2 IMMMMINIIN THESIS Analysis of the Mich. C. R. Bridge Fam sy NV OL AAR Tenn eexcU E,B. SCOTT, | M, A. CHAMBERS. 1914 THESIS — SUPPLEMENTARY MATERIAL IN BACK OF BOOK PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. MAY BE RECALLED with earlier due date if requested. DATE DUE DATE DUE DATE DUE pga, 012002 + C 74 eee Perea ar ereereaerss: Fs PEERS SETI OT TTT TTF > Pany ~F4 Pere ptaters tose cae ane ret eee soe ge eee ae rs mad — ft Seis Parse. rene le Hon 4 a res ‘ j Pa a ree a a SET ST ITS TST TPT PETIT TY o J ig cs oa o PESTT aarore 5 ae A SeTTSIT PEED EERSHS G4 POTERIIIES ESSE SCT ORE SASSER ER CER ET ST yf ry a S| Se ae - om cd ony Lm] ay on Ca aT 4 Pe mae. rw ret 6/01 c/CIRC/DateDue.p65-p. 15 wrt: ~~ ars Pe ine ey we OK ee Paks — L ee ahs, 7 ee awe. Py we: ew ~ w+ y ~ ae ey ay od wr +» eee * vw oa re. a wr ‘ Pare Pte... - ow . +. ~/ileoee rr Ue or Me . eo: fay Ma - pwy-- fee -6 ewe ae =>. we. i 14 z , 2 wos, crs Pe we * Bee . | a ae We ton Owen, ae | fk a Pe we = BPN a oe Pee a oe Whee -. - Pa “ue we. a | en eee rn Pie + Pee wee cad Pern a. Car oa - e... Pres cae | (WN. are Peery. _ we I fA ia, t | AN ANALYSIS of THE MICKIGAN CENTRAL RAILWAY BRIDGE over THE SAGINAW RIV™R at BAY CITY, SICHIGAN PRESENTED IN PARTIAL PULFILLMEINT POR A DEGREF OF BACHELOR OF SUIENCE to THE CIVIL ENGINEERING DEPARTMENT MICHIGAN AGRICULTURAL COLLEGE. -: 1914 :- e, E. B. ScQrT. M. A. CHAMBERS. THESIS PREFACE d The object of this thesis is to analyze the bridge and to become more familiar with the make-up of the dife ferent members, and the general construction of a modern railway bridge, 103752 tay A eee TP ONLY, The bridge runs over the Saginaw River, connecting East Bay City and West Bay City, its direction being east and West. The bridge contains three simple truss spans and a draw span of the Rim Bearing Swing type. An inspection was first made of the bridge in which all different parts of the make up of the bridge were meas- ured and many pictures taken, Some of these pictures will be referred to from time to time and they can be found in- Closed herewith, The two trusses on the west side of the draw are of the Through Pratt type, with rivet connected joints, built by the American Bridge Co. in 1905, The draw span was built at the same time and is of the same general construction, The other simple span, on the east side of the draw, is a pin connected truss of the Through Pratt type. It was built by the Detroit Bridge & Iron Works in 1888 and has been moved to its present location from the site for which it was originally built, The rivet connected simple trusses are made up of seven panels @ 20'-10", having a depth between centers of chords of 30'-00" and a width of 17'-00", These trusses seem to be of heavy construction having end portals made of solid sheets of retal as can be seen from photograph No.1. The old Pin Connected Pratt truss, shown in Frhotos Nos, 2 and 3, contains 4 panels at 20'-00", and 2 end panels at 20'-84", The depth is 24'-00", center to center of pins, and the width of roadway 17'-00", center to center of (2) chords, The truss is much lighter than the others and is made of wrought iron, The draw span is of the Kim Bearing Swing type, con- taining five panels at 20'-9 5/8" on each side of