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Analysis of flexible arches Sled, John James 1959

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ANALYSIS OF FLEXIBLE ARCHES by JOHN JAMES SLED B.Sc. ( C i v i l Engineering), University of Saskatchewan, 1955 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF APPLIED SCIENCE in the Department of CIVIL ENGINEERING We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA September, 1959 i i ABSTRACT A method of analysis of flexible arches under the action of axial deformation, support movements, and fabrication errors by the deflection theory method is presented in this thesis. The elastic moments and dimensionless magnification factors for parabolic hingeless arches with rise-span ratios of 1/8, 1/6, 1/4 and 1/3 are given. Although the data is given for parabolic hingeless arches with a constant EI and one prescribed variation of EI B i t is shown, by numerical examples, that the tables may be used for other arches whose shapes do not di f f e r greatly from a parabola and, by interpolation, to other variations of EI. It is also shown that these solutions for hingeless arches may be used to obtain the solution of one and two hinged archeso It i s shown by theory and by numerical tests that the deflection theory moments are directly proportional to the magnitude of the axial deformation, support movement, or fabrication error. It i s also shown that these moments, when determined separately, may be added to each other and to moments due to load to obtain the correct total moment. The solutions in the tables were calculated by a numerical procedure of successive approximations. The electronic computer, the ALWAC III E ( at the University of British Columbia was used to perform the large amount of numerical work required. In presenting t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study. I f u r t h e r agree that permission f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s r e p r e s e n t a t i v e s . I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. The U n i v e r s i t y of B r i t i s h Columbia Vancouver 3, Canada. Department of 6-.^WfS A ^Uf^ i i i TABLE OF CONTENTS PAGE INTRODUCTION 1 CHAPTER I THE GOVERNING DIFFERENTIAL EQUATION 4 CHAPTER II PROGRAM FOR AUTOMATIC COMPUTATION 17 CHAPTER III NUMERICAL VERIFICATION OF THEORY 24 Part 1 Variation of Moment With Axial Deformation 25 Part 2 Variation of Moment With Vertical Support Movement 27 Part 3 Variation of Moment With Rotation of a Support 30 Part 4 Comparison of Uniform Axial Deformation, Fabrication Error Producing a Horizontal Gap at the Crown, and Horizontal Movements of Supports 32 Part 5 Comparison of a Fabrication Error Which Produces a Vertical Gap at the Crown With a Vertical Movement of a Support 34 Part 6 Superposition of Axial Deformation and Support Movements With Load 39 CHAPTER IV NUMERICAL DATA Part 1 Tables of Magnification Factors 46 Part 2 Deflection Theory Moment Curves 59 Part 3 Influence Lines for Elastic Thrust 76 CHAPTER V NUMERICAL EXAMPLES Part 1 Analysis of a Hingeless Arch 78 Part 2 Analysis of a Two Hinged Arch 85 iv ACKNOWLEDGEMENT The author wishes to express his sincere thanks to his adviser, Dr. R. F. Hooley, for his expert guidance and help. The contribution of his great knowledge and valuable time was greatly appreciated. The author also wishes to thank the staff of the Computing Centre at the University of British Columbia for their assistance in the use of the computer. This research was carried on during the summer of 1959 and was sponsored by the National Research Council of Canada. Their generous financial assistance was gratefully received. September, 1959 Vancouver, Br i t i s h Columbia V NOTATIONS (i F l e x i b i l i t y factor of the arch A Fabrication error of the arch (SH Horizontal deflection of a support 6y Vertical deflection of a support £ Horizontal component of deflection *l Vertical component of deflection 9 Angle which the tangent to arch axis makes with a horizontal line d© Change of 8 due to moment A Coefficient of axial deformation 0 Magnification factor U) Rotation of a support E Modulus of elas t i c i t y f Rise of arch H Horizontal thrust 1 Moment of inertia I a v Average moment of inertia £( ) Linear function L Span of arch MQ Deflection theory bending moment Mjr Elastic theory bending moment P Concentrated load w Intensity of load on arch Uniform loading on arch Vertical shear Horizontal coordinate of arch axis Vertical coordinate of arch axis 1 INTRODUCTION The elementary and most commonly used theory of arch analysis Is the elastic theory. This theory neglects the effect of changes in geometry on the stresses in the arch. The deflection theory takes these changes of geometry into account. In flexible arches the geometry changes, when combined with the axial thrust, produce moments of significant magnitude. The f l e x i b i l i t y of an arch is measured by the dimension-less ratio (i =J £j . The secondary moments due to deflection may usually be neglected in fixed arches with a value of ft less than 1. A load which produces a /3 of 7 is approaching the c r i t i c a l buckling load. The behaviour of an arch under load i s similar to that of a beam under a lateral load and an axial thrust. In both cases the deflections combined with the axial forces produce additional moments which in turn produce additional deflections and moments until the structure comes to equilibrium. In the case of a beam under load i t is possible to solve the differential equation and obtain an exact expression for the stresses in the beam*. In the case of the arch, however, due to the complexity of the differential equation a general solution i s not available in tabulated functions and numerical methods must be used. Lee and Pelton presented tables and graphs to fa c i l i t a t e the analysis of hingeless flexible arches under vertical 2 load. This work is to be extended to include the effects of axial de-formation and support movements. The results are presented as dimensionless magnification factors. A magnification factor i s the ratio of the deflection theory moment to the elastic theory moment. Tables for parabolic hingeless arches having rise-span ratios of 1/8, 1/6, 1/4, and 1/3 are given in Chapter IV. Values of the magnification factors 9 are given for /9 s 3 3, 13 •= 5, and in some cases f or ft *» 7. Graphs of the deflection theory moments are also included. The tables and graphs are given for a constant EI and one particular variation of EI. For other variations of EI interpolation is necessary. Theory and numerical examples are presented to show that the magnification factors are practically independent of the magnitude of the axial deformation or support movements. It w i l l be shown that these moments may be superimposed with the moments due to other loadings. The object of this thesis i s to present magnification factors, not deflection theory moments, since the moments are sensitive to variations in EI while the magnification factors are not nearly as sensitive. Also, for arches whose axes di f f e r l i t t l e from a true parabola i t may be assumed that their magnification factors are approximately the same as those for the parabolic arches which are presented in this thesis, although their elastic moments may d i f f e r . It i s assumed that the designer can find the elastic moments of the arch and choose the proper magnification factors, by interpolation, for ri s e -span ratios and variations of EI other than those given. The program for the electronic computer, the ALWAC III E, 3 used by Lee and Pelton was altered to enable the calculation of the effects of axial deformation and support movements. Details of the modifications are given in Chapter II. 4 CHAPTER I THE GOVERNING DIFFERENTIAL EQUATION The action of an arch under load, axial deformation, and movement of the supports is best studied by i t s differential equation. From the differential equation certain conclusions w i l l be drawn which w i l l be verified by numerical examples in Chapter III. W-C/X M 1- C/M H Figure(1-1) In Figure ( l - l ) an element of an arch, i n i t i a l l y in position A B T deflects under the action of a load w to a position A J B J , 5 The horizontal deflection of point A i s •? and the vertical deflection is £ . The forces are shown acting on the element in i t s deflected position. The moment, vertical shear and horizontal thrust acting on the element are M, V, and H respectively. Equations of Statics 5 > x - 0 Since the load is assumed to act vertic a l l y the horizontal thrust H is constant. ZFy - 0 w dx + dV - 0 or V' ™ -w where ' represents _d_ dx ^ M B I - 0 H(dy • d?) - V(dx + d?) + wdx ~ + dM - 0 After dividing by dx we obtain H(y« + ? ') - V(l + 4») • M« - 0 To combine the preceeding equations the latter must be divided by 1 J y' + ? » - V + M« o 0 1 • «• 1 Differentiating with respect to x we obtain i + S « ( i • s «) z ^ 1+s* ( i + Multiplying by ( l +£') and substituting for V gives M" - M ' + H(y" ••?") .HV' M ' g" - -w(l +^») (la) 1 + £ • 1 + £ » 6 Equations of Axial Deformation The i n i t i a l length of the element AB is ds - / ( d x ) 2 • (dy) 2 The length of the element after deformation i s dsi - J (dx + d?)2 • (dy • d?) 2 Assuming that the change in length of the element due to temperature change and/or axial thrust i s Ads : ds(l • A) «= ds t or J(dx)2 + (dy) 2 ( l + A) - J (dx)2 + 2dxd£ + ( d £ ) 2 + (dy) 2 • 2dyd? + (d?) 2 Squaring both sides gives (dx) 2 + (dy) 2 1 + 2A + A 2| - (dx) 2 • 2dsd£+ (d£) 2 + (dy) 2 + 2dyd? + (d?) 2 or (dx) 2 + (dy) 2 2X+ V - 2dxd£ + ( d £ ) 2 • 2dyd? + (d'?) 2 Dividing by (dx) we obtain 1 + (y') 2] [2 A + A 2J - 2£» + ( E » ) 2 • 2 y » f « • ( »?') 