Ctn4Y a· y JUN 23 1923 U8RA¶3 THES I S. AN ISNVESTIGATION AlND DEVELOPMEN-T OF A CERMTAIN SLOPE DEFLECTION MITHOD AND ITS APPLIZ CAT-ION TO RIGIDLY FRAMED STRUCI TUMRES. 61 Prepared BI A KALGC 1 B by SHIH. To T1: CLYka, vng eeriYg%5 lc5ge. for the F1ulfilment of M.S. Degree in the Department of Civil Engineering of the Massachusettes Institue of Teclhnology, Mass. JiTue, 1922. F, SJue 1, 1921. Prof. A. L. Merrill, Secretary of Faculty, Massachusetts Institute of Technology. Dear Sir; In accordance with the requirement for the degree of Master of Sience in Civil Engineering, I herewith submit my thesis entitled " The investigation and deve.lopment of a certain slope-deflection method for designing rioxidly framed structures." Respectfully submitted, Talean Shih. .C) - Contents. Page Preface and acknowledgement. ........................... 1 Introduction. 1. A review of previous methods on rigidly framed structure. ............................ 2. General outline of the present method. 2 4 .............. 3. Foundamental principals and assumptions of the present slope deflection method.............77 4. Method of getting primary sections of the frame for investigation. .............. 8 Part 1. Theories for getting the characteristics of a given rigidly framed structure. 1. The change of slope of a beam. 2. The change of '1ope of a column. ,..13 ........... .............. 18 ............ 20 3. Ascertaining of moment of inertias. momert factor 4. Cha;racteristic point of Contraflxureq,& 5. The procedure of computation. 20 .............. 29 ..... ., ..... .. 31 Appendix Part ll.Moments and stresses due to vertical loads. 1., Several short-cut methods for finding moments of rigid members 2. Separat•on of moments and their sigrs. ....... ..... .32 t....37 ....... 3. Deter~ination of shear and normal forces. ....... 40 4, Criteria for maximum combined stresses in beams and columns,. ............. 41 Part III, Moments and stresses due to horizontal loads, 1. Derivation of expressions for moments due to deflections. .f.....r......,4 2. Single-story frames. J. K)K)-1 ~3 ) V 5. Two-story frames. .................. 51 4. Three- and more- story frames. .................. 56 5. Secial cases. .................. 60 6. Abstract on theory of determinants. .................. 63 Part IV. Examples. Example 1. Single bent subjected to vertical laod. .......... 68 Check by least work method. Single bent subjected to horizontal load. Check by least work method. Example 2. Investigation of three story reinforced concrete ............. . ........... building. . . .. 71 Example 3. Method of application of theory of determinants...82 Conclusions. .. ** 4* *** 4*** *** *** ***** **.... 04444*. 84 1 Preface. This work is prepared for the fulfilment of M.S degree in the Institute. In doing it a time of about T~o hours has been spent, of which the most part was devoted in reading and researching. Pains have been taken to present the method in unique and simple manner and to developý all the necessary expressions, so that the easiness of the application of the method can be a~chieved. Examples are also given at the end of the work, they serve the purposes of illustrating the method of application, checking up the correctness of this method, and, at the same time, giving a brief comparison of this method with those existing. Oving to the limited time and the large amount of work involved, some points might have been overlooked, although great care has been taken by the writer. The writer wishes to acknowledge his indebtedness to Professor Charles M. Spofford for the inspiring instruction received from him as a student in his class on advanced structures. He par- ticularly wishes to express his appreciation of the valuable instruction and criticism on this work from Professor Hale Sutherland, under whom the writer has taken the course on advanced structural design. Kalean Shih May, 1922. t 1 2 INTRODUCTION 1. A review of previous methods on rigidly framed structure. It is believed that a.wide knowledge of the analysis of rigidly framed structure is generally required in gaining etconomies and in securing effective designs. Notwithstanding the importance, the close investigation of the stresses as they actually occur in this kind of structure is not usually attempted in practice. The reason is simply due to the fact that the existing methods for investigation requires either considerable time to work out formulas for statically indeterminate quantities to suit a particular case in question or laborious work to apply the different expressions already determined to get so many unknown quantities. In view of the ec4onomy of the time element, several very approximate methods are in general use, although they sometimes give results which are serious in error. The methods that have been so far developed can be classified as follow': a) The approximate methods. They are originated by various authorities ( Spofford, Fleming: Smith, Burt, Thayer, etc ). The methods consist. of certain combination. of the following assumptions -. : 1. Point of contraflexure of columns at mid-height. 2 I -Point of contraflexure of beams at mid-length. 3-. Direct stresses in columns are propottional to the distance of the column from neutral axis. 4. The shear on all interior columns is equal and the shear on each exterior column is equal to one half the shear on interior column. 5. Shear is distributed amongst columns in proportion to their moments of inertia. 6. Shear is distributed amongst columns in proportion to area of vertical rectangle of which the column is the axis, As the methods are common in practice and can be found in papers written by respective authors mentioned above, so they will not be given here in further detail. b) The more exact methods. These consistofthe methods based on slope and deflection originated by Dr. CA.Melick and Mr,. .F.Jonson. The original work is laborious and is almost inpractical for buildings several stories high. The further development of the slope and deflection method by W. M. Wilson and several other men ( bulletins 80 and 108, Eng. Exp, Sta of Univ; QOf Ill.) rendered this method generally applicable. been derived. Expressions for various cases have It is considered simple, although in some cases its application is restricted. ( See Conclution, Part 4). 4 c) The exact method. This method is a development by method of least work. The well-known papers regarding to this method are by Prof. A. Smith (Journal of Western Society of Engineers, Vo.l. XX ) and by Dr. Mikishi Abe , ( Bulletin 107-, Eng. Exp, Sta. of Univ. of IIi..) .. The method is exact and the direct stresses can be taken into account.: It is, of course:, very longj. 2. General outline of the present method. The method herein developed and investigated belongs to Class b, the more exact method, of the previous discussion, or, in other words, the slope deflection method treated in more unique manner. The preference of this me- thod will be discussed in the conclusion of Part IV of this work. The work is devided into 4 parts with an introduction at its beginning, in which, besides some general discussion, a section for getting primary sections of a reinforced concrete frame is also given, Several formulas derived from the writer"'s experience are presented, which, when being applied, needs only a single slide rule operation and are deemed simple and helpful. The first part of the work, Part I, is the basis of the wholed All t:he theories and derivations are included in it. They serve in the main to get the characteristics of a ggiven frame. By characteristics of a frameare here meant the following': I V5 Taking each member of the frame separately and assu-ing one end being fixed, while the other end being sub~ad arzv rc••cror.. jected to a unit moment, the expressions for the points of contraflexures of-the memberacan be obtained. The position of the points of contraflexure does not change for any mag- nitude of the accually applied moment. ( Characteristic I) Taking each joint of t'he 4,ame. searately and assuming a unit moment being applied to the joint. Now since it can be easily proven that in the strained position all the members meeting at one point are subjected to the same change of slope, the relation of transmission of the moment, or the moment factors, can be derived. The moment factor does not change for any magn~itude of the actually applied moment. ( Characteristic IT ) At the end of Part I a summary of the procedure for computation is also given. IMost of the theories given in this part are taken from Dr. Strassener's work appeared in the paper " Forscherarbeiten auf dem Gebiete des Eisenbetons, Hefte 26". The expressions for momeat factors for cases where four members meeti$3 at a joint are the writer's own developmen.t from the same principles' ; these complete the missing link of Dr. Strassener's work. The second part of this work, Part II,: treats the method of application of the development given in Part I to the case of a frame subjected to vertical loads. Section 2, the separation of moments and their signs, sho-ws the unique manner of this method, in which Tt lies its superiority. The two sections on several short-cut methods for finding moments of rigid members and Criteria for maximum combined Sstresses in beamse'and columns are collecteda rom various sources and seem to be helpful. The third part of this work, Part lI:,1 treats the method of application of the development gFiven in Part*I 4 loads., to the case of a frame subjected to xerwn4( On ac- count of the fact that the tops of all columns are subjected to a deflection in this case, expressions for moments due to unit deflection are derived. Now since the deflection and the resulted moments are always ý:in direct proportion to each other, advantage can be taken from this fact for deterLmining the moment of each member due to an ass~umed deflection unity of each story of the frame. From these moments the horizontal shear subjected to each column can be found. The sum of the shears of all the columns in each story should mat eaily be equal to the external force applied at that story., which causes the assumed unit deflection. by multiplying the mcment due to the unit deflection by the ratio Acetually appli.ed force External force causing unit defl. the acctual moment of each member can be obtained. Dr, Strassener gives some hint for attacking such problem in his paper mentioned above, but the method has not been treated Ay him in unique and simple manner. The present method is a collection of the developments of various engineers appearing in "Schweizerische Bauzeitung" Sand put up in simple and collective manner by the writer. It should be noted that this method is. eore simple than Wilson,'s method by reducing two equations from each story. The fourth part of the work, Part IV, giveza complete .illustration of the method and conclusions. 