TABLES FOR STRESS CALCULATION THESIS FOR THE DEGREE OF CIVIL ENGINEER JOHN R. LAMBERT St oe SOR ON ON ee THESIS TABLES FOR STRESS CALCULATION AND PROBLEMS IN DEFLECTIONS THESIS FOR THE DEGREE OF CIVIL ENGINEER MICHIGAN AGRICULTURAL COLLEGE JOH R. LAMBERT fem 1914 THESTS This thesis consists of a set of tables for the calcu- lation of stresses in bridge trusses and girders due to Cooper's wheel loads. The tables are from original computations with the following exceptions: Table II was published in the Engineerin;z News of June 31st, 1906, Tables V and VI were computed by if. Clifford Rowe, M.A.C. 1907, and the writer. In addition to the tables there is given a number of problems in which the writer has been interested at various times since graduation in 1906. The problems treat of deflections of beams and trusses and of methods of calculating stresses. 102135 ay —_ —es EXPLANATION OF THE TABLES. Table I gives the static moments of the wheel loads and of the uniform load for every foot up to 233 feet from the head of the firsg engine. In all the tables the quantities are given for one rail of Cooper's E-50 loading, except where otherwise stated. This table was computed by means of the constant first diffference between any two wheels and the constant second difference under the uniform load. Direct interpalation is correct between any two wheels and suffici- ently exact between any two feet under the uniform load. The exact moment may be obtained by deducting 425(x-x*) where is the fractional part of the distance This correction is a maximum when *: 25 its value is then .37/25 Table II shows the proper wheel to place over the common support of two adjacent spans in order te find the maximm concentration. Ifthe maximm moment at any point of a span is desired, the table is entered with the two segments of the span,in fact it may be used to determine the setting of the load for the maximum value og any function which can be repre sented by a triangular influence line. Table III shows the position of the wheels for the absolute maximum moments in girders, and gives the equations for the moments up to a span of 70 feet,and the span limits over wnicm each equation can be used. WwW woe —t. ; 4 : Af x —— W The total load on the span. f Xs 3 t Ww, The load to the left of a given wheel. i an ‘ 4a ” . e . ’ . ’ e . - 7~ “im Moment of the loads to the left of any given load about that load. “a” the distance from the centre of gravity of all the loads to ay given load, « is measured from the centre Of gravity and is negative when measured to the left. The bending moment under any whee, = “ %,— “7% . This expression is a maximm when *-~ since the mum of * ‘7%, is constant, (+2. gabstituting fy for ~ ana ~,, t= MEA) If the moment under the .” wheel is greater than that under the (nvr) Wheel, then 2 +Anss) W faa) — Mn » We —Tln+y Reducing the im@equality by substituting for 7.+) » ma + Wi (anei an) there is obtained, Ww, >W (x + Se") This is the criterion ty which to test for maximm noment. The maximum does not mecessarily occur under a wheel Which is adjacent to the centre of gravity:of all the loads. the equation for moments may ve written, “7-60 + +Cs in which —, 2.x C3 are constants for any give wheel in a given set of wheels, ¢, - “, Cz = Wa”, C3: War rm The equations are written for miccessive mmbers of wheels and solved for the simitaneous values of -, fne limite obseined hy solution may be restricted by the following conditions: mS last span upen which a given set of wheels can be placed in correct position 1@ equal to twice the distance from the mid-point of the distance < to the farthest load used in the calculation of the distance <, aM thegreatest span upon Whieh the wheels can be properly placed ® . a ¢ ‘ . A” t 7 4. tt . . te “—z 1 . ‘ a e: t e - ‘ , { “ ) @ i . a 4 ' . eos . ry “oy, ’ a - we’ é ~ ‘ we, “ ’ “3 ar ~ s.4 . ; eo ae \ - a 1° - | , . . ‘ ’ . oN ° v= a , . * . 1 . + + sy & be . ’ _~ } - « ot] - toy é « ae ’ « 3 a) > ae t o. . « ae . ‘ - + ' ‘ td . : wy -_ Le . ‘ v4 . . —~ = 6 + ve : . ct 1 @ ° a | zi + , _ ¢e + Ree , - tm 4 : V8 . a _ - 4 ran . 4 74 ca x ' --4 oo _ ¢ : fy . : . ' ‘4 ‘ 4 . . -~ ’ -~ re ' * 4 , . .’ ¢ - + ® . abe aA a oad ea a ’ ¢ L.3 ‘ wd." ; e- ‘ eee . : "% wae ‘ > ’ ' . . 7 F 7 . deny 4 - eo. ° . . . 4 « : ‘ ‘> *y “4 » Lk . ‘ . ~ * t wv. ~ * t . ¢ q , . ‘a ~ a. ae is equal to twice the distance from the mid-point of =~ to the nearest wheel off the span. With Cooper's loadings, the uniform load is on the span for lengths above 69-9 feet. The distance of the critical wheel from the centre of the span therefore changes for every span length. Bquations are given for determining the length of uniform load om the spang and thus the other quantities can be founde A= 2 for a critical position M+W2w+~x™ ee — -— W +~~x 4 f x —-¢ #{6*-G.w[M-W4- J X= x 3A Mis the moment of the wheels on the spam about the head of the train, W is the weight of the wheels,~- is the weight per foot of the uniform load, c= 2W---(“~f) The lengths e, ff and ~are ag shown in the figure. The aqations for length ofaniform load given in fable III are wriften for a Cooper’s loading having one pound on the driver. These equations giveg correst results for all of Cooper's loat- ings, since the spacing remains the same and the weights are proportional for each class. Im no case does the absolute max imum moment for spans above 70 feet exceed the centre moment by one per cemt, and the excess is generally mch lesse Im nearly ail cases the absolute maximin moment occurs under the wheel which causes maximim centre monent. Table IIIa gives equations for a Mallet emgine, the present stand:ré of the Chesapeake & Ohio R.R. Table IV gives the maximum moment, the end pang the quarter point shear, and the center shear for girders up to 129 feet in span. The shears at the intermediate points are a 2 . & . ' mt . . a % . . ‘ S. . co . eo8 “ ~ . - * a ° . ‘ . , a * ' 4 e « 4 : Loe . . . Ls 7 ° : > . oy “” ~ , ' > ® . ° ~~ «& £ ce ) a eo .- ’ {7 a? , * . os, : a > ‘ wi. A ve oe * . - a ° — * a,v,ew. . . . er - . e rt ef or) 3 «' aie - t * . i e re . ' “ . a _=- “- $ ', t . - . ok . ° ‘ 6 . ° . ‘ . > , -~ . , « . ~ ~ we . « . . ' e t .. et . ~ yt el . . . . ae ‘ a . , - 4 @ eo a A a i ‘ * eo +4 . : * - de zg - - i) =m oe ee - Me . ‘es o . a , o- t -- -- « «te : : “ . . %. . oy ru 2 o's 2 4 -3 - tad * . e - . ‘» € . & r - d. “oe a « ee ’ 8 * . a ® oN oh . eas . ™ ms. $ . . 8 we . - -~ - wl - - om we - ee, 4 . a oe 6 , : . t ne , - 8 * - ry - ‘ ' - a ° . a ‘ , . . e- a - , , o : be eo -_ * ‘ 4 , wee “ we: - ~ £ . 7 * 4 - “ * ot a he 4 . . > Up . a - es i 24 .¢ . . .- 7 tw 4 . , 7 re - - : -~ . ; - . ; . . yao bes, ae ' t >! _- . aa —_ = a * : - _ae * , ‘ @~ | tes . ‘ « - . , . - « ~ —_ - ‘ode * ‘ l - —- o . a. e on 4 ae £ e , . = re - - « - abe an | ey +. a ' cS af « fe . - Oe eo . oo « os = _ + ote useful for computing rivet pitches. For end shears the first drives of the first engine is placed at the end of the span, ex- cept for spans 2% to 27 feet in length inclusive, where wheel 5 is placed at the od of the span with wheels 4, 3, 2, and 1 on the span: and except for spans 46 to 62 geet in length, where one engine followed ty the uniform load is uded. This arrangemert governs for quarter point shears on spans 61 to 82 feet, and for centre shears @& spans 91 to 12% feet. In figuring the end shears, the pilot wheel is & feet off the span to the left. fo avoid mitiplying: the weight om the pilot by the length of the span plus & feet, in order tosubétract its moment about the right end of the span, the following formule Was used; Bod shear «22540 _ 2500 | /\f, 1s the moment of all the wheels up to the right end of the span, and 7, is the moment of the wheel off the span about the left end of the span, equal to 100 in this case. gnear for seer foot spm = 42,100 — 100. — IQS500 = 17500 TABLE V gives the shear in trusses. These are used for finding wed stresses in trusses with horizontal chords. The were figured ty the usual method, for example the shear in the second panel of a § panel truss, the panels 25 feet in length is found with wheel 3 at the pancl point. Meg = J3520 5% MIG) = 1437.5 I2OF2, FS tIAS? 76700 TABLE VWI gives the bending moment in trusses. When the forward wheels of the loading are off the span, the o~ - , - % - “eo to. ~ d 2 . - - : ¢ ote : , ‘ % as © . . “> s e : ‘ < - . - ” 4 = - “ * . > . ‘ ‘ we =e + - ‘ a . 4. ‘ ; °@ be - _ ¢ . 2 «oe a ‘ t * y eo. . > . ' - - . - eo - = 4 v. . « q@- ‘na: + ae ~ . .. . following method may be used: The moment will be found at the second panel point of a truss consisting of 5 panels of 20 feet. Entering Table II with the segments 40 and 60 :feet, it is found that wheel 12 gives the maximum and that the longer segment is ahead. ookel J Lu lL » GI+tYo: 109 I hog * 2OY SS x Ks Abele Abs . _ —! 