2 Since the change in length of the element w i l l be small, X.2 may be neglected compared to 2 A and we obtain x [ i +. ( y » ) 2 ] - c » + ( g ' ) 2 • • + ( ? ' ) 2 (2a) Equations of Bending Deformation to moment is d(d©) -From Hooke's law the rotation of Bj with respect to A^ due M ds + dsi EI 2 7 or d(dO) M dx 1 • * 7 EI cos 6 A temperature change of 80 degrees or an axial stress of 15,000 p . s . i . in a steel arch gives a value of A of .0005; hence * may be 2 neglected when compared to 1. Dividing by dx gives (de)' - M_ EI cos § Before the load i s applied : tan § ™ dy_ •» y» dx After the load i s applied ' tan (8 + d9) dx • d £ or tan (9 • de) <° v' • *l ' l + e.' The change in slope is d© •» arctan / v' * - arctan (y') 1 + C After substitution we obtain _, » arctan / y* + * \ - arctan (y f) 1 • C M (3a) M" EI cos 8 In summary then, the three basic equations of an arch are' M» S" + H(y" + ? ") - H/y' • I'V" *= -w(l + £•) <la) l + \ 1 + e. *) ( y.)2 ( g ' ) 2 + y» + illi.2 2 2 arctan \ I • s »/ - arctan (y') M EI cos © (2a) (3a) 8 These equations contain 4 unknowns £, ®2 , M and H. The thrust H i s determined from boundary conditions and the other three unknowns may be obtained by the solution of the three simultaneous equations. These equations are highly non linear. Their solution for an arch would correspond to the "elastica" problem of an i n i t i a l l y straight column. In applying these equations to an arch of normal proportions certain approximations may be made. If i t is assumed that the horizontal deflection may be approximated by the expression £ «* £ 0 sin j^p- and the maximum horizontal deflection is —h— the maximum value of C' is -JL- . 2T5£F 200 This i s small compared to 1 and may be neglected in equations (la) and (3a). Assuming the maximum deflections are , the terms ,{ and 0 200 2 ^ i^ *^  are very small compared to the f i r s t power of these terms and may be neglected in equation (2a). In equation (3a), since d8 is small, the tangent of d8 may be assumed equal to d8. Applying these approximations the three equations become M" + M»<£" + H (y" + 9( ") - H(y« + *?•)£" « -w (lb) A.[l + ( y ' ) 2 - £* • y**l* (2b) *l" - M (3b) EI cos 8 To obtain a better understanding of the action of a flexible arch these three equations w i l l be combined into one equation. Rearranging equation (2b) we obtain and C" - 2Ay'y-' - y»<?" - y" Rearranging equation (3b) we obtain M - EI cos 8*?" r ~I " and M" . EI cos 8<? " 9 Substitution into equation (lb) gives [EI COS 8 ? "] " + M' C," + H(y" + ^  ") . H(y» + Jp«) (2Ay'y" - y' V" - y"^') w Multiplying and collecting terras gives [EI cos e ? " ] " + M«£" + H [ l + ( y * ) 2 ] * ? " + H [ l - 2A<y') 2]y" + H [ l - 2A] y ' y " f + H [ y » ^ » ^ " + y"(7«)2] - -w (4a) It i s d i f f i c u l t to show that the term M» £ " is small com-pared to the other terms, however, tests made with the computer show that the effect of this term i s small and may be neglected when the deflections are small. The effect of this term w i l l be discussed further in Chapter III. If the rise-span ratio — is 1/3 and the coefficient of axial deformation is .0005 (a temperature change of 80°), the value of 2 A i s .001 and the maximum value of 2A-(y') 2 is .0018. Both these terms are very small com-pared to 1 and may be neglected. The terms due to axial deformation thus disappear from the differential equation and must enter the solution of the equation only as boundary conditions. The terms containing products of differentials of *} are small compared to the other terms and may be neglected. The differential equation now becomes + + - W - Hy To f a c i l i t a t e the study of this equation the variables w i l l - j-> y = L* A N C* + + _ y y » W 10 T3 yt-l 2 o If we multiply the equation by and let ... - ft *• where E I a v E I a v ft i s a dimensionless ratio defining the f l e x i b i l i t y of the arch, we obtain cos e^"J " + ft1 [ 1 • (y») 2 *)" * ft2 y'y" ^» - - wlP - a 2 y» (4c) E I a y If fi and the deflections are both small, the terms which are products of ft and the derivatives of ^ are small. Neglecting these two terms gives the equation I cos 8 fjn lav wL3 - /32y'' <4d) EIav Equation (4d) is a fourth order linear differential equation with variable coefficients. It is the differential equation of the elastic theory of arch analysis and moments obtained by the solution of this equation are the elastic theory moments. If V and the deflections are large we have a flexible arch and the moments obtained from the solution of equation (4c) are the deflection theory moments. Equation (4c) is a linear differential equation but the structure i t governs i s non-linear. This Is due to the fact that an increase in the load also changes fi and thus the coefficients of the equation as well as the deflections. Because of this change of coefficients, superposition in the ordinary sense is not possible. Tests have shown that the thrust due to load determined by the elastic and deflection theories are almost the same. Thus i t is possible to use the readily obtained elastic thrust In equation (4c). Using this thrust a modified method of superposition, to be described later, i s possible. It w i l l be seen that the larger the value of ft , the larger w i l l be the difference In the moments determined from equation 11 (4c) as compared to those determined by equation (4d). The ratio of the deflection theory moments, determined by equation (4c), compared to the elastic theory moments, determined by equation (4d), are the magnification factors. The term/32?" in equation (4d) represents the effect of horizontal thrust in the arch. If this term is neglected, the equation of an arch whose ends are unrestrained horizontally is obtained : i cos e lav wL3 Flav The fact that an axial deformation enters into the solution of the differential only as a boundary condition indicates that a uniform axial deformation A is equivalent to a fabrication error which produces a gap at the crown of A c «• A.L or a horizontal movement of a support of SH = AL. This was found to be true by tests on the computer. Results of these tests are given in Chapter III. To il l u s t r a t e the principle of superposition consider an arch, shown in Figure (1-2), which is loaded by a variable load w^  (load l ) . It i s desired to combine this loading with a rotation Ct) of the lef t support (load 2) to obtain load 3. To superimpose these two cases the horizontal thrust of each individual case must be made equal to the thrust of the combination. It w i l l be assumed that the elastic theory and the deflection theory thrusts are equal in a l l cases. A uniform load qj which produces a thrust equal to the thrust produced by the rotation must be added to the variable load w^ . For convenience, equation (4c) w i l l be designated as L^) + tf2L2(H) - - _WL3 - / ? 2 y ' (5) EIav 12 4 Load 3 Figure (1-2) 13 Letting the deflection due to the load (wj + q^) be ^a, the differential equation for case A becomes LxWa) + /32L (? f l) - - <„. + qi) J s L - /32r C5a) FI 3v o 2 where Q => "3 L Elav A uniform load q2 which produces a thrust equal to the thrust produced by the variable load must be combined with the rotation to give the correct horizontal thrust. Letting the deflection due to rotation and q2 alone be ^  the differential equation for case B is LltfyJ * /32L2(*?h) - - q 2 _ L L - / ? 2 y " <5b) Elav where ft2 - h 3 l 2 Elav To combine case A and case B 9 the case of a uniform load (qj • q 2) must be considered. Letting f}c be the deflection due to this loading the differential equation i s £l<?C> + # 2 £ 2 ( ? C > - - (qi + q 2) - B2T <5c) Elav where (31 - H 3 L Elav Letting tfd be the deflection due to load 3, the combination of the load w^  and the rotation CaJ, the differential equation becomes Li<?d) • / ? 2 £ 2 < £ d > - - «i _ t L - /? 2y" ( 5 d > Elav where /? 2 - H 3 l 2 Elav Adding equations (5a) and (5b) and subtracting equation (5c) we obtain Ll<fm + H ' ^c> + / ? V ? a + H - 7c> - - w l l 3 - 2 ? " ( 6> Elav 14 The addition and subtraction of these equations is only possible because care was taken to make the value of ft identical in each equation. Comparing equations (5d) and (6) we see that - U + "^b - ^ c In equation (5c) t assuming a parabolic arch <*2> L * 8 f E I a v H3 m (qj + q 2 M )L2 and (i 2 -" a v ( q i + also y • (Lx L 2 - x 2) and y - 4f (x -L x2) Therefore y" - - 8f r In arches which have a shape other than parabolic, the distribution of load which produces the correct thrust and zero bending moment must be used. Substituting these quantities into equation (5c) gives M?"c> + /3 2L 2^c> - - <qi + q2> J L E I a v + (q! • q 2) 8f (5c) 8 f E I a v L or LX< YV • tf2L2(?c> - o For loads less than the c r i t i c a l buckling load the only solution to this equation i s c - 0, hence ? d - 9 a • 9 b and - ?a + ?b From the equation (3b) the moments are proportional to the 15 deflections and Mud - MDa + MDb - M F A x ^ a + M E S MOb MEa MEb - MEa x 0a + MEb * 0b Thus the deflection theory moments may be determined by multiplying the elastic theory moments by magnification factors. The total elastic thrust of the combination must be used in determining the magnification factors. The magnification factors 0b due to rotation are tabulated in Chapter V. The magnification factors 0a due to load are tabulated in the Appendix of Ref. 2. A more d i f f i c u l t problem of superposition is the analysis of a loaded arch hinged at one support. The moments in this arch may be determined by superimposing the solution determined for the rotation of a support of a hingeless arch with the solution found for the loaded hinge-less arch. To obtain the hinged condition the moments produced in the loaded hingeless arch must be superimposed with a rotation of a support which produces a moment at the support equal and of opposite sign to the moment produced by- the load. The rotation is unknown and since the rotation produces a thrust, the magnification factors which must be applied to the elastic moments are unknown. To find the moments in this arch, i t is necessary to assume a rotation and calculate the total elastic thrust in the arch due to the load plus the assumed rotation. Using this thrust, the magnification factors to be applied to the elastic moments at the hinged support are chosen. If the elastic moments due to load and rotation multiplied by the proper magnification factors are not of opposite sign and equal, a new value of the rotation must be assumed 16 and the procedure repeated. When the rotation at the hinged support and the thrust is determined, the proper magnification factors may be applied to the elastic moments at the other points in the arch to determine the governing moments. It was found that i t i s permissible to neglect the thrust due to moment for small rotations. 