¥ Gg ve oe iat detail org/ in Ive igitizec D arch Ww p/w htt THE DESIGN OF STEEL MILL BUILDINGS AND THE CALCULATION OF STRESSES IN FRAMED STRUCTURES « BY MILO S. KETCHUM, C.E. DEAN OF THE COLLEGE OF ENGINEERING AND PROFESSOR OF CIVIL ENGINEERING, UNIVERSITY OF COLORALO; CONSULTING ENGINEER Yo C wv SECOND EDITION, ENLARGED re SECOND THOUSAND TOTAL ISSUE, FIFTH THOUSAND is /NEW YORK THE “ENGINEERING NEWS PUBLISHING CO. ¥ , 1907 ‘ pee u Q t, 1906 Copyrig THE New ERA ee oe PREFACE. This book is intended to provide a short course in the calculation of stresses in framed structures and to give a brief discussion of mill building construction. The book is intended to supplement the elemen- tary books on stresses on the one hand, and the more elaborate treatises on bridge design on the other. While the book is cuncerned chiefly with’ mill buildings it is nevertheless true that much of the matter will apply equally well to all classes of steel frame construction. In the course in stresses an attempt has been made to give a concise, logical and systematic treatment. Both the algebraic and graphic methods of calculating stresses are fully described and illustrated. Each step in the solution is fully explained and analyzed so that the student will get a definite idea of the underlying principles. Attention is called to the graphic solutions of the transverse bent, the portal and the two-hinged:arch, which are believed to be new, and have proved their worth by actual test in the class room. The diagram for finding the stress in eye-bars ue to their own weight is new, and its use will save considerable time ia designing bridges. In the discussion of mill building construction the aim has been to describe the methods of construction and the material used, together with a brief treatment of mill building design, and the making of esti- mates of weight and cost. The underlying idea has been to give methods, data and details not ordinarily available, and to discuss the matter presented in a way to assist the engineer in making his designs and the detailer in developing the designs in the drafting room. Every engineer should be familiar and be provided with one or more of the standard handbooks, and therefore only such tables as are not ordinarily available are given. iv PREFACE, The present book is a result of two years experience as designing engineer and contracting agent for the Gillette-Herzog Mfg. Co., Min- neapolis, Minn., and four years experience in teaching the subject at the University of Illinois. This book represents the course given by the author in elementary stresses and in the design of metal structures, pre- liminary to a course in bridge design. While written primarily for the author’s students it is hoped that the book will be of interest to others, especially to the younger engineers. As far as practicable credit has been given in the body of the book for drawings and data. In addition the author wishes to acknowledge his indebtedness to various sources for drawings and information to which it has not been practicable to give proper individual credit. He wishes to thank Messrs. C. W. Malcolm, L. G. Parker and R. H. Gage, Instructors in Civil Engineering in the University of Illinois, for assistance in preparing the drawings, especially Mr. Malcolm who made a large part of the drawings. The author will consider it a favor to have errors brought to his notice. Champaign, II1., M. S. K. August 17, 1903. PREFACE TO THE SECOND EDITION. In this enlarged edition more than 100 pages of new material, in- cluding many cuts, have been added. The additions include a discussion of influence diagrams; the calculation of stresses in pins; a chapter on “Graphic Methods for Calculating the Deflection of Beams”; data on loads, foundations, and saw-tooth roofs; descriptions of various buildings; and an appendix giving the analyses of 22 Problems in Graphic Statics. The author wishes to acknowledge the appreciation with which the first edition was received by engineers and instructors. Boulder, Colo., Mops KK. June 1, 1906. TABLE OF CONTENTS. INTRODUCTION PAGE. Puente att: UUdINGS,. ks en Do ee oes le aberedaws ews i Steel Mill Buildings with Masonry Filled Walls................ 2 Mane tdinies. with. Masonry Walls... 02.66.05 05.2 cece eden’ 3 PART I. LOADS. | es CHAPTER I. Deap Loans. Weight of Roof Trusses....... Ber GO RPE LSI oe Saw cca ana nek ge 5 Weight SERRANO 2h gd ashi gee ts bus oo be hae cia DE 8 Weight of Covering..... ea artim OT Aes claet SoG BY ccd ve mee ae 8 Oven Or SUuctiife. i... SES RAE MEE IED NP ERR 9 ts Cuapter II. Snow Loans. SE ete Ce rs EAs Pe ada sca aves Stu few eaekbe 10 | Cuapter III. Wrnp Loaps. IRIS cae ote 50) ig a Ns ee i et 12 Weincet ressure on Inclined Surfaces. . 2.56. 040. 2. cence ce devcces 13 Cuapter [V.. MiscELLANEOUS LoADs Live ES USS SI ede ee a PUD pie SAU 17 Oe OEM Qt A OPANOS ie. ay oe ss en ee es ane cewek coe bkeek 18 PCT MIROLEEG CTANCR, Fs oe cl. ee ees eee wenens 18 Weights of Miscellaneous Material............. 0.2.0. cece eee 19 Concentrated Live Loads 2 RS a Rr. | ais 0 sine a ie Res) 2I vi TABLE OF CONTENTS PART Il. STRESSES, CHAPTER V. GRAPHIC STATICS. Pegtaslabriearn «coc oie sceies saa: os hie oo sop le Representation of Forces... 0... ..5 0s bate eee eee ee Force Triangle 0.002 i oes Oo oi ee ee Pile iste Force Polygon «oss. j20% ss a < ola \8 eso ee Equilibrium of Concurrent ‘Forces. :.:.\..\acsaue oe semen ene ai Equilibrium of Non-concurrent Forces... -(..2.0aa.95 ees ee eee Equilibrium ‘Polygons: . i235 .1c 6 c4.'s odes woop 0 eae ee pee Reactions of a‘Simple Beam... Galen et oes we CHAPTER VI. STRESSES IN FRAMED STRUCTURES. Methods of) Calculation (nc5.S Si ee ee ee ee Algebraic Resolution .... 2 (siaac wus fee a eee pie as ee eee eae Graphic Resolution Algebraic Moments Graphic Moments ceeee ee eer ee eee ewe err wre eee err eseevewweeevneseee ee es @ see eee ee we we ee ee wwe ewe ee ee eee ewer wees eeseeeesee CHAPTER VII. SrressEs IN SIMPLE RooF TRUSSES. Loads Se ©, 0 2 06 8 6 OO 6 81S C'S CO. 8. #240 8) OSHS CC. eo (6 1e © 6. 2 ea 6 6 6) 20 Ble a. 250 2) eee eee e006 8 0 6H 6 8. © O86 2 PO 810 E68 €) do) ele) '6 ee oe 8 OU ee ee en eee ow) ae Dead and Ceiling Load Stresses Snow Load Stresses a | e 0 © @ 3 @ 6) 956) Olé 4 € 0) oe a 6, S,:8) Poe Oe. O) Ose: 9S, aoe Le Ce ee ee ee a ee ie a oe Oe sey Ph er er lr ae Do: a ae TF ee ere ey OE ne EN ee ee ee ee ee ee oe ae ee TABLE OF CONTENTS Vii CuHaApter VIII. SimpcLe BeAms. Moment and Shear in Beams: Concentrated Loads............... 56 Moment and Shear in Beams: Uniform Loads.................. 57 CHAPTER IX. Movinc Loaps on BEAMS. ICE PON ITT C10RG5.,0)6 5 ics F-2s uces ahd Cab de blip Rilvie'a 0: 59 PY HOC DAVID. TOA eos. cele scans vs ha busca bad cwae 61 CHAPTER X. STRESSES IN BRIDGE TRUSSES. Method of Loading.............. pie Ra ptene eae Babee ett ae 65 RR OGM re Wi we Aas FV ene b ce Hic Maw ne hp 65 rr TI yl. Sk ac so eidhe wise bs sie cea sae ee 70 TN Be RCM Slog cro og os ok oss ches hs Died wets oN bp les CaS 6 72 Psrapiic Moments. ....6.0..0.3.... FL MINE RO Amity b yb uidaten a a Sate eA 73 MMReL O88, 5.4 kc pane os BOR RN Sadat eo Ra Ne ang 76 MM i os is As sicv dic co Cock ae Voloa Hees veces oh 7 Maximum Moment in a Truss..........0....i0e000008 77 bac yea 11h BL LUSS soos sido ge ca vik a we eee ee bse 79 Masimunt Floor Beam = Reaction, .. <5 0. <6... ee ashes 0% 81 CHAPTER XI. STRESSES IN A TRANSVERSE BENT. Me WL DOAG SITCSSES. ola os cee obec oc ts beeen coweces 83 RI EEO ee 5a a iisyiy dca Lice e/ Cd Se Ja vos vied eee een e's 83 Algebraic Calculation of Stresses: Ogee. Conmne Plinged atthe Base. i... ue ee tae 84 iGase ti. Columns Fixed at the: Base. so. .c0 0.00... 87 SNM RIC CACC AOE eels 5-s ait eb a 6 anit cs deans wage ne QI IESE TEI CVT TS aa 92 racmm wy the Upper Chord and Sides... 0... ee eee . 92 Graphic Calculation of Stresses: | OM PG EE Oo Ce hee ciycic on ov eS dik ese Me avian 93 Case 1. Permanent Dead and Snow Load Stresses....... 94 Case 2. Wind Load Stresses; Wind Horizontal; Columns Fineed. c...6.:. EN OPO N 6 oe ia'e: sp ais, ac Ree See 96 Case 3. Wind Load Stresses ; Wind Horizontal ; Columns PCR APR rier sre. a ok thas ooo 2 Us Space ties the 98 Vili TABLE OF CONTENTS Case 4. Wind Load Stresses; Wind Normal; Columns Hinged i... oc. ss ace o's 0's tga 99 Case 5. Wind Load Stresses; Wind Normal; Columns Fixed at Base... :..4.+.s: site eh IOI Maximum: Stresses... <6 2s c.a:5 bois .sicegatn eee 112 Simple Portal as a Three-hinged Arch................. 114 Case II. Stresses in Simple Portals: Columns Fixed. Algebraic Solution... . .:. 2.s:c7 Waites harem ete sees eater eae 115 Anchorage of Columns; su, eee, Ry Bhalla ae valig gaec laa Para tee 115 Graphic Solution:.: >.<. ds.Ge erases vee eae 117 Stresses in Continuous: Portals, .. .2..20.dsue suis a ewe eee 117 stresses in a Double: Portal. ..ci.5. 40 vaso ee eee 118 CuHaApTer XIII. Stresses In THREE-HINGED ARCH. Introduction 2.00003). 54... cc a nee oo 120 Calculation of Stresses i.:00 0 os sacs wee eee See 120 Calculation of Reactions: Algebraic Method.................... 120 Calculation of Reactions: Graphic Method.................+4-- I2I Calculation of Dead Load Stresses........+--+ecceceeeceece eine $°~". Calculation of Wind Load Stressésio.s3 des oy nose ae 125 CHAPTER XIV. STRESSES IN TwWO-HINGED ARCH. Introduction... 04.05 ss 5 lbs ne ewe pe wa eevee ee 127 Calculation ‘of Stresses... Vi sz seston 2k ee ee 127 Calculation of the Reactions: .4.23535.2 2c ep eee 128 Algebraic Calculation of Reactions. ...............+-..- 131 Graphic Calculation of Reactions...%.5.440) 06 meee oe 133, i ‘ < ‘ ¥ 2 2 ae pM Se ee Ce ee an oc 7 Ae os Pe hs Se oo S PG Swi ES ogee 2 TABLE OF CONTENTS ix Calculation of Dead Load Stresses in Arch..................... 135 Seated Vy ind Load Stressés:in Arch... secs bese ances ees 137 MMPI ROMEO AL L1G oe sn iin he ward aldo cate tine ood os ecw 139 INNER TREASOS sic. nade aaa E eS aS Moule ls dbo oda 140 Mere Oonineed ATH. ci i.e ccc a ce cans acccbes bleteasdages 141 CHAPTER XV. COMBINED AND EccENTRIC STRESSES. Combined Direct and Cross Bending Stresses................... 143 Combined Compression and Cross Bending...............-..--. 145 Combined Tension and Cross Bending... ......0.... 0. eccccacecs 148 miteop ia-bar. Jue to its Own Weight... 02.5000 0.0 cnvenesccce 149 Diagram for Stress in Bars Due to Their Own Weight........... 149 Peemnr ie IM@LGHCONNECHIONS: o.oo ke ees cheese eadecae 152 NS iggy sec wile bebe veadccavey os Ea Gly eth et 154 CHAPTER XVA. GRAPHIC METHODS FOR CALCULATING THE DEFLEC- TION OF BEAMS. Introduction ..... Be eae eiet ey ave. ens-.p: do staves fide ios: bs dias WE CR UV aa eee 158 Graphic Equation of the Elastic Curve.................0000- ,--158 NN NEI Re re cl sao os a oe wb Sac soe Seba epithe Rel 161 a Roe lg Cyn G5 cso a b's alee tyne Ch ORS aa beled 162 Continuous AS SEE ee ei ane ee hirer ray 164 entmunis Beam of # Spans... cs... soc. ese cc ce aveescses 168 IEEE eB oo a ey arary slags since slaw s vas ev BME EROS vee es 168 Reactions of Simple Draw Bridges srerare Wealhaee Neate $029 bs Re ee a 169 Praw Bridge with Three Supports... 0.6660. sss ween 170 Draw Bridge with Four Supports..................0.- 172 PART III. DESIGN OF MILL BUILDINGS. CHAPTER XVI. GENERAL DESIGN. EMC en oy ks abc Sep ened sdeesaceeeweseeraes 175 CuHarterR XVII. FRAMEWORK. NEES oy ole SoG Sig's (as wieis cig'é'ss's%s sees Oe ee se ON eae as 179 Meena es Of TTUSSES. fi... ec ees wes cee eens 186 EUS OSS a SS St ee ra OE eM ae Oy Dear Wanacieice -kS 181 x TABLE OF CONTENTS : Ketchum’s Saw Tooth Roof... ..2., 5ic..e ee suites ne 186 Pitch of Roof... 0.5.0.5. «ives oo QE Pitch of Truss... ...0. 0... + 0/0 ao 192 Economic Spacing of Trusses... .2i 34 eq 9 eee ee eee 192 Transverse Bents..... 0.604. ses et se ote een ee aes Truss: Details . 2.0655 cies ca eos wa ee 197 Coltynns 05. 3. ce ae sel 50 a0 0s Ss oP 202 Column Details: 0.6.0.2... 2.35 ape ee ee 205 Strots"and Bracing... o 2... 2. 26. .s'era le eae nee 211 Purlins and Girts . 23.5 cbs 250s + 0 oe ne 6 ae ee 213 Design of Parts of the Structure... 2.255 gy ease oe 213 Design of ‘Trussés:......:. 3.32% «93 sae ee 214 Design of ‘Columns... 2.\2.55 cs ol cece ie ee 217 Design of ‘Plate Girders:. ..0. 295.42 04 « cane eee 221 Crane Girders 2... 6. oss cw aul Do 2 pe eee 224 CuHapter XVIII. CorruGATED STEEL. Introduction | 20.00. 28 vs we eg oe pee ee 225 Fastening Corrugated. Steel... i.45:.4 dou Pe eae wee 227 strength of Corrugated Steel... 0.5 Sp eeaaweee ee ee 230 Corrugated Steel Details.s;:. «20:5. 22a eae eee ee 232 Anti-condensation Roofing ....;.°. . Gs. sepa ack oe 241 Corrugated Steel Plans... . 2.4 43.4's.055 cate ee ee 244 Cost of Corrugated Steel. . i... sess cameeeen 5 wicca pg kone wan iw CHAPTER XIX. Roor COVERINGS. Introduction 2. .fs ais 1s Jelena e 246 Corrugated Steel Roofing... 2: ... ene ee ee ee ee 246 Slate Roofing .0...0052.. 20.420 ieee 247 Vile Roofing... ic... 5 ecco nla ane ale ne ea 250 te Roots oo... 2. 2 hy Ae eee S315, cate eater oat tor ona 251 sheet: Steel Roofing. :. i v3u.. 5 oe tea ee MAPS bore eae 253 Gravel Roofing ... ..... tjaxass ben ee pene eee eee 254 slag Roofing 3.5.05. 26 .<.¢44 a sik 6 oe ee 250 Asphalt Roofing <... .. 2. ..2. . ting obs one nee dere oer 257 pninele Roofs... ..:..0 ose ee ani Ree PREPARES >... <6. ..5. face lev eee ebasee ve 278 a Allowable Pressures ........... Bec iruict show owen 4's satay Va eas. 279 j CHAPTER XXII. Froors. q premaractsdo0rs—— | ypes Of PIOOTS, 0.5. ooo b ee csae vee ve ees sve 281 E MMMM ERIS DS gon cGy io by ds cords KR Laan Seeds 282 | Pen eee ICCeTe WOES 5 a AUN sn Foose cea bbe eeean 284 SS SRR ig GR ie Na fa ce AER Pet 284 MRT AOR es Win ein ire ve od Cepek ee 4 ada we bas 285 MGs EEE UNE G ss 0 oe ib sa ca Rie do ¥ 6 oe wow wbrhe 288 RNIN SERIE irate We Ui cratuiety ss ka woh ora cs vee auee waned 290 ; NRW eae eats, i. obo 8 oak 's oe) bo Bie ee wars eee 290 SPEER ican tas yt tha o's. oe «v's eos Sw ae PR 290 Mae, TRO LCN 2. SEP pws ov ec Ok eee cee 291 PROG OMEN creck kc seed chase ae sess skate &. eM 292 xii TABLE OF CONTENTS Roebling o.oo... ods ose sie + cielo by elena gn 293 “Buckeye” Fireproof .......... 555 «sp aeneee ae eee 2094 Multiplex Steel. 2.2... .. ui ieee male 295 Ferroinclave: ; ....3 0...) ¢ os¢w. «eee ‘enetaisen opus 205 Corrugated 0.) 06665 cee 0 oe ah aetna 295 Buckled Plates ..... ....... ..'. catenin gee 296 Steel Plate... 0). 52.2404. = oes ole atelier 297 Thickness of Timber Flooring: .39 0 ess eas sees os Fee 297 CHAPTER XXIII. Wunpows AND SKYLIGHTS. GAZING oo bes ng os w sin se 0 wd 1b wg 298 BSS oes 55 Ge GS 3 Suelo a leie oe 3 ve eee ocxaee 298 Diffusion of. Light: 3.0... . snot ie aealy pee oy ee 299 Relative Value of Different Kinds of Glass............. 301 Kind of Glass to: Use. ... so. ess a ot > alee ee 301 Placing the Glass \. ....\.:0:c « ic:sise's-b abs ees ate, 302 Use of Window Shades. . 2.2. 2.:5242.4 5 oe 303 size and Cost of Glass.... : i534. <2 bose ote ee 304 Cost. of Windows. = ..6....00.5 see son bet obs ee ee 305 Translucent Fabric)... 05 6. <0. «cos aniea da 5 Oke hc eee ee 306 Cost of Translucent Fabric. -.. i052.) «a oe eee 307 Double; Glaze --s'.6. vsos vccehiwg eens Sa eee wie 65 enemies 307 Details of Windows and Skylights. ..........2.:..ss49sseeeeees 307 Amount.of Light Required ... ........ <6 «00 0 eee 310 skylights for Trainsheds.....)... os. -siewd see + 5 sialew ace oa 315 CHAPTER XXIV. VENTILATORS. Ventilators 0.00606 3.055 Gelsc ese Ws eres aratans phe ne eee 317 Monitor Ventilators 22.000... ea% + apis cses ares va aeeenne a 317 Cost of Monitor Ventilators... ...... <..5 3.) sae 320 Circular Ventilators... ..... 0. ic ices a viele oles a ee 320 Paneled. Doors 2... 62. ...06< cs cums bap by 6 an go 322 Woodem Doors). 2... ii. ec oNa sae 5 eee Be Ry 322 pteel Doors . 6.6... 5. cen ceeu teem beets) el ge 323 Cost of Doors . . 225.06: shea wee semen se 325 . a, i. o = eo a x A a wea ov «ae a Ff 4 ia a i re” a : a a er, a” ' ae oe ee ees et Cll ie > Shilo Yas Pai er ’ a sn 4 it a ee Pee eT” ere ek ee Aaah ee Eee eh Sw, ee Nee’. le late ae.) = TABLE OF CONTENTS Xiii CHAPTER XXVI. Suorp DrAwINGs AND RULES. Shop Drawings Bere tio nic tte een ee tig hy eet Ais ere 326 NEE CRG cas ho) nop eh oh May ca ek ee deem oh awe ets 327 ID yao ccsnt 5 'a.4 io gle eg eee ky aed Seay cole 6 ae 2s 328 CHAPTER XXVII. Parnts AND PAINTING. ERA cir eb iie ss Sitn wo eas Wenala eins Beene Vee o6ae «390 NM Rar rat 2° So fo weg x) 4,4 a law icgle le Jere cece W's, Av 8 Waihi ow Deere Rcdses 330 MN i608 sig; yo by rd MS Le dein eas Meae no els poe ae 330 Ss ee ESTs A ae epee Raereartieee Nate AINE Yee cores att 331 I in od oe ES eg ae CR Fa TE cd Oa ee 332 0 os Sag 5 SIE a an a eA a EOLA en TR aS Se a 333 Me a Ga Sess hE eS FR, Wola Bd Weta Ske Wide 6 aw a 333 Ne eae aitere id 2G a OR RA SOD 333 MDT MEME TE DANI eco. oie os sits, Hoare A ees wk aie Fe Lose CO eS a a ne ee a Sata tik a aisaly: cas ag Soe MI BD ATIE Yoo ick ces 5b RUNES NS om 3 kn WR de ee Riek ate 334 PE TAU AIC esos vis y Gms yitc ma beaten ed Oey 0s 335 Ree OM SCs ak 35 slater ala vid oc wes Rios ORI 336 eM E MENISCI cma sg hy ed anes neal oe aby a 336 Gost of Painting oe 8o6% ss Se STA ry Gis Seca p Lenya lees $2) CRS SRB Iie ea pean alts Soa hg Dar ee arr SENN ky 2a I OANA fiiiia cea 0k bg didn SB Rae ae hg OS Beles 338 Me gra hie Fe ws av 6S ¢ Mogi Sipeavs-> wl eiv ead swig Bie Ne 339 ER RELI stae esgic Sve cbs Sac, « OM dca Cae 0 9 8a ee ake we 339 Permanent Paints coe. cs ec bo calcd s ev cee nedeceteeees 339 POEM PIRAIAD ROTTEN ALTIC. «5.05 5 wets asa ods SlRibie' a be als wal elles 340 Sermrcnees cal Aiat 2nd Patines ie Se. Cele sos cae pe eee 340 CHAPTER XXVIII. Estimate oF WEIGHT AND Cost RRO POT eto ir g's ew 2 AAA Gj 3000 PPVIA PLZT Weight of Roof Trusses D LEV HA TL Copacity 40 lbs persa ft ra ZAAK. F 2000 Vid w= Z8AL (i+ Ze) we) 22g Compression 15000-55¢ Ibs-persqun} + . < Le Tension 15000\bs per sq:in- S 1000 Fr oa Pitch=4 (6'inl2") > Trt eee CI FrCCLLCLE ne LLLI | be i ei 40) 30 AO 50 60 70 80 gO 100 Length of Spanof Truss Z, Ft. Fic. 2. WEIGHT OF ROOF TRUSSES FOR A CAPACITY OF 40 LBS. PER SQUARE FOOT. The weight of steel roof trusses per square foot of horizontal projection of roof for a capacity, P, of 40 lbs. per square foot is given in Fig 3. It should be noted that W , is the dead load per square foot carried by an interior truss. The actual weight of trusses per square foot of horizontal projection for a building with panels will be W, AG where end post bent (a), Fig. 1, is used, and -W . eee! where end truss bent (b), Fig. 1, is used, assuming that all trusses are made alike. Weight of Light Trusses—Formula (1) gives the weight of mill building trusses and will usually cover the weight of knee braces and ventilator framing. By reducing the minimum thickness of metal and LAL aOR Tia ities sitet ok i ee SEE. Li. Fae ae 7 —T Pra Yas en Ff “I aa. Mil) ‘hehe ST eis 1 Nahin edie ntl hi nt en, Serra herds SeLeseints ecethmbpests Sect akilad » ers ee ‘ ie “ ‘ . a x oe o mings 5 ‘ ’ aC. ba ae 7 ENS St We. a 7 “4 ee Jd im 27s Length of Span of Truss,L, in feet. Fic. 3. WEIGHT OF ROOF TRUSSES PER SQUARE FOOT OF HORIZONTAL PROJECTION FOR A CAPACITY OF 40 LBS. PER SQUARE FOOT. by skinning the sections it is possible to materially reduce the weights. Weight of Simple Roof Trusses.—Simple roof trusses supported TABLE I. WEIGHT OF FINK TRUSSES, SUPPORTED ‘ON MASONRY WALLS, DE- SIGNED FOR A VERTICAL LOAD OF 40 POUNDS PER SQUARE FOOT OF HORIZONTAL, PROJECTION OF ROOF. Distance Distance Span, L, between Weight of Span, L, between Weight of Be ee | merous || Fer | Pree A | in Pounds 30 16 741 65 20 3226 30 14 621 70 20-4; 3951 Sie 16 910 75 20 4564 40 16 1211 75 14 3200 40 14 976 80 20 5160 45 16 1423 85 25 6730 50 16 1865 85 14 4000 50 14 1550 90 25 8010 55 16 2103 95 25 8600 oo. ft 20 2870 100 25 9392 60 14 2120 Weicut oF Roor Trusses 7 % ; -t-] Weight of Roof Trusses per sq-ft hor. proj: 8 AO ws £& 6EEO wWewt-per sqft-hor proj. of roof’ oS - 4 =spanof truss in ft. apr 3 {_|_} A=distance between trusses in fr pele ¢f § 5 : BOG ahead Ss f = 6 ] at f = B 2 Z a 4 - c gage ssesene a eusteserna @2225 a 2 c age S Compression = 15000-5577 Ibs. per Sq: ne + Tension =15000 Ibs.persg in. ai wy Et ve LJ oO Fe Pitch =4 (6"inl@") 8 4 ae 930-40 - 50: «O80: 80. | 00 8 DeaD LOADS on walls will usually weigh somewhat less than the value given by Formula (1). The computed weights of Fink roof trusses without ven- tilators and with purlins spaced from 4 to 8 feet are given in Table I. These trusses were designed by two different bridge companies to serve as standards and represent minimum weights. The trusses with a spacing of 14 feet were designed with minimum thickness of metal 3-16” and minimum size of angles 2” x 13@” x 3-16”. In the remainder of the trusses the minimum thickness of plates was 4” and minimum -size of angles 2” x 2” x 14”. The trusses are all too light to give good service although their use in temporary structures may sometimes be allowable. Weight of Purlins, Girts, Bracing, and Columns.—Steel ‘purlins will weigh from 14 to 4 pounds per square foot of area covered, depending upon the spacing and the capacity of the trusses and the snow. load. If possible the actual weight of the purlins should be cal- culated. Girts and window framing will weigh from 1% to 3 pounds per square foot of net surface. Bracing is quite a variable quantity. The bracing in the planes of the upper and lower chords will vary from \% to 1 pound per square foot of area. The side and end bracing, eave struts and columns will weigh about the same per square foot of sur- face as the trusses. Weight of Covering —The weight of corrugated iron or steel covering varies from 114 to 3 pounds per square foot of area. WEIGHT OF FLAT AND CORRUGATED STEEL SHEETS WITH 2% INCH CORRUGATIONS. Thickness|_ Weight per Square (100 sa-ft-) Gage No. in Flat Sheets Corrugated Sheets _inches_ | Black _|Galvanized|Black Painted [Galvanized /6 0625 250 266 ZTs 29/ /8 -0500 200 2/6 220 236 20 0375 1350 /66 169 182 £2 0313 123 /4/ 138 154 24. 0250 100 1/6 aS A /27 26 0/88 735 a 84 99° 28 0/56 63 79 69 66 WEIGHT OF RooFr CovERING 9 Ip estimating the weight of corrugated steel allow about 25 per cent for laps where two corrugations side lap and 6 inches end lap are re- quired, and about 15 per cent for laps where one corrugation side lap and 4 inches end lap are required. Nos. 20 and 22 corrugated steel are commonly used on the roof and Nos. 22 and 24 on the sides. Weight of Roof Covering—The approximate weight per square foot of various roof coverings is given in the following table: Corrugated iron, without sheathing......... I to 3. Ibs. Felt and asphalt, without sheathing........ 2 i Felt and gravel, without sheathing.......... 8 to Io is Slate, 3-16” to %4”, without sheathing...... 7to 9 + pea Without slieathing 00... cee ck ee i403 Skylight glass, 3-16” to 1%4”, including frames 4 to 10 = “ White pine sheathing 1” thick ............ 3 . Yellow pine sheathing 1” thick............ 4 2 etal Pid ure a Chena Sa Few e eae I5 to 20 7 A ITU SOUGGS chs ec loco sess tebe cscs « 8 10 I0 . Tiles, on concrete slabs... ........3.00600- 30 to 35 . RMMOR ORY ENING solos ok. gees See wc Ve sl eeas 10 _ For additional data on weight of roof coverings, see Chapter XIX. The actual weight of roof coverings should be calculated if possible. Weight of the Structure—The weight of the roof can now be fcund. The weight of the steel in the sides and ends is approximately the same per square foot as the steel work in the roof. A very close approximation to the weight of the steel in the en- tire structure where no sheathing is used and the same weight of cor- rugated iron is used on sides as on roof, may be found as follows: Take the sum of the horizontal projection of the roof and the net sur- face of the sides and ends, after subtracting one-half of the area of the windows, wooden doors and clear openings; multiply the sum of these areas by the weight per square foot of the horizontal projection of the roof, and the product will be the approximate weight of the ‘steel in the structure. CHAPTER II. Snow Loans. The annual snowfall in different localities is a function of the humidity and the latitude and is quite a variable quantity. The amount of snow on the ground at one time is still more variable. In the Lake Superior region very little of the snow melts as it falls, and almost the entire annual snowfall is frequently on the ground at one time; whiie on the other hand in the same latitude in the Rocky Mountains the dry winds evaporate the snow in even the coldest weather and a less pro- portion accumulates. In latitudes of 35 to 45 degrees the heavy snow- falls are often followed by a sleeting rain, and the snow and ice load on roofs. sometimes nearly equals the weight of the annual snowfall. From the records of the snowfall for the past ten years as given in the reports of the U. S. Weather Bureau and data obtained by personal experience, in British Columbia, Montana, the Lake Superior region and central Illinois the author presents the values given in Fig. 4 for snow loads for roofs of different inclinations in different latitudes. For the Pacific coast and localities with low humidity, take one-half of the values given. The weight of newly fallen snow was taken at 5 Ibs. and packed snow at 12 Ibs. per cubic foot. A high wind may follow a heavy sleet and in designing the trusses the author would recommend the use of a minimum snow and ice load as given in Fig. 4 for all slopes of roofs. ‘The maximum stresses due to the sum of this snow load, the dead and wind loads; the dead and the wind loads; or of the maximum snow load and the dead load be- ing used in designing the members. Snow Loap ON Roors II iy LS OG a RR a Ca 30.5 a eS e " — 4orFecitic Coastand ao a. Aria Fegions use orie- AO i} Seat hal) (f 7 ‘Pe he ALUIAT valves Pr £ _— Veecae = Lf ; r na a 40 ny q = ¥G O ce ian s q 3 ae eg & 5 i Pit: T 20 < 2 — h 5 3S a SP El TS asl Pie, S dd F § 10 L if yh 2 Straight Line F=£A,A 2457 q TIN V ii G7 #3=Normal Pressure, lbs-per sq.ft: 4 / LZ Ss. a " ’ id PaHorizontal “ “o « “ / YIRC £ : Slists s A=Angle of inclination of surface Were teenerely_ [a ih “KO —10) F=is¢] 19 =u ; © 5 0 1 2 2 3 35 40 45 50 5 6 6 7 75 6& 8& ; Angle Exposed Roof makes with Horizontal in Degrees .A- 4 Fic. 6. NorRMAL WIND LOAD ON ROOF ACCORDING TO DIFFERENT . FORMULAS. ; Duchemin has also deduced the formula 4 site Dela A (7) _ guretne gs ge Sah Le ‘ where P, in (7) equals the pressure parallel to the direction of the ‘ wind, Fig. 5; and : Beak: P 2 sin A oon A (8) E 1+ sin? A 16 WIND LOADS where P in (8) equals the pressure at right angles to the direction of the wind, Fig. 5. P, may be an uplifting, a depressing or a side pres- sure. With an open shed in exposed positions the uplifting effect of the wind often requires attention. In that case the wind should be taken normal to the inner surface of the building on the leeward side, and the uplifting force determined by using formula (8). If the gables are closed a deep cup is formed, and the normal pressure should be increased 30 to 70 per cent. That the uplifting force of the wind is often considerable in exposed localities is made evident by the fact that highway bridges are occasion- ally wrecked by the wind. The most interesting example known to the author is that of a 100-foot span combination bridge in Northwestern Montana which was picked up bodily by the wind, turned about go degrees in azimuth and dropped into the middle of the river. The end bolsters were torn loose although drift-bolted to the abutments.* The wind pressure is not a steady pressure, but varies in intensity, thus producing excessive vibrations which cause the structure to rock if the bracing is not rigid. The bracing in mill buildings should be designed for initial tension, so that the building will be rigid. Rigidity is of more importance than strength in mill buildings. For further information on this subject see a very elaborate and valuable monograph on “ Wind Pressures in Engineering Construc- tion,’ by Capt. W. H. Bixby, M. Am. Soc. C. E., published in Engi- neering News, Vol. XXXIII., pp. 175-184, March, 1895. Wind Pressure on Office Buildings.—The following specifica- tion for wind pressure on office buildings has been proposed by Mr. C. C. Schneider. + Wind Pressure-——The wind pressure shall be assumed at 30 Ib. per sq. ft. acting in either direction horizontally: 1. On the sides and ends of buildings and on the actually Boe surface, or the vertical projection of roofs; 2. On the total exposed surfaces of all parts composing the metal framework. The framework shall be considered an inde- pendent structure, without walls, partitions or floors. *For a description of the wreck by the wind of the High Bridge over the Mississippi River at St. Paul, Minn., see article by C. A. P. Turner in Trans. Am. Soc. C. E., Vol. 54, p. 31. 7 Trans. Am. Soc. C. E., Vol. 54, 1905. CHAPTER IV. MIscELLANEOUS LOADS. LIVE LOADS ON FLOORS.—Live loads on floors for mill buildings are very hard to classify and should be calculated for each case. Floor loads as specified in the building laws of various cities are given in Table II, and the engineer should govern himself accordingly. . TABLE II. FLOoRr LOADS IN POUNDS PER SQUARE FOOT AS SPECIFIED IN VARIOUS CITIES. New York Chicago Philadelphia Boston | : Upper floors 75 Dwellings ........ tek po en 1200 100 100 Public Buildings.. 90 100 120 150 Light Manufac- PRIS 3\5.s'a'e'o e's 120 100 100 eee eres Warehouses and Factories....... BO BONG UD). 5:05:50 150 and up | 250 and up Sidewalks ........ SU 2.) baw eb gaa slave dia weageeles + 065 Reena Without reference to building laws the live loads per square foot, exclusive of weight of floor materials, given below are about standard practice. PIWRMINOS «oo vices «ds was 70. Ibs. per sq. it. RN oa. 4 5 5:5 0c «6 jade ee.270-t0 100: Ibs." per. sq. it. Assembly halls ......... 120 to 150 lbs. per sq. ft. NWEATCHOUSES: So. covceses 250 and up, lbs. per sq. ft. Factories ............--200 to 450 lbs. per sq. ft. 18 MISCELLANEOUS LOADS Reo) + The weight of floors above ground in mill buildings varies so much that it is useless to give weights. of floors see Chapter on Floors. WEIGHT OF HAND CRANES.—The approximate weight of a few of the common sizes of hand cranes made by Pawling and Har- For a few data on weights TABLE, III. WEIGHT OF TRAVELING HAND CRANES. 20-FOOT SPAN 30-FOOT SPAN Capacity of Distance Crane in ctocof - Weight of |MaximumLoad| Weightof | Maximum Load Tons end wheels Crane on each Wheel Crane on each Wheel ibs. lbs. lbs. lbs. 3 3’ — 0” 4500 4500 5000 5000 2 3’ — 0” 5500 7000 6000 7500 3 1% 3' — 8” 8000 10500 9000 11500 10 8’ — 0” 15000 18000 17000 20000 ; 15 8’ — 0” 16000 20000 18000 21000 af a 8’ — 6” 20000 26000 22000 27000 nischfeger, Milwaukee, Wis., and the maximum load on each wheel when the loaded trolley is at one end is given in Table III. A I NS aa S120) ee PTS (approx}=COPpocrty fons , 7 BR ins sae I BB os sete ge ne 1) st 3 = voy os 0 OS OB Bs SN > 90 —— pH ne re) a = = 80 50 Ton € PEE 70 to ; ed = 40 Ton Crane am oO 60 a= iy ceo 3 507 30 Ton Crane | oO _ ha el iG ed 40 20 Ton Crane — ee o © /5_Ton Crane = 20 eer 70 Ton Crane Tor) aonee i 7 a io} = Lrtrrrrrt 20 30 840 50 60 70 80 Span of Crane in Feet Pic. 7, WEIGHT OF ELECTRIC CRANES.—The maximum load on each of the end wheels for common sizes of electric cranes made by WEIGHT OF ELECTRIC CRANES 19 Pawling and Harnischfeger is given in Fig. 7. Cranes made by different manufacturers differ considerably in weight. The weights and dimensions of typical traveling cranes as given in Table Illa have been proposed for adoption as a standard. TABLE IIIa. TYPICAL ELECTRIC TRAVELING CRANES.* skein in | span. | Wheel Base. Wheel font, 8. v. eae res in Pounds. Plate Girders. | Beams. Bia esesexnccsas 40 8 ft. 6 in. 12,000 10in. | 7 ft. | 401b.peryd.| 40 60 G64 .Q 46 13,000 ‘s te 40 6 40 Maite ed cusses 40 PF Qs 19,000 6 es 45 “ 40 60 9 « § «8 21,000 “6 és 45 6 40 Boosh chase Cass 40 9 « 6 « 26,000 4 oh 50 6 50 60 | 10 * 0 « 29,000 $4 se 50 “6 50 Se 40 | 10 * 0 « 33,000 12in. | 8 ft. 55 “6 50 60 | 10 * 6 *« 36,000 6 ss 55 “s 50 Bee en eatiisies 40 |10* 0 * 40,000 6 6 60 ‘6 50 60 | 10 *-6 <« 44,000 ¢ 6 60 és 50 BOs sacoswstsvesse 40 1.10" 6. « 48,000 és ¢ 70 “6 60 60 | 11 «0 « 52,000 «6 “6 70 ‘6 60 DN wes saves as OO 4°11 0 + 64,000 14in. | 9 ft. | 80 “6 60 60 | 12 0 « 70,000 6 ¢ 80 “6 60 . ae 40 }11«0<« | 72,000 « se gp. ts 60 60 | 12 «0 « | 0,000 «“ « |100 « 60 I. Wheel-load can be assumed.as distributed in top flange, over a distance equal to depth of girder, with a maximum limit of 30 in. 2. In addition to the vertical load, the top flanges of the girder shall withstand a lateral loading of two-tenths of the lifting capacity of the crane, equally divided between the four wheels of the crane. $ == Side clearance from center of rail. v= Vertical clearance from top of rail. 3. The top flanges of the crane girders. shall not be of smaller width than one-twentieth of their unsupported length. WEIGHTS OF MISCELLANEOUS MATERIAL.—The weights of various kinds of merchandise are given in Table IV. For weights of other materials consult steel makers hand books. * Mr. C. C. Schneider in Trans. Am. Soc. C. E., Vol. 54, 1905. 20 MISCELLANEOUS LOADS TABLE IV. WEIGHTS OF MERCHANDISE.* commonty | Menge || commoatiy | ela Wool-in Bales......). 5 to 28 Caustic Soda: .<.ceus 88 Woolen Goods ........| 13to22 /||Barrel Starch....... 23 Baled Cotton..........| 12to43 /|/Barrel Lime..:..... 50 Cotton Goods..... oan. di to 37 ‘* Cement...... 73 Rags in Bales..... Sen cd tO556 ‘e.- Plaster cca) 53 DOADCL Sno vied eaesecacenl 20 40.09 “art Of. 34 WRU: Lae snncis vee NPAs 39 to 44. !Rone i.ncccessct ee 42 CEN on 3 6 ev kes hee ROS 31 Bos: Piisssus reve 278 MRIS be osicGe aca cues 27 Box Glase: ies s.e8 60 Baled Hay and Straw.| 14to19 _ |!Crate Crockery...... 40 Bleaching Powder.... 31 Bale Leather........} 16 to 23 SHAR NG ahs nae sews 62 SiG GF iadse snus’ eens 45 Box Indigo...... ios 43 Cheeses 025-5 v naswens 30 The weights of miscellaneous building materials are given in Table IVa. : TABLE IVa. WEIGHTS OF BUILDING MATERIALS. Material. bi 3 9 per Material. vee per Pav ACK 5 oie sigsecn wage unvess 150 ae ois scan cepuaa-y tere 160 Common building brick...... 120 Snow, freshly fallen ........ 10 Soft building brick............. 100 Snow; Wet 6..i as ivesancone 50 CRAYNE iis. oss ats sapasenuvias ee 170 Spreice® 5 .izscssitewansersees 25 BARONET snsasusaavcannaneuvaes 170 PlewsOGl | op catoveess ecient 25 LEMCRIONE i os cdcsssvekoesedsuen: 160 Wile PINE 6. oi. ccsiceneodeenes 25 SANSONE: . <<, cecsseaavcaanen sven 145 DOURAS THE 5.5 ccsasshasaee vase 30 SSPE Mahe onknscskaaseaesvaspepaeed 40 Yellow pine ..:...026000.Sees< 40 GEOWEN Tg acd Jopwinsiktodaeas dieses 120 White 00k. ..2ica¢conrvenanas 50 SURE sulci caycdeas sce keane 175 Common brickwork ......... 100-120 Sand, clay and earth (dry).. 100 Rubble masonry.............. 130-150 Sand, clay and earth (wet). 120 Ashlar masonry ...........+0: 140-160 DERRY Ss 65 cc geteencuvaqeesaoshs 100 Cast In0R.. horizontal components of forces = 0 (a) > vertical components of forces =O (b) = moments of forces about any point = o (c) Representation of Forces.——A force is determined when its magnitude, line of action, and direction are known, and it may be rep- resented graphically in magnitude by the length of a line, in line of action by the position of the line, and in direction by an arrow placed on the line, pointing in the direction in which the force acts. A force may be considered as applied at any point in its line of action. | Force Triangle——The resultant, R, of the two forces P, and P, meeting at the point a in Fig. 8 is represented in magnitude and direc- tion by the diagonal, R, of the parallelogram a b ¢ d. The combining of the two forces P, and P, into the force R is termed composition of forces. The reverse process is called-resolution of forces. Force TRIANGLE 23 Pe Cc Pe i 77 on , ae : or ad ; P, as) e-" i C- Bet : ee i 7 ee LE : M LE ~ a as | oe < | a Ps; dq P> . (a) (b) (c) Fic. 8. The value of R may also be found from the equation fe Pit PS 4 2 PSP, cos: 0 It is not necessary to construct the entire force parallelogram as in (a) Fig. 8, the force triangle*(b) below or (c) above the resultant R being sufficient. If only one force together with the line of action of the two others be given in a system containing three forces in equilibrium, the magni- tude and direction of the two forces may be found by means of the force triangle. If the resultant R in Fig. 8 is replaced by a force E equal in amount but opposite in direction, the system of forces will be in equi- librium, (a) or (b) Fig.9. The force E& is the equilibrant of the system of forces P, and P,. U7 Gal RB a ee Pe (a) (b) Ric: 6; i aa It is immaterial in what order the forces are taken in constructing the force triangle, as in Fig. 9, as long as the forces all act in the same direction around the triangle. The force triangle is the foundation of the science of graphic statics. 24 GRAPHIC STATICS Force Polygon.—If more than three concurrent forces (forces which meet in a point) are in equilibrium as in (a) Fig. 10, R, in (b) will be the resultant of P, and P,, R, will be the resultant of R, and Pz, (b) Fic. 10. and will also be the equilibrant of P, and P;._ The force polygon in (b) is therefore only a combination of force triangles. The force polygon for any system of forces may be constructed as follows :—Beginning at any point draw in succegsion lines representing in magnitude and direction the given forces, each line beginning where the preceding one ends. If the polygon closes the system of forces is in equilibrium, if not the line joining the first and last points represents the resultant in magnitude and direction. As in the case of the force triangle, it is immaterial in what order the forces are applied as long as they all act in the same direction around the polygon. A force polygon is analogous to a traverse of a field in which the bearings and the distances are measured progressively around the field in either direction. The conditions for closure in the two cases are also identical. It will be seen that any side in the force polygon is the equilibrant of all the other sides and that any side reversed in direction is the re- sultant of all the other sides. Equilibrium of Concurrent Forces.—The necessary condition for equilbrium of concurrent coplanar forces therefore is that the force polygon close. This is equivalent to the algebraic condition that = horizontal components of forces = 0, and 3 vertical components of forces = 0. If the system of concurrent forces is not in equilibrium the resultant can be found in magnitude and direction by completing EQUILIBRIUM OF ForRCES 25 the force polygon. The resultant of a system of concurrent forces is always a single force acting through their point of intersection. Equilibrium of Non-concurrent Forces.—If the forces are non-concurrent (do not all meet in a common point), the condition that the force polygon close is a necessary, but not a sufficient condition for equilibrium. For example, take the three equal forces P,, P, and P,, making an angle of 120° with each other as in (a) Fig. 11. A ~ O 1 | Resultant Moment GF R wh =-FPh { Pi Ps Positive Moment J Moment=+Ph CRS, . P, A \ PR; rs “Hs. (Q) (b) GF ° Negative Moment Moment =-Ph (Cc) Fic. 11. The force polygon (b). closes, but the system is not in equilibrium. The resultant, R, of P, and P, acts through their intersection and is parallel to P,, but is opposite in direction. The system of forces is in equilibrium for translation, but is not in equilibrium for rotation. The resultant of this system is a couple with a moment = — P, h, moments clockwise being considered negative and counter clockwise positive, (c) Fig. 11. The equilibrant of the system in (a) Fig. 11 is a couple with a moment = + P, h. A couple—A couple consists of two parallel forces equal in amount, but opposite in direction. The arm of the couple is the per- pendicular distance between the forces. ‘The moment of a couple is equal to one of the forces multiplied by the arm. The moment of a couple is constant about any point in the plane and may be represented 26 GRAPHIC STATICS graphically by twice the area of the triangle having one of the forces as a base and the arm of the couple as an altitude. The moment ofa force about any point may be represented graphically by twice the area of a triangle as shown in (c) Fig. 11. It will be seen from the precéding discussion that in order that a system of non-concurrent forces be in equilibrium it is necessary that the resultant of all the forces save one shall coincide with the one and be opposite in direction. Three non-concurrent forces can not be in equi- librium unless they are parallel. The resultant of a system of non- concurrent forces may be a single force or a couple. Equilibrium Polygon.—First Method.—In Fig. 12 the resultant, a, of P, and P, acts through their intersection and is equal and parallel to a in the force polygon (a) ; the resultant, b, of a and P, acts through their intersection and is equal and parallel to b in the force polygon; the resultant, c, of b and P, acts through their intersection and is equal and parallel to c in the force polygon; and finally the resultant, R, of c and P, acts through their intersection and is equal and parallel to R in the force polygon. R is therefore the resultant of the entire system of forces. If R is replaced by an equal and opposite force, E, the sys- tem of forces will be in equilibrium. Polygon (a) in Fig. 12 is called EQUILIBRIUM PoLyGoN 27 a force polygon and (b) is called a funicular or an equilibrium polygon. It will be seen that the magnitude and direction of the resultant of a system of forces is given. by the closing line of the force polygon, and the line of action is given by the equilibrium polygon. The force polygon in (a) Fig. 13 closes and the resultant, R, of Resultant Moment | (b) = -Feh BiG. 73. the forces P,, P,, P;, Ps, Ps is parallel and equal to P,, and is opposite in direction. ‘The system is in equilibrium for translation, but is not in equilibrium for rotation. The resultant is a couple with a moment = —P,h. The equilibrant of the system of forces will be a couple with a moment = + P, h. From the preceding discussion it will be seen that if the force polygon for any system of non-concurrent forces closes the resultant will be a couple. If there is perfect equilibrium the arm of the couple will be zero. Second Method.—Where the forces do not intersect within the limits of the drawing board, or where the forces are parallel, it is not possible to draw the equilibrium polygon as shown in Fig. 12 and Fig. 13, and the following method is used. « The point 0, (a) Fig. 14, which is called the pole of the force poly- gon, is selected so that the strings a0, b 0, co, do and eo in the equi- librium polygon (b), which are drawn parallel to the corresponding 28 GRAPHIC STATICS rays in the force polygon (a), will make good intersections with the forces which they replace or equilibrate. In the force polygon (a), P, is equilibrated by the imaginary forces represented by the rays oa and bo acting as indicated by the arrows within the triangle; P, is equilibrated by the imaginary forces repre- sented by the rays o b and co acting as indicated by the arrows within the triangle; P, is equilibrated by the imaginary forces represented by the rays oc and do acting as indicated by the arrows within the tri- angle; and P, is equilibrated by the imaginary forces o d and e o acting as indicated by the arrows within the triangle. The imaginary forces are all neutralized except a o and o e, which are seen to be components of the resultant R. To construct the equilibrium polygon, take any point on the line of action of P, and draw strings o a and o D parallel to rays o a ando D, b o is the equilibrant of o a and P,; through the intersection of string o b and P, draw string c o parallel to ray c 0, co is the equilibrant of ob and P,; through the intersection of string co and P, draw string do parallel to ray do, do is the equilibrant of c o and P,; and through the intersection of string do and P, draw string €0 parallel to ray eo, eo is the equilibrant of do and P,. Strings o a and eo acting as shown are components of the resultant R, which will be parallel to R in the force polygon and acts through the intersections of strings o a and ¢ oa. ; a i . a 7. es : Ls _. ; ' J m4 ‘a ; 7 ie! - = 7 * - L'a a a 5 > AEE” 0 REACTIONS OF A BEAM 29 The imaginary forces represented by the rays in the force poly- ‘gon may be considered as components of the forces and the analysis made on that assumption with equal ease. It is immaterial in what order the forces are taken in drawing the force polygon, as long as the forces all act in the same direction around the force polygon, and the strings meeting on the lines of the forces in the equilibrium polygon are parallel to the rays drawn to the ‘ends of the same forces in the force polygon. The imaginary forces a 0, b 0, c 0, d 0, e 0 are represented in mag- nitude and in, direction by the rays of the force polygon to the same scale as the forces P,, P,, P;, P,. The strings of the equilibrium poly- gon represent the imaginary forces in line of action and direction, but not in magnitude. Reactions of a Simple Beam.—The equilibrium polygon may be used to obtain the reactions of a beam loaded with a load P as in Fig. 15. a p Oe ee ' mths ile Ss Yy aoe P ae ul k ; RO ebeigs2 Ri BS rec age ss Re AP a 0S shall ae nal Ret olan on ~ eg ¥- a ~~ ¢ (a) (b) Pic; XS. The force polygon (b) is drawn with a pole o at any convenient point and rays oa and oc are drawn. Now from the fundamental con- ditions for equilibrium for translation we have P = KR, + R,. At any convenient point in the line of action of P draw the strings 0a and oc parallel to the rays o a and o ¢, respectively, in the force polygon. The imaginary forces ao and oc acting as shown equilibrate the force P. 30 GRAPHIC STATICS The imaginary force a o acting in a reverse direction as shown is an equilibrant of R,, and the imaginary force ¢ o acting in a reverse direction is an equilibrant of R,. The remaining equilibrant of R, and of R, must coincide and be equal in amount, but opposite in direction. The string. b o is the remaining equilibrant of R, and of R, and is called the closing line of the equilibrium polygon. The ray bo drawn parallel to the string b o divides P in two parts which are equal to the reactions R, and R, (for reactions of overhanging beam see Chapter VIII). : | } Reactions of a Cantilever Truss.—In the cantilever truss shown in Fig. 16, the direction and point of application B of the reaction R, are known, while the point of application A of the reaction R, only is known. The direction of reaction R, may be found by applying the principle that if a body is in equilibrium under the action of three external forces which are not parallel, they must all meet in a common point, 7. e., the forces must be concurrent. ‘The resultant of all the loads acts through the point c, which is also the point of intersection of the reactions R, and R,. Having the direction of the reaction R,, the values of the reactions may be found by means of a force polygon. _ The direction of reaction R, may be found by means of a force and equilibrium polygon as follows: Construct the force polygon (b) with pole o and draw equilibrium polygon (a) starting with point A, the OPAL Be. Lone. vee Se! AF DP) e Oe ye ee EQUILIBRIUM POLYGON AS A FRAMED STRUCTURE 31 only known point on the reaction R,, and draw the polygon as prey- iously described. A line drawn through point o in the force polygon parallel to the closing line of the equilibrium polygon will meet R,, drawn parallel to reaction R,, in the point y, which is also a point on R,. — The reactions R, and R, are therefore completely determined in direc- tion and amount. The method just given is the one commonly used for finding the re- actions in a truss with one end on rollers (see Chapter VII). Equilibrium Polygon as a Framed Structure-—In (a) Fig. 17, the rigid triangle supports the load P: Construct a force polygon (a) (b) Fic. 17. by drawing rays a 1 and c 1 in (b) parallel to sides a 1 and ¢ 1, respec- tively, in (a), and through pole 1 draw 1 0 parallel to side 1 D in (a). The reactions R, and R, will be given by the force polygon (b), and the rays 1a, Ic and 1 b represent the stresses in the members I a, I¢ and 1 B, respectively, in the triangular structure. The stresses in I a@ and Ic are compression and the stress in 1 b is tension, forces acting toward the joint indicating compression and forces acting away from the joint indicating tension. Triangle (a) is therefore an equilibrium polygon and polygon (b) is a force polygon for the force P,. From the preceding discussion it will be seen that the internal stresses at any point or in any section hold in equilibrium the external forces meeting at a point or on either side of the section. 32 GRAPHIC STATICS Graphic Moments.—In Fig. 18 (b) is a force polygon and (a) is an equilibrium polygon for the system of forces P,, P,, P;, P,. Draw Fic. 18. the line M N = Y parallel to the resultant R, and with ends on strings o e and o a produced. Let r equal the altitude of the triangle L M N and H equal the altitude of the similar triangle 0 e a. H is the pole distance of the resultant FR. Now in the similar triangles L M N and oea Roy 2 sie and Rr= AY But R r = M = moment of resultant R about any point in the line — M N and therefore M oe The statement of the principle just demonstrated is as follows: The moment of any system of coplanar forces about any point in the plane is equal to the intercept on a line drawn through the center of moments and parallel to the resultant of all the forces, cut off by the strings which meet on the resultant, multiplied by the pole distance of the resultant. It should be noted that in all cases the intercept is a distance and the pole distance is a force. This property of the equilibrium polygon is frequently used in finding the bending moment in beams and trusses which are loaded with vertical loads. a Cetin po ke Bie Ce. wae te ~ ay Oe i OER ONG BENDING MoMENTsS IN A BEAM 33 Bending Moments in a Beam.—lIt is required to find the mo- ment at the point M in the simple beam loaded as in (b) Fig. 19. The Fic. 109. moment at M will be the algebraic sum of the moments of the forces to the left of M. The moment of P, = H x B C, the moment of P, = H x C D and the moment of R, = — Hx B A. The moment at M will therefore be M,=HAxBC+HxCD—HxBA=HxAD=—Hy The moment of the forces to the-right of M may in like manner be shown to be M,=+Hy In like manner the bending moment at any point in the beam may be shown to be the ordinate of the equilibrium polygon multiplied by the pole distance. The ordinate is a distance and is measured by the same scale as the beam, while the pole distance is a force and is measured by the same scale as the loads. To Draw an Equilibrium Polygon Through Three Points.— Given a beam loaded as shown in Fig. 20, it is required to draw an equilibrium polygon through the three points a, b, c. Construct a force polygon (b) with pole o, and draw equilibrium polygon a D’ c’ in (a). Point b’ is determined by drawing through 0b a line DD’ parallel to b, b” which is the line of action of the resultants of the forces to the 34 GRAPHIC STATICS \r Pe [p Pf y \ | a \ i ! gr a Vig oleae | : R.| \S RS heed ; i] —_— — a os | S JAe oe he ra | \ 7 i ce | taal ee Sara ~ LY at se i (Q) Fic. 20. left of b, acting through points b and a. Through o draw oc” and o b” parallel to closing lines a c’ and a OD’, respectively. Point ¢” de- termines the reactions R, and R,, and point b” determines the reac- tions acting through a and b of the forces to the left of point b. Points c” and b” are common to all force polygons, and lines c” o’ and b” o’ drawn parallel to the closing lines of the required equi- librium polygon, a c and a b will meet in the new pole 9’. With pole o’ the required equilibrium polygon a b ¢ can now be drawn. Center of Gravity——To find the center of gravity of the figure shown in (a) Fig. 21, proceed as follows: Divide the figure into elementary figures whose centers of gravity and areas are known. Assume that the-areas act as the forces P,, P;, P, through the centers of gravity of the respective figures. Bring the line of action of these forces into the plane of the paper by turning them downward as in (b) and to the right as in (c). Find the resultant of the forces for case (b) and for case (c) by means of force and equilibrium polygons. The intersection of the resultants R will be the center of gravity of the figure. The two sets of forces may be assumed to act at any angle, however, maximum accuracy is given when the forces are assumed to act at right angles. If the figure has an axis of symmetry but one force and equilibrium polygon is required. as - * eS 7 2 ‘ - . etl me a ies oe ee a oa Jee 2 ia ¥ ee me, oe ee ee bina al . ee ee ee eee ee et _ Ni Cook Sh Wie Dag ter ae - ee ee "™ MoMENT OF INERTIA OF FORCES 35 P. P, Ps ‘ + sangsoo Sepia, 2) a ! ry \ / oR. > “. Pay 1 ad ee. ‘> pee 'p, \ Nv ly =-j-- 2-2 R YW \ L_%G6 Negtt AGS) = \ >y | i ! 7 ‘ \| West - : \ 16 Jeter! >) Bae a “= ae \. ~+ + Pe nee gut \ aes A) 2p0 4 I 4 Ir (ad) ‘ oe b) Fic. 21. Moment of Inertia of Forces.—The determination of the moment of inertia of forces and areas by graphics is interesting. There are two methods in common use: (1) Culmann’s method, in which the moment of inertia of forces is determined by finding the moment of the moment of forces by means of force and equilibrium polygons, and (2) Mohr’s method, in which the moment of inertia of forces is determined from the area of the equilibrium polygon. The moment of inertia of a force about a parallel axis is equal to the force multiplied by the square of the distance between the force and the axis. Culmann’s Method.—lIt is required to find the moment of inertia, I, of the system of forces P,, P., P;, Py, Fig. 22, about the axis M N. Construct the force polygon (a) with a pole distance H, draw the equilibrium polygon a b c d e, and produce the strings until they intersect the axis M N. Now the moment of P, about axis M N equals ED x H; moment of P, equals D C x H; moment of P, equals C B x H; moment of P, equals B A x H; and moment of resultant R equals E AxH. Withintercepts E D, D C, CB, B A, as forces acting in place of P,, P,, P;, P,, respectively, construct force polygon (b) with pole distance H’, and draw equilibrium polygon (c). As before the moments of the forces will be equal to the products of the intercepts and pole distance and the momént of the system of forces represented by the 36 GRAPHIC STATICS Culmannis Method - I of Forces about axis M-N Fic. 22. intercepts will be equal to the intercept G F multiplied by pole distance H’. But the intercepts E D, D C, C B, B A, multiplied by the pole distance H equal moments of the forces P,, P,, P;, P,, respectively, about the axis M N, and the moment of inertia of the system of forces P,, P,, P;, Py, about the axis M N will be equal to the intercept G F multiplied by the product of the two pole distances H and H’, and l=F CxS Mohr’s Method.—lIt is required to find the moment of inertia, I, of the system of forces P,, P,, P3, P,, Fig. 23, about the axis M N. Construct the force polygon (a) with a pole distance H, and draw the equilibrium polygon (b). Now the moment of P, about the axis M N equals intercept F G multiplied by the pole distance H, and the moment of inertia of P, about the axis M N equals the moment of the moment of P, about the axis, = F Gx H xd. But F G x d equals twice the area of the triangle F G A, and we have the moment of inertia of P, equal to the area of the triangle F G A x 2 H. In like manner the moment of inertia of P, may be shown equal to area of the triangle G H B x 2H; moment of inertia of P, equal to area of the triangle H I C x 2 H; and moment of inertia of P, equal to area of the triangle ] J D x 2 H. Summing up these values we have the moment of inertia of the sys- ; oe oe om ee ee Se ee oe eee. ee ¥ Mour’s MrerHop FoR MoMEN’? oF INERTIA 37 (a) ase Mohr Method I of Forces about axis M-N =Area FABCDEUF XCH (b) PIG: 23; tem of forces equal to the area of the equilibrium polygon multiplied by twice the pole distance, H, and I=areaaFABCDEJFx2H To find the radius of gyration, 7, we use the formula | en A In Fig. 23 the moment of inertia, J,, of the resultant of the sys- tem of forces about the axis M N, can in like manner be shown to be equal to area of the triangle F E J x 2 H. If the axis M N is made to coincide with the resultant R the mo- ment of inertia J, of the system will be equal to the area of equi- librium polygon AB CD Ex2H. This furnishes a graphic proof for the proposition that the moment of inertia, J, of any system of parallel forces about an axis parallel to the resultant of the system is equal to the moment of inertia, J, at of the forces about an axis through their _centeroid plus the moment of inertia, J, of their resultant about the given axis. i= Ic. g. + kr? = Ie. g. ab I, It will be seen from the foregoing discussion that the moment of inertia of a system of forces about an axis through the centeroid of the system is a minimum. 