Chapter 8 BAXAL BENDN 8.1 DEFNTON A cross section is subjected to biaial (oblique) bending if the normal (direct) stresses from section are reduced to two bending moments and. enerall oblique bending is accompanied b oblique shearing, when the shear stresses τ are reduced to two shear forces V and V. f the shear forces V = V = 0, we discuss about pure oblique bending. The positive convention of the vectors moment and is presented in Fig. 8.1. From figure we ma conclude that the vectors moments and, which are included in the cross section plan, are positive if acts inversel principal ais, while acts in the positive direction of principal ais. n other words and are positive, when the resultant moment i stretches (tensions) the cross section first quadrant. >0 i >0 Fig.8.1 8.2 ODE OF LOADN There are 2 main modes of loading which produce biaial bending: a. The loads are applied perpendicular to the torsion ais C ~, in two plans which are parallel to the principal plans and.
We assume that for a certain cross section the sstems of ais of the significant centers and C are shown in Fig. 8.2. f.l.(p) T f.l.(p) P1 T C P2 P1 q P2 q Fig. 8.2 n.a.(p)! The forces lines must pass through the shear center C in order to have onl oblique bending with shearing. Otherwise this compound solicitation is accompanied b torsion in that cross section. From Fig. 8.2 we ma conclude: a.1 The vertical forces P will produce the shear force V acting in the shear center C, and the bending moment acting in the centroid. The forces plan is ~ C ~ and the forces line in an cross section is C ~ ais. The neutral ais for bending is ais. a.2 The vertical forces P will produce the shear force V acting in the shear center C, and the bending moment acting in the centroid. The forces plan is ~ C ~ and the forces line in an cross section is C ~ ais. The neutral ais for bending is ais. b. All forces act in a plan that passes through the shear center C (which contains ~ ais), but it is inclined with respect to the principal plans and with an angle (Fig. 8.3).
f.l.(p) P2 q P1 C T T T n.a.(p) Fig. 8.3 The forces plan is inclined with the angle, and the intersection of this plan with an cross section defines the forces line which is also inclined with the angle with respect to C ~ ais. Due to the forces acting in this inclined plan, shear forces V and bending moments will subject an cross section. This case of loading ma be reduced to the first one ( a. ), decomposing from the beginning the forces P with respect to and ais: P = P cos and P = Psin As in the first case of loading, P will produce the shear force V and the bending moment, while P will produce the shear force V and the bending moment. n the first case of loading the ratio isn t constant along the bar, so that the deformed ais will be a skew curve in space. n the second case of loading the ratio is constant in an cross section and the deformed ais will be a plane curve, but situated in a different plan with respect to the forces plan. 8.3 THE NORAL STRESSES N OBLQUE BENDN We consider a simple supported beam subjected to pure oblique bending, on the interval CD from this beam (Fig. 8.4). The forces P, inclined with an
angle with respect to ais, produce between C and D sections onl a bending moment (without shear force), having a constant distribution on C-D interval. For simplicit we assume the cross section is smmetrical double, so C and the sstems ~ ~ ~, are identicall. f.l. A P P P P P D P C a b a B Fig.8.4 n cross section the plan of action the bending moment is A-A (Fig. 8.5), inclined with the angle with respect to ais. Also, the same angle is between the vector moment and ais ( and are principal ais). f.l. A β =C (n.a.) A (n.a.) Fig.8.5
The components of the bending moment, with respect to and ais, are: = cos and = sin (8.1) The produce bending in plans, respectivel. Each moment will generate normal stresses, calculated with Navier s formula: = and = Taking into account the hpothesis used in echanics of aterials, the hpothesis of the small deformations and the material has a linear elastic behaviour, the normal stresses ma be calculated superposing the effects of the two straight bending (from and ): = +, or: = + (8.2) The positive moments and are those which produce tensional stresses in an point from the first quadrant of the principal sstem of ais (positive and coordinates). Replacing and from (8.1) in the relation of (8.2), we get: cos sin = + (8.3) The neutral ais equation is obtained from the condition = 0. As, in (8.3) the bending moment 0, the single solution is: cos Replacing: sin + = 0 sin = cos =, or: = tg (8.4) tg, in (8.4), we obtain another form for the neutral ais equation: = (8.5) Equations (8.4) and (8.5) represent the equation of the neutral ais in case of oblique bending, which can be written shorter: = m (8.6) where m: represent the neutral ais slope m = = tgβ (8.7) Replacing (8.7) in equation (8.6), we obtain: = tgβ (8.8)
This final equation (8.8) shows that the neutral ais is a straight line passing through the centroid (0,0) and being inclined with the angle β with respect to ais.! Note that in case of oblique bending, the neutral ais does no more coincide with the support of the moment vector ( doesn t act around the neutral ais, because tgβ tg ). Eception to this observation, are the cross sections having the moments of inertia = tgβ = tg (as for the square or circular cross sections).! As β, the neutral ais is no more perpendicular to the forces line. To trace the neutral ais we either determine the angle β ( β = arctg ), or we define the second point Q (figure 8.6) through which the neutral ais will pass (the first point is the centroid ), of abscissa 1 and ordinate m: Q(1,-m). 2 Ac Q β n.a. At 2 1 >0 >0 1 Fig.8.6! Note that the neutral ais passes alwas through the quadrant limited b the two vectors moment and. After tracing the neutral ais, two parallels to the neutral ais which are tangent to the cross section contour, are also traced. These two parallels (fig.8.6) will pass through the etreme points 1( 1, 1 ) and 2( 2, 2 ). The normal (direct) stress diagram is drawn considering the reference line (which is even the cross section) perpendicular to the neutral ais and having etreme values, maimum tensile stress and maimum compressive stress, in point 1, respectivel 2 (fig.8.6):
= 1 1 + 1 > = 2 2 + 2 < 0 0 As in practical calculations we have to check the strength capacit of the most subjected cross section, meaning that the strength condition ma R must be verified, we are interested mainl in computing the maimum normal stress (tension or compression, in case of steel cross sections): ma ma = ma + ma R (8.9) ma This means that we shall work in the critical (most dangerous) sections, where and have maimum values. f and aren t maimum in the same section we have to consider 2 critical sections: Section 1: ma. and afferent Section 2: afferent and ma. The strength condition will be: ma (8.10) ( ) R = 1 2 ma, n case of sections that ma be inscribed in a rectangle (Fig.8.7) and if there is material in the corners of rectangle (e: rectangle, double T() shaped cross sections, or an other composed section which form a rectangle) we ma determine that normal stresses without determining first the neutral ais position. The maimum stresses will be, no doubt, in the cross section corners: = ± ± ( - ) ( + ) Fig.8.7