THE ANALYSIS OF THE NON-SYMMETRICAL VIERENDEEL TRUSS BY THE ITERATION METHOD OF LEAST WORK Ching-Ena Teac A THESIS Submitted in partial fulfillment of the requirements for the degree or Master of Science in.the Graduate School, Michigan State College, Department of Civil Engineering June, 1948 THESIS "‘xc. The author wishes to expre§s his sincere gratitude to Professors C. L. Allen, 0. M. Cade and C. A. Killer for their valuable aid in the preparation of this thesis. 200018 Contents Introduction -------- - ------ -------------- ........ Sign convention ----------------- - ................ Part I. Ron-symmetrical parallel-chord Vierendeel truss Derivation of the general formula ----- --------- The general formula --- ---------------------- -- Illustrative example (1) Solution ------------------- — ............. --- Explanation --------------- - ----------- - ..... Special case (1) Symmetrical loadings on even number of symmetrical panels. Illustrative example (2) ------------- ------- Special case (2) Symmetrical loadings on odd number of symmetrical panels. Derivation of formula -- ------ --------------- Part II. NOn-symmetrical inclined-chord Vierendeel truss. Derivation of the general formula -------—----- The general formula ~2------------------------- Illustrative example (3) ----------- --- ....... - Special case (1) Symmetrical loadings on even number of symmetrical panels. Derivation of formula --—-------------------- l 4 13 16 25 38 41 49 52 57 58 64 Special case (2) Symmetrical loadings on odd number of symmetrical panels. Derivation of formula - ------- ------ --------- 66 Special case (3) Triangular panels. Derivation of formula --- -------------------- 68 Introduction A Vierendeel truss is a statically indeterminate rigid frame composed of a series of rectangular or trapezoidal panels without diagonal.members. It can be analyzed by any of the standard methods -- Least lbrk, Virtual work, slope-deflection and moment dis- tribution. However, in all except the simplest cases, the application of these methods is extremely laborious. Various improved'methods1 have been proposed, among thich the panel methodzadapted by Prof. L. C. laugh.is perhaps the best. But all these methods are far from being satisfactory, especially in the analysis of the non-symmetrical Vierendeel truss. ‘ In this thesis, the writer's original method, the Iteration.method of Least work, is used to analyze the nonrsymmetrical Vierendeel truss. Original formulae are derived, and several typical non-symmetrical Vier- endeel trusses are analyzed by this method as illustra- tive examples. This original method, is, to the writer's c; 1. Dana Young, "Analysis of Vierendeel Trusses", A.S.C.E. Transactions, 1937, p.869. Louis Bass, "Rigid Frames Without Diagonals", A.S.C.E. Transactions, 1942, p.1215. Rathbun and Cunningham, ”Continuous Frame Analysis by Elastic Support Action", A.S.C.E. Proceedings, Apr e 1947 e 2. Engineering News Record, Mar. 14, 1935, p.379. best knowledge, the best existing method in the analysis of the Vierendeel truss. A comparison between the writer's simple solution of his illustrative example (1) on page 16 and the laborious solution of the same problem appeared in the Apr. 1947 issue of the A.S.C.E. Proceedings (EXample 6, "Continuous Frame Analysis by Elastic Support action" , by Rathbun and Cunningham) clearly shows the farmer's superiority. This new method involves no new principle. It is simply the application of the age-old principle of iteration to the solution