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