38 GRAPHIC STATICS Moment of Inertia of Areas——The moment of inertia of an area about an axis in the same plane is equal to the summation of the products of the differential areas which compose the area and the squares of the distances of the differential areas from the axis. The ‘moment of inertia of an area about a neutral axis (axis ‘through center of gravity of the area) is less than that about any parallel axis, and is the moment of inertia used in the fundamental formula for flexure in beams x= 3 where ; M = bending moment at point in inch-pounds; S = extreme fibre stress in pounds; I = moment of inertia of section in inches to the fourth power; ¢ = distance from neutral axis to extreme fibre in inches. An approximate value of the moment of inertia of an area may be obtained by either of the preceding methods by dividing the area into laminae and assuming each area to be a force acting through the center of gravity of the lamina, the smaller the laminae the greater the accuracy. The true value may be obtained by either of the above methods if each one of the forces is assumed to act at a distance from the given axis equal to the radius of gyration of the area with reference to the axis, d = /a*+ r*, where a is the distance from the given axis to the center of gravity of the lamina and r is the radius of gyration of the lamina about an axis through its center of gravity. If A, is the area of each lamina the moment of inertia of the lamina will be [=4,? =A, 7°?+4+4,7°=4,0°+T1,, which is the fundamental equation for transferring eennenta of inertia to parallel axes. . ee ee a a !, CHAPTER VI. STRESSES IN FRAMED STRUCTURES. Methods of Calculation—The determination of the reactions of simple framed structures usually requires the use of the three funda- mental equations of equilibrium > horizontal components of forces = 0 (a) > vertical components of forces = 0 (b) > moments of forces about any point = 0 (c) Having completely determined the external forces, the internal stresses may be obtained by either equations (a) and (b) (resolution), or equation (c) (moments). These equations may be solved by graphics or by algebra. There are, therefore, four methods of calcu- lating stresses : Algebraic Method Graphic Method Algebraic Method Graphic Method The stresses in any simple framed structure can be calculated by Resolution of Forces Moments of Forces using any one of the four methods. However, all the methods are not equally well suited to all problems, and there is in general one method that is best suited to each particular problem. The common practice of dividing methods of calculation of stresses into analytic and graphic methods is meaningless and mis- leading for the reason that both algebraic and graphic methods are analytical, 7. e. capable of analysis. The loads on trusses are usually considered as concentrated at the jeints in the plane of the loaded chord. 40 STRESSES IN FRAMED STRUCTURES Algebraic Resolution.—In calculating the stresses in a truss by algebraic resolution, the fundamental equations for equilibrium for translation > horizontal components of forces = 0 (a) > vertical components of forces =0 (b) are applied (a) to each joint, or (b) to the members and forces on one side of a section cut through the truss. : (a) Forces at a Joint—rThe reactions having been found, the stresses in the members of the truss shown in Fig. 24 are calculated as ne yes ——- Re my! nb ) (b) (C Fic. 24. follows: Beginning at the left reaction, R,, we have by applying equa- tions (a) and (b) l-x sin 0—1l1-y sin x =0 (9) 1-x cos 8— 1-y cos x« — Rk, = 0 (10) The stresses in members I-x and 1-y may be obtained by solving — equations (9g) and (10). ‘The direction of the forces which rep- resent the stresses in amount will be determined by the signs of the results, plus signs indicating compression and minus signs indicating tension. Arrows pointing toward the joint indicate that the member is in compression; arrows pointing away from the joint indicate that the member is in tension. The stresses in the members of the truss at the remaining joints in the truss are calculated in the same way. | The direction of the forces and the kind of stress can always be determined by sketching in the force polygon for the forces meeting at the joint as in (c) Fig. 24. ALGEBRAIC RESOLUTION 41 It will be seen from the foregoing that the method of algebraic resolution consists in applying the principle of the force polygon to the external forces and internal stresses at each joint. Since we have only two fundamental equations for translation (resolution) we can not solve a joint if there are more than two forces "or stresses unknown. Where the lower chord of the truss is horizontal as in Fig. 25, we x io 0 o | 2 Ri” eae x (a) (b) (C) Hic, 26: have by applying fundamental equations (a) and (b) to the joint at the left reaction 1l-~=-+ R, sec 0 (11) 1-y =— Ff, tano (12) the plus sign indicating compression and the minus sign tension. Equa- tions (11) and (12) may be obtained directly from force triangle (c). Equations (11) and ( 12) are used in calculating the stresses in trusses with parallel chords and lead to the method of coefficients (Chapter X). (b) Forces on One Side of a Section—The principle of resolu- tion of forces may be applied to the structure as a whole or to a por- tion of the structure. If the truss shown in Fig. 26 is cut by the plane 4 4, the internal stresses and external forces acting on either segment, as in (b) will be in equilibrium. The external forces acting on the cut members as shown in (b) are equal to the internal stresses in the cut members and are opposite in direction. 42 STRESSES IN FRAMED STRUCTURES Applying. equations (a) and (b) to the cut section 3-y +2-3 cos x —2-rsinO@ =0 (13) 2-3 sin x«— 2-xcos0-+ R, —P,=0 (14) Now, if all but two of the external forces are known, the un- knowns may be found by solving equations (13) and (14). If more Fic. 26. than two external forces are unknown the problem is indeterminate as far as equations (13) and (14) are concerned. Graphic Resolution.—In Fig. 27 the reactions R, and R, are found by means of the force and equilibrium polygons as shown in (b) and (a). ‘The principle of the force polygon is then applied to each joint of the structure in turn. Beginning at the joint L, the forces are shown in (c), and the force triangle in (d). The reaction R, is known and acts up, the upper chord stress 1-~ acts downward to the left, and the lower chord stress 1-y acts to the right closing the polygon. Stress I-r is compression and stress I-y is tension, as can be seen by applying the arrows to the members in (c). The force polygon at joint U, is then constructed as in (f). Stress I-47 — acting toward joint U, and load P, acting downward are known, and stresses 1-2 and 2-x are found by completing the polygon. Stresses 2-x and 1-2 are compression. The force polygons at joints L, and U, are constructed, in the order given, in the same manner. The known forces at any joint are indicated in direction in the force poly- GRAPHIC RESOLUTION 43 ee | Scale of Lengths ae ins 0 { O' 2 o' 50° J Lae Joint Us Joint Lo © 4000 8000 12000 ie 3 y Scale of Sirecens ey at {g Joint L 2 P, R X2 4 ' 1 Y- 2 oe i 3 2 P, Fal \ ¥ 2 X2 ) i (j) (k) Joint Uz Stress Diagram Fic. 27. gon by double arrows, and the unknown forces are indicated in direc- tion by single arrows. The stresses in the members of the right segment of the truss are the same as in the left, and the force polygons are, therefore, not con- structed for the right segment. The force polygons for all the joints of the truss are grouped into the stress diagram shown in (k). Com- pression in the stress diagram and truss is indicated by arrows acting 44 STRESSES IN FRAMED STRUCTURES toward the ends of the stress lines and toward the joints, respectively, and tension is indicated by arrows acting away from the ends of the stress lines and away from the joints, respectively. The first time a stress is used a single arrow, and the second time the stress is used a double arrow is used’ to indicate direction. ‘The stress diagram in (k) Fig. 27 is called a Maxwell diagram or a reciprocal polygon diagram. The notation used is known as Bow’s notation. The method of graphic resolution is the method most commonly used for calcutating stresses in roof trusses and simple framed structures with inclined chords. Algebraic Moments.—The reactions may be found by applying the fundamental equations of equilibrium to the structure as a whole. In the truss in (a) Fig. 28 by taking moments about the right reaction we have 3 KK; x60 = 5 Pxae R, = Pian, ALGEBRAIC MOMENTS 45 To find the stresses in the members of the truss in (a) Fig. 28, proceed as follows: Cut the truss by means of plane A A, as in (b), and replace the stresses in the members cut away with external forces. These forces are equal to the stresses in the members in amount, but opposite in direction, and produce equilibrium. To obtain stress 4-x take center of moments at L,, and take moments of external forces 4-x~xa+P,xd—R,x2d=0 _R,XW—Pid_ 4Pad ee a iol aeg 4-x (compression) To obtain stress in 4-5 take center of moments at L,, and take moments of external forces 4-5X 6—2P,X id=0 3 Pid b 45— (tension) To obtain the stress in 5-y take center of moments at joint U, in (c), and take moments of external forces Syxh—R,x3d¢d+3P,d=0 3R,d—3P,d_ 9 Pid h en ee To Determine Kind of Stress—If the unknown external force is 5-y = (tension) always taken as acting from the outside toward the cut section, 1. e. is always assumed to cause compression, the sign of the result will in- dicate the kind of stress. A plus sign will indicate that the assumed direction was correct and that the stress is compression, while a minus sign will indicate that the assumed direction was incorrect and that the stress is tension. In calculating stresses by algebraic moments, therefore, always observe the following rule :— Assume the unknown external force as acting from the outside toward the cut section; a plus sign for the result wili then show that the stress in the member is compression, and a minus sign will indi- cate that the stress in the member is tension. The stresses in the web members 3-4, 2-3, 1-2, are found by 46 STRESSES IN FRAMED STRUCTURES taking moments about joint L, as a center. The stresses in y-3 and y-I are found by taking moments about joints U, and U,, respectively; and the stresses in x-2 and #-1 are found by taking moments about joint L,. The method of algebraic moments is the most common method used for calculating the stresses in bridge trusses with iaclined chords and similar frameworks which carry moving loads, Graphic Moments.—The bending moment at any point in a truss may be found by means of a force and equilibrium polygon as ~ € { 1 i ~ RiP pi Pete ee: I ~ ~ ~ ~ a ee ee 10) i ee ae Lap 1 -—-—— oye ! ele ' \ - Metgien te Pes ! | RP ---7--H----> ee ai a ) . Fic. 20. in (b) and (a) Fig. 29. ‘To determine the stress in 4-x, cut section A A and take moments about joint L, as in Fig. 28. The moment of | the external forces on the left of L, will be M, = — H ¥,, and stress M, Ay, Oe a geen To obtain stress in 4-5 take center of moments at joint L,, and stress 3 M, Hy, baton? haere To obtain stress in 5-y take center of moments at joint U,, and stress M; ys ies acme The method of graphic moments is principally used to explain other methods and is little used as a direct method of calculation. ech ee sie ctr ~— eo tage ee ee dl) ns en 4 4 ; : CHAPTER VII. STRESSES IN SIMPLE ROOF TRUSSES. Loads.—The stresses in roof trusses are due (1) to the dead load, (2) the snow load, (3) the wind load, and (4) concentrated and moving loads. ‘The stresses due to dead, snow, wind and concentrated loads will be discussed in this chapter in the order given. Dead Load Stresses.—The dead load is made up of the weight of the truss and roof covering and is usually considered as applied at the panel points of the upper chord in computing stresses in roof trusses. If the purlins do not come at the panel points, the upper chord will have to be designed for both direct stress and stress due to flexure. The stresses in a Fink truss due to dead load are calculated by graphic resolution in Fig. 30. | 6~ ee ha 1) 2000 4000 6000 Y Pp t | i] } Lo EE, Fic. 30. 48 STRESSES IN Roor TRUSSES The loads are laid off, the reactions found, and the stresses calcu- lated beginning at joint L,, as explained in Fig. 27. The stress diagram for the right half of the truss need not be drawn where the truss and loads are symmetrical as in Fig. 30; however it gives a check on the accuracy of the work and is well worth the extra time required. The loads P, on the abutments have no effect on the stresses in the truss and may be omitted in this solution. In calculating the stresses at joint P;, the stresses in the members 3-4, 4-5 and #+-5 are unknown, and the solution appears to be in- determinate. The solution is easily made by cutting out members 4-5 and 5-6, and replacing them with the dotted member shown. ‘The stresses in the members in the modified truss are now obtained up to and including stresses 6-~ and 6-7. Since the stresses 6-x and 6-7 are independent of the form of the framework to the left, as can easily be seen by cutting a section through the members 6-4, 6-7 and 7-y, the solution can be carried back and the apparent ambiguity removed. The ambiguity can also be removed by calculating the stress in 7-y by algebraic moments and substituting it in the stress diagram. It will be noted that all top chord members are in compression and all bottom chord members are in tension. The dead load stresses can also be calculated by any of the three remaining methods, as previously described. Dead and Ceiling Load Stresses.—The stresses in a triangular Pratt truss due to dead and ceiling loads, are calculated by graphic resolution in Fig. 31. : For simplicity the stresses are shown for one side only. The re- action R, is equal to one-half of the entire load on the truss. The solu- tion will appear more clear when it is noted that the stress diagram shown consists of two diagrams, one due to loads on the upper chord and the other due to loads on the lower chord, combined in one, the © loads in each case coming between the stresses in the members on each side of the load. "The top chord loads are laid off in order downward, while the bottom chord loads are laid off in order upward. a ee ee a. ee oo a ee Snow Loap STRESSES 49 DEAD AND CEILING LOADS me) > tone w ew D------- ae oO OF PR < ' c 2000 4000 6000 l l Fic. 31. Snow Load Stresses.—Large snow storms nearly always occur in still weather, and the maximum snow load will therefore be a uni- formly distributed load. A heavy wind may follow a sleet storm and a snow load equal to the minimum given in Fig. 4 should be considered as acting at the same time as the wind load. The stresses due to snow load are found in the same manner as the dead load stresses. Wind Load Stresses.—The stresses in trusses due to wind load will depend upon the direction and intensity of the wind, and the con- dition of the end supports. The wind is commonly considered as act- ing horizontally, and the normal component, as determined by one of the formulas in Fig. 6, is taken. The ends of the truss may (1) be rigidly fixed to the abutment walls, (2) be equally free to move, or (3) may have one end fixed and the other end on rollers. When both ends of the truss are rigidly fixed to the abutment walls (1) the reactions are parallel to each other 50 STRESSES IN Roor Trusses and to the resultant of the external loads; where both ends of the truss are equally free to move (2) the horizontal components of the reactions are equal; and where one end is fixed and the other end is on frictionless rollers (3) the reaction at the roller end will always be vertical. Either case (1) or case (3) is commonly assumed in cal- culating wind load stresses in trusses. Case (2) is the condition in a portal or framed bent. The vertical components of the reactions are independent of the condition of the ends. Wind Load Stresses: No Rollers.——The stresses due to a nor- mal wind load, in a Fink truss with both ends fixed to rigid walls, are calculated by graphic resolution in Fig. 32. The reactions are parallel ee ae et os Ps L L: | Se “ “ Fic. 32. and their sum equals sum of the external loads; they are found by means of force and equilibrium polygons as in Fig. 15 and Fig. 27, —_ ee es WInp Loap STRESSES 51 The stress diagram is constructed in the same manner as that for dead ' loads. Heavy lines in truss and stress diagram indicate compression, and light lines indicate tension. The ambiguity at joint P, is removed by means of the dotted mem- ber as in the case of the dead load stress diagram. It will be seen that there are no stresses in the dotted web members:in the right segment of the truss. It is necessary to carry the solution entirely through the truss, beginning at the left reaction and checking up at the right re- action. It will be seen that the load P, has no effect on the stresses in the truss in this case. Wind Load Stresses: Rollers—Trusses longer than 70 feet are usually fixed at one end, and are supported on rollers at the other end. The reaction at the roller end is then vertical—the horizontal com- ponent of the external wind force being all taken by the fixed end. The wind may come on either side of the truss giving rise to two conditions ; (1) rollers leeward and (2) rollers windward, each requiring a separate solution. Rollers Leeward -—The wind load stresses in a triangular Pratt truss with rollers ‘under the leeward side are calculated by graphic resolution in Fig. 33. The reactions in Fig. 33 were first determined by means of force and equilibrium polygons, on the assumption that they were parallel to each other and to the resultant of the external loads. Then since the reaction at the roller end is vertical and the horizontal component at the fixed end is equal to the horizontal component of the external wind forces, the true reactions were obtained by closing the force polygon. In order that the truss be in equilibrium under the action of the three external forces R,, R, and the resultant of the wind loads, the three external forces must meet in a point if produced. This fur- nishes a method for determining the reactions, where the direction and line of action of one and a point in the line of action of the other are known, providing the point of intersection of the three forces comes within the limits of the drawing board. 52 STRESSES IN Roor TRussEs Wind Load Rollers Leeward Fic. 33. The stress diagram is constructed in the same way as the stress diagram for dead loads. It will be seen that the load P, has no effect on the stresses in the truss in this case. Heavy lines in truss and stress diagram indicate compression and light lines indicate tension. Rollers Windward.—The wind load stresses in the same trian- gular Pratt truss as shown in Fig. 33, with rollers under the windward side of the truss are calculated by graphic resolution in Fig. 34. The true reactions were determined directly by means of force and equilibrium. polygons as in Fig. 16. The direction of the reaction R, is known to be vertical, but the direction of the reaction R, is unknown, the only known point in its line of action being the right abutment. The equilibrium polygon is drawn to pass through the right abutment and the direction of the right reaction is determined by connecting the ——— — heal Tw ae ee CONCENTRATED LoaD STRESSES | 53 | _ Wind Load ~*~ Rollers Windwarg oe point of intersection of the vertical reaction R, and the line drawn through o parallel to the closing line of the equilibrium polygon, with the lower end of the load line. _ Since the vertical components of the reactions are independent of the conditions of the ends of the truss, the vertical components of the reactions in Fig. 33 and Fig. 34 are the same. It will be seen that the load P, produces stress in the members of the truss with rollers wind- ward. If the line of action of R, drops below the joint P, the lower chord of the truss will be in compression, as will be seen by taking moments about P;. Concentrated Load Stresses.—The stresses in a Fink truss due to unequal crane loads are calculated by graphic resolution in Fig. 35. The reactions were found by means of force and equilibrium poly- gons. The truss is reduced to three triangles for the loading shown. The solution of this problem is similar to that for ceiling loads in Fig. 54 StRESSES IN Roor TRUSSES 31. The moving crane trolley will produce maximum moment when it is at the center of the truss, and this case should be investigated in solving the problem. 0 2000 4000 6000 l r 1 i Fic. 35. The method of graphic resolution is commonly used. for calculat- ing the stresses in roof trusses and similar structures. For examples of the calculations of stresses in trusses by algebraic resolution, al- gebraic and graphic moments, see Chapter X. Ea ees ’ 7, ‘ a a ee ee ee a ae, en Te CHAPTER VIII. SIMPLE BEAMS. Reactions.—The reactions of beams may be found by the use of the force and equilibrium polygon as shown in Chapter V. As a sec- ond example let it be required to find the reactions of the overhanging beam shown in Fig. 36. > eee isn rRa 1 we . mt ek a eS 2 1 P, 5 es x 1 a eee f Pa ore pita. Ra ! Pe ee 1 : i: Ri Pore real | aokS ‘ ; Me - — - --=-H---m 1 te vw y- po pa pepe ey es \7 Fic. 36. Construct a force polygon with pole 0, as in (b), and draw an equi- librium polygon, as in (a). The ray o d drawn parallel to the closing line o d in (a) determines the reactions. In this case reaction R, is negative. It should be noted that the closing line in an equilibrium polygon must have its ends on the two reactions. The ordinate to the equilibrium polygon at any point multiplied by the pole distance, H, will give the bending moment in the beam at a point immediately above it. 56 SIMPLE BEAMS Moment and Shear in Beams: Concentrated Loads.—The bend- 7 ing moment in the beam shown in Fig. 37 may be found by constructing the force polygon (a) and equilibrium polygon (b) as shown. é yA an ! XN t : ; ; V b ig Ra! me oe s Bs a II e Ri PS oa ! ; ! fs a atrb-a ae fo toffee neg 2 ||| Mb) tc) Melee 2 ae . ' ! ' ge" a ' 4: 1 Moment Diagram Reh k---20-H - 41 Ri oot : . 41) ee ; -it-y ' : S (a) i} ! 5 Force Polygon A ' x AG) a Shear Diagram 4°] rRRe nd x ¥ Fic. 37. The bending moment at any point is then equal to the ordinate to the equilibrium polygon at that point multiplied by the pole distance, H. The ordinate is to be measured to the same scale as the beam, and the pole distance, H, is to be measured to the same scale as the loads in the force polygon. ‘The ordinate is a distance and the pole distance is a force. , Or, if the scale to which the beam is laid off be multiplied by the pole distance measured to the scale of the loads, and this scale be used in measuring the ordinates, the ordinates will be equal to the bending moments at the corresponding points. This is the same as making the pole distance equal to unity. Diagram (b) is called a nioment diagram. Between the left support and the first load the shear is equal to ne ‘ ey Oe a, ee a a Se a r* MoMENT AND SHEAR IN BEAMS 57 R,; between the loads P, and P, the shear equals R, — P,; between the loads P, and P, the shear equals R, — P, — P.,; between the loads P, and P, the shear equals R, — P, — P, — P,; and between load P, and the right reaction the shear equals R, — P,— P,—P,—P,= — R,. At load P, the shear changes from positive to negative. Diagram (c) is called a shear diagram. It will be seen that the maximum ordinate in the moment diagram comes at the point of zero shear. The bending moment at any point in the beam is equal to the algebraic sum of the shear areas on either side of the point in question. From this we see that the shear areas on each side of P, must be equal. This property of the shear diagram depends upon the principle that the bending moment at any point in a simple beam is the definite integral of the shear between either point of support and the point in question. This will be taken up again in the discussion of beams uniformly loaded which will now be considered. Moment and Shear in Beams: Uniform Loads.—In the beam loaded with a uniform load of w lbs. per lineal foot shown in Fig. 38, the reaction R, = R, = %wL. Ata distance x from the left support, the bending moment is M= R,x— Oe =F (Le 2?) which is the equation of a parabola. The parabola may be constructed by means of the force and equi- librium polygons by assuming that the uniform load is concentrated at points in the beam, as is assumed in a bridge truss, and drawing the force and equilibrium polygons in the usual way, as in Fig. 38. The greater the number of segments into which the uniform load is divided the more nearly will the equilibrium polygon approach the bending moment parabola. e. The parabola may be constructed without drawing the force and equilibrium polygons as follows: Lay off ordinate mn = n p = bend- ing moment at center of beam = % w L?. Divide a p and b p into the same number of equal parts and number them as shown in (b). Join the points with like numbers by lines, which will be tangents to the 58 SIMPLE BEAMS Load = w Ibs. per lin: ff. aa ie eee GRE cee = ARCOAEEE URC OERE DEA Boe ee Po { - - Ri i RS 4 ‘ H 1 : ' 1 ~~. on, ae a} | Dea a a ie cock cme RT BpRe FE ====S TRO : te! 2 ian. Utd t Iibeeg 4 | Re laste Pers! : BS Cr Rae Ot ie ; H “er - 1 ad Pd pi 2 Bee ae ze : a tMoment seis Sol | Force Polygon i i ! ic ' ¥ » i (c) re Shear Diagram ‘ Fic. 38. required parabola. It will be seen in Fig. 38 that points on the parabola are also obtained. The shear at any point -r, will be S= et Wr = w (5— x) which is the equation of the inclined line shown in (c) Fig. 38. The shear at any point is therefore represented by the ordinate to the shear diagram at the given point. Property of the Shear Diagram.—Integrating the equation for shear between the limits, x = oand + = # we have Ss=S 0 (F-*) = 5 (La— x?) which is the equation for the bending moment at any point, +, in the beam, and is also the area of the shear diagram between the limits given. From this we see that the bending moment at any point in a simple beam uniformly loaded is equal to the area of the shear dia- gram to the left of the point in question. The bending moment is also equal to the algebraic sum of the shear areas on either side of the point. ; 4 : j . a Ne : CHAPTER IX. Movinc Loaps on BEAMs. Uniform Moving Loads.—Let the beam in Fig. 39 be loaded with a uniform load of p lbs. per lineal foot, which can be moved on or off the beam. > octal er ay Pee pasan get » HAT RiR----x ~---> Rea ai) Uniforrn Moving Load 3 t ne . { R, =p Ibs. per lin. ft. ‘ ‘ iJ | “i (a) Maximum Positive Shear op : Load moving off to the right- A bee 8 a Re Moximum Negative Shear ». J. Load moving off tothe left. 3 Fic. 39. To find the position of the moving load that will produce a max- imum moment at a point a distance a from the left support, proceed as follows: Let the end of the uniform load be at a distance + from the left reaction. Then taking moments about R, we have R, eae) ? (15) and the moment at the point whose abscissa is a will be “os lag ie ee ‘ -F ro S's a% 60 Movine Loaps oN BEAMS Differentiating (16) and placing derivative of M with respect to x equal to zero, we have after solving x= 0 (17) Therefore the maximum moment at any point in a beam will occur when the beam is fully loaded. The bending moment diagram for a beam loaded with a uniform moving load is constructed as in Fig. 38. To find the position of the moving load for maximum shear at any point in a beam loaded with a moving uniform load, proceed as fol- lows: The left reaction when the end of the moving load is at a dis- tance + from the left reaction, will be R, = en (15) and the shear at a point at a distance a from the left reaction will be- _ (bay be S=R, —(a—x)p p—(a--x)p (18) which is the equation of a parabola. By inspection it can be seen that S will be a maximum when a = x. The maximum shear at any point in a beam will therefore occur at the end of the uniform moving load, the beam being fully loaded to the right of the point as in (a) Fig. 39 for maximum positive , shear, and fully loaded to the left of the point as in (b) Fig. 39 for maximum negative shear. If the beam is assumed to be a cantilever beam fixed at 4, and loaded with a stationary uniform load equal to p Ibs. per lineal foot, and an equilibrium polygon be drawn with a force polygon having a pole distance equal to length of span, L, the parabola drawn through the points in the equilibrium polygon will be the maximum positive shear diagram, (a) Fig. 39. The ordinate at any point to this shear diagram will represent the maximum positive shear at the point to the same > scale as the loads (for the application of this principal to bridge trusses see Fig. 50, Chapter X). ee nn. ae ae ene ees ee CONCENTRATED Movinc Loaps 61 Concentrated Moving Loads.—Let a beam be loaded with con- centrated moving loads at fixed distances apart as shown in Fig. 4o, To find the position of the loads for maximum moment and the amount of the maximum moment, proceed as follows: The load P, will be considered first. Let x be the distance of the load P, from the left support when the loads produce a maximum moment un- der load P,. _ Taking moments about R, we have R P, (L—2x+a)+ P,(L—2)+P,(L— «—4)4+ P,(L—x—b—0) Coca Tc _(L—4) (A, +P. +P3+P,)+P, a—P3b6—P, (6+) (19) or a and the bending moment under load P, will be M=R,x—P,4a _ x L—z) (Pi + P2+P3tP,)+%(P, a—P; 6—P,(6+¢) ) ou pie —P,a (20) Differentiating (20) we have d M_(L~2x) (P,+Pit+ Pst Pi) +Pra—Ps b—Pi(b+<) _ 9 a1) ax L and solving (21) for x we have _L Pia—P,b—P,(b+0) (22) emt. OCP, +P, +2,4+P,) Now P, a— P,; b — P, (b + c), is the static moment of the loads about P, and 62 Movine Loaps oN BEAMS P, a— P,b6—P,(6+<) Put Ps + hare center of the gravity of all the loads. Therefore, for a maximum moment under load P,., it must be as far from one end 4s the center of gravity of all the loads is from the other end of the beam, Fig. 4o. The above criterion holds for all the loads on the beam. ‘The only way to find which load produces the greatest maximum is to try each = distance from P, to one, however, it is usually possible to determine by inspection which load will produce a maximum bending moment. For example the maximum moment in the beam in Fig. 40 will certainly come under the heavy load P,. The above proof may be generalized without diffi- culty and the criterion above shown to be of general application. For two equal loads P = P at a fixed distance, a, apart as in the case of a traveling crane, Fig. 41, the maximum mament will occur under one of the loads when : Taking moments about the right reaction we have a Pe ( a +) 23) and the maximurn bending moment is aR (SS) (24) os CONCENTRATED Movine Loaps 63 There will be a maximum moment when either of the loads satis- fies the above criterion, the bending moments being equal. By equating the maximum moment above to the moment due to a single load at the center of the beam, it will be found that the above criterion holds only when a< 0.586 L Where two unequal moving loads are at a fixed distance apart the greater maximum bending moment will always come under the heavier oad. | The maximum end shear at the left support for a system of con- centrated loads on a simple beam, as in Fig. 40, will occur when the left reaction, R,,isa maximum. This will occur when one of the wheels is infinitely near the left abutment (usually said to be over the left abutment). The load which produces maximum end shear can be easily found by trial. The maximum shear at any point in the beam will occur when one of the loads is over the point. ‘The criterion for determining which load will cause a maximum shear at any point, x, in a beam will now be determined. In Fig. 40, let the total load Gn the beam, P, + P, +P; +P, = W, and let x be the distance from the left support to the point at which we wish to determine the maximum shear. | When load P, is at the point, the shear will be equal to the left reaction, which is found by substituting + + a for x in (19) to be (L—x—a) W+ P,a— P,b6—P, (6+¢) “ L and when P, is at the point the shear will be een Bees be G+) p> S, =k, Subtracting S, from S, we have P,L— Wa S,— SS, = , te Now 5S, will be greater than S, if P, L is greater than W a, or if PZ got The criterion for maximum shear at any paitit therefore follows: : The maximum positive shear in any section of a beam occurs y “. pk epee ee, ee ee ere ae F ¢ < Per Tah ns ae - YY ey eee ST : at IP ae RE MAT EI Re ee ee Pee eee oe CHAPTER X. STRESSES IN BRIDGE ‘T'RUSSES. Method of Loading.—The loads on highway bridges, and in many cases on railway bridges as well, are assumed to be concentrated at the joints of the loaded chord, and if the panels of the truss are equal the joint loads are equal. The assumption of joint loads simplifies the solution and gives values for the stresses that are on the safe side. Equal joint loads will be assumed in this discussion. Algebraic Resolution.*—Let the Warren truss in Fig. 42 have dead loads applied at the joints as shown. From the fundamental equations for equilibrium for translation, reaction R, = R, = 3 W. Wtand : +6 +10 +12 +l2 +10 +6 - @M az O -1 @M - @ -s Ere Wtan@ Dead Load Coefficients Fic. 42. The stresses in the members are calculated as follows: Resolving at the left reaction, stress in 1-r = + 3 W sec @, and stress in I-y = — 3 W tan @. Resolving at first joint in upper chord, stress in 1-2 = — 3 W sec 9, and stress in 2-+ = + 6W tan@. Resolving at second joint in lower chord, stress 2-3 = + 2 W sec@, and stress 3-y = — 8 W tan@. And in like manner the stresses in the remaining members are found as shown. ‘The coefficients shown in Fig. 42 for the chords are to be multiplied by W tan 0; while those for the webs are to be multiplied by W sec 8. ~ *Also called ‘‘Method of Sections.” 66 STRESSES IN BrIDGE TRUSSES It will be seen that the coefficients for the web stresses are equal to the shear in the respective panels. Having found the shears in the different panels of the truss, the remaining coefficients may be found by resolution. Pass a section through any panel and the algebraic sum of the coefficients will be equal to zero. ‘Therefore, if two coefficients are known, the third may be found by addition. | Beginning with member 1-7, which is known and equals —3; coefficient of 2-7 =— (— 3—3)=+ 6; coefficient of 3-y = — (+ 6+2)=— 8; coefficient of 4-7 = — (— 8— 2) =+ 10; coefficient of 5-y = — (+ 10+ 1) =—II; coefficient of 6-4 = — (— 11 — 1) = + 12; coefficient of 7-y = — (+ 12+ 0) =—12. Taine for Maximum Stresses —The effect of different positions of the loads on a Warren truss will now be investigated. Let the truss in Fig. 43 be loaded with a single load P as shown. Chord Stresses = Coefficients x Ptan9 eed Se) eos ee Th -F -F RaeyP ve GP Coefficients for One Load. Fic. 43. 6 ; es a The left reaction, R, = -7-P, and the right reaction, R, = =, The 6 6 stress in I-y = — > P tan 0, and stress in I-v =-+— Psec® The . 6 ness 1 stress in 1-2 = — —7 P sec @ and stress in 2-3 = — = P sec#, etc. The remaining coefficients are found as in the case of dead loads by adding coefficients algebraically and changing the sign of the result. In Fig. 44 the coefficients for a load applied at each joint in turn are shown for the different members; the coefficients for the load on left being given in the top line. ae ee ee ee re a a EA aR ToT Sr mB PORE ae eS IF OE ea ae a, ae ee a ee ee MaxIMUM AND MINIMUM STRESSES 67 Pran€ a +10 +8 +*6 +4 +2 +10 +42 +e 14 a +*4 g¢ i$ + (8 +24 sis 38 Se ee eee ee em +B + S& + BF +2 +p 3 PsecO 2, ae “A 6 27, 2! +H "eB 7 Ned 7G ~ 6 ss a) Te = -3 ~ pi Of On On OR OF O23 Ty R. -4 ~- -20 -2! 15 -9 -3 3 R es -3 -9 -15 -2i -20 Pe = 4 4 * 2 = -2 -¢6 - 10 -\l4 -\18 -15 - 5 2 a ae BS ee -5_ -7 - 9. -N = x2 ake _>.6 77 _B4 pi : 5 2. 21 « 3 Tr ie may 7 ag Mle lay Ptan® Maximum and Minimum Coefficients Fic. 44. The following conclusions may be drawn from Fig. <¢y. (1) All loads produce a compressive stress in the top chord and a tensile stress in the bottom chord. (2) All the loads on one side of a panel produce the same kind of stress in the web members that are inclined in the same direction on that side. For maximum stresses in the chords, therefore, the truss should be fully loaded. ‘For maximum stresses in the web members the longer segment into which the panel divides the truss should be fully loaded ; while for minimum stresses in the web members the shorter segment of the truss should be fully loaded. _ The conditions for maximum loading of a truss with equal joint loads are therefore seen to be essentially the same as the maximum load- ing of a beam with a uniform live load. | Stresses in Warren Truss—The coefficients for maximum and minimum stresses in aWarren truss due to live load are shown in Fig. 45. These coefficients are seen to be the algebraic sum of the co- efficients for the individual loads given in Fig. 44. The live load chord coefficients are the same as for dead load, and if found directly are zound in the same manner. The maximum web coefficients may be found directly by taking off one load at a time beginning at the left. The left reaction, which may be found by algebraic moments, will in each case be the coefficient of 68 STRESSES IN BRIDGE TRUSSES the maximum stress in the panel to the left of the first load. A rule for finding the coefficient of left reaction for any loading is as follows: Multiply the number of loads on the truss by the number of loads plus unity, and divide the product by twice the number of panels in the truss and the result will be the coefficient of the left reaction. nl 3 © - © - © 2 © 1 © -3 © -z Re Maximum in Webs Ptan® Minimum in webs Live Load Coefficients Fic. 45. If the second differences of the maximum coefficients in the web members are calculated, they will be found to be constant, which shows. that the coefficients are equal to the ordinates of a parabola. 2 nN Om > W a 21 SECOND DIFFERENCES OF NUMERATORS OF WEB COEFFICIENTS. This relation gives an easy method for checking up the maximum web coefficients, since the numerators of the coefficients are always the same beginning with unity in the first panel on the right and progressing in order I, 3, 6, 10, etc.; the denominators always being the number of panels in the truss. It should be noted that in the Warren truss the members meeting on the unloaded chord always have stresses equal in amount, but op- posite in sign. Stresses in Pratt Truss— In the Pratt truss the diagonal members are tension members and counters (see dotted members in (c) Fig. 46) must be supplied where ihere is a reversal of stress. The coefficients for the dead and iive load stresses in the Pratt truss shown in (a) and (b) a A oo aera 8 ee wee tS ee a ee ee ee ee ee eer mE we STRESSES IN Pratr Truss 69 The member U, L, acts as a hanger and carries only the load at its lower Fig. 46, are found in the same manner as for a Warren truss. end. The stresses in the chords are found by multiplying the coeffi- cients by W tan @, and in the inclined webs by multiplying the co- efficients by JV sec @, The stresses in the posts are equal to the ver- tical components of the stresses in the inclined web members meeting them on the unloaded chord. Ui +4 U2 +45 Us +288 U2 +256 Ui __ | ir a * Ko) of. & a A > + t ’ t | Loy Le -4 Ls -256 Ly 16.0 ay =16.0 Hi k~-20-0"- > Dead Load Coefficients Dead Load =8Tons per Joint. Dead Load Stresses Sec 6 =!:28- Tan 6 =0-80 (a) Us +576 Uz +512 Ur A | Lo i Bs Se ee ee ee Lz Be She “5 320-13 320 AR, Live Load Coefficients .and Stresses Live Load =I6 Tons per Joint. Sec@=!28- Tan @=0.80 (b) Ul +768 Uz +864 Us +28-8 Uz +25.6 U} s , XN 7 XN Vv an -48-0 1; “48.0 1, -768 3 -256 L} -16.0 na - 16.0 Minimum Stresses Maximum Stresses (C) Fic. 46. 7o STRESSES IN BRIDGE TRUSSES The maximum chord stresses shown on the left of (c) are equal to the sum of the live and dead load chord stresses. The minimum chord stresses shown on the right of (c) are equal to the dead load chord stresses. The maximum and minimum web stresses are found by adding algebraically the stresses in the members due to dead and live loads. Since the diagonal web members in a Pratt truss can take tension only, counters must be supplied as U, L+, in panel L', L,. The tensile stress in a counter in a panel of a Pratt truss is always equal to the compressive stress that would occur in the main diagonal web member in the panel if it were possible for it to take compression. Care must always be used to calculate the corresponding stresses in the vertical posts. , Graphic Resolution.—The stresses in a Warren truss due to dead loads are calculated by graphic resolution in Fig. 47. The solution is the same as for ceiling loads in a roof truss. The loads beginning with © the first load on the left are laid off from the bottom upwards. The analysis of the solution is shown on the stress diagram and truss and © needs no explanation. From the stresses in the members it is seen (a) that web members meeting on the unloaded chord have stresses equal in amount but op- posite in sign, and (b) that the lower chord stresses are the arithmetical means of the upper chord stresses on each side. The live load chord stresses may be obtained from the stress dia- gram in Fig. 47 by changing the scale or by ee the dead load stresses by a constant. The live load web stresses may be obtained by calculating the left reactions for the loading that gives a maximum shear in the panel (no ioads occurring between the panel and the left reaction), and then con- structing the stress diagram up to the member whose stress is required. In a truss with parallel chords it is only necessary to calculate the stress : in the first web member for any given reaction since the shear is con- stant between the left reaction and the panel in questicn. : r - as 5 — Ee a a als tae GRAPHIC RESOLUTION The live load web stresses may all be obtained from a single dia- gram as follows: With an assumed left reaction of, say, 100,000 Ibs. construct a stress diagram on the assumption that the truss is a canti- lever fixed at the right abutment and that there are no loads on the afl. ao Y ° 1G 32 t i Warren ‘Truss. Span 120-0: : P, =7 Tons. i o 6 Tons 12 Tons | I j Fic. 47. truss. Then the maximum stress in any web member will be equal to the stress scaled from the diagram, divided by 100,000, multiplied by the left reaction that produces the maximum stress. This method is a very convenient one for finding the stresses in a truss with inclined chords. 72 STRESSES IN Bripce TRUSSES Algebraic Moments.—The dead and live load stresses in a truss with inclined chords are calculated by algebraic moments in Fig. 48. The conditions for maximum loading are the same in this truss as in a truss with parallel chords, and are as follows: Maximum chord stresses occur when all loads are on; minimum chord stresses occur when no live load is on; maximum web stresses in main members occur > when the longer segment of the truss is loaded; and minimum stresses in main members and maximum stresses in counters occur when the shorter segment of the truss is loaded. An apparent exception to the latter rule occurs in post U, L, which has a maximum stress when the truss is fully loaded with dead and live loads. a it = Bi 4 *. of bs e ‘\ Max +26.40 . . ue "a teous +19. S& \., 19.12 : - z “7 ~ 4 <--> ° iy Vara Pa “aed | a ~ oe OO ¥ i} eae a “ ie Pod “o oe ¢ ee o | RO See 1 ‘ A } 1 ieee : ei a L-16.097, L—-16.00-7 1-19.20» Max~22.00,, Max —22.99 WSS R, Le Pe L, WA lL, ngs yh Ls s ; Le 2 res ig / 7? i¢-- 20.0 -- > so / J +. % a 3 MaAxiMuM AND Minimum STRESSES 3" z og IN eo bd . "One" / ‘ Came ts Back TRUSS =~ j f BY . a ¢ Re Mae ce oy AuceBraic Moments x . 7 re Dead Load 3 tons per Joint 2 7 Live Load 8 33 393 39 eS ys pa Fic. 48. To find the stress in member U, L, take moments about point 4A, the intersection of the upper and lower chords produced. The dead load stress is then given by the equation U, 1.x 70.7 + R, x60 —W x80 =0 U,L, x 70.7 = — 6x 60 + 3 x 80 = — 120 foot-tons U, L, = — 1.70 tons et _ ae eS ee ee oe ee ris ot eee, 2, Pen ee ee ALGEBRAIC MOMENTS 73 The maximum live load stress occurs when all loads are on except L,, and UL, * 70.7 -- RX 60'='0 ita 8 70.7 ae P x 60 = — 576 foot-tons U, L, = — 8.14 tons The maximum live load stress in counter U, L, occurs with a load at L,, and is given by the equation —U, L, x 62.43 + R, x60—Px80=0 OL x 62.43 == P x 60 — 8x 80 U, L, = — 4.10 tons The dead load stress in counter U, L, when main member U, L, is not acting will be U, L, x 62.43 = + 120 foot-tons U, L, = + 1.92 tons The maximum stress in U, L, is therefore — 1.70 — 8.14 = — 9.84 tons, and the minimum stress is zero. The maximum stress in counter U, L, is + 1.92 — 4.10 = — 2.18 tons, and the minimum stress is zero. ; The stresses in the remaining-members may be found in the same manner. To obtain stresses in upper chords U, U, and U, U,, take mo- ments about L, as a center; to obtain stress in lower chord L, L, take moments about U, as a center. The dead load and maximum live load stress in post U, L, is equal to the vertical component of the dead and live loads, respectively, in upper chord U, U,. The stresses in L, U,, L, L,, L, L1,, Uz, U1, and U, L, are most easily found by algebraic resolution. Graphic Moments.—The dead load stresses in the chords of a Warren truss are calculated by graphic moments in Fig. 49. Bending Moment Polygon.—The upper chord stresses are given by the ordinates to the bending moment parabola direct, while the lower chord stresses are arithmetical means of the upper chord stresses 74 STRESSES IN Bripce TRUSSES on each side, and are given by the ordinates to the chords of the parabola as shown in Fig. 4o. The parabola is constructed as follows: The mid-ordinate, 4j, is made equal to the bending moment at the center of the truss divided by the depth ; in this case the mid-ordinate is the stress in 6-1; if the num- jk --—20-0- X > ] > _- = — ae oe eee cr p— mom ae ee ee ee J +6-Y 2 WEL. 90) Bp 7 20000-% \ ber of panels in the truss were odd the mid-ordinate would not be equal to any chord stress. The parabola is then constructed as shown in Fig. 49. The live load chord stresses may be found from Fig. 49 by chang- ing the scale or by multiplying the dead load chord stresses by a con- stant. Shear Polygon.—In Chapter IX it was shown that the maximum shear in a beam at any point could be represented by the ordinate to a parabola at any point. The same principle holds for a bridge truss loaded with equal joint loads, as will now be proved. - In Fig. 50 assume that the simple Warren truss is fixed at the left end as shown, and that right reaction R, is not acting. Then with all joints fully loaded with a live load P, construct a force polygon as shown, with pole o and pole distance H = span -L, and beginning at point a in the load line of the force polygon construct: the equilibrium polygon a g h for the cantilever truss. | : Now the bending moment at the left support will be equal to , GRAPHIC MoMEN’S 75 TR © ©) © ay © ©) Ss Tr 1 ‘ Pee fem > Ret a oe ee ee ee 1 7% "x 4’ Ca . Ss Jy \ / \ 4 Y re bt “A ‘ fr EY : RS gee 8. Eee Gata eA Ck me pylons / / ry / \ < Ee ordinate Y, multiplied by the pole distance H. But the truss is a sim- ple truss and the moment of the right reaction ‘will be equal to the moment at the left abutment and LAs mee ey and since H = L | tay She Ae NB and Ey se ie Now, with the loads remaining stationary, move the truss one panel to the right as shown by the dotted truss. With the same force polygon draw a new equilibrium polygon as above. This equilibrium polygon will be identical with a part of the first equilibrium polygon as shown. _As above, the bending moment at left reaction is Y, H = Y, L = Rk, L, and Y, = R,. In like manner Y, can be shown to be the right reaction with three loads on, etc. Since the bridge is symmetrical with reference to the center line, the ordinates to the shear polygon in Fig. 50, are equal to the maximum shears in the panel to the right of the or- dinate as the load moves off the bridge to the right. 76 STRESSES IN BRIDGE TRUSSES To draw the shear parabola direct, without the use of the force and equilibrium polygons proceed as follows: At a distance of a panel — length to the left of the left abutment lay off to scale a load line equal to one-half the total load on the truss, divide this load line into as many parts as there are panels in the truss, and beginning at the top, which call 1, number the points of division of the load line 1, 2, 3, etc., as in Fig. 49. Drop vertical lines from the panel points and number them I, 2, 3, etc., beginning with the load line, which will be numbered 1, the left reaction numbered 2, etc. Now connect the numbered points in the load line with the point f, which is under the first panel to the left of the right abutment; and the intersection of like numbered lines will give points on the shear parabola. It should be noted that the line h g - is a secant to the parabola and not a tangent as might be expected. The dead load shear is laid off positive downward in Fig. 50 to tlie | same scale as the live load shears, and the maximum and minimum shears due to dead and live loads are added graphically. The stresses in the web members are calculated graphically in Fig. 50. Wheel Loads.—The criteria for maximum moments and shears in bridge trusses loaded with wheel loads are as follows: (1) Maximum Moment at any joint in a bridge loaded with wheel loads will occur when the average load on the left of the section is the same as the average load on the whole span. (2) Maximum Shear in any panel in a bridge loaded with wheel loads will occur when the load on the panel is equal to the load on the bridge divided by the number of panels. These criteria will be proved by means of the influence diagram in the following discussion. For a more complete discussion of the danes see standard books on bridge design. For the calculation of stresses in simple trusses see problems in Appendix II. ae INFLUENCE DIAGRAMS 77 INFLUENCE DIAGRAMS.—An influence diagram (commonly called an influence line) shows the variation of the effect of a moving load or a system of loads on a beam or truss. The differen¢e between bending moment or shear diagrams and influence diagrams is that the bending moment and the shear diagram gives the moment and shear, respectively, at any point for a fixed system of loads, while an in- fluence diagram gives the moment or shear, etc., at a fixed point for a moving system of loads. Influence diagrams are used principally for finding the position of moving loads that will produce maximum shears, moments, reactions, or stresses, although they may be used for calculating the quantities themselves. For convenience where a number of loads are considered the influence diagrams are drawn for a single unit load. The unit influence diagram may then be used for any load by multiplying by the given load. The unit influence diagram will be referred to in the following discussion. Maximum Moment in a Truss or Beam.—Let P, in Fig. 50a, represent the summation of the moving loads to the left of the panel point 2’ and P, be the summation of the moving ldads to the right. Fic. 50a. INFLUENCE DIAGRAM FOR MOMENTS. The influence diagram for the point 2’ is constructed by calculating a(L —a) ' the bending moment at 2’ due to a unit load, = Z = ordinate Ce en Ee a) NS ne ee “hy 8 SS ae oe ees a | . ~- >.4 78 STRESSES IN BRIDGE TRUSSES 2-4, and drawing lines 1-2 and 2-3. The equation of the line 1-2 is «(L—a) L and the equation of the line 2-3 is a(L— x) E y= Now when «=a the two lines have a common ordinate which is equal to a(L —a) Also when +=—L the ordinate to 1-2—L—da; while when +0, the ordinate to 2-3 is a, as is seen in Fig. 50a. This rela- tion gives an easy method of constructing an influence diagram for moments for any point in a beam or truss. Now in Fig. 50a the bending moment at 2’ due to the lade P, and P, is M=P,9,+ Poe (a) Now move the loads P, and P, a short distance to the left, the dis- tance being assumed so small that the distribution of the loads will not be changed, and M+dM=P, (y:—44,) +P, (Yo + 4 Vo) (b) Subtracting (a) from (b) and placing d M =o, we have dM=—P,dy,+P,dy,=0 Lae | a a But d y,==d # tan a,—=d x end), 0 ene ee —a_. a PA his eee from which P,a—P,L + P,a—o, and (P,+P,) a=P,L INFLUENCE DIAGRAMS 79 Solving, we have Poo Pa, Pe cacy ees (c) From (c) it follows that the maximum bending moment at 2’ occurs when the average load on the left of the section is the same as the average load on the entire bridge. Uniform Loads.—In Fig. 50a, the bending moment at 2’ due to a uniform load pd will be pyd-w in (a). But yd -~ is the area of the influence diagram under the uniform load, and the bending moment at 2’ due to a uniform load will be equal to the area of the influence diagram covered by the load, multiplied by the load per unit of length. For a uniform load, ~, covering the entire span the bending moment at 2’ will be p times the area of the influence diagram 1-2-3. For a uniform load the bridge must be fully loaded to obtain maximum bend- ing moment at any point. It will be seen that the general criterion for maximum moment is satisfied when the bridge is fully loaded with a uniform load. . Maximum Shear in a Truss.—Let P,, P,, and P, in Fig. 50b represent the loads on the left of the panel, on the panel, and on the right of the (7-+-1)st. panel, respectively. It is required to find the position of the loads for a maximum shear in the panel. m With a load unity at 2’ the shear in the panel is — —, and 1-2 is the n influence shear line for loads to the left of the panel. With a load unity n—m—I1 at 3’ the shear in the panel is , and 3-4 is the in- n fluence shear line for loads to the right of the panel. For a load on m n—m—I ; the panel the shear will vary from — — at 2’ to at 3, n n and the line 2-3 is the influence shear line for loads on the panel. The influence diagram for the entire span is the polygon I-2-3-4. It will be seen that the lines 1-2 and 3-4 are parallel, and are at a distance unity apart. The total shear in the panel will then be S=—P,y,+ P2vetPs¥s (d) So STRESSES IN BRIDGE TRUSSES Now move the loads a short distance to the left, the distance being assumed so small that the distribution of the loads will not be changed, and S+dS=—P, (y,—491) +P. (Y2—4 92) + Ps (Ys +s) (Cl) Subtracting (d) from (e) and solving for a maximum dS=P,dy,—P,dy,+P,d),=0 But r dy,=—ds tan a,=—d x — n 1 n—TI d\.—d * ton ag=—ae nl I dy,—=d-s tan a,=dr — nl Fic. 50b. INFLUENCE DIAGRAM FOR SHEAR. and substituting we have dx n—I dx dS=P, ——P,d-x + P,—=>=0 nl nl nl. P,— P, (n—1) + P;=0 P,+P,+P,=P,n Pe wy INFLUENCE DIAGRAMS Si _Pit Pi +P, n P. (£) From (f) it follows that the maximum shear in the panel will occur when the load on the panel is equal to the load on the bridge divided by the number of panels in the bridge. Uniform Loads.—From Fig. 50b, it will be seen that for a uni- form load the maximum shear in the panel will occur when the uniform load extends from the right abutment to that point in the panel where the line 2-3 passes through the line 1-4 (where the shear changes sign). For a minimum shear in the panel (maximum shear of the opposite sign) the load should extend from the left abutment to the point in the panel where the shear changes sign. For equal joint loads, load the longer segment for a maximum shear in the panel, and load the shorter segment for a minimum shear in the panel. Maximum Floor Beam Reaction.—It is required to find the maximum load on the floor beam at 2’ in (a) Fig. 50c for the loads carried by the floor stringers in the panels 1’-2’ and 2’-3’. In (a) the diagram 1-2-3 is the influence diagram for the shears at 2’ due to a load unity at any point in either panel. In (b) the dia- gram 1-2-3 is the influence diagram for bending moment at 2’ for a unit load at any point in the beam. Now diagram in (a) differs from diagram in (b) only in the value_of the ordinate 2-4. It will be seen that the reaction at 2’ in (a) may be obtained from the diagram in (b) for any system of loads if the ordinates are multiplied by d,+d, d, d, . We can therefore use diagram (b) for obtaining the d,+ d, dy dy To obtain the maximum floor beam reaction, therefore, take a simple beam equal to the sum of the two panel lengths, and find the maximum bending moment at a point in the beam corresponding to maximum floor beam reaction if we multiply all ordinates by d,+d the panel point. This maximum moment multiplied by > 7 = gill be 2 hd -4 the maximum floor beam reaction. If the two panels are equal in oe ‘STRESSES IN BRIDGE TR ~~ cath a) eS length the maximum bending moment at the center of the Ae i a plied by > where d is the panel length, will give the maxim beam reaction. Poe oo Fic. 50c. INFLUENCE DIAGRAM FoR MAXIMUM FLOooR BEA’ cm REACTION. = eS aa Me ee ee ae / ile tae eon aha Pom Ie AMS S ae bs =, o aS i Pe 77 as bial eee ee ae ee ee A De Le ; ‘ = < 7 Ll ie ied he Tt el r A es | =a 4. CHAPTER XI. STRESSES IN A TRANSVERSE BENT’, Dead and Snow Load Stresses.