Aircraft Structures 3E Episode 2 pdf

26 Basic elasticity For the case of plane stress they simplify to (1.47) It may be seen from the third of Eqs (1.47) that the conditions of plane stress and plane strain do not necessarily describe identical situations. Changes in the linear dimensions of a strained body may lead to a change in volume. Suppose that a small element of a body has dimensions Sx, Sy and Sz. When subjected to a three-dimensional stress system the element will sustain a volumetric strain e (change in volume/unit volume) equal to (1 + &,)ax( 1 + &,,)by( 1 + EZ)SZ - SxSySz SxSySz e= Neglecting products of small quantities in the expansion of the right-hand side of the above equation yields e = E, + E,, + E, (1.48) Substituting for E,, E~ and E~ from Eqs (1.42) we find, for a linearly elastic, isotropic body 1 E e = - [a, + a,, + az - 2v(ax + ay + 41 or In the case of a uniform hydrostatic pressure, a, = a,, = az = -p and 3(1 - 2~) E P e=- (1.49) The constant E/3( 1 - 2v) is known as the bulk modulus or modulus of volume expansion and is often given the symbol K. An examination of Eq. (1.49) shows that v < 0.5 since a body cannot increase in volume under pressure. Also the lateral dimensions of a body subjected to uniaxial tension cannot increase so that v > 0. Therefore, for an isotropic material 0 < v < 0.5 and for most isotropic materials v is in the range 0.25 to 0.33 below the elastic limit. Above the limit of proportionality v increases and approaches 0.5. Example 1.2 A rectangular element in a linearly elastic isotropic material is subjected to tensile stresses of 83 N/mm2 and 65 N/mm2 on mutually perpendicular planes. Determine the strain in the direction of each stress and in the direction perpendicular to both 1.1 5 Stress-strain relationships 27 stresses. Find also the principal strains, the maximum shear stress, the maximum shear strain and their directions at the point. Take E = 200000N/mm2 and v = 0.3. If we assume that a, = 83 N/mm2 and ay = 65 N/mm2 then from Eqs (1.47) (83 - 0.3 x 65) = 3.175 x lop4 1 E, = ~ 200 000 E (65 - 0.3 x 83) = 2.005 x - 200000 -0.3 200 000 e, = ~ (83 + 65) = -2.220 x lop4 In this case, since there are no shear stresses on the given planes, a, and av are principal stresses so that E, and are the principal strains and are in the directions of a, and cy. It follows from Eq. (1.15) that the maximum shear stress (in the plane of the stresses) is 83 - 65 2 rmax = = ~N/IYUII' acting on planes at 45" to the principal planes. Further, using Eq. (1.45), the maximum shear strain is 2 x (1 +0.3) x 9 200 000 'Ymax = so that -ymax = 1.17 x on the planes of maximum shear stress. Example 1.3 At a particular point in a structural member a two-dimensional stress system exists where a, = 60 N/mm , ay = -40 N/mm' and rxy = 50 N/mm2. If Young's modulus E = 200000N/mm2 and Poisson's ratio v = 0.3 calculate the direct strain in the x and y directions and the shear strain at the point. Also calculate the principal strains at the point and their inclination to the plane on which a, acts; verify these answers using a graphical method. 