3. Foandamental principles and. assumptions of the present slope deflection method. The principles and the assumptions of the method in successive sections given in this work are almost the same as are given in tWilson 's paper ,. Bulletins80 and 108, Eng'g experimental station of University of Illinois. The trzeatment and the applications of the principles and4assumpt-ons are entirely different from those given in Wilson's work and the preference of the present method will be discussed in the conclusions vn P+ 4t-- . The principles and assumptions adopted cmn be summarized as follows": 4 1) The moment at an end of a member of a frame is a func- tion of the changes in the slopes of the tangents to the elastic curve of the member at its ends and of the deflection of one end of the member relative to the other end. 2): In the strained position, all the columns and girders which intersect at one point have been subjected to the same change in slope. Upon these two principles the Part 1 of this work is based'.. 3) The change in the length of a member due to direct strees is equal to zero. 4) The .horizontal deflection of the tops of all columns due to vertical load is equal to zero. Upon these two assumptions the part 11 of this work is based. 5) The horizontal deflections of the tops of all columns of a story due to horizontal load .I are equal.. Upon assumptions 3 and 5 the Part 111 of this work is based. Other minor assumptions are,: 6) The connections between the columns and girders are perfectly rigid. 7) The length of a girder is the distance between the neutral axes of the columns which it connects and the length of a column is the distance between the neutral axes of the ggirder which it connects.. 8) The deflec-tion of a member due to the internal shearing stresses is equal to zero.. 9) The wind load is resisted entirely by the rigid frame. 4:. Method of getting primar-ry sections of the frame for in.vestigation. In all the methods discussed in section 1, except the approximate ones, it is required to know all the sections of the members before investigation. This is, how- ever,,, the hardest task, which the designer has to encounter, if any exact computation is to this fact, it intended to make at all. Owning seems to the writer to be favorable to in- clude this section in this work, Now since the scope of s~trctIura1o. •oditiOcns is so wide that the degree of rigidity varies with all factors, like supports, joints, materials,, etc. , it is impossible to present .t I any method which can be applied to all cases. lowing discussion it is, In the fol- therefore, confined to reinforced concrete frames- which has absolute rigidity and the degree of accuracy of the new slope deflection met-hod herein presented can be warranted. The formulas given below are derived from the writer's experience and are based upon 1 ': 2': 4 concrete ( .fc fs = 16000 and n 650 . 15):, which is the mixture used in almost all casee of building frames.. formulas is = T hough the derivation of the so simple and t heeir frms do not differ much from ordinary ones, yet the merit of a single slide rule operation can hereby obtained. Formula I,, For rectangular beams. ( standard notations) d= -. 4L 1080b This formula is derived from ordinary chart for rectangular beams. For the specified stresses of 1 ' 2 : 4 n1 = 108 1---- and whence concrete bd2 Formula 2, For T,-beams, we have the formula. neutral axis lying in stem. d 35*b' in which b" is the assumed flange width, and according to the specification of Joint Committee a.) the beam.:, It should not exceed-I- of the span length of 4 :j b) Its overhanging width on either side of the web shall not exceed 6 times the thi'ckness of the slab. It e) s-hall not exceed the distance c-c of the beams. This formula is derived by examining a series of Tbeams together with the c-hartAgiven on page 364r-365 of 'Hool and Johnsont's Concrete Eng"'r Handbook. It is found that the design of T-beams is always limited by the s.tress of steel, while the stress of concrete is below its limit. generally far The latter stress ranges from 300 to 400 for -competent design with cost ratio r = 70 approximately. Now let draw a horizontal line on diagram 8, and J.'s page 364H-i.. book, which passes the curve fc= 350 and so place the line that half of the curve is above and half below it it will be found that this: horizontal line corresponds L - = 35 and whence we have the formula. It has been tested bd2 that this formula satisfifes most of the cases, especially for getting primary s:ectionss Formula 3. •.rt getting reinforcing steel of rectangular and T.beams.. As 5 This formula is obtained by substituting j -7 general one A= Formula 41. --- jdfs: For columns centrally loaded. into the P Ac Where Ac= the net area of cross section of the column, P= the centrally applied load and C a constant given in the table below, its value is different with different values of "p",, the percentage of re.inforceme-nt. f1 Ir 1 1"::2:4 (2000-.lb. concrete) n=15. fc=450 for ordinary lateral ties.* iO .! ý o.ca o.oz5 o.oC1o0oo o .o-o45i o5O The values of C in the above table are computed from the expression C=fc[ 1+(n-l)p] , the formula is a general one and requires no further discussion. For beams reinforced with compressive steel and columns subjected to eccentric loading, formulae given in various text books are supposed to be used. * Specification of Joint Committee. Part 1. THEORIES FOR GETTINRG THE CRARACTERIS'TICS OF A GIVEN RIGIDLY FRAMED STRUCTURE4 T-he foundamental theories from which this part of the work is based- are orighated by Prof. W. Ritter in his book " Anweandng der graphiache Statik"' During the last ten years research works have continuously been made in Germany to study the effect of rigidnity of any kind of framed strucature with the intention to obtain a method of so.lution, which should be practical in actual use and, at the same time, will give a degree of accuracy not far from the exactness. In year 1916 Dr. Strassener plblished a long article treating this subject .inForsc-herarbeiten in Gebiete des Elisenbetons" and since. then several articles regarding this work have appeared during the-: years 1917-1921 in Sch1 s work weizerische Bauze.itung' Tn both of Dr. Strassener' and the articles in Schweizerische Bauzeitung the expressions given for moment, factors are confined to joints where only three members meet. .I order to solve problems like rigidly framed buildings; cases alwafs occur where four f laca ADtrs lougbpt, er, A, ke in~Ine i r b ys of th L) u 3A din Il consequent# expressions for these, equations 10 to 14, are I developed by the writer. ! 1. The change of slope of a beam. Let- Ma= the moment appli'ed at the left end of a beam. Mb= " "" " " " " right " " ea= the slope angle ait the left end of a beam b =1 due to Ma = eb = the slope angle at the right end of a beam due to Ma = Mb = 1 $= the slope angle at left end due to Mb= 1, or the same at right end due to Ma= 1 and let x = the. distance of any section from left end of the beam in question and x" the same from right end, then the expressions for the slope angles are.: aa=·'4L *· J' x-' E*dxx I. xY b 1dx. r L2 2EI .. (1) EI. For beams of constant cross section, ea S L .... x'xadx p·•1- * and = =b and . ....... (la IET The derivation of equation (1) can be obtained very easily by following the discussions given in Morlefs Strength Ma:terials , pages 170o-179, of og by Prof. Swain's theory given in Transactions of A" S. C. T. 19184 SThe liit * of interati---------------n n equaton------ is ro The limit of integration in equation (1) is f-r-or --to to0 . r ~31 'I-i kb Er A L. A J#~)(a) dV! (3t 1de (@} (Amy .1m ZCL K= a t, z -[i Wki. ;. f *I: MP s F e, 3 I ! -- In case that the cross section of the beam is not constant and its moment of inertia varies, the integral symbols of the above expressions can be rrplaced by the symbol "E" and by looking the value dx as a part of the beam having a length of x. The solution of each expression can be done graphically without difficulty, In view of practical use, two tables are hereby re- produced from "Forscherarbeiten auf dem Gebiete des Eisenbetons, heft xxvi". In order to facilitate the use of the tables, the following equation is given': e-,2 = ," a El Equation (l b ) -. -b .(1b 6F1E is only applicable to symmetrical beams, which type is used in almost all cases in building construction'. The factors , and L are the slope 2EI, 6E I, angles 6 and 3 respectively for beams of constant moment of inertia I,, and the factors Ta and Pb are constants found either from table 1 or table 11 corresponding to the type of haunches that the beam has:. These constants depend also upon the following items of the beak: KL = length of haunch, and CtI C,= -- , d d-a ds a - or r . ..... . .. .... . . ..... .* . 1 for beams having constant width l (2 ) .. ~1118~08~18~2~B~-·d··r;-*i~~--.rr·· . , • rli :.· · ;·.-~ur;~--r Li- - ----- . ~-__1__1_- -- ------ - . Table I. For beams with straight haunches. 'Pa= upper vmlue, 'b= lower valuel I ..% - - --- ~----· - -----· ----------- ·"· ------------------- - C2 0 0.4 0.8 1.0 -- `~-----p~ .432 .518 61.2 1/3 1 "0 1.0 .742 .843 .621 .758 --':-----UY*-r·i 1/4 . .0 .0 .806 .907 .716 .856 1.4 IT.2 .,,• . "• . .375 .331 .295 .. 460 .41 4 .376 "" -C-~UII~------------.·.·~--11111~-, -. r~uC~--~k~--Llur .583 .729 · 1?·1-····-1-~--- .688 .839 554 705 ·-- r··l·· 3----- .665 .825 1.0 1.0 .845 939 .773 .905 .750 .894 .732 *884 .266 .344 ~l~·r~------YrrC1-··r .511 i 8.48 .813 .803 .706 .869 ~~--~-~~- j717T7J U7LZIF2 ý---4_ .633 -.--- j7TIZ77 3.0 .222 .156 .. 294 . .21 5 -4 .2I·lr··~-~.u·5·_ 4-9 .481 6"69 i-~ .718 .876 3.0 $~O .530 .'86 4 -- -5 1__-· ly$ 1.6 1.6·-W··-II·~ 4------- L. .438 .643 .64 .601 I .611 .. 578 .787 .761 .689 L.859 .663 S842 -.l·l...----~-·l-.sp~__~.II U7CLJ27J jc~ 01 Ob2 L . _ _..... ---- ·- . ... . ,. ... .4 1¾ 1.0 1.0 1/3 1.0 0.8 -- n 1.0 1.2 PA4 .545 .670 .507 .635 .475 .605 1.0 1.0 1C 1 .73 C .820 .592 I-~ .7tl-O .819 .913 1-·4111-·-----·1-~-· .728 .859 -~---~-·U·· .,696 8339 .865 .949 ..796 r772 .917 .905 -- .671 .821' I-- - .892 .. 967 .. 837 ..946 -w~ YCz3 ~--:6- .0-4 1 2 0 I - 3.0 .753 .895 .8T'8 .803 1 3P'9 .. 931 S.832 .793 1, .738 .886 r·· L.,604 .771 .L---·.L - I .724 .703 ..878 .865 --------·- I~ .790 .779 .925 .920 ~7IZEL7 .336 .459 ~--.-·---~-L-.I ,650 .806: i------~--)ll* .406 .535 .449 .5 79 i.~,__~-___~- 1.0 1.0 . - f·~- .........---... -U~Llii-~~C~ __ _1 .. ! t 1/4 _~I_,, ... "~'`~' ""i"~~;"----~"~` "--'~"i^'~`~-~"' ~-"--~~~~~-~~--~~'~'~-^~~'~~~I-"-"~~I ..... ~~~--...~~--...--.I.--~.. -. .-------~--~-------- ~--~ ~-·------------~--·--------- t 1/2 ... . Table II. For beams with parabolic haunches, •a= upper value, 'b = lower value. ·--------- · ·-- ·--C IC ~. !i .. .7862 .911 .557 .73-2 ·~--·I-~··· I .668 .842 --~---~------L--~I.734 .896 jy2IrIyj 2. The change of the slope of a column. Let , M = the moment applied at the top of a column, Md = " " foot " " %c = the change of slope at the top of a column due to Me = Md 1 ed = the change of slope at the foot of a colunr 1 due to Pc = = Md = 1 the change of slope at top due to Md = 1, or the change of slope at foot due to Mc = 1. The slope angles are expressed as (fig. 3 )': l SJ. xedx xd 1 =- and J dx'x dx -.