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CHAPTER 16CURVED BE MS ND RNGS

Joseph E ShigleyProfessor Emeritus

The University of MichiganAnn Arbor, Michigan

16.1 BENDING IN THE PLANE OF CURVATURE / 16.216.2 CAS TIGLIANO S THEOREM / 16.2

16.3 RING SEGMENTS WITH ONE SUPPORT / 16.316.4 RINGS WITH SIMPLE SUPPORTS / 1 6 . 1 016.5 RING SEG MENTS W ITH FIXED ENDS / 16.15REFERENCES/16.22

NOTATION

A Area, or a constantB ConstantC ConstantE Modulus of elasticitye EccentricityF Force

G Modulus of rigidity/ Second m om ent of area (Table 48.1)K Shape constant (Table 49.1), or second polar moment of areaM Bending mom entP Reduced loadQ Fictitious forceR Force reac tionr Ring radiusr Centroidal ring radiusT Torsional m om entU Strain energyV Shear forceW Resultant of a distributed loadw Un it distributed loadX Constant

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Y Constanty DeflectionZ Constanty Load angle

$ Span angle, or slopea No rm al stress6 Ang ular coordinate or displacement

Methods of computing the stresses in curved beams for a variety of cross sectionsare included in this chapter. Rings and ring segm ents loaded norm al to the plane ofthe ring are analyzed for a variety of loads and span angles, and formulas a re givenfor bending moment, torsional moment, and deflection.

16.1 BENDINGINTHEPL NEOFCURV TURE

The distribution of stress in a curved mem ber subjected to a bending mom ent in theplane of curvature is hyperbolic ([16.1], [16.2]) and is given by the equation

MyG= A , r (16.1)Ae(r -e-y)

where r = radius to centroidal axisy = distance from neu tral axise = shift in neu tral axis due to curvature (as noted in Table 16.1)

The moment M is computed about the centroidal axis, not the neutra l axis. Themaximum stresses, which occur on the extreme fibers, may be computed using theformulas of Table 16.1.

In most cases, the b ending mom ent 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 stressesobtained using Table 16.1.

76.2 CASTIGLIANO S THEOREM

A complex structure loaded by any combination of forces, moments, and torques canbe analyzed fo r deflections by using the elastic energy stored in the various compo-nents of the structure [16.1]. The m ethod consists of finding the total strain energy

stored in the system by all the various loads. Then the displacemen t correspondingto a particular force is obtained by taking the partial derivative o f the total energywith respect to that force. This procedure is called Castigliano s theorem. Generalexpressions may be w ritten as

a t / w w f .y i=Wt ^=W1 ^=Wt (16'2)

where U = strain energy stored in structureyt = displacement of point of application of force F1- in the direction of F/

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6 / = angular displacement at T1ty i = slope or angular displacement at moment M1

If a displacement is desired at a point on the structure where no force or momentexists, then a fictitious force or moment is placed there. When the expression for the

corresponding displacement is developed, the fictitious force or moment is equatedto zero, and the remaining terms give the deflection at the point where the fictitiousload had been placed.

Castigliano's method can also be used to find the reactions in indeterminatestructures. The procedure is simply to substitute the unknown reaction in Eq. (16.2)and use zero for the corresponding deflection. The resulting expression then yieldsthe 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.

6 3 RINGS GM NTSWITHON SUPPORT

Figure 16.1 shows a cantilevered ring segment fixed at C. The force F causes bending, torsion, and direct shear. The moments and torques at the fixed end C and at anysection 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. Neglectingdirect shear and noting from Fig. 16.16 that / = r d0, we determine the strain energyfrom Table 16.2 to be

pAPrd* f * r2 r < * 9l 2 E T+J0 F

(163)

Then the deflection y at A and in the direction of F is computed from

> f ;K>*

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TABLE 16.1 Eccentricities and Stress Factors fo r Curved Beams 1

1 Rectangle

2. Solid round

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3. Hollow round

4. Hollow rectangle

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TABLE 16.1 Eccentricities an d Stress Factors fo r Curved Beams1 (Continued

5. Trapezoid

6. T Section

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fNotation: r radius of curvature to centroidal axis of section; A area; / sdistance from centroidal axis to neutral axis; < r / - Kp and TO - Kjr where < r / and a 0 are tfibers having the smallest and largest radii of curvature, respectively, and a are the correon the same fibers of a straight beam. (Formulas for A and / can be found in Table 48.1.)

