CHAPTER 12 MISCELLANEOUS TOPICS 12.1.Bending of beams with initial curvature The bending theory derived and applied in Mechanics of Materials I was concerned with the bending of initially straight beams.Let us now consider the modifications which are required to this theory when the beams are initially curved before bending moments are applied.The problem breaks down into two classes: (a)initially curved beams where the depth of cross-section can be considered small in relation to the initial radius of curvature,and (b)those beams where the depth of cross-section and initial radius of curvature are approx- imately of the same order,i.e.deep beams with high curvature. In both cases similar assumptions are made to those for straight beams even though some will not be strictly accurate if the initial radius of curvature is small. (a)Initially curved slender beams Consider now Fig.12.1,with Fig.12.1(a)showing the initial curvature of the beam before bending,with radius Ri,and Fig.12.1 (b)the state after the bending moment M has been applied to produce a new radius of curvature R2.In both figures the radii are measured to the neutral axis. The strain on any element A'B'a distance y from the neutral axis will be given by: A'B'-AB strain on A'B'=8= AB =®+8-R1+9, (R1+y)1 R22+92-R181- (R1+y)81 Since there is no strain on the neutral axis in either figure CD=C'D'and Ri=R202. 82-81_y02-0) E= (R1+y)81(R1+y)8 and,since 02 R101/R2 y(R1-R2) e= (12.1) (R1+y)81 R2(R1+y) 509
CHAPTER 12 MISCELLANEOUS TOPICS 12.1. Bending of beams with initial curvature The bending theory derived and applied in Mechanics of Materials 1 was concerned with the bending of initially straight beams. Let us now consider the modifications which are required to this theory when the beams are initially curved before bending moments are applied. The problem breaks down into two classes: (a) initially curved beams where the depth of cross-section can be considered small in (b) those beams where the depth of cross-section and initial radius of curvature are approxrelation to the initial radius of curvature, and imately of the same order, i.e. deep beams with high curvature. In both cases similar assumptions are made to those for straight beams even though some will not be strictly accurate if the initial radius of curvature is small. (a) Initially curved slender beams Consider now Fig. 12.1, with Fig. 12.1 (a) showing the initial curvature of the beam before bending, with radius R1, and Fig. 12.1 (b) the state after the bending moment M has been applied to produce a new radius of curvature R2. In both figures the radii are measured to the neutral axis. The strain on any element A’B’ a distance y from the neutral axis will be given by: A’B’ - AB strain on A’B’ = E = AB - w2 + y)e2 - wI + Y)el - (RI + Y)OI (R1 + Y)OI - R262 + $2 - Riel - @I - Since there is no strain on the neutral axis in either figure CD = C’D’ and Riel = R2&. ye2 - ye1 - Y(02 - 01) (R~ + Y)eI .. E= - (R~ + y)oI and, since e2 = Rlel/R2. (12.1) 509
510 Mechanics of Materials 2 §12.1 (a) (b) Fig.12.1.Bending of beam with initial curvature (a)before bending,(b)after bending to new radius of curvature R2. For the case of slender,beams with y small in comparison with R (i.e.when y can be neglected in comparison with Ri),the equation reduces to: (R1-R2)「111 8=y (12.2) R2R1 2=y R2R1」 The strain is thus directly proportional to y the distance from the neutral axis and,as for the case of straight beams,the stress and strain distribution across the beam section will be linear and the neutral axis will pass through the centroid of the section.Equation(12.2)can therefore be incorporated into a modified form of the "simple bending theory"thus: (12.3) For initially straight beams R is infinite and egn.(12.2)reduces to: =成= (b)Deep beams with high initial curvature(i.e.small radius of curvature) For deep beams where y can no longer be neglected in comparison with Ri eqn.(12.1) must be fully applied.As a result,the strain distribution is no longer directly proportional to y and hence the stress and strain distributions across the beam section will be non-linear as shown in Fig.12.2 and the neutral axis will not pass through the centroid of the section. From eqn.(12.1)the stress at any point in the beam cross-section will be given by: o=Ee- Ey(R1-R2) (12.4) R2(R1+y) For equilibrium of transverse forces across the section in the absence of applied end load fodA must be zero. ER-R)dA=EK-R） ·dA=0 (12.5) R2(R1+y) R2 (R1+y)