3§8°55. O G7I-GoI=7 A1o : 130.5 x We FS. Oo 397 ¥3, 0 Bo~ Lg Gh leet, An @Xplanation of this method is given farther on. TABLE VII gives the reactionsof beams contimous over two equal spans,and the reaction of beams continuous over three spans when the middle span cannot carry shear. The equations for two equal spans can be derived as follows: The value of ~-Z< over L the centre support . ee ’ in f3./ << Ef" (ak, k,-2&,) where ‘; is measured Fig. from the centre support. See page “i It is desired to measure « from the left support as awn. substituting /-« for «*, gives fla-— f(k-*) deflection Remove the right mpport. Them the at the right end of the overmanging beam is — “2 (K-«7) . GET With a single load of unity at the right end of the beanthe deflee — tion is twice as large as for a cantilever beam of length -“, that is. 23 3EL Then if three supports are furnished and the toad 0s él aced upgm the beam , the recction at the right end is: _ PL” (K-k3) + 2° > — PUR K*) CEI 3FI of nit Load, LL yy Unit head er, Yor unit loa@ “3° va le, Te Yes Taking moments about ©, &,:/-K+R; “\ - £36. 4 i #y ed Hits e £ - 7 . . 3 4 t ‘ ‘ - 1 ‘ : . , o-t , s 4 sot o . ° a ~ ' @ e “~ e “4 . \ j . ' > om 1 ‘ ‘ 1 » . - 4 ‘ a 1. faking moments about “,, &. -<-2*3 TFT ey In the case of three spans ,partially contimious, Rk, Rs v4 4g assumed that no shear can be carried through the middle span +f; the moment in this span is therefore con- stant. The end deflection when a load is placed upon tne left. span and the support at f removed, is as before ~ £2 (k-k?) The end deflection with a load placed at the end is greater than that for the two span beam; owing to the effect of the bending moment in the middke span 7/.». This moment equals-(,and the 4 deflection due to it - 7" 2 - ee 3 . 7 £3 _ Kaw R, Ro & Ry (for unrt load ) -_ — = (k- k3) + (381 + #7 Y+on 2 { have the Values shown om Table VII. fables VIII,IX,X,and Xl give the bending moments and shears in eight and tem panel swing bridges, for panel lengths Of 20,25,an@Z 30 feet. The influence lines were computed ty meaus Of the reactions given in Table VII. The trusses therefore are asmmed to act like bemmm of constant moment of inertia. The moment influence lines are plotted fo# values of "“% and “7p for the eignt and tem panel bridges respectively. The tangent of th e angle that each segment makes with the horizontal is given, or numbers which are proportionaz to these tangents. The loading is set go that 2/P&-.< passes through sero. * is the load on any Daneleyukes The following mechanical idea is interesting in this connection. TiiT | If each wheel of the series is free to rotate BK and to slide on its vertical rod,while the horizont@l distances between the wheels are maintained constant, and if the series is placed upon an influence dine, then the posi- tion of the wheels when they come to rest is a position for max- » a ’ « “ . at, ‘a =? roe ’ 4 ‘7 “y ’ , _ e , ‘ ~~ ‘ ‘ a a, \ - e ae. "4 oa, x oo imam influence. It is evident that when an influence line is convex towards the base line, as tose for positive moments, critical positions for maximum moment will be obtained only with a wheel at the largest ordinate; but when the influence line is concave towards the base line,as those for negative monents, then critical positions may occur With a wheel at any ordinate. ‘The moments and shears were found by mitiplying the panel concentrations by the cprresponding ordinates. The positive moment Will be found at the second panel point of the ten panel bridge, using 25 foot penels. Referring to Table II ma 4 6s it is found that wheel 12 placed at the maximnn #0. 2¢-2000 O00 e ts 4a simple. Top Ly a4 ss ordinate would cause. ‘greatest moment in a singie spen, the engines are headed toward the right. This position will be tried. |Mevement! oR & ek Ra ee Se © 1 _P [sas 1OVS” §VS | 53.95" 18.98 T205°— 4 82S to tan Xx SOY” fd. S25’ MAY BSD: 256° ft Ptan 27.0 56.8 43.6228 #233.% | 260 | et ; J The rate of change of the moment at the instant the train moves toward the left is found by adding the positive and negative rates for each pened INCREASE TECREASE / 29,0 ° 2. 56.8 3 22.6 ~ uf 26,9 , ny . _ 7 a L, ca 2 eg. G Tat DPecrea 22 ‘ Similarly if the loading moves to the right the rate of change of the moment is: | INCREASE DECREASE / 29,0 2 4 3.@ 33.4 LG. oa ly LOS | Ss 72. — § 1, F —72.G =F, 2. Gleft Dectsaare . teed oe. ‘ 6 . . ‘ oe i. =, ae. ~ ‘ oe ‘ . ’ a) aes : * . . 6 eg t ‘ x 1 ‘ ® . 2 a Wneel 12 is thus seen to give a maximm value of the bending moment. It is possible that wheel 13 may also give a maxima value, because more uniform l0ad comes on the bridge. This may be determined by taking account of the wheels that pass panel points and of the additional uniform load coming on the bridge as the loading advances five feet. Unk = as (.352 —25¢6) = 2.40 Wheel g 1G.2S (424-,352) = ht Urriforrr look F'n 2.5% 504 * 6.70 2.5 %, g.67 —~9J4U4 = 167 This shows that the decrease Of 92 has been changed: to an inorese ee of .67 during the five foot advance of the loading. That is, a point of minimum moment exists between the two positions, but wheel 13 has nearly reached the panel point, having o nly .53 (GT - of a foot to gO. asxsoy °°: It 1s evident from this that wheel 12 gives the larger moment. The change of moment can be represented th us: & ws 7 aX. oF" oY Wheel 13 will be tested py 2 “~~ [de a | P| 90.0 53.78 82.5 73.75, G25 825 6S es 7 fare | so" | sae | way | ase | ase [Plant | 35.3 290 | 436 389 | 265 370 | 229 | ae Mevement fo Left. Movement to ight. Increase. Decrease. Increase. Decrease. / 35.3 27.1 2 ¥Y 3.G 38.9 3 26.5 37,1 y 224 22.9 FS PHF -— 28.8 78.2. 78,7 C6 .O 88. F 78, 2 GC. O 7 Net Decrease. 22.§ Net Decrease, With the loading headed towards the left, the smaller wheels come over the largest ordinate, and while critical pos— itions can be obtained, there is no doubt but that the resultir bending moments would be less than those under wheels 12 and 33 with the loading. neaded toward the right, these moments will be computed. I, 1, 3. f, 5, 19 GPUS 2350 ,25G6 GOr.7 “yf 3725 1622.5 6o8 98 6. G@ GF S395 2326.75 1,032 LYO 3,3 GU 15373,.75 19767, 5 SOY &70.9 119 24130 7 YF &2.5 Bending Moment Gap “ay 1150 2247,5 1256 S7S5.4 49 YSY7hS 1GY4YoO Gog G97. & qa GSES 2363.75 1,037 143974 99 IGIGGUS = =1693,95 504 GYFG 124 2606/,25 YF SSG Bending Moment 73) Column 1 contains the distance of the panel points from the head of the loading;:column 2, the static moments of all the wheels in front of the panille points, these are read from Table I; column 3,the panel concentrations mitipiied by the panel lengt) that is Mis, —2Mn + Mn-, Where./7, is the static moment at the panel point; column % the values of % from Table X; and cohwes 5 gives the products of the quantities in columns 3 and 4, which are the bending moments. It is possible to determine the difference in the moments figured above without actually making both computations, it some tines takes longer however, than to figure the moments. For ex- ample, start with wheel 12 at the largest ordinate and figure the change in the moment for a movement of 5 feet. eae . 10. Wheels 1,2, 3,a0d ¥ 97.5 « & « QS5G + 12,0 te S 235 xy ff « ” x GY 25 84,5 8 yos. '€F Decrease. Wheels 13, 14,15, and IG G2.5 vS 5 S9F 219.80 7, and IS 32,5 *«S& * ,S04Y Shi LUnitterm Lead ~ ad’frerence in area 2.s[(1s x. son) $— Conse: J 78.75 398 Ys Increase, Ltyos, 1 Decrease, 26.73 Net Decrease, Difference from computed moments = 2¢7, The maximm negative moments at all points in the left span are cae a wnen the loading is upon the right span and in the propém position to cause tie maximum uplift at the left sup- porte The exact Placing of tne loads upon these influence lines is a matter of more theoretical than practical interest,since with a little practice the loads can be placed by aye in positinns which will give values close to the maximum, and moreover, these influence lines do not give true bending moments because the truas- es do not follow the three-moment theorem exactly. fable XIZ gives the uplift and centre reactions for deck piate girder draw spans. The derivations of the equation for setting the load for uplifts is as follows: or _K3 , x ann Ry = “Z fhis is the equation of the influeme NT line. latluence Line for Uplift, R3, Le - ' * Ll . . r -# \ . be te *, a . * ww . . . oh . ot . ; : -< . ue < - see . . e 4 x , . } - . 8 o ‘ . . : ° a “ : -- a ~~ : + ; « - : 7 . t ’ « , . ' : a a ' ( a "Nae ~ : . - or. f* ° “a : 1 . eo. °- . . . oe - : ~ . - 8 s -~ i . . Qs. ' vf, . , - - sc @ =< - nr » e . » : , . # - ' ‘ ‘ - ‘ < + -~ 7 > « = . o . - _ 4 . ~ 4. ‘ «J , t e . ‘ ‘ : . ‘ as ae - , o ay , ‘ ue ° ’ os ye ae . c a - * - ~ - wwe fe 8 ~ ‘ o- he mete - ~ . = - -. ‘ - sé ¢ : ‘ : ‘* . s. ? L oa . e . ° ~ i - . { : a 5 \ ‘ . . . - ‘ . 4 : - : o- 4 - - * 4 . 