of a set of earefully formed simultaneous Least-work Equations. .In this method, the total internal work of the Vierendeel truss is first expressed in terms of the statically redundant internal moments. Then the partial derivatives of the internal work with respect to these redundant moments are each.placed equal to zero to obtain the simultaneous Least-Work equations. These equations are similar in form to the equations of the Theorem of Three Moments, and a single general formula is sufficient to formulate all of them. The actual solution of a problem consists of two steps, the formation of the simultaneous equations by a .klm I ,.. irlvr (de‘A1'finuwzl.i‘-§uri . m "IrElrl using the general formula, and the solution of these simultaneous equations by iteration. The greatest obstacle to the general acceptance of the Vierendeel truss has been its reputation for necessitating laborious methods of stress analysis. With the removal of this obstacle by his simple method presented in this thesis, the writer believes the Vierendeel truss is destined to play a prominent role in the bridges of the future. Sign Convention All counterclockwise moments are taken as positive. All clockwise panel shears are taken as positive. Part I. Non-symmetrical parallel-chord Vierendeel truss. Derivation of the general formula. t. MI. , W, E I K”- k E, L", . '—'—* 3‘ ’jr\ Mu” n. (4“ L7 \JM'O H1, \l M“) (D I K! R 1,5]: ‘M J i I"\ q I“ [‘48 Pg 32‘ L‘ \JM" M7 w, __., c 1‘ K. ya); M I“ “b a 5 H‘ >24 ~ 1. L M ,4 (V 4 365M 4 "i ”3* 5'} L' M, ,4; 2* Ea \ MI ”Jam! ”mg ‘ A A’ Figure (1L In the above figure, 1!, and W; are external forces acting on the Vierendeel truss. L3,L.,L1,L”_, 1:3,L;,L’q and L2 are external moments acting on the joints B, C, D, E, B’, C’, D/ and E’ respectively. The problem is to find the internal moments M, mulls, I l I I I I M‘pugsmg9M79M89MQsM/wuqsnjzs MieML9M3amzfsMésMgsm7sM3s I I I u’ ,uwm“ and an. For each joint, one joint equation can be written. The joint equations of joints B, C, B" and C’ are as follows: “1*“ M5 4— 11+ = L5 """""""" "” (1) M511- MG '1- M7 '2 Ls “""' """"""" " (2) Mfi+ M; +- I; = L; """""""""" ' (3) m;- + a; +— M} = Lé ' ---- """"""’”"" ‘4’ For each.panel, one shear equation can be written. The shear equation of panel BC is as follows: 114+ MS + u; + u; := Hh ------------ (5) where H:- panel shear of panel BC, clockwise as positive. h -- panel length of panel BC. Let "8c. '= the internal work of member BC. we’d = the internal work of member Bd. I“: = the internal work of member Cd. atce . i M1 6.7x Then "so “i 2J51 (The axial stress item is neglected.) tv—L ——a M”: M % moment diagram Figure (2) From figure (2), JC. 1! = - I! that MS)? '2 Hence ‘- L7 l "se: ~J2EI[- n + (M +- refit-$01236 -.-_——(M: - M4MS+ M5) 01' I 2.. z. wag cram (Mtr - magi-Ms) also am)“. 3:1: _ t; )J __.-:_E: — . 1 [v — M +- Q 1 MS- I 2M; -16th. *[C [if S>a + similarly, , 1; I 1 2;- "66:051—4w "E‘s "L "5) 'cc. eeK ( c; “cmcfl'uc’ . ’ M~ I 8” ’ ) 3 We” CRIM- HsmIEIW'S M‘*)M*s] BM; 65!», I 2M; >de ’ mm ~—-—.i'- (1M. 9 3M 1 ”‘45 (.El’\ [C e Q>2MS S’ (f! n - . 