13 Deflections of beams 13.1 Introduction In Chapter 7 we showed that the loading actions at any section of a simply-supported beam or cantilever can be resolved into a bending moment and a shearing force. Subsequently, inChapters 9 and 10, we discussed ways of estimating the stresses due to these bending moments and shearing forces. There is, however, another aspect of the problem of bending which remains to be treated, namely, the calculation of the stifiess of a beam. In most practical cases, it is necessary that a beam should be not only strong enough for its purpose, but also that it should have the requisite stiffness, that is, it should not deflect from its original position by more than a certain amount. Again, there are certain types ofbeams, such as those camed by more than two supports and beams with their ends held in such a way that they must keep their original directions, for which we cannot calculate bending moments and shearing forces without studying the deformations of the axis of the beam; these problems are statically indeterminate, in fact. In this chapter we consider methods of finding the deflected form of a beam under a given system of external loads and having known conditions of support. 13.2 Elastic bending of straight beams It was shown in Section 9.2 that a straight beam of uniform cross-section, when subjected to end couples A4 applied about a principal axis, bends into a circular arc of radius R, given by 1M R EI - (13.1) where EI, which is the product of Young's modulus E and the second moment of area I about the relevant principal axis, is the flexural stiffness of the beam; equation ( 13.1) holds only for elastic bending. Where a beam is subjected to shearing forces, as well as bending moments, the axis of the beam is no longer bent to a circular arc. To deal with this type of problem, we assume that equation (13.1) still defines the radius of curvature at any point of the beam where the bending moment is M. This implies that where the bending moment varies from one section of the beam to another, the radius of curvature also vanes from section to section, in accordance with equation (13.1). In the unstrained condition of the beam, Cz is the longitudinal centroidal axis, Figure 13.1, and Cx, Cy are the principal axes in the cross-section. The co-ordinate axes Cx, Cy are so arranged that the y-axis is vertically downwards. This is convenient as most practical loading conditions give rise to vertically downwards deflections. Suppose bending moments are applied about axes parallel to Cx, so that bending is restricted to the yz-plane, because Cx and Cy are principal axes. 296 Deflections of beams Figure 13.1 Longitudinal and principal Figure 13.2 Displacements of the longitudinal centroidal axes for a straight beam. axis of the beam. Consider a short length of the unstrained beam, corresponding with DF on the axis Cz, Figure 13.2. In the strained condition D and F are dsplaced to D' and F', respectively, which lies in the yz- plane. Any point such as D on the axis Cz is displaced by an amount v parallel to Cy; it is also hsplaced a small, but negligible, amount parallel to Cz. The radius of curvature R at any section of the beam is then given by d2v - 1- dz2 (13.2) R * [1 + ( $)2r We are concerned generally with only small deflections, in which v is small; thls implies that (dv/dz) is small, and that (d~/dz)~ is negligible compared with unity. Then, with sufficient accuracy, we may write d2v 1 R dz2 (13.3) - = f- The equations (13.1) and (1 3.3) give d 'v dz2 (13.4) & EI-= A4 We must now consider whether the positive or negative sign is relevant in this equation; we have already adopted the convention in Section 7.4 that sagging bending moments are positive. When a length of the beam is subjected to sagging bending moments, as in Figure 13.3, the value of (dv/dz) along the length diminishes as z increases; hence a sagging moment implies that the curvature is negative. Then (13.5) d2v dz' EI-=-M where M is the sagging bending moment. Elastic bending of straight beams 297 Figure 13.3 Curvature induced by sagging Figure 13.4 Deflected form of a beam in bending moment. pure bending. Where the beam is loaded on its axis of shear centres, so that no twisting occurs, M may be written in terms of shearing force F and intensity w of vertical loading at any section. From equation (7.9) we have d2M - dF - -w dz2 a2 On substituting for M from equation (1 3.5), we have (13.6) -[-El$] d2 = 5 = -w dz2 Thls relation is true if EI vanes from one section of a beam to another. Where El is constant along the length of a beam, d4v - dF dz4 dz (13.7) -El- - - = -w As an example of the use of equation (13.4), consider the case of a uniform beam carrying couples M at its ends, Figure 13.4. The bending moment at any section is M, so the beam is under a constant bendmg