-- h2 1 h X"`dx El (3 El It is generally that the columns are built as shown in fig. 4 and in that case h' d (2h-h') * ---- 2hEFI C H (3a) 2hEIl Sc= . (.-2. 6h 2 EI ) c wher h' is the clear height of the column and h is the distance between the foot of the column and the neutral axis of the adjourning beams With constant I = Ic throughout the column, then h = h', and e = 8c e-=--h-2O &JI dp hence and -= h 6EI3 e•• ,, •, The strengthening of the top part of the column due to beam haunches does not exert an influence so serious as the haunches to the beam itself;, however, the effect depends largely u.pon their,relative stiffness. With strong column and weak beams, i.,e. beams easily subjected to deformation:,, the effect is very small and h' will be almost equal to h. In the contrary, the whole strengthening effect should be taken into consideration to warrant the correctness of the design. * The derivation of equations (31, (3 ) and (3b ) can be in made without difficulty from the references given Section I. The limits of integration of equation (3a ) are from 0 to h'. 3 E.c..an of mome,nt o Q_. !As ne tia m I'n reinforced concrete buildings the moment of inertia for beams and girders is generally not the .same throughout the .e.vb:er,-it is therefore necessary to compute and use the least moment of inertia for a given member.. A chart for determini:ng that for T-sections. is reproduced here from the same book refered above, which can be used for Tcolumns and sometimes for beams and girders. The moment of inertia is expressed as ' • bI Whr "i~is . a c 'on:s 1 .* (4) ." ob.t. "-U Where u is a constant obtained from the chart. By characteristic point of contraflexure is meant the zero moment point in a rigid member resulted by applying a unith.moment at one end. By moment factor is here understood as a facter by which the moment applied at a rigid joint is to be multiplied so as to obtain the- actual portion of the moment resisted by each elastic member meeting at the joint,; Now let the charater-istic point of contra.flexure and its distance from either end of a beam or a column be denoted in the way as shown in fig'. 7a and let a = the c-hange of. slope at the left end of a beam due to Ma = If the change of sl.ope at the rfight end of a beam due to Mb , cc c the change of slope at the top of a column due 1, and to Me ad = the change of slope at the bottom of a column due to Md 1, 6 then from Fid. b, we have ab F. ... = XPRIWI qI I I f. d U. =.1!5 . e .4' .40 .435 .55 .50 • o0 i -4 0 .5 I5 dlo t • uLAxS •Zo *zS Crk fcfincns Norn~nV .o of ,45 lncr o' T-5econ .55 .50 r. 2b~ cS5- 22 The moment M=1 applied at B will produce a moment Ma at A. The magnitude may be obtained by simple proportion, whence, a L-a Now by definition ebis the slope angle produced by moments M=1 both at ends A and B (fig 7a_c_d_e 86c), then from figures it is evident that Ma ab =b a = eb- (1 + L-a SL-a With the same conception and draw similar figures we can derive expressions for ma, ac, and rd. a = and •aa L-b ab =b,- - L- - c C C h h-d L-a ad = 9 d- -h hac Thus we have 8, at left end of a beam at right end of a beam * f, . (C5) at top of a column c , at foot of a column The values of 8 and P are given by previous equations, and mostly, for beams either of constant cross section or of syfmmetrical in form, aa = ab = a . The values of a and b will be given later. To ascertain the moment factors the transmission of the moment should be carefully considered. In order to avoid confusion the following set s of notations are used: na = a fraction of the moment applied at a continuous column, which is taken by the ._.. nb = a ftaction of the moment applied at a continuous column, which is taken by the•_•l•UbJi._e n, =a fraction of the moment applied at a continuous beam, which is taken by the _he•...te_~ ig nI = a fraction of the moment applied at a continuous beam, which is taken by the_Bkpa•,_Jadt_,, nul= a fraction of the moment taken by a column, when the moment is transmitted to it from any member adjourning its ".pjz end. nlu = a fraction of the moment taken by a column, vhen the moment is transmitted to it from any member adjourning its L.oj_ end nrl= a fraction of moment taking by a beam,, when the moment is transmitted to it from any member adjourning its ri•ght end. n Ir= a fraction of moment taking bya-beam-, when the moment is transmitted to it from any member ad-ýjourning its lfi end. For clear understanding of thezse notations figure 16 given in the section on separation of moments will be refered. Let us denote the resulting slope angle common to two rigidly connected bars due to M = 1 by "y" with suffixes . -IC. 5- tc ii ' kM -( c I/ a'l '-(I COL) (c) F;9. T1 cIL or fc S~ry M Lr -rcol or C.. ('r~ (a) \I 'rr T 7~r7 ~4 .i r6 3, ' f'Z-'r: r or l; I \1\ ~--*--- (eb) / , -r , i/ b,, cý, and be representing the angles common to beams, columns, beam and column respectively. Now let the four members given in figjures 7 b andc be arranged as shown in figure 8, first consider the two vertical members, or columns, be connected rigidly and are continuous and a moment M = 1 is incurred at the joint, fig. 8 a, then the slope at the top of the lower column = nb = ( 1-na) a, and of the bottom of the upper column -na d = ( 1-nb) d By assumption II these two slopes should be equal to the common slope angle Yc at the joint, therefore have, ac na + ac 0d d. nb = and - **.... a c ad ac + d (6) we Similarily from fig. 8b we can derive Lb nl=- ab:+ &, Yb - U a• aa+b (7) I· -aB _ s~a+ag~ Again consider the case shown in fig.8c. we have, f-or a moment coming from column to beam, Ybc= nrlab= (1,nrl) "c and for amomemt coming from beam to column, Ybc = nulac = (lnnul) ab therefore., rl n b+i Lc+-b Ybc )= me 71 (b+ac 4.0· - ,** *I.-O.~li (Q\ In the simila:r way we can derive expressions fornnlr, nitetc ey examin'e.the equations (6), (7),and (8),-the following rule holds true': -If a moment M=1 is applied at the joint of two bars, x and y, then the portion of the moment which will come to bar x, at bar = , ,, , , .,. ... ; .. (9) i at bar x* a at bar y and the portion of the moment which will come to bar y n -a at bar x .'-T•• bar x+% at bar y By a it is understood the angle caused by a moment M=1 applied only to the bar in question. n *I Of at *.! at 01 *1 0./0 0 i'. (M~) (L\• I, ~il 77~T6 27 I; -i1S /// d 1 _-c C ---a-- A:Z7-......-7§S - - A- -i .................... - _,.-.5__F • "; i• IQI-----•<:_. .... ----- "' ,/-; 5 s• '•;•<:•: .......... -5; (• ,_• 1' So far we only have considered a combination of two *usbers, for cases as shown in fig. 8c and 8d, equation (9) can still be applied. By looking the two continuous members as a single member and according to assumption (2) we have, r n-r=- Yc+"b nul= - %b +Yc nla n Yb+ ec+Yb In the same way we can obtain the value of n for other comb ination of members.. To find the values of a, b, and c, d, distances of point of contraflexure from ends of beams and columns respectively, as shown in figures 9 and 10 ýgeneral expression is here first derived: Let p = change of slope of the support due to M=l1 and k = change of slope at ends of the member due to subjected load. Using other notations as used before,then refer to fig.(11) it is evident -paa= ka a+ Ga- ) + Mbb -b Mp=kl, + M (Ob-e ) + MaP If the member is only subjected to a moment, then oa + M (ea-) -PaM -P b = + MbB - + MbGOb- ) + Map (ilL :: Now since M=- a ••• b 6 ), by s•ubstitution can -ae find we can find (11) a+pa Similar1'-ily. for expressions of b, c, and d. From equation (11) and kee in mind that p is the angle made by the support due to M=1 the following expressions can be derived': FProm fig. (9) a= b"= b'8L _n _a _L_.L_ ea+nrlb, eb' +nlr a ' cA= h'Bc d= - he c,+nluad, d+nul d,.-,-***** (12) c Where nrl, nul, etc. are those given in equation (8) tif h" c ?, C.' a 'cI+ Yb L . . - * -* *** (13) a+ Ye From Fig. (10) Same expressions as equation (12) except ****(14) arl,nuletc. are those given inequation (10) For beams and columns of constant moment of inertia, the expressions in equations (5), (12) and (14) can be very much , simplified and in that case::,. .=L 6E nd aI, 8=-- = 29I Equation (5) c 3P, = h 6EIc and ec =h__ 3e 2E:Ic is simplified to '(a = p[ 3 Mc = ic[3- L 4] L-b h a b=p ] 3_ 'L ] L-a d = p-L[ 3- .. *** *******(15) .] h.-c h-d and all values of 8"s in equations(12), (13) and (14) can be substituted by 3-times their correspondingp 's. 5. The procedure of computation For investigating an existing building or a,building, of whose members the sections have already been approximately determined by the method given in the introductory, the distances of the points of contraflexure, a,b,c, and d from each end of the meymber have first to be found. In order to do this the following steps have to be taken: (1) Determine the moment of inertia of each member by the aid of the given chart, if necessary. (2) Determine the values of E and P of each member by the aid of equation (1) and (3) and by the given tables, if nacessary. (3) Beginning from the columns of the lowest storT, which are generally fixed or rigid. The distance d will be': Taking care of the strengthening effect of the column head, S3h-.2h' 2h-A' . h' . (16) 3 d=- h.- .**....* and when the same is neglected, (1T) 3 In the first case th:eangle at the top of the column, = h13 ----* 4EIc(3h[h-h']+h'2) and in the second, dc =-- 4EIc or C . - * .(. 3 2 ** (1-) (19 8c (For derivation of equations 16, 11, 18, andl9 see appendix.) In order to proceed the computation of angle'%y' column, it is nacessary first to determine of each ' (equation 6 ) "s in determining the latter , the distsnce"c"o the next upper column has to be first assumed (eqauation 5 ). c=•-.h is in 4 most of the cases close enough. With this assumption and the values of Yec thus found we can determine-, from left to right the distance a , and from right to left the distance b by eqinuation. (13) for the beam elose.to the wall columnP and by equation (14) for beams in interior spans. (4) After knowing the location of points of contraflexure so proceed to determine the:,'distance'd: of the columns in the story next above. In doing that use equation (7) to compute Yb, equation ?