the cen- ter, and a center panel of 17'-00", The tower is 33'-8*" from the center of the lower chord up to the center of the pin which holds the eye-bars connecting the tower with the extending truss, The upper and lower chords of this draw span are all parallel asa can be seen by photo No, 4, The dead loads were computed by using the Cambria Handbook in connection with the measurements taken, For the old wrought iron truss a handbook edited in 1886 by the Phoenix Iron Works, Phoenixville, Pa., was secured and used, For live loads Cooper's H-50 loading was used, altho this bridge should be strong enough to carry E-55 or E-60 to be consistent with modern heavy loadings, WEIGHTS OF SPANS, The bridge inoludes three entirely different spans, for each of which a table such as Plate No, 1 was filled out, This table is made up of the measurements taken on the bridge, and the first calculations to be made, For example: the calculations were made as follows, We found the member U1U2 to be made up of four angles 34" x 34" x 5/8" © 3.98 aq. in. = 15.92 aq. in. and two plates 188 x +" &© 9.00 sq. in, = 18,00 eq. in, Gross Area 33.92 sq. in. As this member is an upper chord and known to be always in f,., fa Dev. ¥ 4 “Bae a = aS alt ee (3) compression, there are no rivet holes to be deducted to get the net area, The estimated weight is found by multiplying the gross section area, in square feet, by the length, center to cen- ter, in feet, the product being multiplied by the weight per cubic foot of steel 33.92/144 x 20.833 x 4907 = 2400#. In order to get the actual weight it is found necessary to add the following:- Lattice- 26 bars - 44" x 4" x 26" © 16.6/ = 433#. Tie plates- 2 plates-21" x 3/8" xl'-10" U49# = 98#., Rivet Heads- 378 (3/4") w 224/C = 83#, Weight to be added 614%, Estimated weight 2400F#. Actual Weight 3014#. The ratio of the amount to be added to the estimated weight is found, in this instance, to be 614/ 2400 = 0.256 = 25.64. The moment of inertia is found about the axis parallel and perpendicular to the plates thru the center as follows:- About axis 1-1 Angles, Ij-1 ® Ig + Bde, = 4 4,334 3.98 x (8.02)* = 1020, Plates, I 1 3 1" > bho. =™2xi1/l12 x34 x TB = 486, Total 1506. (4) About axis 2-2 Angles, Io.g = 4 4,334 3.98 x (8.60)° = 1198. Plates, Ing = 2x(18x¢)x(7.25)7 = 948, Total 2146, The moments of inertia about these two axes should be about the same for the best design but on account of making riveted connections with other members this is not always possible. The above calculations were carried thru for each dif- ferent member of the truss and the weights of all members totaled, So this sum is added the weights of the sway bracing and the end portals which are also caiculated on Plate No. l. The sum of the weights of the trusses, sway bracing and end portals give the total weight of the bridge above the floor. The remainder of the dead load is in the floor and in the girders supporting the floor, A section of one is sliown on Plate No, 2, on which all of the weights are tabulated for one panel, The weight of the floor is found for the seven panels, On this same plate is summed up the total weight of the span, ‘he weights of the gusset plates were added in the last, This should have been included