510 Mechanics of Materials 2 912.1 I i Fig. 12.1. Bending of beam with initial curvature (a) before bending, (b) after bending to new radius of curvature Rz. For the case of slender, beams with y small in comparison with R1 (i.e. when y can be neglected in comparison with RI), the equation reduces to: (12.2) The strain is thus directly proportional to y the distance from the neutral axis and, as for the case of straight beams, the stress and strain distribution across the beam section will be linear and the neutral axis will pass through the centroid of the section. Equation (12.2) can therefore be incorporated into a modified form of the “simple bending theory” thus: Ma ZY For initially straight beams R1 is infinite and eqn. (12.2) reduces to: YY R2 R &=--=- (12.3) (b) Deep beams with high initial curvature (Le. small radius of curvature) For deep beams where y can no longer be neglected in comparison with Rl eqn. (12.1) must be fully applied. As a result, the strain distribution is no longer directly proportional to y and hence the stress and strain distributions across the beam section will be non-linear as shown in Fig. 12.2 and the neutral axis will not pass through the centroid of the section. From eqn. (12.1) the stress at any point in the beam cross-section will be given by: (1 2.4) For equilibrium of transverse forces across the section in the absence of applied end load adA must be zero. .. (12.5)
§12.1 Miscellaneous Topics 511 Neutral axis Centroidal Beom X 0x15 section (a)Initially stroight beam-Linear (b)Initially curved beam-Non-linear stress stress dis封ribution distribution Fig.12.2.Stress distributions across beams in bending.(a)Initially straight beam linear stress distribution; (b)initially curved deep beam-non-linear stress distribution. i.e. 2·dA=0 (12.6) (R,+y) Unlike the case of bending of straight beams,therefore,it will be seen by inspection that the above integral no longer represents the first moment of area of the section about the centroid.Thus,the centroid and the neutral axis can no longer coincide. The bending moment on the section will be given by: M=a.dA.y=E(R-R) ·dA (12.7) R (R1+y) but (R+y) =dA-/ (R1+y) and from eqn.(12.5)the second integral term reduces to zero for equilibrium of transverse forces. ∫0朵)A=dA=A=M where h is the distance of the neutral axis from the centroid axis,see Fig.12.3.Substituting in egn.(12.7)we have: M E(R1-R2).hA R2 (12.8) From egn.(12.4) (R1+》= E (R1-R2) y R2 M=(R+yhA (12.9) M i.e. (12.10) y hA(R1+y) My My or 0= (12.11) hA(R1+y)hARo
512.1 Miscellaneous Topics 51 1 +--f+. Centroidol Beom X axis section ( a) Initially straight beam -Linear ( b Initially curved beam - Non-linear stress stress distribution distribution Fig. 12.2. Stress distributions across beams in bending. (a) Initially straight beam linear stress distribution; (b) initially curved deep beam-non-linear stress distribution. i.e. *dA=O J& (12.6) Unlike the case of bending of straight beams, therefore, it will be seen by inspection that the above integral no longer represents the first moment of area of the section about the centroid. Thus, the centroid and the neutral axis can no longer coincide. The bending moment on the section will be given by: but (12.7) and from eqn. (12.5) the second integral term reduces to zero for equilibrium of transverse forces. .. where h is the distance of the neutral axis from the centroid axis, see Fig. 12.3. Substituting in ean. (12.7) we have: From eqn. (12.4) i.e. or (12.8) (1 2.9) (12.10) (12.1 1)
512 Mechanics of Materials 2 §12.1 Axis of curvature R Neutrai oxis Beam cross-section Fig.12.3.Relative positions of neutral axis and centroidal axis. On the opposite side of the neutral axis,where y will be negative,the stress becomes: My My =-hA(R1-y)hAR; (12.12) These equations show that the stress distribution follows a hyperbolic form.Equation(12.12) can be seen to be similar in form to the"simple bending"equationT. o M y=7 with the term hA(RI+y)replacing the second moment of area 1. Thus in order to be able to calculate stresses in deep-section beams with high initial curvature,it is necessary to evaluate h and Ri,i.e.to locate the position of the neutral axis relative to the centroid or centroidal axis.This was shown above to be given by the condition: y ·dA=0. J(R1+y) Now fibres distance y from the neutral axis will be some distance ye from the centroidal axis as shown in Figs.12.3 and 12.4 such that,in relation to the axis of curvature, R+y=Rc+yc with y=yc+h .'from egn.(12.5) (+h) (Rc yc) ·dA=0 Re-writing yc+h (Rc+yc)-Rc+h=(Rc +yc)-(Rc-h). Timoshenko and Roark both give details of correction factors which may be applied for standard cross- sectional shapes to be used in association with the simple straight beam equation.(S.Timoshenko.Theory of Plates and Shells,McGraw Hill,New York;R.J.Roark and W.C.Young,Formulas for Stress and Strain,McGraw Hill,New York)
512 Mechanics of Materials 2 912.1 IRc -. 1-11 ! iY‘ Beam cross-section Fig. 12.3. Relative positions of neutral axis and centroidal axis. On the opposite side of the neutral axis, where y will be negative, the stress becomes: (12.12) These equations show that the stress distribution follows a hyperbolic form. Equation (12.12) can be seen to be similar in form to the “simple bending” equationt. aM __ - - YI with the term hA(R1 + y) replacing the second moment of area I. Thus in order to be able to calculate stresses in deep-section beams with high initial curvature, it is necessary to evaluate h and RI, Le. to locate the position of the neutral axis relative to the centroid or centroidal axis. This was shown above to be given by the condition: Now fibres distance y from the neutral axis will be some distance y, from the centroidal axis as shown in Figs. 12.3 and 12.4 such that, in relation to the axis of curvature, R1 + y = R, + yc with Y=Yc+h :. from eqn. (12.5) TTimoshenko and Roark both give details of correction factors which may be applied for standard crosssectional shapes to be used in association with the simple straight beam equation. (S. Timoshenko, Theory of Plares and Shells, McGraw Hill, New York; R. J. Roark and W.C. Young, Formulas for Stress and Strain, McGraw Hill, New York)
§12.1 Miscellaneous Topics 513 一.Axi5 of curvoture R R =Rc+yc dy Fig.12.4. (ye+h) ·dA (Rc+yc) =∫+对-- (Rc yc) (Rc+yc) =A-(R-h) 1 ·dA=0 (Rc+yc) A h=Rc- A =Rc- (12.13) (Rc+yc) A A and R1=Rc-h= dA (12.14) (Rc+yc) Examples 12.1 and 12.2 show how the theory may be applied and Table 12.1 gives some dA usefulqio forfor standard shapes of beam cros-section Note Before applying the above theory for bending of initially curved members it is perhaps appropriate to consider the benefits to be gained over that of an approximate solution using the simple bending theory. Provided that the curvature is not large then the simple theory is reasonably accurate;for example,for a radius to beam depth ratio R/d of as low as 5 the error introduced in the maximum stress value is only of the order of 7%.The error then rises steeply,however,as curvature increases to a figure of approx.30%at Rc/d=1.5. (c)Initially curved beams subjected to bending and additional direct load In many practical engineering applications such as chain links,crane hooks,G-clamps etc.,the component cross-sections will be subjected to both bending and additional direct load,whereas the equations derived in the previous sections have all been derived on the assumption of pure bending only.It is therefore necessary in such cases to obtain a solution by the application of the principle of superposition i.e.by resolving the loading system into
$12.1 Miscellaneous Topics 513 .. Axis of curvature 11-1- -.__ --~ - I Fig. 12.4. P1 = A - (R, - h) .dA =O A =Re-- (12.13) A h= R, - (12.14) A A and R1= Re - h = J (Rc dA +Ye) =E Examples 12.1 and 12.2 show how the theory may be applied and Table 12.1 gives some useful equations for J - for standard shapes of beam cross-section. Note Before applying the above theory for bending of initially curved members it is perhaps appropriate to consider the benefits to be gained over that of an approximate solution using the simple bending theory. Provided that the curvature is not large then the simple theory is reasonably accurate; for example, for a radius to beam depth ratio R,/d of as low as 5 the error introduced in the maximum stress value is only of the order of 7%. The error then rises steeply, however, as curvature increases to a figure of approx. 30% at R,/d = 1.5. dA r (c) Initially curved beams subjected to bending and additional direct load In many practical engineering applications such as chain links, crane hooks, G-clamps etc., the component cross-sections will be subjected to both bending and additional direct load, whereas the equations derived in the previous sections have all been derived on the assumption of pure bending only. It is therefore necessary in such cases to obtain a solution by the application of the principle of superposition i.e. by resolving the loading system into
514 Mechanics of Materials 2 §12.1 Table 12.1.Values of dA for curved bars. Cross-section (a】Rectangle bloge d (N.B.The two following cross-sections are simply produced R。 by the addition of terms of this form for each rectangular portion) axis of curvature R (b)T-section R。 oilo.(点）+loe(年a) d11 R (c】I-beom b3+ d3 R。 (t)+e,(+)+与（） (d)Tropezoid -b1+b2 d (e）Triangle As above (d)with b2 =0 As above (d)with b=0 d:2R R 2π{(R+R)-[(R+R)2-R2]/2)
5 14 Mechanics of Materials 2 Table 12.1. Values of s - for curved bars. $12.1 Cross-section (a Rectangle bb I axis of curvoture -fRi (b T-section J? N.B. The two following cross-sections are simply produced iy the addition of terms of this form for each rectangular lortion) As above (d) with b2 = 0 As above (d) with bl = 0