7. U Section

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TABLE 16.2 Strain Energy Formulas

Loading Formula

1. Axial force F F * l2AE

2. Shear force F r j F*lU = 2AG

3. Bending moment M r j m f M2 Torque T = F r I cos B T c * Fr I - cos < / >

dM STDerivatives = r sin B = K 1 cos 6)

or dr

Deflection A = 0 sin f> cos 0coefficients B = 30 4 sin 0 + sin 0 cos 0

Distributed load w;fictitious load Q Moment M = W r2O - cos 0) A/ c - wr\\ cos 0)

Torque T = wr fy - sin 6) T c wr2^ sin 0)

^M arDenvatives 7: = r sin B = r 1 cos B)

d 2 oQ

Deflection A = 2 2 cos > sin 2 0coefficients B = ^ 2 20 sin 0 + sin2 0

been formed, the force Q can be placed equal to zero prior to integration. Thedeflection equation can then be expressed as

wr4 (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

norm al to the plane of the ring segment; (b)

view of portion of ring AB;

force W

is the resu ltant of thedistributed load w acting on portion AB of ring, and it acts at the centroid.

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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 determ inate. The systemshown 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 forcesand reactions can be located by the angles $ measured counterclockwise from A. Bytreating the reactions as negative forces, Den Hartog [16.3], pp. 319-323, describes asimple method of determining the shear force, the bending moment, and the tor-sional moment a t any point on the ring. The m ethod is called Biezeno s theorem .

A term called the reduced load P is defined for this method. The reduced load isobtained by multiplying the actual load, plus or minus, by the fraction of the circlecorresponding to its location from A. Thus for a force Fh 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 thetorque TA at section A, all statically indeterminate, are found from the set of equations

V ^ = S P ,n

MA = ^ P1Y sin to (16.10)n

T A = P 1T I - CO S to)n

where n = number of forces and reactions toge ther. 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.

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Example 1. Find the shear force, bending moment, and torsional moment at thelocation of .R3 for the ring shown in Fig. 16.4.

Solution. Using the principles of statics, we first find the reactions to be

R 1 = R 2 = R 3 = F

Choosing point A at R 3 , the reduced loads are

p --$r R -o P-ljf-o.o

H= l t

P, = ||F=0.5833F

' -I *-- -^

Then, using Eq. (16.10), we find V A = O. Next,

MA = ^ Fjrsin^5

- Fr ( O + 0.0833 sin 30 - 0.2222 sin 120 + 0.5833 sin 210

_ 0.4444 sin 240)

= -0.0576Fr

In a similar manner, we find TA = 0.991F r.

FIGURE 16.4 Ring loaded by the two forces F and sup-ported by reactions RI, R 2 , and R 3 . The crosses indicatethat the forces act downward; the heavy dots at the reac-tions R indicate an upward direction.

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The task of finding the deflection at any point on a ring with a loading like that ofFig. 16.3 is indeed difficult. The problem can be set up using Eq. (16.2), but the result-ing integrals will be lengthy. The chances of making an error in signs or in terms dur-ing any of the simplification processes are very great. If a computer or even aprogrammable calculator is available, the integration can be performed using a

numerical procedure such as Simpson s rule (see Chap. 4) . Most of the user s manu-als for programmable calculators contain such programs in the master library. Whenthis approach is taken, the two terms behind each integral should not be multipliedout or simplified; reserve these tasks for the computer.

16.4.1 A R ing with ymmetrical Loads

A ring having three equally spaced loads, all equal in magnitude, with three equally

spaced supports located midway between each pair of loads, has reactions at eachsupport of R = F/2, M = 0.289Fr, and T = O by Biezeno s theorem. To find the momentand torque at any location 0 from a reaction, we construct the diagram shown in Fig.16.5. Then the moment and torque at A are

M = M1 cos 9 - R1T sin 0(16.11)

- Fr (0.289 cos 0 - 0.5 sin 0)

T= M1 sin 0 - R1T (I - cos 0)(16.12)

= Fr (0.289 sin 0 - 0.5 + 0.5 cos 0)

FIGURE 16 5 The positive directions of the moment and torque axes are arbi-trary. Note that 1 = F /2 and M 1 = 0.289Fr.