—The stresses due to the dead load in the trusses of a transverse bent are the same as if the trusses were supported on solid walls. ‘The stresses in the supporting columns are due to the dead load of the roof and the part of the side walls supported by the columns, and are direct compressive stresses if the columns are not fixed at the top. If the columns are fixed at the top _ the deflection of the truss will cause bending stress in the columns. 4 ‘The dead load produces no stress in the knee braces of a bent of the _ type shown in Fig. 1 except that due to deflection of the truss, which may usually be omitted. The stresses may be computed by algebraic or graphic methods. The stresses due to snow load are found in the same way as the a dead load stresses. In localities having a heavy fall of snow the freez- _ ing and thawing often cause icicles to form on the eaves of sufficient q weight to tear off the cornice, unless particular care has been exercised in the design of this detail. Wind Load Stresses.—The analysis of the stresses in a bent due a to wind loads is similar to the analysis of the stresses in the portal of a _ bridge. The external wind force is taken (1) as horizontal or (2) q as normal to all surfaces. The first is the more common assumption, a although the second is more nearly correct. For a comparison of the 4 stresses in a bent due to the wind acting horizontal and normal, see : Figs. 54, 55, 56 and 57, and Table V. In the discussion which immed- 4 iately follows, the wind force will be assumed to act horizontally. The magnitude of the wind stresses in the trusses, knee braces and 84 STRESSES IN A TRANSVERSE BENT columns will depend (a) upon whether the bases of the columns are fixed or free to turn, (b) upon whether the columns are rigidly fixed to the truss at the top, and (c) upon the knee brace and truss con- nections. Of the numerous assumptions that might be made, only two, the most probable, will be considered, viz.: (1) columns pin connected (free to turn) at the base and top, and (II) columns fixed at the base and pin connected at the top. Columns in mill buildings are usually fixed by means of heavy bases and anchor bolts. Where the columns support heavy loads the dead load stress in the columns will assist somewhat in fixing them. Where the dead load stress plus algebraically the vertical component _of the wind stress in the column, multiplied by one-half the width of the base of the column parallel to the direction of the wind, is greater than the bending moment developed at the base of the leeward column when the columns are considered as fixed, the columns will be fixed without anchor bolts (see Chapter XII, Fig. 61). In any case the resultant moment is all that will be taken by the anchor bolts. The dead load stresses in mill buildings are seldom sufficient to give material assistance in fixing the columns. Unless care is used in anchoring columns it is best to design mill buildings for columns hinged at the base. The general problem of stresses in a transverse bent for Case [ and Case II, in which the stresses and forces are determined by alge- braic methods, will now be considered. The application of the general problem will be further explained by the graphic solution of a par- ticular problem. ALGEBRAIC CALCULATION OF STRESSES: Case I. Columns Free to Turn at Base and Top.—In Fig. 51, H = H? => = horizontal reaction at the base of the column due to external wind force, W. Ee trees dt 25 the wind force, W. = vertical reaction at base of column due to The wind produces bending in the columns, and also the direct — ee CoLuMNs HINGED AT THE BASE 85 stresses V and V?. Maximum bending occurs at the foot of the knee brace and is equal to (H — W,) d on the windward side, and H* d _on the leeward side. These bending moments are the same as the bend- _ ing moments in a simple beam supported at both ends and loaded with a concentrated load at the point of maximum moment. Since the max- ; uy » L c CF Cc Fr 4Nb —Bslb ae 3 7 44 er . arr c H' H' BE Mi (a) v vb) (c) (d) (€) - External Forces Leeward Col. Beam Shear Moment a Fic. 51. imum moment occurs at the foot of the knee brace in the leeward q column, we will consider only that side. We will assume that the lee- q ward column (b), Fig. 51, acts as a simple beam with reactions H* and Cand a concentrated load B, asin (c). The reaction C and load B will 4 now be calculated. 4 From the fundamental equation of equilibrium, summation hori- 4 zontal forces equal zero, we have POSS: 4 B= H?+C (25) Taking moments about b, we have C (h—d)=H'd. ve ae 3 26 C= (26) The stresses K, U and L can be computed by means of the follow- p ing formulas: K = B cosecant m | (27) _ Where m = angle knee brace makes with column; q U = (V+ — K cos m) cosecant n (28) _ where » = angle of pitch of roof; and . 86 STRESSES IN A TRANSVERSE BENT L=C—Ucosn (29) In calculating the corresponding stresses on the windward side, the wind components acting at the points (a), (b) and (c) must be subtracted from H, B and C. The shear in the leeward column is equal to Ht below and C above the foot of the knee brace, (d) Fig. 51. The moment in the column is shown in (e), Fig. 51, and is a max- imum at the foot of the knee brace and is, M = H' d. The maximum fibre stress due to wind moment and direct loading in the columns will occur at the foot of the knee brace in the leeward column, and will be compression on the inside and tension on the out- side fibres, and is given by the formula* Mw -f M y nth eee (30) 3 ey OF where {, = maximum fibre stress due to flexure; f. = fibre stress due to direct load P; A = area of cross-section of column in square inches; M = bending moment in inch-pounds = H? d; y = distance from neutral axis to extreme fibre of column in inches ; I = Moment of Inertia of column about an axis at right angles to the direction of the wind; P = direct compression in the column in pounds; h = length of the column in inches; E = the modulus of elasticity of steel = 28,000,000 ; Pwd 5 io @ iS minus when P is compression and plus when P is tension. The maximum compressive wind stress is added to the direct dead and minimum snow load compression and governs the design of the column. *This formula was first deduced by Prof. J. B. Johnson. For deduction of the formula see Chapter XV, or “Modern Framed Structures’ by Johnson, Bryan: and Turneaure. Se ee, ae CoLUMNS FIxED AT ‘THE BasE 87 Having the stresses K, U, and L, the remaining stresses in the truss can be obtained by ordinary algebraic or graphic methods. For a simple graphic solution of the stresses in a bent for Case I, in which these stresses are computed graphically, see Fig. 54 for wind horizontal, and Fig. 56 for wind normal to all surfaces. Case II. Columns Fixed at the Base—With columns fixed at the base the columns may be (1) hinged at the top, or (2) rigidly fixed to the truss. | | (1) Columns fixed at the base and hinged at the top.—It will be further assumed that the deflections at the foot of the knee brace and the top of the column, Fig. 52, are equal. w, o> | Ww. U 4 ws oe nik c Soe anes (=< Fag oe - 5, Wi heat Mn ' 1 : a? op} 1 ae ! a. | 2 aes Seta, alk— ale Wy po ras ee pees lage pe H Hi H Sar M (a) i! 1b) te) (d) (e) External Forces Leeward Col. Beam Shear Moment Fic. 52. In Fig. 52 we have as in Case I HH =f V and V? are not as easily found as in Case I, but will be cal- culated presently. The leeward column will be considered and will have horizontal external forces acting on it as shown in (c) Fig. 52. For convenience we will consider the leeward column as a beam fixed at a and acted upon by the horizontal forces B and C as shown in (c) Fig. 52, the de- flection of the points b and ¢ being equal by hypothesis. From the fundamental condition of equilibrium, summation hori- zontal forces equal zero, we have | B=H+C | (31) 88 STRESSES IN A TRANSVERSE BENT To obtain B and C a second equation is necessary. From the theory of flexure we have for the bending moment in the column at any point y, where the origin is taken at the base of the column, when y = d dia MAE =B(d—»)-Clh-») sé Integrating (32) between the limits y = o and y = d, we have EIS = [Bay—4P Chyt+ Dr Anis =—c(4a—5 7) + oe (33) Now (33) equals & J times the angular change in the direction of the neutral axis of the column from y = 0 to y = d. When y > d, we have a? .x Scien Hees a 4 M=EISS L(h—y) (34) Integrating (34) we have E feast 0 hyp Sek. (35) Now (35) equals E J times the ae in direction of the neutral axis of the column at any point from y = dtoy=h. | To determine the constant F, in (35) we have the condition that - the angle at y = d must be the same whether determined from equation (33) or equation (35). Equating (33) and (35) and making y = d, we have F,= ae (36) Substituting this value of F, in (35) we have ax C3 ara So CH Pia RE eA 37 Sele PAs ahs Cs : (37) Integrating (37) between the limits y = d and y = h, we have Be fa) ; 5 ee, SS ‘o 26 2y1y=h Pe Ae ae [ + +. ae (es d* 3 m, (ek h a? (38) amy —__— CoL.uMNs FIxep At THE BAsE S9 Now (38) equals & J times the deflection of the column from y = d to y = h, which equals zero by hypothesis. Solving (38) we have es 3d27h—3a? Bn TI ESE TI = Oh Et oak— a? In a beam fixed at one end there is a point of inflection at some point, between y = o and y = d, where the bending moment equals zero. Now if y, equals the value of y for the point of inflection, we have from (32) B (d—) = C (h—y) and C a — Yo 57 Bimal yes (40) Equating the second members of equations (39) and (40) and solving for y,, we have adja h Je = nae AN) To find the relations between y, and d, we will substitute / in terms of d in (41) and solve for 4. h 5 F a=, —~qd 2 4 a@=rh =! 1 ad—h, Fo 5 Solving (31) and (39) for C, we have HT} 3a? Meese Ta ay (FFA) (42) To find the moment M » at the base of the leeward column, we have from (32) M, =Bd—Ch go STRESSES IN A TRANSVERSE BENT Substituting the value of B given in (31) we have M, =Hd+Cd—Ch (32a) Eliminating h and d by means of (41) and (42) we have finally My, = H* yo (43) In like manner it can be shown that the moment at the base of the windward column is 2 My, =H yy — we (43a) _where w equals the wind load per foot of height. To find V we will take moments about the leeward column. ‘The moments M, and M,, at the bases of the columns respectively, resist ” overturning and we have 2 and since St =a a =*[Hi—2Hy, + 2S (2) S 8 Now if ¢ is taken equal to y,, we have after transposing Vo Vis = [2H —wye ] —E= ee (45) It will be seen that (45) is the same value of V and V’? that we would obtain if the bent were hinged at the point of contra-flexure. From (43) and (45) it will be seen that we can consider the col- umns as hinged at the point of contra-flexure and solve the problem as in Case I, taking into account the wind above the point of contra- flexure only. The maximum shear in the column is shown in (d) | Fig. 52. ; The maximum positive moment occurs at the foot of the leeward knee brace and is M, = H (d — y,) ; the maximum negative moment occurs at the base of the leeward column and is equal to M as A ge CoLUMNS FIXED AT THE BASE gI The maximum fibre stress occurs at the foot of the knee brace, and is given by the formula P M ft+h= + ra - . A im Pkt (30a) 10 & The nomenclature being the same as for (30) except h, which is the distance in inches from the point of contra-flexure to the top of the column. (2) Columns fixed at the base and top.—tIn this case it can be ns seen by inspection that the point of inflection is at a point y, = a and we have for this case B= H* --C (31a) E Wes <- = Me (32a) iets at: ay (42a) It is difficult to realize the exact conditions in either (1) or (2), in Case II, and it is probable that when an attempt is made to fix columns at the base, the actual conditions lie some place between (1) and (2). It would therefore seem reasonable to assume the minimum value, y) = + as the best value to use in practice. This assumption is commonly made and will be made in the problems which follow. Having the external forces Ht, B, C and V’ the stresses K, U and L are computed by formulas (27), (28) and (29). The remaining stresses ‘in the truss can then be computed by the ordinary algebraic or graphic methods. For a simple graphic solution of this problem, where the ex- ternal forces B and C are not computed, see Fig. 55 and Fig. 57. Maximum Stresses.—It is not probable that the maximum snow and wind loads will ever come on the building at the same time, and it is therefore not necessary to design the structure for the sum of the maximum stresses due to dead load, snow load and wind load. A g2 STRESSES IN A TRANSVERSE BENT common method is to combine the dead load stresses with the snow or the wind load stresses that will produce maximum stresses in the members. It is, however, the practice of the author to consider that a heavy sleet may be on the roof at the time of a heavy wind, and to design the structure for the maximum stresses caused by dead and snow load; dead load, minimum snow load and wind load; or dead load and wind load. It should be noted that the maximum reversals occur when the dead and wind load are acting. For a comparison of the stresses due to the different combinations see Table VI. A common method of computing the stresses in a truss of the Fink type for small steel frame mill buildings is to use an equivalent uni- form vertical dead load; the knee braces and the members affected directly by the knee braces being designed according to the judgment of the engineer. This method is satisfactory and expeditious when used by an experienced man, but like other short cuts is dangerous when used by the inexperienced. For a comparison of the stresses in a 60-foot Fink truss by the exact and the approximate method above, see Table VI. Stresses in End Framing.—The external wind force on an end bent will be one-half what it would be on an intermediate trans- verse bent, and the shear in the columns may be taken as equal to the to- tal external wind force divided by the number of columns in the braced panels. ‘The stresses in the diagonal rods in the end framing, as in Fig. I, will then be equal to the external wind force H, divided by the number of braced panels, multiplied by the secant of the angle the diagonal rod makes with a vertical line (For analysis of Portal Bracing see Chapter XIT). Bracing in the Upper Chord and Sides——The intensity of the wind pressure is taken the same on the ends as on the sides, and the wind loads are applied at the bracing connection points along the end rafters and the corner columns. ‘The shear transferred by each braced panel is equal to the total shear divided by the number of braced panels. The stresses in the diagonals in each braced panel are com- GRAPHIC CALCULATION OF STRESSES 93 puted by applying wind loads at the points above referred to, the wind loads being equal to the total wind loads divided by the number of panels. The stresses are computed as in a cantilever truss. The brac- ing in the plane of the lower chord is designed to prevent undue de- flection of the end columns and to brace the lower chords of the trusses. All wind braces should be designed for, say, 5,000 pounds initial stress in each member, and the struts and connections should be proportioned to take the resulting stresses. It should be noted that a mill building can be braced so as to be rigid without knee braces if the bracing be made sufficiently strong. GRAPHIC CALCULATION OF STRESSES.—Data.—To il- lustrate the method of calculating the stresses in a transverse bent by graphic methods, the following data for a transformer building similar to one designed by the author will be taken. The building will consist of a rigid steel frame covered with cor- rugated steel and will have the following dimensions: Length of building, 80’ 0”; width of building, 60’ 0”; height of columns, 20’ 0”; pitch of truss, % (6” in 12”); total height of building, 35 0”; the trusses will be spaced 16’ 0” center to center. ‘The trusses will be riveted Fink trusses. Purlins will be placed at the panel points of the trusses and will be spaced for a normal roof load of 30 Ibs. per square foot. The roof covering will consist of No. 20 corrugated steel with 24-inch corrugations, laid with 6-inch end iaps and two cor- rugations side lap, with anti-condensation lining (see Chapter XVIII). The side covering will consist of an outside covering of No. 22 corru- gated steel with 2!4-inch corrugations, laid with 4-inch end laps and one corrugation side lap; and an inside lining of No. 24 corrugated steel with 14-inch corrugations, laid with 4-inch end laps and one corrugation side lap. For additional warmth two layers of tar’ paper will be put inside of the lining. Three 36-inch Star ventilators placed on the ridge of the roof will be used for ventilation. The general ar- rangement of the framing and bracing will be as in Fig. 1 and Fig. 81. 9-4 STRESSES IN A TRANSVERSE BENT The approximate weight of the roof per square foot of horizontal projection will be as follows: PRESSES. fh lds wee eh ee 3.6 Ibs. per sq. ft. Purlins and Bracing...... 9. te Rae Corrugated Steel ........ 2.4 St ee Roof Lining gt rs Oe eae 6708 ‘c bes ae TOtal Seinen = oars 16(03. eee The maximum snow load will be taken at 20 pounds, and the minimum snow load at 10 pounds per square foot of horizontal pro- jection of roof.- The wind load will be taken at 20 pounds per square foot on a vertical projection for the sides and ends of the building, 20 pounds per square foot on a vertical surface when the wind is considered as acting horizontally on the vertical projection of the roof, and 30 pounds per square foot on a vertical surface when the wind is consid- ered as acting normal to the roof. The stresses in an intermediate transverse bent will be calculated for the following: Case 1. Permanent dead and snow loads. aan Casg 2. A horizontal wind load of 20 pounds per square foot on the sides and vertical projecticns of the roof, with the columns hinged at the base. | Casg 3. Same wind load as in Case 2, with columns fixed at the base. Case 4. k--H’'= 5600 Ibs- > e (5.t 3200 see c TZ A d = IS aa | | phceeepepay is] ae 4 omempen sy 4 re b is X)} ao 3 ‘ 3 Hi alts i \ x2 2% f a aw * yr a CASE ad a PP 2 x a ~, , 2240 a al Columns Pin Connected ee a Pg f = Maximum Mom. in Col-=940800 in-Ibs» CMS 3 ” Wind Load Stress Diagram . Wind Horizontal, 20 |bs.persqft- ee re) 4000 8000 Sea 1 lL 1 j re oe nnn ------ -------------------- ai7 : Compression ve Tension — i ae ¢ 9 Fi ‘ 3 ete of a ws ‘ . at 1 rs 7 6 ges 2 6 eae ~< Gs x xX * a v-- B 12 8 4 A Fic. 54. WIND LOAD STRESS DIAGRAM, CASE 2, by computing the stress in 5-6 (component due to stress in 6-7), and substituting it in the diagram, or by substituting an auxiliary member. The stresses in the auxiliary members are represented -by dotted lines and are of no value in designing the bent. It should be noted that the auxiliary members do not affect the stresses in the trusses and knee braces, which are correctly given in the stress diagram. The maximum stress in the knee brace A-15 is compression, and occurs on the leeward side. 98 STRESSES IN A TRANSVERSE BENT The maximum shear in the leeward column below the knee brace is H* = 5600 lbs. ; the maximum shear above the knee brace is 13,100 lbs. The maximum moment occurs at the foot of the knee brace and is H? x 14 x 12 = 940,800 inch-lbs. : Case 3. Wind Load Stresses: Wind Horizontal; Columns Fixed at Base.—This is Case 2 with the base of the column hinged at the point of contra-flexure. In calculating H and V, Fig. 55, the wind 600 _Us ye M4 ‘5 20 1200 Us 7 1200 Uz 6 a Ls 8 cane 1200 U, 4 5 it Ns Wind=8960 Ibs. H=/4 480 Ibs->%«¢H= 4480 | Ae os 1560, be e a ay = 14 cee NF anes Bee Ue STE SS Pa on ¢ i oe Wing €080., SS Seated _ sis a sone ne s> wee b os. a 17 Ps Eee CASE 3 aaa ——> SK... ~ RQ’ B Columns Fixed at Base Cc 2240 Max:-Mom.in Col: = 376320 in- Ibs. VA | Wind Load Stress Diagram Wind Horizontal 20 |bs.per sq; ft 13 15 Fes L 1 l S000 ‘Compression Tension pain i2 8 Fic. 55. WIND LOAD STRESS DIAGRAM, CASE 3. above the point of contra-flexure only (see formula (45)) produces stresses in the bent. ‘The value of fixing the columns at the base is seen by comparing the stresses in Case 2 with those in Case 3, both being drawn to the same scale. Maximum shear in the leeward column below the knee brace is Ht = 4480 Ibs.; above the knee brace is 5230 lbs. The maximum positive moment occurs at the foot of the knee brace and negative moment at the foot of the column, and is H* x 7 x I2 = 376,320 inch-lbs. : * + , ee ee | LET a ee ee ne Sa Sp ee ee ee ee eee ee Sd en SE ee ha ee ee —— - ees Wrnp Loap STRESSES, CASE 4 99 Case 4. Wind Load Stresses: Wind Normal; Columns Hinged. —In Fig. 56 the resultant of the external wind forces on the sides and the roof acts through their intersection, and is parallel to C B in the stress diagram (line C B is not drawn). To calculate V and V+ con- nect the point of intersection, 4, of the resultant wind and the vertical line through the center of truss, with the bases of the columns B and C. z \8 3 % .A MiG oe oe woe % Us 3, OW SF ees Na 960 aoe 2 3/4 ‘ le Sy if ; KP ps | or ee mre oo" AT 16 fa, sana Sere exer ; io Columns Pin Connected °7*"7"" > ws 18 Max.Mom:in Col-924000inlbs} 17 4” x b ped ” a hy in “ +2 2240s __.-----* : BS CASE 4 Wind Load’ Stress Diagram Wind Normal, Roof I8 Ibs- Sides 20 Ibs. sq.ft fe) 4000 8000 l l l l 3/3 13 15 16 Sauer ee '7 Compression : 7 Tension el Fic. 56. WIND LOAD STRESS DIAGRAM, CASE 4. 100 STRESSES IN A TRANSVERSE BENT From A lay off one-half of resultant wind on each side, and from the extreme ends drop vertical lines V and V? to the dotted lines A B and AC. The vertical lines V and V+ will be the vertical reactions, the horizontal lines will be H and H', and R and KR? will be the resultants of the horizontal and verticai reactions at B and C respectively. The stresses are calculated by beginning at the base of the column B as in Case 2. In the polygon a-B-A-b at B, A-B = R, a-B = 22¢0 lbs., and a-b and A-b are the stresses in a-b and A-b respectively. & “ cs % Q A rec) (o) ’ ' ‘ ' ’ 4 % big 0 5 10 15 20 eS) [o) Us L i l 1 j 2 be Y | T / ¥ ne IN “ cS) Us 5 : 8 rs ‘ Ne Sf 4 ) . J i2 NSF ; 960 \ le 2 | 1 y4- pea patos x | Bee “ee % 2060. Prectieeemee ind 4I6QIbs| 2 l Columns Fixed at Base IO See : mR «| i 8 Poca in Col: 361200 in- Ibs 17 ae : se i ~ 7 Pa ie lou Pi Se, OR ak >} ‘ < . 4 a. ; 2240 R CA CASE 5 Wind Load Stress Diagram Wind Normal: Roof 18 lbs-Sides 20 lbs- sq-fr Oo 4000 8000 i L 'B Fic. 57. WIND LOAD STRESS DIAGRAM, CASE 5. WIND LOAD STRESSES, CASE 5 101 The maximum shear in the leeward column below the knee brace is Ht = 5500 lbs., above the knee brace is 12,800 lbs.; the maximum a sd moment occurs at the foot of the knee brace and is H! x 14 x 12 924,000 inch-lbs. Case 5. Wind Load Stresses: Wind Normal; Columns Fixed at Base.—This is Case 4 with the base of the column moved up to the point of contra-flexure The maximum shear in the leeward column below the knee brace is 4300 Ibs., above the knee brace is 5000 lbs. ; the maximum positive moment occurs at the foot of the knee brace and negative moment at the foot of the column and is H* x 7 x 12 = 361,200 inch-lbs. | Maximum Stresses.—The stresses For analysis see Fig. 57. in the different members of the bent for the different cases are given in Table V. The maximum TABLE V. Stresses ina Bent For Dead Load Snow Load Wind Load Casee Case Case4 Case5 +9300 78800 +8200 +7700 +7700 +9300 -8300 t/100 -/200 +2200 -/200 #1100 -Z400 -3600 -3600 #2200 -/200 -7/100 | -4700 - 7/00 -8300 +4800 +4800 t4800: +4800 #/8900 #17600 +/6400 +/5400 4/5400 #18600 -/6600 + 2200 - 2400 +4400 #3700 +4900 + 400 +1400 -6/00 -/9600 + 5700 t+ 500 -6800 +3800 - 600 t+ 500 -4300 -4900 t7500 -6700 7/5200 +2400 +6600 t/4200 +4600 -9000 +22300 ~ 53200 t+/700 +3200 - 8600 #2900 +4000 +1400 +3400 -/900 -8600 +2800 + 500 -4900 t5000 - 600 t+ 500 =JIS00 =22U00 +3600 -3200 +7400 + 800 t4000 +7700 +3000 - 6200 #11000 -21/100 t+/3500 +2100 - 5800 #19400 #15400 #10200 4/0200 - 1400 -/4600 - 3100 t+ 2400 -8500 +7300 - 2600 +2400 - 8200 10800 + 7400 - 6700 +/4800 - 6000 t2200 + 9700 t 500 -~8I00 421500 t+ 53400 + 8000 #+I500 - 6400 t/4900 +14900 #112 00 t/1Z 00 + 2800 — 53600 - 6000 +2400 -6700 +6600 -2600 +2400 -7400 -/0000 +3500 -3200 +7000 -7700 ~ 400 #+2/00 ~/300 -6700 4/0400 +4500 t+ 7600 +4/00 - 2400 102 | STRESSES IN A TRANSVERSE BENT stresses in the different members of the bent for (1) dead load plus max- imum snow lodd; (2)' dead load plus wind load, Case 4; (3) dead load plus minimum snow load plus wind load, Case 4; and (4) a vertical dead load of 40 lbs. per sq. ft. horizontal projection of the roof are given in Table VI. ‘The stresses which control the design of the members may be seen in Table VI. By comparing these values with the stresses given in the last column the accuracy of the equivalent load method can be seen. ; Bal TABLE VI. Maximum Stresses ina Bent For Name}Dead Load +/Dead Load +/Dead Load+Min-: |Vert-Dead Load of [tMax:Snow |Wind Load {Snow LoadtWind |of40 Ibs-per Sq: Piece} Load Case 4 |Load.Case 4 |Ftof Hor: Proj- x-2 | #28200 +24700 | +34300 + 37200 “3 726400 +24Z200 #+IZ000 # 35200 x*-6 | #426400 #/8400 +26600 + 32800 P oa f #23/00 t+/7900 t+ 25600 F30800 x-9 | 423/00 + 6300 +/4000 #350800 x13 | 428200 ~ 2500. + 4000 + 37200 1-2 | ~24900 -/3400 -2/700 -33200 Ea SIO +3500 t+ 4600 + 4400 3-41 - 3600 EGO 2 -/0900 - 4800 4-5 t+ 6600 t+ 9500 ~/1700 + 8800 5-6 | - 3600 - 3800 - SOC? -+800 6-0 |. F-D3Ce +3500 t+ 4600 + 4400 5-8 ~7E00 -/0600 -/3000 - 9600 7-8 -/0800 -/4400 -/8000 -/4A400 8-9 | -/0800 %+ 53800 Ff Zoe -/4-400 9-12 | #6600 ~ 4500 ~2 3500 + 8800. I2-15 | -3600 7/3600 t/2400 - #4800 Y-4 | 2/300 -/3/00 -20200 -28400 Y-8 | -/4/00 - 2500 - 5200 -/4 800 Y-12 | -23/00 + 2600 - 4500 -28400 13-15) -24900 - 7800 -/6/00 -32400 P-l - 8500 - 86500 AcI5 #2/500 -2/500 Avb| 4/4400 t+ 8200 t+/35000 +/9200 C-! | 4/4400 #/2800 +/7600 +/92Z00 PX\IT| 4/4400 t/0/00 +/4900 t+/9200 I5-I6 | 4/4400 = /600 # 35200 +/9200 CALCULATION OF REACTIONS 103 Graphic Calculation of Reactions.—The graphic method for calculating the reactions given in Fig. 56 and Fig. 57 may be proved as follows: In Fig. 57a the intersection of } W and the center line of the bent is at d. Draw A-B and A-C, lay off } W so that it is bisected by the point A, and draw 1-2 and 3-4. Then 1-2 equals V’’ and 3-4 equals V as shown in the following proof: Proof.—To calculate V take moments of external forces about C, and > sW Xb ok ae 4 times area triangle A—4-C “= e : : 34XL But area triangle A—4-C is also equal to ————— . and 4 3-4 XL V = —_—_- = 3-4, | L 3-4 which proves the construction. It may be proved in like manner that I-2=V". oe op ee ee me ee me ee oe ae Wa Bi . Wis sy =>"y li * “ap as B i] ¢! ie re C eo a" ee LEN He 6 - Ns ’ : ; ’ ABS of qn =- H'-—-je--- H---a Eg a een nee ae = TRANSVERSE BENT WITH VENTILATOR.—The calcu- lation of the wind stresses in a transverse bent with a monitor venti- lator is shown in Fig. 57b. The bents are spaced 32’ 0” centers and Ye ee ee er 104 . _ STRESSES IN A TRANSVERSE BENT are designed for a horizontal wind load of 20 Ibs. per sq. ft., the nor- mal wind roof load being obtained by, Hutton’s formula as shown in “ Fig. 6. Sere The point of contra-flexure is found by substituting in formula “4 (41) to be at a point y= 17.0’. The external forces are calculated for the bent above the point of contra-flexure by multiplying the area sup- ported at the point by the intensity of the wind pressure. For example, the load at B is 32’ X 6.75’ X 20 lbs. = 4320 lbs. = pee ‘ SG ; i Sag Se aS Ee ee ee e Reg ae ie ee a Leia aes tet ies Py - 4 -9- SNA Scale Truss =: 2 “ - ~‘ : > * oe {So 1 . ; % = ! i iy eels Cop abe ett . ¥ ts 7 4520" 2th K-H=/3500/b.*-H=15500 Ib-- Me "3 ee fs 4 ret — iss = i. 10880 640". 5 4 Me see we wcewccesscesane 64 | aw eace eeeawcans > Med no oe en = Ih a I-18 9 a m . sae eae Receewaseaan a eae by § a ry t ; a “ey a : ra 3 1 x 7 ; ‘4 a ‘ : 4 . oa iS 4 \ a Relates Sp a 00d = 20 |b. 59. Ft Hox. Wind Load=20 ae ” Yert wal Ay 4 os Ts i A at) 0 5000 10000 20000 t 1 2 } Scale Stresses 8 A WIND LOAD STRESS DIAGRAM Covumns Fixeo (c) Fig. 57b. * ; 5 Satay We ae SUP ST ee ae « ol in eel The line of application and the amount of the external wind load, x W, is found by means of a force and an equilibrium polygon. }W acts through the intersection of the ‘strings parallel to the rays O-B i hs tls . 7 ee hs s 7 ees ee aN ret a tal a cnc ee? 2 BENT WITH SIDE SHEDS . 105 and O-C, and is equal to C—B (line C—B is not drawn in force polygon) in amount. The reactions R and FR’ are calculated by the graphic method as previously described. The calculation of stresses is begun at point B in the windward column, and in the stress diagram the stresses at B are found by drawing the force polygon a~B—A-b-a. The remaining stresses are calculated as for a simple truss. In calculating the stresses in the ventilator it was assumed that diagonals 9-10 and 10-12 are tension members, so that 9-10 will not be in action when the wind is acting as shown. Before solving the stresses at the joint 6-7-9 it was necessary to calculate the stresses in members 7-11, 10-11, and g-h. The re- mainder of the solution offers no difficulty to one familiar with the principles of graphic statics. The stress in post b-a is equal to V, while the stress in 1-c is found by extending 1-c to c’ in the stress diagram, c’ being a point on the load line. The stress in post n—A is equal to V’, while the stress in 19q~m is found by extending 19-m to m’ in the stress diagram, m’ being a point on the horizontal line drawn through C. The kind of stress in the different members is shown by the weight of lines in the bent diagram and by arrows in the stress diagram, one arrow indicating the direction and kind of stress the first time a stress is used and two arrows indicating the second time a stress is used. TRANSVERSE BENT WITH SIDE SHEDS.—Transverse bents with side sheds are quite often used in the design of shops and mills. The calculation of the stresses due to wind load in a bent of this type is an interesting application of the author’s graphic solution of stresses in transverse bents. It is required to calculate the stresses due to a horizontal wind load of 30 lbs. per square foot on the sides and the normal component of 30 lbs. (Hutton’s Formula, Fig. 6) on the roof, the bents being spaced 20’ 0” centers, as in Fig. 57c. The loads are calculated, and by means of a force polygon in (d) and an equilibrium polygon in (a) the resultant wind = W is found to pass through point £, and to be equal to 30,800 lbs. Calculation of Reactions.—The horizontal shear of 25,400 Ibs. will be taken by the columns in proportion to their rigidities, in this 106 STRESSES IN A TRANSVERSE BENT case the rigidities of the columns are assumed equal and the shear at the foot of each column will be 6350 Ibs. The vertical reactions will be due to two forces: (1) a vertical load of 17,200 lbs., which will be taken equally by the four columns, making a load of 4300 Ibs. on each; and (2) to a bending moment of 25,400 lbs. X 9.2 ft. == 233,680 ft.-lbs. (the bending moment about C. G. is also equal to 30,800 Ibs X 7.6 it.), which will be resisted by the columns and will cause reactions varying as the distance from the center of gravity of the columns, (E), as in the case of the continuous oa, Fig. 63. 3 Let v’,, vs, v's, and v’, represent the reactions due to moment in the columns, respectively ; then if a is the reaction on a column at a units distance from the center of gravity we will have v', =—a 40, v', =— a 20, v', = + a 20, and v’',=-+a 40. The resisting moment of each column will be equal to the reaction multiplied by the distance from the center of gravity, and a 40? + a 20? + a 207 + a 40? = 233,- 680 ft.-lbs. from which a = 58.42 lbs. and v’; == — 2340 lbs.; v’, =— 1170 lbs.; v’, =-+ 1170 lbs. ; v’, = + 2340 lbs. | Now adding the reactions due to (1) and (2) we have V , = 4300 — 2340 = + 10960 lbs., V, = 4300 — 1170 = + 3130 lbs., V ,== 4300 — 1170 = + 5470 lbs., V , == 4300 — 2340 = + 6640 lbs. Combining the horizontal and vertical reactions we have R, =a,-A = 6600 lIbs.; R,—A-B=7200 lbs.; R,=B-C=8 400 lbs.; R,= C-D =9g100 lbs. These reactions close the force polygon in (d). Calculation of Stresses—Auxiliary members are substituted as shown by the broken lines. It will be seen that these members are arranged so that all bending is removed from the columns and that the stresses in the truss members are correctly given in the stress dia- gram. The calculation is started at point A at the foot of the left- hand column as in the case of the simple transverse bent, and reac- tions Rk, and R, are substituted as the calculation progresses, the stress diagram finally closing at the base of the leeward column, point D. The stresses are given on the members in (a). The direct stresses in the columns are easily found by algebraic resolution beginning at the its) Oe . i] w bt | 107 v2 i 8 is! Se ev S$PpoO07 ajD2S tT T T ‘ . | oN § 00091 000% 0, pt im ' A ‘ t H . ‘ +> « a bs y cohiehine---$ je 19% en \\ Lf 2) \9¢ te ov eg H-6¢ &. , / Eiibes nae Upside teste he 1S < ye we ats Le ¥ : * ; ve, fo i Pa : oe. ‘eer se re : : ao °° FT att ee ene ve-ty ---: Pe rereaess - ° 9F RM ons ov Tig a 5g % 12-S2 ' ee ‘Y "9 bry “ojnuisoy suoyny Aq ' ae Re oa ‘S ~ JODY 40 JusuodunD pousoy : SX... i ake vant a alas ane a Palgeidl et a a 702 ann7oh Wybs aad ‘g/ gg poo] pulpy, et) + ; %'? D-02 par0ds yuag + soBUS ! ; L0sc9="SR AIT aoe ie eeee VSPeSagAee seca ng MUDGSCLNS0EEENSIiceccesccceccnesesee 2" WS =) . ’ SP. Wal cey: 00) So 1&6 = s. “ ooosel="W Ss . “8 , 1 Gl, ‘\ w 2 ¢ : 4 2 es [Ae eee Sao ee te 1 P “yy, . ; a “eee = ~<-+- oe cee Seer ren Seneenene a is oe #1 ¥ “ope YS SSeeSe 28s 0g.6a ye 4 OS-6a NOSED ='56 EH es 0SFL2 = eer aa : == Sh eS ie’ . = = 2! _ , : S, S22: Riga AW 2 ' SSNAL BOIS . Enno glen en sesageoensacn-o 7 a Tere = a, ae een ee Se ae 108 STRESSES IN A TRANSVERSE BENT foot of the columns where the direct stress is equal to ae ae y, V’,, respectively. The stresses in the leeward side of the main truss very large due to the small depth of truss in line of the member z The stresses could be materially reduced and a considerable saving material obtained by using a main truss of type (b) or (h), Fig. Moment and Shear in Columns.—The bending moment | main leeward column is shown in (b), the maximum moment i the foot of the knee brace and is M, = 274,500 ft.-lbs. The shear d gram is shown in (c), the maximum shear is between the foot of ‘g knee brace and the top of the column and is Sy = 45,750 Ibs. : Nore.—The stresses in a bent with side sheds obtained by th ceding method are approximate for the reason that the assumed c tions are probably never entirely realized. In the exact wae ietormuation of the framework is considered in a manner simil that of the two-hinged arch in Chapter XIV. The approximate tion is entirely adequate for all practical purposes. : Me a “ay = “al CHAPTER XII. STRESSES IN PoRTALS. Introduction.—Portal bracing is frequently used for bracing the sides of mill buildings and open sheds. ‘There are many forms of portal bracing in use, a few of the most common of which are shown in Fig. 58. ACE SN Ae Saat Ge PR GredeR A aes A > I | i : aah si WE id a 4 I | | | - l : h | h | h d d d e—-S—-—- ; <—|- S-—- ; i= f= a. ' ! 1 ‘ | | ‘ I | H H H H KH a: A+ BK. Pina =e Ato vr V4 vA Vt wae Vv (a) (b) (c) a ' tw 6 FOR © ee oe 6 ER : T We ~ ee D a G4 * i ! | | { h h | h e-4+S- > I ar —> | ‘there sat Pecix , ' | | ‘ | | ' i | Ho i» Fo) Hobe y Hod ma + Le re HA 7 “la Re toy (d) (e) (f) Fic. 58. Portal bracing may be in separate panels or may be continuous. The columns may be hinged or fixed at the base in either case. wigs ete) STRESSES IN PORTALS _ CASE L STRESSES IN SIMPLE PORTALS: — 58 are feed to be equal and R — c ren cit H= Hs 5 Taking moments about the foot of the windward column Vt = aie Rh s -may be found by either algebraic or graphic methods. a: Algebraic Solution —Portal (a).—To obtain the stress in memk moments of the cies forces to the dhe be the section aes point F i as a center. odes pe But H = =, and (k—d) sin © = > cos ®, Substituting ‘de : . values in (46) we have Ge oA ee V sec © E E’ and H E’ = 0. S To obtain stress in G D, pass section cutting HG, H E’ ane G ». ; point H as a center. [Th Oe 3 Gade ee To obtain stress in G F, pass a section cutting G F, E F and G CG e and take moments of the external forces to the right of the section about point C as a center. Oe e Peed tare @)+ Ha a ho 8 ; | (49) | ‘ ALGEBRAIC SOLUTION Sy Gt To obtain stress in H G, pass a section cutting H G, H £’ and G D, and take moments of the external forces to the left of the section about the point D as a center. Had HE =— ——— h—a (50) The stress in the windward post, A F, is zero above and V below the foot of the knee brace C ; the stress in the leeward post is zero above and 7! below the foot of the knee brace D. The shear in the posts is H below the foot of the knee brace, and above the foot of the knee brace is given by the formula Hd net og = stressin 7G (51) The maximum moment in the posts occurs at the foot of the knee braces C and D and is M=Hd (52) For the actual stresses, moments and shears in a portal of this type, see Fig. 59. _ Portal (b).—The stresses in portal (b) Fig. 58, are found in the same manner as in portal (a). The graphic solution of a similar portal with one more panel is given in Fig. 60, which see. It should be noted that all members are stressed in portals (b) and (d). Portal (c).—The stresses in portal (c) Fig. 58, may be obtained (1) by separating the portal into two separate portals with simple bracing, the stresses found by calculating the separate simple portals with a load = % FR being combined algebraically, to give the stresses in the portal; or (2) by assuming that the stresses are all taken by the system of bracing in which the diagonal ties are in tension. The latter method is the one usually employed and is the simpler. Maximum moment, shear, and stresses in the columns are given by the same formulas as in (a) Fig. 58. | Portal (e).—In portal (e) Fig. 58, the flanges G F and D C are assumed to take all the bending moment, and the lattice web bracing 112 STRESSES IN PoRTALS. is assumed to take all the shear. The maximum compression in the — upper flange G F occurs at F, and is ora kb D+ Ha 3) ha Beis The maximum tension in the upper flange G F is ‘ a — ge es The maximum stress in the lower flange D C is [Th ~— DC xt 4S ee (SS maximum tension occurring at C, and maximum compression occurring at dD. taken equally by the lattice members cut by a seta, as a a. | Maximum moment, shear, and stresses in the columns are an ; by the same formulas as in (a) Fig. 58. : Portal (f).—The maximum moment in the portal strut J Fi’ in () Fig. 58, occurs at H and G, and is ‘ Ma Hh he (56) The maximum direct stress in H G is + H, and inI H is | Hd : Be DOLL ee weet ees 7) aay a The maximum stress in G F is given by formula (49): The maximum shear in girder J F is equal to V. ‘The stress in G C is — V sec @ and in H D is + V sec @, asin (a) Fig. 58. “ Portal strut J F is designed as a girder to take the maximum moment, shear and direct stress. Maximum moment, shear, and stresses in the columns are given by the same formulas as in (a) Fig. 58. Graphic Solution—T'o make the solution of the stresses atndicallye determinate, replace the columns in the portal with trussed framework : GRAPHIC SOLUTION 113 as in Fig. 59. The stresses in the interior members are not affected by the change and will be correctly given by graphic resolution. x; Hd = 2000 G -2000 F +4000 £ R=2000 . J ‘ st Ih 2 / \ | at ° be a I \ bce ay Hd= \ > Cc ee 192000 in.|bs \ ! !o ‘ ii ‘1 Loy \ -! ‘on RAs Ps \ xO) ae oie. a} , yj i gras *t 1, i+ ae) se a 1 i] \ C= -—5 216-0 -<—e- S ’ i \ ' ! " ‘ 1 z - t e oon B-& H=1000 Wn ly aH = V= 3000 V=3000 Moment Shear Portal Oh Sy Se ge. eS, SOR a eran 6 i a A / ' ON roa 4 ’ % ‘us no CASE | ' \ he tae v2 : t \ Ce ee / t Columns Hinged \ ee ee / ! Stress Diagram ; Rare, SN : ' oo 1000 2000 3000 : \ / I \ 4 t t 1 l j ' \ J i '-z \zs aai8 3% by 4 Compression Tension ———- Fic. 59. As bef R 1 i et ee ee ’ S Having the calculated H, H1, V, and V1, the stresses are calculated by graphic resolution as follows: Beginning at the base of the column 4, lay off A-4 = V = 3000 lbs. acting downward, and A-a = H = 1000 Ibs. acting to the right. Then a-1 and 4-1 are the stresses in members a-I and 4-1, respectively, heavy lines indicating compression and light lines tension. At joint in auxiliary truss to right of C the stress in I-a is known and stresses in 1-2 and 2-a are found by closing the polygon. 114 STRESSES IN PorTALS The stresses in the remaining members are found in like manner, taking joints C, E, F, etc. in order, and finally checking up at the base of the column B. The full lines in the stress diagram represent stresses in the portal; the dotted lines represent stresses in the auxiliary members or stresses in members due to auxiliary members, and are of no con- sequence. ‘The shears and moments are shown in the diagram. b ~3500_ -500 43000 +5500_ F?=2000 > S 1743000 | / Sse / K - = _ QO" ------- Hd - 4 1 / 192000inIbs 13 ! f 5 B zs! ' S| / % b\ a of d ‘ / \ i] bined a an Sa/6'O meme 7 \ \onac-s-h a 24 =H=1000 4 Moment Shear vi CASE! Columns Hinged Stress Diagram © 1000 2000 3000 \ l i j /0 Compression ————. Tension —- Fic. 60. Simple Portal as a Three-Hinged Arch—In a simple portal the resultant reactions and the external load R meet in a point at the mid- dle of the top strut, and the portal then becomes a three-hinged arch cht wane CoLuMNS Fixep, ALGEBRAIC SOLUTION 115 (see Chapter XIII), provided there is a joint at that point (point b, Fig. 60). . | In Fig. 60 the reactions were calculated graphically and the stresses in the portal were calculated by graphic resolution. Full lines in the stress diagram represent required stresses in the members. Stresses 3-2 and 11-12 were determined by dropping verticals from points 3 and II to the load line 4-10. CASE II. STRESSES IN SIMPLE PORTALS: Columns Fixed.—The calculation of the stresses in a portal with columns fixed at. the base is similar to the calculation of stresses in a transverse bent . with columns fixed at the base. The point of contra-flexure is at the point _ &@4fa+a2h 41 an 3 ( 2a +h | ct) measured up from the hase of the column. The point of contra-flexure is usually taken at a point a distance £ above the bases of the columns, The stresses in a portal with columns fixed may be calculated by considering the columns hinged at the point of contra-flexure and solv- ing as in Case I. 7 Algebraic Solution —In Fig. 61 we have R pa — H' = 2 ee R(h— >) s Having found the reactions H and H', V and V1, the stresses in and = VY=—Vi= the members are found by taking moments as in (a) Fig. 58, consider- ing the columns as hinged at the point of contra-flexure. The shear diagram for the columns is as shown in (a) and the mo- ment diagram as in (c) Fig. 61. Anchorage of Columns.—In order that the columns be fixed, the anchorage of each column must be capable of developing a resisting moment greater than the overturning moment M = — ses , Shown in 116 | | STRESSES IN PORTALS (c) Fig. 61. The anchorage required on the wihdoeea side is a max “ ? imum and may be calculated as follows: Let T' be the tension in the : windward anchor bolt, 2a be the distance center to center of anchor = G -I!000 F +3000 E R=2000 Hd =|000 \ A S xia) 3 ee, | i= > ! C27] . go ® a |x ; hale sn Oo igor ce M tz 96000 in-Ibs: Me 8% 3] ok a + Oba : H’=1000_| | H=1000] Bl v'= 2000 v=z000 |” (wa 5s 1G ows oe TH=1000 LA wy Shear Portal 3 Columns Fixed | (a) (b) (Cc) (d) Fic. 61. the leeward anchor bolt we have 2Ta—(P—V)a + 72-0 rad, P—-V (58) 4a 2 i a If the nuts on the anchor bolts are not screwed down tight, there will be a tendency for the column to rotate about the leeward edge of ” the base plate, and both anchor bolts will resist overturning. 5: The maximum pressure on the masonry will occur under the leeward edge of the base plate a will s > Sve ae - = PRET TS Fe CoLuMNS FixeD, GRAPHIC SOLUTION 117 where W = direct stress in post; = area of base of column in sq. ins.; bending moment = % Hd; = one-half the length of the base plate; = moment of inertia of the base plate about an axis at right wo Ra | angles to the direction of the wind. Graphic Solution.—The stresses in the portal in Fig. 62 have been calculated by graphic resolution. This problem is solved in the same manner as the simple portal with hinged columns in Fig. 59. b Lo BEE She . WS ae SQ F KM AFI's “S > ! REA Mader 1 1 +1500 1 =1300 a, 108000 in-Ibs J ! 7] : 4 3) ‘ a! Ms a] fa 9 9 n, NY © H D u — WA < io} 1000 vy 7 1 V 1 ! 1 ‘ ' 7S=16-0 -——> ’ ! ' 1 ; ' ---f— ¥ ¥ Hd _ — 108000 in Ibs. ‘,H=1000 GA ~~ Moment Shear Portal = 19 is x i CASE 2 14 at a to. ri ya Columns Fixed ‘ \ 1 1 Y y) 1 piney ee / ' Stress Diagram ' \ batt Pee a 1 9 1000 2000 3000 1 1S 1 \ / t rn 1 J 8 ‘ tgs es ~ v at “4 RS ee eee & a Compression -—— Tension weenie 4 6 Fic. 62. STRESSES IN CONTINUOUS PORTALS. — The © portal with five bays shown in Fig. 63 will be considered. The columns will all be assumed alike and the deformation of the framework will be neglected. ‘The shears in the columns at the base will be equal, and will 118 STRESSES IN PORTALS R be Sieotig = To find the vertical reactions proceed as follows: Determine the center of gravity of the columns by taking moments. about the base of one of the columns. Now there will be tension in each one of the columns on the windward side and compression in each one of the columns on the leeward side of the center of gravity of the columns. The sum of the moments of the reactions must be equal to the moment ash. “F 1 ! ' ' H h : k- - -S5-~ - Ke -+S4--- 2 --S3--- ae ---Se-- > --S)-- >] I C.G.of Posts ' H H s H H ¥ V. Vs Ht Siz Vs a eae Ve w el a ds =<-<-3e= 5 He nan ee ‘plimcleas y meaeediee oe hao! de--------- BES = = ks me ane di ---------- > Fic. 63 of the external wind load, R. The reactions at the bases of the columns will vary as the distance from the center of gravity and their moments will vary as the square of the distance from the center of gravity. Now, if a equals the reaction of a column at a units distance from the center of gravity, we will have V, = —ad,,V, = —ad,, V, == —a@ d,, Vz=+adV;=-+ad;,andV,= +ad, and the moment M =a (d,’ + d,? + d,? + d? + d,? +d”) = Kh as dks Sd? Having found a, the vertical reactions may be found. Now having found the external forces H and V, the stresses can be calculated by either algebraic or graphic methods. Stresses in a Double Portal—To illustrate the general problem the stresses in a double portal are calculated by graphic resolution in — Fig. 64. In this case . ‘ ee, ee eee ee ee eee - p— DousLe PorTAL 11g and Pk 7 os FEF 25h 1b: The vertical reaction of the middle column is zero. By substitut- ing the dotted members as shown, the stresses can be calculated as in the case of the simple portal. The full lines represent stresses in the portal oy - members. ‘The shear in the columns is equal and is H below, and sis h--ad above the foot of the knee brace. b -2750 +/750 +1200 +5700 _ FR= 5000 er an PDH 11 AY WF AY IM 6 MY KR4 GY IN ' WY 10 “O ») 8 WY 5d) cite Bi pena tf +1500 N 7 ny “(500 SR ’ h ! > Be i wy | Ne . 3 AS ! be \ ou here air a. t : i .! nt i ! or 9 — | ; Jj fy = © ; \ } I) = --S= /6+0"1\--->«}- -S= 16:0" ----- ~ aio = xe) ! ! I = | 1 ' | = ‘| . Hy ! pete 1 2 U ' Hie 14-1000 Leg A Bete ; p Aly. Moment Shear V'2 2250 Double Portal V=2250 ee Bare Ge a et ie 4 (4 x) (64) The horizontal reactions at the crown are the same as at the sup- ports. Reactions for an inclined load may be found by substituting the proper moment arms. Calculation of Reactions: Graphic Method.—Let P, Fig. 66, be _ the resultant of all the loads on the left segment. Since there is no 122 STRESSES IN ‘T‘HREE-HINGED ARCH bending moment at hinge C, the line of action of the reaction R, must pass through the hinge at the crown. ‘This determines the direction of reaction R,, and since the three external forces P,, R, and P produce equilibrium in the structure they must meet in a point. Therefore to find the direction of R, produce B C to d and join d and A. Fic. 66. The values of R, and R, may then be obtained from the force poly- gon. | , 3 The reactions due to loads on the right segment may be found in the same manner. The two operations may be combined in one as il- lustrated in the solution of the dead load stresses in a three-hinged arch, Fig. 67. | 7 Calculation of Dead Load Stresses.—To find the reactions for the dead loads in Fig. 67, the loads are laid off on the load line of the force polygon in order, beginning at the left reaction A, and two equi- librium polygons, one for each segment, are drawn using the same force polygon. The vertical reactions at:the crown, P «> and at abut- ments, P, and P,, are found by drawing a line through pole o of the force polygon parallel to the closing lines of the equilibrium polygons. The load P g at the crown causes reactions R,1 and R,1, and combining ei DrapD LoAD STRESSES 123 reactions R,* and P, at A, and R,* and P, at B, we have the true reactions FR, and R,. ~=% DEAD LOAD Yj | ' Stress Diagram Force Polygon has el f 4 Pa ae ie 3 ' | 4 | oe 7 Pepe of | 7 6 ! 7 Sai i ro ae pis O SS a 8 ! rT Fig ee J \ 9 1 \ j aby lO 4 at hs s : ' ' I e ie 3% 4000 8000 ‘12000 Ps 13 | | ] 14 | aR Fic. 67. Having obtained the reactions, the stresses in the members are found in the same manner as in simple trusses. In Fig. 67 the stresses in the left segment are calculated by graphic resolution. The diagram 124 STRESSES IN THREE-HINGED ARCH is begun with the left reaction x-y = R,. Where the dead load is sym- metrical a stress diagram need only be drawn for one segment, — Rr tere ewer --=- Oo WIND LOAD STRESS DIAGRAM be p: FOR WINDWAR D SIDE ks | nm I7o a me! Z R, cee g P; 15 fy 7 2 Pe 18 IS . 13 14 lak. 8 9 0 1000 2000 3000 L | - 3 Fic. 68. WIND LoapD STRESSES 125 Calculation of Wind Load Stresses——The reactions for wind load in Fig. 68 are found as follows: The reactions P, and P, for the windward segment, considering it a simple truss supported at the hinges, are found by means of force =a. ~ ~ ~_~. —— Wc ak ue in ei ets ive pre e) WIND LOAD STRESS DIAGRAM FOR LEEWARD SIDE: 1000 2000 3000 4000 J ] l a | = er hae eee BP as Pe Ry ae SU Sag FAN re) A Se Wier ae ffm are. ln, Lyy 126 STRESSES IN ’[HREE-HINGED ARCH and equilibrium polygons. ‘The lines of action of P, and P, are D allel to each other and to the resultant R. The line of action of | right reaction, R,, must pass through the center hinge C, and the re: tion P, will be replaced by two reactions R, and ,* parallel to J and R,' in the arch respectively, and the force triangle will be closed by drawing R, in the force polygon. The intersection of force & and reactions R, and R, falls outside the limit of the diagram. : Having obtained the reactions, the stresses in the members u calculated in the same manner as in a simple truss. _ | The wind load stresses must be calculated in both the windy “ and leeward segments. The wind load stress diagram for the wind- ward segment is shown in Fig. 68, and for the leeward segment in Fig. — _ 69, compression being indicated in the stress diagrams by heavy lines — and tension by light lines. Both wind load stress diagrams and the de load stress diagram are usually constructed for the same segment of t arch. By comparing wind load stress diagrams in Fig. 68 and Fig. 6: it will be seen that there are many reversals in stress. The maximu stresses found by combining the dead, snow and wind load stresses a in the case of simple trusses and transverse bents, are used in designi the members. Tee I eT eee = * } CHAPTER XIV. STRESSES IN ‘T'WO-HINGED ARCH. Introduction.—A two-hinged arch is a frame-work or beam with hinged ends which has inclined reactions for vertical loads. The bot- tom chords of two-hinged arches are usually cambered, however, a simple truss becomes a two-hinged arch if the ends are fixed to the abutments so that deformation in the direction of the length of the truss is prevented. The horizontal components of the reactions may be supplied either by the abutments or by a tie connecting the hinges. Jn the latter case the deformation of the tie must be considered in determining the hori- zontal reactions. ‘Two-hinged arches are statically indeterminate struc- tures and their design is subject to the same uncertainties as continuous and swing bridges. | Two-hinged roof arches are rigid and economical, but have been used to a very limited extent on account of the difficulties experienced in their design. The methods outlined in this chapter are quite simple in principle, although they necessarily require quite extended calcula- tions. Two-hinged roof arches with open framework, only, will be considered in this chapter. CALCULATION OF STRESSES.—The vertical reactions in a two-hinged arch are the same as in a simple truss or a three-hinged arch having the same loads and span. The horizontal reactions, however, de- pend upon the deformation of the framework and cannot be determined by simple statics alone. Before the deformations can be calculated, the sizes of the members must be known, and conversely, before the sizes 128 STRESSES IN ‘T'WO-HINGED ARCH of the members can be calculated, the stresses which depend upon the deformations must be known. Any method for the calculation of the stresses in a two-hinged arch is, therefore, necessarily a method of successive approximations. With a skilled computer, however, it is rarely necessary to make more than two or three trials before obtain- ing satisfactory results in designing roof arches. Two-hinged bridge arches require somewhat more work to design than roof arches on : account of the greater number of conditions for maximum stresses in the members. Having determined the correct value of the horizontal thrust, H, the stresses in a two-hinged arch may be calculated by the ordinary algebraic or graphic methods used in the solution of the stresses in simple trusses. Calculation of the Reactions.—In Fig. 70 the vertical reactions, V, and V,, are the same as for a simple truss. The horizontal reactions, H, will be equal and will be the forces which would prevent change in length of span if the ends of the arch were free to move. The horizon- tal thrust, H, will therefore be the force which, applied at the roller end of a simple truss, will prevent deformation and make the truss a two- hinged arch. An expression for H may be determined as follows: In Fig. 70 assume that all members are rigid except the member 1-y, which is increased in length 8, under the action of the external load, W. The CALCULATION OF THE REACTIONS 129 movement of the truss A’ at the hinge L’, will then be due to the change in length, 8, oi the member 1-y. Let h* be the horizontal reaction necessary to bring L1, back to its original position, and let U h* be the stress in the member I-y due to the horizontal thrust h?. Now the internal work, % §h' U, in short- ening the member I-y to its original length will be equal to the external work, %4 h1 A’, required to bring the hinge L,? back to its original position, | RRA = Yth O and AL= 3-7 (65) but 6 = ee where P is the unit stress in the member 1-y due to the external load W, L, is the length of the member 1-y, and E£ is the mod- ulus of elasticity of the material of which the member is composed. Substituting this value of § in (65) we have a PUL (6) 1 where U is the stress in 1-y due to a load a =— unity at L*, Now if each one of the remaining members of the arch is assumed to be distorted in turn, the others meanwhile remaining rigid, the dis- tortion in each case at 1,1, will be represented by the general equation (66) and the total deformation, A, at 11, will be Y i © dr & E A= (67) Let P* h* be the unit stress in the member I-y due to a horizontal thrust h*, then by the same reasoning Ai =3U | (65a) 1 1 but ra fe ee. a PLO‘ an = Ez and the total deformation, A, will be 130 STRESSES IN "T'wo-HINGED ARCH fi PY Oe, PL Gs s= > E oe 1 er (68) Now for equilibrium, the values of a as given in equations (67) and (68) must be equal, and we have, after solving for H PoE H ‘ = az PP OL ns (69) which is an expression for computing the horizontal thrust in any two- hinged arch due to external loads. This formula holds for any system of loading as long as P is the unit stress due to that loading, U is the stress in the member and P" is the unit stress in the member due to a unit load acting at the point at which the deformation is desired, and parallel to the direction in which the deformation is to be measured. The method of finding the correct value of the horizontal reaction, H, is as follows: (1) calculate the stresses in the arch for the given loading on the assumption that it is a simple truss with one end sup- ported on frictionless rollers; (2) calculate the stresses in the arch for an assumed horizontal reaction, Ht =, say, 20000 lbs. on the assumption that it is a simple truss on frictionless rollers; (3) calculate the defor- mation, A, of the free end of the truss for the given loads by means of formula (67) ; (4) calculate the deformation, A’ of the free end of the truss for the assumed horizontal reaction Ht = 20000 lbs. by means of formula (68). The true value of H is then by formula (69) given by the proportion Ms HE aS A = ——— = _H*A _ 200004 (70) A’ A' The calculation of the horizontal reaction, H, and the stresses can be made by algebraic methods alone or by a combination of graphic and algebraic methods. ‘The first requires less work, while the second —- ALGEBRAIC CALCULATION OF REACTIONS 131 is probably easier to understand. The algebraic solution will be given first. Algebraic Calculation of Reactions.