2 From Eqs (1.47) E, = - (60 + 0.3 x 40) = 360 x lop6 200 000 (-40 - 0.3 x 60) = -290 x lop6 1 EY = - 200 000 From Eq. (1.45) the shear modulus, G, is given by - 2oo Oo0 = 76 923 N/mm2 G=- E 2(1+v)-2(1+0.3) Hence, from Eqs (1.47) 28 Basic elasticity (-290 x 1 04, Y Q, (360 x 1 04, ix 650 x 1 04) Fig. 1.14 Mohr's circle of strain for Example 1.3. Now substituting in Eq. (1.35) for E,, E, and -yrY 1 &I = 10- + \/(360 + 290)2 + 6502 which gives &I = 495 x Similarly, from Eq. (1.36) EII = -425 x IOp6 From Eq. (1.37) 650 x 360 x lop6 + 290 x lop6 = tan20 = Therefore 20 = 45" or 225" so that 0 = 22.5" or 112.5" The values of E~, and 0 are verified using Mohr's circle of strain (Fig. 1.14). Axes OE and Oy are set up and the points Q1 (360 x lop6,: x 650 x and Q2 (-290 x lop6, - 4 x 650 x IOp6) located. The centre C of the circle is the intersection of Q1Q2 and the OE axis. The circle is then drawn with radius CQ1 and the points B(q) and A(eII) located. Finally angle QICB = 20 and angle QICA = 20 + 7r. Stresses at a point on the surface of a piece of material may be determined by measur- ing the strains at the point, usually by electrical resistance strain gauges arranged in 1.16 Experimental measurement of surface strains 29 Fig. 1.1 5 Strain gauge rosette. the form of a rosette, as shown in Fig. 1.15. Suppose that are the principal strains at the point, then if E,, &b and E, are the measured strains in the directions 8, (8 + a), (8 + a + p) to we have, from the general direct strain relationship of Eq. (1.3 1) (1 SO) E~ becomes qI and -yxy is zero since the x and y directions have and E, = E~COS 2 8 + cII sin 2 8 since E, becomes become principal directions. Rewriting Eq. (1.50) we have 1 + COSM 1 - cos 28 %=EI( 2 )+EII( 2 ) or E, = $ + + 4 - cII) COS 28 (1.51) Similarly &b = $ (€1 + EII) + $ (E1 - EII) cos 2(8 + a) (1.52) and E, = ; (EI + EI1) + 4 (EI - EII) cos 2(e + a + p) (1.53) Therefore if E,, &b and E, are measured in given directions, i.e. given angles a and p, then and 8 are the only unknowns in Eqs (lSl), (1.52) and (1.53). in Eqs (1.47). Thus The principal stresses are now obtained by substitution of and (1.54) 1 E &I = - (Cq - vq1) and Solving Eqs (1.54) and (1.55) gives E 01 = - - 9 (&I + %I) (1.55) (1.56) 30 Basic elasticity Fig. 1.16 Experimental values of principal strain using Mohr's circle. and (1.57) A typical rosette would have Q = ,O = 45" in which case the principal strains are most conveniently found using the geometry of Mohr's circle of strain. Suppose that the arm a of the rosette is inclined at some unknown angle 8 to the maximum principal strain as in Fig. 1.15. Then Mohr's circle of strain is as shown in Fig. 1.16; the shear strains +yay +yb and 7, do not feature in the analysis and are therefore ignored. From Fig. 1.16 oc= +(&,+Ec) CN =E, -0C =;(E, -E,) QN= CM =~b -OC=E~ -$(E, +E,) The radius of the circle is CQ and CQ = dCN2 + QN2 Hence which simplifies to 1 CQ=~ J( E, - qJ2 + (E, - Eb)2 Therefore which is given by = OC + radius of circle is (1.58) 1.1 6 Experimental measurement of surface strains 3 1 Also i.e. cII = OC - radius of circle Finally the angle 8 is given by i.e. 2Eb - E, - E, tan28 = Ea -E, A similar approach may be adopted for a 60” rosette. (1.59) ( 1.60) Example 1.4 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: E, = 1000 x 1K6, Eb = -200 x where the gauges ‘a’ and ‘cy are in line with, and perpendicular to, the axis of the bar respectively. If Young’s modulus, E, for the bar is 70 000 N/mm2 and Poisson’s ratio, v, is 0.3, calculate