1.! All: tl’ Let N be the total internal work of the whole truss, then W = Wed“- We’d+‘vc¢31— WQD't ---_--- etc. The given unsymmetrical truss is statically inde- terminate to the twelfth degree. M‘,l1_,M4 ,M5-,M7, m3,nw,Mu,M:,M;,M; and Mn (marked red in figure 1) are chosen as the statically redundant elements. From the principle of Least work, the partial derivatives of W’ with.respect to each of the above redundants should be equal to zero. The partial derivatives of all the internal moments, except M5,MG,Mé.and It, with respect to‘Ms are equal to zero. By definition, 3&3; 3M5 From equation (2), .3M6 = ‘( 3M5 From equation (5), f9 ”‘5" .2 -1 BMS From equations (4) and (5), I. z _ y .0 _ / MLtM7‘f’Hh (MffM5+M+)—- LC. M' cw a “v :2 I own; Hence ‘BVJBL', __ - 37?; " GEN. ‘ 2 "5‘ M ’ DV’V‘QQ’ ~ ‘I am; “oE1<’(2M;:+'M) QNLEI __ 5 / 3M;- " oErxg ( Mo- "6) Ewell _. M = --._.- -- O 314$ '_ a M— .— I and 3:1"? ""' 4' :N4 G E KC, 2W __ since 2M5 — O , we have I _ 1M; ‘IV‘4- 1M:“~t++3(M;'-flfb)_ o - - (6) Mr “4 K1, similarly we have, '“TY‘BW = o 3v“ 1 M -r 11 M M 1 11v 5' 1111;.1 7.1 13:;3 or “ti—'- f—-r- L;+ _ - :0--- -------- (7) ‘1 \4 “(a ““3 3w} __ M“ 13-11:) «:11 M W—n‘ or 3C 1 1 t __?-1‘3 ‘320 ................ (3) K4 he K3 Eliminating 119156 and Mg; from equations (2 ), (4), (5) and (6), we have 2th! --(3‘-+--3’- Eh s (—'—+-'-) M’ S'_ M Na) K; Na 41 L 3" i 3 t' I - (Nit-R: M) M++' K6017 - “71.1% - Lg) ----(9) where 1' 3 i t .2: ‘7‘"1" T— 1- ~57- N \lo a, “(It-[It f1 I‘LL! 10 This equation of MS- is a useful general equation. It can be used to find ML,M3,M”,MIL,M Malay and M”. For instance, 2tmg=(§+3—)Hh3(:- (trig) 14+ ’ I - 1.... 3-5))H1- -l7+Lé- L6) - ------ (9)q 2t‘M1-(-r-—-)n‘h 31-51%) M" 2— I - (F1. MK. '8) I TE”: MILK L5) ------ (9)5 I 3 2 131116: (hr 21mm.- amt-R3) M' ‘ I I 3 I I "EV N K3 1; M1+W3(M4' M++L3- l1'3) ..... - (9". where 1 3 3 . “'3 ‘7' i” ”*7- t. k‘ ‘1" -K-; K. H, - shear in panel BC. clockwise as positive. hI '=-— panel length of panel BC. Substraoting equation (8) from equation (7), we have 5%.!sz ._ 2.11;;1’; +£13.39; ....... (10, Eliminating M3,M;,M5 and M' from equations (1), (3), (9): (9).: (9)be (9)6 and ”(10),;0 have {. 30:;3-{3 f1J£[J—K‘r._ )+N(K‘1 t)”}l\/ '{ :Kg)1x[WWr—H-n)§’ —‘r-c-‘.—- ——-—) ' ’ L -—'- r a:- { KLN} r; g Hh m1}— {fing’H LII -1— {3—H (Y; 44%“!!! 4,. -'-.-+-'.,,-- 4—+—'-r .. m K: EM, --1- {WE (M; -1Vl7)+{_’%iT<_:x.g(LQ-L£) I 3’... .. ..................... .. ........ 3|- ’1- C - - . . .- - F 0‘ fi . . I - . ’ .- \ \ n" \ - s -Q ' .- . ‘ "‘ e O 3 an.- . .I Eliminating! 3, @M6,M;,Ms and m; from equation: m. (2). (a). (4). (a). (a). ma. (sub. and (9)“ w- haw . . {71‘0“} ‘) “L :43; A” «3K; W H— z‘j’n + :Qé; " < ”M W =;’-g ifi<fi*%*—lfi*m§ ‘1'" WG-3.- - 73M +2.2:K3C—‘z-If-é‘7- e: M, +21%; 3%- {S'U'f—E’MV 25$ f; if V317 1-57; fi+i aw; Intorchanging all tho L, M and K with the respec- tive L, M, and K, (1.0., ohango nfi inton lg, If into I‘, Ki into Ki, K. into Ki' etc.) '0 get a naw equation, aay,eqnation (15). Adding equation (13) and equation (12), we have, : a Q -'.——‘ ' , s _I_ '7 ‘ K3 2. 1; l H (3 LL? . i .‘L L ' 75° 7‘: T— KKK: 1— 7’: K m) MI ._r_ .—_._ 4. . L Cf K3 2. L __ I _L ” 4 H +"’L, H _ 2 ,c i m, m, w 3 ME, ’1 Hm. + M s __ 1.. .E. H M _ “3 K3 LYN; N N3 N4) 7 ‘ 7 G H ,/ ""fi— + --- ——-——- IV __ L L + 1mg C R; My. *‘W3 17 C 3 a) 9 I 9 _ ______ _(14) K. + LLKQ}. -L[t\q_>Lt" + TIC/(H 24:“; 2:r\+)L :3 Eliminating M%.¢ from equations (11) and (14), '0 have our general formula. (on next page! 15 The general formula. (v’u+ u’v) M4 = G {-Qk-jh - -- u) M 2QB. 13’, + 3R3” fTu) l:+(E v «A u) ITHE V+Au) II; \I where l ._ ,é .f 12; ' '- + Q = 2 (Jur a) r = 11A1- lZfi‘fiéfi- 6 A .= A - fl; B .=---- A + 4 c -- afiWIs/qfir -—-)+11K + 6 D = 43; (J; + 5) - 4+ E = 11-45 - vig+ 6 G =- (Aqu) Hui—{fiffi-‘fiiu - v) H‘hl'i- J J = (v - D) a (La. L3)-t- §¢§3v (L3+L3) wk 11 (LG- L’G) - v (EchfiL’G) u = V: +0 v = Bfi,(l-~§—)1~D c ’51 \ .44: 7::- f’ = 55° {4- "\4_ _ Kc, £3 '— K3 . r r f . ‘9 I . . . o 0 .. V b a. u-a ‘ ‘. ~ ‘ s ' V S e- . ' ‘3‘ ". ‘, Q 1 a o- .u a -1 _ ‘ P J ‘ a , u a , . ‘ ‘ g e ' , w | l ,_ I ‘ ‘ ' ‘ - , . — ‘ .n l a n ' x v .