17 CHAPTER II PROGRAM FOR AUTOMATIC COMPUTATION To check the superposition principles and to produce the tables and graphs presented in Chapter IV, the program for the ALKAC III E electronic computer, which was used by Lee and Pelton, was altered to enable the effect to axial deformation and support movements to be determined. This program finds the deflections, moments and thrust in the arch by the conjugate beam method. The program begins by finding the elastic theory deflections and forces with the deflections assumed to be zero. The deflections theory moments and thrust are found by successive application of the conjugate beam formulae using the geometry altered by the deflections determined in the previous cycle. The program i s set up to analyse an arch divided into 20 divisions not necessarily of equal length. The arch may have any symmetrical shape. The location of the end of each division must be specified. In each division the flexural ri g i d i t y i s constant; however, any r i g i d i t y may be assigned to each division. The loads are assumed to act at the centre of each division and they may have any desired distribution. The moment and deflection at the end of each division, the shear at the crown, and the horizontal thrust is determined by the computer. The complete details of the program are given in Ref. 2. A diagram of the arch is given in Figure (2-1). The steps which the program follows in computing the moment, thrust, and shear at the crown and then the moment and deflections at the 1l », fe ts V , ft i J J - J Forces and Geometry of the Arch IIOR In i)n In Deflections of the Arch 03 Figure (2-1) 19 end of each division, using the geometry altered by the deflections determined in the previous cycle, are as follows: (1) Assuming the arch cut at the crown, the horizontal and vertical deflections and the rotation at the crown due to a unit thrust applied to each half of the arch at the crown are found. (2) Horizontal and vertical deflections and the rotation at the crown due to a unit shear applied to each half of the arch at the crown are found. (3) The horizontal and vertical deflections and the rotation at the crown due to a unit moment applied to each half of the arch at the crown are determined. (4) The horizontal and vertical deflections and the rotation at the crown due to the applied loads on each half of the arch are calculated. (5) The relative horizontal and vertical deflections and rotations at the crown due to the unit loads and the actual loads are determined. (6) The coefficients required for the three simultaneous equations necessary to find the moment, thrust, and shear at the crown are determined. (7) The three simultaneous equations are solved by iteration to find the moment, thrust, and shear at the crown. The computer types out the values of Mc, H,and V"c. By typing the proper code, the computer may be made to perform another cycle of iteration of the three simultaneous equations and type out new values of M c > H,and V"c or to continue on in the program. (8) Having the moment, thrust,and shear at the crown, the moment, horizontal deflection, and vertical deflection at the end of each division of the 20 arch is now computed. (9) By the proper setting of the jump switches, either the moments or the deflections or both the moments and deflections at the end of each division may be typed out. (10) The deflections which have been determined are added to the i n i t i a l geometry for use in computing the moments in the next cycle. The new deflections divided by 2 are added to the i n i t i a l geometry for obtaining the deflections due to the moments. (11) The cycle is repeated until the arch comes to equilibrium in the deflected position. The axial deformation and support movements being boundary conditions, i t is permissible to neglect the small changes in ds which occur. Since i t is the ratio ~ which is used in the conjugate beam method E I of analysis, a temperature change of 80 degrees in a steel or concrete arch is equivalent to a change in the moment of in e r t i a l of .05 percent which i s negligible. The free deflections of the half arches due to axial deformation and support movements must be added to the deflections resulting from the applied loads. To include these deflections the program must be altered in three places. To enter these deflections into the main memory of the computer i t was necessary to add a deflection input routine. This routine is stored in channel 95 with the last few instructions in channel 96. The routine is begun by the command 9500 carriage return. The routine accepts the axial deformation and support movement deflections and rotations in the following sequence: 21 * 1 . *2* » » 9 > » > • * lOL* ^18» » i » » » . *10R. ? 2 » » > * » » » » ^10L» 719» fyl8» » * » > > » » ?10R» (Check Sum) Horizontal deflections t o Z\c\i a r e P o s i t i v e t o t n e l e f t . Horizontal deflections £jg to £JQr are positive to the right. Vertical deflections are positive upward. &>L i s positive counterclockwise. (t)^ is positive clockwise. During input the computer forms the sum of the forty deflections plus 4 times the sum of the two rotations and then outputs the difference between this sum and the check sum. The same storage scheme as that used in the original program is used in storing the deflections in the channels. The deflections £j to C^gL a r e stored in the channel CO. Deflections £^ 9 to CioR a r e s t o r e < * i n channel Cl. Deflections to *?JQL are stored in channel C2 and deflections 1^9 to ^ JQR are stored in channel C3. 6^ L and are also stored in words If and lb respectively of channel 2 5 C3. As in the original program the deflections are scaled 2 and the 27 rotations are scaled 2 No provision was made for consideration of deflections at the supports in the original program. The deflections at the support are assumed equal to zero and^when the geometry i s input, the coordinates of the support points are automatically set equal to zero. Rather than make extensive program changes i t was found to be easier to introduce a deflection at the supports by setting the coordinates of the supports equal to the desired deflection by making "one word changes". The coordinates of the supports are stored in word 00 of their respective channels. The horizontal 22 and vertical coordinates of the l e f t support are stored in channels a4 and a6 respectively. The horizontal and vertical coordinates of the right support are stored in channels a5 and al respectively. The coordinates, representing the deflections, of the supports must be introduced after the elastic moment, thrust, and shear at the crown i s computed but before the moment and deflection of every point is calculated. It was found most convenient to do this by stopping the program after i t has typed out the moment, thrust, and shear at the crown of the f i r s t cycle. The deflections are entered in hexadecimal form scaled 2 2 5 i.e. .10 ^ 00333333 .20 ~ 00666666 1.00 - ~ 02000000 Care must be taken in choosing the correct signs. The program is re-entered by typing the code 8b00 carriage return. The moment, thrust, and shear are typed out again and the program continues in i t s usual manner. The second instance where the program must be altered i s in the computation of the relative deflections and rotations at the crown due to the unit loads and the applied loads (Step 5). The relative deflections and rotations at the crown due to axial deformation and support movements must be included in computing the moment, thrust, and shear at the crown. Since no provision was made in the original program for additional factors, i t is necessary to add the deflections and rotations at the crown due to axial deformation and support movements to those which have been determined for the applied loads. The actual alteration is accomplished by changing word 12 in channel 8a from 818bll00 to 81961107 and the addition of one 23 channel of program operating in channel 96. The third alteration of the program i s in the computation of the deflections at the end of each division (Step 8). The deflections due to axial deformation and support movements must be added to the deflections due to load. This i s accomplished by changing word 05 in channel 8f from 87a485d4 to 81971103 and the addition of one channel of program operating in channel 97. The three additional channels of program with entry codes are as follows: 9504 280a4940 49031718 49031719 00000000 871e5500 8dcl5500 5b0fll60 00000000 5blfll60 5bl7H60 495f6l03 001b0092 484b6103 484b6l03 49035b0b OOlbOOOa 49031708 49031709 1160495b 001900Id 8dc05500 8dc25500 8dc36l03 0019000d 5blbll60 5bl31l60 81961102 0019001c 484b6l03 484b6l03 00000000 0019000c 9604 87db85c0 497d797a 49035blf 00000000 7978614a 615f497a 11606703 8b408d41 497885cl 797e6l5b 871f5blb 8f421100 797c6l4a 497e8fdb H601b00 00000000 497c85c2 83408541 00000000 00000000 7979614a 87420000 00000000 00000000 497985c3 818bll00 00000000 0219058e 797d614a 00000000 00000000 00190086 9704 85d487c0 85d687c2 83408541 8b408d41 55 If784b 55 If784b 87420000 8f421100 606b484b 606b484b 87a485d4 00000000 17848dd4 17858dd6 818fU09 00000000 85d587cl 85d787c3 00000000 00000000 55 If784b 551f784b 00000000 00000000 606b484b 606b484b 00000000 00000000 17948dd5 17958dd7 00000000 OOOaOOOO 24 CHAPTER III NUMERICAL VERIFICATION OF THEORY With the program altered and checked, a number of runs were made to check the ideas set forth in Chapter I. Specifically we wish to check the variation of the deflection theory moments with axial deformation A » vertical movement of a support 6y and rotation of a support (SL> . Tests were made to compare the moments due to axial deformation A , horizontal movements of the supports SJJ and a horizontal gap at the crown £ J I c. The moments due to a vertical support movement were compared to the moments produced by a fabrication error which produces a vertical gap at the crown. Finally, the superposition of axial deformation and support movements with various load conditions are checked. 