—In Table VII the values of the unit stress, P, in each member due to the external loads are given eae ys U, in each member due to a unit horizontal thrust are given in column Yogi 2 Ge & E the quantities in column 9 gives the total deformation, A = .956 in column 5; values of are given in column 6; values of the stress, 8; and values of are given in column 9. The algebraic sum of inches at the point where the unit horizontal thrust was applied meas- ured parallel to the line of action of the thrust. TABLE VII. Simple Truss with Vertical Loads l oot 2B 4 5 6l7 its fs Member|AreajLength|Stress |UnitStress} PL. |No-of} jy |PUL Sqin-jinches | Ibs. | P lbs. | E |Mem. E I-x. | 53 | 252 |[+60000|4 1/320 |t.095| 9 |-090|--086 2-X | 53 | 192 |t4/000\t 7740 |t-050| 6 |-0.80|--040 4-X | 53 | /80 |#67000|\4/2650 |+.076| / |-/45|--1/0 e'-x | 53 | 792 |#4/000\t 7740 |+050| 12 |-0-80|--040 '-X | 53 | 252 |#60000|4//320 |4-095| /5 |-0.90|--086 I-Y | 53 | 2/6 |-25000|- 4720 |--034| 8 |t/-60|--054 3-Y | 53 | 192 |-57000|-/0760 |--069| 4 |#2.05|--/4/ 3-Y | 53 | 192 |-57000|-/0760 |--069| /O0 |t205|--/4/ I-\y | 53 | 2/6 |-25000|- 4720 |--034| /4 |t/-60|--054 I-2 | 20 | /50 |-30000|-/5000 |--075| 7 |t0-75|--056 2-5 | 40 | 204 |#32000|+ 8000 |+.054| 5 |-0.45|--024 3-4 | 40,| /90 |-22000|- 5500 |--028| 2. |t080\--022 3-4 | 40°| /50 |-22000|- 5500 |--028| 3 |t0-80|--022 2-3 | 40 | 204 |432000\+ 8000 |+-054| // |-0-45|--024 -2' | 20 | 190 1\-30000|-/5000 |--075| 13 |t0.75|--056 Total Deformation = FUL = -956 In Table VIII similar values are given for the arch as a truss with an assumed horizontal reaction of Ht = 20000 lhs. The algebraic sum of the quantities in column 9 gives the total deformation, A’ = .574 inches at the point where the horizontal thrust was applied. 132 TABLE VIII. Simple Truss with H=20000 Ibs. | 2 5 4 5 6 y dae ee tees - IMember|Area |Length,L| Stress|Unit Stress PL |No.of v PUL Sain} inches| lbs. | P lbs. | E |Mem. F. I-X | 53 | 252 |-/86000 |-3400 |-:028| 9 |-0.90| .025 em 5.3 192 |-/6000\- 3000 |--0/9| 6 |-0.60| .O/5 4-X | 53 | /80 |-29000|- 5500 |--033| / |-145| .048 2'-X 535 /92 }\-16000 |- 3000 |--0/9| /2 |-080) .O/5 K 533 252 ~=|-/8000 |- 3400 |--028| 1/5 |-090| .025 I-Y | 53 | 2/6 |#32000|#6000 |t-043|" 8 |t/60} .069 3-Y | 53 | /92 |h41000|4+7800 |t.050| 4 |#205| .102 ZY | 53 | /92 |t4l000 |#7800 |4+050| 10 |\t205| ./02 cag IZ 2/6 |\t3Z2000\|t6000 |4+043| 1/4 |t/60| -069 I-2 | 20 | £50 |#/5000|\# 7500 |4-038| 7 |+075| .029 2-3 | 40 | 204 |- 9000 |- 2250 |-.0/5| 5 |-045| .007 3-4 | 40 | 400 |t/6000|44000 |t.020| 2 |#0.80| -0/6 3'4 | 40 | 150 |+/6000|44000 |t.020| 3 |#0.80| -016 2-3 | 40 | 204 |-9000 |-2250 |-.0/5| /1 |-0.45| -007 '-2' | 201 450 14/5000 '\+7500 \4.038' 13 140751 .029 Total Deformation= see = Ife TABLE IX. Simple Truss with Dead and Wind Loads ] 2 3 4 5 6 Tees Member] Area|LengthL|Stress |UnitStress| PL |No.of} |) |PUL Sq.in-| inches] lbs. | Plbs. | E |Mem. E I-X | 53 | 252 |487000|+/6400 |\4/38| 9 |-090|-/24 2-X | 53 |./92 |t72000|4/3600 |\t.087| 6 |-080|-.069 4-X |53 | /80 |#95000|\4/7800 |4/07| 1 |-1.45|-.155 2'-x |53 |-/92 |#52000|\4 9800 |4.063| /2 |-080|--050 ('-X | 53 | 252 |#72000\+/3600 |+//4| 15 |-090)--/03 l=+¥ | 53 | 2/6 |-58000|-/0900 |-.078| 8 |H-60}-/25 | 3-Y | 93 | 192 |-87000|-/6400 |-/05| 4 |#205|-2/5 3'Y | 53 | /92 |-74000|-/4000 |-090| /0 |t2-05|-./85 I"y | 53 | 2/6 |-30000|- 5650 |-04! 4 tl60|-.064- I-2 2.0 | /50 |-36000|-/8000 |-.090 +0,75|2067. 2-3 | 40 | 204 |#28000|}4+7000 |t.048| 5 |-0.45|-022 3-4. |40.| /90 |-22000\- 5500 |-028| 2 |#0.80\-.022 “4 |40 | /50 |-42000\-/0500 |-053| 3 |#080|-042 2-3' | 40 |.204 |#46000|t//500 |4078| // |-0.45|-035 '-2' 120 1 450 \-42000\-2/000 ‘\-105\ 13 \4075\-.079 Total Deforrnation=> cur = (557 STRESSES IN ’[‘Wo-HINGED ARCH - ww. — oe ee ee ey Te GRAPHIC CALCULATION OF REACTIONS 133 The correct value of H is given by the proportion Dg So Gena ak a _ 20000 x .956 H = 574 = 33400 lbs. In Table IX the deformation, A, for the same arch considered as a simple truss and acted upon by dead and wind loads is 1.357 inches, and 20000 & 1.357 574 = 47300 Ibs. HA = Graphic Calculation of Reactions.—In the graphic solution of the horizontal reactions the total amount of the deformations, A and A are found by means of deformation diagrams. Before con- structing the deformation diagrams the quantities in the first seven columns in Tables VII and VIII or VIII and IX must be 5*0'-——» t ' | | K-—2 Simple Truss Vertical Loads. (Q) (b) ic H=20000 Ibs. Simple Truss | aioe Ibs. Stress Diagram , H=20000 Ibsi (d) 134 STRESSES IN T'WO-HINGED ARCH. calculated. ‘The stresses given in column 4 are calculated as shown in Fig. 71. Column 6, giving deformations of each member, and col- umn 7, giving the order in which these deformations are used, are, how- ever, the only values used in constructing the deformation diagrams. Deformation Diagram.—The principle upon which the construc- tion of the deformation diagram is based is as follows: ‘Take the two members a-c and c-b in (d) Fig. 72, meeting at the point c. Assume that a-c is shortened and b-c is lengthened the amounts indicated. It is required to find the new position; c’, of the point c. With center at a and a radius equal to the new length of a-c = a-c’, describe an arc. The new position of c must be some place on this arc. Then with a center at b and a radius equal to the new length of b-c = b-c’, describe an arc cutting the first are in c’. The new position of c must be some place on this arc and will therefore be at the intersection of the two arcs, c’. Since the deformations of the members are always very small as compared with the lengths of the members, the arcs may be replaced - by perpendiculars, and the members themselves need not be drawn, (e) Fig. 72. ; To draw the deformation diagram, (b) Fig. 72, for the arch as a truss with one end on frictionless rollers and loaded with vertical loads, proceed as follows: Begin with the member marked 1, lay off its deformation = ++ .076 inches (Table VII., column 6) to ‘scale and parallel to member 1. Now lay off the deformation of 2 = — .028 inches away, from the joint U, and parallel to the member 2, and lay off deformation of 3 = — .028 inches, away from the joint U’, and parallel to the member 3. Perpendiculars erected at the ends of de- formations 2 and 3 will meet in the new position of L,. The vertical distance between the deformation 1 and point L, represents to scale the change in position of L, relative to the member U, U1,. At L, in the deformation diagram lay off deformation of 4 = — .069 inches, away from the joint and parallel to the member 4, and at U, lay off deforma- tion of 5 = + 054. inches, toward the joint and parallel to the member 5. The perpendiculars erected at the ends of the deformations 4 and 5 DEFORMATION DIAGRAM 135 Deformation Diagram for , Simple Truss te HH = 20000Ibs. F A Ss os 12\ = 4 be / \ Tees / 7 4 \2 yr. 58 4 (C) b (b) Ae eae ey Pe fi i uy ‘ ° , yon ; \ i : f EN Upe? *, a . Ras i Re ae / 195 C2 " é Deformation Diagram /_-><<_\ | Pe ea Me ‘ 4 or \ 4S aA 7 en Wace * , r \ 1 ; / ‘ Spee. ay f ey. Mts painter == ‘ Vertical Loads S10} ‘ Fic. 72. i. determine the new position of joint L, relative to the other points. In ___ like manner perpendiculars erected at the ends of deformations 6 and . 7 determine U,, and finally perpendiculars erected at.the ends of ‘deformations 8 and g determine J,,. The deformation diagram for the 4 right half of the truss is constructed in the same manner. The horizon- tal line joining L, and 1}, represents to scale the movement of the joint L},. ; In drawing the deformation diagram it is immaterial whether plus : deformations are laid off toward the joints and minus deformations ; away from the joints as was done in the preceding problem, or the E reverse. The former method is more common, but the latter is prob- a e s ° D ® ' ably more consistent. The deformation diagram (b) if drawn in the latter way would be turned upside down and inside out. Calculation of Dead Load Stresses in Arch.—In Fig. 71, (bd) is the stress diagram for the arch as a simple truss with vertical loads as shown in (a) ; and (d) is the stress diagram for the arch as a simple 136 SrrESsES IN T'wo-HINGED ARCH truss with a horizontal thrust, Ht, of 20000 lbs. as shown in (c). The quantities for calculating the deformations of the simple truss with vertical loads are given in Table VII, and the deformation diagram is shown in (b) Fig. 72. The quantities for calculating the deforma- tions of the simple truss with a horizontal thrust of 20000, lbs. are given in Table VIII, and the deformation diagram is shown in (c) Fig. 72. The true value of H is found by the proportion H+ 20000: 3.956" tg om H = 33400 lbs. The stress diagram for the two-hinged arch with V = V’ = 42000 Ibs., and H = H = 33400 lbs. is shown in (b) Fig. 73. . The difference in the stresses in the members of a simple truss and a two-hinged arch may be seen by comparing stress diagram (b) Fig. 71, and stress diagram (b) Fig. 73, both diagrams being drawn we es Yat a sca ironies > Xi Two Hinged Arch (Q) Stress Diagram Two Hinged Arch (b) Fic. 73. to the same scale. The stresses in the arch may be found from the stresses given in Tables VII and VIII by adding the stresses in column ao Table? VIL- 46 the corresponding stresses in column 4, Table VIII, multiplied by 1.67, the ratio between the actual and as- DraD AND WIND LOAD STRESSES 137 sumed horizontal reactions. For example, the stress in 1-x% in the arch equals ++ 60000 — 18000 x 1.67 = + 29800 lbs. Stress in 1-y equals — 25000 ++ 32000 x 1.67 = + 28440 lbs. Dead and Wind Load Stresses in Arch.—In Fig. 74, (b) is the stress diagram for the arch as a simple truss loaded with dead and wind loads as shown in (a). ‘Table IX gives the same data for this case as Dead and Wind Load Stress Diagram (b) x 1 ' 1 Sirnple Truss . ! Dead and Wind Loads 4p ne Wee Ee imple Truss Sy ' Simple Tru Ps! L ‘Simple Truss | H =20000 Ibs. Stress Diagram , H=c0000 Ibs. (c) (d) Fic. 74. are given in Table VII for the simple truss with vertical loads. The deformation diagram for the deformations given in column 6, Table IX, is shown in (b) Fig. 75. In drawing the deformation diagram for this case the member marked 1 was assumed to be fixed in position and the other members were assumed free to move. The horizontal dis- tance between L, and L’, will be the total deformation required. 42\ \ 16 is ‘5 Bas aE a y \ if e.° ty 4 \ / Urs KU, fe 75" Yessy 4 at re pe was Seer 0.574 inches------ ahha ails ik Diagram or Simple Truss H=20000 Ibs. (Cc) given in (c) Fig. 73: The true value of H is found by the proportion i] :-20000 2° 35357 2 7e7e = 47300 lbs. | loads and a horizontal thrust, H, of 47300 Ibs. is given in (b) Fig. ag The stresses in the arch for this case may be found from the stresses in Tables IX and VIII by adding the stresses in column 4, Table IX, to the corresponding stresses in column 4, Table VIII, multiplied by 2.865, the ratio between the actual and assumed horizontal reactions. “a es ee A Aap a | fe i ae ey Kea Arcu Write Horizontay, Tie “Two Hinged Arch (a) \ re) ~20000 40000 8 Ll i 1 l j bet Stress Diagram Two Hinged Arch (b) Fic. 76. Be As a check on the accuracy of the calculations the movement at L,” in the arch was calculated in Table X and was found to be zero as it should be. Arch With Horizontal Tie.—If a horizontal tie is used the final deformation of the arch will be equal to the deformation of the tie. TABLE X. Two Hinged Arch with Dead and Wind Loads | 2 Se 4) 9 6 ro) oS Member|Area|Length,L| Stress |UnitStress; PL. | yy |PUL Sain-|inches | lbs. | P Ibs. | E ape I-X 53 252 \t45500| + 8220 |t.069|-090 |-.063 toa A take Ne, 192 \434200\| +6450 |t0A4l/ |- 0.60|--033 A-X 53 /80 \t2@6500| +5000 |t.030|\-/-45|--045 2X | 535 | 192 |tH4200| #2660 |t.0/7|-0-80|-.0/4 1-X ae. 252 \t29500| 9550 |\t.047\-0.90|--042 ty wo 2/6 =|#/8000| 3400 |t.024)t /.60|4.038 ony: 5:5 192 }\#/0000)\ #/890 |#.0/2 |\t2:05|t.025 i a 3:3 192 \t2@5000| 4350 |t.028 |t205|t.057 Py 53 2/6 = |t46000| #8700 |4.062\t/-60 |t.099) - ie 2:0 130 |- 300\- 250 |-.001 |t0.75 |--00/ ES 4:0 204. |t 6500|¢4/625 |tOll |-0.45|--005 3-4 40 150 |#/S800|45950 |t.020|t0.80|t-.0/6 5-4 4.0 /50 -4200|-/050 |-.005|+0-80|\-.004 ao 4.0 204 |t22800|457Q0 |4.059|-0.45|--0/8 I-2' | 20 | 150 \-5000\-2500 1-.0/3\40.75\-.0/0 Total Deformation => Ul = .000 140 STRESSES IN T'Ww0-HINGED ARCH Assume that the joints L, and L,’ in (a) Fig. 73 are connected by a tie having 3 sq. in. cross-section. A force of 1000 lbs. will stretch the tie , 1000 Kae = 35229,000,000 = -0083 inches. The movement for 1000 lbs. applied as H is equal to =a = .0287 inches. ‘The value of H therefore which will produce equilibrium for the arch with vertical loads will be + .0083 H + .0287 H = .956 x 1000 lbs. _ .956 X 1000 ‘i 0370 = 25840 lbs. The stresses in the arch for this case may be found from the stresses in Table VII and Table VIII as previously described. Temperature Stresses——Where a horizontal tie is used and all parts of the structure are exposed to the same conditions and range of temperature, the entire arch will contract and expand freely and tem- perature stresses will not enter into the calculations. Where the tie is protected and where rigid abutments are used the temperature stresses must receive careful attention. The deformation 4’ due to a uniform change of temperature of ¢ degrees Fahr. when the arch is assumed to be a truss supported on frictionless rollers, will be etl, where e is the coefficient of expansion of steel per degree Fahr. = .00000665 ; ¢ equals change in temperature in degrees Fahr.; and L equals the length of the span. For a change of 75 degrees Fahr. from the mean, the deformation will be A’ = + .00000665 K 75 L L oe Pee 1 2000 7) * d he , ‘a an wet ee See Oe Ee ee Sy oe Le ee Ee ETE ye He Me ye Cee eee DesIcN oF ‘T'wo-HINGED ARCH 141 For the arch in Fig. 73 720" A’'=+ 2000 = + .36 inches This will be equivalent to a change in H of + fost x 20000 = + 12540 lbs. The stresses due to temperature in the two-hinged arch -will be equal to the stresses in column 4, Table VIII, multiplied by + .627. ‘The maximum stresses due to external loads and temperature will be found by adding algebraically the temperature stresses to the stresses due to the external loads. If the arch is not erected at a mean temperature this fact must be taken into account in setting the pedestals. Design of Two-Hinged Arch.—In designing a two-hinged roof arch proceed as follows: (1) With one end free to move, calculate the stresses in the arch as a simple truss; (2) with an assumed horizon- tal reaction, H*, of, say, 20000 lbs., calculate the stresses in the arch as a simple truss; (3) calculate the stresses in the arch for some assumed value of H, this value of H may be guessed at or often may be estimated with considerable accuracy; (4) design the members for approximate stresses in the arch; (5) calculate the deformation of the arch as a truss for the approximate sections and stresses; (6) calculate the deforma- tion of the arch as a truss for an assumed horizontal reaction of 20000 Ibs.; (7) determine a more accurate value of H from the deformations as previously described; (8) recalculate the stfesses in the arch, re- design the members, recalculate the deformations, recalculate a new value of H, etc., until satisfactory sections are obtained. The second approximation is usually sufficient. Corrections for horizontal tie and temperature should be applied in making the approximations. The gross area of the sections of all members should be used in determining _ the deformation of the members. If riveted tension members are much weakened, a somewhat smaller value of E, say, 26,000,000, may be used than the 29,000,000 commonly used for the compression members. The method just described is much more expeditious than the usual method of designing the members for the stresses found by con- CHAPTER XV. CoMBINED AND ECCENTRIC STRESSES. Combined Direct and Cross Bending Stresses. — Thus far members of trusses and frameworks have been considered as acted on by direct forces, parallel to the axis of the member. While this is the more common case, it often becomes necessary to design members which support loads as in (a), (b), (c), or (d), Fig. 77, or in which the line of action of the external force does not coincide with the neutral axis of the member, (e), (f), (g), or (h), Fig. 77. 7 Pa oe onion a a x < ---¢-- U | x TE SSE OES ata a tase Se (h) 144 ComBINED AND EccENTRIC STRESSES The following nomenclature will be used: Let P = total direct loading on member in pounds; J = length of member in inches ; I, = length of member in feet; I = moment of inertia of member; y, = distance from neutral axis to remote fibre on side for— which stress is desired ; E = modulus of elasticity of the material ; hae e = eccentricity of P, i.e. distance from line of action of ; P to neutral axis of member in inches; _ | v = deflection of member in inches; A = area of member in square inches ; f, = fibre stress due to cross bending; Be = direct fibre stress; total bending moment ; on art J moment due to deflection, = P v; to S58 | a oO ie) =] es =) - eee. 2) 3 i) =} c Au S @ ct ) oO ~*~ *: =] 2. =F S n a Ta a a4 s es ee V4 W lin (a) and (b) ; % w in (c) and (d); a in (e), (£), (g) and (h) Fig. 77. oa: Now | M=M, 22M, = Pos Mya! 1 But tes 2 chy, in which c is a constant depending upon the condition of the ends, and the manner in which the beam is loaded. Substituting this value of v in (72) we have Pf, P beet eh —10 4 WM =42e cEy, * 1 V1 i, ee ree Pe Sy te COMPRESSION AND Cross BENDING 145 and reducing, the stress due to cross bending is frp pe (73) where the minus sign is to be used when P is compression and the plus sign is to be used when FP is tension. The factor c may be taken equal to 10 for columns, beams and bars with hinged ends as in Fig. 77; equal to 24 where one end is hinged and the other end is fixed; and equal to 32 where both ends are fixed. The total stress in the member due to direct stress and cross bend- ing will therefore be for columns with hinged ends M,y ts hth~ pat (74) yg! BR asian » Details and Lacing-26% =_328 » Total weight,W,=1585 ” Bending Moment due to weight of the rnember,M =§WIsin6® Stress due to weight, fw= tr te wl open TOE ~ 10E Stress due to weight in upper fibre #X1I585 X 358 X-633 X 5-81 = +1100 Ibs. (compression) 2 199.8 - 25300 X 358 10x 28000000 Stress due to weight in lower fibre fw = £44 x 1100 = -1860 Ibs.(tension) 8! Sey a? a ah th ee ent YP ete” re Ds a catcls Y a eS aad +i at ae ie

IN an R =6400 RR=6400 9g) ° aah al fh a. =o g ws ° SFO -! —/ = LY) SX b'e3200 FB! je — 5 =|5'-4" —+% c fi H=32 ee a V=8520 =8520 H'=32 H=3200]) 4 E = ia 12500 Vv= 2500 Portal,Columns Hinged. Portal,Columns Fixed. Windward Pedestal, (a) (b) (Cc) 147 148 CoMBINED AND ECCENTRIC STRESSES Before calculating the stress due to wind moment, it will be necessary to de- termine whether the end post is fixed or hinged. If the end post is fixed,the negative moment developed at the lower pin will be M= Hd = 3200x228 _ 56/600 in Ibs. In order to obtain this condition of fixidity, the stress in the member must develop a resisting moment equal in amount. Therefore the post may be considered fixed if -V H (esnce jaz ua or (28300-8520 ) 8 2 361600 but 3547120 < 561600 and the énd post will not be fixed. (While this is the usual solution ,the resisting moment certainly reduces the - bending moment and the bending stress is less than computed below-) The maximum moment will then occur at the foot of the portal knee brace and will be M =3200 X226 =723200 in Ibs- Stress due to wind moment isa maximum in the [eeward post and is fy = M yi = 723200-X7 Wyo Pl? ~ p75, (95300412500) 358° lOE : 10 X 28000000 fw =+ 22480 Ibs. Stress, fw,is compression on the windward side and tension on the lee- ward side of the member. pression on the windward and tension on the leeward fibres. ‘The maximum fibre stress will come at the foot of the knee brace either on the upper or lower fibres on the windward side of the post, depending upon whether the stress due to weight is greater than the stress due to eccentric loading, or the reverse. In this case the maximum stress comes on the lower fibres of the windward side of the post. Combined Tension and Cross Bending.—The stress due to cross bending when the member is also subjected to direct tension is given by the formula Ai iy eee (75) the nomenclature being the same as in (74). The constant c is taken equal to 10 where the ends are hinged. een, sa Pe ee ee | SS ee Stress IN Bars DUE to WEIGHT 149 ‘Stress in a Bar Due to its Own Weight :— Let b = breadth of bar in inches; h = depth of bar in inches; w = weight of bar per lineal inch = 0.28 DN’; Si f= —— = direct unit stress. bh We will also have ¥, = “Uh; M,=% wW Ps P =f, bh. Substituting in (75) we have Lond A= FR, fybhl? 12. * 10X 28,000,000 Jf 4,900,000 % | Fa + 23,000,000 ( ey ) where f, is the extreme fibre stress in the bar due to weight, and is tension in lower fibre and compression in upper fibre. If the bar is inclined, the stress obtained by formula (76) must be multiplied by the sine of the angle that the bar makes with a vertical line. | | Formula (76) is much more convenient for actual use than for- mula (75). } Diagram for Stress in Bars Due to Their Own Weight.—Tak- ing the reciprocal of (76) we have : Bp 23,000,000 (5)? — 7, 4,900,000% * —4000,000 ==. 4, Ve 1 (7 = : cranes ° 150 COMBINED AND ECCENTRIC STRESSES Fig. 78 gives values of y, for different values of f,, and values of y, for different values of the length in feet, L. The values of y, and y, can be read off the diagram directly for any value of h, f,, and LL. And then, if the sum of y, and y, be taken on the lower part of the 20 : SDSS FORMULA K NIN SK Z 4.900,000h ‘ eS 5 ROSS Aq" ez300n000 ADRS R |__ fe _2300q000(eh NO vay f 4900000h 4900,000h NES AIA AIA = i+ Ye 10 N N N YA y' % = NICD FA 7 Y+ % 9 Z he f oN, N wa Where ae 8 SoS S 7 | f=extreme fibre stress es j x a tN * 4 Vi due to weight. c Nc. oa KA b=direct fibre stress Asi g AN, LAMA V1 Ah=depth of bar,inches S 2 G4Z04z8 /=length of bar, inches} = 5 KY AV Alel2h 5000 —_— 4 > Aid 5, ~ 4 4000 Vv) N . ’ ee ne VIS 3 3 es 4 sa S< 3000 fe] ‘ 4h xX ZZ “ 4 2 : RATA ¢ Ay 6 S Z KAAS AYP] y 2500 = Ho MATZ wave Zils YW) oa ZASAZ TS KEK ss 2 TAZ Ae MIA AL ATS 5 2000 m 7 y 7 eo Vhiyps ZL yeh Al A + ay YS So vy 7% ANZ ANY . a s . 15 ie VIM LANL AL Ae RX Ia Oa" 0 SAV LAM LL ALL GY WAAL KLIS 150 ”) fi GUvaZeweaasa Cava 2 Tag OO ain c: VAAL ELLE AA Vi OIL INS “has AAV OAWONSSASTAVABOKA 4 N ean AV 6200 OLS AV AVAVA 2 1000 -— * 0.9 4 G64 VAW. Pe ZL vii y Y 900 > Los VAAL ALALIE N ™ ! V4 ¥, “A ¥, wa Va y \ 800 VILL 0.6/- Y VY VY a “a rg S4 4 a 600 HAMMAM 05 4 2 500 | 1.5 2 3 4 5 6°78 3'-8 15 20 lall.Depth of Bar in Inches Ill. %¥+Y2 in Tens of Thousandths Fic. 78. DIAGRAM FOR FINDING STRESS IN BARS DUE TO THEIR OWN WEIGHT. | IIL. ExtremeFibre Stress, f, due to Weight of Bar, Lbs. per Sq.ln. ll el ee ee Se, ee ee ee eg mii anasto s i ide tet dll = re, i 4 » 2 a “ e me}: vy, a. i i ee, © ee ee - -- a , 7. en ens a a ok a DIAGRAM FOR FINDING STRESS IN Bars 151 _ diagram, the reciprocal, which is the fibre stress f1, may be read off the right hand side. The use of the diagram will be illustrated by two problems: Problem 1. Required the stress in a 4” x 1” eye-bar, 20 feet long, which has a direct tension of 56,000 lbs. In this case, h = 4”, L = 20 it., and f, = 14,000 lbs. per sq. in. The stress due to. weight, tes is found as follows: On the bottom of the diagram find h = 4 inches, follow up the vertical line to its inter- section with inclined line marked L, — 20, and then follow the horizontal line passing through the point of intersection out to the left margin and find y, = 3.3 tens of thousandths; then follow the vertical line h = 4 inches, up to its intersection with inclined line marked f, = 14,000, and then follow the horizontal line passing through the point of intersection out to the left margin and find y, = 7.2 tens of thousandths. Now to find the reciprocal of y, + y. = 7.2 + 3.3 = 10.5, find value of y, + y. = 10.5 on lower edge of diagram, follow vertical line to its intersection with inclined line marked “Line of Reciprocals” and find stress f, by following horizontal line to right hand margin to be f, = 950 lbs. per sq. in. By substituting in (76) and-solving we get f, = 960 lbs. per sq. in. Problem 2. Required the stress in a 5” x 34” eye-bar, 30 feet long, which has a direct tension of 60,000 Ibs., and is inclined so that it makes an angle of 45° with a vertical line. In this case, h = 5”, L = 30 feet, f, = 16,000 lbs., and 9. = 45°. From the diagram as in Problem 1, y, = 1.8 tens of thousandths, and 4, = 6.5 tens of thousandths, and dy = 555: sino = 1200 x sin © = 850 lbs. per sq. in. Relations between h, f,, f, and L. For any values of f, and L, f, will be a maximum for that value of h which will make y, + y, a min- 152 CoMBINED AND ECCENTRIC STRESSES imum. ‘This value of h will now be determined. Differentiating equa- tion (76) with reference to f, and h, we have after solving for h- after placing the first derivation equal to zero Z ~ A= Bs | (78) in which h is the depth of bar which will have a maximum fibre stress for any given values of / and f,. Now if we substitute the value of h in (78) back in equation (76), we find that f, will be a maximum when y, = 4p. Now in the diagram the values of y, and y, for any given values” of f, and L, will be equal for the depth of bar, h, corresponding to the intersection of the f, and JL lines. It is therefore seen that every intersection of the inclined f, and L lines in the diagram, has for an abscissa a value of h, which will have a maximum fibre stress f,, for the given values of f, and L. For example, for L = 30 feet and f, = 12,000 lbs. we find h = 8.3 inches and /, = 1700 lbs. For the given length LZ and direct fibre stress f,, a bar deeper or shallower than 8.3 inches will give a smaller value of f, than 1700 lbs. 3 Eccentric Riveted Connections.——The actual shearing stresses . in riveted connections are often -very much in excess of the direct shearing stresses. This will be illustrated by the calculation of the shearing stresses in the rivets in the standard connection shown in Fig. 79 and Fig. &o. The eccentric force, P, may be replaced by a direct force, P, acting | through the center of gravity of the rivets and parallel to its original direction, and a couple with a moment M = P x 3” = 60,000 inch-lbs. Fach rivet in the connection- will then take a direct shear equal to P divided by u, where is the total number of rivets in the connection, and a shear due to bending moment M. | The shear in any rivet due to moment will vary as the distance, Me Se rt Ee a en ae ee ae FeO ee ee ee =e * . ] r 1 r, ; = 4 9 é ," : Per ewe eee ae ee ee ee Eccentric RivETED CONNECTIONS 153 and the resisting moment exerted by each rivet will vary as the square of the distance of the rivet from the center of gravity of all the rivets. Now, if a is taken as the resultant shear due to bending moment in a rivet at a unit’s distance from the center of gravity, we will have the relation M=a (d’?+ 42+ 42+ d,? + d,’) =asd? . a ae (7) The remainder of the calculations are shown in Fig. 79. The re- sultant shears on the rivets are given in the last column of the table Direct Shear S = 20000+5 = 4000 Ibs. Moment = 20000.X3= 60000 in: Ibs: = ald? +dé+df +df+d% ) Where a=Moment shear on rivet 3 = 2630 Ibs.- Rivet} cd | d* |Moment| M S R 2.70 | 7.29}19185 | 7100 | 4000} 9300 1.90} 5.61} 9500 | 5000 | 4000} 3200 1.00} 1-00] 26350 | 2630 | 4000] 66350 1-90 | 3.61 | 9500} 5000 | 4000} 3200 2.10 |} 7.29 19185 | F100 | 4000} 9300 axd?= 22.80a=60000 in.lbs.|20000 a= 2630 |bs.=moment shear on rivet 3 M = Shear due to Moment. S = Shear due fo Direct Load,P R = Resultant Shear. . oO 4000 8000 12000 t ! n J Oban — Fic. 79. and are much larger than would be expected. The force and equilibrium polygons for the resultant shears and load P, drawn in Fig. 80, close, which shows that the connection is in equilibrium, - 154 CoMBINED AND ECCENTRIC STRESSES 1” 1@ le Standard Connection 2is6"x4"x5" P= 20000 Equilibrium Polygon -* | Fic. 80. STRESSES IN PINS.—A pin under ordinary conditions is a short beam and must be designed (1) for bending, (2) for shear, and (3) for bearing. If a pin becomes bent the distribution of the loads and the calculation of the stresses are very uncertain. The cross bending stress, S, is found by means of the fundamental } Cc : ; formula for flexure, S = ae where the maximum bending moment M, is found as explained later; J is the moment of inertia, and c is the radius of a solid or hollow pin. : The safe shearing stresses given in standard specifications are for a uniform distribution of the shear over the entire cross section, and the actual unit shearing stress to be used in designing will be equal to the maximum shear divided by the area of the cross section of the pin. The bearing stress is found by dividing the stress in the member —__ by the bearing area of the pin, found by multiplying the thickness of a the bearing plates by the diameter of the pin. es ny « ip Pl.2a'k she : Ss we Pls.\ RCS 2LS.4"x3"x% “ity? u, 2s: Rees : QY © ee NO @ Us STRESSES IN PINS 155 “1 mm > Vert. Shear Diagram: Maximum Shear =165400* =18000* (page : 309 Cambria) =381700" Maximum Shear=165400* Actual Fibre Stress=5750* Allowable" " =9000% »~ + ° AY u Oy oh wu n> Toegege F oT ik Teta Ra, att xg C1) 4, @ e U nee aa 14 5! Ak iC, c "a Membe fe rv +1] a Pl. . . es 2 : aa . Pin Packing. (b) a ° ce) Tt Ke 1986007 & 452k - 1.83" —ke— 123% alae! ‘Al braic M ' “ 5 ‘Algebraic Moments, ¥ t Soe Saga | 4 Pe << sts , cl ‘Horizontal Components. oO o A =< re} fe) Hf _O 0 l i | Moments at oO Fk | 2 Is © nT — 1 02" o ° 13132 (d)3 | 5) 0 cu ro o 29 co! ©) wl 2=165400x.53-87660". - . lal8 9 O} 21 sand4=165400%236 _, o S13 S uy 5 ~99300X1.85=208600" V¥Lx 0 oe OF t-——|\Vertical Components, U,U2330600 516 3 8 ‘Moments at Force Diagram.Stresses UU. 5:07" es 7=126300X1.83=231100. Total Moment at4ands6. e=i26300x3.06 =/208600 *+ 283000* -64100x1 23+283000"" Hor. Mom. Polygon. « | *351600.*" (9) _Fn 4 II | O Olli a Q = AT l A i Force Polygon, »7 fF F 453k — 183"= 4 1.23"44 18" Hor. Comp.,~ ies. te : ~ Ve ge e}> 4 Graphic Moments. Moments. “© 18 i iS |e r -| @V Horizontal Components. , >| ! 2 13 2 (e) 3 =! 5] Max. Hor Moment,at4, 7 Pe LSA > Q D1 7.04% 200000 hae: cs t-200000%- ~ —a, | ils 9° =| =208000*" f-Eavt i ie ists $ “| 3 rs 5 | rede 3 tL O! Nertical Sibert i= Chiy 5 t6 17 iB ‘Components. sigh AP Stk SRE Sea d rea 4 | ‘Max. Vert. Mom. Ores —--—-------- EF l NTTTTTT]TT"] 4] Hat e1.42x200000 oS aaah =] lilo Ne | =284000,*' Ms ean Sy | Flt [+ — H=800000 * ae 45 206000%" Pics mrt i x force Polygon! 2 1 Vert. Comp. if | l = rk 14 | 9 Design of Fin. : iHor. Shear eee | < Maximum Bending Mom. =351600*" > 64 | Z Allowable Bending Mom. “w! | zal \_y _ fore'pin,using fibre stress i] | Fic. 80b. CALCULATION OF STRESSES IN A PIN. 156 COMBINED AND ECCENTRIC STRESSES Calculation of Stresses.—The method of calculation will be il- lustrated by calculating the stresses in the pin at U, in (a) Fig. 80b. In the complete investigation of the pin U,, it would be necessary to calculate the stresses when the stress in U, U, was a maximum, and when the stress in U, L, was a maximum. Only the case where the stress in U, U, is a maximum will be considered. However, maximum stresses in pins sometimes occur when the stress in U, L, is a maximum, and this case should be considered in practice. Bending Moment.—The stresses in the members are shown in (c), which gives the force polygon for the forces. The makeup of the ~ members is shown in (a), and the pin packing on one side is shown in (b). The stresses shown in (c) are applied one-half on each side of the member, the pin acting like a simple beam. The stresses are assumed as applied at the centers of the members. Algebraic Method.—The amounts of the forces and the distances between their points of application as calculated from (b) are shown in (d). The horizontal and vertical components of the forces are considered separately, the maximum horizontal bending moment and the maximum vertical bending moment are calculated for the same point, and the resultant moment is then found by means of the force triangle. In (d) the horizontal bending moments are calculated about the ~ points 1, 2, 3, 4; the maximum horizontal moment is to the right of 3, and is 208,600 lIb.-in. The vertical bending moments are calculated about points 5, 6, 7, 8; the maximum vertical bending moment is to the right of 8, and is 283,000 Ib.-in. The maximum bending moment is at and to the right of 4 and 8, and is 208,600? + 283,000? = 351,600 Ib.-in. 3 ene: Me Substituting in the formula S ears the maximum bending stress is S = 16,600 Ibs. The allowable bending stress for which this bridge was designed was 18,000 lbs. Graphic Method.—The amounts of the forces and the distances between their points of application are shown in (e). The force poly- gon for the horizontal components is given in (f), and the bending moment polygon is given in (g). The maximum horizontal bending a STRESSES IN PINS 157 moment is to the right of 3, and is H & y= 200,000 X 1.04 = 208,- 000 |b.-in. The force polygon for the vertical forces is given in (h) and the bending moment polygon is given in (i). The maxi- mum vertical bending moment is to the right of 8, and is H KX y= 200,000 X 1.42 == 284,000 lb.-in. The maximum bending moment occurs at and to the right of 4 and 8, and is 351,000 lb.-in., as shown in (j). Shear.—The shear is found for both the horizontal and vertical ‘components as in a simple beam, and is equal to the summation of all the forces to the left of the section. The horizontal shear diagram is shown in (k), and the vertical shear diagram is shown in (1). The maximum horizontal shear is between I and 2, and is 165,400 lbs. The shear between 2 and 3 is 165,400 — 99,300 = 66,100 Ibs. The maxi- _mum vertical shear is between 6 and 7, and is 126,300 lbs. The result- ant shear between 2 and 3, and 6 and 7, is 126,300? + 66,1007 = 145,000 lbs. as in (m), which is less than the horizontal shear between 1 and 2. The maximum shear therefore comes between I and 2, and is 165,400 lbs. The maximum shearing unit stress is 5750 lbs. The allowable shearing stress was gooo lbs. Bearing.—The bearing stress in L, U, is 165,650 6 X 1.94 = 14,300 lbs. Bearing stress in U, U, is 165,400 6 X 1.88 = 14,600 lbs. Bearing stress in U, L, is 42,200 + 6 X 0.89 = 7900 lbs. Bearing stress in U,L, is 107,000-+-6 X 17%; 12,400 Ibs. The allowable bearing stress was 15,000 lbs. - CHAPTER XVa. GRAPHIC METHODS FOR CALCULATING THE DEFLECTION OF BEAMS. Introduction.—Algebraic methods for calculating the deflection of beams and the reactions of continuous beams are given in detail in standard works on applied mechanics.. The graphic methods are, however, not given the attention that their elegance and simplicity merit. The underlying principles of the graphic method will be de- veloped and a few. applications of the method will be given in the following discussion. The discussion will be limited to beams with a constant moment of inertia. Graphic Equation of the Elastic Curve.—Load a simple beam with a continuous load represented by the equation yf #, as in (a) " ay ee ee = EQUATION OF ELASTIC CURVE 159 Fig. 1. Assume that each differential load, y dx = fx dx, acts through its center of gravity. Now construct a force polygon as in (c) and an equilibrium polygon as in (b) Fig. 1. ~ Now, in (b) the tangent of the angle between any side of the d equilibrium polygon and the X-axis is tana == fe If the string bo in x (b) is produced until it cuts the vertical line through 3, it will cut off the intercept 3-3’, which is the difference between two consecutive values of d y and therefore equals d? y. Now, it has been proved that the moment of the force acting through point 2 in (b) about point 3, is equal to the intercept 3-31 multiplied by the pole distance H, is equal to 3-3!x* H=d?yH. But the moment of the differential load fr dx, which acts through point 2, about point 3, is fx dx*, and fede dyH and = | (1) It is evident that (1) is the differential equation of the equi- librium polygon in (b). Now, if the loading is taken so that y—fr—WM, where M represents the bending moment at any given point x, due to a given loading, the equation for the equilibrium polygon becomes dy M ; ae (2) dx? H From mechanics we have the relation that d’y M 3 Ses (3) pass: Fh which differs from (2) only in having EJ substituted for H, E being the modulus of elasticity and J the moment of inertia of the given beam. This relation may be deduced as follows: In (d) Fig. 1, let equilibrium polygon 1-2-3 represent the neutral axis of a beam as in and, solving (5) and (6) for R, we have _ substituting 160 DEFLECTION OF BEAMS OCB S B85 ae Now, in a beam as in (e) Fig. 1, the stresses at any point i beam will vary as the distance from the neutral axis, and from Si triangles we have : 7 R ef Met one and RA=cdxr Now, if S is the fiber stress on the extreme fiber, ang E is modulus of elasticity, we have B's S32 ies AE=S dr RS=E£e But from the common ‘theory of flexure we have M cae I, Substituting the value of R in (7) and (4) we have dy M Hee BE for constructing the elastic curve of a ‘tenet: re aa "Construct the bending moment polygon for the given iiéding the beam. Load the beam with this bending moment cae Se and SIMPLE BEAM 161 with a force polygon having a pole distance equal to EI, construct an equilibrium polygon; this polygon will be the elastic curve of the beam. It is not commonly convenient to use a pole distance equal to ETI, and a pole distance H is used, where N H equals EJ; the de- flection at any point will then be equal to the measured ordinate di- vided by N. Simple Beam.—The simple beam will be considered when loaded with concentrated and uniform loads, using both algebraic and graphic methods. Algebraic Method—Concentrated Load at Center of Beam.—The simple beam in (a) Fig. 2, is loaded with a load P at the center. The bending moment diagram is shown in (b) and the beam is loaded with the bending moment diagram in (c) Fig. 2. To find the equation of the elastic curve take moments of the forces to the left of a point at a distance r+ from the left support, and’ Pix P x —Ely= Se 16 12 and 48Ely=P (4a*—3L?x) (8) |? (a) omaart 7 at Fic. 2. Fic. 3. The maximum deflection will occur when += ¥% L in (8), or it may be found by taking moments of forces to left of r—=¥% L to be ri; “AS ET (9) Beam Uniformly Loaded.—The simple beam in (a) Fig. 3, is - 162 DEFLECTION OF BEAMS loaded with a uniform load of w per lineal foot. The bending mo- ment parabola is shown in (b), and the beam is loaded with the bending moment parabola in (c) Fig. 3. To find the equation of the elastic curve, take moments of forces to the left of a point at a dis- tance x from the left support. The equation of the bending moment parabola with the origin of co-ordinates at the left support is y= Z’wL xr— Yw 2’, the area of a segment of the parabola is d—=4wlLl-s«’?—1-6w2*, and the center of gravity measured back from + is a+ (2L—v-2z) 6L—4-r Taking moments of forces to the left of a point x, and reducing, we have 24Ely=w (—at*+2LlLx7—L' xr) (10) The deflection is a maximum when += 1% L, and may be found directly by taking moments, or may be found from (10), and is 4 Ax aoe = (11) 384 El ca Cantilever Beam—Concenirated Load.—The cantilever beam in (a) Fig. 4, has a concentrated load, P, at its extreme end. It will be seen that the cantilever beam may be considered as one-half of a simple beam with a span 2L, and a load 2P, at the center. The equation of the elastic curve may be found as in Fig. 2. Load the beam with the bending moment diagram as in (b) Fig. 4, and consider- ing the cantilever as one-half of the simple beam we have, after re- ducing, 6Ely=3PL2x4—P x (12) : iy' | eae ee ee SIMPLE BEAM 163 The maximum value of A is equal to y when x equals L, and fas Bg ASSET ~ AeeD To find the maximum deflection we may take the moment of the entire bending-moment parabola about the point 1, and Pole 2 e | BrA= x —Z, and 3 9 Bs mee op 3 This method of finding the maximum deflection of a cantilever beam is the one to use in calculations, and will be used in the solu- tion of the problem of the transverse bent. Simple Beam—Graphic Method—In Fig. 5 a simple beam is loaded with a load P, as shown. With force polygon (b), draw - equilibrium polygon (c). Now load the beam with equilibrium polygon as in (c), and divide the area of the equilibrium polygon into segments, which are treated as loads acting through their centers of gravity. Con- struct force polygon (d) and draw equilibrium polygon (e). Now, the deflection at any point having an ordinate y in (e) will be, if proper scales are used, y X HX H ey, In Fig. 5, if P equals 3000 lbs., and the area of the equilibrium polygon and pole distance H* are measured in square-foot pounds, pole distance H in pounds, and y in feet, we will have % y X HX H* X 1728 ea EI =1.88 inches at center, while maximum value of deflection is _ A? == 1.92 inches. Tangents to Elastic Curve.—If strings 1 and 3 in (e) be pro- 164 DEFLECTION OF BEAMS duced, they will intersect at 2 on a line through the center of gravity of the moment-area polygon, and the strings 1-2 and 2-3 will be tangents to the elastic curve at the supports R, and R,, respectively. This gives an easy method of constructing the tangents to the elastic curve without constructing the curve. It is also seen that the tan- = ; ~A S on ao TB ok 3 K- -R = et ae KOrea moment at: Fic. 5. gents to the elastic curve depend only on the amount of the moment area and position of its center of gravity, and are independent of the arrangement of the moment areas. Continuous Beams.—A beam which in an unstrained condition rests on more than two supports is a continuous beam. For a straight beam the supports must all be on the same level. Beams of one span with one end fixed and the other end supported, or with both ends fixed, may also be considered as continuous beams. In Fig. 6a the continuous beam in (a) with spans L, and L, carries a uniform load w per lineal foot. It is required to calculate the re- actions R,, R,, and R3. . i j rs =' r iil ss ie EE Se ee ee Z 7 Se ee ee ae een eee ae See ae ee CoNTINUOUS BEAM 165 The reactions of the continuous beam in (a) may be replaced by the reactions of the two simple beams loaded with the uniform load w in (b), and the reactions and the load of the simple beam with the span L,+L, and carrying a negative load r,’ in (c). The reactions in (a) will then have the following values; R,=r,—r,'; R,=r,+7,'; R, =, — 1,'. Now the upward curvature of the beam in (a) due to the load r,’ will be neutralized by the load above equal to 7,’ which is transferred to the reaction R, by flexure in the beam. The upward deflection of c~Load w per lin. ft. Fi, Fs Rs ea, ee pe + ee (2) 7 Wy 4 '2 3 the beam in (c) at any point will be the bending moment divided by E J at the same point in (d) due to a bending moment polygon with a maxi- mum moment M,=—r,’ X L, =r,’ X L,; and the downward deflection of the beam in (b) at any point will be the bending moment divided by EI at the same point in (d) due to the bending moment polygons for a uniform load w covering the simple spans in (b). But the de- flection of the beam in (a) is zero at the reaction R,, and therefore the bending moment at the corresponding point in (d) is zero. 166 | DEFLECTION OF BEAMS From the above discussion it follows that to calculate the reactions of the continuous beam in (a) by moment areas, take a simple beam with a span equal to L,-+ L,, and load it with the bending moment polygons for beams (b) and (c) as in (d); the bending moment in beam (d) at the points corresponding to the reactions will be equal to zero, and the reactions of beam (a) can be calculated by statics when the M, is obtained. Continuous Beam—Concentrated Loads.—In (a) Fig. 6, a con- tinuous beam of two equal spans of length L, is loaded. with two equal loads P, at the centers of the spans. Calculate the bending 1 Mat. PL? 7p )~Me pttel., PL aa a Cb) R3= > e Fic. 6. moments and load a simple beam with a span equal to 2L, with the bending-moment diagrams due to P in each span, and with the nega- tive bending-moment diagram due to the reaction R,. Then to find M,, the bending moment at 2, take moments of forces to the left of 2, and we nee Moi ee 2 8 6 16 M,=— Spy 16 To calculate R, take moments in (a) about 2, and PL | R, Ls Mies : 2 ee, ee eo eS -s —— Se coe oy mee ConTINvousS BEAM 167 and Beam Uniformly Loaded—lIn (a) Fig. 7, a continuous beam of three equal spans of length L, is loaded with a uniform load equal to w per foot. Calculate the bending moments due to a uniform load of w on each span, and load a simple beam of span 3 L with the posi- tive bending-moment diagrams due to load w, and with the negative bending-moment diagrams due to the reactions R, and R,. The bending moment M, is equal to M,. Now the deflection of the beam is zero at 2 and 3, and the bending moments must, therefore, be zero at these points. Taking moments of forces to the left of 2, we have AU eke oer Se a e552 5dlar XO Ss Po PO See = 7.0.8 ane BS 525 tae = SR eS \ e@ ‘ 3 ae: Rite“ — (b) ue Fic. 7. wh: wit M,I* M, L? + — — ==6 24 6 72 L* Mo=—— — M, IO To calculate R, take moments about 2 in (a), and w L? R,L— —M,=0 2 Bee wl Ky 10 168 DEFLECTION OF BEAMS 3 4 II R,=R,=-wl——wl=—wLl 2 10 10 Continuous Beam of n Spans.—To calculate the reactions for a continuous beam of m spans, equal or unequal, loaded with any system or systems of loads proceed as follows: Calculate the bending moment due to the external load, or loads, or system of loads in each span considered as a simple beam. Take a simple beam having a total length equal to the length of the con- tinuous beam, and load it with the bending moment polygons found as above. Also load the beam with the bending moment polygons due to the reactions. The reactions being unknown, the bending moments at the reactions will be unknown. Now calculate the bending moment in the simple beam at points corresponding to each reaction and place the result equal to zero, for the reason that the deflection at the sup- ports is zero. For a continuous beam of m spans there will be 7+ 1 equations which is equal to the number of unknown reactions. Solving these equations the unknown moments will be found, and the reactions may be calculated algebraically. Transverse Bent.—The problem of the calculation of the point of contra-flexure in the columns of a transverse bent—the algebraic c <— c aC fl ai “K b b ie /7, E } /7, 1 BE Te a A? ! BAN > 2 He, KN “WN, WN. To To (HQ) Ips (C) Fic. 8. solution of which is given in Chapter XI—will now be solved by the use of moment areas. The nomenclature in Fig. 8 is the same as in REACTIONS OF DRAW BRIDGES 169 Chapter XI. It is assumed that the deflection at points b and ¢ are equal. , In (b) Fig. 8, the deflection at b from the tangent at a is found by taking moments of the moment areas below b to be M,d oe \M,dd EIA= ra 24 3 2M, d?—M, d? = (14) 6EI The deflection at c from the tangent at a is found by taking moments of moment areas below c to be M,d 2M, (h—d)? M,d EIAt=—— (h—d/3) — (hi a/a od) 2 6 2 M, (3dh—d?) —M, (2h? —hd ax Mo (34h —d*) — M, ( ) ae , 6EI But A is equal to At by hypothesis, and equating (14) and (15) we have 2M, d?—M, d?=M, (3dh—d?) —M, (2h? —hd) 4 transposing, E M, (3hd—3d*)=M, (2h? —hd—d?) (16) q Now in (c) Fig. 8, it will be seen that M,: M,:: yy: d—%, q and 4 M, (d—y) =M,y% (17) q Solving (16) and (17) for y,, we have 4 d (2h4+d) Tet (18) 2 (h+2d) which is the same value as was found by algebraic methods. _ Reactions of Simple Draw Bridges.—The preceding methods are not adapted to the solution of problems involving moving loads, as in draw bridges. The following method, which is an application of — curved influence lines, is quite simple in theory and application, al- 170 DEFLECTION OF BEAMS - though requiring considerable labor in preparing the diagrams. The solution will first be explained and a proof given later: Draw Bridge With Three Supports—In Fig. 9 a continuous beam with spans L, and L, is loaded with concentrated moving loads represented by P, and P,, as in (a). In (b) load a simple beam having a span L, + L, with a bend-— ing-moment polygon due to the reaction R, (the value of R, is unknown, and any convenient load will do). Divide the bending-moment diagram into segments, construct a force polygon as in (d) and draw an equilibrium polygon as in (c) POMS Sal aie eam Ne bie ees ee a a eT ae ae _— ys. a nah Aan REACTIONS OF DRAW BRIDGES 17! Fig. 9, assuming that the segments are loads acting through their centers of gravity. The pole distance H may be taken as any convenient length, and the pole O may be taken at any point [in (c) the pole has been selected - to bring the closing line horizontal for convenience only]. : . Then in (c) R,a=P,b—P,d and | P,b—P,d Rk, = * (19) R,c=P,m+P.n and 3 r Pon | Rk, = ee (20) Cc R,g=—P,e+P.f and : —P Pg She att of re Proof.—The ordinates to the equilibrium polygon in (c) are pro- portional to the ordinates to the true elastic curve of the beam in (b) when it is loaded with a given load at 2. Now in (e) Fig. 9, if the deflection at 2 due to a load P at 1 is d, then if the load P be moved to 2, the deflection at 1 will be d. This can be proved by calculating the bending moments at 2 and 1 for the conditions, since the deflections are directly proportional to the bending moments. With P at 1, the bending moment at 2 is Pab a ; and with P at 2, the bending moment at 1 is » and the proposition is proved. Now in (c), if the deflection due to a load unity at 2 is m at P,, then the deflection at 2 due to a load unity at P, will be m. If load R, is applied at 2, the work done in making the elastic curve pass through 2 will be R,c; while the resistance due to a load P, will be P, times the deflection at 2 due to the load P,, which is equal to P, m. In like manner the resistance due to P, will be P,n, and 172 DEFLECTION OF BEAMS R,c=P,m+P,n and P,m+P,n R, (20) c To find R, take moments about 3 in (a), and R, (£,+L.) + Rk, L,—P,1,—P.l,=0 and from similar triangles in (c) R,a+R,c—P, (m+ b)—P, (n—d) =o (22) Substituting the value of R, from (20) in (22), we have R,a=P,b—P,d pole | an a and since R,+k,+h,=P,+P, —P,e+P .— +P.f (21) & Uniform Load.—For a uniform load on the beam the areas of the diagram covered by the uniform load will be used in the place of the ordinates as in Fig. 9 (see discussion on Influence Diagrams, Chapter X). For example in Fig. 10 the reactions are given by the following formulas : p (area B, — area B,) R,= : (23) area 4d + area B area C. poet ee as Re p (area C, — area C,) (53 g Draw Bridge with Four Supports—To find the reaction at R, in Fig. 11, proceed as follows: With a load represented by the triangle I-2-4, construct a force polygon (not shown) and draw an equi- q : ; } a Pe ee ee ee ee REACTIONS OF DRAW BRIDGES 173 ar onrers asta eee ee WILL _ 3 SS a (b) | 8 3° * PS ATT] eae ‘ U ' 0 YY en Bie i WML (<) i. vw PIG. 10, librium polygon passing through m-—n—o—p. Now, with a load repre- sented by the triangle 1-3-4, construct a force polygon (not shown) and draw an equilibrium polygon passing through m—o—p. The method of drawing an equilibrium polygon through three. points is explained in Chapter V, Fig. 20. ; (a) Pp R =) ir iS i, ee ates ee a a (4 Then in (c) Fig. 11, R,d=P,a+P,b—P,¢ and P,a+P,b—P,¢ 2S a R, may be found in a similar manner by deveine an * : polygon for a load, 1-3-4, through point n. . most easily be sbiatied by algebraic mania Proof.—With the load, 1-2-4, and full line deflection ¢ cu ur have, as in the case of three supports, . 2 (d-+h) =—R,k+P, (ate) +P, (b+f) +P And with the load, 1-3-4, and dotted line deflection have, in like manner, - R,k=—R,h+Pie+P.f+P, ee) Subtracting (28) from (27) we have tia R,d=P,a+P,b—P,¢ __ P,a+P,b—P,c aie d which is the equation of which proof was required. PART III. DESIGN OF MILL BUILDINGS. CHAPTER XVI. GENERAL DESIGN. General Principles.—The general dimensions and outline of a mill building will be governed by local conditions and requirements. The questions of light, heat, ventilation, foundations for machinery, hand- ling of materials, future extensions, first cost and cost of maintenance should receive proper attention in designing the different classes of structures. One or two of the above items often determines the type and general design of the structure. Where real estate is high, the first cost, including the cost of both land and structure, causes the adoption in many cases of the multiple story building, while on the other hand where the site is not too expensive the single story shop or mill is usually preferred. In coal tipples and shaft houses the handling of materials is the prime object; in railway shops and factories turning out heavy machinery or a similar product, foundations for the ma- _ chinery required, and conyenience in handling materials are most im- portant; while in many other classes of structures such as weaving sheds, textile mills, and factories which turn out a less bulky product with light machinery, and which employ a large number of men, the principal items to be considered in designing are light, heat, ventilation and ease of superintendence. 176 GENERAL DESIGN Shops and factories are preferably located where transportation facilities are good, land is cheap and labor plentiful. Too much care cannot be used in the design of shops and factories for the reason that defects in design that cause inconvenience in handling materials and workmen, increased cost of operation and maintenance are permanent and cannot be removed. The best modern practice inclines toward single floor shops with as few dividing walls and partitions as possible. The advantages of this type over multiple story buildings are (1) the light is better, (2) ventilation is better, (3) buildings are more easily heated, (4) founda- tions for machinery are cheaper, (5) machinery being set directly on the ground causes no vibrations in the building, (6) floors are cheaper, (7) workmen are more directly under the eye of the superintendent, (8) materials are more easily and cheaply handled, (9) buildings admit of indefinite extension in any direction, (10) the cost of construction is less, and (11) there is less danger from damage due to fire. 3 The walls of shops and factories are made (1) of brick, stone. or concrete; (2) of brick, hollow tile or concrete curtain walls between steel columns; (3) of expanded metal and plaster curtain walls and glass; (4) of concrete slabs fastened to the steel frame; and (5) of corrugated iron fastened to the steel frame. The roof is commonly supported by steel trusses and framework, and the roofing may be slate, tile, tar and gravel or other composition, tin or sheet steel, laid on board sheathing or on concrete slabs, tile or slate supported directly on the purlins, or corrugated steel supported on board sheathing or directly on the purlins. Where the slope of the roof is flat a first grade tar and gravel roof, or some one of the patent com- position roofs is used in preference to tin, and on a steep slope slate is commonly used in preference to tin or tile. Corrugated steel roofing is much used on boiler houses, smelters, forge shops, coal tipples, = similar structures. Floors in boiler houses, forge shops and in similar structures are generally made of cinders; in round houses brick floors on a gravel or DETAILS OF DESIGN 177 concrete foundation are quite common; while in buildings where men have to work at machines the favorite floor is a wooden floor on a foun- dation of cinders, gravel, or tar concrete. Where concrete is used for the foundation of a wooden floor it should be either a tar or an asphalt concrete, or a layer of tar should be put on top of the cement concrete to prevent decay. Concrete or cement floors are used in many cases with good results, but they are not satisfactory where men have to stand at benches or machines. Wooden racks on cement floors remove the above objection somewhat. Where rough work is done, the upper or wearing surface of wooden floors is often made of yellow pine or oak plank, while in the better classes of structures, the top layer is com- monly made of maple. For upper floors some one of the common types of fireproof floors, or as is more common a heavy plank floor supported on beams may be used. Care should be used to obtain an ample amount of light in build- ings in which men are to work. It is now the common practice to make as much of the roof and side walls of a transparent or translucent ma- terial as practicable ; in many cases fifty per cent of the roof surface is made of glass, while skylights equal to twenty-five to thirty per cent of the roof surface are very common. Direct sunlight causes a glare, and is also objectionable in the summer on account of the heat. Where windows and skylights are directly exposed to the sunlight they may best be curtained with white muslin cloth which admits much of the light and shades perfectly. ‘The “saw tooth” type of roof with the shorter and glazed tooth facing the north, gives the best ‘light and is now coming into quite general use. Plane glass, wire glass, factory ribbed glass, and translucent fabric are used for glazing windows and skylights. Factory ribbed glass should be placed with the ribs vertical for the reason that with the ribs horizontal, the glass emits a glare which is very trying on the eyes of the workmen. Wire netting should always be stretched under sky- lights to prevent the broken glass from falling down, where wire glass is not used, 178 GENERAL DESIGN Heating in large buildings is generally done by the hot blast sys- tem in which fans draw the air across heated coils, which are heated by exhaust steam, and the heated air is conveyed by ducts suspended from the roof or placed under the ground. In smaller buildings, direct radiation from steam or hot water pipes is commonly used. The proper unit stresses, minimum size of sections and thickness of metal will depend upon whether the building. is to be permanent or temporary, and upon whether or not the metal is liable to be subjected to the action of corrosive gases. For permanent buildings the author would recommend 16,000 Ibs. per square inch for allowable tensile, and 16,000 — 70 Ibs. per square inch for allowable compressive stress for direct dead, snow and wind stresses in trusses and columns; / being the center to center length and r the radius of gyration of the member, both in inches. For wind bracing and flexural stresses in columns due to wind, add 25 per cent to the allowable stresses for dead, snow and wind loads. For temporary structures the above allowable stresses may be increased 20 to 25 per cent. The minimum size of angles should be 2” x 2” x 14”, and the minimum thickness of plates 4”, for both permanent and temporary structures. Where the metal will be subjected to corrosive gases as in smelters and train sheds, the allowable stresses should be decreased 20 to 25 per cent, and the minimum thickness of metal increased 25 per cent, unless the metal is fully.protected by an acid-proof coating (at present the best paints do little more in any case than delay and retard the corrosion). The minimum thickness of corrugated steel should be No. 20 gage for the roof and No. 22 for the sides; where there is certain to be no corrosion Nos. 22 and 24 may be used for the roof and sides respectively. The different parts of mill buildings will be taken up and discussed at some length in the following chapters. CHAPTER XVII. FRAMEWORK. Arrangement.