- .. . . -‘ - - . i . ~- I \ - . a. o - -. . ‘ — 14 In the above general formula, A’ is obtained by interchanging all khvalues with kLvalues, and A, is obtained by using the kbvalues of the panel immediately below the panel under consideration. Thus - / AI=fi+-’£K A. = 4| -— k: The above relation.holds true for all the constants. Note that the values of «L .fi , i” and-{(3 do not change by interchanging k-values with kLvalues. Nets also that'é.is not the stiffness, but is the stiffness ratio. In analyzing any nonesyunetrical parallel-chord Vierendeel truss, there are five steps: First step. Calculate for each.panel all the constants. Second step. Formulate two equations for each.panel by using the general formula. Third step. Solve these simultaneous equations by the principle of iteration. Fourth step. Solve for "Batucetum’mewndfi ”661'ka and ‘53 by using the general equation of [5 shown below: . SWISS:(l&;+3)Hh_(B’*—5)M4 _3<&;+I)Pé-+3 --~ E -— - —«»~> - 6.0J~: - 4.2rxo' — - -1—9' -19.? I 7. ‘ I e]. "' *L ““““ L L 08 e1 “ "L """" "‘ — — ‘J Tan]. -5205 -5302 1.0 " 608 ’4 e " " . . .- .4 ~ *2 - . \. . . — ~ 0. . A O I e' e e -. e "' s '- . 0 ‘ 0 w _ O p 0 o '- 45 Table VIII. Computation of M34 Final moments [: goszicients 0;, Rib M 3(ILM+D§§9 due to 4. 4. 4‘ ‘7'7 Loading B '05 2 -1.5 - 3 1.0 "—fi -41.7 - 72 Ida, 1 O -105 -608 ———_ -2405 “BA 5 -495 ’ - 6 -5205 -23.‘ -6007 -216 "5N 4 - 3 -4.5 -53.2 23.4 -49.4 Thble II. To check the values of Ms-found in table VIII by means of the shear equation. I I BC - 72.0 - 72 checked AB -215.0 -216 checked M found in T‘bl..x= To check the values of Mk_and 5 table VII and table VIII by means of equation (8) on page 9. sum of sum of Panel positive terms negative terms in equation (8) in equation (8) BC 133.2 2132.8 checked AB 325.1 -325.7 checked 46 Table II. Final solution. ‘ Moments due to Final Moments by Member Loading A loading B Homents Amirikian (ft-kips) 15 -12.0 -52.5 -64.5 -64.7 A'E -55.2 -55.2 -55.1 BC -12.0 1.0 ’1100 -1101 95’ - 6.6 - 6.6 - 6.9 BA 12.0 -60.7 -48.7 -49.0 BK -49.4 -49.4 -49.4 06 12.0 -41.7 -29.7 -29.5 dd' -24.5 -24.5 -24.5 1x 64.5 .64.6 11 55.2 55.1 66’ 59.7 60.0 66 56.2 56.3 00' o 0 at 0 0 4'7 Elanation. The calculations are similar to that to Illus- trative mample (l) on page 16 with the following exceptions: (1) For panel BC, . .z 3 r -= 12g4(f(1+1—) A 0 J, , A ‘ 550(04)- 614‘) (1+3?) D -==- 72; (Ar 5) 0 Thus for “50’ r =- 12X1K0.5 (l+%) = 18 A = 0 c = 5.1 (1.5+6xo.5) (1+;%—) = 40.5 D = l (1.5-('3) = 4.5 (2) There are no IL, and In; terms for panel BC. (3) For panel BC, the equation of Us is zans=2fiah - 514+ - 515:5; Thus in table VIII, the computations for ”‘8 are: Coefficient of H h- 2 ééf" 261= 2 Coefficient of M4 = -B’ = - 1.5 Coefficient of 11;”; -3&;'='- -3Xl=-3 .___!_ .. .. '::- mu aw "‘1...9E"72’H’1’5”‘1 5( 6.6)] 41.7 e ‘ e .C — ' . I \ ~ ‘ o. ' . '- ' s . 0 , v- , . C. em I . v. 4- .. . l 48 The last column of table XI gives the moments computed by A. Amirikian on page 130 of his book, "Analysis of Rigid Frames". The check is close enough for slide-rule computa- tiODfl e Special case (2). of symetrical panels. Symetrical loadings on odd number 49 >< 4 f I L l h I / '5'.- .. 7w ‘ ‘75 VT» . E 6 5'1; /~ (LN L K4 (\l K: (\ ’AR A T7 J . / “ 3» cf 5’ M"I "5' "4- M: M) ' L l a? 3 f 45;: {‘5 ‘ C“ B A N 'N/ v' /// r V L“ MEL 7113 “L (r w "W. “V;- TJS l ‘7; "w; W.