25 Part 1 Variation of Moment With Axial Deformation It was found that A enters the solution of the differential equation of the arch only as a boundary condition. Hence i t would be expected that for a constant /3 the moments shall be directly proportional to X. Several runs were made for various values of X for different values of /3 . The results are given in Table (3-1) The results show that, for constant (i , is for a l l XEI practical purposes a constant. The magnification factors 0 are thus independent of X and need only be computed for various values of B . These runs were made with a uniform load applied producing thrust to give values of (i equal to 3, 5 and 7 with the thrust due to the axial deformation neglected. This was found to be more correct than including the elastic thrust due to axial deformation in the determination of 3 . This is due to the fact that an increase in the length of the arch produces a positive thrust when computed by the elastic theory; however, on subsequent cycles the increase In the rise-span ratio reduces this thrust or, in the case of the more flexible arches, the thrust due to an increase in length becomes negative. Therefore, the change in thrust due to axial deformation may be neglected. The vertical deflection at the crown i s : 1.56 XL for /3 - 0 1.58 XL for fS " 3 1.62 XL for 0 - 5 1.70 AL for 3 - 7 -~- DUE TO UNIFORM AXIAL DEFORMATION f - 1, CONSTANT EI L 6 ft- 3 ft- 5 ft - 7 X L A - -.001 A - .001 A - -.001 A - .001 A • -.001 A - -.0005 A - .0005 A - .001 .00 37.57 37.00 31.66 31.23 20.15 20.13 20.12 20.08 .05 28.19 27.86 27.15 26.83 24.24 24.17 24.06 24.00 .10 18.80 18.65 20.48 20.28 23.43 23.35 23.18 23.11 .15 9.84 9.82 12.56 12.47 18.41 18.33 18.15 18.09 .20 1.61 1.68 4.21 4.21 10.36 10.29 10.17 10.12 .25 - 5.64 - 5.52 - 3.87 - 3.79 .70 .68 .60 .60 .30 -11.74 -11.59 -11.12 -10.98 - 9.20 - 9.19 - 9.16 - 9.13 .35 -16.58 -16.42 -17.14 -16.96 -18.16 -18.10 -17.99 -17.92 .40 -20.09 -19.92 -21.63 -21.41 -25.23 -25.13 -24.93 -24.85 .45 -24.21 -22.04 -24.39 -24.15 -29.73 -29.61 -29.36 -29.26 .50 =22.92 -22.75 -25.32 -25.08 -31.27 -31.14 -30.87 -30.77 Table (3-1) 27 Part 2 Variation of Moment With Vertical Support Movement It is impossible for vertical support movements 5y to enter the solution of the differential equation of the arch other than as a boundary condition; hence i t would be expected that, for constant /3 , the moments w i l l be directly proportional to <5y. ^ n e results of several runs, which were made to compare the moments produced by various values of 6 \j for a constant /3 , are given in Table (3-2) The results of these runs show a small degree of non-linearity. The deflection theory moment due to 6yj at a point in the arch, for constant /3 , may be expressed as MD - ( C l • c 2 6-v \ 6v \ L / L or in terms of magnification factors as The factor 0i is a function of the f l e x i b i l i t y /3 of the arch as well as the geometry and distribution of EI and thus is similar to the usual magnification factor. The factor 0 2 i s a constant which is practically independent of /3 but depends mainly on the rise-span ratio and distribution of EI. The term 0j is much larger than the term 0 2 -jY. for deflections of usual magnitude. The 0\B for corresponding points on each half of the arch are equal. The 0 2s for corresponding points on each half of the arch are of equal magnitude but of opposite sign. This non-linearity appears to be mainly due to the term M'£ M of equation (lb) of Chapter I. This non-linearity appears only in cases where there is non-28 symmetry and shear in the arch. At the points in the arch where 0 2 - 0 , i.e. no non-linearity, both the second derivative of the horizontal deflections and the f i r s t derivative of the moments are very small. It w i l l be seen that this term is independent of the f l e x i b i l i t y of the arch. Bearing in mind that 6y =» -.005L represents a support settlement of 1 foot in an arch of 200 foot span, this non-linearity may be neglected in practical cases. A vertical movement is positive upward. The elastic thrust due to the vertical movement of one support is zero and the deflection theory thrust is negligible. M P L DUE TO A VERTICAL MOVEMENT OF THE LEFT SUPPORT 6 V E I f - 1, CONSTANT EI, /? . 5 L 6 X L 6vi - -.005L 6 V L - -.002L <5 V L *" -.OOIL 6 V L - .001L 6 V L - .002L 6 V L - .005L .00 -2.45 -2.51 -2.53 -2.58 -2.60 -2.67 .05 -3.40 -3.45 -3.47 -3.51 -3.53 -3.58 .10 -4.04 -4.08 -4.09 -4.12 -4.13 -4.18 .15 -4.35 -4.38 -4.38 -4.40 -4.40 -4.43 .20 -4.35 -4.36 -4.36 -4.36 -4.36 -4.37 .25 -4.07 -4.06 -4.06 -4.05 -4.05 -4.04 .30 -3.56 -3.54 -3.53 -3.51 -3.50 -3.48 .35 -2.85 -2.82 -2.81 -2.78 -2.77 -2.74 .40 -2.01 -1.97 -1.95 -1.92 -1.91 -1.86 .45 -1.07 -1.02 -1.00 - .98 - .96 - .91 .50 - .08 - .03 - .01 .01 .03 .08 .55 .91 .96 .98 1.00 1.02 1.07 .60 1.86 1.91 1.92 1.95 1.97 2.01 .65 2.74 2.77 2.78 2.81 2.82 2.85 .70 3.48 3.50 3.51 3.53 3.54 3.56 .75 4.04 4.05 4.05 4.06 4.06 4.07 .80 4.37 4.36 4.36 4.36 4.36 4.35 .85 4.43 4.40 4.40 4.38 4.38 4.35 .90 4.18 4.13 4.12 4.09 4.08 4.04 .95 3.58 3.53 3.51 3.47 3.45 3.40 1.00 2.67 2.60 2.58 2.53 2.51 2.45 Table (3-2) 30 Part 3 Variation of Moment With Rotation of a Support A rotation of a support also enters the differential equation of the arch only as a boundary condition; hence i t may be assumed that, for constant /3 , the moment due to the rotation of a support w i l l be directly proportional to the magnitude of the rotation (s). A number of runs were made to determine the moments produced by various values of The runs were made for a constant /? , the thrust due to the rotation being taken into account in the determination of fi . The results are given in Table (3-3). The moments due to a rotation also show a small degree of non-linearity. If the deflection theory moment i s expressed as MD - M E (0x + 0 2 & J ) neither 0^  nor 0 2 are equal for corresponding points on the two halves of-the arch. The factor 0 2 also varies somewhat with the f l e x i b i l i t y of the arch. At points in the arch where the second derivative of the horizontal deflections is zero the non-linearity i s negligible. There i s , however, non-linearity at the points where the f i r s t derivative of the moments is zero. This suggests that the two terms involving the products of the derivatives of the vertical deflections in equation (4a) also contribute to the non-linearity. ^ DUE TO A ROTATION OF THE LEFT SUPPORT Ct>EI f - 1, CONSTANT EI, /3 «=• 5 L 6 X L (t)Lm .03 U)L - .01 cuL - -.003 & i L = -.01 dJl " -.03 .00 -4.95 -4.95 -4.94 -4.94 -4.94 .05 -5.34 -5.32 -5.30 -5.29 -5.26 .10 -5.26 -5.21 -5.18 -5.17 -5.14 .15 -4.72 -4.69 -4.67 -4.66 -4.63 .20 -3.91 -3.88 -3.87 -3.86 -3.83 .25 -2.90 -2.88 -2.88 -2.87 -2.86 .30 -1.80 -1.80 -1.80 -1.80 1.80 .35 - .69 - .70 - .71 - .72 - .74 .40 .35 .32 .29 .29 .26 .45 1.24 1.20 1.17 1.16 1.12 .50 1.95 1.90 1.87 1.86 1.81 .55 2.44 2.38 2.35 2.33 2.29 .60 2.67 2.62 2.59 2.57 2.53 .65 2.63 2.58 2.56 2.54 2.50 .70 2.31 2.28 2.26 2.25 2.22 .75 1.73 1.71 1.70 1.69 1.68 .80 .90 .90 .90 .90 .90 .85 - .12 - .10 - .08 - .08 - .06 .90 -1.26 -1.21 -1.19 -1.17 -1.13 .95 -2.40 -2.34 -2.31 -2.29 -2.23 1.00 -2.43 -2.36 -2.32 -2.30 -2.24 Table (3-3) 32 Part 4 Comparison of Uniform Axial Deformation X, Fabrication Error Producing a -Horizontal Gap at the Crown A H C , and Horizontal Movement of Supports 6H It was found that, for equal changes in horizontal span length, the moments produced by these three effects are practically equal. This MQL confirms the findings of the theoretical analysis. A comparison of «T» 2 A ti. ^ ^ E T ' 3 N (* O*~E"T v a r * o u s v a l u e s °f fi a r e given in Table (3-4). In a l l H cases the thrust due to A , 8R or A^vas neglected in the determination of (3. The larger discrepancy of the moment at the crown due to a gap at the crown is due to the fact that there is a discontinuity in the arch axis at this point because the two sections of the arch meeting at this point are too short. In the case of a uniform axial deformation or a support movement the two halves of the arch retain their parabolic shape. From these results i t may be concluded that any effects which produce or tend to produce changes in the span length of the arch may be represented by a uniform axial deformation which produces a free horizontal deflection of the same amount. A variable symmetrical axial deformation may be considered as a uniform axial deformation which produces the same horizontal free deflection. COMPARISON OF JV;, ^ AND MD L XEI AU^I • SttVI £ «• l s CONSTANT EI L 6 • 3 P - 5 / ? - 7 X L X" - . 0 0 1 A H C •» . 0 0 1 L A - - . 0 0 1 A H C - . 0 0 1 L A - - . 0 0 1 6H " - . 0 0 1 1 A H C » . 0 0 1 L . 0 0 3 7 . 5 7 3 7 . 5 6 3 1 . 6 6 3 1 . 6 8 2 0 . 1 5 2 0 . 1 3 2 0 . 2 3 . 0 5 2 8 . 1 9 2 8 . 1 8 2 7 . 1 5 2 7 . 1 6 2 4 . 2 4 2 4 . 2 2 2 4 . 2 8 . 1 0 1 8 . 8 0 1 8 . 7 9 2 0 . 4 8 2 0 . 4 8 2 3 . 4 3 2 3 . 4 2 2 3 . 4 3 . 1 5 9 . 8 4 9 . 8 3 1 2 . 5 6 1 2 . 5 4 1 8 . 4 1 1 8 . 4 0 1 8 . 3 6 . 2 0 1 . 6 1 1 . 6 0 4 . 2 1 4 . 1 8 1 0 . 3 6 1 0 . 3 6 1 0 . 2 8 . 2 5 - 5 . 6 4 - 5 . 6 4 - 3 . 8 7 - 3 . 9 0 . 7 0 . 7 0 . 6 0 . 3 0 - 1 1 . 7 4 - 1 1 . 7 5 - 1 1 . 1 2 - 1 1 . 1 6 - 9 . 2 0 - 9 . 2 0 - 9 . 3 1 . 3 5 - 1 6 . 5 8 - 1 6 . 5 9 - 1 7 . 1 4 - 1 7 . 1 8 - 1 8 . 1 6 - 1 8 . 1 6 - 1 8 . 2 5 . 4 0 - 2 0 . 0 9 - 2 0 . 0 9 - 2 1 . 6 3 - 2 1 . 6 6 - 2 5 . 2 3 - 2 5 . 2 2 - 2 5 . 2 8 . 4 5 - 2 2 . 2 1 - 2 2 . 2 1 - 2 4 . 3 9 - 2 4 . 4 1 - 2 9 . 7 3 - 2 9 . 7 2 - 2 9 . 7 4 . 5 0 - 2 2 . 9 2 - 2 2 . 7 7 - 2 5 . 3 2 - 2 4 . 9 1 - 3 1 . 2 7 - 3 1 . 2 6 - 3 0 . 4 4 Table ( 3 - 4 ) 34 Part 5 Comparison of a Fabrication Error Which Produces a Vertical Gap at the Crown Ay c With a Vertical Movement of a Support 6y The deflection theory moments due to a fabrication error which produces a total vertical gap at the crown A.yc were also determined. It was assumed that the free vertical deflections of each half of the arch vary uniformly from zero at the supports to ± ^  V c at the crown, i.e. both halves of the arch retain their parabolic shape although their rise-span ratios are different. An error which leaves the l e f t half of the arch higher than the right is considered positive. The moments produced by various values of ^ v c f ° r a constant ft are given in Table (3-5). A com-parison of the moments produced by a vertical support movement and a vertical gap at the crown for /3 » 7 are given in Table (3-6). A comparison of Tables (3-2) and (3-5) and a study of Table (3-6) shows that the deflection theory moments produced by these two conditions are practically identical. The non-linearity of the moments due to a gap at the crown i s less. The moments due to a non-uniform temperature change in an arch may thus be determined by the superposition of vertical and horizontal support movements. The moments due to the horizontal change in length are equal to the moments produced by a horizontal movement of a support of the same amount. The moments produced by a difference in the vertical deflections of the two halves of the arch are equal to the moments produced by a vertical support movement of the same amount. An example of this method is given in Table (3-7). The table is set up in a similar manner to those in Part 6. — ^ w r DUE TO A VERTICAL GAP AT THE CROWN ^ V C E I f - 1 „ CONSTANT EI, =. 