(10) nlu, and finally equation (14) gives the required istance d. 0(5) The location of the points of contraflexure for beams above the second story will then be calculated. The angle m for the column below can be found either by equation (5) or (15). In order to do this it is again nacessary to assume the distance c =1 h of the column in question. In the same way as before we 4 a4b,for the beams. compute Yc and the dista:nces a,a•g (6) Repeat the process mentioned in (4) and (5), we can obtain all the distances a, b, c, and d for all stories of the building. Beginning from top we can finally determine the last unknowjn : i.e. the distancec, which was assumed before for each story of the building. In case that the final value c obtained in (6) varies a great dea-l compared with the assumed value, so the process may be repeated by assuming c = iits•- final value obtianed in the first computationL IN general, however, the first computation gives result sufficiently accurate, while in the second computation the result is almost exact. The degree of exactness to be attained depends:, of course, upon the character of the building and the opinion of the designer. N•t•'l For c = I h and for constant moment of inertia from 4 equation (5),1 YCv cd I = h V•T 6• .,L. or or , d 5 3 31 Appendix. The Derivation of equations 16, 17, 18, and 19. From equation (11) ·¶ !·*~··t i· :· 9 d+ (A) Pd From equation (3a) Od ~-d " '2h- ~ -2h' -- AA 3hr.2j .= 2 6c ih1~* Bc-~~h:2 2hETT 8h"ET also from equation (5) cCd " C h" ** h *..;*** *.- (B) Substituting the values of edand Scin (A),we have d3.(h-Zh' -.. . 2 3h" (2hi-h') + hBIcpd For continuous and rigidlj fixed columns Pd =. 0 , then h,2h' d=- h'- 3* (2h-h') If h = h'. then d = ........ .... h. (17) 3 Similarlj assume pd = 0 and substituting the values of OC, pc and d in (B), we obtain .h'. - 3h2 3h-2h' (h-h"_ c hE6h "I )-6hh'Ic(3h-2h") OEc3(2hh' (18) * * *·~ i"] +h'') 4ETI c (3h[ h-h If h = hr', then 4EI *... * or, since P 3 h Cg 2 Pc. . , * * (19) -PART 2 MOME1MTS AND STRESSES DUE TO VERT'ICAL LOADS. In computing moments and stresses due to vertical loads a general assumption has here been made that the deflection due to vertical load at the. top of the columns is equal ýto zero. The theories followed in this section are mostly given in Prof. W. Ritter's book " Anwendungen der Graphische Statik." It is endeavored, however,in this section to put thdse theories into practical adoption: for the particular use to solve rigidly framed buildings. The use of moment factors developed in part 1 is a special feature, upon which the separation of the moment at each joint is based. This principle can be adopted for any kind of rigidly framed structures. In view of practical use several short cut methods for finding moments of a rigid member are presented here also. ae•k~rs. From equations (11)a, (11)b, and (11) if are not =0 we 'have,: from eq-ua~tion Pa, a-.a , :((11, ' Pb = -eb- b Substitute these values, in (11)a and simplifying ing expressions result': + + a L-a Mb L 'b MaL = the follow- *.a-- -a _a b ka and kb .. b . kb L Refer to fig. (12) and by geometrical relation, we can see at once:, kaa a b S =m .. kb -kb i For members of constant moment of inertia ka and kb =EI -ALLx) L )=. L(,rk 0 6ERI substitute these expressions into former equation-, we obtain, Sa a - 6 L-., a 2 L L L ( Morley's Strength of Materials, page 1 4•6.) and b Sb -. L2 L -b L A ' *- Where A is the bending moment diagram due to the applied load when the member is simply supported. T is the distance of the centroid of the area A from the revre origin A, and (L-r) is its distance from BforSc-, s q and q' are the second factor of the right hand expression of each equation. If we substitute c amd d for a and b respectively into equation (20) , the expressions for Sc and Sd for a vertical member result. All the short-cut methods given below are based on equa-tion (2.0). For beams of varying cross section separate discussions are made in Case A and Case B. -- Ca A. Uniform distributed load . Fig. 13.-- In this case the moment diagram for a simply supported beam is a pa:rabola:, Calling t-he moment at center =M, then, A=2/3 ML, L-'Z=--L hence q=q'=2M. 2 ' From simple geometrical relation of similar triangles it is evident that the graphical solution of Saand Sb shown in fig. 13 holds. The intersectilon of the two crossing lines coinsideswith the apex of the parabola. For bea:ms of varying cross, section , we have for equation _,20)9 , ka=kb 8 = fL dx, I-x'x dx L2B tw (Morley's ,page 183) (Part I, ) EIx By putting these two angles in the form.of a ratio-the factor 34 i1 S-_dl both in the Inumera:tor and in the denominator caacels I 0 x out, therefore the variation of moment of inertia of the member does not effect the values of- Saand :Sbin this case. -Case B.. Single concentrated load-. Fig. 14-A§ shown in the figure, the moment diagram for a simply supported beam under this case is a tri.angle, whose A=1/2 LM, Yfrom A = 1/3 (L+x) and from B 1/3 (L+•i), hence from equation (20) qkL.. q,= ) L and L To find the values of qand q' without calculation, lay off the length L either side from the load, join the ends with the apex of the moment,diagram,and extend the lines as shown, then q and 4'qare obtained, sior. by similar triangles the relation q ":. M = L+x : 'L and q' :M = L+x":L exists. Again dra&- crossing Jines in the way shown, then the values of Sa and Sbcan be found. A s:imilar relation holds, f.e. S a: a and = q:L Sb ":b = q"': L Now in case that the cross section of the beam varies, the s olution of the problem is somewhat difficult here, for the fc+k--or : - J- dx ( compare Case A.) become Ix o and dx iLdx •o IX + x xx in the expression for angle k, in the expression for angle B , and these factor-s will not cancel out when expressed inithe f formof a ratio, therefore the variation of the cross section of the beam does have effect on the values of Sa.and Sb To take care of this effect the work is complicated. In order to facilitate speedy solution, a table is reproduced tere. from "StrassenerJs Forscherarbeiten". The values of S a and Sb can be taken directly from th±p ta2ble. For single con•.... i-P ,b I I·~ Ir - IIr~t 1 b ~C-------t------------I I I Il 191 _ , iC---- _ _ -Fc----. 2. i r ;-I _ i Lk_/ Fi7 I? z. /: Sr~a 6 1~ \ r' bb Me c* C") r' - -t- - >.---+ L I o - r '" N t -h---9----·r I /: -I -\ - 7 N /. - W47 I Ii ti N.K I / - ~" , - i :!, -M~d Fig .15 .9 i cA- Yaloc-: +for hbt< - - - P= , L C (upperfg , - ·;i ~~· *1 -Tu -- -w · I c), .irC) ,ower -- -0 A .Q5 2 ~i-----~ z) 2 0[0o o. \-r. o. 1o o. z'5 0. 110 0.16 0 o .oi0 0o' r .Oto o.oa4 0.117 0.083 oo• o. 83 0. 1z o,ods 0, 0.083 .00 0. I•7 0.084 010~ 00 I -1 ).Z66 0.524 o.0oa 0. P83 l. ci 0.3518 o.B5lz 0 .5to I-3 I 0.396 C o 0.31 /41 5 o.5, o.o083 0.%45 O,6v o,5 3 I Q 0.,J6 0,39e ( 0.D38 3-.94 c , o.z-cq 0o, o, (2Tj o,\bZ,. 0.•4o 0 o o3 0. 0o,i. 0o , 0 0 30: o.58 o- 3qo o 0204 o o.9 0) 0.24 Og3z oBT0I Q.)9l o384 00.04 0.35, oa.s• i .o 9 ,164 I 0.~38 o.•9o I , 0.39o0 a o.o'0t083 o .• 9 --··--- ·u~---··---- D. I•,o o.0o8 0.591 I~ I . . v o ~I 0.38. o.:e9 .. 1 O2.4o o,;oA ----TI- 0.2.40 -r)----lr4o. 24t5 0.06-5 o, t(0- 0.\51 0.,ws0,1t • 0.321 . I . . . , · 0.585 .·594 o.asg - Q0 o I _I 0,36\ o.Ds ! 0.32,4 0.-11 0.o58 o. ,1 I oalj --- . . • o0>,39 0.\(9 ~--·- o.39, •9B o, o.BS• -·- .'239 Ozc23 o.3 0 o.3o 0.5'1-z ------- 0;-3q3 O. \o65 o.IZ9 o0 -1 - 0;.39•3 o.-,rI O.o03 o. 54 I)oz,0 .Q3I 0.39• • 0.301 ---- 0.398 .•,1 O. i2 -r 0.8 . O,9 ' -·YI-·ll- o.l·- a oNl0z4 o.SQ8 0.535 o., LA o.r3 13 I 04 4>1 c. Il-·--1---CI-ll 1 o,453 0,242 Co oka O,ZZ6 0.|65 0' rs 0400je I a 1Z-5 0.o084 q.3rs · o t-o lo,5-[o o 341 -o 5 0.083 04 O18 o .za, oo.-e.Z5 ,5? 0. 165 ÷- o,Z43 1e8 0. 0A\66 O.,<5 Iz 'Z. 0 .4 o ---- { 43- 0.-3z o. zzo. 0.(•o- o.398 0. k05 f~aG o , o 4-o1 ff -- I· - is8 0.1 0 0. i aunCcs. 040$ +5 0i.56 _1 o•,1 !o.\ OAI0 0.08Ct o.Zo 0.3509 o. c -Z,; o.O833 o .402-.o- 0.40\ o.zgg aracbole -- -- -- C---i I o.,. 4G; oJ 5'7 oa ~C_ i -f--~- o.,z'4 o,10 e for 0.315 &5385 0 .310 --- *- 5 4 o,3vzt c. $5 - i----.--+------------ =3roigh yFor o. 14O .4-. -- 7 - I o.89o a 00~ 0.5sS 05,52.9 ~-:-~ --- · - · - 37 centrated load multiply the table value by the load and for severa:l concentrated loads or for a combination of concentrated load and uniform load take each load separately and sum up the moments. Case A and Case B hold also for vertical members. Columns subjected to crane load. Fig. 15. -- qCas_. The applied crane load P causes a moment P'e:, which is equavalent to the moment caused by a horizontal couple H as shown. Lay off CC' eýqual to the moment P'e. Draw line CK parallel to J'D and join LK as shown, then the resulting area bound by the broken lines CK, KL, LD, and CD will be the moment area of the column due to the crane load P with no regard to t'he continuity of the column. The value of A and x can be easily computed. Compute q and q' from equation (20), and lay off the same equal to CC" and DD" in the figure,.then join C"D and CD" , the values of Seand Sd are found. The proof is the same as the first two cases. The moments due to vertical load will, in general, be investigated by considering each single span separately as shown in fig. 1&, in which it brings out the way of the separation of moments onto the columns and beams surrounding the loa.ded central span. If the moments Ma and Mb for the loaded span be determined according to the method given in the previous section, then at the column to the left of the load, the beam will take a moment = nri'Ma and the total moment which falls to NAMa Ma - nrlMa. From this the adjacent columns at left = moment the lower column takes an amount = nb 'AM A and the Similasxrily at the column rest falls onto the upper column. to the right of the load, the beam takes a moment = nr' Mb, the total moment which falls onto the adjacent columns at right = AMb Mb - nlrMb, from which the lower column takes a portion = nbAMb while the upper takes the rest. The tra'sition of moment lines for unloaded spans _ __ b it, A I ---I ~ F "KY. rn(,Py flri , rl A lb 1/-~L P jIMA I P -I-t I J~TJ- 'R9. la I I - , r39 is as followsý: For all spans to the right of t-he loadedi t--heamoment line passes through .the point B, for all spans to the left of the loaded the moment line passes t~hrough the point A, for all the columns abcve the moment line passes through the point C, and for all the columns below t-he moment. line passes thru. the point D. The pointsA, B, C, and D are the points of contraflexure as marked in fig:ure 6, part 1. For the.j.oints in stories next above or below the moment transmitted over will be sepatated again. Let the moment to.6 be reseparated be called McandMd., then at the joint next above the reduced column moment nlu' Mc, the total moment to be resisted by the beams = AM= Mc - nluMc, from which the left beam takes a portion beam.; = nlAM c and the rest falls to the right Similarlly at th.e joint next below t-he reduced column moment = nul'Md, the beams have to resist a moment = AMd= Md- nul Md:, from which the beam at left takes the portion nl'AMd';', while that at right takes the rest. The moment tra.nsmitted and to be reseparated becomes smaller and smaller , the further is the span from the loaded one. At the end columns the moment carried over by the beam should be separated in such a way that it will wholly be taken by the columns, so in that case AMO-Maand AMb= Mb. In the above discussion we ha:ve only considered the absolute magnitude of the moments. Hoever,in ca:se that several spans are lo-aded, t-he moment due to the load in one span may in most c:,ses cancel that due to the load in the other. In order to get the algebric sum of the moments due to the loadings in various spans and to avoid confusion the following conventional signs will be found helpful= .f " The moment is considered positive, if the elastic curvature eof the beam axis at the portion considered slopes downward, and nagativer, if it slopes upward.. The moment is considered positive, if the elastic curvature of the column axis at the portion considered slopes to the right, and negative, if it slopes to the left." In figure 15 the positive moment is drawn below and to the right of the beam and column axes respectively and the negative moment, above and to the left. Keeping the above signs in mind it is very easy to determine the character of the moments by considering the elastic deflection of the member.. 