— as... (61 In figure (6), both the (Vierendeel truss and the loading are symmetrical with respect to 11.!- Hence for panel DE, "5" 11+ "55“ I; Hh=0 1‘6““ L3 Lé=- L’3 K6“- K3 .1! Substituting l§=- M4 into the equation of M5. on page 14, we have I a, 1' {11+ -(Bf 5) 2+ -5(4 +1) {+5(M;-u7+L3-L3) --- (20) Substituting M,:=- M7 and M"- =- E? into the 2.. shear equation of panel CD, we have I 11,11, - (1171-119 =--— 2,5, ---------------------- (21) Eliminating m; from equations (21) and (22), we have 9 . ~/_ - _ i- ------ 2s7==(k+r1)x4-(Qr1)sg+my»: spins, L5) (22) similarly I 4;. M’ (3 _ g , / 2l7- (541-1) 44 4r1)n++(u'+u, -Hih|1-L3-L3) ------(23) Substituting values of M7 and 11,7 of equations (21) and (22) into equation: (11) and equation (14) on pages 10 and 12 respectively, we get two new equations, say, (11)" and (14)”. Eliminating M’ from equations (11)” and (14)”, we 4. get the following modified general equation: I 1.)?) (v’u+u’ v) 1141-0 +Q(2v - — u) I. +€(2v+-—-‘-u) M’ Q: ( where 1A. G J:t=-={(~---u-2v) H‘h,+ J =(v-D)u (Ig~L§)+ €v(Lg;1-L) 9A'(L 3-L;)+ 7(EL3+ E L3) 0 2&(611-19” 2(£+5) D = 34%; 51 All other constants and coefficients have the same definition as that of the general formula on page 13. and M, terms. 7 7 The general equation of MS-for panel DE is, of course, Note that there are no Hih, M l5n==- quwhich.will not he needed since only half of the symmetrical truss will be analyzed. Care should be taken to apply the modified firmula to panel DE only. The original general formula should be used for panels AB, BC and CD. 52 Part II. Non-synetrical inclined-chord Vierendeelttruss. Derivation of the general formula. L.)_ I L.‘ Mn. /) M” “n. M; Figure UL In the above figure, VI, and '2. are external forces acting on the Vierendeel truss. LB’Lé'Lq’Ln.’ 53 Ig,Lé,L; and Lil are external moments acting on the Joints B, C, D, E, B], C’, D, and B] respectively. The problem is to find the internal moments Mi , 111,143,M+,MS—,MQ,MI,M%,Mq,M;.V,MH ,Mil,mf,u1,mg,u’+,mg,1{b,1( , %WMWWJaMM The derivation of the general formula is similar (2' to that of the parallel-chord Vierendeel truss shown in Part Ie Thus we have the same Joint equations: Ma."- M3 -r- M4 = L3 ---- ------------- (24) ms. + MG + M7 = L5 ------------ ”m"- (25) M: +- M} "r- M; = L; - -------- --- '''' " (26) I i I ...... I. - .......... - ----- - 11.5—4— mg -+— m7 -— LC? (27) The shear equation of panel BC is derived as follows: First, we have 9+ + 14+ + ML)- vfl3 - 0 -- ------------ (26) .-.-= .. - i ....... --------- Q+ Q.7 H 114-12;- LC, (29) where Q. =The moment at a section Just above BE, caused by all the external loads above this section, counterclockwise as posi- tive. Q, “The moment at a section just above 00’, 7 caused by all the external loads above this gum-amelockwise as posi- tive. O C -u- 0'. fi ‘ r. _ \ O ‘ - f‘ f. .0 . _ i a. _.‘.— .4“ I — .7 r- - - .— u- L. -- N- 54 £3... length of BB. H = panel shear of panel BC, clockwise as positive. h = panel length of panel BC. V .= vertical component of the axial stress in BC, tension as positive. y v l. . ‘ ’(é a fir—1,65 I ' , ! MS C. M5 Q. % l , H ‘ 5 ‘B 4 {4+ M; £3 \TJ From the equilibrium of BC and B'd as shown in the above figure, we have M+M+MI+M/+V(§2 ”(1') = Hh ----------- (30) + 5' 4 s -3 6 where _ £0 = length 01‘ cc. Eliminating V and H h from equations (28), (29) and (30), we have our shear equation for panel BC: 55 Ms 5 'hOI'O .C L n: ‘ £3 / Again choosing Ml ,ML,M+,MS,M7,MY,M’G,MV,M; ,W+,M7 ’ l . _ / +M + n(M4+ M4) - Q7 --nQ,“:t-Léi-LC9 -- (31) and II". (marked red in figure '7) as the statically redundant elements , and noting that a [L B M I =-———-~—-. 