5 L 6 X L Av - -.005L *c Ayc - -.001L A\JC - .001L Aw " .005L .00 -2.54 -2.55 -2.56 -2.58 .05 -3.47 -3.49 -3.50 -3.51 .10 -4.09 -4.10 -4.11 -4.12 .15 -4.38 -4.39 -4.39 -4.40 .20 -4.36 -4.36 -4.36 -4.37 .25 -4.06 -4.06 -4,06 -4.05 .30 -3.52 -3.52 -3.52 -3.51 .35 -2.81 -2.80 -2.79 -2.78 .40 -1.95 -1,94 1.93 -1.92 .45 -1.01 - .99 - .99 - .97 .50 - .02 .00 .00 .02 .55 .97 .99 .99 1.01 .60 1.92 1.93 1.94 1.95 .65 2.78 2.79 2.80 2.81 .70 3.51 3.52 3.52 3.52 .75 4.05 4.06 4.06 4.06 .80 4.37 4.36 4.36 4.36 .85 4.40 4.39 4.39 4.38 .90 4.12 4.11 4.10 4.09 .95 3.51 3.50 3.49 3.47 1.00 2.58 2.56 2.55 2.54 Table (3-5) 2 2 COMPARISON OF ft^ AND f..- 1, CONSTANT EI, ft ~ 7 L 6 X L 6vL • -.001L A v c - -.001L Uniform .00 3.62 3.61 .05 .51 .50 .10 -2.69 -2.70 . 15 -5.43 -5.44 .20 -7.34 -7.34 .25 -8.24 -8.24 .30 -8.09 -8.08 .35 =6.99 -6.98 .40 -5.12 -5.11 .45 -2.71 -2.69 .50 - .02 .00 .55 2.66 2.68 .60 5.07 5.09 .65 6.95 6.96 .70 8.05 8.06 .75 8.22 8.22 .80 7.33 7.33 .85 5.44 5.44 .90 2.72 2.71 .95 - .47 - .48 1.00 -3.58 -3.59 Table (3-6) 38 SUPERPOSITION OF HORIZONTAL AND VERTICAL MOVEMENTS OF THE LEFT SUPPORT TO GIVE THE MOMENTS DUE TO A UNIFORM AXIAL DEFORMATION OF THE LEFT HALF OF THE ARCH L ~ 200 f t . , f - 1, CONSTANT EI - 40,000 kip f t . 2 , 0 - 7 L 6 Uniform Load Col. 1 Col. 2 Col. 3 Col. 4 Col. 5 X L 5 H l *" -.0005L <5vL - -.0005f 2(2) • (3) A L - -.001 .00 -2.013 - .121 -2.134 -2.135 .05 -2.422 - .017 -2.439 -2.434 .10 -2.342 .090 -2.252 -2.244 .15 -1.840 .181 -1.659 -1.650 .20 -1.036 .245 - .791 - .784 .25 - .070 .275 .205 .208 .30 .920 .270 1.190 1.189 .35 1.816 .233 2.049 2.043 .40 2.522 .171 2.693 2.683 .45 2.972 .090 3.062 3.050 .50 3.126 .001 3.127 3.113 .55 2.972 - .089 2.883 2.871 .60 2.522 - .169 2.353 2.343 .65 1.816 - .231 1.585 1.579 .70 .920 - .268 .652 .651 .75 - .070 - .274 - .344 - .340 .80 -1.036 - .244 -1.280 -1.273 .85 -1.840 - .181 -2.021 -2.013 .90 -2.342 - .091 -2.433 -2.424 .95 -2.422 .016 -2.406 -2.401 1.00 -2.013 .119 -1.894 -1.893 Table (3-7) 39 Part 6 Superposition of Axial Deformation and Support Movements With Load A number of runs were made to check the superposition of axial deformation and support movements with various load conditions. The moments determined separately by the computer due to two effects are added together to be compared to the moments which are obtained by the computer when the two effects are combined. In each of the following tables of comparison two conditions are compared. The f i r s t column gives the location of the point on the arch. The next two columns give the deflection theory moments due to each condition when determined separately with the computer. The fourth column gives the sum of the moments of these two conditions.determined independently. The f i f t h column gives the moments which were determined by the computer when the two cases were combined. In a l l cases the thrust due to A o r ^ was neglected in the determination of /3 . The units of the moments are f t . kips. In a l l cases the discrepancy due to determining the moments of the two effects separately is negligible. It i s thus shown that the conclusions of Chapter I are correct and the sum of the elastic moments of each effect, multiplied by the correct magnification factor, gives the correct total moment. 40 SUPERPOSITION OF A LOAD PRODUCING A MAXIMUM POSITIVE MOMENT AT THE CROWN WITH A NEGATIVE UNIFORM AXIAL DEFORMATION L - 200 f t . , f - _L, CONSTANT EI - 40,000 kip f t - 2 , LL » 1,0- 7 L 6 DL Live Load Placed on 6 Central Divisions of Arch Col. 1 Col. 2 Col. 3 Col. 4 Col. 5 X Load A - -.001 2 (2) • (3) Load and L Alone Alone A - — .001 .00 94.08 -4.03 90.05 90.19 .05 36.70 -4.85 31.85 31.78 .10 -15.47 -4.69 -20.16 -20.43 .15 -52.95 -3.68 -56.63 -57.03 .20 -70.11 -2.07 -72.18 -72.57 .25 -65.14 - .14 -65.28 -65.54 .30 -39.56 1.84 -37.72 -37.77 .35 2.60 3.63 6.23 6.43 .40 45.27 5.04 50.31 50.73 .45 72.90 5.94 78.84 79.40 .50 82.44 6.25 88.69 89.31 Elastic ^ at crown - -.860 f t . Deflection theory 1 at crown - -1.374 f t . Table (3-8) 41 SUPERPOSITION OP A LOAD PRODUCING A MAXIMUM POSITIVE MOMENT AT THE CROWN WITH A NEGATIVE UNIFORM AXIAL DEFORMATION L « 200 f t . , f - 1, VARIABLE EI, EL^ => 80,000 kip f t - 2 , LL - l,/3= 5 L 3 DL Live Load Placed on 6 Central Divisions of Arch Col. 1 Col. 2 Col. 3 Col. 4 Col. 5 X L Load Alone A - -.001 Alone £(2) • (3) Load and X - -.001 .00 260.28 -5.69 254.59 254.41 .05 138.44 -5.74 132.70 132.56 .10 29.95 -5.20 24.75 24.61 .15 - 54.65 -4.09 - 58.74 - 58.88 .20 -103.76 -2.51 -106.27 -106.39 .25 -109.20 - .70 -109.90 -109.96 .30 - 72.41 1.04 - 71.37 - 71.33 .35 - 2.48 2.49 .01 .11 .40 69.05 3.53 72.58 72.72 .45 113.89 4.15 118.04 118.19 .50 129.08 4.36 133.44 133.59 Maximum elastic *j «• -.949 f t . ) ) at crown Maximum deflection theory V "* -1-450 f t . ) Table (3-9) 42 SUPERPOSITION OF A LOAD PRODUCING A MAXIMUM NEGATIVE MOMENT AT THE LEFT SUPPORT WITH A NEGATIVE UNIFORM AXIAL DEFORMATION L - 200 f t . , f - 1, CONSTANT EI = 40,000 kip f t . 2 , LL - 1, Q m 5 L 6 DL Live Load Placed on 8 Left Divisions of Arch Col. 1 Col. 2 Col. 3 Col. 4 Col. 5 X Load X - -.001 £ (2) + (3) Load and L Alone Alone A " -.001 .00 -114.17 -6.33 -120.50 -120.30 .05 - 63.59 -5.43 - 69.02 - 68.89 .10 - 15.48 -4.10 - 19.58 - 19.50 .15 24.82 -2.51 22.31 22.31 .20 53.86 - .84 53.02 52.96 .25 69.80 .77 70.57 70.46 .30 72.07 2.22 74.29 74.15 .35 61.02 3.43 64.45 64.31 .40 37.71 4.32 42.03 41.92 .45 9.87 4.88 14.75 14.69 .50 - 14.64 5.06 - 9.58 - 9.57 .55 - 34.37 4.88 - 29.49 - 29.43 .60 - 48.16 4.32 - 43.84 - 43.74 .65 - 55.12 3.43 - 51.69 - 51.58 .70 - 54.68 2.22 - 52.46 - 52.34 .75 - 46.63 .77 - 45.86 - 45.78 .80 - 31.28 - .84 - 32.12 - 32.08 .85 - 9.46 -2.51 - 11.97 - 11.98 .90 17.31 -4.10 13.21 13.17 .95 46.83 -5.43 41.40 41.31 1.00 76.13 -6.33 69.80 69.66 Notes This load also produces a maximum positive moment at the 1/4 point. This is the load studied in Example 1 of Ref. 2. Maximum elastic » -1.035 f t . ) ) at f - .30 Maximum deflection theory « -1.539 f t . ) Table (3-10) SUPERPOSITION OF A LOAD PRODUCING A MAXIMUM NEGATIVE MOMENT AT THE LEFT SUPPORT WITH A SETTLEMENT OF THE LEFT SUPPORT L . 100 f t . , f - i , VARIABLE EI, E I a v «. 40,000 kip. f t . 2 , LL ~ 1,0 - 5 L J DL 4 Live Load Placed on 9 Left Divisions of Arch Col. 1 Col. 2 Col, 3 Col. 4 Col. 5 X Load 6v L - -.20 f t . £ < 2 ) + <3> Load and L Alone Alone SVL - -.20 f t . .00 -243.26 - .59 -243.85 -244.14 .05 -175.62 .67 -174.95 -175.20 .10 - 99.96 1.84 - 98.12 - 98.32 .15 - 24.36 2.78 - 21.58 - 21.70 .20 40.63 3.34 43.97 43.92 .25 84.93 3.40 88.33 88.35 .30 103.78 3.04 106.82 106.86 .35 98.62 2.41 101.03 101.11 .40 73.42 1.66 75.08 75.17 .45 32.58 .84 33.42 33.53 .50 - 12.34 .01 - 12.33 - 12.21 .55 - 50.20 - .82 - 51.02 - 50.91 .60 - 77.73 -1.64 - 79.37 - 79.26 .65 - 91.73 -2.40 - 94.13 - 94.05 .70 - 89.08 -3.03 - 92.11 - 92.06 .75 - 67.47 -3.40 - 70.87 - 70.87 .80 - 26.95 -3.35 - 30.30 - 30.37 .85 27.60 -2.81 24.79 24.65 .90 88.03 -1.88 86.15 85.94 .95 146.64 - .71 145.93 145.67 1.00 198.01 .53 198.54 198.26 Notet This load also produces a maximum positive moment at x «= .30 L Maximum elastic - -.381 f t . ) ) at 2, . .30 Maximum deflection theory *} - -.821 f t . ) Table (3-11) 44 SUPERPOSITION OF A NEGATIVE ROTATION OF ONE WITH AN EQUAL POSITIVE ROTATION OF THE OTHER SUPPORT L » 200 f t . , f - 1, CONSTANT EI - 40,000 kip. f t . 2 , 0 -7 L 6 Uniform Load Col. 1 Col. 2 Col. 3 Col. 4 Col. 5 X OJL - -.003 (&)R - .003 2(2) • (3) iaJL - -.063 L U)R » .003 .00 - .455 -3.612 -4.067 -4.076 .05 1.290 -2.796 -1.506 -1.521 .10 2.733 -1.529 1.204 1.185 .15 3.674 - .087 3.587 3.568 .20 4.032 1.282 5.314 5.298 .25 3.825 2.384 6.209 6.200 .30 3.139 3.095 6.234 6.232 .35 2.104 3.352 5.456 5.464 .40 .871 3.153 4.024 4.039 .45 - .409 2.539 2.130 2.153 .50 -1.589 1.589 .000 .028 .55 -2.539 .409 -2.130 -2.098 .60 -3.153 - .871 -4.024 -3.990 .65 -3.352 -2.104 -5.456 -5.426 .70 -3.095 -3.139 -6.234 -6.211 .75 -2.384 -3.825 -6.209 -6.197 .80 -1.282 -4.032 -5.314 -5.315 .85 .087 -3.674 -3.587 -3.602 .90 1.529 -2.733 -1.204 -1.232 .95 2.796 -1.290 1.506 1.469 1.00 3.612 .455 4.067 4.029 Table (3-12) 45 SUPERPOSITION OF A LOAD PRODUCING A MAXIMUM NEGATIVE MOMENT AT THE LEFT SUPPORT WITH A ROTATION OF THE LEFT SUPPORT L m 200 f t . , f « 1, CONSTANT EI - 40,000 kip. f t . 