—The common methods of arranging the frame- work in simple mill buildings are shown in Fig. 1, Fig. 81 and Fig. 82. The different terms which are used in the discussion that follows will be made clear by an inspection of Fig. 1 and Fig. 81. 5 . “1% R ~ ms 7 ing- ¢ Furlin pit or a Q & i aS ne Rae : rez i. <1 $ 6 3 mS 18 a -0o Sb oe Ss Ree 5 ire |S TST SAVE NYS ESS sf ~~! Cd NSN : We - = Nae Eee eS) < \ ahi a] Seve Stet a BS S (Ag N € yt BY "cir? « Rea A 8 § S S$CKKE S S ax Os “8 ere? N — — TI rs eee CS eee Vj- Ne N : ” ‘ S N ~ Q "4 KS aS 4 > i: + NG = ra Gro * r L_™ NI a bs: ee ee Transverse Bent Side Elevation L Column t $\Eave Strut &4L.Column T<- y Ve 4 pal, eo 0 Yd as A £Girt Vy s BI ce el ol meh © a 20 7 S x} PS ae lo -s ~~ ~T iH 38 a x 4 ei lS ©) Sienrshut ~ |S eh Strut ep aS er 5 rc ; wil 4 a Pe £ N .S i yy K N = % 3 oN S a 4 4 S 18) iS) Pik % My i % HS ss S S Mel eed 2, See ud = ‘7 & — ie} =Do : Oe BEARS OR. Pf} ERS Bp 2 | 1/ ‘ \ » 8 v x v >< wy “i pe 2 Strut} ‘ Do ” ; yee ae End Framing oY . ‘ ‘ Do Ro Pi 2 . oe Do re Ske «lL Column & , Lave Strutge4 LColumn ua Plan Lower Chord Plan Upper Chord Fic. 81. The three types of mill buildings—steel frame mill buildings, mill buildings with masonry filled walis; and mill buildings with masonry walls—have been discussed in the Introduction. The end post bent, shown in (a) Fig. 1 and in Fig. 81, usually requires less material than the end trussed bent shown in (b) Fig. 1 and in Fig. 180 FRAMEWORK 82, and is commonly used for simple mill buildings. Extensions can be made with about equal ease in either case, and the choice of methods will usually be determined by the local conditions of the problem and Composition Root —4 ny ae PRS = Composition as ae = aN \ A f hee erect Tes dh 2 ag Sate Ve = eS | gr ad ‘53x 23g" Jt 4:2 Portland Cement Finish. yi. ___ Finished Floor Level _ 2 Ci Rm, ane RR, © © iN ; : ZS ra fa i Ca ue eso” Re i 4 i (e) 7 & ww *uoHeAgTy pug _-- —-- ee a a a ” 190 FRAMEWORK a OE Cage PES. CT ree Fic. 85. Locomotive SHop, EASTERN Rattway oF FRANCE. A few of the forms of trusses in common use where ventilation and light are provided for are shown in Fig. 86. The Fink truss with (b) glass glass glass or louvres ; -«y glass | glass a re NS Za N (c) (d) ae glass glass HOSS —s- glass a : : : N fe) (f) (g) | (h) Silk Mill Fic. 86. eS Pitcu oF Roor | 191 monitor ventilator and skylights in the roof shown in (a), is a favorite tvpe for shops; truss (b) with double monitor ventilator is especially adapted to round house construction; trusses (c) and (e) are adapted to shop and factory construction where a large amount of light is de- sired, ventilation being obtained by means of circular ventilators ; truss (d) is similar to (c) and (e), but allows of better ventilation ; truss (f) has skylights in the roof and has circular ventilators placed along the ridge of the roof; truss (g) is the type in common use for blacksmith shops, boiler houses, and roofs of small span. The “silk mill” roof shown in (h) was used by the Klots Throwing Co. in their silk mill at Carbondale, Pa. The spans of the three trusses are 48’ 8” each, with a clerestory of 13’ 9” in the monitor ventilators, which are glazed with glass 11’ 0” high. The monitors face east and west, al- lowing a maximum amount of direct sunlight in the morning and evening, and none at midday. This roof has given very satisfactory results, however, it would seem to the author that it would be necessary to use shades, and that there would be shadows in the building. The trusses in this building are spaced 10’ 6” apart and support the plank sheathing which carries the roof, no purlins being used. The shafting to run the machinery in this building is placed in a sub-base- ment; a method much more economical and convenient than the com- mon one of suspending the shafting from the trusses. Pitch of Roof.—The pitch of a roof is given in terms of the center height divided by the span; for example a 60-ft. span truss with 4 pitch will have a center height. of 15 ft. The minimum pitch allow- able in a roof will depend upon the character of the roof covering, and upon the kind of sheathing used. For corrugated steel laid directly on purlins, the pitch should preferably be not less than 14 (6” in 12”), and the minimum pitch, unless the joints are cemented, not less than \%. Slate and tile should not be used on a less slope than 14 and preferably not less than 14. The lap of the slate and tile should be greater for the less pitch. Gravel should never be used on a roof with a greater pitch than about 1%, and even then the composition is very liable to run. As- 192 FRAMEWORK phalt is inclined to run and should not be used on a roof with a pitch of more than, say, 2 inches to the foot. If the laps are carefully made and cemented a gravel and tar or asphalt roof may be practically flat; a pitch of 34 to 1 inch to the foot is, however, usually preferred. ‘Tin may be used on a roof of any slope if the joints are properly soldered. Most of the patent composition roofings give better satisfaction if laid on a roof with a pitch of % to %4. Shingles should not be used on a roof with a pitch less than %4, and preferably the pitch should be % to 4. Pitch of Truss.—There is very little difference in the weight of Fink trusses with horizontal bottom chords, in which the top chord has a pitch of %, %4, or %. The difference in weight is quite notice- able, however, when the lower chord is cambered; the truss with the ¥% pitch being then more economical than either the % or the 4% pitch. Cambering the lower chord of a truss more than, say, 1-40 of the span adds considerable to the weight. For example the computed weights of a 60-ft. Fink truss with a horizontal lower chord, and a 60-ft. Fink truss with a camber of 3 feet in the lower chord, showed that the cambered truss weighed 40 per cent more for the %4 pitch and 15 per cent more for the % pitch, than the truss having the same pitch with horizontal lower chord. It is, however, desirable for appearance sake to put a slight camber in the bottom chords of roof trusses, for the reason that to the eye a horizontal lower chord will appear to sag if viewed from one side. . In deciding on the proper pitch, it should be noted that while the ¥% pitch gives a better slope and has a less snow load than a roof with \% or &% pitch, it has a greater wind load and more roof surface. Tak- ing all things into consideration % pitch is probably the most econom- ical pitch for a roof. A roof with % pitch is, however, very nearly as economical, end should preferably be used where corrugated steel roof- ing is used without sheathing, and where the snow load is large. Economic Spacing of Trusses.—The weight of the trusses and columns per square foot of area decreases as the spacing increases, while ee =a ee yy Ps r ere ee wee ee a ae a ee * Pitcu oF Truss 193 the weight of the purlins and girts per square foot of area increases as the spacing increases. The economic spacing of the trusses is a func- tion of the weight per square foot of floor area of the truss, the pur- lins, the side girts and the columns, and also of the relative cost of each kind of material. For any given conditions the spacing which makes the sum of these quantities a minimum will be the economic spacing. It is desirable to use simple rolled sections for purlins and girts, and under these conditions the economic spacing will usually be between 16 and 25 feet. The smaller value being about right for spans up to, say, 60 feet, designed for moderate loads, while the greater value is about tight for long spans, designed for heavy loads. (eae 4 i ae a Truss pice pert tt Ei ene % ap ga i 474 NS S aos 4 r "y 5 x Re ee —* ae | Mate 2 = te CBee thcNbesch TUSS_ k- 8g $854 UE Part Framing Plan 160 is é 3 | JZ'0" > u Section A-B Fic. 87. STEEL ROOF COVERED WITH LUDOWICI TILE. Calculations of a series of simple Fink trusses resting on walls and having a uniform span of 60 feet, and different spacings gave the least weight per square foot of horizontal projection of the roof for a spacing of 18 feet, and the least weight of trusses and purlins com- bined for a spacing of to feet. The weight of trusses per square foot was, however, more for the 10-ft. spacing than for the 18-ft. spacing, so that the actual cost of the steel in the roof was a minimum for a 194 FRAMEWORK spacing of about 16 feet; the shop cost of the trusses per pound being several times that of the purlins. Local conditions and requirements --Glass or Louvres ~Glass ye Traveling Crane SD | y (a) (e) -Glass -Glass or Louvres . ae Traveling Crone “Cides eo S a ~ (b) (f) -Glass Glass <-Glass or Louvres (c) (g) - Louvres. -Glass — Comet {d) {h) Fic. 88. ‘TypEs oF TRANSVERSE BENTS, ot a F q d 7. a 4 TRANSVERSE BENTS. 195 usually control the spacing of the trusses so that it is not necessary that we know the economic spacing very definitely. For long spans the economic spacing can be increased by using rafters supported on heavy purlins, placed at greater distances than would be required if the roof were carried directly by the purlins. This method is frequently used in the design of train sheds and roofs of buildings where plank sheathing is used to support slate or tile cover- ings, or where the tiles are supported by angle sub-purlins spaced close together as shown in Fig. 87. TRANSVERSE BENTS.—The proper cross-section for a mill building will depend upon the use to which the finished structure is to —— ee ee ee ne Ok NE Se OR ee Ne ee ee z Skylight Gravel Roof-~ mg : : . " ——_ . , Skylight Skylight | Rate AS ss re > pin spams eae 75*0"--------- tk-------- 75-0%------=--=- Fig == ‘ou ' BK ~- 33°3 =~ SE -38°6"- gray | ees Apiate Rees ! is = ! % Ds “ 26} » 7 ' . . oS m 8 Trusses Spaced (8 0 * ! : 4 | ! Pita ; : ‘ ! | i 48 ? Plank Floor || 4 : YY mee, -- Concrete’ © = Ke eee wee ee ee wee ho) 0 0 ee ee a er ea pa 400'Long Locomotive Shop-Oregon Short Line Fic. 89. be put. A number of the common types of transverse bents are shown in Fig. 88. Transverse bents (a), (b), (d) and (h) are commonly used for boiler houses, shops and small train sheds. Where a travel- ing crane is desired, the crane girders are commonly suspended from the trusses in the bents referred to, although the crane may be made to span the entire building as in (h). ‘Transverse bent (d) was used for a round house with excellent results. Transverse bents (f) and (g) are quite commonly used where it is desired that the main part of the building be open and be provided with a traveling crane that will sweep 196 FRAMEWORK the building, while the side rooms are used for lighter tools and mis- cellaneous work. Transverse bent (c) may be used in the same way as (g), by supplying a traveling crane. Transverse bent (e) is very often used for shops. Cross-sections of the locomotive shops of several of the leading railways are shown in Figs. 89 to 92, inclusive, and the locomotive shops of the A. T. & S. F., and the Philadelphia and Reading Railroads are described in detail in Part IV. For the most part these buildings ———, a a 1 { { t i As t 70 Ton Crane S 2 ied ‘ . 4 + SS Lecomotive Erecting 2 x Wy ; . e A) i S 8 and Boiler Shop Machine Shop 9 f & f *) 4 — 1 NU 4 : N I -peaqe=---—- 72-0" —-----— —-+eR----- - 62-0" ------ oa 0 Bee ri» Nie sh as es i ee ee ISTIO® = wn me ww oe ee ee eek > 540' Long Locomotive Shop - St-L»!-M-& 5S. Fic. 90. are built with self-supporting frames, and have brick walls built out- side the framing. The arrangement of the cranes, provisions for light- SSSR Skylight OVERS LRMABD VPPRAVLARRABABASARARRARA p= SIO = mB , meee ew ee = £69-0O0° «a_weee eee eo SS et ee : ee ee ee ie ~~ -----6 40" kn === — agi ---—o i NSS 486-0 Long Locomotive Shop -Union Pacific: Fic. 1. - eS paw Truss Derats 197 ing and ventilating, and the main dimensions are shown in the cuts and need no explanation. Beene ta tls La pewionst sk seeks r » Skylight Skylight a“ q 1 pCR Oh ge MO ENT a a eS Cot I yl aw |Tooth atom : Her : x x Ass =! Fi w a wil? sn Shaw Tobth iy? es Es ¥ |[K4+----- wee th —~-~70°6"- -- 1 ¥ > ~ . =! N) 1 : . ~ 2 Sy " Ay Q rs . Ny y URE wc les ak lena riiyyy <----- 59°93"----» LJ} J) J) fons 39°93 "—— => Mo--------- 74'5"--------- 4 852-1" Long Locomotive Shop-A‘T-& S-F: Fic. 92. A cross-section and end view of the train shed of the Richmond Union Passenger Station are shown in Fig. 93. Riveted trusses are eu AUN DEAD DEAN USAT 22 2 "EUG LDDs LHL FON SOUND GUDND NOUNS HONE HgUn2 a00E ene eS 1% §: orth Galle < | Platrorm Cross Section, Looking North. Elevation, North End. Train Shed—Richmond Union Passenger Station. Fic. 93. quite generally used in train sheds; a notable exception to this state- ment, however, being the trusses for the new train shed of the C. R. I. & P., and L. S. & M.S. Railways in Chicago. The trusses in this struc- ture have a length of span of 207 ft., a rise of the bottom chord of 40 ft. and a depth of truss at the center of 25 ft. The trusses are pin con- nected, the compression members being built up channels and the ten- sion members eye-bars. The building is described in detail in Engineer- ing News, August 6, 1903. Truss Details.—Riveted trusses are commonly used for mill build- ings and similar structures. For ordinary loads, the upper and lower ee Ue ee th ee eS Te ae al Ai a's Pre tome (, So ory Per )\ lag ome oF a a ae 7 mt | . ics a) aoa 198 FRAMEWORK chords, and the main struts and ties are commonly made of two angles ~ placed back to back, forming a T-section, the connections being made — by means of plates. The upper chord should preferably be made of unequal legged angles with the short legs turned out. Stb-struts and ties are usually made of one angle. Flats should not be used. Where a truss member is made of two angles placed back to back, the angles should always be riveted together at intervals of 2 to 4 feet. Trusses that carry heavy loads or that support a traveling crane or hoist, are very often made with a lower chord composed of two chan- nels placed back to back and laced or battened, and are sometimes made — with channel chord sections, throughout (see Fig. 175). When the purlins are not placed at the panel points of the truss the upper chord must be designed for flexure as well as for direct stress. The section in most common use for the upper chord, where the purlins are not placed at the panel points, is one composed of two angles and a plate as shown in (c) Fig. 96. Ojo OO C000 | e ‘o'eeee * (a) (Cc) ’ Trusses may be fastened to the columns by means of a plate as shown in (a) Fig. 94, or by means of connection angles as shown in (b) and (c). The first method is to be preferred on account of the - ‘ . rigidity of the connection, and the ease .with which the field connection can be made. / t99 Rivets a Holes 7" hud DETAILS OF A STEEL Roor Truss ; La I-PLOEK i ue : w 2-09 28x25 xi x/6-0" 2-08 23x25 xox 18-63 id 1 s6¥ Sof 2-94"= 8°33 PHO" Fx O-N" 15" M111 Sus ois Fills |g ‘S| Ine 2% vu N filh< Wiis Ls L ov 2 4 £ ® re) ead J ! IPLSE RE x VA qe MW 8k" 2-6" SHOP DRAWINGS FOR A STEEL ROOF TRUSS. _ Fic. 95. 200 | FRAMEWORK Trusses supported directly on masonry walls have one end sup- ported on sliding plates for spans up to about 70 feet; for greater lengths of span one end should be placed on rollers, or should be hung on a rocker. ‘Trusses for mill buildings should be made with riveted rather than with pin connections, on account of the greater rigidity of the riveted structure. The complete shop drawings of a truss for the machine shop at the University of Illinois, are shown in Fig. 95. This truss is more completely detailed than is customary in most bridge shops. The practice in many shops is to sketch the truss, giving main dimensions, number of rivets and lengths of members, depending on the See - » © ” oF 7 lé ~-PLIZKPRS‘O” Lavel 3x2 Foy 6 Kee eee Span=60-0: ——Fifthk ---— ip? ee (a) ME ry Cast Lateral. “Lara Froa 412 feelers Sige 73° EL SAS 15 lp ~ —> 21 5335%5°071G> sie gid Sane ASO = CUE RE re Fic. 96. Truss DETAILS 201 templet maker for the rest. In Fig. 95 the rivet gage lines are taken as the center lines. This is the most common practice, although many use one leg of the angle as the center line in secondary members. The latter method has the advantage of reducing the length of connection plates without introducing secondary stresses that are liable to be troublesome. The detail drawings of a transverse bent are shown in Fig. 96. The common methods of attaching purlins and girts, and of making lateral connections are also shown. ‘The fan type of Fink Truss shown in Fig. 96 is quite commonly used where an odd number of panels is desired, and makes a very satisfactory design. The details of the end connection of a 60-ft. span truss are shown in (a), and of a 45-ft. span truss with a reinforced top chord are shown in (c), Fig. 96. The method of reinforcing the top chord shown (c) is the one most com- monly employed where purlins are not placed at the panel points. The method of making lateral connections for the lateral rods shown in (c) is not good, for the reason that it brings bending stresses in a plate which is already badly cut up. The detail drawings of a saw tooth roof bent for the Mathiessen & Hegeler Zinc Works, LaSalle, Ill., are shown in Fig. 97. This building was erected in 1899 along the lines suggested by an experience with a similar saw tooth roof building erected in 1874. The building was de- signed by Mr. August Ziesing, Vice President American Bridge Co., and was erected by the American Bridge Co. The following description is from a personal letter from Mr. Julius Hegeler of the firm of Mathiessen & Hegeler, to the author in reply to a request for plans: “The cast iron gutters are fastened to the pur- lins and roof boards by spikes through holes in the gutters (holes are not shown in the drawing) ; on account of their slope, however, hardly any fastening is necessary. ‘hese gutters are so placed that the gal- vanized iron down spouts are next to the posts, there being two down spouts at each post. The condensation gutters are fastened to the gut- ters and empty into the down spouts. Ice has never caused any trouble by forming in the gutters.” 1 a CA ee ee a a: ~ -y ad ~~ Steer soy ee 8 Oe ; Fs = il iy ree 4 id) ip a4 : a? 4 += ‘ ay ; : 202 FRAMEWORK - on - - prs - - - - get tr 5 - - - - - or Py - as ae 0h we ae Alege chan do's yk bo Ww ar DBL Sais 1a er ect mee ee oe wo mm we we - OM oe ee * : ane ge ” LAasxZni2 * -10-1§ ~--- wo -- += > Pad kD 3 | s = .% > 2 B Holes 1+5"6 toc. Built in mason Fille tile Wane es Fee Rin GRA" Swe kc eMaag au eo S SJ alisha eed . 5 = ; ’ , t H : \ S me = —_—— = = A

Troe) aa, 1. c \ ae a _ a ‘. - 8 Say ut 1 = | + ee ago al oh - “ " ms bs ie 4 oe ee a pe of Ne 7: os, | ome } 2 4 PLOxg” 418 4x3xE ww www ° Details of Cast Iron Gutter Hae ei ~ Fic. 97. CROSS-SECTION OF THE SHOPS OF THE MATHIESSEN & HEGELER Zinc Works, LASALLE, ILL. The original saw tooth roof shop built by this firm in 1874 is still NT in use, and is one of the first, if not the first, saw tooth roofs builtin America. COLUMNS.—The common forms of columns used in mill build-_ a ings are shown in Fig. 98. For side columns where the loads are not — ra excessive, column (g) composed of four angles laced is probably the best. In this column a large radius of gyration about an axis at right angles to the direction of the wind is obtained with a smal] amount Typrs oF Mitz Buitpinc CoLuMNS 203 of metal. The lacing should be designed to take the shear, and should be replaced by a plate, (f) Fig. 98, where the shear is excessive, or where the bending moment developed at the base of the column requires the use of excessive flanges. ‘The I beam column (h) makes a good side column where proper connections are made, and is commonly used for end columns (see Fig. 81). The best corner column is made of an equal legged angle with 4, 5 or 6-in. legs, (i) Fig. 98. Details for the bases of the three columns above described are shown in Fig. 99. | Off lo ° Ho}lo mn " ° : oH No ° 0} {0 t “ ° ou i) ° Ho} }o t ¢ = oy No} ° 0 ° ot #0 ° f O}}O) ew | 2 Channels 2Channels 2Channels 2 Channels 4Z Bars Laced Laced 2 Plates 1 I Beam | Plate (a) (b) (c) (d) (e) oe 4Angles 4 Angles 1I Beam 1Angle Special | Plate Laced I Beam (f) (g) (h) (i) (j) i [ Larimer Gray 4Angles 4 Angles 4 Angles Box Laced Box Laced Starred (k) (I) . (m) fray” (oO) Fic. 98. ‘TyPEs OF MILI, BUILDING COLUMNS. 204 FRAMEWORK Columns made of two channels laced, or two channels and two plates, are used where moderately heavy loads are to be carried. Chan- nel column (a), with channels turned back to back and laced, is the form most commonly used; column (b), with the backs of the channels turned out and laced, gives a better chance to make connections and can be made to enter an opening without chipping the legs of the channel ; column (c) is a closed section and is seldom used on that account. The cost of the shop work on column (b) was formerly considerably more than for column (a), for the reason that it was impossible to use a power riveter for driving all the rivets. A pneumatic riveter is now made, however, that will drive all the rivets in column (b), and the shop cost for columns (a) and (b) are practically the same. Where very heavy loads are to be carried, columns (d) or (e) are often used. Column (d), composed of two channels and one I beam, is a very economical column and is quite often used as a substitute for the Z-bar column shown in (e), for the reason that it can be built up out of the material that is in stock or that can be easily obtained. Con- nections for beams are easily and effectively made with either columns (d) or (e). The special I beam column (j), with flanges equal to the depth of the beam, is now being rolled in Germany by the use of a process patented by an American, Mr. Henry Grey. This column makes an almost ideal column for heavy loads, since it has all the advantages of the Z-bar column with a very much smaller shop cost. The Larimer column (k) is a patented column manufactured by Jones & Laughlins, and is used by their patrons quite extensively. The Gray column(1) is a patented column and is but little used. Columns made of four angles box-laced, are used where extremely light loads are carried by very long columns. The shop cost of column (m) is somewhat less than that of column (n), although with small angles there is no dif- ficulty in riveting (n) with a machine riveter. Column (0) is a very poorly designed column, for the reason that the radius of gyration is very small for the area of a cross-section. of the column. Columns made of two angles “starred” and fastened at intervals of two or three i) am CoLuMN DETAILS | 205 ~ feet by means of batten plates, are quite frequently used for light loads. Column Details.—The details of a 4-angle laced column attached to a truss are shown in Fig. 96; and the details of a 4-angle plate column are shown in Fig. 97. The details of bases for a angle, I beam and angle columns are shown in Fig. 99. . ey “Inq “IN Sf 8 \ ‘IN ss : Nie ae FY SKgXI-28" | pile le, FI 9S§ HI -5 PUIOKE 0" a FY ONG ATU" Fi LEEKS NG 19" LEONE NZI 8 8 2L5 6X52 46 x10" R CLO'X4A B63" “7 a FV AKA 2. 2264) 8 498 “Sade F7/ 494.48. a 4/414 az w-- ~~ /'-/0%----> w= 4-/ 5" =m a2e--- /-7"--~- nce H2K--/'-1'-- 2% = 1-3 % + Litas Nal SLE SPiecel S: HS Alba scr 6 is Shc ihe $1 Wile s_ 2 ef Ot “lle fle @|ije|] ¥) ee (a) (b) Fic. 99 Shop details of a 4-angle column are shown in Fig. 100. This column was designed for a mill building with a span of 60 feet, trusses spaced 16 feet apart. The long legs of the angles are placed out, to give a larger radius of gyration about an axis at right angles to the direc- tion of the wind. The details of a 4-angle and plate column, designed to carry a crane girder as well as the roof, are shown in Fig. 101. The details of a heavy column composed of two channels placed back to back and laced, are shown in Fig. 102; the lacing is heavy and is well riveted. The bent plate connections for the anchor bolts on this column are very satisfactory. ‘This is one of the columns used in the A. T. & S. F. R. R. shops at Topeka, Kas., to carry the crane girders. The shop details of a light channel column are shown in Fig. 103. The single lacing alternates on the two sides of the column. ‘The various details of the columns can be seen, and require no explanation. FRAMEWORK 206 Sidi E2 & = NS UN ID UdIig —-——-— — i) —— Se AKSEMN- ae | =, p . & 2 ! % aa s | > ct ee Re eere nt Say Sos = wae — ~s : : 5; PILE S Z eo wo LN g eu J F j \K---— a0! -3--- f meee ee ~S : 2 WO 72:24 3's fil bes. : PM 124 Fob 67 2- ~ "2/0H, [> [oes ae wales Fa t occ Cte eed ined --4 eneeeree a % fU:04. 249 YF 2 2 SHELLED 672 HOG 40f [OlHOpy aSOG ULUNIOD jo UDig Pe XZ Bawa PES 7 & -45- | 7 , Ey i areas pale i w'nkccisiere SR SE pish SERRE ASSN On re ene On Me aS Soph Oe INOS So 9s 2 hee ES sis ‘ am = = OO. = = eee 87 east Pp s/s oengere - - - -- ee e - el SE ES OS ST ~ * * . . x --lee 6 : eee * $e APO MOT 2 Nee Re MORO es Fic. 101. > 58 358 N ® «--4- ey PT . ° . = 3° a - ua 124 so -- CL6 5 1g 1-2" & Lacing ai ag 10%," foc OLS 404 B5/'5"--~, CF 3.128 «1-2 0NCL ~. MD 1} 4-7 1 ' ‘ 1 1 ' ' ' ‘ ' ' ' ‘ 1 ‘ ‘ ' ' ‘ ‘ ' t] ' ‘ ‘ ‘ ' ' ' ' ' ' ’ ' ’ & 1" AR : ’ ’ ' ' ‘ ! 1 ‘ 1 ‘ ‘ ' ' ‘ ' ' J ‘ ‘ i ' 4 ‘ ‘ ‘ ' ‘ ‘ ' ! ’ ’ ' ‘ ' ' ' ' ' 265331 240°6" 726 0@ bE 16-04" aesee te « sk " ip, Peete Al eae WRN Me Ree ee eens a (ES eee ee eee eer yt oO" 2 *FUEUR «HBF co. NO 1 624% wg But2o7_ CoLuMN DeEraIts law = <2¢0"- ood 707 ee TZT Ct 0.12 -~------ we renee Se ee re * tie ee ee ee -——-—<—— - ee ee ee ee ee Ce ee ed -— -_— ne 2p CF "or Orage. re ' t ne re : ay ed | £2 Past w¥O/202-82, 6,97 2 y ay N Py & are i he ‘ F . 7 j i ig | feeb 4] 4 » A 7 proses tte F098 eee ates och neramennnsnae ccwrmrmcnnccrene see = = 402 me gate gi SiS os Salons was asataini Y in BST wwe Ae Ra aa t a3 “1 1 ' 3 as : : cE ee ee Te eo 5 0 8 $08 oO OsIIS ice Pe ek eee el ee Een oe ee ee 6 ao ‘epee | RS PRSERSS: = rere mead Sie seeeiinns j £35 Sede Ledeen wow n wen serene eS Cea | 3 sy a : wk Sat -2 9.8 OF re re " : ' O-2 ! ed ™ pete " ie 8 * & wy §- In B% “af 0 | ey t eS" 9 NS oe 26 IduZ%ell a = z= 4 y < | ¥ © wu wo 4 ry s 2 r) | Ane 21 | v0 / At 3 x 5 ae x Be a 2 Ral fi : 5 ae ° Fy c ret » : Ss ; Ve) \er 2 OS oe # J a Preven ™ r —= Blew DE = { in -- 20" Knee ew men ne ” Center Roof Truss End Strut $09---~~- = 2 hehe Ee me oy < | \ eel eat || | pk haan atin ae as SSNAL JOOY BPIG AOZ SUTIZIQUUOD 0 s7u I) -2 FRAMEWORK — eee feves-ne- 40,9 ---9n==-9-- 208 ees Smad 2 yt: | | ood 1 1 2 ' ws BL, pO%nele fe Sg ee ae ee By SEU -Y > Ye H Slt u Eu. ' & 3 * H SE TT EVE ETT Ee : or. > © - wu 2] ‘ = — io as k-7@2-- Ny k--- ete emeen-~ - ye/2l we terre creer a Aen yee, Ae Scia= masse OO): este man ee = 3 wore on erm wnnte~ YB) coress roe een------ Gti Ge o 26-0 G oo 3 FE YOu ~4v000 96 8 v oO 8 0 Ries we mlb we ori stk nin wi Se oe ene ssnay pourpnyibu07 g and Forgs Shop Main Column in Bending and Forge Shop. ‘Main Side Column in Bendin Fic. 103b. Fic. 1032. which have been taken from the : and 103d, 103¢c, 5 Figs. 103a, 103b Engineering Record es it | aA & WAN * Holes for 14° DK Bolts. oe bras "18 | Conter Aisle Root 234 Wah ne 3 "1s Runway Girder for 20. ard %0Fon branes. Truss, er aX aka Soe oF \/ ™. | RRR RDP eAeReROs Lede iclooo Sogo OO qoOOOY CoLUMN DETAILS S US Side Aisle j Vibe TASS. H H 43/7" Gall Floor Girdgr. | 5-Ton Crane Rurwa ayes VIG Center ‘Column in Machine Shop. Fic. 103¢. 209 Longitudinal Truss : Roof SS : “ooo oo oo : Section B-B a x SRR omnes [FP EPL ah" b> | Plea of jo — Section O-D. aameng tt IE) sh gen North Aste [|] *. +] connection for Run EL porta Sener B- : 2 tI 5 South Aisle. of of of0'd »: °! : i Knee Brace. oo) t Diagonal Rod. 0} ‘ @¢ Floor Beams, tl: ij i Longitudinal Sint cy ' ‘ for : 5 Bn Crane : oo ‘Diagonal tods, . " D Double : PD} oD a loagitdina! = russ PY " aw ‘ 19) ‘ y : " ‘ o : iy H 10 | ' ' x H § H 5 H ¥ : A ‘ 1 | bear rm OO 08 * 9 0 = 'Main Column No 38, between North and South Aisies Fic. 103d. 210 FRAMEWORK The American Bridge Company’s specifications for lattice bars for single and double lacing are shown in Fig. 104. 76 7. 5 m0 = ‘ fo TES Maximum Distance C for given thickness of bar. SINGLE LACING t=35 | DouBLE LacinG t= THICK, | DISTANCE | DISTANCE | THICK, t c c r Q-10 1- 3 fe_| 1- 03 1- 63 16 7 1- 8 1-103 i6 1- 53 2- 24 16 3 1- 8 q i6 1 - 10+ fi 5 32-1 eet ae HI hf Single lacing should make an angle of not less than 60 degrees, and double lacing, riveted at the center, not less than 45 degrees with the axis of the member. These specifications are standard. . The properties of angles, I beams and channels, and of Z-bar, Larimer and Gray columns are given in the manufacturers handbooks. The moment of inertia of two channels placed back to back and laced, as in (a) or (b) Fig. 98, about an axis parallel to the webs and through the center of gravity of the section, is given by the formula IT=2I'+2 Ad where J’ = moment of inertia of one channel about an axis through its center of gravity and parallel to the given axis, A = the area of one channel, and d = distance from the center of gravity of one channel to the center of gravity of the column. The lacing is omitted in find- ing the moment of inertia and the area of the section. ‘The moment of © inertia of the column about an axis perpendicular to the webs is equal STRUTS AND BRACING 211 to twice the moment of inertia of one channel, which may be found in the table of properties of channels given in the handbooks. Having the moment of inertia J, the radius of gyration of the column is given by the formula rat A With channels placed back to back and laced, the radii of gyra- tion about the two axes are equal when the clear distance is equal to about 3 inches for 5-in. channels, and 10 inches for 15-in. channels. A common rule is to space the channels about eight-tenths the depth. With channels placed with backs out and laced, the radii of gyration about the two axes are equal when the clear distance is about equal to 5 inches for 5-in. channels, and 13 inches for 15-in. channels (see Cambria Steel, 1903 Edition, p. 217). The moment of inertia of a 4-angle laced column, about an axis perpendicular to the lacing and through the center of the post, is given by the formula I= 4!l'+ 4Ad? where J’ = moment of inertia of one angle about an axis through its cen- ter of gravity and parallel to the given axis, d = the area of one angle and d = the distance from the center of gravity of the separate angles to the center of gravity of the column. The moment of inertia about the other axis is found in a similar manner. STRUTS AND BRACING.—Eave struts are very commonly mide of four angles laced, made in the same way as the 4-angle posts, Fig. 100. Eave struts made of single channels are more economical, and are equally as good as the laced struts for most cases. End rafters are commonly made of channels. The sides, ends, upper and lower chords are commonly braced as shown in Fig. 81. The bracing in the plane of the lower chords should preferably be made of members cap- able of taking compression as well as tension. The diagonal bracing in the plane of sides, ends, and upper chords is commonly composed of rods. Initial tension should always be thrown into diagonal rods by 212 FRAMEWORK screwing up the turnbuckles or adjustable ends. Stiff bracing should be made short, and should be brought into position for riveting by using drift pins; to accomplish this there should be not less than three rivet holes in each lateral connection. A connection for lateral rods to the chords of trusses is shown in Fig. 105. Lateral Connection Fic. 105. Cast lateral lugs for connecting lateral rods to the webs of I — beams and to heavy plates are shown in (a) and (b), Fig. 106. — Cor- = LT ii “= tN : Cast Lateral Lug (Q) Cast Lateral Lug (b) Fic. 106. Where rod bracing in the ends and sides of buildings interferes with windows and doors, or where the building is to be left open, portal bracing is used. In the latter case the bents are usually braced in pairs, ee — DESIGN OF PARTS OF THE STRUCTURE 213 although the portal bracing is sometimes made continuous. Stiff brac- ing is often placed between the trusses in the plane of the center of the building and materially stiffens the structure (see Fig. 175). PURLINS AND GIRTS.—Purlins are made of channels, angles, Z-bars and I beams, Fig. 111, where simple shapes are used. Channel and angle purlins should be fastened by means of angle lugs as shown in Fig. 107. I beam purlins are very often fastened as shown in the A. T. & S. F. R. R. shops, Fig. 175. Z-bar purlins are bolted direct- ly to the upper chords of the trusses. The channel purlin is the most economical, and the I beam purlin is the most rigid. Girts are made of channels, angles, and Z-bars, and are fastened as shown in Fig. 111. Where the distance between trusses is more than 15 or 16 feet the pur- lins and girts should be kept from sagging by running 3 or 14-inch rods through the centers to act as sag rods, the ends of the rods being fastened to the eaves and ridge (see Fig. 81). Purlin Clip Fic. 107. Where the columns and trusses are placed so far apart that the use of simple rolled shapes is no longer economical, purlins and girts are trussed. DESIGN OF PARTS OF THE STRUCTURE.—The methods of determining the sizes of the various members in a mill building will be illustrated by a few examples. For a more detailed treatment of this subject, see “Modern Framed Structures” by Johnson, Bryan and Turneaure ; “Roofs and Bridges” by Merriman and Jacoby; and other standard works on bridge design. Manufacturers of structural material issue handbooks which con- 214 FRAMEWORK tain tables that give weights, areas of sections, positions of centers of gravity, moments of inertia, radii of gyration, etc., for the shapes manufactured by the different companies. Tables are also given for the resisting moments on pins, the shearing and bearing values of rivets, standard bolts, eye-bars, bridge pins, standard connection angles, bear- ing plates, minimum size of rivets, spacing of rivets, and many other useful tables. The handbooks best known are as follows, the popular name being given in brackets: Cambria Steel (Cambria), issued by the Cambria Steel Company, Johnstown,. Pa.; Pocket Companion (Car- negie), issued by the Carnegie Steel Company, Pittsburg, Pa.; Stand- ard Steel Construction (Jones & Laughlins), issued by Jones & Laugh- ins, Limited, Pittsburg, Pa.; Steel in Construction (Pencoyd), issued by A. and P. Roberts Company, Philadelphia ; and Structural Steel and Iron (Passaic), issued by the Passaic Rolling Mill Company, Pater- son, N. J. These books can be obtained for from 50 cts. to $2.00. The American Bridge Company issued, in 1901, a book entitled Standards for Structural Details, for use at its various plants. The Carnegie handbook was formerly very generally used in de- signing offices, but recently the supply has been limited so that the Cam- bria handbook has taken its place in schools and in many offices, and for this reason references will be made to Cambria in obtaining weights, properties of sectiohs, etc. All references to Cambria will be to the 1903 edition. Design of Trusses.—The method of determining the proper sizes of the truss members will be illustrated by designing a few of the members of the truss in the transverse bent of the mill building shown in Fig. 53; the stresses in which are given in Table VI. The secondary members will be omitted from the truss in the design, as they were in obtaining the stresses. | The material will be assumed to be medium steel and the allow- able stresses as given in Appendix I, will be taken. The allowable stresses are as follows: ; Pension > visa sesoreee es 16,000 lbs. per sq. in. Compression iss aa.ivewete 16,000 — 701-~r lbs. per sq. in. a a ES a ee ~~ DESIGN OF TRUSSES . 215 where / = the length of the member in inches, and r = radius of gyra- tion of member in inches. Mewereeand Fins, Dearing. ... . xsis as o's 60ce eesiee a 22,000 lbs. per sq. in. CAEN FS BUCS Te OL ou a pas ged gM Le PEO eee eorims, bending on extreme fibre................ BEC i oe Pee Plate Girder webs, shear on net section......... SOD pois 8s cme Compression Members. Piece 4-2. Maximum stress = + 34,300 lbs. The upper chord will be made of two angles with unequal legs placed back to back, with the shorter legs turned out, and separated by 3g-inch gusset or connection plates. Try two 4” x 3” x 5-16” angles. From table on page 187 Cambria, the least radius of gyration, 7, is 1.27 inches. The unsupported length of the member is 8.5 feet, and ]—-r = 102 + 1.27 = 80. The allow- able stress per square inch = 16,000 — 70] ~r = 16,000 — 5,600 = 10,400 lbs.. The area required will be 34,300 + 10,400 = 3.3 sq. in. The combined area of the two angles is 4.18 sq. ins. (page 170 Cam- bria), which is somewhat large. Try two 3%” x 3” x 5-16” angles. From the table on page 186 Cambria, ry = 1.10 inches; then / — r = 93, and allowable stress is 16,000 — 70 x 93 = 9,490 lbs. per sq. in. Required area = 3.62 sq. in. The area of the two angles is 3.88 sq. in., so the section is sufficient. To make the two angles act together as one piece it is necessary to rivet them together at intervals, such that the two angles acting singly will be stronger than the two angles acting together. On page 170 Cambria, the least radius of gyration of a 344” x 3” x 5-16” angle about a diagonal axis is 0.63 inches. ‘The angles must therefore be riveted at least every 0.63 x 93 — 58.6 inches. It is the common practice to rivet angles in compression about every 2% to 3 feet. The truss will be shipped in two parts and in order to avoid a splice, and because the difference in the stresses is small, the entire top chord will be made of two 3%” x 3” x 5-16” angles. 216 FRAMEWORK Tension Members. Member 1-2. - Maximum stress = — 24,900 Ibs. The net area required is 24,900 — 16,000 = 1.56 sq. in. The gross area of the section must be such, that there will be a net area of not less than 1.56 sq. in. after the area of the rivet holes in any section has been deducted. Try two 3” x 3” x 4” angles. It will be necessary to deduct the area of one rivet hole from each angle. The diameter of the rivet hole deducted is taken 4% inch larger than the diameter of the rivet before driving. Assuming the rivets as 54 inch, it will be necessary to deduct 0.19 sq. in. from each angle (page 310 Cambria). The net area of two 3” x 3” x %” angles is 2.88 — 0.38 = 2.50 sq. in. The section is somewhat large, but will be used, for the reason that ane much smaller than these will be deficient in rigidity. The angles will be riveted together about every 3 feet to make them act as one member. Member 5-6. Maximum stress = — 5,000 lbs. The net area required is 5,000 -- 16,000 = 0.32 sq. in. The gross area of the section must be such that there will be a net area of at least 0.32 sq. in. after the area of the rivet holes in any section has been deducted. Try two 2” x 2” x 14” angles — the minimum angles that can be used under the specifications. Deducting the area of one rivet the net area is 1.88 — o. 38 = 1.50 sq. in. The section appears to be exces- sively large and one 2” x 2” x 14” angle will be tried. Where angles in tension are fastened by one leg the specifications require that (para- graph 35) only one leg shall be counted as effective, or the eccentric stress shall be calculated. The net area of the one 2” x 2” x %4 ” angle when fastened by one leg, will then be % 0.94 — 0.19 = 0.28 sq. ins., which is insufficient. One 214” x 214” x 4" angle will have a net area of 0.40 sq. in., which will be sufficient. However, since it is preferable to make tension members of symmetrical sections, the member will be made of two 2” x 2” x 14” angles. DESIGN OF COLUMNS 217 Alternate Tension and Compression.—Where members are subject to alternate tension and compression the specifications require that they be designed to take each kind of stress, (paragraph 32). Member 4-y. Maximum stresses = (— 21,300 and + 2600 lbs). Try two 3” x 3” x 4%" angles — the same as member 1-2. The net area required for tension is 21,300 -- 16,000 = 1.34 sq. in. The net area of two 3” x 3” x 4” angles is 2.88 — 0.38 = 2.50 sq. in. which is ample for tension. The least radius of gyration is r = 0.93 inches (page 185 Cam- bria). Length = 108 inches, and /— r= 117. The allowable stress per square inch = 16,000 — 70 x 117 = 8,810 lbs. Required area = 0.30 sq. in. The section appears to be large, but it can not be made much smaller without exceeding the maximum limit of 125 for ] ~ r. Two 3” x 2%” x 14” angles will be found by a similar calculation to be sufficiently large, and will be used. Member 3-4. Maximum stresses = (— 10,900 and + 13,600 lbs). Try two 2” x 2” x 4” angles. The area required to take the ten- sion is 10,900 — 16,000 = 0.69 sq. in. The net area of the two ~ angles is I 88 =..01 38 = 1.50 sq in., which is ample for tension. The least radius of gyration is r = 0.61 inches (page 185 Cambria). The length is 108 inches, and 1+r= 177. This is greater than the max- imum allowed of 125, and a larger section must be used. Try two 3” x 2” x 44” angles, with short legs out. In this case 1+ yr equals 108 - 0.89 = 120. The allowable stress per sq. in. is 16,000 — 70 x 120 = 7,600 Ibs. The required area for compression is 13,000 -- 7,600 = 1.79 sq. in. The area of the two angles is 2.38 sq. in., which is ample. The section is sufficiently large to take both ten- sion and compression, and will be used. | Design of Columns.—Columns must be designed to take the stress due to direct loading, to eccentric loading, and to wind moment. The method of column design will be illustrated by the design of the leeward column in the transverse bent shown in Fig. 56; the stresses for which are given in Table VI. 12 218 FRAMEWORK Direct stress in A-17 = 14,900 Ibs.; and bending moment 924,000 inch-lbs. A 4-angle laced column will be used. : Try four 4” x 3” x 5-16” angles, long legs out, and a depth of 18” out to out of angles; 3%” lacing and connection plates will be used. The radius of gyration of two 4” x 3” x 5-16” angles with the long legs out, is found on page 189 Cambria to be 1.93 inches. The moment of inertia of a section of the post about the shorter axis is Il=4l + 4A = 4x 1.65 + 4x 2.09 (9.00 — 0.76)? = 574.24 : | and the radius of gyration is ae = 8.3 inches. The maximum fibre stress will occur on the windward side of the post and will be found by substituting in formula (30a) to be 14,900 | 924,000 x 9 fuls ts ae pe 34 14900 X 2408 | 280,000, 000 gt = 1780 + 14,560 = 16,340 lbs. per square inch. rae The allowable stress per square inch for direct loads is 16,000 — 70 1 +- r = 16,000 —°70 x 29 = 14,000 lbs.; and since the wind mo- ment comes only occasionally we will increase the allowable stress for direct loads by 25 per cent when wind loads are considered, making an allowable stress of 14,000 x 1.25 = 17,500 lbs. per square inch. The section chosen is therefore sufficiently large. The direct load will have to be carried by the column, and it will be necessary to investigate the column about its longer axis. For this case ] + 7 = 240 -- 1.93 = 125, which is allowable under the specifica- tions, and the section is ample. | The lacing will be designed to take the shear, which is 5, KG Ibs. below and 12,800 Ibs. above the foot of the knee brace. ‘The maximum stress in the lacing will be 12,800 x sec 30° = 14,700 lbs. ‘The al- ~ lowable tensile stress per square inch will be 16,000 x 1.25 = 20,000 TABLES 219 TABLE XI. RIVET SPACING IN ANGLES. =, 1 Ga Ge Leg. G Max. Leg. Rivets. = ey Ca Bivors Inches. | Inches. | Inches. Inches. | Inches. | Inches. | Inches. 8 41g % 8 3 3 % ES 4 “ 4 21g 3 “ 6 314 «“ 6 214 216 “ 5 3 be 5 9 1% rT: 4 24 « 31g 2 i Where 6” Angle Exceeds %’. 8 =. 2 144 4 6 213 2 % 144 % % TABLE XII. MAXIMUM SIZE OF RIVETS IN BEAMS, CHANNELS, AND ANGLES. I-BEAMS. CHANNELS. ANGLES. Depth | Weight | Size | Depth | Weight} Size | Depth | Weight | Size | Length] Sixe | Length Size of per of of - per of of per ot of of of of Beat. | Foot. | Rivet.) Beam. | Foot. | Rivet./Channel| Foot. | Rivet.| Leg. | Rivet.| Leg. Rivet. Tnsk’s. | Pounds. |Inches.jInches. | Pounds.| Inches.| Inches. | Pounds. |Inches.| Inches.|Inches.| Inches. Inches S15 e513 | 159142.0) 3¢ 3}4.0| 3% | #4 1Mi 241 % 4] 7.5| % | 15|60.0) x 4} 5.25) % |1 %| 241 BK 5|9.75| % | 15180.0] % 516.5|%11%]%| 3 6 12.25] 5 | 18|/55.0] % 6|8.0| %113%|/% | 3% 1 7115.0 | 5% | 20}65.0) 1 7| 9.75] % 11% | 3%4 4 1 8118.0 | 3 | 20/80.0} 1 8 |11.251 % |1%1%| 4% 1 9 |21.0 | % | 24/80.0) 1 9 113.25) 3% |1% | % | 5 1 10 |25.0 | x 10 |15.0-| % |2 | 3% | 6 1 12 (31.5 | % 12 |20.50] 4 |2¢ | % | 7 1 12 140.0 | x 15 133.0 | 3% |2,4 | ¥ 220 FRAMEWORK TABLE XIII. RIVET SPACING. iy Minimum — [ at ——— aig in|Distance from Edge of Piece to 0 $ 0 anges 0 Rivet. Piteh. a Chords xe Gird’s. Genter of: Rivet Hola. embers, Inches, Inches. Inches. Inches, Minimum, Usual, Inches. Inches, 4% 34 eeee Ce . % 1% cove ee % 1% eeee eree * eee 8 1% 2% 4 +3 1% 34 2% 3 4 1% 1%: 7 278 3% . I; 134 1 3 + 4 114 2 TABLE XIV. SHEARING AND BEARING VALUE OF RIVETS IN POUNDS. Diameter of ——— aC Rivet Area | Single Bearing Value for Different Thicknesses of Plate in ¥ in | Shear Inches, at 22,000 Pounds per Square Inch. z Inches —_|Square|at 11000 on Inches.} Lbs. = hes j Frac- | Deci- % ti ¥% iG % is % 16 34 = tion mal , : ¥% |.375].1104| 1210 2060)2580|3090 % |.500].1963| 2160|2750 3440[4130/4820 5500 5 —|.6251.3068| 3370 |3440/4300)5160/6020] 6880] 7740} 8600 % =|. 750|.4418] 4860 41306160 6190 7220 8250] 9280] 1032011340} 12380 % |.875|.6013| 6610/4810 6020 7220|8430| 9630| 10840 12040113040 14440 1 |1.00|.7854| 8640 |5500|6880 sasipeso 11000] 12380] 1375015130 504 All bearing values above or to right of upper zizgag lines are greater than double shear. Values below or to left of lower zigzag lines are less than single shear. lbs. ‘The required net area for tension will be 14,700 +- 20,000 = 0.74 square inches. The gross area of a 3” x 3@” bar is 1.125 square inches (page 388 Cambria) and the net area after deducting for one 34" rivet is 0.795 square inches (page 310 Cambria) which is sufficient for tension. DESIGN OF PLATE GIRDERS ; 221 The allowable stress for compression is (16,000 — 701--r) 1.25. The moment of inertia of a 3” x 3%” bar is 0.0135, and the radius of gyration is 0.11 inches. The ends of the lacing bars are practically fixed, and it will be assumed that the length c. to c. of rivets will as a result be shortened by one-half. Then +r = 75, the allowable stress will be 13,340 Ibs. and the required area 1.10 square inches. ‘The lac- ing bars are therefore sufficient to take either the tension or compression, Lacing bars 214” x 3%” will be found sufficient below the foot of the knee brace. The allowable shear on each rivet in the lacing will be(Table XIV) 4860 x 1.25 = 6,075 lbs.; and the allowable bearing will be 6,190 x 1.25 = 7,740 lbs. The stress in the lacing bars below the foot of the knee brace is 5,500 x sec 30° = 6,300 lbs.; the 34-rivets are all right for bearing but are not quite large enough for shear, however it is so near, that they will be used. Above the foot of the knee brace it will be necessary to increase the thickness of the lacing bars and put two rivets in each connection as shown in Fig. 102, or use a solid plate. In designing the bases of columns hinged at the base, part of the stresses may be assumed to pass directly to the base plate if the abutting surfaces have been milled; but in columns fixed at the base all of the stresses must be transferred by the rivets. The rivets must be designed to take the direct stress and the stress due to bending moment; the so- lution is similar to that for anchorage (Fig. 61) and will not be given. Design of Plate Girders—The maximum moments and shears are found as described in Chapter X. If the plate girder were de- signed by means of its moment of inertia, as in the case of rolled sec- tions, about % of the web would be effective as flange area to take the bending moment; or deducting rivets about % would be found ef- fective. It is, however, the common practice to assume that all the moment is taken by the flanges, and that all the shear is taken by the web; and this assumption will be made in the discussion which follows. Flange Stress—The stress, F, in the flanges at any point in a plate girder is F=M—+h (80) 222 FRAMEWORK where MM = bending moment in inch-pounds, and h = the distance between centers of gravity of the flange areas (effective depth), (a) Fig. 108. The net flange area, 4, will be A = F + f where f = the allow- able unit stress. The tension flanges of plate girders are designed as above, and the compression flanges are made with the same gross area. & pf o 0 0 0 0 0 0 O fol oOo o } ° $ lo s Flange Angles ¢ 3 < ¥ 2 Web Plate & Pe : ° & 0 4 = i ° er q ° Flange Angles ° , Oo} 0 ° ©1680. 8. 0 6 oO |} (b) Q, ° ==: x $ l | bo | | $35 ot lo ° 9 geet {o Y | 3 S Web.—tThe web plate should not be less than 5-16 of an inch in thickness although %4-inch plates may be used if provided with suf- ficient stiffeners. ‘The shear in the web is commonly assumed as uni- formly distributed over the entire cross-section of the plate. Siiffeners—There is no rational method for the design of stif- feners. If they are placed at distances apart not exceeding the depth of the girder, nor more than 5 feet, where the shearing stress is greater Z ‘ a = 5 DESIGN OF PLATE GIRDEKS 223 than given by the formula — allowed shearing stress = 12,500 — 90 H, where H = ratio of depth to thickness of web plate, the stiffeners will be near enough together. Where the shearing stress is less than given by the above formula, stiffeners may be omitted or spaced as desired, Stiffeners are commonly designed as columns, free to move in a di- rection at right angles to the web, with an allowed stress P = 12,000 — 55/7. Stiffeners should be provided at all points of support and un- der all concentrated loads, and should contain enough rivets to transfer the vertical shear. Web Splice ——In the plain web splice shown in Fig. 108, the rivets take a uniform shear equal to S — nu, where S is total shear, and n is number of rivets on one side of splice, and a shear due to the shearing stress not being applied at the center of gravity of the rivets. This is the problem of the eccentric riveted connection, which has been dis- cussed in Chapter XV. If the web is assumed to take part of the bending moment there will be an additional shear due to bending moment. Rivets in the Flanges.—In Fig. 108, let S = the shear in the girder at the given section, ’ = distance between rivet lines, p = the pitch of the rivets, and r = the resistance of one rivet (7 is usually the safe bearing on the rivet in the web). Then taking moments about the lower right hand rivet, we have Sp = rh’, andp=rh’ —-S © (81) Where the rivets are in double rows as shown in (d), the distance h’ is taken as a mean of the distances for the two lines. . The crane loads produce an additional shear in the rivets, (e) Fig. 108, which will now. be investigated. We will assume that the rail dis- tributes the load over a distance of 25 inches; this distance will be less for light rails and more for heavy rails. The maximum vertical shear on one rivet will-be Pp — 25 = 0.04 Pp. The horizontal stress due to bending moment is r = Sp ~ h’, and the resultant stress from the two sources will be (0.08 Pe)? +(32)' and solving for p p= ¥—— (0.04 P)? +( =i Crane Girders——The maximum moments and shears in ¢ : ders are found as explained in Chapter X. For small cranes | 2 girders are commonly used, and are designed by the use of thei ments of inertia. Plate girders are designed as previously desc In designing both rolled and plate girders care must be used to ly support the girder laterally. ia a . J “ a CHAPTER XVIII. — CorRUGATED STEEL. Introduction.—Corrugated steel is made from sheet steel of stand- ard gages, and is either galvanized at the mill or is left black. The black corrugated steel is usually painted at the mill and is always paint- ed after erection. Paint will not adhere well to the galvanized steel until after it has weathered unless a portion of the coating is removed by the application of an acid. The common standard for the gage of sheet steel in the United States is the United States Standard Gage, and this should be used in specifying the weight and thickness. The thickness and weights per square of 100 square feet, for black and gal- vanized sheet and corrugated steel are given in Table XV. The weights of the corrugated steel given in the table are for standard corrugations, approximately 214 inches wide and 54 of an inch deep. If black sheet steel is painted, add about 2 lbs. per square. _ TABLE XV. WEIGHT OF FLAT, AND CORRUGATED STEEL SHEETS WITH 2)4-INCH CORRUGATIONS. lee TThickness| Weight per Square (100 sa-ft-.) Gage No. in | __ Flat Sheets Corrugated Sheets inches Black _|Galvanized|Black Painted |Galvanized /6 0625 250 266 re fe 29/ 18 -0500 200 2/6 220 236 20 O37 /50 /66 169 /82 22 03135 129 /4/ 138 154. 