2 ~11 3 4, a "1'4 we have 3w __ 3M5 —. / / - . ’ or 3M5 -M+ _ ,ZMS -M+ f ,$£'*‘e“Me) ___ 0 «mm (32) .. 'I ' K4. K4 V\g .29."- .= o 8M4 I ’ . I ~’ - «M - _w —:~ I» . —V' U“ 3 or 1M,” ‘5 -- 1M5... Mvm- Lm‘” -. ‘b'm’ "BK—fl?" 0" (33 m, M N 3 22!. = am, 0 , / v ,/ -' -. .m’ «4 or 3““‘4'MS- EM-Ymi 71,+ 1MQ’LMB K- L 34-. 3 ~=0---(54 Kg "4 Kr. K3 Eliminating Méfllé and M2 from equations (25), (2'7), (31) and (32), we have the general equation of M5: Q IS=(2&;r5)(Q7-nQ++L6'1-Lg) - Baggan ~ng 114 I I I. / _ _ ’ ........... - [(2§+3)n+£4]u4r3(n7 M7i-Lé Le) (35) -1 “6 where {4 == ‘7:; ‘;l :2 K6 A I, \ Proceeding in the parallel-chord the same manner as in the case of Vierendeel truss, we can derive the following general formula: (on next page) 5'7 Theieneral formula. (v’u-i—u/ v) M4: G +§\€<3[niv - where a»: Q. 112:3 [11, V " 1%.,” (A: “(+43: )] “.1 '1- [(E’n+3 A") v win] M7+BE n+3 A ) v “hi! M/7 (A‘ n,+i§, )1 MI dL-r— £4 + 474: {t = 2 (okf 3) I r: 8(.L+£41€4)+ 6 2 -== 34 1- 4K44é1, A = 45+ - £4: = 3 (fit 2%) + 841.5; fi;(3l+ 2&312) 44(0H- 4) == 4 (zfig-ELH 6 6= [Au+(rn+$)v] (07- nQ++Léleé) +§fi3(%-'u - ”(of n.Q'+L3+ L13) + J J ---— (1%,(1 - %)u(L3-L;) +§€3v(13+Lg) +J. u(L6-L’G) - [_(s’m-si’ )LQ+(En+ 3A)Lé]v II II B C D E G u = 9&3 +rn"+Bn + c a... ,. é - v @533” - .5) An1—D .19. ‘1"- N} in. 434 M . __Kt'a. «2,- K n: r: Illustrative exgple g 3 2 58 K? 5 IS F I Q K=z 5’ . 1. f \L" K: \. D/ 2 A/ \g #9 ‘3 3‘ ‘Q. 5‘ if ‘0 x x 'r ‘p K=32 Kn. K‘a A ’1?" C 5 77977 I I 3 ms = 45-0" j 5 Kit» (’[0 KC? 5. Figure (8) . ihe solution is given in. tabular form below: Table XII. Computation of constants. . _ q ' ' . Q4 n _QT 71.10“ 46+ (L Q ( g Q’fi} A B ”CD 1.333 -.667 43.3 75 .625 46.8 3.333 12.67 54 20.7 16.9 I, 2 .667 39.3 or! A 'Bc. ’ 150 l 75 1.5 3 12 48 18 12 0 ; 36 'B’c' 'AB 1 -0.5 28.5 0 1.6 -l50 2.5 11 38 13.5 8.25 IA’g' 1.5 0.5 25.5 1 ‘k Ar . _ L -_ .-‘.i _. _ ._--__-._.»._.. ‘ 7‘ . ‘V 4 I | 0 | ‘ e A. . ‘ O ‘ ' A O V v - \ - < . J r 0 O 0 e 0 s O . — e s . . . O O 59 Table XII. (continued) 1 1 I c ffi i t ' r C D i E u ! V f 1233:? c i? M M” i i I ”if“ ““3 QWWF ' ' nnb 49.4 14.67 8.67 114.4325.5‘ 1204 -597 - 86 877 4510 ”6'25 54.6 9.76;16.7 97.1117.8 1055 -501 -109 708 ? ! i 1‘ Ma; I i '3361 250 970 36.0 10.50 12.0il32.0116.0 4220 1055 7 m J - 48 -554 365 6. I MAB 52.2 9.75 8.0 185.2 18.8 1500 .__ -—— -—— 5740 MAIB’. 22e5 6.50 14e0 168.7 14.0 3.122 """’ "— ___ -»————..—- I) -, a. - - 1 . ._ .-..( -. ._ (Li.__.._..-_‘._.ll_ - ,. L. -. -.. 11-..--.“ .L.-i.____.1_ Table XII. (continued) Coefficient. First -Iteration factors °’ -~ G I I M M A rox. M M M M 7 7 pp I I 7 7 Men ~——- -— 26500 6.2 -.0200 .204 -——- -—- “60 129700 50.8' .0595 .250 -204 588 -.0485 ..159 Md! 86500 20.5 -.0840 .0865 MAB - 9 671 ~195000 ~54.0 .—~— -———- -.0016 .117 ngfld ~262 756 -168500 :29.5 J -*—:H’_:::::_L-.O456 .152 *r 60 Table XIII. Iteration. i "A3 ”115’ u 04: 14 8’6 “CD “0515 Iteration new" .0573 .0365 .230 43+ -1076 .132 .,0200 16+ .117-.0456.20+20253 :0485 .137 .139 -.0$85' First approx. '34'.0 ~29'.3 30.8“ 20.5] 612 6&0 I ; \9 \9 ' I Iteretien 1.-----9---” - 2.0M- - 2.9 g : u . k-“ F“ - 6e7 (:1 205 y l ' " 03 I 08 +"‘*"‘J : OBJ : "' .3 ‘(fi “-‘.“ “1' ~— 0 3.0 *“""‘—""‘/k"" ““““““ 3"- .5}[~. 