3. De-.e..ai Leq~t.. . ra For all members of a building , if their moments at joints are known, we can find the shear and normal forces very easily. For rectangular construction l:dvantage can be taken from the fact that, just as the shear is perpendicular to the normal force, the column axis is perpendicular to the beam axis, therefore we can find the normal force of a column from the shear of a beam and conversely the normal force of a beam from the shear of a column. Let Q be the normal force or the shear required, Qobe the same corresponding to a simply supported member, and, as before , Ma and Mbthe moment at the. left and the right end of a fixed beam, and Mcand Md the moment at the top and the bottom of a fixed column respectively, then, taking positive shear as that directed to the top at the left section of a beam, and. that directed to the left at the lower section of a column, we have : For the shear of the beam or normal force of the column in = Q0- question * * (21) - For the shear of the column or normal force of the beam in question a = Qo + .(22) c-d h For columns , since they are generally free from external forces:, then Qo= 0 , and -.--. S. A .. ** (23) h The derivation of the above expressions follows immediately by taking EM = 0 of the member considered. Both the maximum stresses in beams and columns occur from the source of maximum bending moment and direct stress resulting from normal force. For columns, especially those of very slender section, the deflection, which we so far assume to be ziero, may cause an additional stres4, but the stress due to maximum moment is , in almost all cases , greater than that due to maximum deflection and these tiro will not occur simultaneously. If we load alternate bays of the structure so as to give maximum bending moment to the column in question, (seF d-0iscussions given later) the deflection ca:used by the load on either side is of compensating character and the stress resulted therefrom, if any, is generally small and negligible. From figiure 16 it can be easily seen the following criteria for maximum moments and hence the maximum stresses of the beams and columnst For beams maximum positive moments are caused by loading all the odd number of bays in both directions from the bay in question, counting the latter zero . The negative moments are caused by loading all the even bays in the direction and all the odd bays away from the direction of the beam end in question, counting in the same manner the loaded bay zero. For interior columns maximum momemts are caused by loading the bay adjacent to the column in question for the full hight of the structure, and, where possible y loading alternate bays in both directions from the bay. For exterior columns loadings causing maximum moments are similar to interior columns. For corner columns the moment may be considered as being introduced into the column by the two girders meeting at right angles to each other, and the moments may then be combined t-b give a diagonal resulting moment. The maximum stresses in all members will be found by combining the stress caused by the maximum momemt with the normal force found by methods given in section 3. PART 3. MOMTENTS AND STRESSES DUE TO HORIZONTAL L ADS. In high building design the stresses caused by the wind load applied in horizontal direction play such an important r6le that a close investigation always seems to be justified. For buildings, of which the width is small compared with its height, the wind stress- is always a great factor. In this part of the work the general theories given in Prof. W. Ritter's and Dr. Strassener's books will be followed. It is endeavored, however, that the theories are extended and put in unique manner for the practical solution of high buildings due to horizontal loads. In dealing with horizontal loads several approximate methods given by various authorities have been known, and they are generally deemed as simple and practical. Nevertheless , none of them seems to be consistant with the actual condition. The exact method given in bulletin 80 of engineering experimental stat;;on of the University of Illinois and sevetal other known methods derived from the theory of least work are all too laborious and they are, therefore, restricted in almost all cases in its practical adaptibility. In the preceeding parts we have made one assumption, that is, the horizontal deflection of a rigid joint due to vertical loads is equal to zero. This assumption, though it does not conform to the theoretical correctness, is rea- sonable for the solution of the moments and stresses due to vertical loads, since the effect is small and negligible. In dealing with heor'~Khal loads this assumption will no more hold, we have here two kinds of moment to consider: a. Moments caused by loads considering the rigid joints not being subjected to any deflection. b. Additional moments caused by resulting deflection. The present method will be described under various sections given below,, from one-story--frame to the general case of many-story frames'" 1. Derivation of expressions for moments due to deflections.. --,~-~-~~~~-~------------------ -~-i I UI / >I 'I /M Sif CS 8 /·" ~ i i' i -~n~,··:--···-? P -- 7i'/~~i 1/i j j M4 "FL.f] ]:s• .\' . Considering the case as shown in figure (1i) the column has an elastic s-ufport at its lower end, the upper end is to move and subjected to a horizontal load free H, then the resulting deflection is Hh2 d A-= H'+ and the angle, through which the top of the column is turned, is ,= where X' = h2(x ' + Hhjd d - Pc) and a'= had For expressions of ad and Pc refer to Part 1. By substitution, we have A= Hh2 (Oc+Od- 1c ) and ,r= Rh(xd+ Pc) Deviding the former expression by the latter, we have 45 =h- ad Od Part 1 But from equation (-), S----- ad + ?d -d .. therefore hd .. or, . h--ý The moment at the foot, of the column due to the deflection A caused by the horizontal load H is then Md6= - - h= A ad+ 0 d (From equation X) -d or, since ad.+kd. d Md 5 * * ** * (24) (h-d)hpci Equation (24) is the general expression for moment due to deflection, when the deflecting end is free to move. Again refer to figure 18 the column has elastic supports at both ends and theck-clu•;i distances c and d,ai of the member, .i.e. already fou.nd according to the method given in Part 1. First assume the top end of the co.lumn is free to move and is subjected to the shear Q, then we have the moment. h 2 pc Similarly by assuming the lower end of the column being free to move and subjected to the horizontal shear Q, we have for MA--^ . M , = h cA hBc Now in the actual condition of the column the ends does not move freely. As soon as the deflection A occurs at one end, there produces moments Mc and Mdat their respective ends; although these end moments will offset a part of the angular turning of the column ends, they,. however, do not have any effect on the value of the moments Sc and Sd, since by hypothesis of Part 1, point C is the zero-moment point of the column due to a moment acting at the bottom end of the column, and point D is the corresponding point of the same due to a moment at the top end. From above discussion we can have the following ex-6 pression to find the distance zc, distance of zero-moment point from the top end. S -Sd zc - c h-zc-d z = d + h-d-c Sd 81 S By substituting the values of Scand Sd in the above expression we have ch c+d which gives the location of the zero-moment point. Now at point C the moment Sc also equal to Q(z- c), from which relation we obtain = A. h2pc c c- C From the figure it is evident that Mc= Q'ze and Yd=- Q( h - zc ) By proper substitution the equation for moments is expressed in the following general form: Mc= m'C Md= m'd Where a is a constant ........... = - (25) . Pch(h "-d) In computing the moments of a building frame the value of A in equation (25) can be assumed as unity. After Mcand Md due to unit deflection are found, we compute the shear Q of each column from the following equation: h for moments of opposite signs, and S. .... ....... (26) ( Md - Mc ) h for moments of the same sign. The sum of the internal shearshould be equal to the external load 1. The actual moments are found by multiplying the values of Mc and Md due to unit deflection by the factor - The separation of the moments will be done by thea aid of moment factors as described in Part 1 and 11. In order to make the method clear the application of the method to various cases tions following. are given in different sec- 2, .S9igie-1toryfraue4 V-5.; V--~ 7 7/77 Let us assume the case given in figure 19, a single story, two span frame acted upon b y a horizontal force H at the top girder. The columns may be either hinged, elastic', or fixed. The horizontal force H causes a deflection of the columns at the top end in the direct-ion of the applied horizontal load. The deflections at all columns are the: same, because the top ends of the columns are connected by a con; tinuous girder and the effect of normal stress can be neg.e. lected.. The magnitude of the deflection due to the applied horizontal load is unknown, we can, however, first assume a unit deflection and by the aid of equation (25) the moments due to that deflection can he solved. From the shear calcu- lated therefrom we can determine the force the assumed deflection. coefficient H_ , By multi'plying the moments by the the required moments due to t-he horizon- tal load H can be found. foll ows: i.which produces The procedure of computation is as f' 777 '147- (a) (b) F9 2 r I A *11 A i 'I ~iITL·_ / A IL' i U0) / '9~zz (Al) A~-- 'i F Si! FT ti 1 7-7 (Ca) 77747 774-; 77 (6) 1. Compute the characteristics of the frame, i.e. distances a, b, c, and d, and the moment factors of the corresponding members by the method given in Part 1. 2 . Treat each column separately and assume the column in question either as fixed or elastic, as the case may be, and the rest as free to move. ( Figures 21 a, 22 a, and 23 a.) Compute the moment of each member in each case by equation (25) in mhich the value of A is taken as unityl. The reason for taking each column separately is for the convenience of separating the moments by moment factors to each member. The moment diagrams in each case are shown in figures 21 b, 22 b, and 23 b. In actual computation these diagrams need not be drawn and the work can be done mentally-. 