 3 + 6d 2 y ) 20d 3 σx= and the total end load = d ∫ σ 1 dy −d = x pl 20d 3 d )dy 0 ∫ (−10 y + 6d y= 3 2 −d Thus the stress function satisfies the boundary conditions for axial load in the x direction. Also, the direct stress, σ y , is given by (see Eqs (2.8)) px ∂ 2φ = 3 ( y 3 − 3 yd 2 − 2d 3 ) 2 4d ∂x σy = Solutions to Chapter 2 Problems 23 When x = 0, σ y = 0 for all values of y. Also at any section x where y = –d σ y= px (−d 3 + 3d 3 − 2d 3 = ) 0 3 4d and when y = +d px 4d σ y =3 (d 3 − 3d 3 − 2d 3 ) = − px Thus, the stress function satisfies the boundary conditions for load in the y direction. The shear stress, τ xy , is given by (see Eqs (2.8)) p ∂ 2φ [5(3x 2 − l 2 )( y 2 − d 2 ) − 5 y 4 + 6 y 2 d 2 − d 4 ] = − ∂x ∂y 40d 3 τ xy = − When x = 0 p [−5l 2 ( y 2 − d 2 ) − 5 y 4 + 6 y 2 d 2 − d 4 ] 40d 3 τ xy = − so that, when y = ±d, τ xy = 0. The resultant shear force on the plane x = 0 is given by ∫ d −d p 40d 3 − τ xy 1 dy = pl 2 − [−5l 2 ( y 2 − d 2 ) − 5 y 4 + 6 y 2 d 2 − d 4 ]dy = −d 6 ∫ d From Fig. P.2.4 and taking moments about the plane x = l, = τ xy ( x 0)12 = dl 1 2 lpl l 2 3 i.e. τ xy (= = x 0) pl 2 6d and the shear force is pl2/6. Thus, although the resultant of the Airy stress function shear stress has the same magnitude as the equilibrating shear force it varies through the depth of the beam whereas the applied equilibrating shear stress is constant. A similar situation arises on the plane x = l. S.2.5 The stress function is φ= w (−10c3 x 2 − 15c 2 x 2 y + 2c 2 y 3 + 5 x 2 y 3 − y 5 ) 40bc3 24 Solutions Manual Then ∂ 2φ w = = (12c 2 y + 30 x 2 y − 20 y3 ) σ x 2 3 ∂y 40bc ∂ 2φ w = (−20c3 − 30c 2 y + 10 y 3= ) σy 2 3 ∂x 40bc ∂ 2φ w = (−30c 2 x + 30 xy 2 ) = −τ xy ∂x ∂y 40bc3 ∂ 4φ =0 ∂x 4 ∂ 4φ w = (−120 y ) 4 ∂y 40bc3 ∂ 4φ w = (60 y ) 2 2 ∂x ∂y 40bc3 Substituting in Eq. (2.9) ∇ 4φ = 0 so that the stress function satisfies the biharmonic equation. On the boundary, y = +c w 0 σy = − τ xy = b At y = —c = σ y 0= τ xy 0 At x = 0 = σx w (12c 2 y − 20 y 3 ) 3 40bc Then c w (12c 2 y − 20 y 3 )dy 3 −c 40bc − c w = [6c 2 y 2 − 5 y 4 ]c− c 40bc3 =0 i.e. the direct stress distribution at the end of the cantilever is self-equilibrating. The axial force at any section is c c w = σ x dy (12c 2 y + 30 x 2 y − 20 y 3 )dy 3 −c −c 40bc w = [6c 2 y 2 + 15 x 2 y 2 − 5 y 4 ]c− c 40bc3 =0 i.e. no axial force at any section of the beam. ∫ c ∫ = σ x dy ∫ ∫ Solutions to Chapter 2 Problems 25 The bending moment at x =0 is ∫ c c w (12c 2 y 2 − 20 y 4 )dy 3 −c 40bc w 5 c = [4c 2 y 3 − 4 y= ]− c 0 3 40bc ∫ = σ x y dy −c i.e. the beam is a cantilever beam under a uniformly distributed load of w/unit area with a self-equilibrating stress application at x = 0. S.2.6 From physics, the strain due to a temperature rise T in a bar of original length L 0 and final length L is given by L − L0 L0 (1 + α T ) = = αT L0 L0 = ε Thus for the isotropic sheet, Eqs (1.52) become 1 (σ x − vσ y ) + α T E 1 εy = (σ y − vσ x ) + α T E εx = Also, from the last of Eqs (1.52) and (1.50) γ xy = 2(1 + v) τ xy E Substituting in Eq. (1.21) 2 2 ∂ 2σ y  ∂ 2σ x  2(1 + v) ∂ τ xy 1  ∂ σ y ∂ 2T 1  ∂ 2σ x ∂ 2T     = − + α + − + α v v E ∂x ∂y E  ∂x 2 ∂x 2  ∂x 2 E  ∂y 2 ∂y 2  ∂y 2 or 2(1 + v) ∂ 2τ xy ∂ 2σ y ∂ 2σ x ∂ 2σ y ∂ 2σ x v v = + − − + Eα∇ 2T ∂x ∂y ∂x 2 ∂y 2 ∂x 2 ∂y 2 From Eqs (1.6) and assuming body forces X =Y =0 ∂ 2τ xy 2 ∂ 2σ y ∂ 2σ x ∂ τ xy = − 2 = − ∂y ∂x ∂x ∂y ∂x ∂y 2 Hence 2 ∂ 2τ xy ∂ 2σ y ∂ 2σ = − 2x − ∂x ∂y ∂x ∂y 2 (i) 26 Solutions Manual and 2v ∂ 2τ xy ∂ 2σ y ∂ 2σ x v = −v − ∂x ∂y ∂x 2 ∂y 2 Substituting in Eq. (i) − ∂ 2σ x ∂x 2 − ∂ 2σ y ∂ 2σ y ∂ 2σ x = + + Eα∇ 2T ∂y 2 ∂x 2 ∂y 2 Thus  ∂2 ∂2  2 0 +  ∂x 2 ∂y 2  (σ x + σ y ) + Eα∇ T =   and since = σx ∂ 2φ ∂ 2φ (see Eqs (2.8)) = σ y ∂y 2 ∂x 2  ∂2 ∂ 2   ∂ 2φ ∂ 2φ  2 0 +  ∂x 2 ∂y 2   ∂y 2 + ∂x 2  + Eα∇ T =     or ∇ 2 (∇ 2φ + Eα T ) = 0 S.2.7 The stress function is = φ 3Qxy Qxy 3 − 4a 4a 3 Then ∂ 2φ = 0= σ y ∂x 2 3Qxy ∂ 2φ σx = − = 2 2a 3 ∂y 3Q 3Qy 2 ∂ 2φ = − = −τ xy ∂x ∂y 4a 4a 3 Also ∂ 4φ ∂ 4φ ∂ 4φ 0 0 = = = 0 ∂x 4 ∂y 4 ∂x 2 ∂y 2 so that Eq. (2.9), the biharmonic equation, is satisfied. Solutions to Chapter 2 Problems 27 When x = a, σ x = –3Qy/2a2, i.e. linear. Then, when = σx 0 y 0= 3Q −a σ x = y= 2a −3Q +a σ x = y= 2a Also, when x = –a, σ x = 3Qy/2a2, i.e. linear and when = σx 0 y 0= −3Q −a σ x = y= 2a 3Q +a σ x = y= 2a The shear stress is given by (see above) 3Q  4a  τ xy = − 1 − y2   , i.e. parabolic a 2  so that, when y = ±a, τ xy = 0 and when y = 0, τ xy = –3Q/4a. The resultant shear force at x = ±a is ∫ 3Q  y2  1 − 2  dy 4a  a  a = − −a i.e. SF = Q. The resultant bending moment at x = ±a is = = a ∫ σ y dy −a ∫ x a 3Qay 2 −a 2a 3 dy i.e. BM = –Qa Solutions to Chapter 3 Problems 27 Solutions to Chapter 3 Problems S.3.1 Initially the stress function, φ, must be expressed in terms of Cartesian coordinates. Thus, from the equation of a circle of radius, a, and having the origin of its axes at its centre. φ= k ( x 2 + y 2 − a 2 ) (i) ∂ 2φ ∂ 2φ dθ + 2 = −2G F= 2 dz ∂x ∂y (ii) From Eqs (3.4) and (3.11) Differentiating Eq. (i) and substituting in Eq. (ii) 4k = −2G dθ dz or 1 dθ k= − G 2 dz (iii) From Eq. (3.8) T =2 ∫∫ φ dx dy i.e. dθ  −G T= x 2 dx dy + dz  A ∫∫ ∫∫ y dx dy − a ∫∫ dx dy  2 2 A (iv) A where ∫∫ A x2 dx dy = I y , the second moment of area of the cross-section about the y axis; ∫∫ A y2 dx dy = I x , the second moment of area of the cross-section about the x axis and ∫∫ A dx dy = A, the area of the cross-section. Thus, since I y = πa4/4, I x = πa4/4 and A = πa2 Eq. (iv) becomes T =G dθ π a 4 dz 2 or dθ 2T T = = 4 dz Gπ a GI p (v) From Eqs (3.2) and (v) dθ Tx ∂φ G x= τ zy = − = −2kx = ∂x dz Ip (vi) 28 Solutions Manual and Ty ∂φ dθ = −G − τ zx = 2ky = y= ∂y dz Ip (vii) Substituting for τ zy and τ zx from Eqs (vi) and (vii) in the second of Eqs (3.15) T ( xl + ym) Ip = τ zs (viii) in which, from Eqs (3.6) l= dy ds m= − dx ds Suppose that the bar of Fig. 3.2 is circular in cross-section and that the radius makes an angle α with the x axis. Then. = m sin = α and l cos α Also, at any radius, r = y r= sin α x r cos α Substituting for x, l, y and m in Eq. (viii) gives τ= zs Tr =( τ ) Ip Now substituting for τ zx , τ zy and dθ/dz from Eqs (vii), (vi) and (v) in Eqs (3.10) ∂w Ty Ty 0 = − + = GI p GI p ∂x (ix) ∂w Tx Tx 0 = − + = ∂y GI p GI p (x) The possible solutions of Eqs (ix) and (x) are w = 0 and w = constant. The latter solution implies a displacement of the whole bar along the z axis which, under the given loading, cannot occur. Therefore, the first solution applies, i.e. the warping is zero at all points in the cross-section. The stress function, φ, defined in Eq. (i) is constant at any radius, r, in the crosssection of the bar so that there are no shear stresses acting across such a boundary. Thus, the material contained within this boundary could be removed without affecting the stress distribution in the outer portion. Therefore, the stress function could be used for a hollow bar of circular cross-section.