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CHAPTER 16 CURVED BEAMS AND RINGS Joseph E. Shigley Professor Emeritus The University of Michigan Ann Arbor, Michigan 16.1 BENDING IN THE PLANE OF CURVATURE / 16.2 16.2 CASTIGLIANO'S THEOREM / 16.2 16.3 RING SEGMENTS WITH ONE SUPPORT / 16.3 16.4 RINGS WITH SIMPLE SUPPORTS /16.10 16.5 RING SEGMENTS WITH FIXED ENDS / 16.15 REFERENCES/16.22 NOTATION A Area, or a constant B Constant C Constant E Modulus of elasticity e Eccentricity F Force G Modulus of rigidity / Second moment of area (Table 48.1) K Shape constant (Table 49.1), or second polar moment of area M Bending moment P Reduced load Q Fictitious force R Force reaction r Ring radius r Centroidal ring radius T Torsional moment U Strain energy V Shear force W Resultant of a distributed load w Unit distributed load X Constant Y Constant y Deflection Z Constant y Load angle $ Span angle, or slope a Normal stress 6 Angular coordinate or displacement Methods of computing the stresses in curved beams for a variety of cross sections are included in this chapter. Rings and ring segments loaded normal to the plane of the ring are analyzed for a variety of loads and span angles, and formulas are given for bending moment, torsional moment, and deflection. 16.1 BENDINGINTHEPLANEOFCURVATURE The distribution of stress in a curved member subjected to a bending moment in the plane of curvature is hyperbolic ([16.1], [16.2]) and is given by the equation My G= A , r (16.1) Ae(r -e-y) ^ ' where r = radius to centroidal axis y = distance from neutral axis e = shift in neutral axis due to curvature (as noted in Table 16.1) The moment M is computed about the centroidal axis, not the neutral axis. The maximum stresses, which occur on the extreme fibers, may be computed using the formulas of Table 16.1. In most cases, the bending moment is due to forces acting to one side of the sec- tion. In such cases, be sure to add the resulting axial stress to the maximum stresses obtained using Table 16.1. 76.2 CASTIGLIANO'S THEOREM A complex structure loaded by any combination of forces, moments, and torques can be analyzed for deflections by using the elastic energy stored in the various compo- nents of the structure [16.1]. The method consists of finding the total strain energy stored in the system by all the various loads. Then the displacement corresponding to a particular force is obtained by taking the partial derivative of the total energy with respect to that force. This procedure is called Castigliano's theorem. General expressions may be written as at/ w w f . yi= W t ^=W 1 ^=Wt (16 ' 2) where U = strain energy stored in structure y t = displacement of point of application of force F 1 - in the direction of F/ 6/ = angular displacement at T 1 tyi = slope or angular displacement at moment M 1 If a displacement is desired at a point on the structure where no force or moment exists, then a fictitious force or moment is placed there. When the expression for the corresponding displacement is developed, the fictitious force or moment is equated to zero, and the remaining terms give the deflection at the point where the fictitious load had been placed. Castigliano's method can also be used to find the reactions in indeterminate structures. The procedure is simply to substitute the unknown reaction in Eq. (16.2) and use zero for the corresponding deflection. The resulting expression then yields the value of the unknown reaction. It is important to remember that the displacement-force relation must be linear. Otherwise, the theorem is not valid. Table 16.2 summarizes strain-energy relations. 16.3 RINGSEGMENTSWITHONESUPPORT Figure 16.1 shows a cantilevered ring segment fixed at C. The force F causes bend- ing, torsion, and direct shear. The moments and torques at the fixed end C and at any section B are shown in Table 16.3. The shear at C is R c = F. Stresses in the ring can be computed using the formulas of Chap. 49. To obtain the deflection at end A, we use Castigliano's theorem. Neglecting direct shear and noting from Fig. 16.16 that / = r d0, we determine the strain energy from Table 16.2 to be pAPrd* , f*r 2 r<*9 ^l "2ET + J 0 "^F (163) Then the deflection y at A and in the direction of F is computed from >-f-;K«>*<sf№ The terms for this relation are shown in Table 16.3. It is convenient to arrange the solution in the form Fr 3 (A B \ ,_ y = -T (EI + GK) (16 ' 5) where the coefficients A and B are related only to the span angle. These are listed in Table 16.3. Figure 16.20 shows another cantilevered ring segment, loaded now by a dis- tributed load. The resultant load is W = wrfy a shear reaction R = W acts upward at the fixed end C, in addition to the moment and torque reactions shown in Table 16.3. A force W= wrQ acts at the centroid of segment AB in Fig. 16.26. The centroidal radius is ?