3 .Taking the algebraic sum of the moments, it gives the combined diagram figure 24 b. The resulting moment in each member is that due to the assumed unit deflection. 4, Qa, Apply equation (26) and find the values of Q1, Q2, and fig. 25, then the sum of the shears must be equal to the load P, the load which causes the unit deflection and produces the moments, or P = Q, + Q, + Q, 5. Find the coefficient of Ji with which the moments in figure 24 b are to be multiplies to give the required moments of the members of the frame due to the horizontal load H. Fig. 26 shows a general:case for the srparation of moments due to deflection b. moment factors. 3. Twow-storv frames. For two story frames the moment of each member are due to the combined effect of horizontal loads H. and H2 , fig. 27, Through t-e influence of theseloads the frame changes its shape as shown in fi;. 28, i.e. all the joints deflect horizontally in the, direction of the load. Now since the magnitude of the horizontal deflection is not known so we have to get at it in some indirect way -as before, by the aid of an assumed deflection and the moments therefrom. resulted 5:t C __ ---1 i ~4_--- rr .,,,- 'I ' /T ' --- -1 - I -;i- 2; I1 T4 rr I I 7 C)t C4 (C',) TP' !4 ) 1 Ii7f (b) FTg9o %M F , I , -Sdr 's r,,, ~( g 1 -- ---. -f.r _ -,--+ F~ N . tg3 l is ?_ ---------- PI F=.35 Fi~ 9.~ - - ----- ---- - ------ --- ~-------------- For two story frame we have to consider two conditions of deformation. For the first condition assume the top girder of the frame is subjected to a deflection A=unity, while the lower girder remains in its original position ( fig. 29 a.). The moment of each member can be computed in the same way as for one story frame, in this we again consider each of the three columns separately and compute moments by equation (25). The moment for each case is separated and the algebric sums for the three: cases are found. The external forces acting on the frame, which produces the deformation of condition 1 are (fig. 29 d) On upper girder P, = Q. + Q& + R, =6 Q + 2q + Q8 + Qa On lower girder al + Q2, + • and on the footings S, = Qz + a, + Qs The suffix given to P, R, and S denote the number of the con- dition of deflection considered. For the second condition a deflection of A = unity at lower girder will be considered. The ujper part of the frame is free to move, so as soon as a deflection is given to the lower girder, the upper girder will have a deflection greater than A. The latter will be assumed as being restricted from producing a deflection greater than A, so that the relative position of the upper part of the girder to the lower part does not change. ( The deflection of the upper girder cannot be assumed as being restricted to zero, as in that case another moment due to deflection in opposite direction will be resulted, which violets the condition of moment separation ). After the moments in three cases are combined and shears found, the external forces will be ( fig. 30c ) " On uppe"girder, R,2 =Q, + Q1 + Q* on l.ower girder, P = , + ~Q + Q, + 1 + Q2 + QS and on the footings, S, = Q1 Now we have two loacL:: + Q2 + Q 3 diagrams on hand, figures 29 c and 3o c The first produces a deflection unity at the upper girder and the second produces the same at the lower. diagrams is also shown the elastic behaviia In both of these of the frame against the acting-loads. From the moments resulted from these assumed deflections we can determine the maentsdue to loa4 any horizontalby some simple modification. Let t-he moments caused by the loads in figi 29 c be called am and those caused by the loads in fig .30 c be called a,:, then by addition we have a combined diagram with forces PL- R, and acting on upper girder, " loker F,R, " - " on o, the supports.! Fig 31 The combined moments are then = al + im2 . It should be kept in mind that this combined load. diagram gives a unit deflection both at upper and lower girders. Now in order to conform the actual deflection for a specific case let the load5t.. in diagrams29 c and 30 c be multiplied by the coefficients x and y respectively, the combined loads,fig(32),will now be : 6PI - xR, on the upper girder, xP, . 7R, on the lower g.irder. and -yS, 2 + x S2 on the supports. Simila.r'ily tne moments = xm i + ya, It is evident that for the applied loads Hi and H2 =R,Hz P,- and xP, gR~ = tH, From these two expressions the coefficients x and y can be found and the moments computed from M = xm, + y2 In practice in order to find the maximum moments it is aenerally required to treat the horizontal loads as moving load and in that case the values of influence moments have to be computed, i. e. t-he moments due to a horizontal load unity at each panel point separately,, thus the conditional equations are as follows': For a load unity acting at panel point 1,( fig.- 34) P R, = 0 xP, and M- "owergirder. = R, - 1 xm 1 + ylam 2 For a load unity acting at panel point 2, (fig.35) at upperl girder, -Pt - xz Rt = 1 Xo and t P1 -I TR = 0 i = x 2 m1 + y7 MR 2 For the load condition shown in fig. 33, M= + l , 2H frames 4, Three-and- more-stor~ The method of computation given above can be extended to the solut.ion of three and more story frames. There are as many conditions of deflection as are number of stories. From the load diagrams the elastic behavior of the frame against the influence of horizontal loads will be given. For any specific caze each condition of deflection gives a load diagram and a corresponding sets of coefficients.. There are always as many conditional equations as are unknown coefficients' To make the method more evident it -. •., .,,, Lor a toerL .,, Ll suory fibE Iram , Cf, r,,, ,~,, is given below f LL~,· h1nUa.flectZio[n conUUiLUS W.imn I'hlr corresponding moment, and load-diagrams (figures 32 a-c) and the conditional equations for the determination of coefficients for each case of loading: For the specific case s'hown in figure 40 the con, ditional equations are?: xP, - yR, + ~R NP, wR" - zR," -, zP + xRI and the moments M = -H, - H', xmi + yMa, + za 3 the If the horizontal loads are taken as moving load, then~conditional equations and expressions for influence moments for a unit load applied.at each panel point separately are as follows': For a load unity acting at first floor girder,Fig..37, the conditional equations are": 7- _ _ _ __ 42 . e-o H..I -4-·- mi~ -- · 7 r 7T T~ .38 9 =a) Ii, W-1i r'1:~ 0o -b (bA £'1 t4 7 17-4- V;4~Ac 9L."3 L w I5w 'C rc\4' 4i .2 77J -P--;·1 - p \i ii ! <I. · 3, 77,r,- ,,,,----- 3 -5+o rj-YFra me __ Locz4 5.tMrent- 4 K- *-- _~i_ ~T F -, 7p. / n .(P) -J1 r I I xlP, r. yiR, + agR s = = 0 aR," - 4P, -•R"j =31 R ' - y, Z- P, *· From which the coefficients xyy, and z, can be solved, The moment expression is M = x ai + y.mi.. + Z.as For a load unity acting at second floor girder:,, fig.- 38, the conditional e.quations are-:x 2 P,, P2 .+ zR - = 7-Z 2 R x22 R 0 "=1 z P, + 1R1 ' -, RS2"=0 and the moment express:ion is M2 = x 2l., + y,2,m2 + z a0 s For a load unity acting at the top girder; fig, 39, the conditional equations are: x,P, - y•,R' + z-,R"'= 1 - xqRa023s = as2`P z , +PxsRI'"~ 3 + R2 ' = 0 and also the moment expression is M, = x.,m,1 + y m. +zm3, s For the specific case; fig. 4po:,. the expression for total moment due to the combined effect HI, 'H M' HzM +'H2 and:H, is given as. M- +÷'HMs In all these cases the signs of the moments should be taken into due consideration. 1 In computing the moments it is eecommended to use t~e 4Follow;vy +fabLac f 0 r-v,; p-uzW rim TaTobux Lorrm aQ PoIn 0±ember a cnvc for Conm4#prfng --L n+ I Slome~i M 4 or H il~+ 9 kMOac~i~th of Tcrce - ry -Frame - I I I I I; I I . Y. 3m +v + .F..r lone~r +ol9ýr 9 c-V Momcaorh a: --- 1 ` r 9 Coy fO;Ld icy t e 3pe; 2d -- ·--------- ~---I-,.-.. - --- ~ M --- 5. eial caseThe. usefulness of the methods is n.ot only limited to cases discussed abbve, but it can be adopted in almost all kinds of framed structure with some aodification. of this method it By means is especially simple and exact for the s.o- lution of truss-formed frames and truss-bents,, Let take the case s-hown in fig. 41, a truss4f ored frame., -It is evident that this truss can be considered as a frame of the same type formerly considered except it is layed in a horizontal position and, instead of one side, both sides are fixed-, i.e. fixed at both supports-.: The deflection con- dition can be so considered that:, while one vertical member is subjected to a deflection A due to any vertlcal load, t-he ends of other members joining this vertical are all subjected to the same deflection, whi:le the rest verticals are kept in their original position by some imaginary external forces. HRere again the moments of all members due to the deflection of each indevidual member for each deflection condition in question are to be determiined and these moments combined to give Ia moment diagram for each particular deflection condition. Figureha 43 gives the three cases of deflection condi- tions with their respective moments and load diagrams; For the case the truss being subjected to the loads W1 , W•, and Wa (fig. 4z) the conditional equations for solving the coefficients x:, Y, and z are as follows': I - I V-- I-_j71-L I--- I· "`f .~...I KA Fe .4 SI Ii- )--~I- - T f;7 ... i. ..• -t"-': / I, ; _ '\ -1 \ AI i, ii - --. ,I i AF.. ....5I--.._...T .. -1 g.. ÷ i.. -4i,... ... ..... ..~L.! . ..., •. .lz. m I ...... ': .. :.. F1. I 1 - /9 - _____ IL 43 VT 1 _ _ --I-- IT A~ }F ~__ ~_I_ -- "7 1ý"P5 j1Pz1B 1~ - A I- Iti n7T· 4 - * -TI.- ~_ __ - -- ---- -- -- ----- -- --------------- xP 1 yR- + zR' zR," =ZTh -xR;"+yP, xR 1 I w yR' + zP, = W and the moment expression is M = xm i + Ym2 + z•aI If the loads are to be treated as moving load;, a unit load can be plased as moving from one panel point to the other and the three sets of coefficients can be solved from similar equations given above. To find the reactions A and B at the supports the f,ollowing equations -hold': xA1.'- yA2 + zAs = A xB, - yB, + zB , = B and A + B = ~I + W2 + Ta If the truss and the loads are quite unsymmetrical then there will be an imaginary horizontal force H acting at the upper chords for each load diagram, fig.4:4. This force is equal to the algebraic sum of the shears in the verticals, or the unbalanced force in each case. We have now an additional deflection condition to consider, fig. 45. In figi 45 the value P4 is evidently equal to the algebrkic sum of F's in fig. 44, and the values of Ri', R4 " and R" are equal to the direct stress of the corresponding verticals, or the end shear of the corresponding upper chords. The four coeffi-.. cients can be sdlved from the following four conditional equations': yR2 ' + zR '"+ vR 4 ' = WI xP2 1 xR`"+ jP,- .zR xR1M -yR" -x1 + vR" = VW 'I + P + vR + yR - 2+ zH3 + VP =0 4 and the moment expression is M= X2 1 + 'i2 +z 3 s + v1i, In case that there is accually a horizontally applied load H acting on the upper chordi, then in solvingthe coefficients t:he left hand expression should be put equal to H, the applied load, instead of 0. If the truss is connected with two posts in the form of a bent as shown in fig. 46, then another deflection condi-. tion,fig. 47 , has to be taken care of and the conditional equations will be raised to five in number. of determinants. 