on Plate No. 1, which gives the total weight of steel above the floor, One half of the total weight is placed on each truss and divided among the panel points. The weight of the truss was divided equally between the upper and lower panel points, the weight of the floor taken entirely by the lower panel points. The calculations were made so as to bring the load FLATE B A,” ad ry wa 7 Ye NAS z oY Pa ee =e ISS kar Mae Re chs Ar QRy ae LEE LT} 2 Bt on [11 ae ia 2 EEE Pte id Cd ae i Cs ~ — Pes aA 74H ey ab I¥5 2 = — a Dd Fle 4 el j WeiGHT of Floor For ome PANEL 1) a a | alla ME det hae Co A a da) 3060 fae LT Me ae er Pte LA) Fa Vg el LPS eee heed 3330 Bas 644A F 120-10" tied) it Gas BA3BAGX I-39" i lceiiia >) (>> ED a COD 9 Sols at ee ve 137 Fa De La al 2 ‘— at a a a (215 4“ EXE KE AIS 6" 4360 Web SIZA FX an ene Bale ee Tha | eae ees bi A 4 Fills Soe 246” o ae Cy Ne 2 MUE KISS Pay oe ee te a Py a7 RF70 OP mA Pome? ed PEEL T:) Fee ae on soeale SEN SF Pee al VEE A] Pe tt ae a ae S083 Total werght of thor for oe pare/ 7A GOS37 = ISSES FSF” Sor srer IA Me CMY ae De ak hd . SRRAIA = Jolel Wl of SPA. TS ITT hie Met 400R 45. for say ne) 487% Lpper Gussel Pls. J676 Lower Gussel PL. rs 6 9673 Weight on each oe ae 1 es ee re ee y co) Oo 'GSC& Vatal Ae Fees (‘EGP A- X 2Zf-F” 165 FSS Fe Pa ry 4¢ eC 10 a 7 ‘i 1 {TT , x i P - = L4—2 +e . att ee HH why cA 4¢ $ Be \} R P Sind Mi le Vs —6 2 * ZA = Zz, u ‘ e {aie | Base aCe ie | \ we Ne oe ee , } Seam 2 | rr CxXGxE ee i ae ee ee Vatal we of floor ee OF eak a Dae 7 Oto “we ae Va ES ea ae at tae rs pe ee ee oe IRMIMAXIVIWIIY Pd 456 es eho ELL $0454) 600 rw 6900 Ve) /40z0~ £370 6370 ig Ary) | fe RA 3k IMC A Gage Oe Ld 4906 * Werghi of Floo~ girder (/parel) NL aes, SECTION of FLOOR (semnA/ Con ae xe 6 od Pri rr (Te te Pe al WEIGHT of FlooR For LE | ye ae ae ee AD Fi ONS EA AE KUK Pe OP ae ae a tes 6XILK 5 AIF" | LORS 4 ae Fe ek oe he ORL hd Werptt, of ¢ r7og¢ Hearn odd i] ae eee GS 6EXIFA LR XO’, FH 2 Y, eT F VE TU AA eae tee Ad Tres 18 (skrox14) yellow pwe @ a Pda Meta i ak CXEX20. oe a cand wr orre per7e/, added for rivels jin Gir-d ers and rod xaskerrers Jolal wi ef Alootr-(ene peret) eI Ms hed se J0FIOXE*9G3I4FO Kfoor Yor Wx paaeds Zi Whadded fot-esl-a ef" on each end, he IRM | LE ea a 6 Dal Laka Sl ie ML ae eel Le ee a Rae A Me aL oo SLL Me A a I hn dd aaa IT weigh! of Sper ee PO LTA ol) A oe (5) on the end panel points equal to one half of that on the other lower panel points as shown under the division of weights on Plate No. 3. The dead load stresses for span "C*" are shown on Plate No. 4. The shear was first found for all panels, This value, when multiplied by the secant of the angle made by the diagonal with the vertical gives the stress in each web member, The shear times the tangent of this angle gives the chord increment which is added to each chord to give the stress in the succeeding one, The stresses were cal- culated in the above manner and tabulated, for spans "C* and "A", on Plate No, 4. The dead load stresses for span "5b", the swing span, were computed for both swinging and closed position, Plate No. 5 shows the graphical solution by which the stresses were computed when the span was in the swinging position. she results are tabulated on Plate No. 10-b,, In order