2 , LL - 1, /3 - 5 L 6 DL Live Load Placed on 8 Left Divisions of Arch Col. 1 Col. 2 Col. 3 Col. 4 Col. 5 X Load cVL - -.003 X<2) • (3) Load and L Alone Alone auL°> -.003 .00 -114.17 2.95 - 111.22 -111.56 .05 - 63.59 3.18 - 60.41 - 60.65 .10 - 15.48 3.11 - 12.37 - 12.49 .15 24.82 2.80 27.62 27.61 .20 53.86 2.32 56.18 56.26 .25 69.80 1.73 71.53 71.67 .30 72.07 1.08 73.15 73.32 .35 61.02 .43 61.45 61.64 .40 37.71 - .18 37.53 37.71 .45 9.87 - .70 9.17 9.31 .50 - 14.64 -1.12 - 15.76 - 15.66 .55 - 34.37 -1.41 - 35.78 - 35.73 .60 - 48.16 -1.55 - 49.71 - 49.72 .65 - 55.12 -1.54 - 56.66 - 56.70 .70 - 54.68 -1.36 - 56.04 - 56.10 .75 - 46.63 -1.02 - 47.65 - 47.73 .80 - 31.28 - .54 - 31.82 - 31.90 .85 - 9.46 .05 - 9.41 - 9.48 .90 17.31 .71 18.02 17.98 .95 46.83 1.38 48.21 48.18 1.00 76.13 1.99 78.12 78.10 Note: This load also produces a maximum positive moment at the 1/4 point. This is the load studied in Example 1 of Ref. 2. Maximum elastic f » -.861 ) ) at £ - .25 Maximum deflection theory V " -1.364 f t . ) Table (3-13) 46 CHAPTER IV NUMERICAL DATA Part 1 Tables of Magnification Factors Curves of magnification factors due to axial deformation and support movements are given. The elastic moments due to each effect are also included. The magnification factors due to axial deformation were obtained by subjecting the arch to a small negative uniform axial deformation and applying a uniform vertical load to obtain the desired value of /3 „ the thrust due to the axial deformation being neglected in the determination of /3 . It was shown in Chapter H I that these moments and magnification factors are applicable to any effect which changes or tends to change the span length of the arch. The elastic moments are exact for any condition which produces a change in horizontal length of the arch of S H •=» XL. The elastic moments and magnification factors are symmetrical thus the values for one-half of the arch need only be given. The moments and magnification factors due to a vertical movement of a support were obtained by subjecting the arch to small vertical movement of one support and applying a uniform load to give the desired value of 13 . The non-linearity discussed in Chapter III was neglected. The average of the magnification factors for corresponding points on each half of the arch are given for one half of the arch. The elastic moments for corresponding points on each half of the arch are equal and of opposite 47 sign. It was shown in Chapter III that these moments and magnification factors are applicable to any effect which produces or tends to produce unequal but uniform vertical deflections of the two halves of the arch. The moments and magnification factors due to a rotation of the l e f t support were obtained by subjecting the arch to a small negative rotation of the lef t support and applying a uniform load to give the desired value of /3 , the thrust due to the rotation being neglected in the determination of ft . Variation in Rise-Span Ratio Magnification factors are given for four rise-span ratios. For other rise-span ratios interpolation is necessary. This may be done by plotting 0 versus ^. However, since the difference in 0 i s small between consecutive ^ ratios, a straight line interpolation w i l l result in a very small error. Variation in Distribution of EI The magnification factors have been determined for two distributions of EI B a constant EI and a variation of EI as given in Figure ( 4 - l ) . In arches with a variable EI the average EI tconsidered along the horizontal axis of the arch, has been used in the determination of Q . For low values of ft the tables show that the change in the distribution of EI has only a small effect on the value of 0. For larger values of ft the difference in 0 is larger. With judgment the designer may interpolate between the two for other distributions of EI. 48 3.0-2.0-£7 i.o 2.7 OO .00 I. 7 Variable El 1.2 In-constant El 0.8 0.6 .10 20 .30 .40 L .50 Distribution of EI in Tables Figure (4-1) Variation of F l e x i b i l i t y Factor Magnification factors are given for/3 a 3,13 a 5, and, in some cases, for 13 «• 7. The magnification factor for 13 •» 0 i s , of course, 1, For other values of $ interpolation is necessary. This may be done by plotting 0 versus (3 or 13 2 . It is found, in most cases, for 0>1 that the curve of 0 versus Q may be approximated by the curve 49 0 - sec k/3 Where 0 <1 i t i s found that the magnification factors may usually be approximated by the curve 0 - 2 - sec k/3 In both cases k is a dimensionless variable which is nearly constant at each point in the arch. These two curves are symmetrical about the 0 - 1 axis for equal values of k. Hence these curves need only be plotted on one side of the axis. Plots of these curves for various values of k are given in Figure (4-2). Values of 0 for 0)1 are given to the l e f t of the values of 0 for 0 (1. MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER THE ACTION OF A UNIFORM AXIAL DEFORMATION X X MEL XEI 9 for 0 - 3 0 for 0 - 5 0 for/3 - 7 L Const. EI Var. EI Const. EI Var. EI Const, EI Var. EI Const. EI Var. EI — i | OO 0 M-> .00 .05 .10 .15 .20 .25 .30 .35 .40 .45 .50 56.29 39.96 25.35 12,46 1.29 - 8.17 -15.90 -21.92 -26.22 -28.80 -29.66 84.70 63.68 44.86 28.26 13.87 1.70 - 8.26 -16.01 -21.54 -24.86 -25.97 .94 1.00 1.05 1.14 2.19 .88 .99 1.02 1.04 1.05 1.06 .92 .96 1.01 1.07 1.17 2.11 .89 1.02 1.06 1.08 1.09 .80 .96 1.13 1.43 5.04 .56 .92 1.05 1.12 1.16 1.17 .71 .83 .98 1.19 1.56 5.05 .55 1.02 1.18 1.26 1.29 .53 .85 1.28 2.05 11.49 - .22 .74 l . l l 1.31 1.41 1.45 .25 .50 .86 1.43 2.54 13.26 - .55 .95 1.51 1.79 1.88 B .00 .05 .10 .15 .20 .25 .30 .35 .40 .45 .50 40.18 28.40 17.87 8.57 .51 - 6.31 -11.89 -16.23 -19.33 -21.19 -21.81 60.76 45.60 32.02 20.05 9.67 .89 - 6.30 -11.88 -15.88 -18.27 -19.07 .93 .99 1.03 1.15 3.17 .89 .99 1.02 1.04 1.05 1.05 .91 .95 1.00 1.07 1.18 2.57 .90 1.02 1.06 1.08 1.08 .79 .96 1.15 1.47 8.27 .61 .94 1.06 1.12 1.15 1.16 .69 .82 .98 1.20 1.60 6.66 .58 1.03 1.19 1.26 1.29 .50 .85 1.31 2.15 20.26 - .11 .78 1.12 1.30 1.40 1.43 .21 .47 .86 1.47 2.68 18.53 - .47 .98 1.54 1.80 1.89 Table <4-l) MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER ACTION OF A UNIFORM AXIAL DEFORMATION A i I L .00 .05 .10 .15 .20 .25 .30 .35 .40 .45 .50 .00 .05 .10 .15 .20 .25 .30 .35 .40 .45 .50 MRL ? CET 0 for B - 3 0 for 0 - 5 0 for 0 - 7 Const. E I Var. E I Const. E I Var. E I Const. E I Var. E I Const. E I Var. E I 23.89 36.46 .92 .89 .76 .65 .41 .05 16.71 27.24 1.00 .95 .97 .80 .86 .38 10.29 18.98 1.07 1.01 1.19 1.00 1.44 .88 4.63 12.70- 1.17 1.08 1.56 1.26 2.48 1.66 - .28 5.39 -1.19 1.21 -6.62 1.73 -21.36 3.25 - 4.43 .05 .93 17.60 .73 63.05 .12 219.27 - 7.83 - 4.32 1.00 .92 .98 .68 .87 - .35 -10.47 - 7.72 1.02 1.02 1.07 1.06 1.17 1.07 -12.36 -10.15 1.03 1.06 1.11 1.20 1.32 1.64 -13.49 -11.60 1.04 1.07 1.13 1.26 1.39 1.90 -13.87 -12.09 1.04 1.08 1.14 1.28 1.42 1.98 15.89 24.43 .91 .87 .71 .58 10.99 18.16 1.00 .94 1.00 .79 6.61 12.55 1.08 1.02 1.26 1.04 2.75 7.59 1.19 1.11 1.67 1.35 - .60 3.30 .49 1.24 --.98 1.90 - 3.43 - .33 .98 - .54 .88 - 5.26 - 5.75 - 3.30 1.02 .96 1.04 .79 - 7.55 - 5.61 1.03 1.04 1.09 1.11 - 8.84 - 7.26 1.03 1.06 1.10 1.22 - 9.62 - 8.25 1.03 1.07 1.10 1.26 - 9.87 - 8.58 1.03 1.07 1.10 1.27 Table (4-2) MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER THE ACTION OF A VERTICAL MOVEMENT OF A SUPPORT cSy MEL2 X <5yLEI 0 for ft - 3 0 for (3 = 5 0 for /3 «, 7 L Const. EI Var. EI Const. EI Var. EI Const. EI Var. EI Const. EI Var. EI .00 -5.613 -6.429 .84 .72 .48 .07 - .55 -2.33 .05 -5.052 -5.786 .92 .80 .71 .29 - .03 -1.85 .10 -4.490 -5.144 .99 .88 .94 .55 .63 -1.10 .15 -3.929 -4.500 1.05 .97 1.15 .85 1,37 - .02 .20 -3.368 -3.858 1.10 1,06 1.35 1.18 2.14 1.39 a .25 -2.806 -3.215 1.14 1.14 1.52 1.50 2.89 3.02 .30 -2.245 -2.572 1.18 1.20 1.67 1.78 3.56 4.67 .35 -1.684 -1.929 1.20 1.25 1.78 2.02 4.13 6.15 .40 -1.122 -1.286 1.22 1.29 1.87 2.19 4.56 7.34 .45 - .562 - .643 •1.23 1.31 1.91 2.29 4.81 8.09 .50 .000 .000 .00 .00 .00 .00 .00 .00 .00 -5.352 -6.190 .84 .72 .48 .04 - .67 -2.82 .05 -4.817 -5.572 .93 .80 .72 .28 - .10 -2.29 .10 -4.282 -4.952 1.00 .89 .96 .57 .63 -1.42 .15 -3.747 -4.334 1.06 .98 1.17 .87 1.45 - .14 .20 -3.212 -3.714 1.10 1.07 1.36 1.20 2.28 1.53 .25 -2.676 -3.095 1.14 1.14 1.52 1.53 3.08 3.45 a .30 -2.141 -2.476 1.17 1.20 1.64 1.80 3.77 5.33 .35 -1.606 -1.857 1.20 1.25 1.74 2.03 4.35 7.04 .40 -1.070 -1.238 1.20 1.28 1.81 2.19 4.75 8.35 .45 - .535 - .619 1.21 1.30 1.85 2.29 5.02 9.19 .50 .000 .000 .00 .00 .00 .00 .00 .00 Table <4-3) MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER THE ACTION OF A VERTICAL MOVEMENT OF A SUPPORT 8 V L2 X T " 5v LEi 0 for Ii - 3 0 for (3 = 5 Li Const. EI Var. EI Const. EI Var. EI Const. EI Var. EI .00 -4.776 -5.639 .85 .71 .48 - .03 .05 -4.298 -5.075 .94 .81 .77 .24 .10 -3.821 -4.511 1.02 .90 1.02 .56 .15 -3.344 -3.948 1.06 1.00 1.21 .91 .1 df> .20 -2.866 -3.384 1.09 1.08 1.36 1.27 • .25 -2.388 -2.820 1.11 1.15 1.45 1.59 V 11.1. 1 .30 -1.910 -2.256 1.12 1.19 1.51 1.85 .35 -1.433 -1.692 1.12 1.22 1.55 2.03 .40 - .955 -1.128 1.12 1.24 1.57 2.16 .45 - .478 - .564 1.12 1.24 1.58 2.22 .50 .000 .000 .00 .00 .00 ..00 .00 -4.218 -5.075 .86 .71 .53 - .14 .05 -3.796 -4.568 .98 .82 .88 .19 .10 . -3.374 -4.060 1.04 .93 1.12 .57 .15 " -2.952 -3.552 1.07 1.02 1.24 .98 .20 -2.530 -3.045 1.06 1.10 1.27 1.37 B .25 -2.109 -2.538 1.04 1.14 1.24 1.67 H l u l l .30 -1.687 -2.030 1.02 1.16 1.17 1.93 .35 -1.266 -1.522 .99 1.16 1.09 1.98 .40 - .844 -1.015 .96 1.16 1.02 2.03 .45 - .422 - .508 .95 1.15 .97 2.05 .50 .000 .000 .00 .00 .00 .00 Table (4-4) MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER THE ACTION OF A ROTATION OF THE LEFT SUPPORT <0L MEL 3 - 7 X fc)LEI 0 for fi - 3 0 f or IS = 5 0 for L Const. FI Var. EI Const. EI Var. EI Const. EI Var. EI Const. EI Var. EI .00 -8.377 -12.064 .89 .85 .66 .52 . .09 - .50 .05 -6.759 -9.731 .95 .89 .84 .64 .45 - .27 .10 -5.282 -7.610 1.01 .95 1.02 .80 .92 .11 .15 -3.946 -5.700 1.07 1.01 1.21 1.01 1.51 .73 .20 -2.751 -4.002 1.13 1.09 1.43 1.28 2.27 1.70 .25 -1.696 -2.516 1.20 1.18 1.72 1.62 3.41 3.21 .30 - .782 -1.242 1.36 1.33 2.33 2.22 5.94 6.22 .35 - .009 - .179 21.22 2.53 79.82 7.11 333.57 32.65 .40 .623 .672 .89 .84 .49 .25 - 1.75 - 4.31 .45 1.115 1.311 1.04 1.05 1.07 1.11 .76 .41 r-<|CC .50 1.466 1.738 1.09 1.11 1.30 1.40 1.79 2.17 P .55 1.676 1.954 1.12 1.15 1.35 1.55 2.38 3.20 .60 1.746 1.958 1.13 1.16 1.49 1.62 2.76 3.80 .65 1.675 1.750 1.14 1.16 1.52 1.62 2.98 4.07 .70 1.463 1.330 1.14 1.14 1.52 1.55 3.08 4.02 .75 1.110 .698 1.13 1.07 1.48 1.31 3.02 3.38 .80 .617 - .145 1.09 1.68 1.36 3.38 2.73 8.45 .85 - .017 -1.200 3.15 1.18 8.38 1.68 18.66 3.92 .90 - .792 -2.467 1.15 1.11 1.56 1.44 3.03 3.12 .95 -1.708 -3.945 1.10 1.07 1.38 1.29 2.47 2.53 1.00 -2.764 -5.635 1.06 1.03 1.23 1.16 1.99 2.06 Table (4-5) U l MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER THE ACTION OF A ROTATION OF THE LEFT SUPPORT CtJi MEL X <tfLEI 0 for p - 3 0 for fi - 5 0 for /3 - 7 L Const. EI Var. EI Const. EI Var. EI Const. EI Var. EI Cons t. EI Var. EI .00 -7,952 -11.536 .88 .83 .62 .46 .10 - .80 .05 -6.410 --9.302 .95 .89 .82 .61 .33 - .54 .10 -5.006 - 7.270 1.02 .95 1.04 .79 .91 - .08 .15 -3.734 - 5.442 1.08 1.02 1.25 1.03 1.64 .66 .20 -2.595 - 3.816 1.14 1.10 1.49 1.32 2.59 1.86 .25 -1.591 - 2.392 1.23 1.20 1.81 1.70 4.02 3.76 .30 - .720 - 1.171 1.40 1.35 2.50 2.36 7.26 7.60 .35 .016 - .153 -10.35 2.77 -54.54 8.42 -219.21 45.58 .40 .618 ,663 .89 .85 .48 .22 - 2.35 - 5.62 .45 1.087 1.276 1.04 1.05 1.08 1.11 .63 .10 .50 1.422 1.687 1.09 1.12 1.32 1.43 1.86 2.30 8 twM .55 1.622 1.896 1.13 1.16 1.46 1.59 2.61 3.63 .60 1.689 1.901 1.14 1.17 1.54 1.67 3.11 4.46 .65 1.622 1.704 1.15 1.17 1.58 1.69 3.45 4.93 .70 1.420 1.305 1.15 1.15 1.59 1.63 3.63 4.98 .75 1.085 .703 1.14 1.09 1.57 1.41 3.63 4.50 .80 .616 - .101 1.11 1.87 1.47 4.07 3.48 9.94 .85 .013 - 1.108 - 1.38 1.20 - 6.31 1.76 - 11.01 