moment. Equation (13.5) gives d2v - dz2 EI- - -M On integrating once, we have dv EI - dz = -Mz +A (13.8) 298 Deflections of beams where A is a constant. On integrating once more (13.9) 1 2 EIv = Mz’ + AI + B where B is another constant. If we measure v relative to a line CD joining the ends of the beam, vis zero at each end. Then v = 0, for z = 0 and z = L. On substituting these two conditions into equation (13.9), we have 1 2 B = 0 and A = -ML The equation (13.9) may be written (13.10) 1 EIv = -Mz(L - Z) 2 At the mid-length, z = U., and (13.11) ML 2 v=- 8EI which is the greatest deflection. At the ends z = 0 and z = L/2, (13.12) dv- ML dv ML (iz 2 EI a!? 2 EI - at C; - = at D It is important to appreciate that equation (13.3), expressing the radius of curvature R in terms of v, is only true if the displacement v is small. Figure 13.5 Distortion of a beam in pure bending. Elastic bending of straight beams 299 We can study more accurately the pure bending of a beam by considering it to be deformed into the arc of a circle, Figure 13.5; as the bending moment M is constant at all sections of the beam, the radius of curvature R is the same for all sections. If L is the length between the ends, Figure 13.5, and D is the mid-point, OB = 4-j Thus the central deflection v, is v = BD = R - \IR2 - (L2/4) Then v=ii Suppose WR is considerably less than unity; then which can be written 1 v = -1 I+-+ L2 L2 8R 4R But and so v = ML -il+- M~L~ + I 8EI 4(E42 (13.13) Clearly, if (L2/4Rz) is negligible compared with unity we have, approximately, which agrees with equation (1 3.1 1). The more accurate equation (13.13) shows that, when (Lz/4R2) 300 Deflections of beams is not negligible, the relationshp between v and A4 is non-linear; for all practical purposes this refinement is unimportant, and we find simple linear relationships of the type of equation (1 3.1 1) are sufficiently accurate for engineering purposes. 13.3 Simply-supported beam carrying a uniformly distributed load A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 13.6; it carries a uniformly distributed lateral load of w per unit length, whch induces bending in the yz plane only. Then the reactions at the ends are each equal to %wL; if z is measured from the end C, the bending moment at a distance z from C is 1 12 2 2 M = -WLZ - -WZ Figure 13.6 Simply-supported beam carrying a uniformly supported load. Then from equation ( 13 S), d2v 1 1 dz2 2 2 El- = -M = WLZ + - wz2 On integrating, +A dv WLZ 2 wz 3 EI- = +- dz 4 6 and +Az+B (1 3.14) wLz3 wz4 12 24 EIv = + - Suppose v = 0 at the ends z = 0 and z = L; then B = 0, and A = wL’I24 Cantilever with a concentrated load 301 Then equation (13.14) becomes (13.15) wz EZv = - [L’ - 2Lz2 + z’] 24 The deflection at the mid-length, z = Y., is (1 3.16) 5wL4 v=- 3 84 El 13.4 Cantilever with a concentrated load A uniform cantilever of flexural stiffness Eland length L carries a vertical concentrated load Wat the free end, Figure 13.7. The bending moment a distance z from the built-in end is M = -W(L - z) Figure 13.7 Cantilever carrying a vertical load at the remote end. Hence equation ( 13.5) gives d2v dz2 EZ- = W(L - Z) Then EIk = w(Lz - iZ2) -L A (13.17) dz and 302 Deflections of beams EIv = W( ;Lz2 - iz3) + Az + B At the end z = 0, there is zero slope in the deflected form, so that dv/dz = 0; then equation (13.17) gives A = 0. Furthermore, at z = 0 there is also no deflection, so that B = 0. Then wz 2 EIv = - (3L - Z) 6 Atthe free end,^ = L, WL 3 VI. = - 3 EI (13.18) The slope of the beam at the free end is 0, = (2) ?=L 2EI (13.19) - WL2 When the cantilever is loaded at some point between the ends, at a distance a, say, from the built-in support, Figure 13.8, the beam between G and D carries no bending moments and therefore remains straight. The deflection at G can be deduced from equation (13.18); for z = a, (13.20) wa 3 v, = - 3 E/ and the slope at z = a is (13.21) Wa 2 2EI eo = - Then the deflection at the free end D of the cantilever is Figure 13.8 Cantilever with a load applied between the ends. Cantilever with a uniformly distributed load 303 Wa ’ 3EI 2EI VL = - wu3 + (L - a) - wu2 (3~ - a) (13.22) - 6EI 13.5 Cantilever with a uniformly distributed load A uniform cantilever, Figure 13.9, carries a uniformly distributed load of w per unit length over the whole of its length. The bending moment at a distance z from C is 1 2 M = w (L - z)’ Then, from equation (13.5), d‘v 1 1 dz’ 2 2 EI- = -W (L - z)’ = -W (L2 - ~Lz + z’) Figure 13.9 Cantilever carrying a uniformly distributed load. Thus EI- = -w L’z - Lz2 + -z3 + A * dz 2 7 3 7 and 22 1 I’ 3 12 ’1 1 EIv = -w -L’z’ - -Lz3 + -z4 + Az + B At the built end, z = 0, and we have 304 Deflections of beams &-O and v=O CL ThusA = B = 0. Then 1 24 E~v = -W (6L2z2 - 4Lz3 + z4) At the free end, D, the vertical deflection is (13.23) WL 4 VL = - 8EI 13.6 Propped cantilever with distributed load The uniform cantilever of Figure 13.1 O(i) carries a uniformly distributed load w and is supported on a rigid knife edge at the end D. Suppose P is the force on the support at D. Then we regard Figure 13.10(i) as the superposition of the effects of P and w acting separately. Figure 13.10 (i) Uniformly loaded cantilever propped at one end. (ii) Deflections due to w alone. (iii) Deflections due to P alone. If w acts alone, the deflection at D is given by equation (13.23), and has the value WL 4 v, = - 8EI If the reaction P acted alone, there would be an upward deflection PL 3 v* = - 3 EI [...]