to compute the dead load stresses in the swing span when closed, the reaction at the end must be found from which the stresses can be figured by the ordinary method, This reaction is caused by a force applied at the ends in an upward direction and should be great enough to take out the deflection of the end when swinging or at least enough to counter-act any negative reaction, or tendency to 1ift off from the support that may be caused by a moving load on the other end. To be on the safe side, about 10,000f should ve added to these negative reactions, There did not seem to any provision made for taking care of the nesative reactions 4La4/L 3 op eae ive Pers QSELESF | _S8S-S-FE Vee aA 7 -ha hod eae cy Peary 3 FS-EZBL cP aD ew Cea Ee = Med Lan ht Pat el ears eee Pees eA | a AA aaa 4 wags s ad Ia aah eee om ee SOY pecan OE eats ce oS cceg, Os ma Ram aN Ma OWT (O62) ( ap fe AM) ae di Vt a A PATA Xe ve sete 72 YN" saves fone y Poth nd /2-a@ LOM OF PUO co? i tea ia Ey] Seka LY vewe7 Of PL07| 4999 %7 LCT & NZS ° a, ant ef LL = AMSG FLATE. F DEAD LOAD STRESSES (SPAN # ) Pe tt ee AEE iad ee oe) et wp Eis Otter Fanele { ee eer hl var, Vee Vio sel ee ae ed Si aga Verticals Tad : YL gh ht Vato aks (AIZaD) OM ace (Ci are Y eae ya fodgz &38f70 14,4 \t2 te Ce eee cog A 24289 F/GZ20 exes eo rey aa VU ae §0%6 /06H0 ae a i VOLTA ae g carole te a me G- R es “ > Pie Pa Pe Ps oe ae < DEAD LOAD STRESSES (SPAN c) seca =/2/77 1a77O0 =0.60%A2Z oe “eT aok ered {a ey 4 “4 ree 1g 77) A = ee ra a co y 3 J SR/te 3zJ4o0 34400 32400 rae rer SY ae Le pester ee ee a ee A ak @) Bvtadated eee ee Cz) VAL LL 32400 §2040 §&2040 Lt, Se i heea Rte ee is JZO36h 24693 (56753 YGI%*LL; 43th 39582 47955 Lb, PIF eee AL - aes (ae SLATE DEAD LOAD STRESSES 74 y= CYA A aL ARIA, SeCce =/. 56/4 ie od de he T a gorals Shears Stresses Veri? mY ae I apne ee ee C aie lew eal pee a (kips} | (kips} | | Co ae A a ed me ee Da me ae ae | | Re ee ALA a | Lp eS | 50.0 ae ie | kI-3 /A#FO 4/2. TL | | Lat, Are ee ae errs kaa Tt ie Je ee Ee Ee * a L 5-4 Cae eee, : ata 5 Va A aye Beas Paco te es ek | | (SWING SPAN Clad Pe es ae Top il rt Bd Cae anak a eae (47 sec om Metcale eee Chords LoG 14,7 | 20,0 wars 294. 338.0 7356 LLV42L Ta” TF 7Q0 vet ps ae) Ve VA A eC SS sss eae 7 - —— Via iA eet hak iY ee U, ‘sed ies, /42.0 SS he n. SS — 71. hk — war Ganial Wea var, hie b36a0 rae a4 nein a ae : t Se een | ee 68.9 |Z j- i =e Ow, igi Aae a eee oe” a en an eae Te oe pe 200 0 O/ a he >) : cs . ES a eA : i 7 a S Spr07 P2727 Y of 2Mp sosted]> tl Pa sd TOD hd have bhakti 7 he i ey Os y | os ; nos & Rie e Bel i Eve A 71-1 @ Sa al gppod fouwol! Jamo )Y , REELS Ri rs i vA t 7 iu |, yi ace ie: a Yi S ela gy v : dt me A] v | 7 J F 3 ae LP ui Cd i o i i i P’, To phd Lhe oat e194 Cm |, Semel (6) in this bridge, because when a load came on one end, the other end was scen to jump up and down considerably so only enough reaction to take out the deflection will be used in our case, Innorder to find the force necessary to overcome the Geflection, the deflection must first be found. This was done by considering a truss as a simple cantilever beam and h the formula used to find the deflection due to each panel point load was f= PLI( 2 - 3k + k3) / 6 EI. The computations are shown on Plate No. 6. The