24 0250 /00 1/6 /// 127 26 :0/88 73 Sf 84. EA 28 0/96 63 79 69 66 Corrugated steel is also made with corrugations 5, 3 and 1% inches wide approximately. Corrugated steel with corrugations 14 226 CORRUGATED STEEL inches wide and 3% of an inch deep is frequently used for lining build- ings. Corrugated steel with 114-inch corrugations weighs about 4 per cent more than steel of the same gage with 24-inch corrugations. Cor- rugated sheets are commonly made from flat bessemer steel sheets, by rolling one corrugation at a time. Iron and open hearth steel corrugated sheets can be obtained, but are very hard to get and cost extra. The standard sheets of corrugated steel with 214-inch corrugations, are 28 inches wide before, and 26 inches wide after corrugating, and will cover a width of 24 inches with one corrugation side lap, and ap- proximately 2114 inches with two corrugations side lap, (c) and (a) Fig. 109. Special corrugated steel can usually be obtained that will cover a width of 24 inches with 114 corrugations side lap, (b). Cor- rugated steel should be laid with 6 inches end lap on the roof and 4 inches end lap on the sides of buildings. Corrugated Roof Steel Side Lap 2 Corrugations —— CoversClé - ee — Covers 212" - —— $1 | | Ul “a e x w22"- b- 28 wide before corrugaring eat 4 ee ee gs a ” (a) Special Cor. Roof Steel Side Lap Iz Corrugations —— Covers 24"— >< — Covers 24" — —— ORS NT em “23'>) 30" wae before corrugaling L/F" 4» after ” Lrha@ Lap for Froof 6" (b) Corrugated Siding Steel Side Lap | Corrugation —— Covers 24"— >< - Covers 24" —-— CONS omer ~--F2 a” | “ - * ie 23-5 C8 wide berorecorrugaring K?6" » offer ” End Lap for Sides 4" (C) Fic. 109. FASTENING CORRUGATED STEEL 2247 Stock lengths of corrugated steel sheets can be obtained from 5 to 10 feet, varying by one-half foot. Sheets of any length between 4 and 10 feet can usually be obtained directly from the mill without extra charge. Sheets from 48 to 5 inches long, cost from I-10 to % cents per pound extra. Sheets from 10 to 12 feet long are very hard to obtain and cost extra. Sheets cannot be obtained longer than 12 feet. Stock lengths of sheets should be used whenever possible as odd lengths often delay the filling of the order. Bevel sheets should preferably be ordered in multiple lengths and should be cut in the field. Sheets to fit around windows and doors should be cut in the field; no part of a sheet less than % the width of a full sheet should ever be used. Wii Mi = == == = SECTION A-A ae o 3s 6 S12" ee ee a a | (==> Fic. 110. For cutting and splitting corrugated sheets in the field the rotary shear shown in Fig. 110 is invaluable. It will make square or bevel cuts, or will split sheets without denting the corrugations. The shear shown in Fig. 110 is one made by the Gillette-Herzog Mfg. Co., Min- neapolis, Minn., and was used by the author in the erection of a steel stamp mill in Northern Michigan, while in the employ of the above named company. ‘The shear is not on the market, but can be made in any ordinary machine shop at a comparatively small cost. Fastening Corrugated Steel—Where spiking strips are used, the corrugated steel is fastened with 8d barbed roofing nails 34 to 2% inches long, spaced 6 to 8 inches apart. The 2%4-inch barbed nails should be used for nailing to spiking strips and to sheathing whenever possible. For weight of barbed roofing nails see Table XVI. 228 CORRUGATED STEEL TABLE XVI. NUMBER OF BARBED ROOFING NAILS IN ONE POUND, Size | eokes | Ne” | opel: |) SEP eee ee 4d 1% 13 339 20d 4 6 6d 2 12 205 30d 4u 5 8d 2% 10 96 40d 5 4 10d 3 9 63 50d 5 3 12d 3 8 52 60d 6 2 16d 3% 7 38 The common methods of fastening corrugated steel directly to the purlins and girts are shown in Fig. 111. Nailing pieces should pref- erably be used where anti-condensation roofing, Fig. 127, is used, or where the sides are lined with corrugated steel. The clinch nail is prob- _-fvers and clinch nails go aor! Wrough top of corrugations ( at eat ee Methods of Fastening es Corrugated Steel to Purlins oe ail, roo! ’ Kiel gral” Table of Clinch Nails “™ eet LPuriinieg] 3 | # | S| oO] 7 of Length . BT abe ns hi) a No.per Ib. | 32 | 29 P ges Z2/ 18 C Purlin leg Bile ee as ww Pie at shar Length OO 17or84 9" 4 SOFT No.per ib. | 29 | 2/ /8 1/6 | /4 Fic. 111. METHODS OF FASTENING CORRUGATED STEEL TO PURLINS AND GIRTS. ably the most satisfactory fastening for the usual conditions. The side laps are fastened together by means of copper or galvanized iron clos- RT nN eet ee ee a, ee ; a 4 4 WEIGHT OF COPPER RIVETS 229 ing rivets, spaced about 8 to 12 inches apart on the roof and about 2 feet apart on the sides. Clinch nails are made of % inch or No. Io soft iron wire and are clinched around the purlin. The usual sizes and weights of clinch nails for different lengths of angle and channel purlins are given in Fig. 111. Care should be used in punching the holes in the corrugated steel for clinch nails and rivets to get them in the top of the corrugations and to avoid making the hole unnecessarily large. Clinch nails are spaced from 8 to 12 inches apart. ‘Two clinch nails are usually furnished for each lineal foot of purlin and girt. i Straps are made of No. 18 gage steel, 34 inches wide, and are TABLE XVII. NUMBER OF COPPER RIVETS IN ONE POUND. ere Length of Rivets in inches. r Meee os fh Pel ag ise cl eh me | wh wl le) nel me 3 70 4 78 5 85 64 60] 53| 48] 46} 44] 39] 36) 32 6 | 180 105] 100} 96] 90] 74] 68| 61| 56] 54] 50) 46 7 | 368] 211] 180] 171| 160] 150] 140] 132] 110] 97] 91] 79] 72] 63 8 | 417| 266] 248] 227] 200| 172] 157] 147| 136] 116| 100] 93| 88] 71 9 | 600} 365] 336] 261) 248] 228| 220) 184] 169] 156] 133] 124] 113) 99 10 | 820} 411] 376| 336| 305] 257] 249| 223] 206] 180| 162 11 | 944] 416] 400] 360] 338) 320 | 12 | 1167] 545| 475| 400] 342] 325] 308] 292] 257) 221] 190 13 | 1442] 799| 640| 547| 502] 448] 400] 392] 316 14 | 1620/1040] 995| 816] 784] 616| 550| 528 i ae ee) fastened with two 3-16-inch stove bolts 34 inches long. Straps are spaced 8 to 12 inches apart. One strap and two bolts are usually fur- nished for each lineal foot of purlin and girt. One bundle of heop steel for making straps contains 400 lineal feet and weighs 50 lbs. Clips are made of No. 16 gage steel, 114” x 24%4”, and are fastened with two 3-16-inch stove bolts 1%4 inches long. Clips are spaced from 8 to 12 inches apart. One clip and two bolts are usually furnished for each lineal foot of purlin and girt. 230 CORRUGATED STEEL Copper rivets weighing about 6 pounds per 1000 rivets have com- monly been used for closing rivets; but galvanized iron rivets made of very soft wire and weighing about 7 pounds per 1000 rivets are fully as good, and cost 7 cents per pound in 1903 as compared with about 25 cents per pound for copper rivets. The weight of copper rivets is given in Table XVII. Strength of Corrugated Steel—The safe load per square foot for corrugated steel supported as a simple beam, for sheets with 24-inch corrugations and of various gages is given in Fig. 112.. This diagram is based on Rankine’s formula _ 4 She Oi? (8 where W = safe load in lbs.; S = working stress in Ibs. ; h = depth of the corrugations in inches; b = width of the sheet in inches ; t = thickness of the sheet in inches; 1 = clear span in inches, fi taza we 4 Shot (Rankine) W=safeload ~ ._.. 3° S = Working Stress =15000 Ibs, 99 FN h = depth of corrugation-ins. g = b = width of sheet in inches § 80 ENS t = thickness of sheet in inches ‘B 1 = clear span in inches. a ¢ - 95 »t 9 2 wENS as 5 £ Sas <5 SS 2 w $b ESN BE 40 ae = 550 - 24 3 io = ane +k + ye AS RE OL MT aE “2 ‘3 4 rhe S pan,Z, in Feet. Fic. 112. SAFE UNIFORM LOAD IN POUNDS FOR CORRUGATED STEEL FOR DIFFERENT SPANS IN FEET. A summary of experiments to determine the strength of corrugated steel made by the author’s assistant, Mr. Ralph H. Gage, is given in STRENGTH OF CORRUGATED STEEL 231 Table XVIII. ‘The coefficient C in column 8 depends on the angle that the metal makes with the horizontal axis and varies as follows: angle Bre. °C = 0.2783 45°, C = 0.203; 60°, € = 0.312; and for 90°, C = 0.393. TABLE XVIII. SUMMARY OF EXPERIMENTS TO DETERMINE THE STRENGTH OF COR- RUGATED STEEL.* 1 2 3 A. 5 6 i ' 8 9 10 Width Thick-| Angle Tensile Gage’'s Actual} Rankine’s No. a of sii a — aoe i er Formula peed Formula. orru- Ss. per a _lgations with sq. in. W = CSh4.| “yy |W= is: SAdE Ins. Ins. Ins. Axis Ins. Ib Y lbs. Z Ss. lbs. 1 | 2.50 |0.6025) .0588/39°11'| 44.0 | 58,000 643 630 597 2 | 2.50 10.612 | .0568/39°10’| 44.0 | 58,000 632 630 587 3 | 2.50 0.625 | .066 |39°30'| 44.0 | 58,000 745 720 692 4 | 2.50 |0.606 | .0655|/40° 42’| 44.0 58,000 725 700 670 5 | 2.88 10.650 | :0367/36° 0'| 43.25) 67,000 505 500 475 6 | 2.88 10.650 | .0366/36° 0'| 44.0 | 67,000 494 490 465 7 | 2.50 10.630 | .0366/36° 0’| 44.0 | 50,000 358 350 335 8 | 2.50 |0.61 .0365|36° 0'| 44.0 50,000 344 340 324 9 | 1.25 |0.27 .0365 36° 0'| 24.0 | 50,000 281 ie 262 10 | 1.25 |0.27 .0365|36° 0’| 24.0 | 50,000 281 29 262 21°} 1.25 (0:27 .0293/36° 0'| 24.0 50,000 225 200 211 12h 4.25-10.27 .0293/36° 0'| 24.0 | 50,000 225 195 211 13 | 1.00 |0.18 .0291/36° 0'| 24.0 | 50,000 298 310 279 14 | 1.00 |0.18 .0291/36° 0’| 24.0 | 50,000 298 300 279 15 | 1.00 |0.18 .026 (36° 0'| 24-0} 50,000 266 280 250 16 | 1.00 |0.18 .026 |36° 0’| 24.0 | 50,000 266 260 250 The actual breaking load agrees in most cases more closely with Gage’s formula than with Rankine’s, although the latter is more often on the safe side. Purlins are commonly spaced for a safe load of 30 Ibs. per square foot as given in Fig. 112; if the purlins are spaced farther apart than this, the steel will deflect a dangerous amount when walked on, and will leak snow and rain. Girts should be spaced for a safe load of about 25 Ibs. per square foot. From an inspection of Fig. 112, it is evident that corrugated steel lighter than No. 24 is of little use for mill buildings. *For details of experiments see article by Ralph H. Gage, in the Technograph, No. 17. ne 232 CORRUGATED STEEL Corrugated steel of No. 26 or 28 gage is so thin that it soon rusts out and should never be used unless for lining cheap buildings. Corrugated Steel Details—Ridge Roll.—The ridge roll most commonly used is made from No. 24 flat steel, and has a 21-inch roll and 6-inch aprons. It comes in 96-inch lengths and should be laid with 3 inches end lap. Plain and corrugated ridge roll are used (see Fig. oo Se fea, yT ——— pail eat CORRUGATED RIDGE ROLL. Gable Cornice Eave Cornice ————, ie" PLAIN RIDGE ROLL. 6 Flashing for Stack y- Flat Sieel te" 4 J Flashing . . CORRUGATED SIDE FLASHING. Outside Corner Finish Fic. 113. FLASHING 233 | 113). Ridge roll is fastened with rivets or nails spaced 6 to 8 inches apart. Flashing.—F lashing is used where the roof changes slope, around chimneys and openings in the roof, and over windows and doors, and should be of sufficient dimensions and so arranged that at least 3 inches vertical height is obtained between the edge of the flashing and the end of the corrugated steel roofing. Vertical and horizontal seams of all flashing should be closely riveted. Flashing is made from flat sheets Bin. JIN. 3 6iNn. y Sin. : H 41N. aes | % 3am. aeeesenenene sweeter 2'/21N. BAIN: 4.1N N a EAVES TROUGH HANGERS Fic. 114. CO ge ee ES ie os 234 CoRRUGATED STEEL of the same gage as the corrugated steel, and can be obtained up to 96 inches in length. Flashing is made both plain and corrugated (see Fig. 113). Corner Finish—Corner finish is made in various ways, three of which are shown in Fig. 113. Other methods are shown on the suc- ceeding pages. | el f 6te° 4 5té 0? Ag tT A ae _ Bol cal CIP LA 2 i es 3 / 4 ; / "20> ! ACHICA /6 | sé | j Ni f y Adjustable hanger \, ai every 4-0". i Hanging Gutter Hanging Gutter Fic. 115. F ote Ping” Wg 1 > Cees. ai Slope qbast{” .. a Clinch EAI in 1071015 FECL) A I Sie ge Oe a7 Basso Jal Valley Gutter DIC PEs, Conductors are made plain round or square, and corrugated round or square. Corrugated conductors are to be preferred to plain conduc- tors for the reason that they will give when the ice freezes inside of them, and will not burst as the others often do. Common sizes of round pipe are 2”, 3°, 4”, 5”, and 6” diameter. Common sizes of square Pipe are 134” x 214”, 236” x 344", 234” x 414” and 334” x 5S’, equal to 2”, 3” and 4” round pipe, respectively. Conductor pipes are fastened with hooks or by means of wire. Design of Gutters and Conductors.—The specifications of the American Bridge Company for the design of gutters and conductors are as follows: 236 CoRRUGATED STEEL Span of roof. Gutter. Conductor. up to 50 6” 4” every 40’ 50 to 70’ y ia ‘ce 40° 70’ to 100’ 8” ied eaten Hanging gutters should have a slope of at least 1 inch to 15 feet. The diagram in Fig. 118 for the design of gutters and conductors was described in Engineering News, April 17, 1902, by Mr. Emmett Steece, Assoc. M. A. Soc. C. E., City Engineer of Burlington, Iowa, as follows :— 6r 8 ef Ss | E oe eR sfisil ‘ ose 35 aig: 3gx4 vd ton € 20 Sie : Liz 15 ” eae Te Azo At ob: 2 ges |] Fed a s »| ! eS ‘| 0" te: 2 25/— » +4 4 x3 eo slee : <}| 9 Sores 8 8 8 8 BE 8 os Area in Plam, Square’ Feet. Eng.News Fic. 118. “The curves are for 4 pitch or flat roofs, to full pitch or domes. The areas are reduced to plan as shown. The minimum sizes of circu- lar and commercial rectangular conductors are given on the left side of the diagram and the sizes and the minimum cross-sectional areas of _ square gutters are given on the right hand side. To use the diagram: Assume an area of roof, say 30 x 100 ft., or 3000 sq. ft., 14 pitch and one conductor for the whole area. Note the intersection of the vertical over area 3000 and the curve of % pitch; following thence the horizontal line to the left it strikes a diameter of 5 ins. for circular, or over 3% x 43% ins. for commercial size. The next larger size would be used. ‘The minimum cross-sectional area of gut- ters is shown on the right to be about 30 sq. ins., and the side of a square conductor about 4.5 ins.” This diagram was based on a maximum rainfall of 1.98 inches per hour, ee ee ae Pas - a> ii a ee . CoRNICE 237 English practice is as follows: Rain-water or down-pipes should have a bore or internal area of at least one square inch for every 60 square feet of roof surface in temperate climates, and about 35 square feet in tropical climates. They should be placet not more than 20 feet apart, and should have gutters not less in v).dth than twice the diameter of the pipe. The practice among American architects is to provide about one square inch of conductor area for each 75 square feet of roof surface; no conductor less than 2 inches in diameter being used in any case. Cornice ——There are many methods of finishing the gables and eaves of buildings. A gable finish for a steel end, and for a brick end - Froof Steel Clinch Fiver; en i "Ou tlooker le EOE eee TE nap ae ei aay, Nl | Chnch ; t Fish Angle : Z ~Outlooker ‘ 7-Ofa-<8 Partin? _. shea /i0’=— -Ena Wall Gable Finish for Steel End Gable Finish with Brick Wall Fic. 119. wm“ 4 Ne er PEA olf “0 Vl as an xi | J ely a \ Cea Zz (pe Flashing - Clinch Friver - — eel Hiding --> Flasting vsvally \ "24 Gal Crimped Hee/ \ Flashed Finish Fic. 120. 238 CORRUGATED STEEL as used by the American Bridge Company, are shown in Fig. 119. The steel end may have a cornice made by bending the corrugated steel as shown, or a molded cornice. The flashed finish shown in Fig. 120, is used by the American Bridge Company ; it is quite effective and gives a very neat appearance. The corrugated steel siding should preferably be carried up to the roof steel. The cornice and ridge finish shown in Fig. 121, designed by Mr. .H. A. Fitch, Minneapolis, Minn., is very neat, efficient and economical. Section at Ridge Fridge Cap P 0 put ya “ib Cal Pivets SE, #22-4°x4" SES oD p . ‘€ EE aa Flashing #22-4"%g 7 Slashing - 2°X%4> ir 4 2°x4"- Section at Gables Section at Eaves, Fic. 121. The galvanized rivets are much cheaper than copper rivets, and are preferred by many to the copper rivets. The detail shown was for a small dry house in which the eave strut was omitted. | pn eae — Nailing strip on .-G cave sITUlZ 6ragpria} | Navling strponend 3 a t------44 Offer or eri T1TUSS Section} through Gable ; Girt * Corner Finish vv tv — Fic. 122. CORNICE 239 In Fig. 122, the eave cornice is made by simply extending the roof- ing steel, while the gable cornice is made by bending a sheet of cor- rugated steel over the ends of the purlins and nailing to wooden strips as shown. Section through Gable Zep 3 & SAS LA ae i \ : KO 4 Corner Finish ‘ EKO Section through Eaves Fic. 123. Sheets heavier than No. 22 should not be bent in the field. The corner finish is made by bending a sheet of corrugated steel. In Fig. 123, the eave and gable cornice are made of plain flat steel bent in the shop as shown. The eave cornice is made to mitre with the gable cornice, thus giving a neat finish at the corner. The corner finish is made by using sheets at the corners in which one-half is left plain. Natling strip,on MS ond rarter : ‘i b Section through J “find, ’ Sable Ne os Cs + je Oh ~ Corner Finish q N Section through Eaves Fic. 124. aad oo c ‘eae ih a ee Oe Oe athe" 240 CoRRUGATED STEEL In Fig. 124, the eave strut and gable cornice are molded. The twe cornices are so made as to mitre at the corners, the mitres being made in the field. A plain corner cap is put on as shown, after bending the corrugated steel around the corner. ‘Nalling strpon end VIVE Of C07 T7055. Grackets /8e-c: Poke) Ci’. Carruce of 24crimpea Ga/.steel -Crimpsé*is. Gutter of *24gGa steel fostered ro corniteand ¥ rooting - Cor. wood filler ---7% Siding of "C2 gal-cor: _ steel 12 Corrvaations. Section through Cornice Oo0o0000 - *20Gd/ cor steel "24 crimped gal. 7 Steel. Clip to 3 ios hd and rivet to rooring - Section through Gable Fic. 126. ANTI-CONDENSATION ROOFING 241 In Fig. 125, an eave purlin is used and a channel is placed along the ends of the purlins. Spiking strips should always be used as shown, and the eave purlin should be fastened to the rafter by means of angle clips. The finish shown in Fig. 126, was used by the U. S. Government and needs no explanation. Anti-condensation Roofing.—To prevent the condensation of moisture on the inner surface of a steel roof, and the resulting dripping, the anti-condensation roofing shown in Fig. 127 and in Fig. 129 is fre- quently used. The usual method of constructing this roofing is as follows: Galvanized wire poultry netting is fastened to one eave purlin LA I | oS en a dls oo Sangster tae: sos Asbestos ~~ =@¢ Tar Faper— gs Hi 0) > Galvanized Wire Cor 57ee/->F ij | i j 2 ~ Poultry Netting Anti-Condensation Roofing Bic.t27. and is passed over the ridge, stretched tight and fastened to the other eave purlin. The edges of the wire are woven together, and the net- ting is fastened to the spiking strips, where used, by means of small staples. On the netting are laid one or two layers of asbestos paper 1-16-inch thick, and sometimes one or two layers of tar paper. The corrugated steel is then fastened to the purlins in the usual way. Stove bolts, 3-16” diameter, with I x % x 4-inch plate washers on lower side, are used for fastening the side laps together and for support- ing the lining (see Fig. 129). The author would recommend that pur- lins be spaced one-half the usual distance where anti-condensation lin- ing is used ; the stove bolts could then be omitted. Asbestos paper 1-16- inch thick comes in rolls, and weighs about 32 pounds per square of 100 square feet. Galvanized poultry netting comes in rolls 60 inches wide and weighs about 10 pounds per square. The corrugated steel used with anti-condensation roofing should never be less than No. 22, and the purlins should be spaced for not less 242 Fic. 128. CoRRUGATED STEEI, PLANS FOR A TRANSFORMER BUILDING. ee a eee CORRUGATED STEEL ‘pe vt see eeecececee wo oceen woe = — = —/-—- 1 1 A & 1 oO # NSS aia | yk-- ee ee meee y Spamgeae 1 $ Ako 305% oh JOS-4\63OF3/3O@5H\5|8 1 t*®#——e- ee e ~“<—<—<—-—s + - ==} 3S 7 ‘ as ¥ ! ‘ % » WS _ A zes's" LE} 7e9:a"_ i 2S Fg 4/0 3@ 521410) SINR | yf alae - ple a Sa : t | 64@4°/0" | |4@4'9 4484-10 2h ed 404°-3|'61@4+10 1 “Sing ERE x cm h Beas: A i AT) i & — ex@oto"ll 8. eed ‘Hea 8 5! lle:e6:0" ap Peay eS TEE Le 3) SERRE 6 ERS (or) aaa ‘ ‘S ‘ 110" ox iz a Sh Lipoais £ ~ : L@ ‘101 1S FOF S | pre saeeri = } a7 i ee | 7 ee en BETS aee ard in Ya geet d MBNA 2 to ima we Pet eae ee ss zs © be ~//-0"~ nla 7-2 Ge 5 Och ‘fu 56ntbe 7! 6c 110% > be— //* £0 me H1-0"- me — — 16:0" he — 11-0 One - -//-0-> Kiss Rees 60-0—---—-------- End Elevation Lanes eee te ne aa i - © ===sssS= roa sod i eepepeperrenper Mat SS TG 926" O@9*6: 1 ‘ Jae ee Sal ng. Se i ; : ar) Rae Re arte nett li SO En lan ES ETRE | I, "Speier pope tele ppbaiete eC papige® erin SRO eRT ox ves ryt” ; bsg 4e4 Sleies'1404'S hes Opies SealuionessHieriess} 26] S| “Tr LS en ees BOE aoe oe 2S grass ry TT s ? ilel me Ae 3 3! 460] 2 1 Lebo’ | ao] & +! 42@6°0 3 1 6: s I eel Oe fammlameel OD Lncpiad st funda Oe pumiemat Sl incall oe shad BEM phates FEE seat Oe e Sol er en yy ‘2 ! ‘0" 1 £ oes ajo PETE aoe t =4.-FE=2 is ee =-J-=-2_¥ 1 ey i L457! “6 shea 15-74 6 “nef Fu 0-868 ri Opt 554 756 tcf Saf 5e-7 6A BA o Len = 5 J650 inte a 0” ~- > = -/6* 0% mek 2/60 oe 68 Side Elevation CoRRUGATED STEEL PLANS 243 Corrugated Steel List for Building Rectangular Sheets _| Beveled Sheets as per Sketch No.|U-:SSGlLenath| Marks __|No.|U-S5S.GlLength| Marks JIS|\*#22 | 45/0" + |*24|7-12"|2% NC#IR 2 4 On OE det Oe 4) *® | 5°982* 212"ZR Ye) I io Bo 4 » | 4+5212* 3}24# 3R 58| » | 6'3" 4) » 1348" |2#412* 4A (90|.» | to" 3| * |6B-lae Zaeen 48 |#24. | 4'-0" 8| » |4-8"|4* 714*7R 62| » | 449" 8| *” | 3°4"l4# gl44gR 87| » | 44/0" 8| ” | 2+ 0°l4* 9/4*9R ah PoE ie be 4 ” 1/0°0"|2#/0|2* 0R 7| » | 55" 4| * (18+8"\2# 11 |24/ 12h? aD a 4 » | 744 l2*/Z\Z2*/2R 67| ” | 60" Fo Se tiet BS dae 67| * | 9'-/0" ry iis.) I Fridge Fro// in eae We Fiat Fee/, Pte 100 lineal tech « 9° 2" k-— Length — >| Flashing oF Ginn #2? Fla? Stee! “ YU sheer steel Feart- ed ore Coat Fred Lead. JI squares Asbestos. Jheets 26 "wide. 1300 lin. tt-60'Foultry Netting Corrvgations 22" Corrugated Steel on Sides, No.l4 Black, Feurtred, /Corrugation Hide lap atria 4"Eind lap. Carrugated Sree! 017 Frook Wale Black, Feursred, 2 Corrugations Side lap ard O' Lina lap Poultry Netting \ . e440 eotetestezes Asbestos Fortin Purliv - Purlin Method of Fastening Steel and Lining on Root 1 |- Walls, ~ +, Method of Fastening Steel on the Sides _-# Cor. Steel : a ‘phe Oo: -Asbhestos ATahAA No 22 cae Steel We ei? F > f° —WireNetting Wire Netting galt on See A Ce #22Cor Stee! pert -fA a a 5 or ve tos ‘i a Ye en N by nee oO topes ~>Wire Netting : ue Z Vx- oN : Mz 12’ u/ ‘x Sx” nS ys pA A Ai P/ Washer ' . #205teI zs ’ ee “ ~ ' pS IO ie ~ Gable Rose a Ah | % Frame #16> Q\+Lovvers No.20 NN , \ ¢_! ; * Cornice \° OER se é xe t : ~ ae oy <¢- i” ae 53 ' teas 302! AA Caly. Steel’ gt * rAl e dsy } 1 oe Sy roc! | ” Lae v OP OR EE ( B~gee- SP - epee * i 8 SA & oe Epi NES + x ‘ NX : a: Eave Cornice Finish at Corner Detail of Louvres Fic. 129. CoRRUGATED STEEL LIST AND DETAILS FOR TRANSFORMER BUILDING. 244 CorRRUGATED STEEL. than 30 pounds per square foot. A less substantial roof will not usually be satisfactory. An engine house with anti-condensation lining on the roof and sides has been in use in the Lake Superior copper country for several years, and has been altogether satisfactory under trying conditions. The covering and lining of roof and sides are fastened by clinch nails to angle purlins and girts spaced about two feet apart. A transformer building designed by the author and built by the Gillette-Herzog Mfg. Co., at East Helena, Montana, has anti-condensa- tion lining on the roof as shown in Fig. 129, and is lined on the sides with one layer of asbestos paper, and 14-inch No. 26 corrugated steel. The black framework, the red side lining, and white roof lining made a very pleasing interior. This building after several years is giving en- tire satisfaction. Corrugated Steel Plans.—The shop plans, list of steel and details of the corrugated steel for a mill building are shown in Fig. 128 and Fig. 129 (for the general plans and a detailed estimate of this build- ing see Chapter XXVIII). Corrugated steel sheets should be ordered to cover three purlins or girts if possible. Bevel sheets should be ordered by number, and sheets should be split and reentrant cuts should be made in the field. All sheets should be plainly marked with the number or length. Sheets No. 22 or lighter can be bent in the field, heavier metal should always be bent at the mill. In preliminary estimates of corrug- ated steel allow 25 per cent for laps where two corrugations side lap and 6 inches end lap are required, and 15 per cent for laps where one corrugation side lap and 4 inches end lap are required. Cost of Corrugated Steel—Galvanized steel in 1903 is aia at about 75 per cent off the standard list, f. o. b. Pittsburg; list price of flat galvanized steel being as follows: No.0: to: 16: Spelusive:c so nsw bean se ee 12c. per Ib. No. 17-40 2F “inclsiven to. cys awe awe E300 es No.. 23,40 “aa inelitere 5s es cee eee EG No. 25 to 26 inclusive ...... se GACH Stine Liens so: No. 2p SoG sian sis a ois RE eee eens TOG a 4 be .. a, = Bb mm - ue “9 = ty i. ‘ — qi e. i 2% — >, a =i 4 oa Oe ee Pe ee Se ee a 7 < \ 7 r ; - ; " ary os eve ' : * ae \ ’ Ve ; vi ae ot ” ; oa aT ae ON ha Fe OL ee ee SS ee eM Mme eS ees Cost oF CORRUGATED STEEL, 245 The net cost of corrugated galvanized steel is found by adding .osc. per pound to the net cost of flat galvanized sheets. The standard card of extras used in 1903 is given below. These extras are to be added to the net price of flat black or galvanized sheets to obtain the cost. These extras are not subject to discount. CARD OF EXTRAS FOR BLACK OR GALVANIZED SHEETS. eC UPATIN owed sw kekSs dba eo ec .o5c. per Ib. For painting with red oxide ................ SSF : For painting with Dixon’s graphite.......... ROE SF “cc “cc For painting with Goheen’s carbonizing coating .30c. For all trimmings, etc., flashings, ridge caps, BG IOUVTCS ei yee cele ces hvales cae ebs I.00c. For flat sheets rolled from reworked muck bar .5oc. For sheets rolled from iron scrap mixture.... .25¢. Me ENESE Ce he re ec is bed eas keke on bes {AGG P= Black corrugated steel in 1903 is quoted about as follows, f. o. b, Pittsburg: PIG SO RO FO WNCIIRIVE Sic cde es ces 2.2c. per lb. MO Oe (0 OS MICTUSIVE 6 iS ee ces 7k a NG. 24°10 20"ineltsive 63... es 2. 9¢. CHAPTER XIX. Roor COVERINGS. Introduction.—Mill buildings are covered with corrugated steel supported directly on the purlins; by slate or tile supported by sub- purlins; or by corrugated steel, slate, tile, shingles, gravel or other composition roof, or some one of the various patented roofings sup- ported on sheathing. The sheathing is commonly made of a single thickness of planks, 1 to 3 inches thick. The planks are sometimes laid in two thicknesses with a layer of lime mortar between the layers as a protection against fire. In fireproof buildings the sheathing is com- monly made of reinforced concrete constructed as described in Chapter XX. Concrete slabs are sometimes used for a roof covering, being in that case supported directly by the purlins, and sometimes as a sheath- ing for a slate or tile roof. The roofs of smelters, foundries, steel mills, mine structures and similar structures are commonly covered with corrugated steel. Where the buildings are to be heated or where a more substantial roof cov- ering is desired slate, tile, tin or a good grade of composition roofing is used, or the roof is made of reinforced concrete. For very cheap and for temporary roofs a cheap composition roofing is commonly used. The following coverings will be described in the order given; corrug- ated steel, slate, tile, tin, sheet steel, gravel, slag, asphalt, shingle, and also the patent roofings ; asbestos, Carey’s, Granite, Ruberoid and Fer- roinclave. The construction of reinforced concrete roofing is de- scribed in Chapter XX. | Corrugated Steel Roofing.—Corrugated steel roofing is laid on plank sheathing or is supported directly on the purlins as described in Chapter XVIII. For the cost of erecting corrugated steel roofing see Chapter XXVIII. - SLATE RooFING 247 Corrugated steel roofing should be kept well painted with a good paint. Where the roofing is exposed to the action of corrosive gases as in the roof of a smelter reducing sulphur ores, ordinary red lead or iron oxide paint is practically worthless as a protective coating; better results being obtained with graphite and asphalt paints. Graphite paint has been quite extensively used for painting corrugated steel in the Butte, Mont., district. ‘The corrosion of | corrugated steel is sometimes very rapid. In 1898 the author saw at the Trail Smelter, Trail, B. C., a corrugated steel roof made of No. 22 corrugated steel and painted with oxide of iron paint that had corroded so badly in one year that one could stick his finger through it as easily as through brown paper. The climate in that locality is moist and the smelter was used for re- ducing sulphur ores. Galvanized corrugated steel is quite extensively used in the Lake Superior district. F Slate Roofing.—There are many varieties of roofing slate, among which the Brownville and Monson slates of Maine, and the Bangor and Peach Bottom slates of Pennsylvania are well known and are of excellent quality. Besides the characteristic slaty color, green, purple, red and variegated roofing slates may be obtained. The best quality of slate has a glistening semi-metallic appearance. Slate with a dull 248 Roor CovERINGS earthy appearance will absorb water and is liable to be destroyed by the frost. : Roofing slates are usually made irom % to % inches thick; 3-16- inch being a very common thickness. Slates vary in size from 6” x 12” to 24” x 44”; the sizes varying from 6” x 12” to 12” x 18” being the most common. ‘ Slates are laid like shingles as shown in Fig. 130. The lap most commonly used is 3 inches; where less than the minimum pitch of % is used the lap should be increased. | The number of slates of different sizes required for one square of 100 square feet of roof for a 3-inch lap are given in Table XIX. The weight of slates of the various lengths and thicknesses tequired for one square of roofing, using a 3-inch lap is given in Table XX. The weight of slate is about 174 pounds per cubic foot, The weight of slate per superficial square foot for different thick- nesses is given in Table XXI. The minimum pitch recommended for a slate roof is 4; but even with steeper slopes the rain and snow may be driven under the edges of the slates by the wind. ‘This can be prevented by laying the slates in slater’s cement. Cemented joints should always be used around eaves, ridges and chimneys. Slates are commonly laid on plank sheathing. The sheathing should — be strong enough to prevent deflections that will break the slate, and should be tongued and grooved, or shiplapped, and dressed on the upper surface. Concrete sheathing reinforced with wire lath or expanded metal is now being used quite extensively for slate and tile roofs, and makes a fireproof roof. Tar roofing felt laid between the slates and the sheathing assists materially in making the roof waterproof, and prevents breakage when the roof is walked on. The use of rubber- soled shoes by the workmen will materially reduce the breakage caused by walking on the roof. Roofing slates may also be swpported directly on laths or sub-purlins. The details of this method are practically the same as for tile roofing, which see. TABLES 249 TABLE XIX. NUMBER OF ROOFING SLATES REQUIRED TO LAY ONE SQUARE OF ROOF WITH 3-INCH LAP. 70 j No. of paar No of ; No. of e =a Slate in Size ast Slate in Size in Slate in Square. Inches. Square. Inches. Square. 6x12 533 8x16 277 12x20 141 7 12 457 9 16 246 14 20 121 8 12 400 10 16 221 Il . 23 137 9 12 355 12 16 184 12. 22 126 10 12 320 9 18 213 14 22 108 12 12 266 10 18 192 12 24 114 7 14 374 11 18 174 14 24 98 8 14 327 12 18 160 16 24 86 9 14 291 14 18 137 14 26 89 10 14 261 10 20 169 16 26 78 12 14 218 11 20 154 TABLE XX. THE WEIGHT OF SLATE REQUIRED FOR ONE SQUARE OF ROOF. Weight in pounds, per square, for the thickness. Length in : Inches. 1" 1 ” yy" 36" 1gn 5g" ay" 1” 12 483 724 967 1450 1936 2419 2902 3872 14 460 688 920 1370 1842 2301 2760 3683 16 445 667 890 1336 1784 2229 2670 3567 18 434 650 869 1303 1740 2174 2607 3480 20 425 637 851 1276 1704 2129 2553 3408 22 418 626 836 1254 1675 2093 2508 3350 24 412 617 825 1238 1653 2066 2478 3306 26 407 610 815 1222 1631 2039 2445 3263 TABLE XXI. WEIGHT OF SLATE PER SQUARE FOOT. Thickness—in.. ........| % | vs | & 36 Vg 5 4 1 Weight—lIbs. .........0.. 1.81 | 2.71 | 3.62 | 5.43 | 7.25 | 9.06 | 10.87] 14.5 250 RooFr CovERINGS When roofing slates are laid on sheathing they are fastened by two nails, one in each upper corner. When supported directly on sub- purlins the slates are fastened by copper or composition wire. Gal- vanized and tinned steel nails, copper, composition and zinc slate roofing nails are used. Where the roof is to be exposed to corrosive gases cop- per, composition or zinc nails should be used. Slate roofs when made from first class slates well laid have been known to last 50 years. When poorly put on or when an inferior qual- ity of slate is used slate roofs are comparatively short-lived. Slates are easily broken by walking over the roof and are sometimes broken by hailstones. Slate roofing is fireproof as far’as sparks are concerned, but the slates will crack and disintegrate when exposed to heat. Local conditions have much to do with the life of slate roofs; an ordinary life being from 25 to 30 years. First class slate 3-16 to %4 inches thick may ordinarily be obtained f. o. b. at the quarry for from $5.00 to $7.00 per square; common slate for from $2.00 to $4.00 per square; while extra fine slate may cost from $10.00 to $12.00 per square. | An experienced roofer can lay from 1% to 2 squares of slate ina day of 10 hours. In 1903 slater’s supplies were quoted as follows: Galvanized iron nails, 2% to 3 cents per lb.; copper nails, 20 cents per lb.; zinc nails, 10 cents per lb.; slater’s felt, 70 to 75 cents per roll of 500 square feet; two-ply tar roofing felt, 75 cents per square; slater’s cement in 10-lb. kegs, 10 cents per Ib. Trautwine gives the cost of slate roofs as $7.00 per square and upwards. ‘The costs of slate roofs per square is given in the reports of the Association of Railway Superintendents of Bridges and Buildings, as follows: New England, $9.00 to $12.00; New York, $9.00 to $10.00; Virginia $4.10 to $5.00; California, $10.00 to $10.50. Tile Roofing.—Baked clay or terra-cotta roofing tiles are made in many forms and sizes. Plain roofing tiles are usually 1014 inches long, 614 inches wide and 5% inches thick; wéigh from 2 to 2% pounds each and lay one-half to the weather. There are many other forms of 3 - o on rat Ks 1 ; : : F Fh > , P ag Viet a 2 1 be 2 sors eee t . ee eee ew ae a Peal — ou “ » * J — ge ee a a Ce eed 2 Me eee 1s ogee . The pressure at A will be 1 W,6WS ; PePyp +P, + —— (83) and at B will be : W 6Wé , a Pah Py a (84) as shown in (c). Now if P, is made equal to P, the pressure at B will be zero and at A will be twice the average pressure. Placing P, = P, in (84) and solving for b, we have b = Y% 1. This leads to the theory of the middle third or kern of a section. If the point of application of the load never falls outside of the middle third there will be no tension in the ma- sonry or between the masonry and foundation, and the maximum com- pression will never be more than twice the average shown in (a). i If the point of application of the load falls outside the middle third (b greater than ¥% /) there will be tension at B, and the compres- sion at A will be more than twice the average. But since neither the masonry nor foundation can take tension, formulas (83) and (84) will give erroneous results. In (d) Fig. 133, assume that b is greater than % /, and then as above, the load W will pass through the center of pressures which will vary from zero at the right to P at A. If 3 a is the length of the foundation which is under pressure, then from the fundamental con- dition for equilibrium for translation, summation vertical forces equals zero, we will have . W='Y%P3a and pPu=2W (85) 3a PRESSURE OF PIER ON FOUNDATIONS 274 Pressure of a Pier on Foundation.—In Fig. 134, let W = resul- tant of the stresses in the column and the weight of the pier, / = length, ¢ = depth and u = the breadth of the footing of the pier in feet. The bending moment at the top of the pier is M = — % H d and at the base of the pier is M, = —H (%d-+ c). Now the pier must be designed so the maximum pressure on the foundation due to W and the bending moment M, will not exceed the allowable pressure. The maximum pressure on the foundation will be see e0 It will be seen from (86) that a shallow pier with a long base is most economical. To find the relations between / and c when the maximum pressure is twice the average, place W 3 H(@+ 20) Ln n ~ pa 2 €4+2 0) (87) W W Pes and For any given conditions the value of / that will be a minimum may be found by substituting in the second member of (87). To illustrate the method of calculating the size of a pier we will calculate the pier required to fix the leeward column in Fig. 57. The sum of the stresses in column A-17 is a minimum for dead and wind load and will be (Table V) equal to 4800 + 4500 = 9300 lbs. Try a pier 3’ 0” x 3’ o” on top, 6’ o” x 6 oO” on the base and 6 feet deep, weighing about 16,700 lbs. Substituting in (86) we have p = 26,000 , 3X 4300 (14 +12) 36 6 X 36 ee (22 + 1553 278 FOUNDATIONS This gives tension on the windward side which will not do, and so we will reinforce the footing with beams and make / = Ito ft., and increasing weight so that W = 40800 Ibs. P =680 + 559 = 1239 or 1214 Ibs. per square foot, which is safe for ordinary soils. If it had been necessary to drive piles for’this pier, a small amount of tension might have been allowed on the windward side if the of the piles had been enclosed in concrete. Design of Footings.—The thickness and length of the offsets in a concrete or masonry footing are commonly calculated as for a beam fixed at one end and loaded with a uniform load over the projecting part equal to the maximum pressure on the footing. If p = projection of the footing. in inches ; # = the thickness of the footing in inches ; P = pressure on foundation in pounds per sq. ft.; and S = safe working load of the material of which the footing is made in pounds per square inch, by substituting in the fundamental formula for flexure and solving for p, -0 8 @ The values of S in common use are: firct class Portland cement concrete 50 lbs.; ordinary concrete 30 Ibs.; limestone 150 Ibs. ; granite | 180 lbs.; brickwork in cement 50 lbs. The projection and thickness of the footing course is sometimes calculated on the assumption that the footing course is a beam fixed at the center, in place of as above. ‘This solution hardly appears to be justified. Pressure of Column on Masonry.—The following pressures in pounds per square inch are allowed by the building laws of New York.—Portland cement concrete 230 Ibs.; Rosendale cement concrete — 125 lbs.; Rubble stonework laid in Portland cement mortar 140 lbs. ; brickwork laid in Portland cement mortar 250 Ibs.; brickwork laid in lime mortar 110 lbs. ; granite 1000 lbs. ; limestone 700 Ibs. It is very com: mon to specify 250 Ibs. per square inch for bearing on good Portland cement pedestals, and 300 lbs. per square inch is not uncommon. i ta ee a ALLOWABLE PRESSURES 279 Allowable Pressures.—The following unit pressures have been proposed by Mr. C. C. Schneider in “ Structural Design of Buildings.” * 1. Foundations.—Pressure on foundations not to exceed, in tons per square foot: Gan SESE Gov aa a apnan eaece Rai IC ES Olin hia vin'a co's $4 oh 00 be nee I Ordinary clay and dry sand mixed with clay................ 2 MEY HAN ABO CET CIAY (Sig vers cee ewe ken Sues oss caneessees 3 Pear clay and Mirth, CORREC SONU woe. Geese Scene See reese ceases 4 Pitt, \COATHE. BANG “ANG BTUUEII Stoel Sika ges s ces ece esas cas 6 2. Masonry.—Working pressure in masonry not to exceed, in tons per square foot: Common brick, Rosendale-cement mortar.................-. 10 Fy re Orta Cement ANOTCAL oc isc cls yes cas eons 12 Hard-burned brick, Portland-cement mortar......... ..... 15 Rubble masonry, Rosendale-cement mortar ................. 8 = < Portland-cement mortar ...........-..e08- 10 Coursed rubble, Portland-cement mortar..................-. 12 First-class masonry, sandstone .......... Miaiulae dsl nars enna a's 3 20 e . * limestone ....... Patate Sst tic oa aioe care ae e 25 - ir sf OMNES POSS A cosa tor eee ib sks sobs 30 Concrete for walls: Portlantl Cement 295.5 ooscis ccev ca ccenscccee eS Sat 20 6 ERM nee e tik vo gcc oe Ooh ere Ne awe 25 3. Pressure on Wall-Plates—The pressure of beams, girders, wall plates, column bases, etc., on masonry shall not exceed the fol- lowing, in pounds per square inch. On brickwork with cement mortar... RS Ce a he oath g wets eauhe’ 200 “rubble masonry with cement mortar...........0...0000- 200 Tite OMERDRICLMEMPAGTIE COTICTOLE oo chs aks cc cv cS bee ogc edacn see ses 350 Wr EMERISAM” BETOLOTG 7s x oie occas S vin Fob ew co Cos ee OovoNes .400 se eae RR CT os ee PaleeUn gala w bes 500 “ ec “ce - Rae lance ox 5 65 ain's do Ae oleae Sev wie Rea Oo 600 4. Bearing Power of Piles—The maximum load carried by any pile shall not exceed 40,000 lbs., or 600 Ibs. per sq. in. of its average cross-sections. * Trans. Am. Soc. C. E., Vol. 54, 1905. 280 FouNDATIONS Piles driven in firm soil to rock may be loaded to the above limits. _ q Piles driven through loose, wet soil to solid rock or equivalent bearing shall be figured as columns with a maximum unit strain of 600 lbs. per sq. in., properly reduced. | The minimum distance between centers of piles shall be 2% it. 5. Loads on Foundations——tThe live loads on foundations shall be assumed to be the same as for the footings of columns. The areas of the bases of the foundations shall be proportioned for the dead load only. That foundation which receives the largest ratio of live to dead load shall be selected and proportioned for the combined dead and live loads. The dead load on this foundation shall be divided by the area thus found, and this reduced pressure per square foot shall be the permissible working pressure to be used for the dead load of all foundations. 6. Reduction of Live Load on Columns.—For columns carry- ing more than five floors, these live loads may be reduced as follows: For columns supporting the roof and top floor, no reductions; For columns supporting each succeeding floor, a reduction of 5 per cent of the total live load may be made until 50 per cent is reached, which reduced load shall be used for the columns supporting all remaining floors. a a a CHAPTER XXII. FLoors. Introduction.—The requirements and the local conditions govern- ing the design of floors for shops and mills are so varied and diversified that the subject of floor design can be treated only in a general way. Floors will be discussed under the head of (1) ground floors and (2) floors above ground. GROUND FLOORS.—Types of Floors.—There are three gen- eral types of ground floors in use in mills and shops: (1) solid heat conducting floors as stone, brick or concrete; (2) semi-elastic, semi- heat conducting floors as earth, macadam or asphalt; (3) elastic non- heat conducting floors of wood or with a wooden wearing surface. (1) Floors of the first class have been used in Europe and form- erly in this country to quite an extent in shops and mills, and at pres- ent are much used in round houses, smelters, foundries and in other buildings where the wear and tear are considerable or where men are not required to stand alongside a machine. Floors of this class are cold and damp and make workmen uncomfortable. ‘The wooden shoes of the continental workmen or the wooden platforms in use in many of our shops which have floors of this class, overcome the above objec- tions to some extent. The gritty dust arising from most concrete floors is very objectionable where delicate machinery is used. The noise and danger from breakage and first cost are additional objections to floors of this class. (2) Floors of this class have many of the objections and defects of floors of the first class. ‘These floors are liable to be cold and damp unless properly drained, and give rise to a gritty dust that is often in- tolerable in a machine shop. 282 | FLoors | Earth and cinder floors are very cheap and are adapted to forge shops and many other places where concrete and brick floors are now put down. Floors of this class should be well tamped in layers and should be carefully drained. ‘Tar-concrete and asphalt floors are more elastic and conduct less heat than any of the floors above mentioned, but the surface is not sufficiently stable to support machinery directly, and floors of this class are very much improved by the addition of a contin- uous wooden wearing surface. (3) Floors of wood or with a wooden wearing surface appear to be the most desirable for shops, mills and factories. Wooden floors are elastic, non-heat conducting and are pleasant to work on. ‘They are cheap, easily laid, repaired and renewed. They are easily kept clean and do not give rise to grit and dust. The most satisfactory wearing surface on a wooden floor is rock maple 7% to 1% inches thick and 2% to 4 inches wide, matched or not as desired. The matched flooring makes a somewhat smoother floor and is on the whole the most satisfactory. ‘The wearing floor should be laid to break joints and should be nailed to planking or stringers laid at right angles to the surface layer. The thickness of the planking will depend upon the foundation and upon the use to which the floor is to be put. The different classes of floors will now be briefly discussed and illus- trated by examples of floors in use. Cement Floors.—The construction of cement or concrete floors is similar to the construction of cement sidewalks, the only difference being that the floor usually requires the better foundation, The foun- dation will depend upon the use to which the floor is to be put, and upon the character of the material upon which the foundation is to rest. The excavation should be made to solid ground or until there is depth enough to allow a sub-foundation of gravel or cinders. Upon this base a layer of cinders or gravel 6 to 8 inches thick is placed and thoroughly rammed. The cement concrete base, made of 1 part Portland cement, 3 parts sand and 5 to 6 parts broken stone or gravel, is then placed on CEMENT FLoors 253 the sub-foundation and thoroughly rammed. ‘The cement and sand should be mixed dry until the mixture is of a uniform color, the gravel or broken stone is then added, having previously been wet down, and the concrete is thoroughly mixed, sufficient water being added during the process of mixing to make a moderately wet concrete. The con- crete is of the proper consistency if the moisture will just flush to the top when the concrete is thoroughly rammed. The concrete should be mixed until the ingredients are thoroughly incorporated and each par- ticle of the aggregate is thoroughly coated with mortar. The wearing coat is usually made of 1 part Portland cement and one or two parts of clean sharp sand or granite screenings that will pass through a %4-inch screen. The thickness of the wearing coat will de- pend upon the wear, and varies from ¥% to 2 inches thick, 1 inch being a very common thickness. The mortar for the wearing surface should be rather dry and should be applied before the cement in the concrete base has begun to set. Care should be used to see that there is not a layer of water on the upper surface of the base or that a film of clay washed out of the sand or gravel has not been deposited on the sur-. face, for either will make a line of separation between the base and the wearing surface. The mortar-is brought to a uniform surface with a straight edge, and is rubbed and compressed with a float to expel the water and air bubbles. As the cement sets it is rubbed smooth with a plastering trowel. Joints should be formed in the floor making it into blocks about 4 to 8 feet square. Cement floors are said to be a failure for railway round houses for the reason that they flake and crack after they have been used a short time, on account of the varying changes to which they are subjected. Cement floors vary in cost, depending upon the thickness of the floor and upon local conditions. In central Illinois a cement floor hav- ing a I-inch surface‘ coat and 3 inches of concrete laid on a cinder foundation 6 to 8 inches thick can be obtained (1903) for about 12 cents per square foot. A very substantial concrete floor can usually be obtained for about 20 cents per square foot. 284 FLoors Tar Concrete Floors.—The following specifications for tar con- crete floors are given in circulars Nos. 54 and 55 of the Boston Man- ufacturer’s Mutual Fire Insurance Co., and are reprinted in Engineer- ing News, March 21, 1895. “The floor to be 6 inches thick, and to be put down as follows: The lower 5 inches to be of clean coarse gravel or broken stone, with sufficient fine gravel to nearly fill the voids, thoroughly coated with coal-tar and well rammed into place. On this place a layer 1 inch thick of clean, fine gravel and sand heated and thoroughly coated with a mixture of coal-tar and coal-tar pitch in the proportions of 1 part of pitch and 2 parts of tar. This layer is to be rolled with a heavy roller and brought to a true and level surface ready to receive the floor plank. No sand or gravel to be used while wet. “A floor of the kind above specified should siwers be protected by. a floor of wood over it, and the plank should be laid and bedded in the top surface while it is warm and before it becomes hard. “For light work the thickness of the lower layer of concrete may be reduced one or more inches if upon a dry gravelly or sandy soil. For storage purposes where the articles stored are light and trucks are little used, the following specification has been found to give a satis- factory floor: “The lower layer being mixed and put down as above specified, the top layer will be of fine gravel and sand, heated and thoroughly mixed with a mixture of equal parts of coal-tar, coal-tar pitch and paving cement, so that each particle of sand and gravel is completely coated with the mixture, using not less than one gallon of the mix- ture to each cubic foot of sand and gravel. This layer should be well rolled with a heavy roller and allowed to harden several days before be- ing used.” Brick Floors.—Brick floors are recommended as the most satis- factory floors for round houses. Round house floors on the Boston & Maine R. R. are made as follows: *Brick is laid flat on a 2-inch layer of bedding sand on well compacted earth, gravel or cinders, Joints are left open 3% of an inch and are swept full of cement grout. Round house floors are made on the Chicago, Milwaukee & St. Paul R. R., as follows: *Vitrified brick is laid on edge on a layer of sand 1 *EKighth Annual Report of the Association of Railway Superintendents of Bridges and Buildings. : : WoopEeNn FLoors 285 to 2 inches thick on a cinder foundation 6 inches thick. Fine sand is broomed into the cracks after-the brick are in place. The cost of this floor per square yard is about as follows: Material. Firebox cinders cost nothing.............. $00.00 aie nei och iosi saya so Lbae g's Wiese ee wo 0.50 Labor. Preparing the foundation ..............005 0.20. Laying the brick ....... Peele Fiske widn devia cate 0.15 Opal GOSt Der SGUATE yard... cc. ee cea ee $0.85 Total cost per square foot............... 9% cents. The cost of brick floors as given in the reports of the Association of Railway Superintendents of Bridges and Buildings varies from 9% to 13 cents per square foot. The Southern & Southwestern Railway Club—Eng. News, Jan. 16, 1896—recommends that round house floors be made of vitrified brick laid as follows: Make a bed surface of slag or chert about 18 inches thick, then put a coat of sand over slag, lay brick on edge and level them up by tamping. After this is done a coat of hot tar is applied which enters the space between the bricks and cements them together. Wooden Floors.—Coal-tar or asphalt concrete makes the best foundation for a shop floor. If Portland cement is used, the planking will decay very rapidly unless the top of the concrete is mopped with coal-tar or asphalt. A floor laid by Pratt & Whitney Co., of Hartford, Conn., is described as follows: “In laying a basement floor about 18 years since of 10,000 square feet, 8,000 square feet were laid over coal- tar and pitch concrete in about equal proportions, and about 2,000 square feet were laid over cement concrete. ‘The latter portion of the floor was removed in about ten years, the timbers and the plank being com- pletely rotted out; while the other was in a perfect state of preserva- toin and has continued so until the present time.” The floor with tar | concrete foundations was constructed as follows: “Excavation was made about one foot below the floor and six inches of coarse stone 286 FLoors was filled in, then five inches of concrete made of coarse gravel, coal- tar and pitch, and finally about one inch of fine gravel tar concrete. Before the concrete was laid, heavy stakes were driven about three feet apart to which the 4” x 4” floor timbers were nailed and leveled up. The concrete was then filled in around the floor timbers and thoroughly tamped. A layer of hot coal-tar was then spread on top of the concrete and the flooring was laid and nailed to the timbers. It is very es- sential that the gravel be perfectly dry before mixing and this is ac- complished by mixing it with hot coal-tar. What is known as dis- - tilled or refined coal-tar must be used as that which comes from the gas house without being refined does not work in a very satisfactory manner.” The following paragraph is abstracted from Report No. V., Insur- ance Engineering Experiment Station, Boston, Mass: “Floors over an air space or on cement are subject to a dry rot. Asphalt or coal-tar concrete is softened by oil, and the dust will wear machinery unless the concrete is covered by plank flooring. Floors made by laying sleepers on 6 inches of pebbles, tarred when hot, then 2 inches tarred sand packed flush with the top of the sleepers, and coy- ered with a double flooring, have remained sound for 37 years. Double flooring at right angles’can be laid on concrete without the use of | sleepers. It is usually preferable to secure nailing strips to stakes 4 feet apart each way and driven to grade, concrete flush to top of strips, and lay 14-inch flooring.” | The floor shown in Fig. 135 was laid in an extensive shop on the Boston & Maine Railway. The earth was well compacted and brought to a proper surface and a 4-inch bed of coal-tar concrete put down in --/ Plank pe nhiad 4 ae, ” RE ae VF Footin ing Pitch ~~ Compacted Earth Fic. 135. WoopEN FLoors 287 three courses. The stones in the lower course were to be not less than 1 inch in diameter. Stones in each course were well covered with tar before laying and were well tamped and rolled afterwards. The third and finishing course was composed of good clean sharp sand well dried, heated hot and mixed with pitch and tar in proper proportions. This was then carefully rolled and brought to a true level to fit a straight edge. On the finished surface of the foundation was spread a coating \4-inch thick of best roofing pitch put on hot and into which the lower _ course of the plank was laid before the pitch cooled. Care was taken to have the planks thoroughly bedded in the pitch and after laying, the joints were filled with pitch. If vacant spaces appeared under the plank, they were filled up with pitch by boring 3 through the plank. The cost of this flooring was about 18 cents per square foot, using spruce lumber. A cheap but serviceable floor may be made as shown in Fig. 136. The soil is excavated to a depth of 12 to 15 inches and cinders are filled in and carefully tamped. The flooring planks are nailed to the sills which are bedded in the cinders. The life of the plank flooring can be increased by putting a coating of slaked lime on top of the cinders. ls op iia maple NY, Cinaers 5" x 4 vy Pp. ae Stryp ip Fic. 136. The floor shown in Fig. 137 was used in the factory of the Atlas Tack Company, Fairhaven, Mass., and needs no explanation. -/° Maple Longitudinal xe eS Vletnlock Diagonal ‘s e Ae “ — ransverse UMS WN Mh: VA 1 — we SSF Z ZNO SS LYNNE AME tte pre 5: ie By RS: oe ten as by Tor. Pitch ¢ ea 5 *s4g" "Portland Cernent Contr: Pic: 137. 288 FLoors A very good floor for mills and factories is shown in Fig. 138. Aa 8" Matched /aple Longitudinal er, Se: « Trarsverse “S'Fine Concrete of tar or asphatt Fic. 138. The pitch or asphalt will prevent the decay of the plank and will add materially to the life of the floor. Maple flooring makes the best wearing surface for floors and should be used if the cost is not pro- hibitive. The floor shown in Fig. 139 was constructed as follows: T'wo-inch plank, matched and planed on one side, were laid on 3” x 3” chestnut joists. ‘The surface of the cinders was kept 2” away from the wood and Pe eX. i -2" Plank eS See TAS DSCRRRR AEE NF SZ GIS et ees. Bee j SSS 5 asa - Fees ae iF) AN this space was filled with lime mortar. After the surface of the cin- ders had been graded, the 3” x 3” joists were held in place by stakes nailed to the joists about three feet apart. The lime mortar was then filled in around and slightly above the surface of the joists to allow for shrinkage. Before laying the floor a thin layer of slaked lime was spread over the surface. This floor in an eastern city cost about 85 cents per square yard; and has a life of 10 to 12 years. Examples of Floors.—The floor of the Locomotive Shop of the A. T.