... method of analysis, malung use of A and Z, is known as the method of moment-areas; it can be extended to deal with most problems of beam deflections When the section of the beam is not constant, equation (13.60) becomes The slopes at the ends of the beam are then given by and It is necessary to plot five curves of (Mdl), (l/l), (do, (;/l), (M,@)and to find their areas As an example of the use of equations... greatest deflection occurs at dv/dz W,LZ* -w * L 3 - - + - + 11 180 3 wG4 wG5 -+ 24 = WG3 6 120L 0, i.e when W#4 - 24L or when + 60(;)3 - 120(;)2 + 22 = The relevant root of this equation is z/L =OS06 which gives the point of maximum deflection neai to the mid-length The maximum deflection is Deflections of beams 324 , v 7.03 w,L4 wo L4 - 0.0195360 EI EI i This is negligibly different from the deflection. .. calculate the end deflection on application of the load TakeE = 200GN/m2 (RNEC) Problem 13.1 Solution (1) The second moment of are of the cross-section is I, = T - (0.050)4 64 = 0.307 x m4 The deflection at the end is then (ii) Let T = tension in the wire; the area of cross-section of the wire is 4.90 elongation ofthe wire is then e = - - - EA (200 x x m2 The T(3) 109)(4.90x The load on the end of the cantilever... constant Deflections of beams 326 This requires that Z - - constant - I or I " 2 Thus, 1 12 -bt' ot z 1 or t " z 3 Any variation of the form 1 t = to ($7 where tois the thickness at the built-in end will lead to bending in the form of a circular arc Problem 13.8 The curve M , below, represents the bending moment at any section of a timber cantilever of variable bending stiffness The second moments of area... 109)(4.90x The load on the end of the cantilever is then (1000 - T), and this produces a deflection of v = 3(200 (1000 - fi(2)3 x 109)(0.307 x Deflections of beams 306 If this equals the stretching of the wire, then (1000 - 71(2)3 3(200 x 109)(0.307x 1O-6) This gives T v = (200 x T(3) 109)(4.90x 1O-6) 934 N, and the deflection of the cantilever becomes (66)(2)3 109)(0.307x 1O-6) = 3(200 Problem 13.2 - x = 0.00276... employed, when a curve of (M/I) must be taken as the starting point instead of a curve of M Problem 13.7 A cantilever strip has a length L, a constant breadth b and thickness t varying in such a way that when the cantilever carries a lateral end load W, centre the line of the strip is bent into a circular arc Find the form of variation of the thckness t Solution The second moment of area, I, at any section... greatest deflection occurs at the point Deflections of beams 320 1 - z = [(a/3) - a)I2 and has the value (2L vmax = -(2L wa 9LEI -4 a) (L - a) If a < %L,the greatest deflection occurs in the range z > a;in this case we replace a by (L - a), whence the greatest deflection occurs at the point = ,/-,andhasthevalue z = q - - 9LEI ,v , (L2 - 3 13.11 Beam with end couples and distributed load Suppose the ends of. .. Cantilever carrying any system of lateral loads 13.14 Beams of varying section When the second moment of area of a beam varies from one section to another, equations (13.58) and (13.59) take the forms - dV dz -A-LJ? E Cantilever with irregular loading 325 and v = Az+B-1 E r+ , The general method of procedure follows the same lines as before If (M/I) and J(M/l)dzare integrable functions of z,then (dv/dz)and v... load W at a distance a from C Then the reactions at C and G are v, = E (L L u) vc = Wa - L Figure 13.1 1 Deflections of a simply-supported beam carrying a concentrated lateral load Now consider a section of the beam a distance z from C; if z < a, the bending moment at the section is Deflections of beams 308 M = vcz = V , Z - Mz - a) and ifz > a, M Then E- d2v I = = -vCz z . (1000 - T), and this produces a deflection of (1000 - fi(2)3 v= 3(200 x 109)(0.307 x 306 Deflections of beams If this equals the stretching of the wire, then (1000 - 71(2)3. section of a beam to another. Where El is constant along the length of a beam, d4v - dF dz4 dz (13.7) -El- - - = -w As an example of the use of equation (13.4), consider the case of. the radius of curvature R in terms of v, is only true if the displacement v is small. Figure 13.5 Distortion of a beam in pure bending. Elastic bending of straight beams 299