the values of T and P. and E, = -300 x Substituting the values of E,, &b and c, in Eq. (1.58) €1 = -(lo00 - 300) +- J( 1000 + 200)2 + (-200 + 300)’ 2 Jz which gives E1 = 1202 x Similarly, from Eq. (1.59) EII = -502 x Now substituting for and sI1 in Eq. (1.56) 70 000 x CTI = (-502 + 0.3 x 1202) = -80.9N/mm2 1 - (0.3)2 Similarly, from Eq. (1.57) cII = - 10.9 N/m2 Since Q, = 0, Eqs (1.1 1) and (1.12) reduce to 32 Basic elasticity and (ii) respectively. Adding Eqs (i) and (ii) we obtain 01 + a11 = a, Thus ax = 80.9 - 10.9 = 70N/mm2 For an axial load P 2p P ax = 70N/mm = - = A n x 502/4 whence P = 137.4kN Substituting for a, in either of Eqs (i) or (ii) gives 7.1. = 29.7N/mm2 From the theory of the torsion of circular section bars Tr T x 25 J ITX 504/32 T,~ = 29.7N/mm = - = from which T =0.7kNm Note that P could have been found directly in this particular case from the axial strain. Thus, from the first of Eqs (1.47) a, = EE, = 70000 x 1000 x lop6 = 70N/mm2 as before. 1 Timoshenko, S. and Goodier, J. N., Theory of Elasticity, 2nd edition, McGraw-Hill Book Company, New York, 1951. 2 Wang, C. T., Applied Elasticity, McGraw-Hill Book Company, New York, 1953. P.l.l A structural member supports loads which produce, at a particular point, a direct tensile stress of 80 N/mm2 and a shear stress of 45 N/mm2 on the same plane. Calculate the values and directions of the principal stresses at the point and also the maximum shear stress, stating on which planes this will act. Problems 33 Am. or = 100.2N/mm2, 0 = 24" 11' a11 = -20.2 N/IIUII~, 6' = 114" 11' T,,, = 60.2N/mm2, at 45" to principal planes P.1.2 At a point in an elastic material there are two mutually perpendicular planes, one of which carries a direct tensile stress at 50N/mm2 and a shear stress of 40N/mm2, while the other plane is subjected to a direct compressive stress of 35 N/mm2 and a complementary shear stress of 40 N/mm2. Determine the principal stresses at the point, the position of the planes on which they act and the position of the planes on which there is no normal stress. ar = ~~.~N/IIuII~, e = 210 38' aII = -50.9N/mm2; 0 = 111" 38' No normal stress on planes at 70" 21' and -27" 5' to vertical. P.1.3 Listed below are varying combinations of stresses acting at a point and referred to axes x and y in an elastic material. Using Mohr's circle of stress determine the principal stresses at the point and their directions for each combination. a, N/m2 ay N/m2 7.u N/m2 (i) +54 +30 +5 (ii) +30 +54 -5 (iii) -60 -36 +5 (iv) +30 -50 +30 Am. (i) aI = +55N/mm2, arI = +29N/mm2, a1 at 11.5" to x axis. (ii) or = +55 N/mm2, aII = +29 N/mm2, aII at 11 .5" to x axis. (iii) or = -34.5N/mm2; arI = -61 N/mm2, aI at 79.5" to x axis. (iv) aI = +40N/mm2, aII = -60N/mm2, aI at 18.5" to x axis. P.1.4 The state of stress at a point is caused by three separate actions, each of which produces a pure, unidirectional tension of 10N/mm2 individually but in three different directions as shown in Fig. P.1.4. By transforming the individual IO N/rnrn2 IO N/mrn2 IO N/rnm2 IO N /mm2 IO N/rnrnz IO N/mrn2 Fig. P.1.4 34 Basic elasticity stresses to a common set of axes (x, y) determine the principal stresses at the point and their directions. Ans. aI = aII = 15N/mm2. All directions are principal directions P.1.5 A shear stress T,~ acts in a two-dimensional