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