4 ‘ * 7 : : wee . - . - ve. - - -- ~~ ‘© on . aos ‘ . o . ' aE ARs | 1-3k~ ak this is the tangent at any point of the influence line. For maximnm influence make 2 we (1% “’) = oO Substituting for K its value for each load, 4%,°-5 . X.: 7 efc, and solving the equation for ZX, X= NEY - Svat a i See Saw S ep oe saw 3 = tu or x+ /E- A — 6 ) where A qnd B qre constants. The values of A and B for Cooper's loadings are given in the table. For the centre reactions the loading is placed as fer maximum reaction for two single spans. Both uplifts and reactions were scaled from plotted influence lines, shown in table XIIlI. Table XIV gives the bending moment at the support for beams fixed at both ends. Mo — PEK (I-k) = This equation is treated sin ilarly to the one of Table XII, and the condition for maximun moment optained. Table XIV gives also the bending moment at the support for veams fixed at one end. The oconditianal equation is similar to that for uplif*. If a beam is fixed at one end and partially fixed at the other,the moment at the fixed end can be interpolatdadda between the two given values. Tables XV and XV1 give the exact equivalent uniform loads for Cooper's E50 loading,and the curves plotted from the same. These curves are given in Johnson's "Modern Framed structures * and their use in connection with influence lines is explained. s *¢ ‘ i. 4 --¢ . i ‘ ro? et fo vs “a. , ‘ed i a er Sy o " r be &, . to -} f t ~ , > sf : t wea * ' @7 * > 28 - ~ . a s . ‘ . o ° so - a ‘ ‘ * «a _ 4 ~ ° . ‘ \, my ot t ‘ t ry, £7) ot é , +4 . , ‘ , 6 s ° e a . ~_= . . te . + ‘ ~— ow ° . £ . ‘ . , ¥ 4 - 4 we , . -? ad ; ‘ . - , ' 4 ee woos »~.# - oh - a a y } S : 1 * 5 1 a -” . 4 LE Oe ee oJ give (AIL aL which ies ‘ ee ey ae os rt a ee aa * s oH RY r+ A i hy 58 is) a ao Pe " - Bs vu oszor(?) . 1% OS'2ZE v - ns 5 ae y 3 oszo/S} 4 SZI%0E * s HT S * \; Nee ooosz(%) = A RS BY ‘| voosz() a Les j Gon ovosz(&) . 1 oonz ry 4 Pree = Os1z N . iS 1 re el] o = SES re Ae =e he aay iS s eA ae S219) iy i cy ary Cee As Ae a | (cee ah 5 ot tA ne Jarl vi E 9 (| G vo0sz(®) ai * 4 | S z Prete ice S'ZI S| Table Il. wet red PP . Absolute Vite ee Span o , Position of Loads. tlc Equaty ons tor Coopers £50, CECE rr 71 ne Bx che en Md LORS 147 | 127504 — 125000 sx Sac 18.7 0 foloKonon DTD de ees mee TT a nana Z7.e yy" 2 O600. i ed Py ae ao pete LLL ae re ET aa 5. be) CONS FEM ClS Ke) EL IL Se A tee aan ee eT = EE ad To ay cer - a (oS) ore! (ohio ge! 4a3| 362504 4 -——p- -630625 J3T? = oo So mmisa 48.3 ree 20 OOO ooo |,%| 4o31256+22YF - #33750 = 52.5 PALE: QO © OOO) OBO. | 5) 443750 + —Q— - 1.045.000 for 2’ vay ‘at bad G58 Q_9 OOGO 090 00/5,| versnst +7tF™ 209780 mea a Re a a ie 184 O99 oRox AO Hc MoM OM oN PANIES 27-5 ay allel ea hod darted RYLTEI Reine Pat Tae a al Thaw load On Spar. Ted 08-70" eR ee to Pl a rf x Je* + 0.0[e4F- 35) — 292.85] —G * Vee) ©, Cs ie C+1G.8 -0.1( 4-35) A es aL aoe 2 ae | .Of'= 92 i 2 Yo" 45d atk aba a —-C a ga t+) ad - sz2| _._@® | G *+/8.1-0,/(F-35) x:495 te 15,13 §3 Ke A ie U Folin We eta bd haa 389.70] — 99 me if; C: eee 35) C2 i) ae eA oan x» 6*+0.6(1(F- 40) ~ 566.70] ~¢ o.3 C=23.4--0.1(4-40) BR:¢56 fo (2.14 10 bespees 7m 107 ed ys 444 = Loe Z o- /20 ee EMA Retr te rm Par cae) ed cee td os °.3 C274 -0.107-45) 392'- 4 5 ee ee ee A] mo ° 5 ea Si ~ roy RH rate H ae renee eae —&18.Z0] —G oh} C+ 2e4-0/ (4-45) WAVE Ha. War al s to Chesa, CSO, ake ek Oe) PR. ae ‘We imed j be Specifications of 19th of 20000/bs, 56,000 on each driver, WT ee TTY Tal ea a ee are tor One Track. e as fate) fi] ZR00c0of a aa —/~0000 Th seloles 1¢7| 420004—- 280000 2 Ta felolero! YAEL 1 eel oe ALLL) a nz 30.6 Q ag00° CUR eee a - - 90000 2s ie EY farelec: WA alee ar re Re erly, ce 47 ss A e00 6000. a1 eae me TLE a WT 0006* 0000 sa7| 11200044 2809900 _ 200000 ee eg Oooo ¢ PE 2 5. eee el EOL , ’ ae eye) ; 4 2000 6000 oo oes| 1320004 + oa — 3900000 Tu Sem Ts yay OOOO Of CARE} rae a ne ALE a 2000 0600 0000 | Fy, ee 791| /S2000f + p — £20000 ~ iS eC 5 A/a Ce eae aed ae ys naettarst End. Quarter Point. Genfre. me A ER faa AP J a 14+ ae Table V. Shears in Trusses. E50 Loading OTA Lee ros Eicon , the deta engine eet eel ae Shears are tata Le aa nee yh pm lar da ee Pears ai Panelsf2* | SF ie MA Ez eh ZO zi | 22 OR) AS A an Wa (a 2 a CO 2 a 2 GA 3|\% yoo |\* @47 |? va7 |? sas |% 143 |* 150 |? ss5e |? 7e2 | a 7-7 aka oN Sk a i Po oe he a ee ee De ee ee ee ee A ee ri eee el eT al) ee Cy foal ae) 2° oe a ba PI La 2 2 D mS iG : el dN eh Z 345! ye et 7/207 6 She ee My a a as A a a | ee | ae a a CO ee 5 ic 2 azit a¢(|? «e3|* #9 |? 93 ze CoS eer ae 3/977 |? 2ZOs@ |* 213.7 nM IRarife. ae 2 ae a ee Petie * 2) Le oe Oh AU ee oe Pt <1, Seo aoe eae Laer oe a esi ss a ee See ae EET A ee. ae GPA* 42:3 / Bose ee Ls +— — ———— Ss = Va = a ’ - aR Sens & ms had % ee he “a ad =. ae 7 Se bs 2 ——_——_}+——++ mea E ee Ne: | Aanel ee ae eae 3 Rnels\font| 23 tae At es 27 ia) ee le a x i Pa aes i eels Ea ae ie C5 in) Gas ; De a A ee ee ee 5 4|? sas |? s2¢ |? 139.4% “ gi pee ed Oh a bay A oY ll et cl ere ly WN aed AD YS pl dg A Z|? 125.4 |? 130.7 |? 134.5 |" 139.0 |* 143.5 |* 148.0 F 6\3 2 729 7 757 ? re P9252 s43 7 £70 le a I be 2 MO gee? age Fo ee ET a ;] si? 97 |* vor |? vos |? vog |? ¢23 |* 0497 ae inl 9 A cl Mkt ee % ZSRSH |* SSS * 252.7 \* 2G02 |" 267.6 |* 275.0 7a a ek A Ak Oo oa ee M2 A A Ac a a x ke oe a ue el Le a Fe ys ae ss 2 ee Se ee a de 2 2 ne 2 ee ee ae Te 2 A Sd a 2 |? 1447 |? 148.3 |* zouq |* zine |” zie.a |* zz50 |* 2346 |* 238.2 | Ee A a 2 eG xO YE a ee a > 7%e |? 980 |? 1043 Cie) 2 Lee) A oe Le A is ne s | _ 52s [> s¥z [* SG4 |* sa% |* Gos |* G27 |* Gute ee 6 % ear Me 26.) 7 Fa) 7 aed 7 Aas 7] 29.3 7) ey “ Prony oer ET i ST eX 2 Ga YS A 7ab/e VI. Reactions of Bearns_ on Three OY] oy lat ae Le] = RTI 17 ae TE RCA Pry | MAGS O |—. nee ces WITT LAE ¥ jt re 7 Wid te eS ra og ay MALT CaCK Lee [eat 24769 -0405/ Cyt ae A ee eo Re = k-Z2R3> 32784 —.0sZ2F%l (BOT —.05859 an Bn H41@FI —-.0G560 43GS50 —.0G82S5 H&@I4T -.OT4+0T7 S36I4 | —.0P OST eit —.08400 Be) £48397) 0349 —.08T4G oe el 2 x3 “4OGZS GkETSO -.09375 zee ee Ba ert Cr: 2 eg Jax ere eee yl + a ae Bera -77Z00 —.09600 72 27979 | .£IF42 —-.09521 | 2 = had lt AI A es dt cK 7 Wissen ees “Sq | 4 Fe2S |) 8892) -.08T46 | ame Le ee SAF-S4O | P3AI4Z \-.07GEZ iE A hao 4230! | 96065 ~.06360 Te ae Tee OT7FT73 Fr Ak Tall bal ik aan (enn nS Th et aX cele OSTZS5 48SSO ~-. O4275— ote ee {4 00000 ie CT ae lala Ry aan ; laa PO Ene PLIEe eg ae 7 eS} 4 « as mE a a ese | La R2 3 a | rae 2 i Real Le ete a R, >= 1-k +Ra e 2 J . - = bs = ee ye ae eee Te oe bes Sh 20 ate tae ee eee eR 7 ee ee a ee ee 47? nts cia wert Y Praree a t-pounds for Coop 0, on - — ae . er Gry ° ene | ey Ere ae el eT nee ial dial Pee eT Med PoPeta of engines, loading Lee LT ee Ed aa are_not separated from the drivers. These Pelt [idk taka oe? on Le three tables Er Cz ae Tae ra 4A cs baa, Y Ke Pi emed we 31.5 el aC i ay a Va) ga ci ks 2 ey, +2394! Ge 315 —Z2045)| @y a Z0|- 2123 4 PTA Ee an Oe oe fe 30 | — 4836 |+(9) KE) an pao rites fo these influence lines Cary ante Pea LAL a i RHE or 1024. The tangents eee la the numerators of yf = whose derrominators are 256. ~ ay an << - v eb ~~ ~™, = ~ Y” > ¥: “ye Wen.” * — ie " . : - 2. - : .e ec: ° ~ : > - . . ri Wee . %4 - ae F _™ et . . . ES eT OE oe hy . re et ae, Se -~ . rs ; oe a a Swe MTT Cc. ¥ ot ) . _* woe - me ° oe c . are Cg Ft. a 6- VET Ga ee Eight ler Swing 5 Bridge. Shears are gh y ousands o curds for 3 oni ie oi TL SS , oe 3p erry me See ee A a Be ES aC a — /$.3| (s» 4] + /07.1| (> a Se eS 25\= 792\@_ |. ee eee ant 1283 \*O 1 a4 + e - zi Py ied PE a H } ve) ao | Ae ee ay bg ' N NY ro Magee a Sore ee it ae ee tt 38t|-@ 2 red eu ood SS XAOS EE 22OF _ | ios + 45.1, <3) Z\ 755 a aan) S rai mt ae ee oc 233] @> {431 an Hq pare’ 3 SS ess es sa al AY <4) a 0 A Ee od or RO eee en 2 Ss we < rc eto —s opts aK 5 bi ooo eS re Ns | i bR = pS = Vv a ae eas ene \ i a >) ele ane a— aa? A — ee = a Se bi Sek cd hd 4 4 va 4 So be j r ey 86718 zo| gon% |“) 2 — [25] 3527 | @ Contre SON SOT MITT Nt — Maximum , Goncentration. ZO-~ ZY8h.G 4¢1.ZSE ¥.970* FOF eM. ha en he ae Ee | te ee ee 2 7 rv : 5 a a + farels@p__ _ |, Pee ot Leta eta te en eee aa Zable X. Bending Menem eC ard Bridge. Tem Meas et [zo] 2255 [+ ¢ [ez] vate ek eC) ees 30| ¥935 |S) ft | 8s7 Se A as ae aes rs ak a 20| 3/66 | @- 2 |.¢08 a = = a a SS eo Ee ti i) ede £5) 48S | Or 2) 201 Ss ‘a ee 30| G02 | @» 2) 803 eat $ft0 Pc) 0 . SJ Ped os o = or Z0| 2790 | +2) ms 7 a RE as. rear. 25| 4293 | (2) Fa | ~ 30| 5984 4) 3. 106 PO . Eat i Se a) ee s Ss S 4 rE eee ZO /090 a Bahk er ae a ee SA ed 3 ak a 7 : Ree 30| Z¥00 (5) Pa hae Zo *-‘#o ZO| 3Y¥49 | eo) 3 ES a a kg Be et ae ee Romero © 4 a as Ms DTA ES oh i OA > ae ee Ce eg c i) ny N ee Fk pee wer 20] 1971+ 6 | 5 ida ed , O < ordirates aig aes bee eras elt a od ZY4.1 | <4) 30| 28.0\%) §| CET S.C cl a a aT gM TAC A Bridge. Goopers E50 TA era A alicia Puen 7 ¥ ee JT 1 . ae __ 5 Panels @ p_ oe a 4 Z Py 570 72! / bE j T4.3 Ya 25. os = a ee oe PEARS ces ees he 2 ——— TG 21% 116 128 712 24 12 ae ee ibe ie atl TF ee al neo alt me W771 Rr 743 5 ZS Zao 788 176 (28 ae FIco ar = 4277 sat” ele G17 & ae to ee n i H H 1 H es ee Pee eee +) ee 7) ee es HH: eed] J Pa ss N bo i = a Uy ’ ' ' ' rm te are a aed aon de 20| 361.8 | WW) Centre 25| 433.7 '@) Gentre Kf) a Me OMA Lara Ht hts 9 ZO'- 296.4 es, ¥.977 = 3GAE a i ete AS ee DS ee | 30'- $06.7 x%f25 «.99S* SOSG Table XI. Reactions tor Deck Plate Girder Swing Bridges. Peo £50 erred - One Feri, Va ame ee Ta as ae ae Pd be ae = Place a a oo ae a ee OY 2 z+ /t£-A -B trl ay Cae a ee G B88 154 Span. whee x. Uphith- nT See Fo 30° 5 65' 000 bbs. FSS 2a" ¢o' 7 48 9800 G 2058 246 ry eee 2 ake Meer ty JO 2596 305 7S a, i+ TD AS Ce qa IE 78S “Yesee a EL | Fs eS Ao 9 /G200 [x “S929 oe (¢ GS69 376 TS. F332. F4t ‘(6 F098 467 ay 17 @F592" SOf hg 989.5 $2.6 , Centre Max.Conc. yee Reaction, TweSimple Spars. Te 129,000 + /07800 x ry ae Ae TT /GZ2500 = /35000% " + 