30;}“1 2.4 (M .1.0 4: “““““““ " ° "’ --"A#" 4.2 m - 0 £75 ; 5 L 1 n ' , i : ~—-—-—»«9—---—----> .1 .2 . b- - "' " " + 05 0‘ " .2 i . : " 02 6 05 '”"""" l .5 1 - .1:$fi*-'“-‘r"-“ { i f 7 *——— .1 «so ------ —-4~ ~~~~~~ ‘ . -—- .1 3:. .1 Total -31.6 -27.2 23.5 21.0 110.0 1 9.3 61 Table XIV. Computation of Is. Coefficients of Q _ Final )an I+ F; 1 "a. M+ Bug-191.1512) momenta “p" 7 -5e04 -6037 e 1 10.0 _ 18.8 ' 46.3 5.67 “105‘ -4087 9.3 — 15.9 23.5 '2.1 15.4 6 -4.50 -7.50 75 21.0 2.1 15.2 6 -8e60 -1101 ' -3106 -705 -3004 '150 5 ’6e50 ‘ 9 '27e2 7e5 '25e5 Table IV. To cheek the valnee of [5 found in table XIV by means of the ehear~equatien. I Panel |‘+l5+n(l4+n’+) 0:; n Q} CD 46.3 46.8 «hacked 80 75.1 75.0 checked AB -149 .9 '150 .0 checked T8131. XVI e To check the valuee of l4_ 5 62 and M found in table XIII and table XIV by means of equation (34) on page 55. eun of an. of Panel Positive terms Negative terms in equation (54) in equation (54) CD 51.4 -5l.4 checked BC 42.2 -42.1 checked AB 66 e3 -66e3 0h.°k°d Table XVII. Final eolution. Moment Ilenber (ft-hips) -3106 2% -27e2 BC 25.5 5d 21.0 CD 10.0 dd 9.5 BA "30e4 H! “2505 08 15.4 dd 15.2 DC 18.8 dd 15.9 AA’ 51.6 n 27.2 88’ 6.9 85 4.5 00’ -25.4 d0 -24e5 Dd -18.8 ab -15.9 64 Special case (I). Symotrieal loadings! on even number of symmetrical panels. 24>,- Ki/ \{J V ‘99" l ‘I In figure (9), both the Vierendeel truss and the loading are symmetrical with respect to DD’. Hence for panel CD. I 1‘5'" Lé— o ‘5'" M7 I: - ll5‘ ll’7 Substituting 115. -=- 117 into the equation of l5. on page 55, we have -g I7=(2fi1+5’)(07- 5,931.61: Lg) - kzfimm €£4JI4 - I32fi+3)n+%4]l;+5(lfl-u7) -------------- -- (56) _ a 65 similarly ' A; n’:= (2£++5)(Q_'- n Q++Lé+1£) - [wag-5):: £11; 7 - [(2fi+5)n+fijn+ r “MT-1’7) -------------- (57) ’ from equations (36) and (5'7), we have Eliminating M 7 I __ - ' l - l ----- 2.111?- 215,497 n 0+) H2fi4n .fl4)u4+(2n+1){;u+ (58) similarly‘ I._ , 2&M7--2 6.4:)7- n Q4) +(2fz4n -.(Z;)l4+(2n+-l)£+u4 ------ (:59) minimting "7'“; and M; in a manner similar to that of the symmetrical parallel-chord Vierendeel truss shown on page 39, we get our general formula which is the same as that shown on page 57 except that there are no [7 ,l;,L6 and 132 terms. Also six of the coefficients Y , S, A, B, C and D each take a new dafinition: r = ekfizurifl $ =2 .150.»- in A = 0 B = afififiuf) = r c =-%1(§A+2€4)(1+i-) D .934. 1 I Note that A is no longer equal to 4+ ~33. — _ The new equation of I; is: I ’ 4 A I 2Au5=2£4<0f :1 Q4) - (2744:: 44):; (2n+ “£7.4- Note that these new formulae apply to panel CD only. The original general formula and the original formula of II; should be used for panels AB and BC. cow-.. 1 \ N i - W ' .- ‘ ~ ~') ‘ I s- 4 9‘ \ . . a- 0 ,.. a. 7 '7 ' . ' a... v x V- . '- 5 O \ O \ . _ 4 t - 4 i ' ' 1 ( ~ ‘ o . e 66 gpecial case (a). Symmetrical loadings on odd number of symmetrical panels. In figure (10), both the Vierendeel truss and the loading are symmetrical with.respect to xx. Hence for panel DE, I, I: I ,I. ' 0" if. r v Q I. . - . la a _ t . II D . ., t... O O 71 4 s . . . o ‘ w 1 .. pa . 4. . .PfO. . a. . I o . ? ‘ O ’ \ . 1 A A a — . 4. .fl. _. n o s . l g . \n I.» V . . ' y e n \. !- a d 67 From equation (29) on page 55, we have I Q7-nQ++Lé+Lé=Hh=O ‘ Proceeding in the same manner as that of the parallel- chord Vierendeel truss shown on page 50, we have the following modified general formula: . 