6. Abstract on theory ------------------From the above discussion it is seen that for each For story of the frame there is an unknown coefficient. buildings more than fo•r- or five stories high, the solution It seems of these coefficients may be felt handicapped. to the writer that the use of the method of determinants The following abstract on this method is is preferable. taken from "Hawkes Advanced Algebra." and will be found helpful in most cases. a) Definations': Let us solve the equations ax + bly =C and a 2 x + b2 Y = c 2 64 . Multiply the first equation by b2 and the second by b1 , we obtain x + b b,2 ab -ubtracting, we obtain or if = azb 2 x + bzb 2 y aib, , a.b L x (a,b 2 b,c2 = bze = - a,.b.)x b 2,c -bc,. OI b 2 cz azb 2 and • 1 y= ab. ac, c a ,-c1 aazb, a 2 bb We note that the denominators of the expressions for x and y are the same. This denominator we will denote syn, bolically by the following notatiod: a z b 2 -a 2b b = a a2 b2 The symbol in the right hand member is called a "Determinant". Since there are two rows and two columns, this determinant is said to be of the "second order/ The left hand member of the equation is of the determinant.. called the " Development" The symbols a ,ba,a 2 ,.b - are called " Elements "of the determinant, while the elements az and b2 are said to comprise its Principal Diagonal". For a determinant of the third order, we may combine the terms as fo.llows'-. a1 bz c1 a2 b2 c2 = a1 (b cz-bd •c-bc) c) - a,(bcc-, bec) + a (b a3 b9 cs b2ý c Sb, 3, 3 bb a 'bs3 e2 c bi ci Cb . b a bt c 2 65 We observe that the coefficient of a1 is the determinant ~that we obtain by erasing the row and column in which a1 lies;. A similar fact holds fEor ýthe coefficients of a, and a3 . The determinant obtained byýsing the row and column in which a given element lies is called the ".'Minor" of that element. 0- c Thus . is the minor of a,, 1 b3 c8 We notice that in the above development by minors the sign of a given term is + or- ac- cording as the sum of the number of the row and the number of th'e columnof the element in that term is.even or old. Rules.: 1. The development of a determinant of the nth order is equal to the algebraic sum of the terms consisting of letters following each other in the same order in which they found in the principal diagonal but in which the subscripts take on all possible permutations. A term has the positive or the negative sign according as there is an even or an odd number of inversions in the subscripts. 2. For the development of a determinant by minors write in succession the elements of any row or column, each muItiplied by its minor. Give each term a + or a, sign according as the sum of the number of the row and the number of the If in a series of positive integers a greater integer precedes a less, there is said to be an inversion, Thus in the series 1 43 2 there are three inversions. 006 column of the element in that term is even or odd. Develope the determinant in each term by a similar process until the value of the development can be determined directly by nultiplication. 3. The value of one of the variables in the solution of n linear equations in n variables consists of 1.o1 r ction + whose den minator is the determinant of +he s s % CL. .A' LIJ' JIJLJ Ud . I LA U-L U.LL A U b tem and whose numerator is the same determinant, except that the column which contains the coefficients of the given variable is replaced by a column consisting of the constant terms. Principles': 1. If every element of a row or a column is multiplied by a number m, the determinant is multiplied by m. 2. The value of a determinant is not changed, if the columns and rows are interchanged. 3. If two columns or two rows are interchanged, the sign of the determinant is changed-. 4. If a determinant has two rows (or two columns) or any row (or column) being m times any other row (or column) its value is zero. 5. If each of the elements of any row or any column consists of the sum of two numbers, the determinant may be written as the sum of two determinantsi 6. The value of the determinant is unchanged if the elements of any row (or column) are replaced by the ele- I ments of that row (or column) increased or diminis-hed by a multiple of the elements of another row or column. A numerical.example for illustrating the use of determinants- for the solution of the coefficients is given at the end of Part IV. S.. 6br high building frames TV e -5 ,ey+ened : ro'- 'ro y; 'd Th 1ro%\cm vo fcr- ser-ve-= .e r -Xcp sk-o0 n eLIkeL2nr\te ra rn upc e r I )Cg ' e Y-,- ou a 41.e rs &td'A Lbef da c V$ r,.A q A- fr10) 1the gveTX c±;t eC N+E-r- cfrre-crdief xte l'zoA a pped vesv;+5 i T+f0 ftrmevvibcrWt -irr vrc-aJ 4 & wneksr G ~ ·~: -I? i : r-5 *l se-ý8.: ý- L=-a : Co 1.. CSe b6c<7, : b~~~ u IC Vo CV.p rpenlc I ia r51 e~cr: "*-f ~ o~- ~:·1:.4 !~1':% L·rtl\p:l ~~c~L~I· 31.4 in , =-= : -JThv~ -p· c IL··t;c:2:'S ~yMU 2: F-`J . ~- ~I3, f r3 =~ 6x40 i1)r u&(o cci= Cl;!t 5 -- I B-16 So jeo 3 teq. t L-3 - 13-~):= e3 .rs-~s 6 -. 3O . Lb to ~t ::2 =Irru4-Q t=*-l + '=73 '3 = 136.11/ 4~.3 5 " 4- I ~r~e+ve~:irtA'~oX ver+iezdz a; iX " - •-"rer Z- 7i 71o f, C45 - 15. y(6) . - 6,- -t~t ok a3 c~nv-r oF a rod 7" x :i• Jo llg3b~ ~~zw:u3 ·139.. ouT) I 3o .. J.ru1 oner o. = C+ Q . e-45V'~ I rT.3z P . t? t k*5,VNCC: I-S ;c5- Srdc-- -tk s drrec bj cf-rrdr5 ~siteshor .. 1, a cZoa vvr r .e x i5; tox 3h17.4 Mr . -7 "1 rl + f kc~ Z5x 3 P foo) =,09 6 40-0-,1%+ 144 J8354) 6 ost.8-i x 1 L c: -~,,E d~t-~.;~ed?=rorrl formcrl.aj .9 a no YH meod wor ;ea3r··obr;; Sgawood's .5?. SLf tocd tocit c2. ~\tC6e~arrJ cx .41.86j Y%-3 ,41-85 rtnr: cro~ t~scr e Lee 1on a =1 a ey a ppljM d; ·2to S•to x I6. z ~x C•l + o4 6 83 .)goixe40s\ • SOI4• \> 48.3 t$ .M,= a?-d 5 n.Cz o ) ov (them --p \4-q G--~ ' aI3c1: .oo-zG+.oo*St a z+t44o14 bj c\C' 1-14> .) oe o L re to..cyC ;Sjrdo% ovrw- v,1;Sg "tleZ oa &. 0 gso'~fb'j"t-ue loc +o e<- We vwers ee = i 5.oy%c - 6r il y 54 ,., = I-to -(. -v & zb2~.o) or cxLY .rtwww' sanw~te -- 4s1 ) Coc- %d Lo Nb 3P{ltr 5Mb = - - o.o zft - OL Co3 . a+ 13 l lbbot'ls_ Co Iu xveds a= -. a. uses duck? wc: to . Ra~ve4le (\ 30 albo, ow.& (o8 .o ;4 P ) e .-1 O 4:,431 •i.-+o S~yrcar 42 ,oo 8 13s ,rk - +So . + 4 1of45 .'i f ) zSa cn -- - 4,'" irebetj. 6 n o4.kt r\ . -To y, -- J LC eie. . didrance = ke > Tte t.c -44 58 < x . .. c- o \-.6 0x- . 2- -13 aw. cu ,'.Derr- c. 513 H o = onS \ -O- -- 1. ~H.- ~3rtzgrjv.n · (a \raer .5195t "'.5 ci b cc 'noe • ie or mtr c .t mtrN(j, ob'rne oI Prok Idro ke ~agi cc?;:~v~q 3;,c41 ~jl;lc~ \.L7- j IY L1 .5. -4oo0 ',oo.+ 15 14 - 3 ' x +" ~2 ~E~l~~-3~~:L ' 1 ·. zL3.x~x Lar 1A IS 3 It =.554 .kL : x2o t .os 4 1 + 5 3 1 h -_ +c"4, T Ii,= t.5 :5," . =1 4 sX 0 5 3,rz -~:=3 SEc5-5~r333C~Q ~~-~4 -. O5 *H 5.s +ct_-t~ i=itc:c~Ic e "-." ~the arablemn Cov'\cre ;.3o Q- SF d; 0V thegypioxtoo "Thc sct tovw-s 4 In th \em. vv Y=, orgiYalv-c e inode - 'o vy;tkin \\k Wt Te bcrScme rcau +s a- v 60. -2. L-oad The re'w\F tJ5+ato'- of 1 av yerm RoFgrevlOu ~RCmebers3 various red~ aC~f aken.; I, ig* mad~e invforceC our--se '1\Sa rem POSs' ab\e Ipa as y- 8 -s ?le -time~lthe OF 4yoe J Acvecifd 09 lrd;c bm ver3 oC a meR~d de1-0n veaice 'smomentS nexgCt cv·ll~reinfavrt-emrk `f ro~vL iasTL ew;:; 1c~ 4cs mesign. '0Y je -dfZtYn of nd ve=Heajt tLc- 5- centh-r or;4;M"\ s -s+oryr 4-Y\or vaacrn detCJgnc bkm -r ag, kavwted ics at.e -91T\ keeY hacVe buoad~n 0. buCt Inive a buý yvok-~ -saour vw ed; 16& ~ahto r n~:J E icsor F,,. 401i ,25; i~t~-;s.fi. ~3S·~t~ryC~. Comv~p, -~2TN^n.on C 0 u z-,S = on\ . .30*e 1 Th-< oJ~e"r Sc n.~ " 09 Covcreir Xs ass..rm c-i #oCi ?ro ." d 4, -_3_8 =S 3.to 1-4 To~a 3,, -- of Jo10 or'0 r- S5 Jt =~z o Cl)TcoCfo "o+j. 5 o~s sq3 o omeai av . 4 4 i - x c .c~~ 4 I= ( Zl.i 2+ 4:x4j. 2'to a;i7SaceioF 4ce rz +-,-o,, T. RA.•v•e ed or .52 -t' (j~) . 1c 2 .7-8 x )4 = ..in4 or 4. 1655 in4 or".o- o- - 0 0 C i· _ i aT ce- iT e LcY i I1r I- i x T--! ~--I~ _ ____ __I r I' - 0_ 6' 'r ?' 64 I Spac.cd ;or'c-c \aierdl- IJ -0" - 4-r ' Jr/ 19Z * 1r s "if II·' aper- . ~i~------ L'rLJ TiY' *- i') r ~1 ~-1· 1 jL21 -----Al ' in ** * a a* L Nzd - spays- ,r )J s~ ?9: = Z43 o o-ca .. G5 D 6f ox 718545'1 1o.5164 Svinc . of .4oor In *rnvme~it o4- inel the re'g;w -orc~emenns~ ,-fe end Sic-;o 0r or cveer 1.0 54 Varix vvoverabts i/ inert'a 09 a beam Cor are workcd oZtI and S-acer ;tA, 71250 l193o rdrc ; or .53 . in o0 1n= 4rdoo x 10 y l L'O7 z or 'i - on bow,~ a3Ccouv cKert- fhe ýa-tIcvo vaue uS eS-- o be Lm!dc. Se*=4'on. - F =,. o& 9.AA= Zcom iTItop , Cx- 1;73) ! ý5 o- ro y- race.5 = -30.4 4 (x.-3.3)x 1 - 2%I = o 4 oCo' 5 5 = x S '.·. = enrd -f \tT **<:00 ';? , + I'5A S54,s5G x-- 15 00, b15 c = orc.8 Z.7 54os5Z 5(at ,-As -).zo2 -P3(3 14 u z.35(v(>L--.5) t 14 4 ,(x-3.5) )ST.e co -CT - - X = c, oF . rovi bolt-tom -t (5 =l(0a50 in <airder 15 42 * Total floor· in. 6.1 -4547.7 4.1 3.4'= xýg Cr-z3-4.5.-)ws (s19.0 - 3.5 - x.) \.z oz(, (4- '-"= 150 .405 1. -r) t zo- 94 'in. s = * 10.94 4815 j5I14( 15 "Toht ( I O.- sr 44c+4r 5.4 -15 X-35)2.rIS xo a •:ooC G-C=der ca,.,%r ect'-'orn x -f.0 )n (•-zo ud X1545 Or o.255 ;t ((J4'o-x) x" . :: = t...5(o a 55 o0 L48 x T. (3 =2 85.4, t oJ Tota = 3,140 ,re.15( o Ld ( Tnbeuse -Fo4-S erl-e x(16. -Z)4 r 4nd n Jc • •-.5t54 -3VoI Ia ·i 14 Iz54z , 15 . 1.32/4I x.42) Tote( .-. Ft.- ) - ( 16.5- 1.5 -x s=>G= (1o.5-3.-x) 82.9 3 s•- [ ,6- . '* . x TG55 .. = useva), .,. 3- = 5.o,7,',.• l.4 x. - t-,9 - o = 7ý xro x 5.C0 - (");-ob - x:, = e"(•rown pop) = [51.5 I. 54,2 xZ.-tS• or "t. ".(-ow botto.l = +.883644.t") A-SO =X-ox 4.4 ;i* or .?.fo ;4 aon COmpUitjaiKt S+rcC e To compue o•- an -raov'e i- SyvmvCdr'4a I\ (5Crform = diSkC• v~, 4 So-- .3 o ,Szo 4.54 a.oE E- r-Crbc '- ~c~mp·Jfornh ,xc tca b . aracr; c b<=\cv . o8 ,o09 . . , b~e5 oc, ezAd. ,151 . , 4.50 ,15 4.50 o53 S.4ok •.o o 9.5 for d;vr • eC. gO:. -of c. z.o c . Yo%- = . 3--- •o= = 3:r Z.Z4 -c = Edes =3 x4.5 (pe ~4.o5- s 7A5 (re19C o0 (eq!.4) . -rr 9.-00 \. 92% E£2 . - 4.T oc4.5+ Z.52. •5-4 E4 4.5o 1 5 (e 4: +24 3r45+ 2.5 E r = -- 3x: = .o#, A = - t .;5 -I-.I - -t- . . .. • 55sm .ne , ca. 3ws3'95j.5, -.- -- 11 =S 5 3c 30o = 3V'.o V5 =W ' I4.3• + S4,5 ~-7T. = t L=- as bq- 9 = Zo.4 - 4_so 4.503 63I80 d tot 40--;mare fot--arf 1h ie ed. I\. cre = \o to 0 .I + 14,9 ' ,L "', =4.50 i (3 ic~-r =~a + 4.•9 (p.Ac = C1 o == S .4.4 . "~IJ .. =.5,o5 1 o= 1250>x M.40 E = •, >2 h,,-d,, tlZ..4 a 3.6 2.tx50 8.34 .2o.4 .59 . ,0 Ll4-a-s,a •, .25 8.9 55 _ = BE )8,12 4 b, 1,4 C 1 = L.5 *- )1 x424.5+brlea o E atAsumed The h.T-C•,, (-$.4.• i5s de 4, 0O.4 x 2 = - z 41 .5 ,$59 ai -fh:oe 5Nformrb . ... -4 obt~arined. -cl'S= \z.