4.75 .90 - .724 - 2.318 1.18 1.13 1.64 1.51 3.53 3.80 .95 -1.595 - 3.730 1.12 1.08 1.45 1.34 2.92 3.09 1.00 -2.600 - 5.345 1.07 1.04 1.28 1.20 2.31 2.50 Table (4-6) ON MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER THE ACTION OF A ROTATION OF THE LEFT SUPPORT (t)-. MEL 4>7ET 0 for ft " 3 0 for (3 - 5 0 for/3 =7 L Const. E I Var. E I Const. E I Var. E I Const. E I Var. E I Const. E I .00 -7.038 -10.359 .84 .79 .46 .29 - 1.06 .05 -5.664 - 8.346 .94 .86 .77 .49 .38 .10 -4.410 - 6.514 1.03 .95 1.08 .75 .73 .15 -3.275 - 4.865 1.11 1.04 1.39 1.07 2.28 ,20 -2.260 - 3.398 1.19 1.14 1.71 1.46 4.38 .25 -1.364 - 2.113 1.29 1.24 2.16 1.96 7.62 .30 - .588 - 1.011 1.51 1.42 3.18 2.83 15.78 .35 .068 - .091 -1.65 4.00 -11.52 15.18 -102.25 .40 .606 .647 .90 .86 .38 .09 - 6.41 -<l<t .45 1.024 1.202 1.05 1.06 1.09 1.12 .48 .50 1.322 1.576 1.11 1.14 1.39 1.51 2.15 .55 1.502 1.766 1.15 1.17 1.57 1.72 3.84 .60 1.561 1.775 1.17 1.20 1.69 1.84 5.14 .65 1.501 1.601 1.19 1.21 1-.78 1,91 6.19 ,70 1.322 1.245 1.20 1.20 1.84 1.91 7.04 ,75 1.024 .706 1.20 1.15 1.87 1.77 7.72 .80 ..605 - .014 1.18 5.36 1.84 14.22 8.28 .85 .068 - .917 .92 1.25 1.11 2.02 13.62 .90 - .589 - 2.003 1.22 1.17 1.89 1.74 6.31 .95 -1.366 - 3.270 1.16 1.12 1.68 1.52 5.76 1.00 -2.261 - 4.720 1.10 1.07 1.44 1.33 4.44 Table (4-7) MAGNIFICATION FACTORS FOR PARABOLIC HINGELESS ARCHES UNDER THE ACTION OF A ROTATION OF THE LEFT SUPPORT (eJL x L MEL Const. EI Var. EI 0 f or Ii - 3 Const. EI Var. EI 0 for /9 - 5 Const. EI Var. EI •Ho 8 .00 -6.179 -9.217 .05 -4.962 -7.416 .10 -3.851 -5.778 .15 -2.846 -4.303 .20 -1.946 -2.990 .25 -1.153 -1.841 .30 - .465 - .854 .35 .116 - .030 .40 .592 .630 .45 .962 1.128 .50 1.226 1.464 .55 1.384 1.636 .60 1.436 1.645 .65 1.382 1.492 .70 1.222 1.176 .75 .956 .697 .80 .585 .055 .85 .107 - .750 .90 - ,476 -1.718 .95 -1.166 -2.848 1.00 -1.961 -4.142 .77 .93 1.06 1.17 1.27 1.40 1.71 . .64 .91 1.07 1.14 1.18 1.22 1.25 1.27 1.29 1.30 1.28 1.29 1.24 1.15 .72 .83 .95 1.07 1.19 1.32 1.55 10.19 .87 1.08 1.16 1.21 1.24 1.26 1.27 1.25 .60 1.32 1.24 1.17 1.10 .17 .65 1.15 1.65 2.18 2.89 4.76 .8.81 .11 1.08 1.50 1.78 2.00 2.19 2.37 2.52 2.64 3.05 2.36 2.19 1.81 . .04 .26 . .66 1.14 1.73 2.48 3.84 57.64 . .26 1.10 1.64 1.95 2.19 2.36 2.48 2.50 1.25 2.51 2.21 1.91 1.61 Table (4-8) CO 59 Part 2 Deflection Theory Moment Curves For convenience, curves of the deflection theory moments are also included. In a l l cases they are the moments which were found in the determination of the magnification factors. In the case of uniform axial deformation and vertical movement of a support the moments determined with a constant EI are plotted to the le f t of the centreline and the moments with the variable EI are plotted to the right. In the case of the rotation of the l e f t support, the moments for constant and variable EI are plotted on separate sheets. In a l l cases the horizontal scale i s 1" => .10 L. • urn t+PF :|±ix TP tit -FFH SHE i - i - H H - h -t i . c r ttit •nit ±t+ r 1 - i ; TXT: x r i x -pp-t -l-!-r' tin St i t i t S t © i l l ii:!3t IRISES. .11-x n x t TT-t tttfi P±F± Part 3 Influence Lines for Elastic Thrust The influence lines of the elastic thrust of the eight arches studied in this thesis are given in Table (4-9). The elastic 2 thrust due to rotation "^ is given in the last line of the table. INFLUENCE LINES FOR ELASTIC THRUST f -L I 8 f -L 1 6 f -L 1 4 f -L 1 3 L Const. EI Var. EI Const. EI Var. EI Const. EI Var. EI Const. EI Var. EI .025 ..033303 ..019014 .025377 .014574 ..017502 .010170 .013558 .007955 .075 .155391 .097737 .118131 .074700 .081089 .051832 .062549 .040351 .125 .367761 .259174 .278746 .197370 .190166 .135959 .145871 .105182 .175 .632067 .500280 .477554 .379426 .323609 .259180 .246677 .199009 .225 .915796 .804759 .689856 .607817 .464451 .411549 .351837 .313434 .275 1.191869 1.139876 .895451 .857639 .599355 .575904 .451410 .435120 .325 1.438289 1.463575 1.078212 1.097717 .718115 .731960 .538132 .549182 .375 1.637834 1.738404 1.225720 1.300752 .813194 .862685 .606919 .643708 .425 1.777814 1.936749 1.328945 1.44687 5 .879326 .956108 .654418 .710711 .475 1.849876 2.04043 2 1.382008 1.523130 .913193 1.004653 .678629 .745348 I. 10.000000 10.000000 7.500000 7.500000 5.000000 5.000000 3.750000 3.750000 HL2 -56.29 -84.70 -40.15 -60.77 -23.89 -36.46 -15.89 -24.43 Table (4-9) 78 CHAPTER V NUMERICAL EXAMPLES Part 1 Analysis of a Hingeless Arch To il l u s t r a t e the use of the tables presented in Chapter IV and the tables given in the appendix of Ref. 2, the moments and magnification factors at a support of the Rainbow Arch Bridge 3 w i l l be determined. The moments and magnification factors due to live load, a temperature drop of 60 degrees, a horizontal support movement of .20 feet, and a support settlement of 1.0 feet w i l l be found. This arch has a span of 950 feet and a rise of 150 feet. Each rib i s subject to a uniform live load of 1.38 kips per foot and a concentrated live load of 26.7 kips. The dead load of the arch i s not uniform; hence the arch i s not a true parabola. The greatest deviation of the arch axis from a parabola which passes through the crown and the supports i s approximately 1.75 feet. The arch supports 24 equally spaced columns which in turn support the deck of the bridge. For the purposes of analysis Hardesty et al considered the arch divided into 24 divisions of equal horizontal length, the mid-point of each division coinciding with a column. Since our tables are set up for the analysis of an arch divided into 20 divisions of equal horizontal length, we shall consider the arch as having 20 equally spaced columns and subject to the same live load. The live load per panel now becomes 1.38 x 2^ 9. ~ 65.5 kips 20 79 Since the shape of the arch i s quite close to that of a parabola the moments and magnification factors which have been determined for parabolic hingeless arches w i l l be used in the analysis. The table on Page 86 and the graph on Page 119, both of Ref. 2, show that live load should be placed on the 8 divisions of the arch closest to the support to produce a maximum negative moment at that support. The concentrated live load should be The moment of inertia at the mid-point of each division is given. The value of E i s , of course, constant and w i l l be taken as 30 x 10^ psi => 4.32 x 10° kip/ft, . The average moment of inertia along the horizontal axis is 94.5 f t . \ The distribution of EI of the arch is given in Figure (5-1). The distribution of EI which was used in the tables is also shown. placed at = ,175. 2..7Q 2.0 — /.70 t.50 1.0 Constant / f / - ^ .325 .80 .85 .60 0.0 .00 JO 20 .30 .40 .50 Distribution of EI Figure (5-1) 80 This figure shows that the variation from constant HI of the EI of the Rainbow Arch Bridge is approximately 1/4 as much as that for the variable EI used in the tables. For interpolation in the tables i t w i l l be assumed that the moments and magnification factors are composed of 75 percent of the value determined for a constant EI and 25 percent of the corresponding value found for the variable EI. The dead load thrust, as determined by Hardesty et a l , is 6,062 kips. This corresponds to a uniform dead load of 8.06 kips/ft. on a parabolic arch. Since the ^  ratio i s small (1/5,8) the thrust due to live load may be determined with sufficient accuracy by using Table (4-9) which is for parabolic arches. By a straight line interpolation for a rise-span ratio £ = ^ 0 and the. assumed distribution of EI ( an influence line for elastic thrust as given in Table (5-1) is obtained. .025 .075 .125 .175 .225 .275 .325 .375 .425 .475 .024 .115 .276 .484 .716 .949 1.140 1.347 1.459 1.520 .375 = 5.051 .025 Influence Line of Elastic Thrust Table (5-1) A live load of 65.5 kips/panel and a live load of 26.7 kips at » .175 produces a thrust of 5.051 x 65.5 • .484 x 26.7 » 344 kips. The total thrust becomes 6,062 + 344 = 6,406 kips. The f l e x i b i l i t y factor 81 for the arch under this load i s thus Q M /jfi£ . / 6406 ^  95Q2 ~ _ ° J EIav J 4.32 x 10° x 94.5 The influence line of elastic moments at the support due to load may be determined by a two way interpolation for a rise-span ratio — «• 150 and the assumed distribution of EI using tables on Pages 75 and 86 L 950 of Ref. 2. The ordinates of the influence line for the divisions which are loaded are given in Column 2 of Table (5-2). Live loads of 65.5 kips/panel and 26.7 kips at • .175 produce an elastic moment at the support of L -.3679 x 65.5 x 950 - 22,900 -.0716x26.7x950 - 1,800 24,700 f t . kips Col. 1 Col. 2 Col. 3 Col. 4 M E MD L PL 0 PL .025 -.0203 1.03 -.0209 .075 -.0522 1.06 -.0554 .125 -.0685 1.09 -.0746 .175 -.0716 1.14 -.0816 .225 -.0645 1.16 -.0748 .275 -.0498 1.23 -.0614 .325 -.0308 1.30 -.0400 .375 -.0102 1.54 -.0157 -.3679 -.4244 Moments at x •=» .00 L Table (5-2) The magnification factors for a value of /3 ° 3.779 a rise-span ratio j - m A?0 flnd t h e a s s u t n e f j distribution of EI are obtained by 82 interpolation and the use of Figure (4-2). The tables on Pages 75 and 86 of Ref. 2 are again used. The magnification factors are given in Column 3 of Table (5-2). Multiplying the ordinates of the elastic influence line by these magnification factors gives the ordinates of the deflection theory influence line which are shown in Column 4 of Table (5-2). The deflection theory moment at the support due to the same load thus becomes .4244 x 65.5 x 950 - 26,400 .0816 x 26.7 x 950 - 2,100 28,500 The overall magnification factor of the elastic moment at a support due to this load i s thus 28.500 « 1.15 24,700 By straight line interpolation in Table (4-l) for the rise-span ratio and the assumed distribution of EI a value of M g L » 49.4 is AEI obtained for the elastic moment at a support due to a uniform axial deformation. Assuming a temperature drop of 60 degrees and a coefficient of expansion of .0000065 the value of A is -60 x .0000065 - -.00039. Hence the elastic moment at the support due to a temperature drop of 60 degrees is ME - 49.4 <= 49.4 -.00039 x 4.32 x 10 6 x 94.5 950 •=> 8,300 f t . kips The magnification factor for Q « 3.77, ~ « and the assumed distribution of EI is obtained by interpolation from Table (5-l) and with the aid of Figure (4-2). The deflection theory moment at a support due to the temperature drop becomes Mjj - x 0 - -8,300 x .87 « -7,200 f t . kips 83 A temperature drop of 60 degrees produces the same moments as the r i b shortening due to an axial stress of A E » .00039 x 30 x 10^ = 11,700 psi. It was shown in Chapter III that the moments due to axial deformation and horizontal support movements are practically identical. The elastic moment due to a horizontal outward support movement of .20 f t . is ME - -49.4 6 n E I L 2 - -49.4 FF20 x 4 .32 x 1Q 6 x 94.5 950 2 - -4,500 f t . kips The deflection theory moment is MD - ME x 0 - -4 ,500 x .87 - -3 ,900 f t . kips By straight line interpolation in Table (4-3) a value of MnL2 A"1 - ' m -5 .61 i s obtained for the elastic moment at a support due to the DyEI vertical movement of a support. The elastic moment due to a vertical settlement of 1.0 f t . of the far support i s ME » - 5 . 