live load stresses in web members for span "C" and span "A" are tabulated on Plate No. 7. All stresses were found by placing the wheel loads in that position which would Cause the maxirmur stress, LIVE LOAD STRESS IN Lol} eae This is a web member and the criterion is followed out that;~- The shear in any panel of a bridge is largest when the load on the panel is equal to the total weight of the live load on the bridge divided by the number of panels, or V=zl/mxw where: V= shear. m= number of panels, Ws total weight of live load. Wheel No. 2 was placed at the right hand end of the panel or at lL) and the load on the panel is found to vary from 12.5 to 37.5 kips by moving the wheel a very emall dis- tance in either direction, The total load on the bridge is 415 kips and since "n" equals 7, the shear should be ATIONS FOR REACTIONS SWING SPAN. (Closed) ae . n AY nN \ N N ; f AN N IS iv 'y tn te ) ‘3 La is hb ‘a he x Va a ¥ oe Pag Y M| j f ¥ y i a F7 ee Rs a Pa 1 ae N os CIAOG” = x A - a S A _Lth.alo* i. 3 g ee, a iss | ee eae : ‘ ‘y ; f 7 = 6 i A oy al daa aes ere the VA ea ola Te eee ‘ a ag Ce Pay 4 yk ae he Me Md ho ci Battom I ae s pe pileig ve (A ch.) Ah a eae coP yoy vA Ca ya AP ee Tae Tle eee a Coe ,, Tae ant a ee eee ae cain _ £4.bb —X ed IJ—Z26.5%= 25x > A =I 266 KT eee a) Ges ha OF a fia! ST = Fi is 4 Fae Lek Thole eh eye) AP ae te ee M266 X12 F669/ X ZG Q/00000 |= SIL SVT yrotes* Saas ie ee ee oe rey s) ee er =s39 7ZF sprchesé ay. yo ff de LLL 700A de =fES?AS * Pa Ae i eT Tole, / A pe yf (Giana ETA ee Tle SSEZATIFIZEFAIO?I ER KUTT £ mer” add aaah Ea ag Sofre Lack a laa am LF «: ee ee ee al Yo 47°29 ha aa te [7 = ZBEIZ4F7S§FO2 0000 _ Px 2%26%¢0/00000 Vay aa 2 ae tone hole) ~ “ZOLIAA/OOO0IVIO Mani ee aA a ee Fat 2, FF ee WEMES TEESE S | bin Lena) Cy Tae A and C Vath haa >: 2440/6 |) 690.0 /68-0 RABE \ ke ; | Es Vea Ait ane poe aan Ie ; . eae WAL 6357 a A a a a | a a ae A ee oe ya EW eae AL Cee — 32.0 UL, mA Ly : TAG P i i fd A) AEM) VEL e ee I iL Team ee | c. PRP er ae 2 sO ae ee fo ae eg or ae ee 3 Vas x Aes ae oa vg MN Nico i daa We AP 72 | AIPA NGEs | F 8745" | /00.0 $€Z |-58Z |/5¢ Or Fe dan APP FLATE 8 CHOMTD STRESSES DUE T0 LIVE LOL LL ols ea oe ae 2 fs 9 5 S 2050 | 5250 |2/F%O B 0 ay OMe Le PL) 50 PY Gis FO 60 VOR RETA x0) sola (7) 415/7 or 59.3. As this value does not lie between 12.5 and 37.5 the condition mentioned above does not satisfy the cri- terion, Wheel No. 3 was then tried, The load on the panel varies from 37.5 to 62.5 and the total load is 427.5. \ie found that 427.5/7 = 61.2 lies between the required limits, satisfying the cruterion and the shear due to this loading is found. Wheel ‘io, 4 also satisfied the criterion but whea Wo, 5 did not, The largest shear of those caused by wheels No, 3 and 4 at Ly was usec which gives tne stress when mul- tiplied by the secant of the angle, the same as in the case of dead loads, This method was followed out for all of the wev members, Yor the chord members, the criterion P' = n'/m x W where Pp' = gum of loads on left side of panel thru which section is taken, mw’ = number of panels on the left side of the center of moments. m = number of panels in truss, W = total weight on bridge. was used to find the wheel position which gives the maximum