V. DEMENKO MECHNCS OF MTERLS 015 1 LECTURE 13 Strength of a Bar in Pure Bending Bending is a tpe of loading under which bending moments and also shear forces occur at cross sections of a rod. f the bending moment is the onl force factor acting in the section while the shearing force is absent, bending is called pure. n most cases, however, shearing forces occur as well as bending moments at cross sections of a rod. n this case bending is called transverse. The eamples of structural elements subjected to plane bending are shown in Figs 1 8. Fig. 1 bridge with movable girder Fig. Beam of wide-flanged shape loaded b distributed load (b) (c) (a) (d) Fig. 3 Non-prismatic beams in bending: a) street lamp, b) bridge with tapered girders and pears, c) wheel strut of a small airplane, d) wrench handle Fig. 4 tall signboard supported b two vertical beams consisting of thin-walled, tapered circular tubes
V. DEMENKO MECHNCS OF MTERLS 015 Fig. 5 Vertical solid wood and aluminum posts support a lateral load P Fig. 6 pontoon bridge consisting of two longitudinal wood beams (balks) that span between adjacent pontoons and support the transverse floor beams (cheeses) Fig. 7 Plane bending of a beam BCD developed b the force P applied to the cable Fig. 8 Simpl supported wood beam Pure Bending Let us consider an infinitesimall small element of a beam d (see Fig. 9). (1) The hpothesis of plane sections holds true for bending as well as for tension and torsion. Section B, which was a plane before bending, has remained such after bending, but turned through a certain small angle dθ. () The plane B continues to be perpendicular to the eternal surfaces of the beam. Therefore we have not the shearing stresses t.
V. DEMENKO MECHNCS OF MTERLS 015 3 (3) The upper fibers of the beam became stretched, i.e. increased in length from ab to a'b', whereas the lower fibers became contracted from ef to e'f'. However the fibers don t press each other. Therefore we haven't the normal stresses in perpendicular direction to the ais. (4) There should be the line cd between the upper and lower fibers, which does not change its length on bending and is called the neutral laer. Fig. 9 Note: ll the laers of the beam which are parallel to the neutral laer, are stretched or contracted and it is true the following relationship n Fig. 9 the radius of the neutral laer is denoted b. σ = Eε. (1) 1 Determination of Strains Let us consider the deformation of particular fiber (laer) ab of the beam, which is at a distance from the neutral laer cd. The relative elongation of the laer ab is ab ab ε =, () ab
4 V. DEMENKO MECHNCS OF MTERLS 015 where ab= d= dθ = cd = cd ' ', ab ' ' = ( + d ) θ, whence ε dθ + dθ dθ dθ ( ) = = = 1 k. (3) Note: the angle dθ is small, the arch a'b' can be determined with a good accurac as ab ' ' = ( + d ) θ. Eq. 3 means, that the strains are distributed linearl between the laers. Determination of the Neutral is Position Due to tension-compression deformation of the fibers, the stresses can be found using Hooke's law: whence it follows after substitution: ( ) = ( ) σ ε E, (4) σ E k. (5) ( ) = = Thus in pure bending the stresses in the cross section var according to a linear law with the proportionalit factor k. The locus of points in the section which satisfies the condition σ = 0 is called the neutral ais of the section: Fig. 10
V. DEMENKO MECHNCS OF MTERLS 015 5 Let us determine the position of the neutral ais. t ma be recalled that the sum of the projections of all forces in a cross section onto the ais is equal to ero, since there are no normal forces in the bending of the beam. The elementar normal force acting on an elementar area dn = σ d= E d. (6) Summing over the entire area, we get: N E = d =0. (7) Noting that the constant E is other than ero, it follows: d = 0. (8) This integral represents the static moment (first moment) of the section with respect to the neutral ais. Since the static moment is ero, the neutral ais passes through the centroid of a section (see Fig. 10). Otherwise M = σ d =0, (9) M E E E = d =0, d = 0, = const, = 0= = 0 d. Hence the and aes are the principal aes of the section. 3 Determination of Stresses For numerical determination of stresses, it is essential to find the radius of curvature of the neutral laer of deflected beam. Elementar moment relative to the neutral ais: dm = σ d.
6 V. DEMENKO MECHNCS OF MTERLS 015 Summing the elementar moments over the cross-sectional area and substituting σ = E, we obtain: M d d d d E E E = σ = = = = =, whence we find the curvature of deflected ais of beam: 1 M = E. (10) Substituting the epression of curvature into the formula for σ, we finall get σ M M M = E = E = ; σ =. (11) E The maimum bending stress occurs at the points most remote from the neutral ais: Mma σ =. (1) ma The quotient / is called the section(al) modulus in bending and is denoted b ma W : W =, [m 3 ]. (13) ma Thus ma = M σ. (14) W The diagram of bending stresses distribution along vertical ais is shown in Fig. 11:
V. DEMENKO MECHNCS OF MTERLS 015 7 Fig. 11 Formula (13) is fundamental in the analsis of rods subjected to bending. For a rod of rectangular section with sides b and h 3 bh =, ma = h, 1 bh W =. 6 For a circular section 4 πd = 64, ma = d, 3 d W = π. 3 4 Condition of Strength in Pure Bending Condition of strength is reall an inequalit in which maimum working stresses and allowable stresses for structural element material are compared: M ma σ = [ σ ]. (15) ma Considering the strength conditions, it is possible to solve three important engineering problems: (1) problem of checking the strength. For specified loads and geometrical dimensions of a cross section the maimum stress in the section (called the critical section) is determined using the formula W
8 V. DEMENKO MECHNCS OF MTERLS 015 σ ma M = W ma and compared with the allowable stress [ σ ]: σma [ σ ]; (16) () design problem. For the specified loads and allowable stresses the crosssectional area of a beam is determined b the formula: W = M ma [ σ ] ; (17) (3) problem of allowable load: [ M] = W[ σ ], (18) where [M] is the allowable load determined for the critical section of a beam. Eample 1 Design problem in pure plane bending For a cantilever beam select the following sections: a) circular section; b) rectangular section with h/ b = ; and c) -section, using [ σ ] = 100MPa. Bending moments diagram shows the constant value of internal bending moment Fig. 1 M =10kNm. To solve the problem, it is necessar to calculate the section modulus ma W. Using condition of strength (14) we obtain 3 6 6 W M /[ σ] = 10 10 /100 10 = 100 10 m 3. n the case (a) (round section) according to epression (14)
V. DEMENKO MECHNCS OF MTERLS 015 9 W 3 πd = = 100 10 d = 10 10 3 6 m, = 80 10 4 m, n the case (b) (rectangle) bh W = ; h/ b = ; 6 b ( b) 6 6 = 100 10 b= 5.3 10 m; h = 10.6 10 m; = 56. 10 4 m. n the case (c) W W = 90.3 10 6 = 118 10 6 m 3, No14, m 3, No14, = 18.9 10 4 m, = 1.5 10 4 m, Note, that > > (80> 56.> 1.5) Fig. 13 Which of those sections is more advantageous from the viewpoint of strength-toweight efficienc? Evidentl, the -section is the most efficient shape of cross section. For a beam to be efficient, most of the beam material should obviousl be put as far as possible from the neutral ais. Remember, that for an section, the normal bending stress in a cross section of a beam is directl proportional to the distance from the neutral ais of the beam.