& S. F. R. R., at Topeka, Kas., is as follows: ‘The floor foun- dation is formed of 6 inches of concrete resting on the natural soil well tamped. On the concrete are laid 3” x 4” yellow pine stringers at 18- inch centers, the whole being covered with 2-inch No. 1 hard maple, surfaced on one side and two edges and milled for 4” x 1” pine splines. bh UF fee ae ee ; A > POP SENSI alg cepa aT, gl EXAMPLES OF FLoors 289 The flooring in the Great Northern Shops at St. Paul, Minn., is 3 x 12” plank on 6” x 8” sleepers bedded in 18 inches dry sand filling. ) The floor of the locomotive shops of the Philadelphia and Reading R. R., is made of bituminous concrete on which are a solid course of 3 x 8” hemlock and a top wearing surface of 11%” x 4” maple. The machine shop floors of the Lehigh Valley R. R. at Sayre Pa., are of concrete with a maple wearing surface in the high grade buildings and yellow pine in the others. The Southern Railway has a vitrified brick floor in a round house at Knoxville, Tenn., which is giving good satisfaction. The cost of this floor was about $1.00 per square yard. Shop floors for the American Locofnotive Works, Schenectady, N. Y., are described in Engineering Record, May 30, 1903 ,as follows: “On a sand fill was laid from 4 to 6 inches of 24-inch broken stone, rammed dry and then flushed with about one gallon of hot tar for every square yard of floor. This course was covered with 2 inches of -hot sand and tar mixed to the consistency of dry mortar, shoveled into place and thoroughly rammed to a level surface. Spiking strips made of 3” x 4” timbers were imbedded in the sand at about 3-ft. centers. To these strips were spiked 2-in. rough hemlock planks, which were in turn covered with transverse tongued and grooved 7%-in. maple boards 4 inches wide.” Cedar Blocks form a neat, clean and durable floor. Care should be used where heavy jacking is to be done on wooden block floor that the blocks are not forced down through the plank foundation. A cedar block floor in the Chicago Ave. round house of the C & N. W. R. R., laid on planks on a gravel foundation cost about II cents per square foot. *A cedar block floor on the C. & E. I. R. R., laid directly on a gravel foundation cost 8 cents per square foot. *A floor constructed of 6 to 8-inch blocks sawed from old bridge timbers, set on 24nch hemlock plank, which in turn rested on 3 inches _*Reports Association of Railway Superintendents of Bridges and Buildings. 290 FLOors of dry gravel or sand, has been used on the Ashland division of the C. & N. W. R. R., and cost about 4 cents per square foot exclusive of the cost of the old timbers. This floor has proved to be quite satis- factory. FLOORS ABOVE GROUND.—The type of floor used for the upper stories of mill buildings will depend upon the character of the structure and the use to which the floor is to be put. In fireproof build- ings the floors should preferably be constructed of fireproof materials, although there is comparatively little risk from fire under ordinary conditions with a heavy plank floor. Where the load on the floor is very heavy some form of trough or buckled plate floor is very often used. : | Timber Upper Floors.—Where steel floor beams are used the floor is often made by placing 2” x 6” or 2” x 8” planks on edge and spiking them together, the wearing surface being made of hard wood boards. Where there is much danger from fire this floor can be fire- proofed by plastering it below with wire lath and hard plaster and by putting a layer of cement or lime mortar between the plank and the wearing surface. The upper surface is also sometimes finished with a wearing coat of cement or asphalt. The standard floor recommended by the Boston Manufacturer’s Mutual Fire Insurance Co., for mill buildings constructed of heavy timbers, calls for a layer of spruce plank, generally 3 inches thick, laid — to cover two floor beam spaces and breaking joints every 3 feet; on this . are laid 3 thicknesses of rosin sized paper, each layer being mopped ~ with tar. The top floor is 1%-in. hard wood, preferably maple. The main beams are spaced 8 to 10 feet. “The floor is smoother if laid across the line of the plank in the under floor, but traveling loads are better distributed when moved in and out of the store house if the top floor is laid parallel to the lower plank.” | Brick Arch Floor.—The brick arch floor shown in Fig. 140 was formerly much used in fireproof buildings and is still used to some ex- tent in mills and factories. The arch is commonly made of a single yaa we yh ate. Ks i ite ee ee a Nala a i —Te — es. ee il ata Os aa BG AEA SEAL IEG SRE EAT SEV Pe tare ups tY a reel la FIREPROOF FLooRS , 291 Brick Arch Construction Fic. 140. layer of brick about 4 inches thick, with a span of 4 to 8 feet and a cen- ter rise of preferably not less than % the span. The space above the brick arch is filled with concrete and a wearing floor is nailed to strips imbedded in the surface of the concrete. The most desirable span is from 4 to 6 feet. Tie rods are commonly placed at about % the height of the beam and are spaced from 4 to 6 feet apart. The thrust of the arch per lineal foot can be found by the formula 1.5 W L? Rk T= where J = thrust of arch in Ibs. per lineal foot; W = load on arch in Ibs. per square foot; L = span of the arch in feet; R = rise of arch in inches. The weight of this floor is about 75 lbs. per square foot. Corrugated Iron Arch Floor.—The corrugated iron arch shown in Fig. 141 makes a very strong floor for shops and mills. The cor- rugated iron acts as a center for the concrete filling above it, and in connection with the concrete makes a composite arch. The corrugated iron or steel is ordinarily the standard 24-inch corrugations, and the gages are Nos. 16, 18 or 20, depending upon the load and the length of span. The rise of the arch should not be less than 1-12 the span and ea gy A A ERE UREN Oe SEB DO. Bee 7e “strats : by. Corrugated Iron Arch Fic. 141. 292 FLoors should have a thickness of from 2 to 4 inches of concrete over the cen- ter of the arch. Beams are spaced from 4 to 7 feet apart for this floor, and tie rods are used as in the brick arch floor. Expanded Metal Floors.—The floor shown in Fig. 142 is con- structed as follows: A wood centering is suspended from the beams, with the upper surface of the centering about 1 inch below the top of the beams, a layer of expanded metai is stretched across the beam in sheets and the concrete is spread over and tamped so that the ex- panded metal becomes imbedded in the lower inch of the concrete. The Fic. 142. concrete is usually made of 1 part Portland cement, 2 parts sand and 6 parts cinders, and weighs 80 to go pounds per cubic foot. Beams were formerly spaced from 5 to 8 feet apart, but have recently been spaced much farther; spacings as wide as 18 to 20 feet having been successfully employed. Expanded metal with 3-inch mesh cut from No. 10 gage sheet steel is commonly used for floors, which are made from 3 to 5 or 6 inches thick. There is at present no rational method for the design of expanded metal floors—descriptions of floors in actual use, and tests of floors may be found by consulting the volumes of the Engineering News, the Engineering Record, and other technical papers. The companies controlling the patents will erect floors for specified loads under a guarantee. Expanded metal floors are also made as shown in Fig. 143. This type is adapted to very heavy floor loads. ‘The arch should have a rise Swill. ND TEN DR AE ul Doe aD LT eUL Gk eee ens Ud Lf Tye, . Sr RI ce mag rin te — ~ te med ARTRIT a Nees * és ™ be i ROEBLING FLOOR 293 of I-10 to % the span. Tests have shown that the steel work in ex- panded metal floors is not ordinarily affected by the cinder concrete. If care is used when erecting the floor to coat the expanded metal with a coating of Portland cement mortar before the metal has become rusted. the protection against corrosion will be almost perfect. Roebling Floor.—The floor shown in (c) Fig. 144 consists of a wire cloth arch, stiffened by woven-in stiff steel rods 3 to I-inch in diameter, at about 9-inch centers, which is sprung between the floor beams and abuts on the seat formed by the lower edge of the floor beam. “BUCKEYE” FIREPROOF FLOORING, MULTIPLEX STEEL PLATE FLOOR (a) (b) Roebling Fire-proof System (d) ) Fic. 144. On this wire centering Portland cement concrete is filled in and is fin- ished with a wearing coat of cement or a wooden floor as shown. The beams are held in position by 34 or %-inch tie rods, placed from 4 to 294 FLoors ft 6 feet apart. The concrete is commonly composed of 1 part Portland cement, 214 parts sand and 6 parts clean cinders, and is laid with a thickness of not less than 3 inches over the crown of the arch. The — _ weights and safe loads for floors constructed on this system are given in the manufacturer’s catalog. The Roebling Construction Company also makes a floor with flat construction as follows: A light framework is made of flat steel bars set on edge and spaced 16-inch centers, with %4-turn at both ends where the bars rest on the steel beams; braces of half round iron are spaced at intervals to brace the bars. The Roebling standard lathing with 14-inch steel stiffening ribs woven in every 7% inches, is then applied to the under side of the bars and laced to them at each inter- section. On this wire lathing cinder concrete from 3 to 4 inches thick is deposited. | “Buckeye” Fireproof Flooring.—The steel flooring shown in (a) Fig. 144 is manufactured by the Youngstown Iron & Stee! Roofing Co., Youngstown, Ohio. This floor is made in two sizes, one for bridge floors and the other for building floors. The flooring shown in (a) is for buildings, is made in sections of four triangles each in lengths up to 10 feet, and will lay a width of 21 inches. Each triangle is 5%4 inches wide and 214 inches deep. The flooring when complete with a concrete filling and a 114-inch wearing surface will weigh from 32 to | 35 pounds per square foot. The weights of the metal troughs laid in place are given in the following table: | cA WEIGHTS OF METAL, TROUGHS 24% INCHES DEEP BY 514 INCHES WIDE, INCLUDING SIDE LAPS. | ; GALVANIZED TROUGHS. BLACK IRON TROUGHS. No. 16—386 lbs. per 100 square feet. No. 16—363 lbs. per 100 square feet. ING; S700. 8 OR 8 INO, BRO RaE te Se eee : No. 18—313 ‘ te ” a “ No. 18—290 ‘“ § “ = a ey No. 19—278 ‘“ se “e $458), ee No. 19—254 “ « “ “ No. 20—241 ‘“ 8 a ae ” No. 20—218 ‘“‘ * ak SOE ho bey No. 22—204 ‘ hc 6s s *“ No. 22—181 “ “ “ “e “ No. 24—168 «e 6 6 6s “ce No. 24—145 e ‘ce “e 66 : “ MIScELLANEOUS FLOORS 295 The safe loads in addition to the weight of the floor as given in the manufacturer’s catalog are given in the following table: SAFE LOADS FOR “BUCKEYE” FIREPROOF FLOOR. Span. No. 18 Gage. No. 20 Gage. No. 22 Gage. No. 24 Gage. 3 ft. 0 in. 1050 lbs. 800 Ibs. 580 lbs. 450 lbs. an. 6 * 820 * 570“ 425 “ $302. ea. Qe" 675 *“ 425“ 315 “ 230.“ et, & St 570 ** 320 “ 240 * 1905-28 CO 475“ 250 =“ 136:;.“ 135 “* aay, a 200 * 200 * 140 ‘“* Multiplex Steel Floor—The steel flooring shown in (b) Fig. 144, is manufactured by the Berger Mfg. Co., Canton, Ohio. ‘This floor is made with corrugations from 2 to 4 inches deep and of Nos. 16, 18, 20 and 24 gage steel. The triangles are filled with concrete and the floor is given a cement finish, or is covered with a wooden wearing surface. Tables of safe loads for Multiplex Steel Floor are given in the manu- facturer’s catalog, and are larger than for the “Buckeye” Flooring for the same span. Ferroinclave.—Ferroinclave is fully described in Chapter XIX. It has been used by the Brown Hoisting Company for floors as well as for roofing and siding. It should make a very satisfactory flooring where light loads are to be carried. Corrugated Flooring.—Corrugated flooring or trough plates shown in (a) and (b) Fig. 145, are used for fireproof floors where ex- tra heavy loads are to be carried in mill buildings, in train sheds and for bridge floors. The troughs are filled with concrete, which is given a finishing coat of cement and sand or is cevered with a plank floor ; the planks being laid directly on the plates or spiked to spiking pieces imbedded in the filling. The details, weights and safe loads for corrug- ated plates are given in Pencoyd Iron Works’ handbook, in Carnegie Steel Company’s handbook, and in Trautwine’s Pocket-book. Details _ of corrugated plates are also given in the American Bridge Company’s “Standards for Structural Details.” 2096 FLoors Corrugated flooring or trough plates are usually very hard to get and the Z-bar and plate floor shown in (c) Fig. 145, and the angle and Pencoyd Corrugated Flooring Z Bar Floor Angle and Plate Floor (a) (b) (c) (d) Fic. 145. plate floor shown in (d) are substituted. The details, weights and safe — loads for Z-bar and plate flooring are given in the handbooks above named. Angle and plate flooring is made of equal legged angles and plates, and the safe loads are not given in the handbooks but must be ~ : calculated. The moment of inertia, J, of a section of flooring contain- ing two angles and two plates is given by the formula I = 2I' + 2Ad? + 2]” where J’ = moment of inertia of one angle about an axis through the center of gravity of the angle parallel to the neutral axis of the flooring; A = area of one angle; = distance from center of gravity of the angle to the neutral axis of the flooring; | I” = moment of inertia of the plate about the neutral axis. The properties of the angles required in the calculations may be obtained from the handbooks, and J” is equal to one-half the sum of the moments of inertia of the plate about its long and its short diameter— since the sum of the moments of inertia about any pair of rectangular axes is a constant. Buckled Plates.—Buckled plates are made from soft steel plates from 3 to 5 feet wide, and are from 14 to 7-16 inches thick. Buckled plates are made in lengths having from one to I 5 buckles or domes in one plate. Buckles vary in depth from 2 to 314 inches, however, dif- ferent depths should not be used in the same plate. Buckled plates are usually supported along the edges and the ends and are bolted to the floor beams. ‘The details, weights and safe loads are given in the hand- books named above and in the Passaic Steel Company’s handbook. The oe a ay Ne RE ce ree eee Nee SPO iy STEEL PLATE FLOORING 297 ? buckled plates are covered with concrete, which is given a finishing coat of cement or is covered with plank flooring. Steel Plate Flooring.—Fireproof floors around smelters, etc., are often made of steel plates. Flat steel plates do not make a very satis- factory floor for the reasons that the plates will bulge up in the cen- ter when fastened around the edges, and because they become danger- ously smooth. : Neverslip Wrought Steel Floor Plates are made from 24” x 72” to 36” x 120”, and from 3-16” to 1” thick. These plates are designed to take the place of the cast iron checkered plates formerly used for floors, weigh about 50 per cent less and last much longer. The stock sizes, weights and safe loads for Neverslip floor plates are given in the Stock List of the Scully Steel Co., Chicago. These lists are issued about six times a year and will be sent free upon request. Thickness of Timber Flooring.—The thickness of white pine and spruce flooring for different spans and loads is given in Table XXXIIIa. The following fibre stresses were used in calculating the thickness: Transverse bending, 1000 Ib.; end bearing, 1000 lb.; bear- ing across the fibre, 200 Ib.; shear along the fibre, 100 Ib. TABLE XXXIIIa. THICKNESS OF SPRUCE AND WHITE PINE PLANK FOR FLOORS. 3 ; THICKNESS, IN INCHES, FOR VARIOUS LOADS PER SQUARE FOOT OF PLANK. an in Meet. lb. | Ib. | Ib. | Ib. | Ib. | Ib. ) Ib. | Ib. | Ib. | Ib. | Ib. | Ib. | Ib. | Ib. | Ib. | Ib. | Ib. 380 | 40 | 50 | 75 | 100 | 125 | 150 | 175 | 200 | 225 | 250 | 275 | 300 | 825 | 350 | 375 | 400 daw a «cabs 0.9 | 1.1) 1.2) 1.5 | 1.7} 1.9 | 2.1) 2.2 | 2.4 | 2.5 | 2.7 | 2.8 | 2.9 | 3.1| 3.2 | 3.3 | 3.4 Bi axtinahenns 1.2 | 1.4) 1.5|1.9| 2.1) 2.4 | 2.6 | 2.8 | 3.0 |-3.2 | 3.4| 3.5 | 3.7 | 3.8 | 4.0) 4.1) 4.3 Scotts ces 1.4|1.6| 1.8 | 2.2 | 2.6} 2.9 | 3.1) 3.4 | 3.6 | 3.8 | 4.0 | 4.2 | 4.4 | 4.6 | 4.8 | 4.9 | 5.1 Peichndesve: 1.7/1.9 | 2.1 | 2.6 | 3.0 | 3.3 | 3.7 | 3.9 | 4.2 | 4.5 | 4.7| 4.9 | 5.2|5.4| 5.6/5.8 | 6.0 aR 1.9 | 2.2 | 2.4| 3.0) 3.4) 3.8 | 4.4/4.5 | 4.8) 5.1/5.4) 5.7|}5.9| 6.1 Beats Matick. ikees Bi ek ee AS | 400 | 961) 5.4 1 5B] GLE bevcccchisse ds lacconslicvane|ecdestlavane MUR Seve etics Dee dad | Ook acd 1, Ged | 48 1-502 1500 | O.0 [oo coc ficacecdeskc ccd elesscfocasoofeoases Bike ite invents AEA AD coe Meet ME LOE [ER a cles wads feo wce<[acion'ss |acas’sleseese looses te tiactatenac BoP id 1 SIO | Sew: loss va lceccee hee: (SSSR ee 3.1 | 3.6 | 4.0! 4.9 5.6). TPR ONS! ARR d FERS Et ieee NR Ae ae ae 3.4|3.9|4.3|5.3}6.1]......[.:. | bei tolp For yellow pine use nine-tenths of the above thicknesses. CHAPTER XXIII. WINDows AND SKYLIGHTS, | Glazing.—For glazing windows and skylights, two substances, glass and translucent fabric are in common use. GLASS.—The principal kinds of glass used in windows and sky- lights are (1) plane or sheet glass; (2) rough plate or hammered glass; (3) ribbed or corrugated glass; (4) maze glass; (5) wire glass—glass with wire netting pressed into it; (6) ribbed wire glass; and (7) prisms. (1) Plane Glass—Plane or common window glass is technically known as sheet or cylinder glass. It is made by dipping a tube in molten glass and blowing the glass into a cylinder, which is then cut and pressed out flat. Without regard to quality sheet glass is divided ac- cording to thickness into “single strength” and “double strength” glass. Double strength glass is %-inch thick while single strength glass is about 1-16-inch thick. In mill buildings, lights larger than 12” x 14” are usually made of double strength glass. With reference to quality sheet glass is divided into three grades AA, A, and B. The AA is the best quality, the A is good quality while the B is very poor. The B grade is suitable only for stables, cellars, etc. For residences, offices, and similar purposes nothing poorer than AA should be specified. ‘The A grade does very well for ordinary mills, although the AA grade should be used if practicable. (2) Plate Glass—Plate glass is made by casting and not by | blowing, and is finished by grinding and polishing on both sides until a smooth surface is obtained. It is usually 1% or 3-16 inches thick. The price depends upon the size of the plate and the quality of the glass. The rough plate glass used in mills is not finished as carefully as for glass fronts, and it may contain many flaws that would not be allow- k= them Slate bite tinal om Neha red SNL ATS ha hae Geer ent = ; KINDs oF GLASS 299 able in the former case. The roughened surface of the glass prevents the entrance of direct sunlight and does away with the use of sun- shades. The only value of rough plate glass is in softening the light, the loss of light in passing through it being very great. (3) Ribbed or Corrugated Glass—Ribbed or corrugated glass is usually smooth on one side and has 5, 7, II or 21 ribs on the other side, fa) big. 146,:: It varies in thickness and shape of ribs. “Factory ribbed” glass with 21 ribs to the inch is distinctly the most effective of the ribbed glasses. RIBBED FIGURED SHEET PRISM (A) VCD) (a) Fic. 146. (4) Maze Glass—Maze glass has one side smooth and has a raised pattern on the other side roughening practically the entire sur- face, (b) Fig: 146. It is quite effective. (5) Wire Glass—Wire glass is made by pressing wire netting into the molten glass. It is made either plane or with ribs or prisms on one side. Wire glass is injured but not destroyed by the action of fire and water, and is now accepted by insurance companies as fire- proof construction. (7) Prisms—Prisms are made in small sections which are set in a frame of lead or other metal, or are made in sheets as shown in (c) Fig. 146. Luxfer sheet prisms, manufactured by the American Luxfer Prism Co., Chicago, will be cut in any size desired up to 84 inches wide (parallel with the saw teeth) by 36 inches high. Diffusion of Light.*—The light entering a room through a win- dow or skylight comes for the most part from the sky and has, there- *Report No. III. Insurance Engineering Experiment Station, Boston, Mass. 300 WINDOWS AND SKYLIGHTS fore, a general downward direction, varying with the time of day and the position of the window. The portion of the room which receives the most light ordinarily is the floor near the windows, but if we inter- pose a dispersive glass in this beam the light will no longer fall té the floor but will be spread out into a broad divergent beam falling with nearly equal intensity on walls, ceiling and floor. There is of course no gain in the total amount of light admitted, the light being simply redistributed, taking up from the floor that which fell there and was comparatively useless, and sending it where it is of more service. Experiments have shown that the diffusion of light in a room lighted by means of windows or skylights depends upon the kind and position of the glass used. The relative intensity of the light admitted in per cents of the light outside the window for plane glass, factory ribbed glass, Luxfer and canopy prisms is shown in Fig. 147%, 5 50 cz Variation of Light 8 240 with '=% %, Angle of Skylight 5 2 fs No Direct Sunlight Bo e eon S,. a £ a © & 5 20 NY XN = r) ox &, 0, — SNe aS % log Ss Cc is, VV YS © ec £ 5 lass aw fe) . BOE AO 60s BO TOs Or Angle skylight makes with horizontal Fic. 147. Fig. 147 shows a great increase in efficiency of factory ribbed glass and prisms as the sky angle diminishes. *Report No. III. Insurance Engineering Experiment Station, Boston, Mass. : oe he 7 : +, ee 2 . re at = : / ‘ ° aliek : ~ . : wads ae eee i dia De ’ : oe a ( TT Pe ee oe ee me foe yee .. Pee te pe ee ee eh ee gee mY te ese ae ea ae, ; ae i eS terres ee ie ren sar ate Me ns ti jee. asia ee S CD on at i ? RELATIVE VALUE OF DIFFEREN’? Kinps oF GLASS 301 The equivalent areas required to give the same intensity of light with the kinds of glass shown in Fig. 147, are given in Table XXIV for sky angles of 30° and 60°. TABLE XXIV. EQUIVALENT AREAS FOR DIFFERENT KINDS OF GLASS. vee te a” Ter Te oa Angle Skylight makes with the Horizontal. Kinds of Glass. 30° 60° Plane 100 sq. ft. 100 sq. ft. Factory Ribbed . 25 “ 6&6 40“ « Luxfer Prisms 1 ly Mage A 30 “6 « Luxfer Canopy Prisms pT eit The American Luxfer Prism Co., recommends that Luxfer prisms be set at an angle of about 57 degrees with the vertical when used in skylights. | Relative Value of Different Kinds of Glass——Ground glass is of little value except as a softening medium for bright sunlight. It be- ‘comes opaque with moisture and makes an undesirable window glass. Roughened plate glass has very little value as a diffusing medium. Of the ribbed glasses, the factory ribbed glass with 21 ribs to the inch gives the widest and most uniform distribution and is distinctly the best. There is no apparent gain in corrugating both sides. Ribbed wire glass is about 20 per cent less effective than the factory ribbed glass. When a glass of a slightly better appearance than the factory ribbed glass is wanted the maze glass is the best; the raised pattern imprinted on the back of this glass giving wide diffusion, especially in bright sunlight. _ The prisms are very much more effective than any of the glasses men- tioned above, but their cost prevents their use under ordinary conditions. Kind of Glass to Use.—Where the amount of skylight is large and the light is not obstructed by buildings plane glass is very satisfac- tory. Where a superior light is desired, or where the skylight area is less than ample, use factory ribbed glass in skylights and in the upper 302 WINDOWS AND SKYLIGHTS panes of windows. Where the skylight area is very small, the light is obstructed, or a very superior light is desired, use prisms. Wire glass should be used where there is danger from fire and in skylights, where - it removes the necessity of stretching wire netting under the glass to protect it and to prevent it from falling into the building when broken. Placing the Glass.—Factory ribbed glass is somewhat more ef- fective if the ribs are placed horizontal, but the lines of light deflected from the horizontal ribs may become injurious to the workmen’s eyes and it is now the custom to set the ribs vertical. Ribbed glass should have the ribs on the inside for ease in keeping it clean, and where double glass is used the ribs should face each other and be crossed. Care should be used in setting thick wire glass in metal frames; the lower edge must bear directly on the frame, but the top and sides should fit loosely so that the differential expansion of the glass and frame will not crack the glass. Plane glass and small panes of other kinds of glass are set with glaziers’ tacks and putty. In skylights and large windows some method must be used that will allow the glass to expand and con- tract freely and at the same time will be free from leakage. Several — methods of glazing skylights without putty are shown in Fig. 148. Skylight bar (a) manufactured by Vaile & Young, Baltimore Md., is made of heavy galvanized iron and lead. ahs i | | a | a o \ (a) (b) (Cc) ta) | Fic. 148. Bars (b) and (c) are made of zinc or galvanized iron, supported by a steel bar. Bar (c) is adapted to small panes of glass and is made of galvanized iron; it is made water tight by the use of putty. The skylight bars in Fig. 148, all have condensation gutters to catch the moisture that leaks through or forms on the inner surface of the glass. r" ‘ Li > «3 We ed Me a A aes ell Reet aia - Use oF WINDOW SHADES 303 Glazing Poirrt TT aoe =f) pang . | Miles ated or Hammer | \ ob Speer lad S\N cB ie 2 ete is opart Vertical Section through Ends. ‘ay (b) Fic. 149. The glass in a large greenhouse at Edgely Pa., was secured to the sash-bars as shown in (a), Fig. 149. It will be seen that the glass is im- bedded in putty on the under side only, and that any water that can pos- sibly leak through between the bar and the glass will be caught in the drip trough “a”, and be carried to the eaves. ‘The lights are 16” x 24” and the sash-bars are spaced 2434 ins., c. toc. The lights are held in place by two patent glazing points per light, driven in such a way as to prevent the glass from moving. The lights overlap but I-16-in., the leakage having been found to be smaller and less liable to occur with this than with a larger lap. The Paradigm system of glazing is shown in (b) Fig. 149. This system is in use in a large number of shops, among which the: steam engineering buildings for the Brooklyn Navy Yard, described in Part IV, is one of the best examples. The patents for the Paradigm skylight are controlled by Arthur EF. Rendle, New York. Skylights are of two types; (1) box skylights covering a small area and placed on a curb raising the glass above the roof, and (2) continuous skylights usually placed in the plane of the roof. The glass used for skylights varies from %4 to % inch thick and should preferably be wire glass. The glass used for skylights usually comes in sheets about 20 inches wide and up to 8 feet long. The details of a box skylight manufactured by Vaile & Young, Baltimore, Md., is shown in Fig. 150. Use of Window Shades.—Where factory ribbed glass is- placed so as to throw light on the ceiling, screens or shades are seldom required, Fic. 150. however, under ordinary conditions shades are necessary when bright sunlight strikes the window. ‘The glass used in factory ribbed and rough plate glass as made in England is somewhat opaque, and the atmosphere is somewhat hazy, so that the use of shades in their shops is in most cases unnecessary. ‘The glass made in this country is so clear and our atmosphere is so translucent that it has been found nec- essary to use shades where windows and shades are exposed to direct sunlight. The most effective and satisfactory shade is a thin white cloth, which cuts off about 60 per cent of the light. Size and Cost of Glass——The regular stock sizes of plane glass varies from 6 x 16 inches by single inches up to 24 x 30 inches, and above that by even inches up to 60 x 70 inches for double strength glass and 30 x 50 inches for single strength glass. The weights of different thickness of glass, assuming 156 pounds as the weight of one cubic foot of glass are given in the following table: WEIGHT OF GLASS PER SQUARE FOOT. Thickness——165 oy. cee % 3-146 % % % %& % 1 Weight—lIbs. ..........00. A Apa pee 1.62 2.48 8.25 4.88 6.50 8.13 9.75 18 The cost varies with the quality and the size, being about twice as much to glaze a given area with 30 x 36-inch lights as with 10 x 12- inch lights. The discounts given from the standard price list vary so ee Cost of WINDOWS 305 much that prices are of very little value except to give an idea of the rel- ative cost of different sizes of glass and to serve as a basis for estimates. In 1903 American window glass was quoted about as given in Table XXV. TABLE XXV. _ Cost OF WINDOW GLASS IN CENTS PER SQUARE FOOT. Sine of Lighta Single Strength. | Double Strength. in Inches. AA A B AA - B 10 X12 5.1 4.3 4.0 | 6.8 6.0 5.5 14 x 20 5.4 4.5 FS a Ga ae 6.6 6.1 16 X 24 5.8 4.8 4.5 | 8.3 7.3 6.7 20 X 30 6.0 .| 5.1 4.7 | 8.9 7.9 7.4 24 X 36 | 6.4 5.5 | 4.9 9.4 8.3 7.6 In 1903 the different kinds of glass were quoted in small quantities at the factory about as follows: My here lass 4 ANC WICK. 6s fe es es 23 cents per sq. ft. | Factory ribbed glass % inch thick........9 “ “ “ “ Maze glass % inch thick................ ia «ee Maze glass 3-16 inch thick.............. 1 cA tea a a: Prismatic glass from 25 to 50 cents per sq. ft. Refrax glass (sheet prisms) made by the Union Plate Glass Co., Limited: Pocket Nook, St. Helens, England, was quoted in 1903 as follows at the factory: Ordinary refrax glass %4” thick with 5 prisms to the inch, cut to any size up to 60” x go”, 20 cents per sq. ft.; wired refrax glass 5-16 inch thick with 5 prisms to the inch, cut to any size up to 40” x go”, 25 cents per sq. ft. Maltby prisms, made by Geo. K. Maltby, Boston Mass., 3-16” thick with 6 prisms to the inch costs about 25 to 30 cents per sq. ft. Cost of Windows.—Windows with frames for mill buildings will cost from 15 to 25 cents per square foot, depending upon the size and quality of the sash, the size of the opening, and cost of glass and frames. In 1900 the cost was about 16 cents per square foot for D. S. 306 WINDOWS AND SKYLIGHTS glass with box frames and sash, and 9 cents for S. S. glass with plank frames and sash. In 1900 skylights cost from 23 to 30 cents per square foot with D. S$. glass. Windows are commonly estimated at 25 cents per square foot and skylights at from 40 to 50 cents per square foot in making preliminary estimates. | The American Luxfer Prism Company manufacture sheet prisms for factory purposes that can be cut to fit any opening up to 36” x 84”. The cost of sheet prisms to fit ordinary windows is about 40 cents per square foot. The improved skylight prisms made by this company cost about $1.50 per square foot. 7 TRANSLUCENT FABRIC.—Translucent fabric consists of a wire cloth imbedded in a translucent, impervious, elastic material, prob- ' ably made of linseed oil. The fabric may be bent double without cracking and is so elastic that changes due to temperature or vibrations do not af- fect it. If a sheet of translucent fabric is suspended and a fire applied to the edge, it will burn up leaving a carbonaceous covering on the wire. But if the edges are protected it will burn only with great difficulty. — Live coals falling on skylights of this material will char and burn holes but will not set fire to the fabric. It is therefore practically fireproof. Translucent fabric will not transmit as much light as glass, but makes a most excellent substitute therefore. It shuts off sufficient light so that the lighting is uniform threughout the shop and makes it possible for men to work directly under it without shading. Where one-quarter of the roof is covered with the fabric the lighting is prac- tically perfect. The fabric should be washed with castile soap and warm water occasionally, and should be varnished every year or two with a special varnish furnished by the manufacturers. It is said to become less opaque with age. When properly cared for the fabric has been known to give good service for ten years. The fabric is man- ufactured in sheets 3’ 3” wide and in lengths from 4’ 6” to 9’ oO. The framework for translucent fabric is the best made of wood. A - standard frame for sheets 3’ 3” x 6’ 3” is shown in Fig. 151. The fab- ric must be stretched tight and carefully nailed around the edges of the sheet. The capped joint with metal cap shown in Fig. 151 is very — eS ee ee ee ee eee oe, Oe eee ee ee ee ee ee ee ee ee Cost oF TRANSLUCENT Fapric 307 satisfactory as it holds the fabric tight, and will give slightly to accom- modate changes in temperature. Cost of Translucent Fabric—The fabric costs from 13 to 15 cents per square foot at the factory at Quincy, Mass. The framework, freight and cost of laying will probably be as much more, making the entire cost of skylights from 25 to 30 cents per square foot. Translucent fabric has been quite widely used and has given uni- formly good results. It has been used recently in the A. T. & S. F. R. R. shops at Topeka, Kas. <-~-- = — 6-0" ----- > Sade < — -3°0"——> =| jeeps “= 3 "S=<2/¥ Wood % | ||4%2 Woo : : ‘ 2 “No. 10 Gal. Wire " _ : RDP ale we Se 2 "=-2's13 Wood i Sai ZigWood ‘& "--Wo./0 Gal Wire ; Le 2 xf Wood ____ == [ dete ie } Dimensions giver are for Lock Yoint Detail of Detail of Detail of Method of Fastening Lap Joint LockJoint Capped Joint Ends of Wire Fic. 151. Double Glazing.—The condensation on the inner surface of glass can be prevented by double glazing the windows and skylights. Build- ings with double glazing are also very much easier te heat than those with single glazing, the air space between the sheets of glass acting as an almost perfect non-conductor of heat. Details of Windows and Skylights.—The details of windows in use in different sections of the country vary a great deal on account of the varied conditions. In buildings that have to be heated and ventil- ated through the windows at the same time, it is necessary to provide ’ some means of opening and closing the windows quickly and easily ; while in many other cases the sash can remain fixed. The author would call especial attention to the saving in fuel by the use of double glazing ; the loss of heat through a double glazed skylight has been shown by 308 WINDOWS AND SKYLIGHTS experiment to be only about one-half what it is through a single glazed skylight. Details of windows for use in ordinary brick and stone walls can be found in books on architectural construction and will not be given here. A few of the best designs available for windows in buildings : with corrugated steel, expanded metal and plaster, and similar walls, | have been selected and are given on the following pages. 3 | The different types of windows for buildings covered with cor- rugated steel siding as used by the American Bridge Company are shown in Figs. 152 to 156 inclusive. anit ge cecil til Tilia - wy c > (| I i | ii) Vi Car Stee/ ss h | il 4 : i Hy II i Peurtin _iimersNails ‘ HQ Yai i il it In| ii 4 ial ast e ba : Top F tt e Pot S a 1 3 S hs ‘ < STurvtire \.t, ‘y. Dimdbsions ok sas 1 ed ruppiber ara size oN lights #28 Gal Stee!” Glass Be a? 5 -~----— Add 2"tot/ for purlin spacing - ---» ms ee ee ee EE ee oe Stile ~.. we /1=6 “O° Max. --= ote Glass # “3 |Slashing x ! +2) ole g a me «2/2 Screw _ Bottom Fai/*\sis va 3 ” T : IMT MUNN CT i! bt) \\ aT TTT A Rib ¥-2°x6" fit ii | ih Hill il] | hl | i V4 Found Wp : Sil j UU £°x23'Bolr 4} | RPE go ELEVATION pes bee pet “rin : tt Bron < 12 tase - Ly Glass Ss se 2 ae waded <--—--- W=Glass,Murtins&43'---— > PLAN Fic. 152. DESIGN FOR A CONTINUOUS FIXED SASH WINDOW. DETAILS OF WINDOWS 309 The sash frames are constructed of white pine and are glazed usu- ally with A qualityAmerican class. The common sizes of glass used in these windowsare 10” x 12”,12” x 12”, 10” x 14” and 12” x 14” single strength. For lights larger than 12” x 14”, double strength glass is used. The window shown in Fig. 152 is used where light is desired without ventilation. This detail is used principally for monitor ven- tilators or for windows placed out of reach. Where it is desirable to obtain ventilation as well as light the window frame with sliding sash shown in Fig. 153 is used. «Cor: Steel $ ) $ flash ai H TO in Lag Icrew a x/F “ey 2 sesabrlcaseg Ay wate 8 Bee eh * in Nap Nw i *oy * reg ‘ iN = Tee * Al! 1 “ME = He N I. mit Su yon of sash detennned a SES by rurnker ana ¥/ze of lplys N S Ni i : A Muntins| N S het DN tT S errs 8 2 FT Os s NN 5 ae NES mh N N i] y zZ § & ry t q% 3k maa 4 ef AOllers |. x ! 4h 12 N Le Go7 Aad Fe | 3 1 ‘a ——}——— _ Tet oh. CATT Frofes a ‘wT Drip - EBT | tae fa Furtin | /'4 Round S Sy - Cor Stee/ Lag Jcrew, © ak S 2°x22 N ie R DW | ~ ELEVATION . -FProller 13" wit ‘ EF Bcd 3 ee nar i 181955 +3 ciysis o£ Vp FRound 6 - W=Glass,Murins t4$"- PLAN Fic. 153. DESIGN FOR WINDOW FRAME WITH SLIDING SASH. 310 WINDOWS AND SKYLIGHTS — es, Flashing Ki doee sine on BANE WTS SONETES 3 ow 2 x9 ee yates = yi 1S Fu ‘~7Qp \\Frait by a 1 2 § ¥3 = 8 2t- cx % /4urtin | . -/Z % i 8 x 1% ; 1 edb ; Botton Frail E 469 | ===> === == Yost 2 Jt : L Furlin et Drip? WF <2" x8 "Sill - Cor: Stee/ 14 Pound? 8/2 Lagscrew3 ELEVATION . SECTION ek ry SIE XE Stop "x7" unt <7 ae -_-————— =———<—<—==s -———<— <= = AES" > - EX ld “4 Z "KE F + (ig stripe hat f “3 Four 4 be. a 1" c nek Glass a S6hzss dGlass we k- - W= Glass, untins +4$"---* PLAN Fic. 154. DESIGN FOR WINDOW FRAME WITH COUNTERBALANCED SASH. Amount of Light Required.—The amount of glazed surface re- quired in mill buildings depends upon the use to which the building is — io be put, the material used in glazing, the location and angle of the windows and skylights, and the clearness of the atmosphere. In glazing — windows for mills and factories in which the determination of color is a necessary part of the work, care should be used to obtain a clear white glass for the reason that the ordinary commercial glass breaks up the light passing through it so that the determination of color is difficult. Amount oF Licu?r REQUIRED 311 Car. Stee/ — Slashing Fun “X18 "LAGSCIEW = = = saakkaaaeoad, Gee ae aT Fade lack Hie mis Ul. Wo im A 2 ie oa 0 *~79p \|Arai/ “ es y | /loriised > 8 i Culley eee S 1 k03-=4 N ‘ re ||| es ¥ i 8 iin? < Meeting\ Fail S! teh W Rens , ete 8 Linens: Vota aerermnyned | t « | & Ly nurnker arr she of highs. ; ae x some CaN Oh | | + sity} 8 ae ) | Botton Frail tag ! 1 8) | le 30 BW Sie-s7 Cor: Stee/ 1-4 FRound/¥f z $512 Lagscrew ELEVATION SECTION a) nie . # xe Leattiog strip: | BxX4g- -Bx4g” ete ke ‘23° ie ne” rah it 5 =I i [#204840 Glass 18 Gass Glass ati viele i} W= Glass, Muntins +43" — PLAN Fic. 155. DESIGN FOR WINDOW FRAME WITH WEIGHTED SASH. _ It is common to specify that not less than 10 per cent of the ex- terior surface of ordinary mill buildings and 25 per cent of the exterior surface of machine shops and similar structures shall be glazed. One- half of the glazing is usually required to be in the roof in the form of skylights. With translucent fabric it has been found that the lighting is good where 25 per cent of the roof is glazed. The present tendency in shop and factory design is to make as 312 WINDOWS AND SKYLIGHTS mn ft Th in a lite wit Ha a —— | M2 yV Stuntins: eae aii A cia eG | 3)" Si Hi ; 1! j VES 1 2 k-~ “Adal 72 70H for Purlin Spacing -» Riga eS ee Alb ns” ld” lg 5" damb SECTION : | Pee, fr | ; 43 ————4 a aS Fai Its 1°y Found *% PLAN Fic. 156. DESIGN FOR WINDOW FRAME WITH SWINGING SASH. a much of the side walls and roof of glass as possible ; the danger of leak- age around and through skylights has prevented many from making use of skylights, although with the present methods of glazing there is no reason why any leakage should occur. The shops of the Grant Tool Company, at Franklin, Pa., shown in Fig. 157, is a good illustra- tion of side wall lighting, while the steam engineering buildings for the Brooklyn Navy Yard, described in Part IV, is a good illustration of side wall and skylight lighting. The A. T. & S. F. shops described in Part IV, is a good illustration of side wall, skylight and saw tooth roof lighting combined. The Central Railway of New Jersey hace at Elizabeth, N. J., have skylights made of translucent fabric in the different buildings in a ; 1 ; ; 3 4 ,* Amount oF Licut RrQuirED 313 92) 4 .4°Spiking Strip, 4 Y Y ao kA ZZ A “ey ; us ; i > S ——<—S H . 14 Ly! Za H 4 I desist Lt (ee "wa a i Pe es im My —-= +--+ -476"Conter fo Center « 4 ee? 1 2k8" Joists 4 FA ie PEt rR ; Half Cross Section, 3 amr —— Cement Fioor, in OU cs CAE A, fi Sa ae NAN Da BP en oe = H+ -— oe _- eey Sew acc ure <== + + Heo - = == -+ 468 Sm ee ne tw wee mw cw wee oe = ~ = = = ~~~ 99 2"Center to Center Fic. 157. SHOPS OF THE GRANT TOOL WORKS, FRANKLIN, PA. per cents of the entire roof surface as follows: Blacksmith shop 30 per cent; machine shop 30 per cent; and paint and repair shop 55 bet cent. 7 The Lackawanna and Western Railway blacksmith shop has 13 square feet of skylight per 100 square feet of floor area. In the Great Northern Railway shops at St. Paul, Minn. Railway Gazette, June 16, 1903—all skylights have Y4-inch ribbed glass, below which is double strength window glass. Suitable drainage is provided for the moisture which collects on the upper surface of the latter. Wire netting is stretched under the skylights to prevent broken glass from falling into the shops. ‘The walls are supplied with windows set at 12 feet centers, 55 panes to each sash. The skylights of the A. T. & S. F. R. R. shops at Topeka, Kas., are made of translucent fabric, about 20 per cent of the roof surface being fabric. | The skylights of the machine shop of the Chicago City Railway are made of wire glass, about 35 per cent of the roof being glass. 314 WINDOWS AND SKYLIGHTS The machine shop of the Lehigh Valley Ry., at Sayre, Pa,, will have the side windows of plane glass. The locomotive shop will have factory ribbed. glass in the side windows and wire glass in the root and monitor skylights. About 25 per cent of the roof of the St. Louis Train Shed is sky- light. In the American Car and Foundry Company’s shop at Detroit, about 27 per cent of the exterior surface is ribbed glass. Fully 60 per cent of the exterior surface of the Steam Engineer- ing Buildings for the Brooklyn Navy Yard is of glass. Ff sreatning ange a a ' x ‘ a oe -} Fic. 157a. Dertarts of Saw TootH WIinDows For P. & L. E. R. R. Suops. SEE Fic. 84d. SKYLIGHTS FOR TRAIN SHEDS 315 Experience with Skylight Construction for Railway Train Sheds.—In 1904 a committee was appointed by a prominent eastern trunk line to investigate and report on train shed skylight failures and their remedy. The committee examined the Broad Street Station train shed, Camden and Reading Terminal train sheds, and other structures in Philadelphia; the train shed at Jersey City; the North German Lloyd Steamship piers at Hoboken; the N. Y. C. & H. R. R. R. train shed, the Macey Building, and other structures in New York; the South Union, North Union, and Back Bay stations, and the Charlestown Navy Yard in Boston; the Union Station, and the Pitts- burg & Lake Erie station in Pittsburg; and the Westinghouse Shops in East Pittsburg. A complete report is printed in Engineering News, April 21, 1904. The conclusions of the committee - were as follows: 1. That gas and smoke from locomotives, because of its influence on the metal framework of skylights, is the primary cause of the breakage of glass. 2. That the contraction and expansion of the metal frame is also a serious cause of breakage where the glass is tightly fitted in the frames. 3. From the testimony elicited and from personal observations, we find the percentage of breakage in ribbed, hammered, and wire glass is about equal. We do not find that the breakage of wire glass results from any internal stress being set up by the contraction and expansion of the wire within. 4. The larger sizes of glass break more readily than the smaller. 5. Glass set horizontally, or at an angle, breaks more readily than glass set vertically. 6. Wire netting hung under glass, from the effect of gases upon it, is unreliable. 7. Wire glass is most desirable, because when peliceiece the wire will generally hold it in position until repairs can be made. 8. Steel bars, such as are used in skylights at Broad Street and Jersey City train sheds, because of the effect of the gases on same, are unsatisfactory. 9. Wooden bars, such as are used in skylights at the Jersey City train shed, are desirable, being unaffected by gases. We recommend a zinc cap in the place of the wooden cap. 316 WINDOWS AND SKYLIGHTS _ eke, 10. We recommend that a zinc expansion bar be used with brass bolts in preference to wooden bars. We are led to this conclusion from. n shed, which show no deterioration from effect of gas. 11. We recommend that the sizes of glass used in skylights should not exceed 24 XX 36 inches. 12. We recommend for future construction and, Sree train sheds, where same can be adopted, a monitor form of skylight, prefer- 5 ably placed parallel with the tracks, of large dimensions; set far enough apart so that one monitor will not obstruct the light of an-— other; small sizes of glass, set in wooden or approved metal frames, fa frames set loose enough to overcome the contraction and expansion of . the metal work of the shed. “ee 13. We recommend ventilating the monitors at the top, the open- ing being covered with an umbrella shelter, also by putting on each side of the monitor an opening under the eaves above the glass the” entire length. CHAPTER XXIV. VENTILATORS. Ventilation— Mill buildings are ventilated either by forced draft - or by natural ventilation. Natural ventilation is usually sufficient, al- though forced ventilation is necessary in many factories and mills. The problem of ventilation is too large to consider fully in this place and the natural method of ventilation only will be discussed. ‘The amount of air required depends on the use to which the building is to be put; a common specification for the ventilation of mill buildings being that ventilators shall be provided and located so as to ventilate the building properly, and shall have a net opening for each 100 square feet of floor space of not less than one-fourth square foot for clean machine shops and similar buildings; of not less than one square foot for dirty ma- chine shops ; of not less than four square feet for mills; and not less than six square feet for forge shops, foundries and smelters. Ventila- tors in high buildings are more effective than in low ones. The follow- ing table will give an idea of the effect of height on ventilation.* Height above ground. 20’ 30’ 40’ 50 Machine shop, sq. ft. per 100 % 34 8 ¥Y round vents. Mills, ih ain ke ke em 4 6 5 4 Louvre vents. ‘ce ‘ec ‘ec “ce 9 8 7 6 Louvres or | open vents. Monitor Ventilators——The openings in the clerestory of monitors Forge shops, are fitted with louvres, shutters or sash, or may be left entirely open. Louvres are made in many different ways, the Shiffler Louvres shown in Fig. 158, and the Berlin Louvres shown in Fig. 159, are in common use. The details of these louvres as made by various firms differ some- what. | *Mill Building Construction, H. G. Tyrrell. 3318 VENTILATORS Froof Steel ~~. Strap -~ 44 Use angle uprights at splice joints of lores xq ' f 238 brackets 0 .. at splice joints Gauge of metal ae: ap xe * 22 unless specitied ~~ : 2g “Strap at fouls -— sy (Tax length of louvres r+ H'wide-Z"holes in uprights, 22 \ = forz oval screw head bolts 7a 2 "fang - --¥ iF Louvre Block * le Long Fic. 159. BERLIN LOUVRES. The details of the Shiffler louvres shown in Fig. 158, and of the Berlin louvres shown in Fig. 159 are those adopted by the American Bridge Company. The details of the louvres are shown in the cuts and need no further explanation. a9 N ry : § it é ¥ Wh of 4a z |g) 2 5x24g 1 . N \ ! £ 5 ee ' 8 ! Q] SN : ' s|* : ; oa ae ~ a mig} 1 on ¥ x . % ~ +S 3 we! g , = re- Shatt $ Re ; jot-b? Th 2 Ll) i! ; St mig] 1) S , a we y e; ‘ : : . Doe BN nd ae : it ecoelever 2 SI | ! B | bbec . i “oe 4 ; ye! ' | Steel Spring to opposite side ’ “Ae Lexx ! ees —YOM — - tT. ' (a re ee ate: Ls he Siecczeoness v = ' = (4"Flashing ger | \-Fpe ° uc \ ! @ 4 + 1 ' J se =o Fic. 160. HINGED MONITOR SHUTTER. Details of a hinged shutter are shown in Fig. 160. The angle iron frame is covered with a corrugated iron covering. The shutters are made from 6 to Io feet long, with two hinges for shutters 8 feet long and three hinges for shutters more than 8 feet long. Where shutters are to be glazed they are hung as in Fig. 156. The lever gear shown by the dotted lines is used in the better class of structures. This device can be used where the shutters are glazed if care is used in operating. In smelters the clerestory of the monitor is often left entirely open or is slightly protected by self acting shutters. In the latter case the shutters are hinged at the bottom and are connected at the top with each other and with a counter-weight so that the shutter will ordinarily make 3.20 VENTILATORS an angle of about 30 degrees with the vertical. A wind or a storm will close the windward shutter and open the leeward shutter wider. The eaves of the monitor are made to project, so that very little of the storm enters. Cost.—The shop cost for louvres is ordinarily about 1 cent per pound. ‘To this must be added the cost of the sheet steel and the cost of the framework and details. In 1900 louvres without frames cost about 25 cents per square foot. Circular Ventilators.—Circular ventilators are often used fon ven- tilating mill buildings in place of the monitors, and on buildings requir- ing a small area for ventilation. They are made of galvanized iron, copper or other sheet metal, and are usually placed along the ridge line of the roof. ith TUN OMA INONMNTTATCT Acorn Ventilator. BUCKEYE VENTILATOR. Fic. 161. CIRCULAR VENTILATORS. w=. oe — ee ee a , 7 ss eee a a CIRCULAR VENTILATORS 321 There are many styles of circular ventilators on the market, a few of which are shown in Fig. 161. The Star ventilator made by Mer- chant & Co., Chicago, is quite often used and is quite efficient. It is made in sizes varying from 2 to 60 inches. In 1903 Star ventilators made of galvanized iron were quoted about as follows: 12-in., $2.00; 18-in., $6.75 ; 24-in., $10.00; 40-in., $45.00. The Globe ventilator made by the Cincinnati Corrugating Com- pany, Cincinnati, Ohio; the Garry ventilator made by the Garry Iron & Steel Roofing Co., Cleveland, Ohio; and the Acorn and Buckeye ven- tilators made by the Youngstown Iron & Steel Roofing Co., Youngs- town, Ohio, are quite efficient and all cost about the same as the Star . except the Garry ventilator, which is cheaper. Home-made circular ventilators can be made that are quite as sat- isfactory as the patented ventilators and are much less expensive. In 1900, ten 36-inch circular ventilators cost $12.25 each, and two 24- inch circular ventilators cost $9.25 each in Minneapolis, Minn. The cost of the 24-inch ventilators was large on account of the small number made. CHAPTER XXV. Doors. Paneled Doors.—For openings from 2’ 0” x 6’ 0” to3’ 0” x 9! o” ordinary stock paneled doors are commonly used. The stock doors vary in width from 2’ 0” to 3’ o” by even inches and in length by 4” to 6” up to 7’ o” for 2’ o” doors, and 9’ o” for 3’ o” doors. Stock doors are made 13 and 134 inches thick, and are made in three grades, A, B and C; the A grade being first class, B grade fair and C grade very poor. Paneled doors up to 7 feet wide and 2% inches thick can be obtained from most mills by a special order. Wooden Doors.—Wooden doors are usually constructed of matched pine sheathing nailed to a wooden frame as shown in Fig. 162 and Fig. 163. Section A-A La a 4... oo BF ie ape ig he eo ee Hin eal high LAOS P cc Cie / vt K--*/70N F106 = * 2» ss (ee Wed ee ae Bey Lehr Baw, sacs : LY ,¢ Top Frail 6 ala “4 . 1 fo Ld «---Up fo 3'--- > FFF ear, a Pail 3 Ly ! Poe ke ie 7 4 ‘ 1 i) Y y+ a ! ! rts) 4 ae ' i te oe Daas ! A LZ “1A I F 7 ! oe 6) y ; sy | ; ' ASA ie ‘ ae : Fae y, ye : by My 4 N 1 | ae | AnaaleRal Gk | RS 2 a Ogee ke eal al Pe “4 { £ ! ob, ee baa : NS ; ia a = : re { ) a Y ae 1 'is t 1 e ¢ =) ee Pas ' Py a J Rage oe Vo Bat oe y Vr . ‘ii YL y ba) “ ‘ a we i177 / in bee oe an eed 1 ‘f l i , Peal t iv PS be oe TS ora tar IF Ok LOS OR ae lt a | ier AE ee \ |Borrim fall Bg \ 1 1 | /Bortdn Aral lOxle’ } 5 ele e i } EY RES TA a i ¥ PAS SLE EY EY BE SRE ES Eo s Mew me ee eee eK 7-0O Swing Wooden Doors Sliding Wooden Door . Fic. 162, Fic. 163. DETAILS OF Doors 323 Designs for wooden swing doors are shown in Fig. 162, and for a wooden sliding door in Fig. 163. These doors are made of white pine. Doors up to four feet in width should be swung on hinges ; wider doors should be made to slide on an overhead track or should be counter- balanced and raise vertically. Sliding doors should be at' least 4 inches wider and 2 inches higher than the clear opening. “Sandwich” doors are made by covering a wooden frame with flat or corrugated steel. The wooden framework of these doors is com- monly made of two or more thicknesses of 74-inch dressed and matched white pine sheathing not over 4 inches wide, laid diagonally and nailed with clinch nails. Care must be used in handling sandwich doors made as above or they will warp out of shape. Corrugated steel with 1%4- inch corrugations makes the neatest covering for sandwich doors. For swing doors use hinges about as follows: For doors 3’ x 6’ or less use 10-inch strap or 10-inch T hinges; for doors 3’ x 6’ to 3’ x 8’ use 16-inch strap or 16-inch T hinges ; for doors 3’ x 8 to 4’ x 10° use 24-inch strap hinges. Steel Doors.—Details of a steel lift door are shown in Fig. 164. This door is counterbalanced by weights and lifts upward between ver- Vole tor 4 sree! cable tor ho/sting ~~~ . Pu yw + \ O-n+ : ! No. 22 Car Steel fas - a | 1 ened ro this side by means le : EVI iron barr 122 and & rivers t Sil 5 shown. -. RE ist Se Boss Mate Eg i108" PUNO GT Se ‘ shel o V2 hl. Shop rivers f and BSS t 7 , 2 an this side 104 MiG. - NI 1 rei 0 v tobi a SHIN x ¥ THIN vy 1 : N 5h “ale me ¥ eg S ww aw we we ee eee 7*O ee oe Steel Lift Door Fic. 164. 324 Doors tical guides. This door was covered with corrugated steel with 14- inch corrugations as described in the cut. Details of a steel sliding door are shown in Fig. 165. ‘This door is made to slide inside the building and swing clear of the columns. Where the columns are so close together that there is not room enough for the door to slide the entire length of the opening, it should be placed on the outside of the building. The track and hangers shown —& Bar Hangers 5 *i*5 abt Wheels about 5 aia. ~»{\ 4 Tre rors £. - Hole ingirt to which \P aoor nay be locked trom [he side ~. SXF ZL -3 "Me Corrugated tron to Le oct torrame Cor. fron by riveting wilh ¢° All ress ~ rivers '0¢ Counter| \ OUTER x e sunt onthe outside \\ WAE0Ex! N us N tugatle. , ¥ as mS = 3 x ‘x “ate AZ AZ WY * Hoy wh ie Nee ae +, Poe »- Lock here from wrsrae Steel Sliding Door Fic. 165. make a very satisfactory arrangement; however there is a tendency for the wheels to jump the track unless the grooves in the wheels are made very deep. There are quite a number of patented devices on the market for hanging sliding doors. ‘The Wilcox trolley door hanger, track and Cost oF Doors 325 7 bolt latch shown in Fig. 166, are efficient and are quite generally used. The prices of the door fixtures shown in Fig. 166 are about as follows: door hangers, $2.25 to $3.00 per pair; steel track, 10 to 25 cts. per ft.; clips, 15 to 25 cts. each; door latch, $1.00, f. o. b. the factory at Aurora, Ill. Discounts for this and several other well known makes of door fixtures are given each week in the Iron Age, New York, and the list prices are given in the manufacturer’s catalogs. L Wilcox O. K. Steel Track Wilcox Gravity Door Bolt and Latch Fic. 166. Cost of Doors.—Stock panel doors cost $1.50 to $5.00 each, depend- ing upon the grade, size and conditions. The details of steel doors vary so much that it is necessary to make detailed estimates in each case. ‘The shop cost of the framework is often quite high and may run as high as 3 or 4 cts. per pound. The wooden frames for sandwich doors cost from 20 to 25 cts. per square foot. The cost of hinges, bolts, etc., required for doors can be found by applying the discounts given in the Iron Age to the list prices given in the standard lists (see Chanter XXVIII). * CHAPTER XXVI. SHOP DRAWINGS AND RULES. SHOP DRAWINGS.