963 50’ /95,000 = IG1000 x « » 969 ro 237,000 + 9/500 x " « 990 7 Z7T¥500 = 22/300 ~ * » 992 FP 306000 = Z2¥8G00 » * « 98S “ AT Tee Lp) i 7 at Mae i 2] ; a iL s NS 571 = (a8 ge 22 oe Lg DE ef ao ag ee EEL HK FFL Table XIV. ear Moments for Beams with Fixed Ends. AR P aan of eae a! fi (= 4k +k?) Seep rae Me = —P# kci-k)? tA fa > —PF ktU-ky SERA rT Line for /We. R, = P (-k)* (2k +1) Re = P-k*(3-Zk) = X My 7] f eee i fi eo all dee haa of a Calon of corcentrated loads for eee adasie? Me make Z P(1-¢¥k Loo lel Oo, ] : eae | ton fame 4 ra Moment Mo| ae Crd Ee er oy Or a ae aie eet ire : Tp eA ne ae 74 3 2:0 Meat dee yD 3 Nx? Se Nb)x +N4*-¢4ia +3Ea Ce a 2 a-s' NZ ua Ae TraAs SG7I90 | GF.0 Gx? +(30-84)x +Zt?7_20f +975:0 | r a ae arte me B Pa a CT N:3 413'3 Tat 692 | xt 490-120) x +342 -GOF 4375+0 I¢ | 3 207 8G600 G30 Em 2 el be 4 Ane 223 | 412700 an a a » Ss < tals Momerts for Beams with One End Fixed. J4 mae: Oy Plead Sa) bah te Ie et a da Nad A Sani - ar (/-3k*) z-3k We La = . R2: P Eee é a Fe I Fh tet le Fined En oy F ead ; wl aes foes Gondition for eae Moment at Laer : Fixed End r mais ae} Pe fale Pe ua iP ie edie = P¢ B hs Ea ae LO Fee | aT aren ai 4 334 bay Ae] e) ; ers PAC MET ee eee N-=2 a:§ 7 ae i ie St i 13°63 | 1.30) sfogGoo0 | yee MF ged tage 1 ee Be [aes a ee 1§ |3 | zG¥)| 148500 | 1@ |3 | 329 | 166700 nf ’ ‘ é a —m_e IE ILe ae lat Te Unitorm Loads, Goqpers E50 Loading- One Kail. eyed End .t Point Ze ye a er iaae Toe cya 4940 4670 “667 4587 W556 4O" wWtTIS ae ao oo yey 4360 Oe a 3972 Eek cP af ke Ga’ ICE Be) cE ais 3730 3697 3643 3597 fon ces Tah | 3563 3497 34973 3428 em cts 64 3590 cra eA 3385 ceed 3375 ree Cran) cere] 3460 39353 3304 aK Loco ET e] pW A ao 33409 ie ae 32Z0 ke oD 4 3473 cared cies eo 3189 4 36/8 clas Be Be a 3295 3ZG3 3192 /30' 3560 3364 Paes i a 323/ 3/80 /50' 3456 3290 3200 Wa 4 a] 3136 ZO00' 3263 wo} 3027 Ae | a A] 29G5 vA?) eer ERR a LAC 2396 2372 aE To | 300° 3039 2914 ZBE4O 2924 2966 27i/ 350 2966 Va a ath 2786 Pa ZGVS ZGSS 400" 2916 al hh 2751 2709 Z2G48 Pao 450 2873 2753 Z7I8 ZGSE8 2GIT VAL A 5 -)¢ Mae oS 2723 ZG8G 26270 2595 2576 “a BS ' eve orem - 4. ]8Y 2U0- 057 Sladoo7 - spoo7 iiLiofiuy) fllajeaney “By aqy 27 THE CALCULATION OF WHEEL LOAD STRESSES FRQL INFLUENCE LIWES. mea eon Let A2,DeCyde@, DE an influence line consisting of a momber of straight segments. Let M, M. etc. be the static momert of all the loads to the left of a,b,c, etce respectively. The load is headed toward the left. fhe tangents of the angles which the segments makeg with the horizontal are 7, Ta, 73, ete, " Mn a Mi Ma M3 My M, Pw K ——— + “Os Q O 2 2 | a © 3 ~ i Fa - ( ie S 7 2-2 \ Fig. f, “ Fg, 2, In fig.2 the influence of the load is ly fy = Px tana — Py, (lr. x + f~9) > Af, tain — M, (ta.« + &-3) Similarly the total influence in fig.l. is Meg T, — My (1 + Ta) + M3 (Tr +72) — Ma (73 + To) + 1,74 The form of the influence line determines whether the succeeding products are to be added or subtracted. Wn Ms <--@ - 4- R - -- U-k) K Fig, 3. The bending moment at any point of a beam is P7,k — Ma (K4I-K) + Mi (1- &) > M,k- MM, +M,(-4) ve 30 If no loads are off the span to the left, the bending The pier concentration for two unequal spans is \15 pl, Mr , AG - Me _ AL - Me fh (| * tn z.) + 4 td 4, Af; — 2 Ai, + + fl, when “4: the concentration is é — Fig.5. shows one arm of a single track swing bridge. Let it be required to find the stresses due to E50 lomig in the chord 3 and in the diagonal T2 wnen the arm is acting as a simple span. Use the panel length as the unit for cal- culating the tangents of the influence lines Wheel 12 at the pa:.el point with the train headed towards the left is the critical position for the bending moment. Some of the wheels are off the bridge to the left. Eso Migg eas? GLITZ per rar'l, N19) S3SS Mya eo a 3S4Y¥S.6 Stress in Chord 3 : ZN ee 2og-e" 3 = | r e eS pus ‘ z Nat | luerce Line for Bending Moment. The maximim stress in the < crea xd ~F995 1% 43SEC KS 423 ,¢00 /bs, vz % Ss - a 7 ef 497% Ale She {3 ~1 » S q itp 347 bs 7 ' 1. I cr i ' “@ P —~ | ' 2s I | 272 ? Vicf ge \274) i— i ' } ,77* J_ Inf | uence Line for Vertical Component of Stress in Te. diagonal T, occurs when wheel 3 stands at the panel point. Alo” = YOG3S Secan{ es 2h C775 29.5 ALH > LIT SF BI.V1 LE2 | Stress ia Ty? [vccgs x25 — 259.5 4239¢] —2e -~:#+ 206400 LIS K2G/LFK292 TaaSsSuS IN ThUSlasS AVD FLOOR TUE TO LOADS ON A CURVED TRACK. centrifugal force stresses in the trusses and laterel system cam be found cirectly from the vertical load stresses in the truscese Stuwressec due to the eccentricity of tne load on a curved trz.ck are usually found from the panei locds computed from an equivelent uniform a0oad, using tr.c average eccentri=- city in each panel. 32 A method ty waich these stresses can pe more easily found follows. Inner Truss. i Oufer Truss, Assume that the path of the centre of gravity of the locds is a pera- bolae This is perhaps as true as tO assume tnat it is circulcer. Fig.t. Figure 2 shows the portion of a uniform load of +» poundsber re: 8 lineal foot, which rests upon tre - Fig. Guter truss,the curve 1s a poratoli Fircure 4% snows a [Eurabola whose rid- sr oe , gle ordinate is aT » its cquaticn referred to O as an origm \ { | | | | e - eet f-_ IT | is ; Ft aa x S — = ard ar. influence area bounded by ' angen of gle + wg ® traignt linés.of indefinite length, Fig. 3, upor the influence area over the tenzsent of whose ironed angle is 3S, The effect of the paravola the @istance XK igs , 4 kb _ Ff, \ t L ! | NU The moiuent factor for the outer fam: | an: | an: t- “ tan: IK truss igs fovnd from figure % which I1'g. 4. snows the infiuence line for bending VOMnente 33 ree yt ava £7 T%y-k)?- Yk) Moment from Parebolic load =(/-kK) 20% (CF HERD FATE f [Fe OC’ J _ AL Mowert from load Vouwent fom uniform load in Figure 2. = ——;-44(/-*) Z f (kK- k*) a f° (8) 8 “2 Q. 7 Factor for Bending Moment - At 7 efs4 (/+k-k ). t Le] Tris is tne factor by wricn to mitipiy tre norent dte to locds on a straight track, to obtain the noment at eny point in tne outer truss when the track is on a curve. The varictle 4 port of the factor (/+4-*') 48 @ weximaa for A> 3 1 a Wher. A is zero the factor becones | (%* +) , tris is the factor for end snecr. For K-37 | the moment factor => 2(F te | - (GAO 1) + poke ot Snear factor - ee . he Ke - f : — > 2 2” - Oo 4 2 4K -KT) FC | | a 2Gay4 c) L tz J When kK>7, factor * 2 “3 _ . # J Fig. 5, The factor for end sneer in trusses is _ Ub ake (A.) that is for stralgnt track - 4 (K- kK) at CC) the same as tnat for ionment at the \. \ first panel pointe For intcrmedicte | de tk panel, q c mt pene shear factors figure to tne }<__— — > MN middle of the distance d,, ~— gh Fr¢g G a, = ay) P Ficsures 7 & 8 shows the valuerofr 3 (tk -k") and 2( 3 k-k") + 2 3 ° 3 an ¢ Both curves are paerccolas.e F, 1¢. 7, * oN 3 7 N nel 3 Fie ¢ For én outer stringer, the left reacticn is it Outer String Tw. I J 4 Mee — wf gt +7%e, + Zen Y ( 3 ) , | H and tne rignt reaction is ~ Fre.7, wP i 3 t€ + ze, t wes ) a [—-_ ~ XN a _——— > v $3 Lead or +N he terms in parontheses are the bo he Over fer Striv ger. Nu y Co sheer factors to be used when the t | kK py stresses are calculated from wheel rere, “ loadse The centre moment in the outer stringer is “L. BL 42g tte ye ap (© + 2ea PvP (e,-% Pap C4 ten) 7 Ye « FE axr EntvEG 76 tr Panel loads on truss, Outer Truss Toad at cs : TC Et 880 4 “wee ae oud Leak at We? ( Ct 26, “be. eo) YN G6 Symbolsi- 9+ > -~ _ “, “AAR “3 ZS wf ( 30 4€, + 4%e, + 63) The outer Penel load + U7, locz.d per lineal foot of treck panel longtnh Gistar.ce centre to centre of trusses Gistance centre to centre of stringers ‘eccentricities of centre line of load,measurdd positive outwards The following illustrates the method of computing the correct= jons in stresses due to whneek loads. 4 § Fanels w2s5':200' . Path ot Col G. of Laat, 4 ~ FC = —{4 op Ess Yr r a Led This problem IS treated on page 277, ,, Z Johinsens SF treuctures Q 9th £of'tion Rewrtten . fart J, g 4 sy K= ,9738 oY 670 SSC Yor ’ ' ™ o 4 Middle cr dinate jor @ curve ,Z00 " hord = 5.3 Firctors for Chord Stresses it Qyfer Truss . 4 [tC tk -k*) +cJ 2 ft Z (1 +h kt) a FY¥O 5,3 = = > 7YO x 3.72 +670 = [oe ** = feta 1% Pane/ Pt 792 me “4,20 fo ([O7° § ge 1090 a" Aveo §23 do 4.3G du /1,0°6 ole 4,106 g7t ole -§33 fo 4.4 / rte Wtf a At Center. f 76f0rs tor Web Stresses tet Outer russ #40 ( FROM) +c] 2 (fk-k") y oer should be, 74t ¥ 5.9% 3.93 +670 = 1064 x 7 * 10603 12 fone, EFU ao 4.5% do IL 22 ao 1,122 a7f ote 58 ao “,7/ do IGE Vo LACT gL as PSE Ae “S54 ao InZY fo AY Yth do 740 te) G.IG ado Jo.6C ao 1,006 sth ay 36 MAXIMUM TENSION IN VERTICALS OF COUNTER C2RERE BRACED TRUSSES WITH CURVED UPPER CHORDS. 