§12.2 Miscellaneous Topics 515 its separate bending,normal(and perhaps shear)loads on the section and combining the stress values obtained from the separate stress calculations.Normal and bending stresses may be added algebraically and combined with the shearing stresses using two-or three-dimensional complex stress equations or Mohr's circle. Care must always be taken to consider the direction in which the moment is applied. In the derivation of the equations in the previous sections it has been shown acting in a direction to increase the initial curvature of the beam(Fig.12.1)producing tensile bending stresses on the outside (convex)surface and compression on the inner (concave)surface.In the practical cases mentioned above,however,e.g.the chain link or crane hook,the moment which is usually applied will tend to straighten the beam and hence reduce its curvature.In these cases,therefore,tensile stresses will be set up on the inner surface and these will add to the tensile stresses produced by the direct load across the section to produce a maximum tensile (and potentially critical)stress condition on this surface-see Fig.12.5. (b) Tensile Bending stresses 6 Pe Compressive Direct lood stresses 60 Totol stress A. (a) ,c。DD Fig.12.5.Loading of a crane hook.(a)Load effect on section AA is direct load P'=P plus moment M=Pe: (b)stress distributions across the section AA. 12.2.Bending of wide beams The equations derived in Mechanics of Materials I for the stress and deflection of beams subjected to bending relied on the assumption that the beams were narrow in relation to their depths in order that expansions or contractions in the lateral (z)direction could take place relatively freely
$12.2 Miscellaneous Topics 515 its separate bending, normal (and perhaps shear) loads on the section and combining the stress values obtained from the separate stress calculations. Normal and bending stresses may be added algebraically and combined with the shearing stresses using two- or three-dimensional complex stress equations or Mohr's circle. Care must always be taken to consider the direction in which the moment is applied. In the derivation of the equations in the previous sections it has been shown acting in a direction to increase the initial curvature of the beam (Fig. 12.1) producing tensile bending stresses on the outside (convex) surface and compression on the inner (concave) surface. In the practical cases mentioned above, however, e.g. the chain link or crane hook, the moment which is usually applied will tend to straighten the beam and hence reduce its curvature. In these cases, therefore, tensile stresses will be set up on the inner surface and these will add to the tensile stresses produced by the direct load across the section to produce a maximum tensile (and potentially critical) stress condition on this surface - see Fig. 12.5. P': P t 1 P Fig. 12.5. Loading of a crane hook. (a) Load effect on section AA is direct load P' = P plus moment M = Pe; (b) stress distributions across the section AA. 12.2. Bending of wide beams The equations derived in Mechanics of Materials 1 for the stress and deflection of beams subjected to bending relied on the assumption that the beams were narrow in relation to their depths in order that expansions or contractions in the lateral (2) direction could take place relatively freely
516 Mechanics of Materials 2 §12.2 For beams that are very wide in comparison with their depth-see Fig.12.6-lateral deflections are constrained,particularly towards the centre of the beam,and such beams become stiffer than predicted by the simple theory and deflections are correspondingly reduced.In effect,therefore,the bending of narrow beams is a plane stress problem whilst that of wide beams becomes a plane strain problem-see $8.22. For the beam of Fig.12.6 the strain in the z direction is given by eqn.(12.6)as: E(0:-vax-vay). 7HH2◆2 T一b Beam cross-section Fig.12.6.Bending of wide beams (bd) Now for thin beams oy =0 and,for total constraint of lateral (z)deformation at z=0, ex=0. 1 0=Ea:-o,） ie. 0:=0x Thus,the strain in the longitudinal x direction will be: 1 Ex=E(ox -voy-vo:) 1 ox-0-(o》 =-加 (12.15) = (1-v2)My (12.16) E 1 Compared with the narrow beam case where er=ox/E there is thus a reduction in strain by the factor(1-v2)and this can be introduced into the deflection equation to give: =a-尚 d2y dr2 (12.17) Thus,all the formulae derived in Book 1 including those of the summary table,may be used for wide beams provided that they are multiplied by (1-v)