= 2rsini6/21 (166) 6 TABLE 16.1 Eccentricities and Stress Factors for Curved Beams 1 1. Rectangle 2. Solid round 3. Hollow round 4. Hollow rectangle TABLE 16.1 Eccentricities and Stress Factors for Curved Beams 1 (Continued) 5. Trapezoid 6. T Section fNotation: r » radius of curvature to centroidal axis of section; A » area; / « second moment of area; e « distance from centroidal axis to neutral axis; <r/ - Kp and <T O - Kjr where <r/ and a 0 are the normal stresses on the fibers having the smallest and largest radii of curvature, respectively, and a are the corresponding stresses computed on the same fibers of a straight beam. (Formulas for A and / can be found in Table 48.1.) 7. U Section TABLE 16.2 Strain Energy Formulas Loading Formula 1. Axial force F F*l 2AE 2. Shear force F rj F*l U = 2AG 3. Bending moment M rj m f M 2 <** J 2EI 4. Torsional moment T 1*1 2GK To determine the deflection of end A, we employ a fictitious force Q acting down at end A. Then the deflection is at/ r f* ,, 9M . 0 r f* 3T , D „, _. y = de = ^i M 3c " e+ ^i r ae " e (16J) The components of the moment and torque due to Q can be obtained by substitut- ing Q for Fin the moment and torque equations in Table 16.3 for an end load F; then the total of the moments and torques is obtained by adding this result to the equa- tions for M and T due only to the distributed load. When the terms in Eq. (16.7) have FIGURE 16.1 (a) Ring segment of span angle (J) loaded by force F normal to the plane of the ring. (b) View of portion of ring AB showing positive directions of the moment and torque for section at B. TABLE 16.3 Formulas for Ring Segments with One Support Loading Term Formula End load F Moment M-Fr sin B M c = Fr sin <f> Torque T = Fr(I - cos B) T c * Fr(I - cos </>) dM ST Derivatives — = r sin B — = K1 — cos 6) or dr Deflection A = 0 — sin <f> cos 0 coefficients B = 30 — 4 sin 0 + sin 0 cos 0 Distributed load w; fictitious load Q Moment M = Wr 2 O - cos 0) A/ c - wr\\ - cos 0) Torque T = wrfy - sin 6) T c - wr 2 ^ - sin 0) ^M ar Denvatives —7: = r sin B —— = r( 1 — cos B) d(2 oQ Deflection A = 2 - 2 cos 4> - sin 2 0 coefficients B = ^ 2 — 20 sin 0 + sin 2 0 been formed, the force Q can be placed equal to zero prior to integration. The deflection equation can then be expressed as wr 4 (A B \ „,<>, y= -T U + OT) (16 ' 8) FIGURE 16.2 (a) Ring segment of span angle (J) loaded by a uniformly distributed load w acting normal to the plane of the ring segment; (b) view of portion of ring AB; force W is the resultant of the distributed load w acting on portion AB of ring, and it acts at the centroid. 16.4 RINGSWITHSIMPLESUPPORTS Consider a ring loaded by any set of forces F and supported by reactions R, all nor- mal to the ring plane, such that the force system is statically determinate. The system shown in Fig. 16.3, consisting of five forces and three reactions, is statically determi- nate and is such a system. By choosing an origin at any point A on the ring, all forces and reactions can be located by the angles $ measured counterclockwise from A. By treating the reactions as negative forces, Den Hartog [16.3], pp. 319-323, describes a simple method of determining the shear force, the bending moment, and the tor- sional moment at any point on the ring. The method is called Biezeno's theorem. A term called the reduced load P is defined for this method. The reduced load is obtained by multiplying the actual load, plus or minus, by the fraction of the circle corresponding to its location from A. Thus for a force F h the reduced load is P^F 1 (16-9) Then Biezeno's theorem states that the shear force V A , the moment M 4 , and the torque T A at section A, all statically indeterminate, are found from the set of equations V^=SP, n M A = ^ P 1 Y sin to (16.10) n T A = ^ P 1 T(I - COS to) n where n = number of forces and reactions together. The proof uses Castigliano's the- orem and may be found in Ref. [16.3]. FIGURE 16.3 Ring loaded by a series of concentrated forces.