chord stress, The wheels were moved so as to satisfy this criterion and the stress found by taking moments about the center of roments for each chord, sor LoL the center of moments is taken at U5. Wheel Ho, 6 was tried at L, and found to satisfy the criterion, as P' varies from 112.5 to 123.75 kips and % © 121 kips, which falls between the required limits. With this load the more (8) ent about the left support is 31016, which, divided by the length of the span, 145.833', gives the left reaction as 213 kips. The moment about Us of this reaction is then 213 x 41.6 = *8900. The moments of t:.e loads on the left about Uo = “2050 and the difference gives the moment about the center of moments which is resisted by LsL3. 8900 = 2050 = 6850, 6850/30 = 228 kips which is the live load stress for L513. this method is used for finding all of the chord stresses, The stress in the verticals is the maximum sanear found in tie section, The end verticals, as U,L,, were found by placing the wheels so as to satisfy the criterion Ps 2p! where P # the load in both panels, Lob Pts the load in Lol and then substituting values in the formula Rp = (uM, - 2%, )/P where RR, = reaction at 1, or the etress inly1, M cg = moment of the loads about Lo Ls moment of the loads about L, P = panel length, The live load stesses in the swing span were found graphically as is shown on Plate tio, ll, Stresses were found for all of the chord members and shears for all of the web members for a load of unity at each panel point and the influence line drawn for each member, The loads are then placed so as to cut the maximum ordinates, the sum of these ordinates, each multiplied by its wieel load gives the stress or dhear desired, In getting moments and shears for these (9) diagrams, a book, The Designing of Draw Spans, by wright was used, As the members in the center panels are very light the case was taken of no shear transferred across the ce center panel. Coefficients for reactions for this case can be taken from this book, page 111.Take for example mem-| ber UjUo:- ‘ake center of moments as Lo. Now for unit load at lL, the left hand reaction is 0,760. Taking mom- ents about L, for a load of unity at 1} My = (0.760 x 41.66) - (1 x 20.8) - U,U, x 24.66 U Uo" (0.760 x 41.66) -20.8 XE 24,66 © which is laid off to some scale under Ly where the load was, = 0.473 Points are plotted in the same way for a load ay each point and a straight line joining each of these points gives the stress influence line for the member UUs. The same is Calculated for each member, The stresses are found and tab- ulated, The diagrams give shear for the web members, this must be multiplied by the secant of the angle to give the stress, In the panei next to the center there are two dia- gonal members and the method of moments had to be used, From the shear influence line for this panel it is found that the maximum shear occurs when wheel No. 4 is at L, with the wheels moving toward the center. Take moments for loade- ing about the same points “a”, the intersection of the other two members, as shown on Plate 10. hy Similar triangles Uge * Ushs : abs 22.8% 33.66: aLs oUs 9 aLe= 85.5' distance to center of mom- ents, ir a GAAPHICAL = S0lvT/on a a a aM Lae FOR LIVE LOAD cand 7 