6 15yEI L2 " - 5 * 6 1 LO x 4 .32 x 10 6 x 94.5 950 2 - -2,500 f t . kips By applying the proper magnification factor the deflection theory moment is obtained MQ - Mg x 0 - -2 ,500 x .78 - -2 ,000 f t . kips The moments at a support of the Rainbow Arch Bridge are summarized in Table (3-3) . 84 EFFECT M E 0 Uniform LL - 1.38 Cone. LL - 26.7k J -24,700 -28,500 1.15 Temp, drop =• 60° - 8,300 - 7,200 .87 H ° -20 - 4,500 - 3,900 .87 v - 1.0 - 2,500 - 2,000 .78 Moments at a Support of Rainbow Arch Bridge Table (3-3) 85 Part 2 Analysis of a Two Hinged Arch The use of the tables w i l l be further illustrated by the determination of the moments and magnification factors at the 1/4 point of the two hinged arch which is also discussed in Ref. 3. As before the arch has a span of 950 feet and a rise of 150 feet. Each ri b is subject to a uniform live load of 1.375 kips/ft. and a concentrated load of 26.7 kips. The coordinates of the arch axis are not given; however, in a l l calculations except the determination of the dead load thrust i t w i l l be assumed that the arch axis is a parabola. In the analysis by Hardesty et al the uniform live load was placed on the 10 divisions closest to one support of the 24 division arch. The concentrated live load was placed at •» .229. For our analysis the uniform live load w i l l be placed on the 8 divisions closest to one support of a 20 division arch. The concentrated live load w i l l be placed at j» 950 _£ - .225. The load per panel due to uniform live load becomes 1.375 x -rr- ra X* 65.4 kips/panel. The moment of inertia of the arch is given as 116.14 f t . ^ at the crown and 120.24 f t . 4 at the 1/4 point. It w i l l be assumed that the arch has a constant moment of inertia of 116.14 + 120.24 - 118.2 f t . 4 2 The dead load elastic thrust determined by Hardesty et al is 7,126 kips. This corresponds to a uniform dead load of 9.50 kips/ft. on a parabolic arch. The thrust due to the live load w i l l be determined by the use of Table (4-9). Interpolating for a rise-span ratio £ => , the ordinates of the influence line for elastic thrust as shown in Table (5-4) 86 are obtained. The thrust due to a load of 65.4 kips/panel and a concentrated L P .025 .075 .125 .299 | .375 .175 .225 2 - 5.122 .275 .959 1 .025 .325 .375 .425 .475 Influence Line of Elastic Thrust Table (5-4) load of 26.7 kips at - .225 i s 5.122 x 65.4 + .739 x 26.7 - 355 kips The total elastic thrust becomes 7,126 • 355 - 7,481 kips. Neglecting thrust due to rotation of the supports the f l e x i b i l i t y factor 3 due to this load i s /3 - / HL2__ - / 7.481 x 950 2 - 3.64 J E I a v J 4.32 x 10 6 x 118.2 To find the solution of a two hinged arch, having solutions for load and support rotation of a hingeless arch, i t i s necessary to combine rotations of the supports with the load so that the moments at the supports of the arch are equal to zero. The elastic solution w i l l be found f i r s t . By interpolation for £ - i|£ of the tables on Pages 75 and 86 of Ref. 2 the ordinates of the elastic influence lines of a hingeless arch 87 for moments at the two supports are obtained. These are given in Column 2 of Tables (5-5) and (5-7) respectively. The ordinates of an elastic influence line of a hingeless arch for moment at the 1/4 point is obtained by interpolation in tables on Pages 80 and 91 of Ref. 2. These are given in Column 2 of Table (5-6) The elastic moment at — - .00 in a hingeless arch due to the live load i s ME - -.3315 x 65.4 x 950 - -20,600 -.0565 x 26.7 x 950 - - 1.400 22,000 f t . kips The elastic moment at ~ ~ .25 due to the live load i s % - .1836 x 65.4 x 950 - 11,400 .0491 x 26.7 x 950 - 1.200 12,600 f t . kips Col. 1 Col. 2 Col. 3 Col. 4 X p M E n MD L PL V PL .025 -.0197 1.03 -.0203 .075 -.0497 1.06 -.0526 .125 -.0632 1.10 -.0695 .175 -.0642 1.13 -.0725 .225 -.0565 1.17 -.0661 .275 -.0430 1.20 -.0516 .325 -.0264 1.27 -.0335 .375 -.0088 1.48 -.0130 -.3315 -.3791 Moments at j- - .00 Table (5-5) Col. 1 Col. 2 Col. 3 Col. 4 2> ME 0 L P L P L . 0 2 5 . 0 0 1 2 1 . 2 8 . 0 0 1 5 . 0 7 5 . 0 0 6 0 1 . 2 6 . 0 0 7 6 . 1 2 5 . 0 1 5 6 1 . 2 4 . 0 1 9 4 . 1 7 5 . 0 3 0 0 1 . 2 1 . 0 3 6 3 . 2 2 5 . 0 4 9 1 1 . 1 5 . 0 5 6 5 . 2 7 5 . 0 4 7 8 1 . 1 7 . 0 5 5 9 . 3 2 5 . 0 2 5 8 1 . 2 5 . 0 3 2 2 . 3 7 5 . 0 0 8 1 1 . 5 0 . 0 1 2 2 . 1 8 3 6 . 2 2 1 6 Moments at — - . 2 5 Table ( 5 - 6 ) Col. 1 Col. 2 Col. 3 Col. 4 x p M E 0 & L P L P L . 0 2 5 . 0 0 1 6 1 . 1 3 . 0 0 1 8 . 0 7 5 . 0 0 7 0 1 . 1 4 . 0 0 8 0 . 1 2 5 . 0 1 6 0 1 . 1 5 . 0 1 8 4 . 1 7 5 . 0 2 6 1 1 . 1 6 . 0 3 0 3 . 2 2 5 . 0 3 5 5 1 . 1 7 . 0 4 1 5 . 2 7 5 . 0 4 2 7 1 . 1 7 . 0 5 0 0 . 3 2 5 . 0 4 6 7 1 . 1 7 . 0 5 4 6 . 3 7 5 . 0 4 6 9 1 . 1 6 . 0 5 4 4 . 2 2 2 5 . 2 5 9 0 Moments at - - 1 . 0 0 Table ( 5 - 7 ) 8 9 The elastic moment at — - 1.00 due to the live load i s L M E - .2225 x 65.4 x 950 - 13,800 .0355 x 26.7 x 950 900 14,700 f t . kips By interpolation for i . - 150 in Table (4-5) the elastic moments 950 due to support rotations of a hingeless arch may be determined. The elastic moment at a support due to a rotation of that support i s found to be M E - -8.04 <J>EI The elastic moment at the 1/4 point due to a rotation of the closest support is M E - -1.61 cam The elastic moment at a 1/4 point due to a rotation of the far support i s ME - 1 . 0 9 ^ E I The elastic moment at a support due to a rotation of the other support is The sum of the elastic moments at — - 1.00 must also equal zero; hence -2.64 £gl -8.04.2Jf!i • 14,700 - 0 Solution of these two simultaneous equations gives L -3,760 L uV R E I 3,050 L 90 The elastic moment at the 1/4 point of the two hinged arch i s thus Mg - 12,600 - 1.61 (-3,760) • 1.09 (3,050) - 22,000 f t . kips Hardesty et al found this moment to be 22,594 f t . kips By interpolation in Table (4-9) the thrust due to support rotation of a hingeless arch i s found to be H - -43.58 MM - -43.58 (-3,760 + 3,050) 950 - 35 kips The rotations at the two supports being of almost equal magnitude and of opposite sign results in a negligible thrust due to rotation. To obtain the deflection theory solution of the two hinged arch, the deflection theory moments due to load and support rotations of the hingeless arch must be determined. They are obtained by multiplying the elastic theory moments by a magnification factor corresponding to fi «• 3.66 and j- - ^-Q> -The magnification factors are found in the same tables as the corresponding elastic moments. The magnification factors are given in Column 3 of Tables ( 5 - 5 ) , (5-6) and ( 5 - 7 ) . The ordinates of the deflection theory influence lines due to load are shown in Column 4 of Tables ( 5 - 5 ) , (5-6) and ( 5 - 7 ) . The deflection theory moment at j- • .00 in a hingeless arch due to live load is MQ - - .3791 x 65.4 x 950 - 23,500 - .0661 x 26.7 x 950 - 1,700 25,200 f t . kips 9 1 The deflection theory moment at £ - .25 due to the live load i s MD - .2216 x 65.4 x 950 - 13.800 .0565 x 26.7 x 950 - 1.400 15,200 f t . kips The deflection theory moment at *• - 1.00 due to the live load i s Mp - .2590 x 65.4 x 950 - 16,100 .0415 x 26.7 x 950 - 1,000 17,100 f t . kips The deflection theory moments due to support rotations are found by multiplying the elastic theory moments by the correct magnification factors. They must be determined for — - and ft » 3 .66. The deflection theory moment at a support due to a rotation of that support is MD - .8 .04 x .83 - -6.65 4L£i L L The deflection theory moment at the 1/4 point due to a rotation of the closest support i s MD - -1.61 ^ Ei x 1.33 - -2.14 ^ The deflection theory moment at a 1/4 point due to a rotation of the far support i s MD - 1.09 ^ F i x 1.25 - 1.36 £EL The deflection theory moment at a support due to a rotation of the other support i s MD - - 2 . 6 4 x ^ x 1 . 1 2 - -2.96 ^Ei As before, the sum of the deflection theory moments at both supports must be equal to zero; hence -6.65 — ~ — - 2.96 — £ — - 25,200 - 0 9 2 and 2 . 9 6 — ~ 6 . 6 5 — | 1 7 , 1 0 0 - 0 Solution of these two simultaneous equations gives L ^ R E I - -6,160 - 5,310 L The deflection theory moment at 1/4 point of the two hinged arch thus becomes MQ - 1 5 , 2 0 0 - 2 . 1 4 ( - 6 , 1 6 0 ) • 1 . 3 6 ( 5 , 3 1 0 ) - 3 5 , 6 0 0 f t . kips Hardesty et al found this moment to be 3 7 , 1 2 3 f t . kips. The overall magnification factor due to this load is 0 - 2& - 3 5 . 6 0 0 - 1 . 6 2 " E 2 2 , 0 0 0 The overall magnification factor found by Hardesty et al is 0 - 3 7 . 1 2 3 - 1 . 6 4 2 2 , 5 9 4 By a similar analysis of the same arch with a hinge at one support the magnification factor for the maximum moment at the 1/4 point is found to be 0 - MD - 2 3 . 4 0 0 - 1 . 3 8 M E 1 7 , 0 0 0 If this were a hingeless arch,the magnification factor of the maximum moment at the 1/4 point i s 0 . - " 0 - 1 5 . 2 0 0 - 1 . 2 1 M E 1 2 , 6 0 0 93 BIBLIOGRAPHY Timoshenko, S.: "Strength of Materials", D. Van Nostrand Company, Inc., New York, (1955) Lee, R. W. M.: "Analysis of Flexible Hingeless Arch by an Influence Line Method". Thesis submitted in partial fulfillment of the requirements of the degree of Master of Applied Science in the University of British Columbia (1958). Pelton, T. E.: "Magnification Factors for Hingeless Arches". Thesis submitted in partial fulfillment of the requirements of the degree of Master of Applied Science in the University of Bri t i s h Columbia (1958). Hardesty, S., Garrelts, J . M., and Hedrick, I. G.: "Rainbow Arch Bridge Over Niagara Gorge". Transactions, American Society of C i v i l Engineers, vol. 110 (1945). 

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