0 en In order to determine the proper position of the loading to get the maximm temtiom in a vertical adjacent to a counter, it is convenient to trace the cycle of stress in the vertical as it passes through its maximum tension. The truss chosen has eight panels of 25 feet, the depth at the hips is 28 feet, at the second vertical 33 feet, and at the centre 36 feet. The dead load assumed ig 18000 lbs. mm the lower panel points and 7000 lbs. on the upper panel points. The influence lines, and dead load stresses are given On page and the curves of stress on page The cycle of stress for /2/2 is as follows: the train, Cooper's E50 enters the bridge from the left. When te pilot wheel is 33.82 feet from the left support, the dead load compression in (,42 , —2/ $07 is reduced to zere: yng * 2% 20S, 7PO X0095S2 +2I,P OO live load stess. When the pilot is 39.77 feet from Z.the dead load tension in t,43, vertical component = +2/7°° 18 reduced to zero: 3 O7F,000 x,0070§ = ~21900 live load vertical component. The truss members acting are as shown, Ve Lh The stress in /.. at this instant may be found from influence iines(2) or (3) in the former case a dead load compression of - 2190° matbe added and in the latter case a dead load tensia of +2/00 mst be added. (2) 3, O75 000 K 00955 > + 29500~ 2) 9700+ +7700 (3) 3,975,000Kx,0097FZ:= + 5606 4 2700: #9700 © os 37 As the live load advances the counter “3:42 is called into action “aaah tne pilot wheel is 74.0 feet from 4. the truss is again in the condition shown in the figure above. May * GSES P7417 M 1, 0 97,585,000 x 0070% — |/5 0000 aa, ~ ~ 2/900 ay - The stress in (.d. may be found from (2 jor (3 ) (2) GS§S 000 x 0O9SS —1/50000K,O4Cf => + BF SAO — 2IPOO = +/E700 (3) 9 5§5 000 %,0O0O/82L—//50 000K 00 2ZY2Z = -/tf4G00 Ft Z/OO = + 1@ 7 OO The stress may be found from the bending moments at “24 or l’zL3 in wnich case the upper panel load of 7000 lbs. mst be taken into account. -Live load Bending Moment at /s- 579724 GSES x % ~ --—-Dead load * ° 7£ * 25000 ¥2S -- 9 Yo8T Ss eens J0OGC78.1 10678,/ x 2 = 4+23700— 7000 := +/16700 3Gx 2S , ; ’ This is the maximum tension that can occur in l,42, As the load advances 4/;comes into action, the stress in “4/4. quickly falls to zero and passes into compression and remains in compression during the remainder of the passage of the tgain. tf U.Ls could take compression and “3/4: were omitted the maximum tension in JU./2 would occur with wheel 32 at Aa. 7120000 x 00955 —2§7 500x046) > +5Y P00 -2LIF00 > +35 000 The curves of stress shown are made up of short straight lines, a break occurs whenever a wheel passes a break in the influence lines, or whenever a wheel comes on the span. The cycle of stress for U;L; is as follows; the train enters the bridge from the left, when the pilot is 17.2 feet from L. the dead load compression in “4/3, -3970e0 is reduced to zero. te 35 $5900? K 00709 = +370° Jive load When the Bead of the train is 35.5 feet from 4-, U3L4 passes out of action, 2,500,000 x cos = -/2500 Jive load shear in panel /;3/,., At this instan’ the stress in 3s may be found from in= fluence lines (22 ) or (‘“%a) (24) 2500000 «00709 = +/79700 -3900 = +3900 (a) 2,500,000 x,00208: 4+ 5200 +7600: +/3800 at 39e77 feet /. 243 goes out of action. (3a) 3 095,000 x 005 = _ 155004 30500 = 4/5000 (42) 3,095 000x 00208: +6400 + FG0O = 4/5000 The stgess in /44; now decreases to a minimum and then increases. It is possible for this exetovonce in the stress curve to go pelow the zero line. If this happens, /;/; nas the following changes during a single passage of the load; com>ression, tension, compression, tension and compression. Tlmws the nec= essity: for careful detailing of the pin joints of such verti- cals is snown. At 74.0 feet /3l.goes out of action. The stress in 4343 is (32) a5%s5000 x,005 — 1/50,000 x.0% » — 19004 30500 = +2600 (4a) 955000 x00209 = +19900 + F600 > +2600 The stgpess in “;43 now increases at a nearly uniform rate until “,L3 yoes cout of action. This occurs when the head of the train is 110.6 feet from Le . 21,026 500 ¥ 005-2513 500 %,04% + 197 500K 0% » 12500 The stress ism U3zLlz is; (24) 2/026 500%.00709- 2.513 500x,04339 + I9q500 X OY 005 = +7900 -3900 ~+YY000 (ta) 2/,O26,5 00 ~ 00209 —~2,513,500%00333 = + 3BSYOO+F 600+ +000 ) s 6: 37 This is the maximum tension which “3 43 can receive. Y;3L3now takes compression as the load advances and remains in compression during the remainder of the passage of the load. If the disgonals are stiff members, /; L3 gets a mich larger tension, as shown in the stress curve. Most specifications add an impact stress to the live load stres, this is equivalent to increasing the weight of the live load, and this may be done in finding the tension in a vertical,taking the impact factor as constant over any panel. Less load will be on the bridge when the vertical gets its maximimy stress, By dividing the load a somewhat larger tension can be obtained, but as this complicates the question of impact, it is not gener ally donew for 4242 the calculation is as follows: The static moment of panel loads 1 and 2 about 4. mst be a max imille Wheel 4 at 4. gives this condition. 6 Live Load Bending Moment af 2 > 7 BIT@— 600+* 5527.5 Dead —" . " " 3750 Tota | Moment at 3 2: GNIZB,90¢ Total G2 7). $ Wow enough load mist be added on the right end of the bridge to make a oa so that no diagonals will be acting in panelg/-/;. 9793.95 +4 — 727745 + 9M 3¢ " 33 _. Mz 3199 TAIs requires an = sore teet of wnsforrmn load. and the maxXimim tension is 17,400 . Similarly with wheel 4 at 43 and 79 feet of uniform load, maximm tension in Us4s ~ ¥7 F700, This is Merriman’s method, given in Part II of /pridge texts. fhe rule for placing an undivided load is; have as mich load on the bridge as possible,without calling into action the main diagonzl adjacent to the vertical and between the vertical andh the centre of the span. anes aT a Dah as cL ' the Stress in Uel 3, a DL. + 21900 Stress in Uzlz 2 J See Uzl3 ac ting. BL. --Zteoo (a Stress in Uzlz 3 2 — ee , a > 4H 7:4 ee .. e PT Te aching, ““O1P2 Se 000G9 Ss z rd DL, +2100 Re a oe : c . ‘tae i Ry a al faye, ee) atm — O.L. Shear + +/zs00 . . WA ead - ; YF ee CP ee , JAP eee Stress in Usl3 « TT eS a : eo se ee a orem OW Re ee Re ear DL, + — 39700 Roy eer, aa UsL3 (3a) eS EY mayo Me id ae) - al erage) le D.L.* #30500 (4a) he i ae Ss — sa = F esate in Ue ie) Pe ei? i ee U2lh3 and UyL3 a ee oN hee ge SE Pras: 2op eee EE] aie ” h (Oey Oe (011 a = OLIGO Roms Oke HOnO! | & i yy Stress in Uzl2. i ovet] Pe A a alle ee el stiff main diagonals, eae “YA” are for counter-braced diagorrals. pee. Ed Pt uv WN \ iw le tr” fa. ibe. “te ae rm at Stress in U3L3. — — ~ j37oo sce ] a 29 ae eee | Pa aN aL A a nly y (/-K) X for lett segment. t | g, Er ct “(- kK) 4c, wo, ¢, = Elx ReI-® RK " Where < is the slope of the elastio line at the left support. 3, El ag = (/-k) E+ Gash C2 for viel g 2 eye E22, * EL Ta » U- Kk) x- (¥- K#) for right segment, x* 2 §. ELSE: UWE -(Z-KO) + G3 Make x= 4 in 2 and 5 since these equations give identical slopes at the load. KrO™ _ pte C . Ela — KES & Elx = —~ (Qa (oO) 4 3 <3 * z s 5 KK Re + Cy y) Ely UE ~ (> 2. ) +£LIX ® 2 {. Make x: k# in 3 and 7 since these equations give identical | Kif3 deflections at the load, and cy= -3 To determine LI« make x<-* and y= Oo in 7, and El: fo (ak k?—2k) Then the equation for deflection in the left segment, that is, when --€ is less than «1s; a. 2 72 Eling s UKE + AE™ (3K 2 B) Make x =m, 43 7 ak —-aktekiym +¢0-kymi ¢ GEL [ ( Changing the sins, in order that the deflections will appear positive,and mitiplying by P gives equation A. In the same way equation 8B is found from 7. In equat bons A and B, m and kK age intercmanged. This is proof of Maxwell's Reciprocal Theorem for beams. The Theorem is, that if a beam supports two wuni@ loads, at any two points ag a and bd , the deflection at a due to the load at b equals the deflection at b due to the load at a, or 24- Peta. a Equations A and B can be used to find the maximum deflection of a beam like that shown in the figure, as follows: Select the gsction which prokably contains it, write the coeffi- clients of m , m*, ard m? Which apply to that section, and add the like coefficients. Let Cc, , €z, and C3 be tie sums of the coefficients of mm, :m% and ni Aum of 93) 26, 44.5,m4+3C ym? respectively. Then for a Maximum, — 7) ¢t2egmtaegm =O Solve form , and if m£ falls in the section assumed, the maximm oer ot ¢ v , “% YS deflection is at that point. Find the deflections due to each load at pointm¢ from equations A and B and add them A full unfform load may be combined with he concentrated loads, the deflection is; +> Wwe WOE - Eee “J W «aw tosal uniform load. In this a cubic equation mst be solved. Cc, t 2C2um -t- 3C3am*+ 4 Cy tn3 =O Cy: The following table gives values of the bracketed parts of equations A and B, Maximin deflections for combinations of con- centrated loads can be found approximately by use of this table Without solving any eqpations. DEFLECTION OF A BEAM SUPPORTING A NUMBER OF LOADS. GEOMETRICAL METHOD. c f - ——— -——— CT To find the point of maximum deflection and its amount. M s/f 7 t 2 + a Plot the moment diagram. Working from the supports call the static moments of the portions of the moment diagram up to amy point, about the left and right supports IM, aM. Locate the panel at the right end of which “% >/Me and at the left end of which Me), To make %=%e the condition for a horizontal tangent, the moment of the part “x about the left support mist be added to M, | and the moment of “x about & the right support mst be subftracted from Me. ax(d+ B) AGIA (+ $a) 4M) = Mg [4 x(Fte-B) OBE Fre FV] kr ae ‘ , atx + Ze 