M O M E N T A X I S

T O R Q U E A X I S

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Neglecting direct shear, the strain energy stored in the ring between any two supports is, from Table 16.2,

>Castigliano's theorem states that the deflection at the load F is

f = i / > f ^ r ^ f * wFrom Eqs. (16.11) and (16.12), we find

^ = r(0.289cos0 0.5sin0)orr

^ = r(0.289 sin 0 0.5 + 0.5 cos 0)or

When these are substituted into Eq. (16.14), we get

Fr3 A B \ / , r i c \y = Y EJ + GK) 1615)

which is the same as Eq. (16.5). The constants are 7E 3

A = 4 J (0.289 cos 0 0.5 sin 0) 2 dQ

(16.16)cn/3

V B = 4 (0.289 sin 0 0.5 + 0.5 cos 0) 2 d0

These equation s can be integra ted directly or by a com puter using Simpson's rule. If

your in tegration is rusty, use the com puter. The results are A = 0.1208 and B = 0.0134.

16 4 2 Distributed Load ing

The ring segment in Fig. 16.6 is subjected to a distributed load w per unit circumference and is supported by the vertical reactions 1 and R 2 and the moment reactionsM I and M 2. The zero torque reactions mean tha t the ring is free to turn at A and B.The resultant of the distributed load is W = wr; it acts at the centroid:

TSiTlMi

l

By symmetry, the force reactions are R I = R 2 = W 2 w r 2. Summing momentsabout an axis through BO gives

IM(BO) = M2 + Wr sin M 1 cos (n < |> ^ sin $ = O

Since M1 and M 2 a re equal, this equation can be solved to give

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FIGURE 16 6 Section of ring of span angle $ with distributed load.

M1 = WtI1 - **-*** 2 )**+] (16-18)

L 1 - COS (J ) J

Example 2. A ring has a uniformly distributed load and is supported by threeequally spaced reactions. Find the deflection midway between supports.

Solution. If we place a load Q midway between supports and compute thestrain energy using half the span, Eq. (16.7) becomes

W 2r (2

, _ 3M , _ 2r [2

3T ,_ imy = =TiI M-* QdQ+GKl TtQdQ (1619)Using Eq. (16.18) with ty = 271/3 gives the moment at a support due only to w to be M I= 0.395 w r2. Then, using a procedure quite similar to that used to write Eqs. (16.11)and (16.12), we find the moment and torque due only to the distributed load at anysection 0 to be

Mw = wr2 M - 0.605 cos 0 - sin 0 J

(16.20)Tw = wr

21G - 0.605 sin 0 - - + ^ cos 0 j

In a similar manner, the force Q results in additional components of

MQ = Q j- (0.866 cos 0 - sin 0)

(16.21)

TQ = - (0.866 sin 0 - 1 + cos 0)

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FIGURE 16.7 (a) Ring segment of span angle (|)

loaded by force E (b) Portion of ring used to com-pute moment and torque at position 6.

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Now multiply Eq. (16.26) by EI and divide by r; then substitute. The result can bewritten in the form

(T1 sin 9 + MI cos 9 - R1T sin 9) cos 9 dQ

+ Fr J (cos y sin 0 - sin y cos 9) cos 9 d QY

E I f r *+ ] [-7\ cos 0 + MI sin 9 - R^(I - cos 0)] sin 0 dQ

GK l/o

f 1- Fr (cos Y cos 0 + sin Y sin 0 - 1) sin 0 dQ [ = O (16.27)

T

Similar equations can be written using the other two relations in Eq. (16.2). Whenthese three relations are integrated, the results can be expressed in the form

a n i2 13 T1IFr ^ 1

2 1 a22 23 M1IFr = b2 (16.28)

a31 a32 033 J [R1IF J \b3_

where

11 = sin2 (J)-- sin2 ^ ) (16.29)LrK

2 1 = ( < > - sin < |> cos (|>) + ($ + sin ty cos < |>) (16.30)GK

EI31 = (0 - sin (() cos $) + ((() + sin cos < |) - 2 sin (()) (16.31)

GK

EI 1 2 = (< > + sin (|) cos (|>) + ((]) - sin (|) cos (|>) (16.32)

GA.