field in which the maximum allowable shear stress is denoted by T~~ and the major principal stress by aI. Derive, using the geometry of Mohr's circle of stress, expressions for the maximum values of direct stress which may be applied to the x and y planes in terms of the three parameters given above. cy = 01 - rmax - ddax - <y P.1.6 A solid shaft of circular cross-section supports a torque of 50 kNm and a bending moment of 25 kNm. If the diameter of the shaft is 150 mm calculate the values of the principal stresses and their directions at a point on the surface of the shaft. A~S. aI = 121.4~/mm~, e = 31043' aII = -46.4~/m~, e = 121043' P.1.7 An element of an elastic body is subjected to a three-dimensional stress system a,, ay and a,. Show that if the direct strains in the directions x, y and z are E,, and .zZ then ay = Xe + ~GE,, a, = Xe + ~GE,, a, = Xe + ~GE, where uE A= (1 +Y)(l -24 and e = E, + + E, the volumetric strain. P.1.8 Show that the compatibility equation for the case of plane strain, viz. may be expressed in terms of direct stresses a, and cry in the form P.1.9 In Fig. P.1.9 the direct strains in the directions a, by c are -0.002, -0.002 Ans. = +0.00283, = -0.00283, 8 = -22.5" or +67.5" P.1.10 The simply supported rectangular beam shown in Fig. P. 1.10 is subjected to two symmetrically placed transverse loads each of magnitude Q. A rectangular strain gauge rosette located at a point P on the centroidal axis on one vertical face and +0.002 respectively. If I and I1 denote principal directions find E~., qI and 0. Problems 35 It Fig. P.1.9 Equal distances QAQ Equal distances // QwQ "i- / I/ I Fig. P.1.10 of the beam gave strain readings as follows: E, = -222 x &b = -213 x lop6, E, = +45 x The longitudinal stress at the point P due to an external compres- sive force is 7 N/mm'. Calculate the shear stress T at the point P in the vertical plane and hence the transverse load Q. (Q = 2bdr/3 where b = breadth, d = depth of beam) E = 31 000N/mm2, v = 0.2 Ans. T = 3.17N/mm2, Q = 95.1 kN [...]... 1 E = -(cy - uc,) Ey rxy = 2( 1 E + u) T x y We find that although E, exists, Eqs (1 .22 )-(1 .26 ) are identically satisfied leaving Eq (1 .21 ) as the required compatibility condition Substitution in Eq (1 .21 ) of the above strains gives 2( 1 , + v)- & axay a2 = -(ay ax2 a2 - vu,) +-(a, - vuy) aY2 From Eqs (1.6) and Adding Eqs (2. 2) and (2. 3), then substituting in Eq (2. 1) for 2a2rXy/axay,we have or The alternative... above for rzx, and dO/dz rzy dw 2Ty T (a2+b2) dw - 2Tx 7rab3G+G m3b3 y 7 % =-E - - dx T (a2+ h 2 ) X 7ra3b3 or (vii) Integrating both of Eqs (vii) T(b2- a2) T(b2- a2) yx+fib), w = X Y +f2 ( X I 7ra3 G b3 nu3b3G The warping displacement given by each of these equations must have the same value at identical points ( x , y ) It follows thatfi(y) =f2(x) = 0 Hence W = T(b2- a2) (viii) XY nu3b3G Lines of... Determine also the extent to which the static boundary conditions are satisfied ax PX =[5y(x2 - z2) - ioy3 ay = Txy = 20 d3 E( 4d3 + 6dZy] - 3yd2 - 2d3) -P -[5(3x2- Z2)(y2- d2) - 5y4 + 6y2d2- $1 40d3 The boundary stress function values of T~~do not agree with the assumed constant equilibrating shears at x = 0 and 1 P .2. 