4x" > My — PN Atcr+t = af ' | , a \ . e . . - e ‘ ~ oS | . | . ’ : ' . - ‘ “ . ‘ | : —™ . , ’ e . : 4 , | e . | . @ | | —# a> | : . - ° a , . : . ; @ , ° | - - - - ° . - -_ +. — = - - « ° oo -e . - -— *+-@ . - «= - ~ - - see i .. “ . e ‘ . . * : : ’ . ‘ . . ob . eo * oo ‘ . , . ' - . : . | . ) ‘ . “ . ~ ( - : ' . ate 1 | - - t ‘ . if ae aa yet any aa +f a, Oa PTL r,. an i | ] | } ra O03'197 iy To eile dk dete E Rares O16T/ cole a ay RL LA 02042 OZOS3 feyAsle) 4 | 01913 Re ee ] + | eee > 7 none eed | 00872 OOS59Fl Si ah le Ves | OOOF2Z $ | + } i. 0122 001G0 | tre Caml Rysre 00279 00ZIEg t ,003// | + 00319 | TET a Ryan an 50 eye a O0GIT 00909 1183 01432 redler $0 01830 O1967 0Z053 RO at Oe) errre 01967 Wee EC Ee yeu ae MOP IE 4 0/432 O11F3 Rel a] 006/17 Tocca ood ri oT ea Mala) Pee trata ind EE 8 i K vs a constant whose va og ow given ns eee 4 the table for values of m and k varying , O05. al OE PTT zo | 2s | go |.35 | #0 ys al Weyer iea Cra era 00238 |.00272 00296 | 00311 [00379 | tela Rceehe Ras O0#4GE7 00534 | Roles ee | Peyelaes 00630 | iS ee yore ae Peed i coll Ee Were as a 003953 rr eoededt heed a4} Reeres 00467 | 0OG7S | 00853 |.00794 | 01097 | O11GS erry 25 |.0027% reer: ora 00944 IIEz 01305 ELE Focal loo l¢ 4: eee 6) Rk eae a RoE O1470 | O1S84Y | O1GSO Creel ee rf Wea es [01395 O1S84 |017925 |.01811 POMP TIT MN YLT LTC LT ae tT eT ad eee Eo 4S | .003/9 | eked 00928 cee Rerasy yaa baked PELE Ii mereke | rae: 00909 |.O1/83 dea 01/650 | 0/830 | O1967 Weer PLD Ezra .0113@ | 01378 | 0/592 |.01772Z 0/913 | Re eee La Re ah PT Lea ee TRE oe Ic 0074S 00977 Ez, 01378 ore es ey eciy L. Ce rae errr rere eed ALL te koe CL A da ee Reread eI |009// |,O1l0S9 Ia) 01296 rome OTT MRE oa ce, Rec TELE Ree Rte 00/22 | 00Z42)|.00358 Eccccm aks ARAL aod 90 Reed eral rrr WaT A et ee Ra [00041 |,00082 |.00122|.00160 |.00195 |.00227 | .00Z55 | LALA Pa ee Cee ee ee a [ Fo os ee CLERC AP LEST Lae PYIr roe ED TES eda cede Cele gelr) 00450 | 00386 ated ai OOZ4Z |.00163 Ed EX atcs ae eeehe cee eure Eccetideti lll eer eckekadidica Pret 01067 | 00917 | TEC races 100613 | 00469 CE 25 |01378 REZ Te Cie meee cL FexaeR Peer Lerce .30 |.01592 |.01500 | 01378 | 01230 | acca ll Racal cl 00666 eke 3S eae Rony el ee 2 Ears ace 00748 | 00506 aa) | 01913 rzEn arard Rene ae) 6] Era cca Reed Rees reas IEIE, | KES Wrra | E 01378 01136 Rt Ke eee La era meer k TL, cURL 01650 rere Ae 2 Rr kcal erm errr PIPLIaT TT aT ae Pate ede oe TEs 00630 aC) AIEEE Zee art ore4y Peesr Reco add hel GS O184S mele aw Kei a LE fo FOE ce Rec) chek keke cro TL ATMR SRT TT iE EGER CF MRL LAR TE rs U7 meL Ice T Al 01395 ozo EXILE ee gO rem |.0/Z00 ea 01097 | ner BS 009728 | 00926 | 00907 “T Compute X,complete the static moment about elther supprt up to the section and find the deflection. Stifle Moment . PIKAK Deflection - a Om = EL GENERAL EQUATION FOR THE DEFLECTION OF A BEAM SUPPORTING A NUMBER OF LOADS. . a Fos we - . bos 4 tL F is the left reaction og all the loads. fe t. Re, is the resultant of R and P,. K2 is the resultant of R,B, and Pz. ty 1g the distance of the line of action of F; R, to the left of of R. 4 =—°S— Of, is the dishance of the, /ine ZL. Ra, + Prana. [of action K, to the left of R. 2 > PPro Let@ be the inclination of the elastic line at the left support, a negative quantity. EZ and I are Wung's modulus for the mater ial and the moment of inertia of the bema respectively. 3 Ela: LL°(3k,~ kak) + BO (3k. -ky th) eh, = L(FE -£f2 _ 23 Pa) The summations cover the whole span. This equation was derived on page ¥Y, The series of constants was established by integrating for a beam supporting two loads which gives the constamts ©, fo C3 & KM, 0X3, gE " Ele , 7 Ky : ° 2.3 = Gy- K > rola _ oe 2. G Pr. 3 ca. C2- fad K3 > K. + ae Cn El — zs Kae 5 fe First Section: EI. 2 ery - Rx* + Cy ci te? +o,X + Ky Second Section: FIZ% = Ry (x+ 4) Elst. R,(% + 4x) +C2 Ely = R, (% t+) 4 Cant Ka All other sections are similar to this. The general equations are ty, then as follows: FI £4 = (R- = py(xt oles) te 5 Pye 4 xv sha t+hlo — 2 a. = (WR - a +2 fa pI = (R- 247 Fa" Eng: 2(R- sP)= 74 2 sha +[ Ela — sfx +2 6 The summations include the loads to the left of the section. This equation is general and may be applied to a simple beam loaded in any manner with uniform and concentrated loads; in the case of uniform loads covering a part or all of the span, the sumiation signs must be replaced by integral signs. The de» flection at the quarter-point of a beam covered with uniform load will Ve found, this will furnish a check on the correctness of the general equation, since the equation for uniform load, _ rm mi? +o) “ ¢ SFY (4 Zz Y IF rer J wz048 Ef ' the equation was arranged to give deflections positive downward. £4 [oe Hf 3 oe ~—~ Liew [xo - at arr 2 the lornt Por Ine VT cenit. gives, positive in this case, sinee ae Ks eo 4 Sh a a ty Z fa a only a a gt fe _ elt © : “ a 4 216 ft. un “x = 192 SfQ4 fi + x rAx- 7 OU 2 sf te. ot at $274 pal" Ely (= -# sey 132042 7 ~ R49 Yo ait - vil” act” _ tk! + a = —_ 19 acl* ° 1540 + To14 4536 CNY 2048 “9 If there are no loads to the left of the section, the general 3 equation becomes: f / ys Ax +FJaxX ; thus for a 7 fe? 1K 1€ — _ L£a 2 53 ; beam with a single load at the centre: a % , Ely-- Le” If equal loads are balanced about the centre of the beam, the fat general equation becomes; £/,, - 2s fa +[ Fla —s fa- a yx+1s Se for middle sections. frus for loads at the quarterpoints T } — | 7 the equation of the middle section is: Pe x tee Pe Pe” Ely: 7 2 +f E LE] + Sy Px —_— pen, + PEF ~ S £ BSF 3 . & - e: 50 SECONDARY STRESSES IN THE RIVETED TRUSSES of a RAILROAD BRIDGE. o—-——— itn “=e The method of solution is that of MOHR with an adaption of the Williot displacement diagram for finding certain angles. The development of the equation of the elastic line for a bar acted upon by moments at the ends is as follows: —— ay CL a fr ny —— “yen, ~" ‘es 7 a | + Mo | - «---€-....- £1 Li 01 (Me tm)? = Me uy a. z / FL 72 = (Do th), —~/ShKue+e (2) Elta = a — [/, xe tO 4+G 3 Ig = (Mott) ~ Plo G w+, (3) When X-o0 1 oY “OO, and wher x=, fH oO. - (LP lo —Ph)-*€ OC ° O anal / ’ c- Co Fut X-=O ia (Z) ee To a (2 fo —/i,) Cs) > L Pad x4 in (A) SBT = Ze (2M Plo) &) J5/ When a truss deflects,all members,in gen-ral,revolve througn some angle,and every panel point rotates through some angle. Calling the former angles Wand the letter $ and subdstituting Go—-f for To and $,-f for 7, 4 , original position, M,= 2Ff (2%.+G6-3¢4) (<) “vere _ + cy" ‘oe (t+) : 9 ! 4 ome FTI) ! yy, = 2EL (26 460-39) (7 Fina ( pesttiog The value of ¥ for each member can be Obtained most easily by scaling from the displacement diagram the net movement of one end of the member with respect to the other end and at a right angle to the menber, and dividing the movement by the length @ f the member. Qn pages s¥ and ss these dlagrams are: given, on page 53 the data from which the diagrams were constructed is gives On page sS@ are shown the values of the moments at the ends of all the truss members expregsed in terms of # and ~/ , agd also the final values found by substituting the values of ¢ solved from the equations on page 57, These equaticns are formed by equating the sum of the bending moments around each panel point to zero. The influence lines for secondary stress and the ratio which the secondary stress bears to the primary stress are given on page §/, For exaxodle, the end bottom chord has a maximum primary stress of +/&O for full load. The secondary bending moment at the end “o- for this loacing is —/.s522, the negative sign indicates the direction of the moment,in this case it causes tension on the to» of the vottom chord. The urit stress on the extreme fibre ecuiis PO, Are ~ e . ' \T 7 -L522 x O12 , The urntt Primary stress eguals Lf . TF Then the ratio sought ig (DLS. , fe F2ZQW KOMDKNAY 2 290 1,.€0O $2 rte 20-10": 125-0" —— ; F ~ Zz a @ 7 2 ies Pea es a a i fan @=.718 ’ aC. : BI r) ji 5 $ hy ie ie tak iT Me Tae) Load at @) eM ae) Member, Length, Sie Baer) if ? “Sires. Y Stress i ¥ O2 428" 3732" —foz2 -—Ih/ +091 ~.82 -89 +,107 -.G2Z -G8 +.097 ae " ! - 20 - 22 ~,029 -.44 -4%5 -,0SG Of 250° 1942 + GO +47 4204 +49 +62 4.1536 +.36 t%6 +,//0 or’ S x + /Z +45 -.0392 +24 439.4 ~.071 ‘te sf 18.02 +14.00 +19.3 +,.049 +0 20 4.07% 20 ee) + 063 + a a 40 40 -022 #20 Peery es 250 19.42 + GO +77 -.008 +.48 +@2 +136 + 3G +44 47/08 ee " " + IZ +45 098 424% 43.4 -.070 23 aS ZLIG ='20 = 39: +.021 +/§2 dV a aN a a de ren " " 4.20 4399 -023 +4) +8) -0¥S Z4 250 383 — 4€§ -3.1 ~00% -96 ~-63 1/2 -=72 -4q7 +)44 rae . a —.24 “1G -048 -4% -3) 09% 34 34S 1942 wT +3.0 +019 +.33 +$9 +037 -$0 -90 +036 ied ee a -./7 -3.0 -013 =33 -S9 -02! 3S . 