516 Mechanics of Materials 2 $12.2 For beams that are very wide in comparison with their depth - see Fig. 12.6 - lateral deflections are constrained, particularly towards the centre of the beam, and such beams become stiffer than predicted by the simple theory and deflections are correspondingly reduced. In effect, therefore, the bending of narrow beams is a plane stress problem whilst that of wide beams becomes a plane strain problem - see 98.22. For the beam of Fig. 12.6 the strain in the z direction is given by eqn. (12.6) as: 1 E - -(az - wax - way). Z-E tY t’ Beam cross-section ‘I Fig. 12.6. Bending of wide beams (b >> d) Now for thin beams cry = 0 and, for total constraint of lateral (z) deformation at z = 0, E, = 0. i.e. 1 E 0 = -(az - wa,) a, = wax Thus, the strain in the longitudinal x direction will be: 1 E E, = -(a, - uay - wa,) 1 E = -(a, - 0 - u(uax)) (12.15) 1 E = -(1 - w2)ax (1 - u2) My - ___.- E I (12.16) Compared with the narrow beam case where E, = a,/E there is thus a reduction in strain by the factor (1 - w2) and this can be introduced into the deflection equation to give: d2Y M - = (1 - 2)- dx2 EI (12.17) Thus, all the formulae derived in Book 1 including those of the summary table, may be used for wide beams provided that they are multiplied by (1 - u’)
§12.3 Miscellaneous Topics 517 12.3.General expression for stresses in thin-walled shells subjected to pressure or self-weight Consider the general shell or "surface of revolution"of arbitrary (but thin)wall thickness shown in Fig.12.7 subjected to internal pressure.The stress system set up will be three- dimensional with stresses o1(hoop)and o2(meridional)in the plane of the surface and o3 (radial)normal to that plane.Strictly,all three of these stresses will vary in magnitude through the thickness of the shell wall but provided that the thickness is less than approximately one- tenth of the major,i.e.smallest,radius of curvature of the shell surface,this variation can be neglected as can the radial stress (which becomes very small in comparison with the hoop and meridional stresses). Rodius Arc d82 in vertical plane g1(hoop】 Arc de in horizontol plane Rodius r (b) 2 (mer idional Fig.12.7.(a)General surface of revolution subjected to internal pressure p:(b)element of surface with radii of curvature ri and r2 in two perpendicular planes. Because of this limitation on thickness,which makes the system statically determinate,the shell can be considered as a membrane with little or no resistance to bending.The stresses set up on any element are thus only the so-called"membrane stresses"and o2 mentioned above,no additional bending stresses being required. Consider,therefore,the equilibrium of the element ABCD shown in Fig.12.7(b)where rI is the radius of curvature of the element in the horizontal plane and r2 is the radius of curvature in the vertical plane. The forces on the "vertical"and "horizontal"edges of the element are oitds and oztds2, respectively,and each are inclined relative to the radial line through the centre of the element, one at an angle de1/2 the other at de2/2. Thus,resolving forces along the radial line we have,for an internal pressure p: 2(din d d02 Now for small angles sin de/2=de/2 radians =pds1·ds2
§12.3 Miscellaneous Topics 517 12.3. General expression for stresses in thin-walled shells subjected to pressure or self-weight Consider the general shell or "surface of revolution" of arbitrary (but thin) wall thickness shown in Fig. 12.7 subjected to internal pressure. The stress system set up will be threedimensional with stresses a) (hoop) and a2 (meridional) in the plane of the surface and a3 (radial) normal to that plane. Strictly, all three of these stresses will vary in magnitude through the thickness of the shell wall but provided that the thickness is less than approximately onetenth of the major, i.e. smallest, radius of curvature of the shell surface, this variation can be neglected as can the radial stress (which becomes very small in comparison with the hoop and meridional stresses). Fig. 12.7. (a) General surface of revolution subjected to internal pressure p; (b) element of surface with radii of curvature rl and r2 in two perpendicular planes. Because of this limitation on thickness, which makes the system statically determinate, the shell can be considered as a membrane with little or no resistance to bending. The stresses set up on any element are thus only the so-called "membrane stresses" al and a2 mentioned above, no additional bending stresses being required. Consider, therefore, the equilibrium of the element ABCD shown in Fig. 12.7(b) where rl is the radius of curvature of the element in the horizontal plane and r2 is the radius of curvature in the vertical plane. The forces on the "vertical" and "horizontal" edges of the element are altdsl and a2tds2, respectively, and each are inclined relative to the radial line through the centre of the element, one at an angle d(}l/2 the other at d(}2/2. Thus, resolving forces along the radial line we have, for an internal pressure p: .d(}l .d(}2 2(al t dsJ .SIn ~ + a2 t ds2 .SIn -= p .dsl .dS2 2 " Now for small angles sinde/2 = de/2 radians ( del 2 (11 t dsl .-+ (12 t dS2 ~ ) = pdsl .dS2 2 2