4 A two-dimensional isotropic sheet, having a Young’s modulus E and linear coefficientof... Integrating Eqs (iv) and (v) and noting that displacements, we find E, (v) Y2) (vi) and sY are partial derivatives of the Px2y +fib), v=- uPxy2 + f 2 ( x ) 2EI (vii) 2EI where fi ( y ) andfi(x) are unknown functions of x and y Substituting these values of u and Y in Eq (vi) u= px2 2EI 'fi +-+-( Y ) 8y vpY2 2EI ah(x) ax P (b2 - 4y2) 81G 46 Two-dimensional problems in elasticity Separating the terms containing... (b) Fig 2. 2 (a) Required loading conditions on rectangular sheet in Example 2. 2 for A = B = C = 0; (b) as in (a) butA = C = D = 0 the coefficients are related in a certain way Thus, for a stress function in the form of a polynomial of the fourth degree Ax4 $=-+- Bx3y 6 12 Cx +-+ -22 g Dxy3 Ey4 6 ' 1 2 and Substituting these values in Eq (2. 9) we have E = -(2C +A ) The stress components are then a24 ox... 0 and from Eq (x) C= It follows immediately that PI2 2EI 2. 6 Bending of an end-loaded cantilever 47 and, from Eq (viii) D = - PI2 Pb2 -2EI 8IG Substitution for the constants C, F and H in Eqs (ix) and (x) now produces the D, equations for the components of displacement at any point in the beam Thus u= (i x) vPxy2 Px3 v= + 2EI 6EI P12x 2EI PI3 +-3EI (xii) The deflection curve for the neutral plane... $=-+- Bx3y 6 12 Cx +-+ -22 g Dxy3 Ey4 6 ' 1 2 and Substituting these values in Eq (2. 9) we have E = -(2C +A ) The stress components are then a24 ox = -= Cx2 aY2 g - -8 4 =A E'- ax2 #q5 x - + Dxy - (2C + A)y2 + Bxy + Cy2 B 2 5= -2 axay - DY2 2cxy -2 The coefficients A , B, C and D are arbitrary and may be chosen to produce various loading conditions as in the previous examples The obvious disadvantage... first of these T~,, -A = BY2 2 = b 0 at y = f2 giving Bb2 A= 8 From the second bI2 J dy = P (see sign convention for T , ~ - -b /2 or dy = P -b /2 7-yy) 2. 6 Bending of an end-loaded cantilever 45 from which The stresses follow from Eqs (ii) l2Pxy b3 a,= - Px I Y u,,= 0 (iii) where I = b3/ 12 the second moment of area of the beam cross-section We note from the discussion of Section 2. 4 that Eqs (iii) represent... T = -2G- de (a2 + b 2 ) ( 'a2 J l d d y + ~ l l y 2 ~ d y - J l d x d y ) dz *b2 J The first and second integrals in this equation are the second moments of area Iyy = 7ru3b/4and I,, = 7rub3/4,while the third integral is the area of the cross-section A = nub Replacing the integrals by these values gives dB m 3 b 3 T=Gdz (a2 b2) + 3 .2 St Venant warping function solution from which (see Eq (3. 12) ) J=... Fig 2. 1 40 Two-dimensional problems in elasticity A t f t t t ' 4 4 _I -7- A I -c i l +- ux =22 c +I - I X * rxy=-B , J + S t t + Example 2 2 A more complex polynomial for the stress function is Ax3 4= -+ 6 Bx2y 2 Cxy2 +- +2 Dy3 6 As before &4- a~ ax~ay2 ay4= O so that the compatibility equation (2. 9) is identically satisfied.The stresses are given by We may choose any number of values of the coefficients . (1 .21 ) of the above strains gives &,, a2 a2 2( 1 + v)- = -(ay - vu,) +-(a, - vuy) axay ax2 aY2 From Eqs (1.6) and Adding Eqs (2. 2) and (2. 3), then substituting in Eq. (2. 1). Bx3y Cx2g Dxy3 Ey4 6 ' 12 +-+- $=-+- 12 6 2 and Substituting these values in Eq. (2. 9) we have E = -(2C + A) The stress components are then a24 aY2 ox = - = Cx2 +. then a24 aY2 ox = - = Cx2 + Dxy - (2C + A)y2 x + Bxy + Cy2 84 E'- ax2 g =A DY2 5= axay 2 2 2cxy - - #q5 - B2 The coefficients A, B, C and D are arbitrary