250 32.26 af +3.7 =OZG +.96 24% = e a ee, oe a. ha ; j a qe a Te Ve | | | 7 a aeiiee ee a 4 : | * ee er oe : F : : ‘: Can oan ere as ol ed 4 zo | ” Pas sere as | cars NS SH a Le en y Vid A J 4 a 1 \ | | a / ! ! ie N a a f | ie . \%S. 7 . , J “ a . a i , i a sig $ / a , * - | \ hg g | a R ; A f i } eer Y - & . a} j pi. } 4 } an eas) oF x aa \ , P Rid Ay re om = ee a i : { ae os . y ; AL 3 g = . —\ | / i sy fee / ! Ay Sg uw S \ 1 Q) rn \ LJ i ae ) vi " ae Py) ne oS Ne As aa wo ¥ i) Z & i a : ’ Mag : hee . ae Os | “ " Pa nee : ; Ae e oe. ee SE ee ak S iS t et asa ---+ +++ Oe BS To Se eT ee i cota eed =| NY <3 | v w | az} i7 for Unit Loads. Vitae ae Ata os aA Ar Eres Diagr 3 a ee ey el cee ee oi ee Fae7 A 4s es rey ts Ns) Ww be Ck) dit) han AO KR aw iS) wf ~~ Ma a ‘2 ONES co iy Nm & et y ATE a wo 1S re ee hee > aia Bs ik a) aa : “ mn hink nS Re ie dd S @yveyz ses 3 SS Oly eek ie ee) ee ae 2 aS » a . + i Bs ek Diag. 4 OF |\8Zl-\tho-| oF | SEI'-|Zh0'-| ZO8'1+ hhO=| HE/'+ pOT'+ E07'~ plo--| A) ML Salo) Mell en ee ae a ee / $£20'+ OE - Eos —- LZE— ELE - | bhI— | bz0'-| Pl eel a ie tae 9250'+ Zho0°- shol# HIE+| F624 Oho 1+ EbE'+ BEZ'+ shor ZhE'4| F10'- \on0y- 09074 | ZZI'— P22. Shed Yai [Tea zbe—| Zi Z'- | i cL ies rake LLE-| b8L'+ beo'+ 9bh'-| L&E* - Hlb'+ ofb'+ OLe'+ Obb4 Sab'+ gre’ | ] LL eee A dA th 1 el Ae ae one'-| £9/"-Lbo'- |8e- Zhe~ $zI-| Ot ie hal el) ee A ed sto'+ LEI + £80'+ sSoo- sEeo't od ea Pr en) ted) a ee 901*—| 840 -peo'= | ZOl- fL0— 8/0'- 90/' 4, mZ/'4+ (tO - | zol'+ #80'+ bo0'+ Part att a ae ee 8Ol'= $80'+p0$ '1- ZEZ— SAl'—|o7b' u Ee Be i] PA 491 - bat —- LLI'-| | soO- Pan aT eel ee Ms On ORE Cla CMa ED) See LI al) Pad eee] i aaa Me ena) . ed? ates A st ee ee A) any) as A ee I] ATA Sh onzi+ Poem =esp = EAL "6 BIL + Bbse = tn ei bd ZhWW aT. ea aet 7 (EW = na id a BS az = 2 4W A ay oT adh LAZ — *$ g081 + bHOb * Oty |\n00°- ee ae 2 Be ee de LEZ + At il oh Silla +€/° a Lit’ + OLL' — oy ies nO*T - rt Bee oe Ae obz- ) a VE, E&-— | 7 a ie ere + nes lozz' + | ons’ + bbe* - ae oe alge a lane ee 9 ir A 7 Obs’ Pao Zshno'+ LHe- 7) he oi ah a rae] Ad ey a aay) Paes a] = hew Le od tte od Lan] ahd arid 24 = Boy As ie lh ee - L9€0°— + Ss O=9L1t — "$SRl'ieot ro a a O ae ae Ee wil ne . oy ee vWEL pin ey ‘ “ ery La hee Be ‘ : —o rn y [ot Al Lsi* a hleo't 2bto'* L2to0'* a} 4 Pelee eee ee Ed et es ue) “gs wr a) *Z45D'Zi ea edad ee as | ca) *g bo’ + '$ oo's eee ea ee PT a ae ae 2 a tae enue oa) a Cam *@3D' + ‘PS LEID + BC O'S a ey ee *rzAOL + ‘Soos + S$ gO'RZ ee ea i en he ie bee 2] A a) ye 7 ee es ed *) ” ea “SF » a ‘S Ps eh ed A Be Ae | "p Py + fp ” + ¥ a t od D i? i oe Ae ed A es ST ee 2 oe a Ye 0 a) rr as ” a K ao a vf " Ef 00'S +*6 8D + ‘S$ LEYZ+ 6 CO's ss ae A + % “ S *Sr0b + Goos + Y 808z Calne wey | Os i be aed 8 ss a a sa I | Bes Influence Lines for lat aTaAc1as ied RT ELiay, Stresses. is a) re RO ae a 2 ad 7 N . 7a ° aa ae: ae Ba fat ae ercentage of Maximum Frimary aan oe Stress. es i : Bae 20%o fs pee SS LIL Ba ee s . Wee Le uae Le Y ee Lae Bi a ar ray A Fi) a YOMEE Toa a ase 4 a ee A pare 1 oe G7 eS See? Se Ls 5 had FON ry ray pe ee ee ap [5 //1///; a A 2 > 70954 | : ‘ es 217. eee = ve. ' . Sew ole + > O52 Sl a L y Pre eee ee ie! al LTA aes Bone) i pe See ens Fig. fee La Lae ee — G7. ns 7, - ee G2 CE eka ct mn cats . 27. Re hee oe ~ 92 Ne af rata {ie 4 | oe 23 a S c1y a os ee —— oz * yeas: ; 7] i sa ae T mer oA ¢ er rt t +,.0922 2 4 ye 4 5 +.IZS57T 8 799 -,002Z ¥h2 ACESS 4.0659 * LZET +.089¢ +4649 1.9139 ye | kr Vy a ee | Fath hehehe ries +.08G3 » .797* + 0670 +,.0992 « 1.2F -.01G1 © 1.287, —,.0207 beget « i797. + .09463 +5112 f.5251 at 1a . i 3 . R3 fora Er Mn eA a LA rR Coe R, (fer balanced loads) —(i-k) cr : o 4 AS Reactions by Elashe analysis, Reactions by the Beam Forrutla. Ry Ra Le) Ri Kz Le) k> tye +.708 +335 ~p4zs +OIIF! +,36718 -os859 oe, an) eo) een PY +.40625 +.68750 —-.09375 J A>% _ _ ht end Lies 41d Ta TT PA — 1&1 PLEA Z2I437 AEM en 29.4 To Ve ame Mee a thre |e eee a ya a A ama am Sen Centre Bearing Draw Bridge. 246 Feet Long. UT da ee i pig Heol Tha vs 91.58" 31,58°" AE al - reer rr ya 4 rere Tare fem Teed ; e Unit Stresses due a ey m re Pa a ee Ae ae = By AG ry ae 9) Fad Po ere eee A Sd he rs} BX *. aA #9 () . “4 e A: oi i Ae i Aa a 26.46°" Z6.46"" ; rrr, @ -24¥37 €3) ~09G6/ =I22/ i} —/29) ul _5. Panels @ 24.0": 123.0 A a ee es a 6z = .3Z23f 3 34/788 Ce otf a we baa 2 at et Ld) R3 AS) Prd Parr) Balanced BT iP PPT re mae eee |) ie re | lye aL O797 * .39238 + +,0Z39> +0725 eae We Lee 0293 ~0853 ~-—07Z0 OO8S x 1.178% 2 —.0100 1366 y PYE4 = +. 1/59 ST) ish ides eT] eT 2 2) rai OT eWay Lacie 0) iar alt bead OF OS * 1.1988 = +,0977 1AGEG * .8484 = +.1430 OYOS * 1.178 « —.O47 OFOl « gyey ? +,.0340| .a5q7 Real 4 1, 1084 TL aed -,O7G/i -,of74oa IASZ we tT8E * +4958 2 Wee ae ee ee WsS2 w 4.7788 = — 1358 0100 » .twad * ~ 0085) . 5993 8 §86GES! hee 1247 -0376 -=.0480 Ab ee ee. 1029 tg ee = > ! 0676 © 1.1788 * + .0797 3057 — 3000 :0O239 » 4842 = .0703 0676 * 1.1788 + -.0797 _ 0036 G8IS 24750 OZ293 * 1.1788 > + 034 O730x F484 = + .0GIG) OST — 3008 - 1.9% ar TY i and ites Gentre Moment by the a ee ee ae Peet A Rim Bearing Draw Bridge, 436 feet Long. 48.0°" oe tied ee ee a 7 oe 3 fo W129” 4+? (4 “ e ° Pb mre yd a AS es i TS aes ke iia 1 5 Fy Ss MI [= te oe G N SKS a £ a a) Pe . a Nd tls [~) he q e . #2 VG <> a y Ba J aa) as 3 a wr a os co fe, P, a" 0% v rv ae om ty Aol fe : ¥ P oa a3! o” gout eo o” seog a | Cllr 2 id rea i rims 5 $406 haa Peau —. 886 etc eee fs Sarr Tt) OL erry -2.98G -.0273 —.0273 er cond -.0s88 ~,oss? rere 3 Gross Areas, Gotangents, Stresses and Unit Stresses due to 1* Load OG2Z2 «GF: S598 OS52 *% + “IG ke ea | 7o Nona) ae Ae ee eee > +.03 490 + 0626 «.4906 Ce IP esd at —~,0006 *.9q952 ee eoe Cd +, OF OOF ¥ $2.25 = hie3ae 4/1736 a oe Lod [i anne e-) + ,OSGY ¥/,0048 Cn gee 22 +.0/92 « 99S2 2 +.01911 %.2753 + .OF7S5O « .COYF + + 07536 +, 4£1793 FOOT eee WE Ls 16,295! a ae Ce ee %.OZ42 “4/4232 = + 094% + 0689 *-,1/723 = — .0118 Ua Yh tie aaa + OFF 2 « &€230: nh ah A a ee ee 2 ee = eA pe aed + .231F + .3723 « $2.25 +19.4527 35.7478 —,OS&2 «.€230= =— ,O¥7Fo — .COLHY *£AISYO* — .0025 sd ne eh ae I ee Le 4 a + .OoZ4!l + «3723 + ,39GY x 26.12ZS= (0.3558 ot oie) he ot Oe ane 2 ee a ee wae eS Poa oh ar @ +.024%0 x 44292 -,oof9/ aL ie it ee eee ek rare A + ,37 64 pamri eee a 4k 2 ie Le $7.313€ 7 aL wd CAT] aed ae la a 4 ba os > -14979 +507 i rd ‘hal Fae tere ee Reactions Ruy ag Ry og Siagle aly Taa) Balanced iF mura 0107 ~=.os7 ee ye eC), ey I444 ISS 04a F Cae ey, en) ls i ee ee 7 2 6237 0632 .05/9 7 Boe4 0383 .0772 ¥F 572% Cee iE ees RE Cham The ad and Gentre Moment OM Forrnula are 23.9% less oP ee ee GS DEFLECTION OF PEAMS OF VARIABLE MOMENT OF INERTIA 0 The deflection of any point of a beam from the tangent at any other point equals [ls the origin being taken at the point whose deflection is required. If the tangent is hopizontal, then the vertical deflection of the point is found. The [ines is the static moment of the moment diagram about the point whose deflection is sought, up to the tangent. With moment diagrams consisting of triangles and parabolas, this integral can be easily evaluated by this methode In the case of plage girders with constant depth @& web and varying mumber of cover plates, the moment of inertia, I, is constant over definite lengths of the girder, and the exact value of the static moment divided by E I can be found. When the we¥ Plate varies in depth and the miter of cover plates also varies, the moment of inertia of the girder does not follow any regular law, and an approximate vaiue of the ceriections must be used. The change in angle between the MAx EPs Having found these angular changes for sections taken at shout tangents to the elastic line at any two sections is ¢ distances apart along the beam,the deflections may be found by Suismning up the ang@MBar changes as in the case of a trtsse The method used will depend also upon whether the deflection of every point of the veam is wanted, or whether the end defeet- ion only is wanted. These methd@s are used on the following two blue prints to find the true reactions of a plate girder draw span and to find the deflection of an engine turntable. In the case of the dréaw-span, the clrect solution by means of tne ecuction shown at the top of the rege gecms to be simpler. GG > ae) | = ma ae lal ita tae He Girder 1 Draw ala ale 100 i org ‘a a — — — ———4 ran? py is a7 a ee at Bd K@x2ZZ | ‘ a 2 Co a J P yard - a 1 os vad l te -Z) % Bw o a 3 ae ye Ty Eso : SG gos o Dee t, ee ZF 1 5 ereen 4 |Z. | f# » Bg x fo sy i) Ex rs ae a] } Ax yaa aH ai ry 3277 T + : a - aa : 4 id oi ary al O 3Z77 ae : t po ae ee er PA ee WEL er See) fo 4 2 73971 | eS a a. ZiG| 1/64!) s145\1517 | 2488 | 1540s aye dee ™ = = + = > a = = = Pear ra eer T MEUALaL ese | TEAL } Feed Tae Ea 14490 | 221790 ks 3 wt atria I | =| Pes tae ea) 25000| 67210 4/7 |280250) with constant I if ce / Par G50| 20G0o | oa : | 85/84) ae 00 | | PALA 5.7 | 259 es | Error 2226 = 4.16 xT 32/,072 *, inet aes ae aaa yaaa ae LT at Pee aia Soe . pees bs2a ie ee 2) A - A cy ee ey 216 | 270 1.32 x 9.60 Ah “a / A A ee ee eh 332680 _66SH = 1.96 x4713 1LGESEO_ i PaeT oars So034#o , ah get aL =a TOF eK Be) NITer er a = 96e | I2F0 am 32.59" 3 , naan 1§ + feoox 3 * SY ‘8S +4392 x S 3 oy a A a a. a at ty 2, IP. = 4,86 * ¥7 aie ie a Pore Ea as he wg NB — 4 T $0 ane} oh. ~ ©4849 Reaction ee RTRs ae 24656 “ t T.O6 ¢ ‘eS + f00 Ne F2 50% 25 =1Z50 ie 32 xe cee 1849 ¥ 50 > ea a ad 63 x» 3405: /70966 Era 257 ae EEK: ee tli L4¥6 56 raw rt Deflection of an Engine ent = 7) ; Depth of id 50g atends 2*tt! Ss ays soso SOs ed ia) Bae Flange Gh Ct ete a sir es F epee Cover Plates 13% % - § at centre. ty FA 5 Roy, ee 5 oe ad ng Om aw as ag ee As yA p 2. : } 1 SF a PR ! i i] j | eS | | | 4 pes 4s00000 inch ra ah. (10000 /bs, at end.) Ee: cache Eee nt I rex 2 ra “275 Tan ee ee ee g 3. ae AJ C) © 3825 TL mad Cer ee CT) 2 ede a - bt ceed Wks eh ae Et A Oe ie ak aT 2925 Ee 2230 | 7500) 4771/0} 029 Os ae so 2230 973°9| 27440) ov¢ ort ea ae Tree Ex Z4GO | {2/90 | 39630) oy yy ee a (2730 | 14920)| 54550)| oFF 1016 ae? ya Z24¥li0 | 17330| 7/8FO)| WIG It @ *3,0° 348 G75 14 2170 | 19500), 9/380! 