•.•3, 48 C~, = I- -2- F C~= f b96(3 3.-o) - (~ 3.5 3 The su~m~ed var~~u c192~jK010 Tkc (2., -- = ee l Llrmed va, ves revis;oDrc t'he Page 36, of:.C'ic-, C-r valves, low bu+;f1the we uc ketp c e-, bu+t ?4xott5e C1ov g r2v-3i v.lAJfor Vc x ke vror i r. rj ver order% -\ 14.2w '+ 4- 3Y II -- .•, 32.24 c,% cn e= Jl.00') 9G -24 -··= me+t--W OF v·e- EI=95(d-e4a 6 +• ,9 ," -3 =2,- IrLe -. 9o II qo4J IO.v.1x7.2•8 422 .- 8 1 G.42, 'E~d ci= = -6 ed ybEa - .de a 4[ x 38.3 cT= h,- >2.5 = 44.o4~-) :-4 .44 1.53 --- = .- e bs ...1P, kte ,eCketsbov = J1.t--:.col = 4L 4 C 3co ;a5 zj Val uc 0$ Co = 3B.05 ' va.1s, 25.4(3-' , Cl a 4 a= or E4 -5 3 = 31.(0 -- v 14-.o 's.o 41 = 0 3O.490 o_i+= = R' -- = E(c_= E- 3.90 t- S\#zr C Iz arc Goveaww= y S eV;:ýe. bC; Yg -,• ~ o'owu s'ct vcout sw aCcorchrato~c t~~Sh-d-erwe 3.. omp-tYd vueze oC r), C , c oveik ' L11 3 v eni-wet tCUe, c- te_ or' 'TMS;J V~~at cA-2 aTk; 3or ~T~v -- V% V=nct~ T0:Yv 4 (3ynPJ eceS n e~rror 1-s \gtitt o s0A akn derZ rr tr-u (L±>'Q YX o da 4.-715 . (4.T•-) f3 Cb' ~ r . C:9~ o:rA?&Ve L~a1c) ' C3,5s3 5C ý4.7(6 - T=4.Z•4 z 4&5 1( 4.To Vo5C268 (. = o. Q5~1j 41~j K3-45 c.-5 d.55) 4 zQ.~:r4,45_ io)~ ~i o. .2: bAq r% Ci o)'71 ·c;; 5-- 514-17- isC%5Ate. Co,-6o x 1!5B = 5sj E1~~i. ~~.;~ vtuL') carm.`?t' CUD ot E`I" ý40 . = G.'ba (G,) Z:3 C¼., ••. C42.1?) az= 4.33 3'L (~,.C ~6 jr< (43) El~f " -a r c po%,a a c 14.M! - C . a __c 36 Zo' oe e) v a 4;a -a~r C.-4-,, 45o 8 ( I•?l ) $<1 z.5 W4 .c l t.2z5 + 4.It -P - E' I ao. __\J * 7.,4 -C,-. - (Zia 0 77!-. 2x.o4x 4 7S St4.t 4 ~F~-to-7 = C00) p 5o N- nC -- - / -756 a_.•, ,u = - =p I = K"w3 i am = C3.6r e) c , = (2.60 Check CYthe Com'put-tZovi 1 cr-vaALues ' bc-"-a),Ucvevcd, czr'c . S(,c?-1 tK~Yr tkhr 7iT Au ?\$Xkt on;5 V-% C.vtJ;co J'YVarYlrj 431 vxicc ;J e r-t err-or-g o5ome:~,cz1Ycze \fcry z .L LeVe(uL 5.0.5 K 8195 ) _i r :: .. - I L 1 !·r?· o 'to -i I' rc5 0o (-1 .d { c·) a5=0 'fl 'J. 8 0 rr 7 VakecS+o rll '' I r+j -~··L- a,6 cta t2) VYO vl S ( 0,x 5l) . j - -- 0 oon o GOmp.A 4 TornmeA- . inkvertce ue-bo veri-icai o 'I order +0 gac+te ,nC\uenc= Yanomerv,= cad a % j+%4fTo cL. ov-ttd c Yloac4d and, bC \oYvn (lvx;\ u- Ccce rYomrve, on yvo'(to) 844 .-Z = vu-% Sjoar-a-t on 1 445.T.5" -4:5 4t-" = V=, Pa + =.5 7 5 - - .< - s 73- .61· T-.+4 - -4 16L 3. ..6 4 s.- -o -i 14-4a~.96e,. 15 7 - . 31- " ct --. o - ~P? 20.4 yZo.4" "=•-5 _1 3 e .t .P = ,4- . 5- 3, - . -- J.9 oIC Stge ai4r4 =cr d; Co rrago (er ,o,•2, 0 .- lvC .eo -- , 4-5.t$ -5."p1M34S t E Succmsty,,k Span , to.4 due kv -e8,c-nae or+ke r-e% e ecS = S= -- o.4dc Qor~ a C\.· wk..ewtr Y%gir: 3W om eniL attt Canke·tr Apt 3 eua rob-c . L3c rr-cer czrr; e twwo conce Set~cc oca-k+ed, load. M tE _ - T .t A ?7 -A*F.T2 r2~ ~I2~ , -x+ .. .44 . - m -r -.I Mq~~~~i±, " " -V) , (x)6-4)5~~u -3,5 x X ,% 4 54 3-Z .4.•o -•.,I M Iq"•= -4-.•.,• aq . :7 - -4- 44 44 44 x•44 vb F1.15= 4.0o a+ (X) '.34 'r~ I ft tpa t Cci r reor ~o .x4 .- 47o0 Sa - 4 --. . AQ Ix • o., X zo. e a vvws - - S - 5.03 jy . - .3 a4a 9..41.3o`3 5 = SG ~44j13" xSG- 5ý07--ý C05 X4.3 S84.jr ."& Q3-00~' = o.2 - -. ceon S i rs a . > -,a =- 3a- + - iS . Y-) 1.0 , rA- . COO -5 . 6£ j ekeor aS gr G;ox-Rhi~44= .51. zIr = +.s-• . )"I % = - 1-5 , 5~ti14 - + .Sr = 4-4 - , -iI ' '3 6 = Srlp~6. ,. o i$5 = 4--* =$ ±· -+ 3. Ir 5 2 31A& 4.31 ~lj r-* + ;.Sz 8 t,. 46,2~I P" r- 1 -4 - •Z 3-,O 8,34 t ±.,2 336 - -7.-1 yr es= +- 7j1 -1 4 47 · 9 +g47 - + ~.. \P (r) .14. IS2 SS- == ;+" 6030 -~-t.C t > >+ 1rE+ .7 + = +.L4o l) -.4t -. 4. - =d -M:b:-LL = I C) 08 =, - A,,•. ,5 "4* s~r~;oY"· of J\IIHtD j4&9 ±cifx~ A4o.I +5.1 ~3~~4 Thci +-3*ý4 Y''I4 + x •. S. = _.. 3 = '.A e 2 - 441±.ý 1.56~ +- --+5. -4-4- +lr-sgý Z 4 2~.'~Sx IC-'j~ - 4'T.I .-'>1 -I, 8 4c•t s _- x.6 513x-c~:=Q 341VC4,= - w = =- -. C,• C-) t6.,x 4 30 Sct·:~ =~C, 5-r = -r.a- (2'ý) Z.·t~ir \ 2 tO4Zi I o. I4, M+, " .' = 4pi M00t% PIT rA - , 14zc". $.ToS~ ~z~f. I .36 >5S~ =3Q7. oS ~~63: S~,,, 3·-6 JYI~ rcbclC~ (. ) of -t.~za -+2r18 3; C69o SC4 = I ( . - jl_(~.6-2IS)~s-2+ 4- w J"•4 - -'•78 _1 1 SF-xxo.512 tQ·l~t.··13 rc 6 -I F58) 4-2-a8 _=-_. = 4t~. - - ½.s+. xo I4 P,-D ) =-.16 -L-:6 T.1~ ~:96 = -*.I e ,,1 =*jIt ~a .T S5 x-6 t5 d ( Y\3. +.29. et = .· '} =--. o4 od o"C. spaac L ord Takt; -joo're o ca = a oS oor to"-d o -ýnoor ID2". 4 a .j4J1 romed +er-Toro3Y\. c -- •, . -Z" -W %.aT-2-,.o - dupph17d84.3 inZ aX , -- - xto.2 -• 5 of 6( 30.2zz , Co7-> 3--Z 1. +.) \~:~-2-t t~ i, o*~ = 72) = - 10 .5(o '-4 1 0.5'I ----- 4'3 .&o8 M -) .58 -I.q 1- (t4x I 4G - ~z18 i - =+.(Z •- · -,* i 1 12. , = - 3 -. =) = -. - t --t3 r3.7 0 - - .e.3 -T. 3 1t - FMPo 71i = -+ $, - (C Lo- I, = T.=Z- . 7 1 t • ÷'S _04 ,j -, >5.03 (5.3 .59) = -,4G = +.2S04 3x.+C + .11 - .•• Ac L -.I=z) 11 F- *=-+ = .9.S77 -14 MCL - -8+.3. x 14 .07 14 f r T~o. t5egar 3 t T.2- -.- •-1 ( -5 (0 1.5+ 14 .8 ;MEte ~-+.cfl = +.(i. 13-- •o ·-~~ , ~~~( la 1.5 S '38 *9' -,r: = -+.5- I2,-•"6:o 15 74, .-L -oo.alo "Li "Sepea 'or ofAMu; *SSr . . ) x5a M4 ks - (13 .3-4.,17z p wi = . .4 . - ; mowctpr~r· Ihom-;S Yl 1,3.3 = T1 (x) I1,.3 S- o- ;3 .10o I2.-, 5, 3.9 2,7e0 q 1+q'. 3-9, Live ocdo rre fl~c \!;J pxnc\ \ocad cour;j+s: c -r Vlre Y'L -t ed e dc r ov .,eCc ew ch+al r ye I1~~~arS-t40a 7:o. rooespc a Do4cxa ~4baskeen War~ 1,ocQ*O, ;~s~r~orc C omuejht. 1...-'.c\al-t ~ r ot"a eor reduard Zna we W1em* i erk o wSclv V0 .ae w - re ae4 b:O cre8 M-;\ v Vv W ar le -ticqe re $v s I'k he s)t p Cv ov40aud b -sorakn-· o z .sm 4=r4 c zc~tci4 \Y:i~k"TC+m~\ W:v~e ±hc: A+ se;o ~;rr~3d Upn~~ 40 mr J·;LoF:5ni-t~ ovainc = twccky.F of 'the \:A -- QCC4, pan\, i4-oyjr roc- \ana-rk 1989 + ifk ye 2-,u on.e,. ~=. In tr.ence Tab eeor g.Wi. . ...... 0,510 , _-- a . a. .+ o i 4~ o -i--;T--- ---x --.----- ---- ; I---- . .,. , .. .. .___,_... _,. - .... -. --... ÷ . . ...... -- - --4 -+- , + • 8 3 to yn6 aer , • . - + .3 , -V-- , ------- _-- 1. -_ -. 4X -.- i - i7 4 . U, in _ , 4- x3 I-. ll - a-' ._Lo __.---M , .t .-- x." - I I- - 5. . , i-5 Vii - •-,3 -O ,A _50___ L.,4.t-5 . -- 1-+3 15..-, ...- 13 - _ r- xo -,l.oo ,. -) , -,.2: ,1. i-.\1 i 16---.-4 I , "I -r - -Zo{rc K .7,~ , 15, o, - - • cc--.. s -•-__ -1_. 5 , 6. -4,4 5 -os ___ _ - . .. o \ 43ok -,08 trI ___ k.I_ z t- -9.4. i+ -4ra -3.+o Ic----... .. .-. s , j -. _,, ... I" . ... 17.-" .. . . + 1413 +oi Oa 245 . .--- -- T . 2 - -- -- -± _ - -r-- 4- - ------ 4--- 2.- -- - - - -- - i C--·- -- ------------ 1------I-t i~------ ___ - 2-' -÷ - --- ~ 5-.3 o 1 ... \ - .1 +.21 +.--. .. ,_ ._. ..'.Ia, 5. -to o l . 9-.j" -. ;•4 (0 .A " - 4(.z-3 to.4 3.541 -\\.AG l4 +z<93( x - - --.- 1 5 3 •,j ___ I . I-q •... 10 . ,-,, + .. -5,te -z.5 -.M.t•+657 " o C a -314.. 'z. I -4 o-,o . 4 Ksp I . .< - o " 5.,7o -•- , 1 : . o.i-.-.8-iC.o -..----- 3 +.(08 e -- - .> +.to .... -- z - z_ ... , . . . .+ , t.-.60-_ s 5 -,.S ____ _ _ _ _ _ L. . , C-,.. u .7 __ - --- - ' ----- o4.423 .. 4 ,. A\ , -- , _____ I + Z .)'3 k, - - - cid o0 13 L+. .' T1 L -0 -7hrgO; 2)? SJ--, 0 •. - " ~ u . _.. ¢ 40 'J I .'- 3 >Jf4o ·S! '! o~- o~uc3) QJS Cj Ii 0 t• f1 -4- 0 Ti t' -. _ '-b rr) I -t- w t4. 1- 0 I' ;~ - -.i-i~-·i~iL·~~-~-· -·(--·-~i I··~~~-l···~·--i··~i~;ky-~ I II) " L -t o 0Z* ,",i .7-; qj 5 Q O~' L Ci 0 • *13 0 • rr 5 Lln i-n O o F o- rit4L JrI •. o L'1> LL 3• -,) 2 cI1 7r ' 1t b 0 1f '"I +1 + t ft -I- •n r. 4- -f- ~ld -~ rPtr' 1t 2ka'2k * -r ,4•,ir t" + I -9rr + 1 i -4 0 i -t I 1 U - -44 -It -4W 0 ' t I 0 4144 r,~fn ý,j Q 74 -fri -t ·1rr" -,tj ---~pi~*IB~YL~L3L~L-~yl~i·ZLYIULyqijl~se o 4t J J + • -I4-) U LI:r '4 cs_5: rz-o- -r o 4. I -ti -e -.t- "- -- +1 -• I 4-g + • •- ;Qr: ~ .... II " P t. P t'B' ~·----i~pi-ii~i--iii)~-·=LWII~--I·YY---- 0 1\ I' •4 vr tI r~~ 21• :----I- -) 110 .. ... -l -I-f dT Pt I,:C rn 0i· F t- -ir- ±l TNs -4- F4lg I4 0 I0 6) ~~~ 1~~~'1111~1%1~1 n Yle cayv%?Qýý4-10y-- CL4 oVr of rre -4- 4 - I +-T, f-i- 3- + 4 -~ + o I i -~ -t5 + - o+4 - I - C 'C 4.!i"+ -z ~ +-\- S0 5 4 +- 2'\ - -1 -C 1' -i 4 +3 4 - - ýj ',t-,=5 Yv t-ý ý+;p + - 4 ~ 4- 13 5 a 43 'v. Ctm 4Kc '3 3 4Je- ree -I.. 3 -i-I 4- \3 3' 6 so 43 4.-- to 43 43G 42 I?3 - o.c, -vV\e It 43 0 43 43 43 (d t .j -13 v Pim-)wkai Y<f"he J =- I 1i..5 a vvac r vyi 4 -i2 --T +t6 vNe KAbve C 3'4 35 - - 43 +-1. 43 S. b -3 4-, 43 Z. -l -\164'1-~ 7 ·t,~i ist ~34; +•4 -at7-- ~+4-5 -1' -3 caxr\ \:~e~ir9c~ 1~ ea x. , YAz 2z , r p. , Tyc <ii: c4San bc 84 CON0CL ~ ITNS. T As the least work method has the special merit for its exactness, the method herein described possesses the preference for its simplicity. The good features of this method can be sum- marized as follows; 1. In applying this method it is not necessary to remember any formula, as soon as the general principles involved are understood. 2. The computation can be done with single slide rule operations. There is no long expressianito be computed, this lessens the liability of making numerical errors. 3. As soon as the so-called characteristics have been comouted for a given framed structure, the moments can be found for any kind of loading. ease. Influence table can also be prapared with graat With the same amount of work it does not seem to be possible to do that in applpin~ both of the least work and the ordinary slope deflecti on method. 4. The moments of all members and at all points of a member can be comnuted at one time. This is illustrated in the example of the three story-building. 5. Irregular structures and structures subjected to eccentric load,-,as crane load, can be investigated without the least difficulty, as the comnutation of characteristics and the seParation of moments can be handled in usual manner. 6.- The amount of labor to be put in for investipgation can vary according to the degree of accuracy required, as the separation of moments is quIte elastic. It can be carried from a fairly ac- curate result to one, which is almost exact. 7. The variation of cross section of a member can be taken care of, as it only involves the additional labor to multiply the slope ang-le by a constant. The writer wants to add a few words on the necessity of a close investigation for buildings by this method. It is a well- known fact that for hiýgh buildings. on account of wind load and the econony of the structure, a close investigation is always necessary. For ordinary buildings up to four to five story high such investigation has never been attempted in ordinary office practice, because the engineer a do not deem it to do it lustified to snend the labor and time by means of exact methods Iknown to them. this method appears, it Now as simnle as seems to the writer that it does pay the small amount of labor involved in this method, for it eliminates the "'uesswork which is p'enerally madei in desiýninF. wall columns or columns subjected to crane load for ordinary office endm_ mill buildin-5s. BITI 0• •R•EY. Tulletins #lo7, #108 and #80, Experimental Station of University of Illinois. Morley, Strength of Materials. V. RL-Utter, Die Anwendung der Graphische Statik. Strassener, Forscher-Arbeittmg auf dem Gebiete des Eisenbetons. Strassener, Die durchlaufende Bogentraeger. Schweizerische Bauzeitung, 1917-1920. Hool & Johnson, Concrete Engineers' Handbook. Huette, Des Ingenieurs Taschenbuch.