22 = n (16.33)

~ F I 3 2 = sin2 0 + - - [2(1 - cos < |> ) - sin2 < | ) ] (16.34)

GK

13 = ~32 16.35)

23 = -31 16.36)

EI 3 3 = - (< |> - sin (|) cos < |>) - (3(|) - 4 sin $ + sin ty cos (|>) (16.37)

GA

EIb 1 = sin Y sin ty cos > - cos Y sin2 $ + (0 - Y ) sin Y + 7 77 [cos Y sin2 tyGK

- sin Y sin fy cos < |) + ( < | ) - Y)sin Y + 2 cos ty - 2 cos Y ] (16.38)

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b 2 = (Y-(^) cos Y ~ sin Y + cos Ysin < |> cos < |> + sin Ysin 2 0

7+ [(Y - ( |>) cos Y - sin Y + 2 sin ty - cos Y sin (|) cos ty - sin Y sin 2 c|>] (16.39)

GA

3 = (y- ( |>) cos Y- sin Y + cos Ysin ty cos (|) + sin Y s in2 ty

+ [(Y- ( |>) cos Y ~ sin Y- cos Ysin ty cos ty - sin Ysin 2 < |)GK

+ 2(sin (|) - (|) + Y + cos Y sin (|) - sin Y cos $)] (16.40)

For tabulation purposes, we indicate these relations in the form

F T F Ta 'i = Xi i + Hk Yii bk = Xk+ GK Yk (16 41)

Programs for solving equations such as Eq. (16.28) are widely available and easyto use. Tables 16.4 and 16.5 list the values of the coefficients for a variety of span andload angles.

TABLE 16.4 Coefficients a tj fo r Various Span Angles

Span angle < />

Coefficients 3r/2 * 3w/4 2* /3 w/2 ir/3 * /4

an X11 1 O 0.5 0.75 0.75 0.5Y1I -1 O -0.5 -0.75 - -0.75 -0.5

a2i X2i 4.7124 T T 2.8562 2.5274 .5708 0.6142 0.2854F2i 4.7124 T T 1.8562 1.6614 .5708 1.4802 1.2854

A31 JT3 1 4.7124 T T 2.8562 2.5274 .5708 0.6142 0.2854Y31 6.7124 T 0.4420 -0.0707 -0.4292 -0.2518 -0.1288

a n X12 4.7124 T 1.8562 1.6614 .5708 1.4802 1.2854Y12 4.7124 T 2.8562 2.5274 .5708 0.6142 0.2854

a22 X22 I O 0.5 0.75 0.75 0.5Y22 -1 O -0.5 -0.75 - -0.75 -0.5

a 32 X32 1 O 0.5 0.75 0.75 0.5K32 1 4 2.9142 2.25 0.25 0.0858

a n X13 -1 O -0.5 -0.75 - -0.75 -0.5Y13 -I -4 -2.9142 -2.25 - -0.25 -0.0858

23 X23 -4.7124 - -2.8562 -2.5274 -1.5708 -0.6142 -0.2854X23 -6.7124 -TT -0.4420 0.0707 0.4292 0.2518 0.1288

a 33 X33 -4.7124 -TT -2.8562 -2.5274 -1.5708 -0.6142 -0.2854K33 -18.1372 -ST -3.7402 -2.3861 -0.7124 -0.1105 -0.0277

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TABLE 16.5 Coefficients b k for Various Span Angles (J) and Load Angles y in Terms of ty

Coefficients, Span angle 0load angles I I I I I

7 37T/2 T T 37T/4 27T/3 7T/2 7T/3 7T/4

, X1 2.8826 1.6661 0.2883 -0.0806 -0.4730 -0.4091 -0.27800 > Y1 2.8826 -1.7481 -1.4019 1.0806 -0.4730 -0.1162 -0.0396

- h X2 -1.3525 -2.3732 -2.1628 -1.8603 -1.0884 -0.4051 -0.18494 2 Y2 -5.2003 -2.3732 -0.4727 -0.1283 0.1462 0.1022 0.0535