518 Mechanics of Materials 2 §12.4 Also ds1 =r2 de2 and ds2 ri de o1tds1 .ds2+ot ds2 rI ds =p.dsds2 r2 and dividing through by ds ds2 t we have: 4+2= (12.18) r r2 t For a general shell of revolution,o and o2 will be unequal and a second equation is required for evaluation of the stresses set up.In the simplest application,i.e.that of the sphere,however,r=r2 =r and symmetry of the problem indicates that o1 =02=o. Equation (12.18)thus gives: a=pr 2t In some cases,e.g.concrete domes or dishes,the self-weight of the vessel can produce significant stresses which contribute to the overall failure consideration of the vessel and to the decision on the need for,and amount of,reinforcing required.In such cases it is necessary to consider the vertical equilibrium of an element of the dome in order to obtain the required second equation and,bearing in mind that self-weight does not act radially as does applied pressure,eqn.(12.18)has to be modified to take into account the vertical component of the forces due to self-weight. Thus for a dome of subtended arc 20 with a force per unit area g due to self-weight, egn.(12.18)becomes: +=±gcos9 (12.19) Combining this equation with one obtained from vertical equilibrium considerations yields the required values of o and o2. 12.4.Bending stresses at discontinuities in thin shells It is normally assumed that thin shells subjected to internal pressure show little resistance to bending so that only membrane (direct)stresses are set up.In cases where there are changes in geometry of the shell,however,such as at the intersection of cylindrical sections with hemispherical ends,the "incompatibility"of displacements caused by the membrane stresses in the two sections may give rise to significant local bending effects.At times these are so severe that it is necessary to introduce reinforcing at the junction locations. Consider,therefore,such a situation as shown in Fig.12.8 where both the cylindrical and hemispherical sections of the vessel are assumed to have uniform and equal thickness membrane stresses in the cylindrical portion are 1=0H=P" and o o whilst for the hemispherical ends pr σ1=02=0H=
518 Mechanics of Materials 2 $12.4 Also dsl = r2d62 and ds2 = rI d81 .. 01 tdsl . and dividing through by dsl . ds2 . t we have: (12.18) For a general shell of revolution, al and a2 will be unequal and a second equation is required for evaluation of the stresses set up. In the simplest application, i.e. that of the sphere, however, rl = r2 = r and symmetry of the problem indicates that (TI = 02 = a. Equation (12.18) thus gives: P‘ 2t u=- In some cases, e.g. concrete domes or dishes, the self-weight of the vessel can produce significant stresses which contribute to the overall failure consideration of the vessel and to the decision on the need for, and amount of, reinforcing required. In such cases it is necessary to consider the vertical equilibrium of an element of the dome in order to obtain the required second equation and, bearing in mind that self-weight does not act radially as does applied pressure, eqn. (12.18) has to be modified to take into account the vertical component of the forces due to self-weight. Thus for a dome of subtended arc 28 with a force per unit area q due to self-weight, eqn . ( 12.18) becomes: (12.19) Combining this equation with one obtained from vertical equilibrium considerations yields the required values of a1 and 02. 12.4. Bending stresses at discontinuities in thin shells It is normally assumed that thin shells subjected to internal pressure show little resistance to bending so that only membrane (direct) stresses are set up. In cases where there are changes in geometry of the shell, however, such as at the intersection of cylindrical sections with hemispherical ends, the “incompatibility” of displacements caused by the membrane stresses in the two sections may give rise to significant local bending effects. At times these are so severe that it is necessary to introduce reinforcing at the junction locations. Consider, therefore, such a situation as shown in Fig. 12.8 where both the cylindrical and hemispherical sections of the vessel are assumed to have uniform and equal thickness membrane stresses in the cylindrical portion are whilst for the hemispherical ends Pr t 01 = 02 = OH = -