1 2M. _ (v/u-ru v) M48 G+Q[2n'v - E-(A‘ n'.1-£:)J I, +5£2n.v+%b(A’,nfl-£._)] l" A: 5:06-5:11 - 210(04- n,Q,+Ig1'1-;) H J : ch - '%)H(L3 -31" )1. Q V(L31"L13) '1' 9AV(L3'L;) .. .Enfi’- 73 ”ran. i-(llg - 74%;”)1.) u = 8 1- 2Q(6.L+19)+2(£+5) v= € (1"?)1'Qfiq. All other constants and coefficients have the same definition where as that of the general formula on page 57. Note that there are no 117,155, and (Q7 -nQ,+ +L 65+]; ) terms. The general equation of M5 for panel DE is, of course, M5=- .4, which will not be needed since only half of the symmetrical truss will be analyzed. Care should be taken to apply the above modified formula to panel DE only. The original general formula should be used for panels AB, BC and CD. 5: 68 Special case (5). Triangular panels. Figure (ll)_ (A) For panelGH, we have n h: 0 an= 0 ((62:60 M 5 Dividing equation (55) on page 55 bylflwe have . I I 1-M5-— Lér L6 the equation of M5 for panel GH: 2111,4114; -§+m;+2é;(L6tL;) ------------ (40) h ,. ___; W ore 4+ '3 '.\+ I f '= “#7 5‘. 69 .4 Dividing the general formula on page 57 by ks , we have the general equation for panel GH: (v’u+u’v) M = 2443(2461u - v)(Q - n Q-rLTLI) 4' {I 1 + ’ l 5 3 2 Ac 1" 2di§3[n‘ V "T(A' nfiflfi] M, 1- 28.1% [n‘ v ,- ;t’i: (A1 n+1“ )] M‘ l ’3 1' .- 2?. -4’2. - ’ *- 4&fiv(Lé+L®)+ 211.3(1 i, )u(L3 L3) 7) / ' 1." 24\&3V(L31‘ L3!) where i "14 '-'- 'fi; ta; :- ‘37" M . z ___1 ”“3 {:3— o u a: 2.1515 Mara-k 'r-2 (:4) - I v== 2551 "g.- )7— «141. All other coefficients have the same definition as that of the general formula on page 57. Thus I a I A :- "l4, " An. ""1 h 75"" a! a K3~ 1 ”-7 31/ Al .. ’51 ~ “I , ‘ I ,5: ‘2 1({sf’kl 1‘3) \ etc. I 1"! Note that '5‘1-a'ff; and 4..., .775 . ‘9 44 9'- .V '70 6*! (B) Still refering to panelhof figure (11), and noting n z 0 and (Cg-00, equation (55) on page 55 becomes: 4.1+(2u+ ‘1‘!) c£3[2(L3-ul-n+)- (L;- dig-1%)] ----- (41) where ’ “‘1. K4 I {3 2 K3 Eliminating M‘ from equations (40) and (41), we have (1:42. a m “7.1;: +2 an...“ M, --(L3 'ML) ------------------ (42) similarlyufl 1’. .{4 $3 I I I I (T+2 “t )M+'=-(—CI +2 3) "Si-2(L3'M1flb‘i‘L‘) ‘ “(L3 .“1) .................. (43) Comparing the notations of panel GH of figure (11) with that of panel AB of figure (12) below, we find w' t ‘. C‘ ‘b hi ‘ cl. .1 . e. . .8 ‘ a . or; .‘ o ‘ C ‘p‘. '9 I...‘ n.‘ 4 \. ’ n ' . ...9 .. n. _ . . o .o I. ”I _ _ .a1l fl . V a ._ l! ._ ,. _ u n ,l... :9. 4 u .n .r a“ I .J _ t._ a v a .4 71 K3 corresponds to K6 corresponds to M5 corresponds to Hg 33: corresponds to M+ mxw +~¢ corresponds to M4 3 corresponds to M7 5. corresponds to M7 corresponds to L6 t"_t" u‘w corresponds to L t-q 6 corresponds to L3 1% corresponds to L3 Hence equation (42) can be transformed into the following equation of M for panel AB: 5 (K Ms. .. R 1141-0174117) t“(2L6 L6) i» 2(L3I'Ig) (44) Eliminating m+, M4 from equations (40), (42) and (43), and transforming the notations of panel GH into that of panel AB, we get the following general formula for panel AB: (figmfigs’r M m: «2&3 431’ )(u -L6) I 1» (% Hr 2K1? )(m’ -Lé)+ (2&3-2fi:d+i§s’+2fl)(L 3+L3) ------- -------------- --(45) In both equations (44) and (45), . _/ K6 ,_ K6 n’fi 1'" “£14" 72'; " 4+ K4 ”L ‘ £4 I ‘ e .. a. ' :Z* R. "' 2(.:+u)~.f§,-t S = (1'&+*3;7+1)R '72 (C) The equation of Us for panel BC in figure (13) is the same as the general equation of I: on page 55. Figure Q3) 1 Proceeding in the same manner as in the derivation of the general formula of page 57, and using h equation (44) as the equation of Hg. for panel AB, we can get the «and following/formula for panel BC: (v’uru’wm 8 G + eggs‘fi‘vf it)”: 1'? 4- . ,- +[ce’mm’w-«L a1M7+BEnHAW+Ml M7 £3 5" (V- A“) M" where G J a: 2. {4.3 (Rf-K.)»(L.+ L1,) ~LELCR,’+R.)V(L°*LS) 4. [(I~1R, ~R,’).u.+-(t+a,’~zfl,)vj €33 L3 +£