, X3 -1.3525 -2.3732 -2.1628 -1.8603 -1.0884 -0.4051 -0.1849^3 Y 3 -13.0342 -5.6714 -2.0455 -1.2699 -0.3622 -0.0544 -0.0135

h X1 3.1416 1.8138 0.4036 0.0446 -0.3424 -0.3179 -0.21801 r, 3.1416 -1.1862 -1.0106 -0.7817 -0.3424 -0.0839 -0.0286

^ , X2 O -1.9132 -1.8178 -1.5620 -0.9069 -0.3346 -0.15223 2 Y2 -4 -1.9132 -0.4036 -0.1307 0.0931 0.0706 0.0373

, X 3 O -1.9132 -1.8178 -1.5620 -0.9069 -0.3346 -0.1522b* Y3 -10.2832 -4.3700 -1.5452 -0.9536 -0.2692 -0.0401 -0.0099

h X1 2.3732 1.5708 0.4351 0.1569 -0.1517 -0.1712 -0.12031 Y1 2.3732 -0.4292 -0.4379 -0.3431 -0.1517 -0.0372 -0.0127

^ X2 1.6661 -1 -1.1041 -0.9566 -0.5554 -0.2034 -0.09222 b2 Y2 -1.7481 -1 -0.2311 -0.0906 0.0304 0.0286 0.0154

, X3 1.6661 -1 -1.1041 -0.9566 -0.5554 -0.2034 -0.0922* Y3 -5.0463 -2.1416 -0.7395 -0.4529 -0.1262 -0.0186 -0.0046

16.5.2 Deflection D ue to Concentrated Load

The deflection of a ring segment at a concentrated load can be obtained using thefirst relation of Eq. (16.2). The com plete analytical solution is quite lengthy, and so aresult is shown here that can be solved using computer solutions of Simpson'sapproximation. First, de fine the three solutions to Eq. (16.28) as

T1 = C1Fr M1 = C2Fr R1 = C3F (16.42)

Then Eq. (16.2) will have four integrals, which areo

AF = [(C1 - C3) sin G + C2 cos 6]2 d Q (16.43)Jo

BF = \ (cos y sin 0 - sin y cos 9)2 d Q (16.44)Jo

CF= \ [(C 3- C1) cos 0 + C2 sin 9 - C3]2 d Q (16.45)Jo

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FIGURE 16.9 A portion of the ring has been isolated here to determine the moment andtorque at any section D at angle 6 from the fixed end at A.

These equations are now employed in the same manner as in Sec. 16.5.1 to obtain

h H f s ^ i= te i i6-5 )21 cos < |) + - 2 sin (16.52)

Y I = ( j) - 2 sin < |) - -^ - sin2 < |) - sin cos ty + (l + cos (|)) (16.53)

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TABLE 16.6 Coefficients bk for Various Span Angles and Uniform Loading

Span angle < />

Coefficients 3ir/2 T T 3ir/4 2v/3 v /2 * /3 * /4

, X} 9.0686 3.1416 1.0310 0.7147 0.3562 0.1409 0.0675D{ Y1 9.0686 3.1416 1.5430 1.0572 0.35 62 0.0602 0.0156

, X2 10.1033 0.9348 0.4507 0.3967 0.2337 0.0716 0.0263 2 Y2 8.1033 0.9348 -1.7274 -2.0102 -1.7663 -0.9750 -0.5810

X2 = --2(1- cos ( |>) - -| sin < |> cos (|> + sin2 < |> (16.54)

Y2 = - - - 2(1 - cos ( |>) + - sin ( |> cos < |> - sin2 < |> + < |> sin < |> (16.55)

Solutions to these equations for a variety of span angles are given in Table 16.6.A solution for the deflection at any point can be obtained using a fictitious load

Q at any point and proceeding in a manner similar to other developments in thischapter. It is, however, a very lengthy analysis.

REFERENCES

16.1 Raym ond J. Roark and Warren C. Y oung, Fo rmulas for Stress and Strain, 6th ed.,McGraw-Hill, New Y ork, 1984.

16.2 Joseph E. Shigley and Charles R. Mischke, Mechanical Engineering Design, 5 th ed.,McGraw-Hill, New Y ork, 1989.

16.3 J. P. Den Hartog, Advan ced Strength of Materials, McGraw-Hill, New Y ork, 1952.