Se SO a la LL SUB! eTer T se Ba Ue he eet er ee so v U ee? Ly ——— x =) + ag o me?) > £) o ms o > a 4 — ar " ve 4 v ce EE ~* ; r ro" r rs Y — = sae — os - 3 id ¥ =o COOPERS £-50 MEEET-Wo 00 Ms slo oW U0 0.098 0'o'> 0 mam on ® | (10) Alao | UE, FU. = ab: al... 24.66 : 33.59 = abs: 85.5 ab @ 62.5' gives the distance at which U,Ls acts. With these distances know, an influence line can be drawn for both U,U, and Y,l,, always taking the moments of the Loads to the left of the section. After all the stresses due te live load, dead lead and impact were calculated a stress sheet such as plate 10 was made eut for each different span. These stresses were summed up av-d the unit stress computed. A column of allowable stresses was calculated and placed alongside. These two results were then eompared. This is far from a complete analysis but it has served the purpose for which it was intended, namely te go a little more inte construction detalis than we have had time for in the reguls course. Conclusions We found that the old truss is stressed in some members nearly te the elastic members. This is not in kecping with the best modern practice. In time a permanent set will prob- ably take place. The results were not altogether umsxpected. It was thought that the bridge was in this condition on account of ecttain unverified reports that its use was to be discontinued. Zt has carried heavy loads for some time and will probably eontinue €o do the same in the future, but we would advise replacing the old span with a more modern structure. We RY ELLA 44704 1H Ta Teal Val i ce igh Wal , N ene e a2 eae) Se” Naar ROY) Za 4 4 % Te Ky . 3 hg N » Ws ~ x y ark r: i Pa e oA Zs oe tg ra Z = a 7 fanels @ 22-10" = 145-10" k: es Viel Viel had es fed i ros % a TS hy ‘a3 Pog PBs Se Y ey Stress | Stress| Stress Ce EI oa a eae Oa pitted: Vea ea bid ye Bs Ce Gees, IA ce AAR a eee LL4 ¢ITS.H WT80 \+/2£0| /5:0 | 6.5" | 24.3 Te ete eh ee eee LLa| 9 |\*FO8\+560\ sso | 165 an oe ee yee ok ole hn Ae A nk eS £42 YOROF WHS |\497/7 | 7500 | 167 | #2 Lihs WIO-7S 226.9 |n/5¢G | (50 | /8 CO | Z5/ L3h4hGFTOF KGZO \4/86.7 | (50 | /8.0 | Z43 UL, Ue | 130.73 £REO |/SCE |/3.3 |/54# | 340 | 75076) AFF ee eS AOE A LyV4 [O#OS | 2629 \-/867 |\/3.3 | /54 | 397 | 75076 FIG LL ¢ 3h AO | GII\|+ S7/\| (500 | (48 | 7 C42 +4636 |-G970\—709| 9.7 | jaz | 706 |\/2%600\ (73 | Le Le et) fe teal A Ec © 23) STRESS SHLET SKIN B i a al 4 a a 5k , = i" as . Vig Gr Te 7 i Ue Te Ze vs Z Z, Z KR x 4 G K vay rs ra - ae Le ria Z, - 4 fanels @ 22-94 =//4A—L dete”) rey rs Po vig Bef Pe ros 1S ¥ x VE ZF PELs Te: Fis pe ey Stress | Stress Se 7 eA Cynon Ieee Cia ia aaa Mer be oe an CAL Unt | frea L* freous&y LZ. \- f00 TET AAV Lae /6h 000 AC eC a ee Ce LL 1 2 AO Le oe Lzlg\* FOO \+680 |\+ 5% |(5000 (0069 | /5-¢0 ry Lote \FFOO \4/ZGO |1//6.3 \16009 | /3F00) £43 4 ‘ Lhehs\ fE4S 2122 | SO \[ROO0 | (350 | ASF \l6KL006 473 | VA ed 4 el tO el oh hd u | ae Le Ee ad i el a A ad Oa Ae A ee A eee FRO | Ll \- GLO \=[6/8 \1/7.G | 13000 | 7360°| 250 | 765°0| GZo | VT ak LN Ld eed Ete LAS ae Lc 4. oe ac SA e aS 2 LI ee eee AAs SA oo ad kk Vy ek 2 ae A hl hate WS MA lc Ue a a Rha a ys / a Mae ee Ui" Ps U” i” Ue v9 F i \e Yi a us a WE y U Z Te ‘ " x y Al / - ~ = a f < FV y ” * > ps Pe % 2a- aes Ph. VA 1 a