147 fed viele ta a er; aed & {Z10 | 2Z0710|//Z090 | 0 ark 20710 112090 i “S57 ¥ 2.054720 e P . ~sel 7 leas | tad A aie Py et ee .% BM Ta) | oe a a ST | |e eae cz 3X o ‘Naa ee . ry es ee PA eee | aT a y, & et pen ae 5 pene 7 /0000 x10 , * rf Ps } } H OA Die sae sa Oe CO aed gi 2 i rE , Detlections = 10> X a vies y, “ ree ee fis 2%, Oe eee ray rs Fens oa at End. (For a load of 10000 /bs,) , a aero Pe te Ph Paro = oe Bi Jf _ /@4 2825 wi V4 Rel tL r, me a ey ee INFLUENCE LINES FOR THREE HINGED ARCH. pcmcia pecicicneapaiee The two blue prints following give a set of influence lines for © three—ninged arch of 200 feet span and 4o feet rise. They are made by combining the effects of the vertical afd horizontal components of the reactions. The curve of the lower chord #s 4 parabolac: which causes the following re- lations between various parts of the influence lines. The maxinum ordinates of the:vertical componentg infiuence line and of the horizontal component influence line are equal for the upper chord members. The comcression and tension Mmmkexx areas are equal for the upper chord and diagonal members. The coupression area of the verticals exceeds the tension area by the amount of the load at the top of the vertical in question. The stresses due to Cooper's Loadings may be easily found by means of the equivalent loads of Table XVI Chord & Egat Load. rea Moment Stress 273 3 24 a27e I 76.G / 762 600 4+ CoYaD 127.3 Bei ge BY 50 of 2 O28 ee a Oe 72-7 Or the stress may be fégured directly from the inflvence lines using the static moments of the wheel loads and the tangents of the influence iines. é ft ' Wheel & 2375has % 2 — boo x = 219642 232.5. +60,4Y¥0 Influence Lines for ZOO Foot Three Hinged GlaclaP 8 Panels @ 25°: 200' Influence Lines for ES ale Ye et .¥ ; Bending Vee Ole mee (Max. Ti alse.) Chord b. Ola Ae al eh Z. Oe Ae Chord ¥#. AN Aye eg we IE Pm nora ee ss, bl eet oe Ne thesebl igo. a deed Lae Gomp, of Chord /,) Influence Lines for ZOO Foot Three Hinged Flat _ 16.03 Z ee tee ia Influence Lines napa for Vertical Component of er yt GC Ty aan = eT Ta: ad vi ke qW24 he a aE il tetaa shoe weet lan + Post lap re PO nay cee Rte Post P2. ‘ag la ae) 72. On Tables Vill, IX, X, and XI near the influence lines are given the ratios of the areas of the continous vpridge influence lines to the areas of the same influence lines drawn for simple spans. To the right of the moments and shears are given the ratios of the wheel load stresses for the two cases. The ratios of the moment areas are seen to be very nearly the same as the ratios of the actual wheel load moments, and can be used to find moments. The ratdos are given on the following blueprint for paneled and unpariled continuous bridges &m@ for a paneled partially continuous vridgee As an example of their use the stresses due to the positive moments in a 316' partially continuous swing bridge having 12 panels of 25 and a centre panel of 16' will be found from the stresses for a simple span. These stresses are given on page 85 of Johnson's structures. f nm! 7 ae ~ e+ ap > AE 2 28 Ratios / FHAr n(n-*) (ar 58 — f 3s / /- fas : ITS / 7.28 GC U6S¢ ' Loy f - 457 + S73 te Ce y Soa f- .2 > Td 3 ~ p-~ .3lt * CFE ( An / /-~ C7 > B74 Stresses scoled from tat lence lines. Simple Span. av 290900 ¥* §75 : 237 oe ta 78 ee “42 5 U0 74/ 3262 90 sare 2740G@O0 397% 2 79,400 ff ° 32¢ 42 §§ $02 FYF + 3927900 , 3 , 434 cue 343 SU? A 1 235 Cuv The ratios for shears do not give such good results tut may be used to get approximate stresses or to check stresses found by someéther method. 73 ae 4 7, lar eae XT Moments. el aes we ip: — # r# of’ ry i Ta aS 50 Oita v | ltd en — el 4 od 7 RATA ses et age a oe rea of Itluence Line for ae for a faneled bridge, Zo” roe Grea tor Pye Rye ECE2) Grea ae La Le ae Ck eo See oe Sees? 1 nn ae Ce Ty Wate Ait 3, A ay Ta gr ine Re iY ele CES ad ay Ratio = 1- ¢ HL Pe a, 4 eed at fol Pe ye Bo x f= x) rc Areas Shr ry =) Ysa dae as fre Canes) ig ee a og Eg ea melts - ha a £*) ra 4 (a ea Pi a Grea tor continuous span= — eX 71 aad [a Paes Ry TT a wa ea 7 ee a ee AYE Td 8 an or = (shear under full load on left span) - ar peor Se ese Ratio ae AS? BTS 28 A Grea aa Se ae aa ” Grea Bod a jae mace?) ad ay Time ra es fe a a eas x -x)p - ye et eae a 2 i An Divide oy at ala oe Io fartially Continuous, area tor negative shear — ogee ee ate TCD) ae Positive Shear» (5'-x)p wee p—aegative shear I 9S - STRESSES IN RINGS AND HOOKS - = Two blue prints following give the derivation of the equations for bending moments and axial stresses ina ring supporting two loads as shown, and the application of the equation to some special cases. The solution is made accord ing to the ordinary theory of bending as applied to straight pieces. fhe more exact theory of bending in curved pieces gives bending moments very nearly the same as the approximate metnod, but the distrivbdution of the stress over the cross= section is different. The following, Bach's equation for the stress at any point in the cross-section of a curved bar : P St _4 S$ E t Fr (7 , sey) P is the axial force at the section, positive when causing tension. ¥ is the area of the section. M is the bending moment, positive when causing compression on the inside edge of the bar. Y is any ordinate from the gravity axis of the section, posi- tive when measured outward. Z is ~ = Ty tr and may: be found by approximate rey integration using Simpson’s rule or by actual integration circular in the case of regular Sections. For a simiter section Ze or*_7_ Sr irea* where a is the radius of the at cross section and/ris the radius of curvature of the bar. The development of the above equation for stress in a curved bar can be found in Bulletin #§1é of the UWivessity of Illinois Engineering Experiment Station. The third blue-print following gives the calculations of the stresses on the dangerous section of a 10 ton wrought iron crane hook designed by Townes: forma. The solution is,.made both by the above forma and by the ordinary method. The maximum tension on the inside of the hook is found to excedd that given by the ordinary method by 39 per cente Q page 508 of Unwins Machine Design" it is stated that the stresses as usually determined are 40 or 50 per cent, too small. Se PsP ee a ae ya =IM Me=M,+2Hr ih i -Pr ee oe eS O gine. rier ye if" ee ie Meosn du = 0 yrs + cose) fe Meosx da =o ei ete oi Gin ae +Hy) Cee a if Qr (cosa —cos) — Pr (sin@ — sinx)] cose dix 5's M) cose ches —Pr sina cosadea + Hr cosada +Hrcosta du = ae a ca BS ie = y ee ak Pe eee has +Hr oJ -Or coskdu + Qrcos@cosu du —Prsirr 8 cosada +/rsina cosa du = e Pe ’ sin eka ae oe ro) - 9r 2 * lea aaa + Oreos @sine —Prsin?e@ + Pr oO Pr ' Sf i) Tene sin ae arty ae ee ee — By 7 ° i SiO cos? _ p, sin @ = -Ptan ore . H'= Ptane $ sn Be) + QrcosO@ sina — Prsin @ sin«x + Pr Py. del Zz -Phanoar ae af ee s | — Planar ee fl lela 3 cata al ‘he Or (cosa -cos@) — Fr (sin 8 - -sina)] ee y ig. EE -Pesinadx +Hrda +H rcosadu= WZ +Prcosa +Hrat+Hrsine a aa a ee oe 5 he T cosuda + Or cosOdu —/rsinOdu + Prsinudu = (a @ ’ ‘ [ -Qrsine +@Qrcos@«% —Prsin@a —Praosx = -Or sin@ +Qrcose@—Prsin@ 8 — Prcos@ + Pr = -Pr tan 6 sin@ + Prsin@@ — Prsin@® -Prcos8+Pr = —Pr fan @sit@ — Preos@ +PPr cise at) fe ee at A Le ae ae + ee) + al , | ee = fr. | + cos @ SES Ee ater for P, the alee ae a cae aS ae or la Pye rere A OR eer a a ZF Le md < eae yeaa = £E(1 +cas6'+ tend sin8) — Hr This is the max. positive moment for 8 < 60% Max. negative moment occurs where fana=— a M: 4 + Hr (1+ cosa ) ea rd Qxial stress at this section = - sina - Hceosa CR ea ye rare ee tl aD, PLT ea te a kcal fe eee ee 6 = 22) | ad M, he Mes FHR-D oP Axial Stress « oa Pt ae P-(2-4 ld a Pr (#-4£) P - 2 Tl ea Ce ee ee 3 z ' mM, = Fe (Gp - cos 6 sin 8) a ld aaa —cos 6) eae Stress = Psi cos @ | or Pecos @ ET SS Se eee designed by Vln ee aes Ee CE ta ae ada Wrought aad a “ae ad rty ia 00 rere) 37S oro) .@o | ee Bin _ BEA - 3.825 =\% of ZG0 G6GS5O9 078 FGF rod, area =8.296" ia 80 292 6,825 22 9.25 2.975. Go 268 61/00 4147 7.97 card 2s ae ee 7 pa yee lai) 2.3% §250 210 ¥97 Ee q q tw 120 2.0% 4925 249 3.47 = soe ‘ i el a a | 2 eo eS Nlé 3975 342 48 ee ee da a 47 3.550 3/0 .20 ae | a0 P.M 3 ey ,00 " a i 3 Areas tor whole cross section by raed Rule. a ie ee see oe Oe Ta ree BS bs) Distance of oe a! eae ry 2/925" . 2 (oF KS Se PS F rw, -—f= ra) * 1.7096 1 OT : 5 . Py Sd r Ld Oe CR A Peta Prats af r Cee} . : OA eres Bending Stress at front ieee res ( orererr® i ha ee el tte) -/o of Beading Stress, ei at back = ae oa ssid ott = -/%,70° tie Ee Led 5 Max, tension = +29 400 + es Te al Max.comp.= -—/%700+ de = —/G6300 Ordinary theory tor straight pieces. iN ee re Tee) } Stress af front = 01,000 »J25 = + 20500 + z¥00 = + 22,900 rE es At 1) peese 7 —2¥700 + Z¥00 = — 22,300 | —— a ———ay ———alligg tl = oa THN NA ee oe 030