—The rules for making shop drawings in use by the American Bridge Company are given in their Standards for Structural Details, and are reprinted in part, in Roofs and Bridges, Part III, by Merriman and Jacoby. The following rules are essentially those in common use by bridge companies, for mill buildings and ware- houses. _ Make sheets for shop details 24 by 36 inches, with two border lines, 1% and 1 inch from the edge, respectively. For mill details use special beam sheets. ‘The title should come in the lower right hand corner, and should contain the name of the job, the contract number, and the initials of the draftsman and checker. " Detail drawings should be made to a scale of 34 to 1 inch to the foot. Members should be detailed as nearly as practicable in the po- sitions in which they occur in the structure. Show all elevations, sec- tions, and views in their proper positions. Holes for field connections should always be blackened. Members that have been cut away to show a section, may be either blackened or cross-hatched. Members, the ends of which are shown in elevation or plan, should be neither black- ened nor cross-hatched. Holes for field connections should be located independently, and should be tied to a gage line of the member. When metal is to be planed, the ordered and finished thickness should be given. In making shop drawings for mill buildings two methods are in use. . The first method is to make the drawing so complete that templets can be made for each individual piece, separately on the bench. The second method is to give on the drawings only sufficient di- mensions to locate the interior of the members and the position of the ERECTION PLAN ; 327 pieces, leaving the templet-maker to work out the details on the laying- out floor. The first method is illustrated in Fig. 95 and the second in’Fig. 96. In the second method sufficient figures should be given to proper- ~ ly locate the main points in the truss; the interior pieces should be lo- cated by center-lines corresponding to the gage lines of the rivets, the centers of gravity lines or the outside edges of the pieces, as the case may be. The drawings should always indicate the number of rivets to be used in each connection, the size of rivets, the usual rivet pitch, and the minimum pitch allowed. Erection Plan.—The erection plan should be made very complete. All the notes that it is necessary for the erecter to have, should be put on the erection plans; how much of the structure is to be riveted and how much bolted, whether it is to be painted after erection or not, whether the windows and doors are to be erected or not, etc. Center line drawings are usually sufficient for the erection plans. The name and the size of the piece should be given and every piece should have a name. The following method was used by the Gillette-Herzog Mfg. Co., for mill buildings, and was very satisfactory: If the points of the compass are known, mark all pieces on the north side with the letter “N”, those on the south side with the letter “S”, etc. Mark girts N. G. 1; N. G. 2; etc. Mark all posts with a different number, thus: N. P. 1; N. P. 2; etc. Mark small pieces which are alike with the same mark; this would usually include everything except posts, trusses and girders, but in order to follow the general marking scheme, where pieces are alike on both sides of a building, change the general letter; e. g. N. G. 7 would be a girt on the north side and S. G. 7 the same girt on south side. Then in case the north and south sides are alike, only an elevation of one side need be shown, and under it a note thus: “Pieces on south side of building, in cor- responding: positions have the same number as on this side, but prefixed by the letter “S” instead of the letter “N.” Mark trusses T. 1; T. 2; etc. Mark roof pieces R. 1; R. 2: etc. ? 328 SHop DRAWINGS AND RULES The above scheme will necessarily have to be modified more or less according to circumstances ; for example, where a building has dif- ferent sections or divisions applying on the same order number, in which case each section or division should have a distinguishing letter which should prefix the mark of every piece.- In such cases it will per- haps be well to omit other letters, such as N., S., etc., so that the mark will not be too long for easy marking on the piece. In general, how- ever, the scheme should be followed of marking all the large pieces, whether alike or not, with a different mark. This would refer to pieces which are liable to be hauled immediately to their places from the cars. But for all smaller pieces which are alike, give the same mark. For architectural buildings adopt the following general scheme of marking: ‘The basement “A”; first floor “B”; second floor “C”; then mark all the pieces on the first floor B. 1; B. 2; etc.; columns between first and second floors B. C. 1; B. C. 2; ete. It will greatly aid the detailing, checking and erection if small sec- tions are made showing the principal connections, such as girt connec- tions, purlin connections, etc. The erection plans of a mill building drawn in accordance with these rules are shown in Fig. 167 and Fig. 168. CHOICE OF SECTIONS.—In designing, it will be found eco- nomical to use minimum weights of sections, and to use sections that can be most easily obtained. As small a number of sizes should be used as is practicable where material is to be ordered from the mill, if good delivery is to be expected. The ease with which any section can be obtained in a mill order, depends upon the call that that particular mill is having for the given section. If there is a large demand for the section, it will be rolled at frequent intervals, while if there is little or no demand for the section, the rollings are very infrequent and a small order may have to wait for a long time before enough orders for the section will accumulate that will warrant a special rolling. The ease with. which sections can be obtained will, therefore, depend upon the mill and the conditions of the market. The standard and permissible Se “aku So fat Se ea ae | hs We Saal ~ Sl WSR e - v CHOICE OF SECTIONS 329 sizes of sections in use by the American Bridge Company, are given in the following table. Standard Angles. Permissible Angles. 6” x ae ray x 4” yaa x ge 6” x 3% 4” x ’” ” 4 3% ” ” x Sf wr x 32" 3” x 34” r x 3 24" x 24" 3u% x 21%" a x te 37/2" x a “at x 2! ci x 2” IZ" tZz"" / 2 I 2/2 s 2Y/2 33" - 2" 4 Standard Channels.. Permissible Channels I - 8” 9” ew 6” as Io” wy - i Standard I Beams. _ Permissible I Beams. 20” Io” : 24” I Q” : Q” 9” dA ” ye 1 6 cn ‘Permissible Tees. cep a ae tes oe he Tacs 3 84) Permissible Zee Bars. 6” Re . 4” a Standard Flats. I yy" a 6” es I 4" 3/2 uw b, 14” 2 4 : Ad TZ ” 274 ” 47 2, 5 10 Standard Rounds. 5" 4” RR” te 14” 14” Standard Squares. 4” wi ut ao rT yy" I yy" Other sizes than those specified may be obtained, but the time of delivery will be very uncertain unless the order is large enough to war- rant a rolling. Deck beams, bulb angles and special section Z-bars are hard to get unless ordered in large quantities. Flats 14” thick and under are very hard to get. Flats under 4” should be ordered by 14” variation in width; flats and universal plates over 4” should be ordered by 1” variation in width. CHAPTER XXVII. PAINTS ee PAINTING. Corrosion of Steel—lIf iron or steel is left exposed to the atmos- phere it unites with oxygen and water to form rust. Where the metal is further exposed to the action of corrosive gases the rate of rusting is accelerated, but the action is similar to that of ordinary rusting. Rust is a hydrated oxide of iron, and forms a porous coating on the surface of the metal that acts as a carrier of oxygen and moisture, thus pro- moting the action of corrosion. If nothing is done to prevent or retard the corrosion of the iron and steel used in metal structures, the metal rapidly rusts away and the structure is short lived. Wrought iron is affected by corrosion more than cast iron, and steel is affected more than wrought iron. The corrosion of iron and steel may be prevented or retarded by covering it with a coating that is not affected by the corroding agents. This is very effectually accomplished by galvanizing; but on account of the cost it is impracticable to use the process for coating anything but sheet steel and small pieces of structural steel. ‘The most common methods of protecting iron and steel are by means of a coating of paint, or by imbedding it in concrete. PAINT.—The paints in use for protecting structural steel may be divided into oil paints, tar paints, asphalt paints, varnishes, lacquers, and enamel paints. ‘The last two mentioned are too expensive for use on a large scale and will not be considered. OIL PAINTS.—An oil paint consists of a drying oil or varnish and a pigment, thoroughly mixed together to form a workable mixture. “A good paint is one that is readily applied, has good covering powers, LINSEED OIL, 331 adheres well to the metal, and is durable.” ‘The pigment should be inert to the metal to which it is applied and also to the oil with which it is mixed. Linseed oil is commonly used as the varnish or vehicle in oil paints, and is unsurpassed in durability by any other drying oil. Pure linseed oil will, when applied to a metal surface, form a trans- parent coating that offers considerable protection for a time, but is soon destroyed by abrasion and the action of the elements. To make the coating thicker, harder and more dense, a pigment is added to the oil. An oil paint is analogous to concrete, the linseed oil and pigment in the paint corresponding to the cement and the aggregate in the concrete. The pigments used in making oil paints for protecting metal may be divided into four groups as follows: (1) lead; (2) zinc; (3) iron; (4) carbon. | Linseed Oil.—Linseed oil is made by crushing and pressing flax- seed. The oil contains some vegetable impurities when made, and should be allowed to stand for two or three months to purify and settle before being used. In this form the oil is known as raw linseed oil, and is ready for use. Raw linseed oil dries (oxidizes) very slowly and for that reason is not often used in a pure state for structural iron paint. _ The rate of drying of raw linseed oil increases with age; an old oil be- ing very much better for paint than that which has been but recently extracted. Raw linseed oil can be made to dry more rapidly by the addition of a drier or by boiling. Linseed oil dries by oxidation and not by evaporation, and therefore any material that will make it take up oxygen more rapidly is a drier. A common method of making a drier for linseed oil is to put the linseed oil in a kettle, heat it to a tem- perature of 400 to 500 degrees Fahr., and stir in about four pounds of red lead or litharge, or a mixture of the two, to each gallon of oil. This mixture is then thinned down by adding enough linseed oil to make four gallons for each gallon of raw oil first put in the kettle. The addition of four gallons of this drier to forty gallons of raw oil will reduce the time of drying from about five days to twenty-four hours. A drier made in this way costs more than the pure linseed oil, so that driers are very often made by mixing lead or manganese oxide with 332 PAINTS AND PAINTING rosin and turpentine, benzine, or rosin oil. ‘These driers can be made for very much less than the price of good linseed oil, and are used as adulterants ; the more of the drier that is put into the paint, the quicker it will dry and the poorer it becomes. Japan drier is often used with raw oil, and when this or any other drier is added to raw oil iu barrels, the oil is said to be “boiled through the bung hole.” Boiled linseed oil is made by heating raw oil, to which a quantity of red lead, litharge, sugar of lead, etc., has been added, to a temper- ature of 400 to 500 degrees Fahr., or by passing a current of heated air through the oil. Heating linseed oil to a temperature at which merely a few bubbles rise to the surface makes it dry more rapidly than the unheated oil; however, if the boiling is continued for more than a few hours the rate of drying is decreased by the boiling. Boiled linseed oil is darker in color than raw oil, and is much used for outside paints. It should dry in from 12 to 24 hours when spread out in a thin film on glass. Raw oil makes a stronger and better film than boiled oil, but it dries so slowly that it is seldom used for outside work without the addition of a drier. | Lead.—_W ute Lead (hydrated carbonate of lead—specific grav- ity 6.4) is used for interior and exterior wood work. White lead forms an excellent pigment on account of its high adhesion and covering power, but it is easily darkened by exposure to corrosive gases and rapidly disintegrates under these conditions, requiring frequent re- newal. It does not make a good bottom coat for other paints, and if it is to be used at aJl for metal work it should be used over another paint. Red Lead (minium; lead tetroxide—specific gravity 8.3) is a heavy, red powder approximating in shade to orange; is affected by acids, but when used as a paint is very stable in light and under ex- posure to the weather. Red lead is seldom adulterated, about the only substance used for the purpose being red oxide. Red lead is prepared by changing metallic lead into monoxide litharge, and converting this product into minium in calcining ovens. Red lead intended for paints must be free from metallic lead. One ounce of lampblack added to one pound of red lead changes the color to a deep chocolate and increases the PIGMENTS 333 time of drying. This compound when mixed in a thick paste will keep 30 days without hardening. Zinc.—Zine white (zinc oxide—specific gravity 5.3) is a white loose powder, devoid of smell or taste and has a good covering power. Zinc paint has a tendency to peel, and when exposed there is a tendency to form a zinc soap with the oil which is easily washed off, and it therefore does not make a good paint.. However, when mixed with red oxide of lead in the proportions of 1 lead to 3 zinc, or 2 lead to I zinc, and ground with linseed oil, it makes a very durable paint for metal surfaces. ‘This paint dries very slowly, the zinc acting to delay harden- ing about the same as lampblack. Iron Oxide.—Iron oxide (specific gravity 5) is composed of anhydrous sesquioxide (hematite) and hydrated sesquioxide of iron (iron rust). The anhydrous oxide is the characteristic ingredient of this pigment and very little of the hydrated oxide should be present. Hydrated sesquioxide of iron is simply iron rust, and it probably acts as a carrier of oxygen and accelerates corrosion when it is present in considerable quantities. Mixed with the iron ore are various other in- gredients, such as clay, ocher and earthy materials, which often form 50 to 75 per cent of the mass. Brown and dark red colors indicate -the anhydrous oxide and are considered the best. Bright red, bright purple and maroon tints are characteristic of hydrated oxide and make less durable paints than the darker tints. Care should be used in buying iron oxide to see that it is finely ground and is free from clay and ocher. Carbon.—The most common forms of carbon in use for paints are lampblack and graphite. Lampblack (specific gravity 2.6) is a great absorbent of linseed oil and makes an excellent pigment. Graphite (blacklead or plumbago—specific gravity 2.4) is a more or less im- pure form of carbon, and when pure is not affected by acids. Graphite does not absorb nor act chemically on linseed oil, so that the varnish simply holds the particles of pigment together in the same manner as the cement in a concrete. There are two kinds of graphite in common use for paints—the granular and the flake graphite. The Dixon Graphite Co., of Jersey City, uses a flake graphite combined with silica, 334 PAINTS AND PAINTING while the Detroit Graphite Manufacturing Co., uses a mineral ore with a large percentage of graphitic carbon in granulated form. On account of the small specific gravity of the pigment, carbon and gra- phite paints have a very large covering capacity. The thickness of the coat is, however, correspondingly reduced. Boiled linseed oil should — always be used with carbon pigments. Mixing the Paint——The pigment should be finely ground and should preferably be ground with the oil. The materials should be bought from reliable dealers, and should be mixed as wanted. If it is not possible to grind the paint, better results will usually be obtained from hand mixed paints made of first class materials than from the ordinary run of prepared paints that are supposed to have been ground. Many ready mixed paints are sold for less than the price of linseed oil, which makes it evident that little if any oil has been used in the paint. The paint should be thinned with oil, or if necessary a small amount of turpentine may be added; however turpentine is an adulterant and should be used sparingly. Benzine, gasoline, etc., should never be used in paints, as the paint dries without oxidizing and then rubs off like chalk. Proportions.—The proper proportions of pigment and oil required to make a good paint varies with the different pigments, and the methods of preparing the paint ; the heavier and the more finely ground ‘pigments require less oil than the lighter or coarsely ground while ground paints require less oil than ordinary mixed paints. A common rule for mixing paints ground in oil is to mix with each gallon of lin- seed oil, dry pigment equal to three to four times the specific gravity of the pigment, the weight of the pigment being given in pounds. This rule gives the following weights of pigment per gallon of linseed oil: white lead, 19 to 26 Ibs.; red lead, 25 to 33 Ibs.; zinc, 15 to 21 lbs.; iron oxide, 15 to 20 lbs. ; lampblack, 8 to 10 Ibs.; graphite, 8 to 10 Ibs. The -_ weights of pigment used per gallon of oil varies about as follows: red lead, 20 to 33 Ibs.; iron oxide, 8 to 25 lbs.; graphite, 3 to 12 Ibs. Covering Capacity.—The covering capacity of a paint depends © upon the uniformity and thickness of the coating; the thinner the coat- COVERING CAPACITY 335 ing the larger the surface covered per unit of paint. To obtain any given thickness of paint therefore requires practically the same amount of paint whatever its pigment may be. The claims often urged in favor of a particular paint that it has a large covering capacity may mean nothing but that an excess of oil has been used in its fabrication. An idea of the relative amounts of oil and pigment required, and the cov- ering capacity of different paints may be obtained from:the following table. AVERAGE SURFACE COVERED PER GALLON OF PAINT.* ! Lbs. |Volumeand Square Feet. Paint. vo? lor Pig-| Weight ; 5 ment.| of Paint. ery Cokie: : . Gals. Lbs. Iron Oxide (powdered) | 1 gal 8.00] 1.2—16.00 600 350 « “ (ground inoil)..| 1 24.75) 2.6—32.75 630 375 Red Lead (powdered)..| 1 « 22.40) 1.4—30.40 630 375 White Lead (g’rdin oil).| 1 « 25.00] 1.7—=33.00 500 300 Graphite(groundinoil).| 1 « 12.50] 2.0—20.50 630 350 Black - Girt 23 +235" _Ge4 P4avPs ’ = . z phen RS TCR 955 to o% ~ od { ~A" 2 Purtiry IL@63" ! == | 4 erin ” , Peds “<> Ih | ae. RANE: < > TT \ ~ ae p XN 7 ~~ oe x Sa 7 mies nn ” AX ~ > | I ; — = 1 Ue S 4-Purtin Lb x3 %5° EP; a Sil ‘wl 2 y Wave Strut FL @/34 Gl. t ae 3 I Se. aed Girk #°7@ 524 ‘ ‘ “il ye . I 2 > Va O55 GS \ Z took | { ae See RN ‘al | aN “| [3 ie * 6 3 6 a i138 Re oe |= aR Nes fap ce S S$ @ x 4 \ eal | YS BN By SG 80. 9 = 7 reBret | 4 : : \ Girti4# C@ 5 "ra s* \ H S i] ane, 1s x I | ssakeod apbea air dies SSS SSH _-_—_t= ie a od - U “3°! 7-6 4° 3714-3” 7*6"_ | 4:3 15°63) 8-62" 3-04 45 7-6" 50 45° 7:6" 1473" BS 2°57 ASIST Zo LIOR aH CANFAS 726% ASL FS 76> Ae Pais Ba ae, 16:0" 16-0" 16:0" D Sat een I le in Sapien oO cn Ceri wl gE WE ae om ph See oe MO ca re ANS aay & eae ole (oe Ou a 16 on Ae 1 G7 “ ie Girt 4 LO@53 -7 S | Gio | Cl 4] Git Window Fost Door Fost 4 LO 5g * 2 { i u 7 it U u q U ———— pas =a tor b xe: '26l 5:92" 14-621 5°32" A AO ote. 7'6= _ BS F788 iF Os, F $a ———- + = = = pc, END ELEVATION Fic. 168. SIDE AND END ELEVATIONS OF STEEL, TRANSFORMER BUILDING. 344 EstIMATE OF WEIGHT AND Cost ESTIMATE OF WEIGHT Steel Frame Transformer Building ,60°0" Wide , 80°O"Long ,20°0" Posts. Pitch of Roof 3 - Covered with Corrugated Iron. umber Lenath |Weight| Weight Details of |Shape] Section Z per = 1 By bn we Pieces | Feet |inches| Foot |Members| Details|fempers| Weight 4 Trusses each thus:- 4 | is | 38x2sxg 1I9 | 53] 60] 463 4] 3ex25x 2 18 | O% | 49 |] 353 4/55 3x 2x8 i | 5:0 222 4 Ls 3xX2xt 10 10 4:0 Be 2 Ls Bx 2xd 16 2 4.0 259 8 | bs 2x2xz 4 9 3-2 70 4 Ls 2ax2xXZ o Be me 3-6 84 8 Ls 2x2xs 5 I 32 130 4 | 1s Bx 2X 10 | 5% 4.0 167 4) 6 2ex2*5 1) | 10 36 | 170 8 Ls 2x2xzZ 10 9 3.2 271 i ty Bx2xz 19 2 4.0 77 4/658 2x2xt 6 fe) 32 77 Sle 2x2xZ 5 | 73 32 90 2 Ls 2x2xzZ 4 73 32 29 4 / 15 OX 2K i 14 32 186 2 | Pls. 18x 5 2|4 9.13 89 2 | Pls 7X s 72 595 7 2 Pls 45xZ 93 36) 6 6 Pls 7x 4 L fe) 595 36 i Pis 74% 3 84 616 42 4 | Pls 8sxe i re) 7:22 29 2 | Pls 8ExXz 1 2 7:22 17 4 | Pis i2xzZ ! 10 10.2 73 2 Pls 165%x3 ! 8 13.61 46 ! Pls 24x35 5 2 204 65 4 | Pis 2xz 1 53 255 15 12 Pls 65x+ 7 553 42 2 Pls 9gxt 7 818 9 2 Pls Sixt 1 rf) 446) 9 2]|b 23x27% 3 93) +0 6 16 Ls DEx22"5 7 4.9 45 2]. 33 x25 x 3g} 49 3 1348 Rivet Heads Per 100 9.95) 13+ 56 Washers Per 100 BOOT eos a. 2621 | 684 | 246 . Total Weight of 4 Trusses = 3505X4= © 14020 8 Posts each thus:- | 4/6 33 25x45 19 Ng | 49 | 391 1 PI 1044 2 fo) 8-71 16 a Ls 33*23%*% 7% 4. 59 15 ’ Pl IBxzZ 1 23 | '105 13 1 a | 10x 2 t i 21-25 41 2 Ls Sued 10 11-6 19 2 Ls 6x4 x 93 12-3 21 17 | Bars 2xZ 1 73 1-7 41 160 2 Rivet heads Per 100 9-95 ; 16 351 | 188 | 48. Total Weight of 8 Posts = 579*8 = 4632 | j 18652 ee ee oe BA EstIMATE OF WEIGHT 345 Number Lenoth [Weight] Weight Details in of |Shape| Section S per FMamn 4 Br aver Total Pieces Feet |Inches| Foot Merober| Details | Members | Weight 4 Posts ,each thus:= Brought Forward 18652 1 I 3"@ 2i# | 35 24 21 740 8 is 32xX32x% 53 ca 26 2 is 6x4x3 63 12.3 13 2 Ls 6X 33 x3 9 11.6 17 ! Pi 9x gx 2 ! 5 19.13 27 88 3° Rivet Heads | per 100 ne; eae aa Cae ee 740 | 92 Total Weight 4 Posts =|a3s7x4 = rt aa 4 Posts,each thus:- I I s"@ 21* 27 103 2) 586 6 is 32 x33 x2 53 7 20 ! i 6x 6x3Z 9 14.8 i 2 Ls 6x4x3 63 12.3 13 2 is 6 X 35x 9 11.6 17 ' PI 9x ! S 1913 27 80 g’ Rivet Heads per 100 995 _8 566 | 96 16.4 Total Weight] + Posts = 682x4| = 2128 4 Posts,each thus:- ! L 6x6 x2 20 82 | 148] 306 10 Pls yo Ie 4 63 595 33 ! PI 5x4 93 | 425 3 ’ PI 1Ox 3 10 21.25 18 Bee Bes: 5 x33x 5 be Bt 9 3 is 5 x 32x ‘i 8 8.7 17 ? no 35x35x ° 5 sa 3 ee: LS 6 xX 33 x 93 1-6 18 100 z Rivet | per 100 to 10 306 V1 36.3 Total Weight 4 Posts = 417*4 |= 1668 4 End Rafters ,each thus:- 1 c 7@ 93* 37 53 | 975] 365 Con 8 and Pils ieee Bie. : 565 7 220 Total Weight, 4 End Rafters = 444*4/= 1176 2 Eave Struts,each thus: - 5 & 9@ 134 16 fe) 1325] 1060 Total Weight 2 Eave Struts = 1060x2 |= 2120 Bottom Chord Bracing :- 8 is 3x 3x4 18 9) 4.9| 706 10 Is 7'@15* 15 2 15.0 | 2275 36 | 6x4 x 5 12.3 184 4 & BX 3BXxZ 5 49 8 128 "Rivet fea per 100 995 15. | 298i | 205 7.0 Total Weight Bottom Chord Bracing = 3186 Purlins :- 30 & 5@e 63* 32 74| 65 | 6355 14 & » 15 112 6.5 | 1452 6 | & » 16 73 | 65 | 650 6] 5x3xz 32 73| 62 | 1604 2 is ets 15 wz &2| 262 2 is af 16 Pe eeees Tere 10595 Total Weight of Purlins = 144053 3.46 EsTIMATE OF WEIGHT AND Cost Number Length Weight} Weight _ etails in| +404 f =) - Percent picces|noPe] Section rece [inches| root [OtaiBersPetails [ef Main] Weight Girts:— . Brought Farward 44053 Cs | 4°@ 54% | 1664 |Inft,| 525] 8736 Ls Sx3xz° SGI... 4.9} A70 Ls 2ex2sxy S6]. -}| 40}. 384 Total Weight 9590 Rods .— : 8 | Rods] 3.9% 28 0 | 20] 448 16 | Rods $°? 18 0 | 20] 576 1G | Rods Z¢ 23 0 20 | 736 32 _| Bolts} Anchor Zo \ 6 2.0 96 4 . \ ) 2.0 8 18 | Rods | Louvre 3 % 3 0 1.0 S4 ISB | Bolts 2 ¢ 1.5 9 18 Spring Catches 0.5 9 36 | Pls [Anchor 4x5" 4 4 50 56 | Pins |Cotter Fees Ss 13 74 . Total Weight FER SOG Ue 2060 Total Weight Steel Framework | 55703 Corrugated lron.- 84 Saquvores No 22 as per list, per sq. 138 | 11592 70 ot ne Pl ee ee whe it 7770 Ridge Roll -« 22 Black 250 Flashing ~ ae * 414 Cornice . i : ane Louvres . and Nol 100) 193621 4264 Totel Weight Corrugated \ron 23626} Total Weight Steel 79329 Summary — Trusses {t284| 2736) 24.6 14020 AL. Posts 3128] 1504] 481] 4632 IB ° S304) 752} 14.1 6056 L - 1224) 444] 36.3 1668 End Rafters 1460} 316] 22.0: 1776 Eave Struts 2120 0.0 2120 Bottom Chord Bracing 298i ZOOS! = 70 3186 Rods 1760} 300) 17.0 2060 Weight Excluding Purling ond Girts [29261] 6257] 220] 35518 Purliins 10595 10595 Girts 9530 9590 Total Frome work 49446] 6257] 130 | 55703 Corrugated tron __ 23626 Total Weight 79329 The weights and per cents of three other buildings, are shown in Table XXVI. The estimates for these buildings were made from the shop drawings, and were checked with the shipping weights. These buildings are of light construction with end post bents, (a) Fig. 1. eS oe ae ee eee Sa on aa, ih EstIMATE oF Cost oF Mitt, BuILpING ESTIMATE OF COST 347 Classification of Material Cost_of Material Cost of Labor ; Weight {Price} Amount |Price}| Amount Riveted Trusses 14020 |*1.60 } 22432 |*1.00 #4020 Latticed Columns 4632 | 1-60 74.1 1 1.00 4632 I Beam ” 6056 | 165 9992 -50 3028 L ” 1668 | 1-60 2669 -50 834 C Struts 3896 1-60 62.34 26 9.14 I Beam Bracing 2480 | 1-65 40.92 25 6.20 L Bracing + 706} 1-60 11-30 25 1.77 C Purlins 10595 | 1-60 169.52 A5 15.90 C Girts 9590 | 1-60 15344 415 14.39 Rods 2060 | 1-80 37.08 1-00 20.60 Corrugated Iron No.22 11592 | 2-60 301.59 ” 7770 | 2.70 209.79 Ridge Roll,Louvres, Ete: 4264 | 2-50 106-60 1.00 42.64 Asbestos Mill Board 1760 | 2-50 44.00 Poultry Netting 400 32-50 8-d Barbed Roofing Naits 32 | 3.00 96 40-d Wire Nails ba 2-50 3 540 Stove Bolts -3"x |" (per 100) 5 | 36.00 1-94 540 Washers I"x,"x4" 76] 6.00 - 4.56 540 Cut Washers 3" ‘2 1.00 A4 800 Wire Staples 5} 4.00 20 Copper Rivets #8 -2"long 6 | 25.00 1-50 1200 Carriage Bolts 3°x23"(per 100) 210 | #110 13.20 200 ” 2°x53s" 40 46] 150 3.00 260 Wood Screws #14-Z" —» 10 | * 60 1.56 54 Steel Butts 32"x35" 60 | * 800 4.32 .@Mortise Door Locks 10 | 75.00 3.00 18 10'T Hinges 40 12.00 216 2- 10" Foot Bolts 5 50.00 1.00 2- Chain Bolts 5 | 50.00 1.00 96 Window Weights 1440 2.00 28.80 24 ” Locks 16 15.00 3.60 2asC gs ee | ea Ma Fi at a ms Ter a singe Wap ‘ Cost oF Mii, Deratrs 351 8. Coping, ordinary beveling, including cut- ting to exact length, with or without punching, including the riveting or bolting of standard connection angles.......... 35 9. For painting or oiling one coat with QTUMELY OF OF PRINEs ios 6 5- aca bv oe a a0 .10 10. Cambering Beams and Channels and other shapes for ships or other purposes. . 25 11. Bending or other unusual work........ Shop rates 12. For fittings, whether loose or attached, such as angle connections, bolts and sepa- PACTS CHES, CU ss ee ib ate ewok ee ee ete a4, The above prices are per 100 lbs. of steel. In ordering material from the mill the following items should be borne in mind. Where beams butt at each end against some other member, order the beams % inch shorter than the figured lengths ; this will allow a clearance of %4 inch if all beams come 3% of an inch too long. Where beams are to be built into the wall, order them in full lengths making no allowance for clearance. Order small plates in mul- tiple lengths. Irregular plates on which there will be considerable waste should be ordered cut to templet. Mills will not make reentrant cuts in plates. Allow % of an inch for each milling for members that have to be faced. Order web plates for girders %4 to % inch narrower than the distance back to back of angles. Order as nearly as possible every thing cut to required length, except where there is liable to be changes made, in which case order long lengths. | It-is often possible to reduce the cost of mill details by having the mills do only part of the work, the rest being done in the field, or by sending out from the shop to be riveted on in the field connection angles and other small details that would cause the work to take a very much higher price. Standard connections should be used wherever possible, and special work should be avoided. The classification of iron and steel bars is given in Table XXVII. The full extra charges for sizes other than those taking the base rate are seldom enforced; one-half card extras being very common. 35 2 EstIMATE OF WEIGHT AND Cost TABLE XXVII. Iron Classification. Adopted Dec. ee by National Bar Iron Association. Adopted March 16, 1899, by Eastern Bar Iron Manufacturers’ Association? Rounds and Squares. Extra. Extra. is wcdeemeesaeavensrcmancwe oe agp ft CO p ccs ccc cecs Onn Ue eC bene bso ne cP esos I¥5 sets: we enee none eee wenn es woes 4, tO xy --------- ----------------- 16 303% ------------------------ #0 A eiwan ynsis hewn nao eames re | 39610 4% oc. ci eee % Ce | Qn epeaer emt venereal 9a ¥ te 1 436 10 BG | nas sents. ae eee Ic f° £9 anual cee ceeata es eeaae Ye | 43 08 (--cc.. .)...5. 135 to ys sg ish stern fin is Sab ss eee as alee as on a io yr \% to 6° sag actors. ne. sla Sal rei eles I ¥% to 5 canna dia imate ie cae eee ts | 0% tO6% 2. eee a MotO FE -2--o ess schon enna ip Ee eee ele 275 L Oval Iron. ; eo 4 to Ye ates | to Hx ae Si Fe sae et Se ee I : FG tO: 98 wee ok a I 3 WO y S-sancanckan casas seeye eres Te. | 3 tO SO sae xc nt ee eee i$ a % to 75 X 5 Been ee: Sa ‘% % to1y% Hares Tee SS < ee ese See ts , Half Oval and Half Round. 4 “ ae es Extra. we wast Babac on aeed Meepereee 4 WY '98 es cece tence ee 4 MPR POR 2 4.| io 10 fe as one no aoe gc nee sends Sp he watts coca a ts MO fe 2005 oc ees ivy 18 tag °o.--. 5.2.2.5 ccc eee Half ova's less than & their width iu thickness, extra price. Flats. Hto yx Yto x nth |4¥to 6 x Xt Y to ye. % tO sh 20 See I 4 x WO: ¥¢° avecaunus %to x X& to ee ... lc. | 4% to 6 x HXtor . we eet ee % %to ~x Kto «ich seats tae ee SO Spe 0% eee ts. 5 to ix YW toe eee mato 6 =x <%t04 ° occu kd 5 to x Seto (MH Scare te [4K to 6 xaktog ....ke ee to x Kto x .------.--- ts | 6% to 64x Kto. ys -------.. rs Yto ix Kto ¥ ssl. 4 by to 8 = tO. ye cicue a ZY to 2} 36 tO SAA vw |6% to 8 x Hto r¥ ---.----- xs a5 5 Se tO % occ. cicasee fe POM 10-8 2 2 8. ee —- r% tory x HKtor --.-.--.-.--. ts | 6% to S x @X%to's v.22 Ic. $4 10°64 ot ME ae rece fr | 8%tor0o x Kto P ....----. 1% to4 x *%toc -___Basenoextra. | 8%toro x %to FT ---.----- to 1% to4 x Ip, tor --.. -.--.-- ft | 8% toro x Ito 1% -.--.-..-- Ys Q: %0 4 SERA Sees tf | 8% toro x r%to 2) .-.-...-.. Ic. @ tO 4 SH 19:9 osc to Flats z4 thick ;,c. per ib higher than ¥f to +}; thick. Bevel edge Shaft Iron ;yc. higher than same size of Flats, All round edge iron ;;c. per tb extra. Horse Shoe Iron all sizes tc. extra. Light Bands. Extra. Extra. % x Nos. 10, 1r and 12._..--- 1; } x tox ,x Nos. ro, rrandzg ..__. ts % x No. 9 to.4% .-.-.5_2.....- Ip [ Eto 3 = No. 6 tye c eee yg to % x Nos. 10, 11 and 12------- 175 | 1% toq4 x Nos. 10, Irand1rZ ---- ys to % x No. g to x; --_.-------.-- 17%, [x1% tog x No. Q to #; ---_..-.---- ys to 56 x Nos. 10, Ir and 12.-.. -_- I7p | 4% to6 x Nos. 10, Irand 12----- 75 zs to 544-x No. g to 35; ---.---------- ic. | 4% to 6 se NGS.'@ 10 fee enc cance ts 4 to % x Nos. 10, rr and 12-_-_-.- a | 6% to 63 x Nos. ro, 11 and 12----- 44 to 4% x No. g to 7 -------------- ts | 6% to6% x No. gto #;._----------- % t% to % x Nos. ro, 1m and 12...---- wy |7 to8 x Nos. ro, rand 12--..-. Ic. 4% to % x No. g to 3% ------------.- tw |7 to8 xNo.9 to-; ... ........ % Bevel Edge Box Iron same as Light Bands of same sizes. Beaded Band Iron 1 inch to 2 inch 7, extra. Sand Band Iron yyc. above same sizes of Light Bands. ET SL TRIG ST | nel) a ed a , F i fi Gr eae . ‘a a a — a NA Sele. in rem hehee rens Synth nde dindig “oe 0 tt the cocbeloht - Fi ail eat Noa lig « j SHop Cost — 353 Shop Cost.—The shop cost of the various classes of work is a variable quantity, depending upon the equipment and capacity of the shop, the number of pieces made alike, the familiarity of the shop men with the particular class of work, and with the cost of labor. The costs given below are the average costs for a shop with a capacity of about 1000 tons per month that has made a specialty of mill building work. The costs given are based on a charge of 40 cents per hour for the number of hours actually consumed in getting out the contract. This charge is assumed to cover the cost of management, cost of operation and maintenance, as well as the cost of labor. The cost of management in a small shop is very small, but in a large concern it may amount to as much as 35 to 40 per cent of all the other charges combined. For this reason small structural shops can often fabricate light structural steel for a less cost than the large shops. The prices given are about an average of those used by the agents of the company above, and have been checked against actual costs for the greater part. Columns.—In lots of at least six, the shop cost of columns is about as follows: Columns made of two channels and two plates, or two channels laced cost about 0.80 to 0.70 cents per Ib., for columns weigh- ing from 600 to 1000 Ibs. each; columns made of 4-angles laced cost from 0.80 to 1.10 cents per lb; columns made of two channels and one I beam, or three channels cost from 0.65 to 0.90 cents per lb.; columns made of single I beams, or single angles cost about 0.50 cents per lb.; and Z-bar columns cost from 0.70 to 0.90 cents per Ib. Plain cast columns cost from 1.50 to 0.75 cents per Ib., for col- umns weighing from 500 to 2500 lbs., in lots of at least six. Roof Trusses—In lots of at least six, the shop cost of ordinary riveted roof trusses in which the ends of the members are cut off at right angles is about as follows: Trusses weighing 1000 Ibs. each, 1.15 to 1.25 cents per lb.; trusses weighing 1500 Ibs. each, 0.90 to I.00 cents per lb.; trusses weighing 2500 lbs. each, 0.75 to 0.85 cents per ib.; and trusses weighing 3500 to 7500 Ibs. 0.60 to 0.75 cents per Ib. Pin connected trusses cost from 0.10 to 0.20 cents per Ib. more than - riveted trusses. 354 ESTIMATE OF WEIGHT AND Cost Eave Struts——Ordinary eave struts made of 4-angles laced, whose length does not exceed 20 to 30 feet, cost for shop work from 0.80 to 1.00 cents per Ib. Plate Girders——The shop work on plate girders for crane girders — and floors will cost from 0.60 to 1.25 cents per lb., depending upon the weight, details and number made at one time. | Eye-Bars.—The shop cost of eye-bars varies with the size and length of the bars and the number made alike. The following costs are a fair average: Average shop cost of bars 3 inches and less in width and 34 inches and less in thickness, is from 1.20 to 1.85 cents per lb., depending on length and size. A good order of bars running from 214” x 34” to 3” x 34”, and from 16 to 30 ft. long, with few variations in size, will cost about 1.20 cents per lb. Large bars in long lengths ordered in large quantities can be fabricated at from 0.55 to 0.75 cents per Ib. To get the total cost of eye-bars the cost of bar steel must be added to the shop cost. 7 Cost of Drafting.—The cost per ton for making details of mill buildings varies with the character of the work and the tonnage that is. to be fabricated from one detail, so that costs per ton may mean very little. The following will give an idea of the range of costs. Details for headworks for mines cost from $4.00 to $6.00 per ton; details for church and court house roofs having hips and valleys cost from $6.00 to $8.00 per ton; details for ordinary mill buildings cost from $2.00 to $4.00 per ton. ‘The details for all work fabricated by the Gillette-Herzog Mfg. Co., with the exception of plain beams and com- plicated tank work, were made in 1896 by contract, by Mr. H. A. Fitch, now structural engineer for the Minneapolis Steel and Machinery Co., Minneapolis, for $2.06 per ton. This price netted the contractor a fair profit. Actual Costs of Detailing—The details of the building for which the estimate is made in this chapter cost $3.60 per ton. The details for the Basin & Bay State Smelter, Basin, Montana, containing 270 tons of steel cost $2.00 per ton. =a ee oe ee CU a i a i en brah A = at rll a , ~ “a _ Pa 7 ee , i . ae —yTs = e oS so i SHop Costs 355 Actual Shop Costs.—The following actual shop costs will give an idea of the range of costs: The shop cost of the transformer building ° of which the estimate is made in this chapter was about $20.00 per ton including drafting; the Carbon Tipple building at Carbon, Montana, weighing 86 tons, cost $18.60 per ton. The shop cost of the structural work of the East Helena transformer building, the estimate of which is given in the next to the last column in Table X XVI, cost $21.80 per ton including details. The shop cost of the Basin & Bay State smelter, weighing 270 tons, was $17.20 per ton including details which cost $2.00 per ton. The shop cost of six gallows frames made by the Gillette-Her- zog Mfg. Co., varied from $21.80 to $41.80 per ton, with an average of $32.20 per ton including details. Cost of Erection.—With skilled labor at $3.50 and common labor at $2.00 per day of g hours, small buildings like those given in Table XXVI will cost about $10.00 per ton for the erection of the steel frame- work, if trusses are riveted and all other connections are bolted. The cost of laying corrugated steel is about $0.75 per square when laid on plank sheathing, $1.25 per square when laid directly on the purlins, and $2.00 per square when laid with anti-condensation roofing. The erection of corrugated steel siding costs from $0.75 to $1.00 per square. The cost of erecting heavy miachine shops, all material riveted and in- cluding the cost of painting but not the cost of the paint, is about $8.50 to $9.00 per ton. Small buildings in which all connections are bolted may be erected for from $5.00 to $6.00 per ton. The cost of erecting the East Helena transformer building (next to the last building in Table XXVI) was $12.80 per ton including the erection of the corrugated steel and transportation of the men. ‘The cost of erecting the Carbon Tipple was $8.80 per ton including corrugated steel. The cost of erection of the Basin & Bay State Smelter was $8.20 per ton including the hoppers and corrugated steel. The cost of erecting 6 gallows frames in Montana varied from $11.20 to $15.20 per ton, with an average of $13.00 per ton, all connections being riveted. COST OF MISCELLANEOUS MATERIAL.—In making an estimate for a mill building the engineer needs to be familiar with the 356 EstIMATE OF WEIGHT AND Cost costs of building hardware, lumber, etc. Prices of building hardware are usually quoted at a certain discount from standard lists. These standard lists and the discounts can be obtained from the dealers, or the method described in the following paragraph may be used. The following method of obtaining costs. of building hardware and other miscellaneous materials, has been found very satisfactory: Obtain standard lists from the dealers, or a very complete one entitled “The Iron Age Standard Hardware Lists” may be obtained from the David Williams Co., New York, for $1.00, postpaid. To find the cur- rent discount, consult the current number of the Iron Age, or a simi- lar publication—the Iron Age is published by the David Williams Co., New York, at $5.00 per year, or Io cents per single copy, and gives each week the current hardware prices and discounts. By applying the discount to the prices given in the standard lists the current price of the material can be obtained. The standard lists of machine bolts, carriage bolts, nails, and turnbuckles are given for convenience in es- timating and to illustrate what is meant by lists. To illustrate the method just described the current prices (1903) will be obtained for a few items: 3 2-in. Barbed Roofing Nails.—The base price of nails is $2.65 per keg of 100 lbs., and from standard nail list (Table XXVIII) we see that 2-in. barbed roofing nails take $0.35 per 100 Ibs. advance over the base price making the price $3.00 per 100 Ibs. Carriage Bolts —2%" x Y%" carriage bolts are listed (Table XXVIII) at $3.00 per 100, from which a discount of 60 and 10% is allowed. A discount of 60 and 10% is equivalent to a discount of 64%, making the price of the bolts $1.08 per 100. Machine Bolts —4" x Ys" machine bolts are listed (Table XXIX) at $4.90 per 100, from which a discount of 65 and 5% is allowed. A discount of 65 and 5% is equivalent to a discount of 6634%, making the price of the bolts $1.36 per 100. The weight of machine bolts and nuts is given, from which the price per Ib.-can be obtained for any size of bolt. ing preliminary estimates: le thal | 5D ee es a | le STANDARD HARDWARE LISTS The following approximate prices of materials will assist in mak- rivets, $2.25 to $3.00 per 100 lbs.; boat spikes, $2.25 to $3.00 per 100 lbs.; washers, cut, $6.00 per 100 lbs. ; washers, cast, $1.50 to $2.00 per 100 lbs.; sash weights, $1.00 to $1.50 per 100 lbs. ; pins $3.50 to $4.00 per 100 lbs.; pin nuts, $4.00 per 100 Ibs. ; wire poultry netting, $0.50-per square ; asbestos felt, 32 lbs. to the square, $2.75 to $3.00 per 100 lbs.; turnbuckles, 50% discount. Other prices and costs will be found in the descriptions of the various articles on the preceding pages. TABLE XXVIII. LIST ADOPTED DECEMBER 1, 1896. Common Carriage Bolts. 357 List Feb’y r, Price na STEEL WIRE NAILS. Length 9 16 inInches. 14 516 88 716 1-2 &58 84 20d to 60d........ Base. 14% $1.00 $1.20 $1.60 $2.20 .. og Deskes Cidnaions 10d to 16d........ $0.05 extra per keg. 1% 1.04 1.25 1.68 2.29 ea gene Fence ’ 8d and 9d ........ aS sae 2 1.08 1.30 1.76 2.38 ts paw and 6d and7d......... ae a . 24 1.12 135 1.84 247 sae Brads. 4d and 5d......... | fants e 358 1.16 1.40 1,92 256 $300 $5.20 $7.20 oh AR Perr RES. “ 934 120 145 200 265 311 5.87 743 BO sGie ssc e eek Tare) Le - 3 1,24 150 2.08 2.74 $22 6554 7.66 844 1 28 155 2.16 288 $388 5.71 7.89 Barbed Common and Barbed Car Nails, 15c. advance over B46 182 160 224 292 3844 5.88 8.12 “ ge eee ge ie ae ; : A R ; i PAD SUM Sova capsudadWoredictocectadlaves curdecs 3d 2a 414 1.44 1.75 2.48 3. 19 3.77 6.89 8.82 MEI DEON 5 so5s ss ccd antandaeccdecssonsaonces 20.50 100 413 1.48 1.80 2.56 38.28 3888 6.56 904 434 1.52 1.85 2.64 8387 38.99 6.73 9.27 Casing and Smooth Box Nails—(Extra per keg). by 4 byt : po My . 6.90 a : ; f ; ; 71.24 9. en ee ee 6 172 210 304 883 454 758 1042 $0.15 615 615 B85 BK «HOBO «01.00 6% 180 2.20 3.20 4.00 4.76 7.92 10.83 4 7 1.88 2380 336 4.18 498 826 1134 Finishing........ 20d 12 & 16d 10d 8& 9d 6&7 5d 4d 3d 2d 7% 196 240 8.52 436 520 8.60 11.80 8 2.04 250 3868 4.54 5.42 8.94 12.96 _ (Extra per keg).80.9 2 .2 .8 45 .65 .65 .85 1.15 a 2.12 2 00 84 472 5.04 28 12.73 : : ‘ j » 3 13 18 ote Ggaptange hed Ai, BOE She aia 9% 228 280 416 5.08 608 996 13,64 (Extra perkeg), $0.30 .40 .50 60 .70 .8 1.0 10 2.36 2.90 4.32 5.26 6.80 10.30. 1410 ll 2.52 3.10 464 5.62 6.74 1098 1502 Clinch .......00+ 20d 12&16d 10d 8&9d 6&7d 4&5d 3d Od - “P+ pe ‘? a e = 15.94 : * . - % 12.384 16.86 (Extra per keg).20 385 .35 .85 .45 55 65 .85°4.05 14 &.00 8.70 5.00 $70 8.06 18.08 17.78 : i v * ‘ s 13. 18.70 Pee Ronis, 8 WK 3 16 8.32 4.10 624 7.42 894 14.38 19.62 (Extra perkeg) 90.35 .45 .45 .55 .60 .65 .% 17 3.48 4.30 6.56 7.78 9.88 15.06 20.54 | fie te Se ie te oe oe Wire spikes, all sizess.scisssscesssessceuee svseese oe. s10C, Extra. : “ . 16.42 2238 ecpcauesie = 20 896 4.90 7:52 8.86 10.70 17.10 23,30 LENGTH AND APPROXIMATE GAUGE OF COMMON WIRE NAILS. 24 8d 4a 5d 6d 7a 8a 9a Nii: 3 1 1 2 2 2 2 ois if ie Oh om 6 Olek fue SPIKES, NAILS AND TACKS. 10a 12d 16d 20d 80d. 40d 50d 60d, STANDARD STEEL WIRE NAILS. . STEEL WIRE SPIKES.° 3 in, a4 ws . £s 5 gi $ sak Common. Finishing. om Sizes, Len iam., | No. per| Diam., | No. i ; oa ery cok hichee Peslg Length. ec pode, Turnbuckl af | [al 108 | 98) |S [asl urn u Ss. * e * 7 di | ab [Sat] |S OB | 4 | Be : : : ; : : 1 0 5 4h’ |. . a . ee adh £3 3 aie 2 | 83 3 “| .0808| 210 | .0641} 350 | 5” | .2576| 13 Or OA | WH | 94 7 i 0858} 160 | .0641| 315 | 54” | .2803] 11 8d | 24” | 0935] 115 | .0720| 214] 6” | .2893] 10 % | $0.40 || 13% | $1.38) 256 | $5.00 9d | 22” | .0963} 93.0720} 1 64" | .2249| 7 1s | -42|| 14] 1-50) 2% | 5-50 10d | 3” |.1082| 77 |.0808| 137| 7” | .2249| 7 an ip 1 ee BRO eee Bd 124 | 34” |:1144| 60 | :0808] 197 | 8” | ‘3648| 5 te | 48] 1% | 2.00] 3° | 650 16d | 3” | .1 48 90 | 9” "|.3648| 43 1 % - 1% = 3 75° 4” |.1620| 31}.1019] 62 . 2 a KR -75|| 2% | 3.10] 336] 9.00 30d | 43” | .1819| 22 I 88 || 2% | 3.50] 3% | 10.00 40d | 5! | -2043) 17 1% | 1.00|| 23% | 4.00] 33% | 15.00 50d a aaa _ 1% | 1.25|| 2%] 4.50] 4 20.00 : 258 ESTIMATE OF WEIGHT AND Cost Fes Re hee ae ae Se te WG en Pee TABLE XXIX. MANUFACTURERS’ STANDARD LIST OF MACHINE BOLTS WITH SQUARE HEADS AND AVERAGE WEIGHT OF SQUARE HEAD SQUARE NUTS. FINISHED POINTS, i ADOPTED SEPT. 20, 1899, TO TAKE EFFECT OCT. 1, 1899. n, DIAMETER. ince] + | fe] & | ve | & [tees] t | & | 2 Length. FF |T-70)2-00|2.40|2.80| 3.60] 5.20) 7.20|10.50|15.10 - |e l[ael ele] sw | «| « | wi] 4 2 |1.7812.12/2.56/3.00] 3.86] 5.58] 7.70/11.20|16.00 94 |1.96/2.24/2.72/3.20] 4.12] 5.96] 8.20/11.90/16.90 1% | 4.0] 6.8] 10.6) 15.0] 33.9) 40.5) 70.0 $ |1.94/2.36/2.88/3.40| 4.38] 6.34) 8.70|12.60]17.80 Ie) Oe Feel aca] sen eel weal on 8} |2.02/2.48/3.04/3.60] 4.64) 6.72] 9.20/13. 30/18. 70 2 -4| 12.6] 18:2] 27:7] 47.0] 79.3] 2. oe 4 |2.1012.60/3.20|3.80] 4.90] 7.10) 9.70|14.00/19.60 a 8/192] 29.0] 49:2 | 2-4 | 12005 200071, 44 2.1812. 72/3.36/4.00| 5.16] 7.48]10.20/14.70|20.50 3% 0 | 171 ace | cal Gos | eel eel ee 5 /2.2612.8418.52/4.20] 5.42] 7.86)10.70/15.40/21.40 3% | 6. 0 | 25:2 | 347] 87:9 | 95:0 | 137-4 | 1980 BE |2.34/2.96|3.68/4.40| 5.68] 8.24/11.20/16. 10/22. 30 4 : °4| %.2| 37.5] 62:8 | 101-2 | 145.8.| 207.0 6 |2.42/3.08)3.84/4.60] 5.94) 8.62/11.70)16.80)/23.20 % | & Ba Sal sol ool eee lies lome 6} |2.50/3.20/4.00/4.80] 6.20] 9.00)12.20|17.50)24.10 % | 9 21.4 | 31:2] 45.7| %.4 | 120.0 | 176.1 | 240.0 7 |2.58)3.32/4.16)5.00] 6.46] 9.88]12.70|18.20/25.00 10. 22:8 | 83.1] 48.4] 79.8 | 126.2 | 184.6 | 251.0 74 |2.86)3.44/4.32/5.20] 6.72) 9.76]13.20/18.90/25.90 4 |i il Bol eee | ict aerate 8 |9.74/3.56/4.48|5.40| 6.98/10. 14]13.70|19.60/26.80 y | 12. o7 | 39.1 | 96:2] 92:9 | taec0 | 200.9 | Bono 9 |2.90)3.80/4.80|5.80] 7.50|10.90/14. 70/21 .00/28.60 48:1 | 20.8 | 29:5 | 41.0 | 59.4] 97.2 | 151-2 | 218.3 | 295.0 10 5.06 ee cat ae 8.02}11. 66/15. 70)22.40|30.40 GE KAP YAM Rede ee eat} 11 |8.22/4.28/5.44/6.60| 8.54/12.42/16.70|23.80|32.20 40.4 |.53:0 | 75-8] 128.5 | 198-7 | 273.9 | a00.0 12 |8.38/4.52|5.76/7.00) 9.06/13. 18)17. 70/25. 20/34.00 44.0 | 57.0] 81.8] 182.2 | 201.0 | 290.0 | saz. 18 |....|....|6.08]7.40] 9.58/13. 94]18. 70/26. 60/35.80 - 86.9] 190.7 | 213-4 | 00-7 | 44-0 44}. .J6240l7!80|10:10!14:70|19.70/28.00/37.60 7-7) 157.6 | 2583 | Sat; io ib [cc fccfeus ol aah ar fe: ia || eed) ie a7 LA. 11 266|16:98129: 70/32 20/48-00 119:5| 101:8 | 388.1 | 408°9 | $98 18 |: *2I/IN]g'18|17!74]28/70/88;60/44:80 125.0] 200.0 | 900.6 | 420:8 | S580 19 |. ++ +|+++.|12.70|18.50/24. 70/85.00]46,.60 = 18.22}19. 26 7 70 186 -£0148.40 was | 1.4] 2.2] 8.6] 4.0] 5.5] 8.6 | 12.4 | 16.9] 92.0 BO ins] actos sclanaateviales fox 27.70/39.20/52.00 ? sf adel aed Hau Rid Geert d Ube reed pipes te APPROXIMATE WEIGHT OF NUTS AND BOLT. 25 |: viese [sss es[80.70/48.40157.40 HEADS, IN POUNDS. 27 “IEEE geitolde:solet.o0. Dlam-of Botetn inches, |< | we | 3 | de | | | 29 | STEELS ee roldo ooled:e0 — anlPaeag'e™s#2" 82%} or | ose | .oo7 | 100 | 128 aor | a8 THR RAS oR, RNS RE SR eet 85.10:50.40168.40 Weight of Square Nut!) gi | .o19 | .069 | 120 | .164 |.820 | .58 The ro owing extras ‘are to be understood as a part Of the §< °. tae er ee Beets oF exec Has ty ls Diam of Bolt in inches. | % : 14 | 1% | 1%) 3 1H ex: Joint Bolte with Oblong Ruta, Belts with ‘Tee Heads, Adkew Weight of Hexagon Nut}) 731 1.10 | 2.14 | 8.78 | 56 | 8.75] 17.0 Heads, and nga Heads, 10 per cent extra. Wek of Bauare. Nat Special Bolts with irregular Threads and unusual dfmensions eight, of Square Nut) 6 | 1.81 | 2.56 | 4.42| 7.0 |10.5 | 21 of Heads or Nuts will be charged extra at the discretion ~and Head -....----.-- manufacturer. ne PART. FY, MISCELLANEOUS STRUCTURES. The descriptions of the different structures given in this part have been abstracted from descriptions published in the Engineering News, Engineering Record, etc. Figs. 171 and 172 are from the Railroad Gazette; and Fig. 169 and Figs. 172 to 181, inclusive, are reproduced from originals kindly loaned by the Engineering News. STEEL DomMeE For THE West Bapen, Inp., Hore..* The dome of the new hotel at West Baden, Ind., is remarkable for its size. ‘This dome has a steel framework and is larger than any other ever built, its span exceeding by about 15 feet that of the Horticultural Building of the Chicago Exposition of 1893. : The dome is about 200 feet in outside diameter and rises about 50 feet above the bed plates. Its frame consists of 24 steel ribs, all con- nected at the center or crown to a circular plate drum, and tied together at the bottom by a circular plate girder tie. Each rib foots at its out- side end on a built up steel shoe, resting on a masonry pier. The rib is connected to the shoe by a steel pin, and the outside plate girder toe is attached to the gusset plate at this point, just above the shoe. The shoes of all the girders are constructed as expansion bearings, being provided with rollers in the usual manner. The dome is therefore virtually an aggregation of two-hinged arches with the drum at the center forming their common connection. Their thrust at the foot goes into the circular tie-girder, and only ver- tical loads (and wind loads) come upon the shoes and the bearing piers. At the same time any temperature stresses are avoided, since the ex- pansion rollers under the shoes permit a uniform outward motion of the lower ends of all the ribs. The outline of the dome and part of the dome framing are shown _ *Engineering News, Sept, 4, 1902. MISCELLANEOUS STRUCTURES 360 "THLOH “GNI ‘NAGVd USHM YO AWOd TWALG “OO “O1T > peBaojug *‘Burroeg pun sujop jaa} 4d [1018q UO}4I9EG OIL440/ "UOWOABT pO!L4OY *ssnay 40 WOE ‘SauiDL4 SSO1) ae fee ce or s 1 “UDI 4404 Fx 02 An’ BIOL "B42/f 14 ' Ky ‘oy 4 ~ , i ches I~ 7! Se + NS \ Om hes K 2 2ll-= rhe, BOI 5 901-7 ey Z iW, S41sl0g RY Ma y Z oe a 2 \ \ \ aN Ps ‘| H \ + 1 1 bs Se ee a a ee bel . i . eT a STEEL Dome For West BapeNn Horer 361 in Fig. 169. The rise is between 4 and % of the span. The outline of the top chord approximates an elliptical curve, and the bottom chord is parallel to the top chord throughout its length, except in the three end panels on either side; the depth of the arch being Io ft. back to back of chord angles. The web members are arranged as a single system of the Pratt type, with substruts to the top chord as purlin supports. In the end sections the arrangement is necessarily modified, the sharper curvature of the chords being allowed for by more frequent strutting. The maximum stresses in the different members of the arch are given on the right half of the rib in Fig. 169. They are obtained by properly combining the dead load stresses with the stresses due to wind blowing successively in opposiie directions in the plane of the rib in question. The loads used in the calculation were a dead load separate- ly estimated for each panel point, a variable snow load, heaviest at the center of the roof, a wind ioad of 30 lbs. per sq. ft. on a normal surface reduced for inclination of the roof to the vertical. The makeup of the members is given on the left half of the rib in Fig. 169. In the plan part of the dome the method of bracing the ribs is fully shown. Suc- cessive pairs of ribs are connected by bays of bracing in both upper and lower chords. In the upper chord the I beam purlins are made use of as struts, angle struts being used in the lower chord. The bracing consists throughout of crossed adjustable rods. At the center, these rods are carried over to a tangential attachment to the central drum, so as to give more rigidity against-twisting at the center. The central drum, 16 ft. in diameter by 10 ft. deep, has a web of 3% inch plate, with stiffener angles to which the ribs are attached. At top and bottom the drum carries a flange plate 24 ins. x 3-16 ins. for lateral stiffeners. In addition it is cross braced internally by four diametrical frames intersecting at the center. The outer tie-girder, which takes the thrust of the arch ribs, is a simple channel-shaped plate girder, 24 ins. deep, as shown on the plan. The weight of the dome complete, including framework and covering was 475,000 lbs. This makes the dead load about 15 lbs. per square foot of horizontal projec- tion of roof surface. Mr. Harrison Albright, of Charleston, W. Va., was the architect of the building and the design of the steel dome was worked out by Mr. Oliver J. Westcott, while in charge of the estimating department of the Illinois Steel Company. ‘The structural steel was furnished by the Ulinois Steel Co. Tue St. Louts CoLiseumM.* The St. Louis Coliseum Building is a rectangular brick building 186’ 2” x 322’ 3”. The steel framework is made entirely independent of the masonry walls and consists of three-hinged arches properly braced. The Coliseum has an area of 222 x 112 feet clear of the curb wall. Ordinarily there are seats for 7,000 persons on the main floor and the galleries, but for convention purposes with seats in the arena the num- ber can be increased to 12,000 persons. The steel framework consists of a central arched section adjoined at each end by a half dome formed by six radial arched trusses. The main arches forming the central section have a span 178’ 6” c. to c. of shoe pins, are spaced 36’ 8” apart, and are connected by lateral bracing in pairs. The pins at the foot of the arches are 4 7-16” diameter, and at the crown the pin is 2 5-16” diameter. The rise of the arches is 80’ 0”, the lower chord points being in the curve of a true ellipse. | ‘The end radial trusses correspond essentially to the semi-trusses of the main arches except for their top connection, where their top chords are attached to a semicircular frame supported by the end main trusses and designed to receive thrust, but no vertical reaction, as shown in Fig. 170. The roof covering of asphalt composition is laid on 13£-inch boards, resting on 2%4 x 16-inch wood joists, 3 feet apart and ceiled underneath. These, in turn, are carried by the steel purlins of the structure, which are spaced about 16 feet apart. The gallery floor beams are carried on stringers of 8-inch channels spaced 3’ 8” center to center, carried by girders running between, and supported by the arches. The rear string-_ er is a plate girder; the front one is a latticed girder, the gallery beams running through the latter and cantilevering out 5’ 4”. The main floor beams, supporting the lower tier of seats, consist of g-inch I beams, ‘spaced 3’ 8” center to center, which are similarly carried on girders, and their lower ends rest on a brick wall. *Engineering News, Aug. 10, 1899, and Engineering Record, 1899. Tue St. Louis CoLIssEUM 363 ———. "1 | | 2 - fj IAT “WW ~ ‘ Purlins 4 Froof ' fy, & Framing 47 - ~-f---e eed he & +298°2"-+ < - -K-+ -- eum Won, aa a 1 uN {! ’ \\ : vO + et WY 1 Lat: |.N ra?) 1 , K- = <= —- VIO-G= ode conn > | +: Nes “learra \ Floor Fraraing \ iy ) ! Gallery eseLe\S 7 : . x3*6 H Framing, EL a ; Wars ec apgriet ite tir a tee Selva SOS Se a ae a a Sectional Plan Figure |. General Plan of Trusses& Framing —— = cs _ 2FY 2LI5xS el3 z OCG, SAMOA WA f Q DX Bracing between Plan of Half Ring Connecting Main Arches Radial Trusses to Main Trusses . Main Arches , Figure 2 - Cross Section Showing Construction of Main Arches Fic. 170. The loads, in accordance with which the trusses were figured, are : 3 as follows: i Case I. | Wooden deck and gravel of roof........17.5 lbs. per sq. ft., vertically eh Re ee te SNe a rou GEV rk “ Sau dd tWid Sues cues eek es eo 85000 8 8 SS " ; “ce “é ‘ce “é “ec OLS Cen RRS ee Sey cart oy WO! ER te th 364 MISCELLANEOUS STRUCTURES Add for floors, viz.: Main floors, banks and galleries........ 105.0 30 ees Attic foots. vicky ea gece eee 66,0750) See Case IT. Wooden deck and gravel of roof....... /.17.5 Ibs. per sq. ft., vertically SCE) i Lents 8b Kees oe OAT eens TPM eee eae ef STOW A :c'n <4 vie bo pre ois Gers \e alata ae LOO ber are é Total i cassie ge cakes Coe eee dae ES te es Wind pressure over entire elevation of wall and roof Of 8... (Wo Saeesaeees 30.0 lbs. per sq. ft., horizontally * LOADS ON PuURLINS. Wooden deck and gravel of roof......... 17.5 lbs. per sq. ft., vertically Steel os sak cows cow ee ee eens sae eG Ce oe aaaee ‘ Snow and Wwitid ....<.asesneeenes sees ngs iT ee 2 Total: ca cicineecaslcepageen eae eee AG aes a Loaps ON Froor Beams, GIRDERS AND CoLUMNS OF MAIN FLoors. Banks and galleries; -beams-v.5.c0s vce eek e ake aes 140 lbs. per sq. ft. Banks and galleries, girders .....45...ss5<0000ses 112.“ * 4o ee Banks and galleries, columns ........2s+e0sesse0. 105 Attic floors beams, columns and girders............ 60°" "> eae For the main trusses, in addition to the stresses of Case II., there was added the stress due to the wind bracing between these trusses. For the radial trusses, in addition to loading of Case II., there was assumed an additional load of 50,000 pounds supposed to act up or down at the upper point of truss; this load being what was assumed probable in case there was slight unequal settlement of the footings. _ For the half ring connecting the tops of the radial trusses there was another case assumed, beside Cases I. and IJ.—viz., a thrust of 50,000 pounds at any point of the half ring; this being the thrust of a radial truss under its full live and wind load. All the material used was of medium steel, excepting the rivets, which were made of soft steel. Both material and workmanship con- form to manufacturer’s standard specifications, ‘ce ‘ce ce “ct 4 = ? 3 ; Et Lee ea ee ee PP edly Att ain ini ea, te ‘ - . aye , eee = ae ere a Pe eee : ne eo pa Ee | ee a ee NS ee Te eR a nae na ne —- . -, + oh ae. k es Oc bd Pam eee NS THE St. Louts ConissuM 365 Unit STRAINS. (ETS ee gu Fe reds 2 a a elt a 16,000 Ibs. per sq. in. Compression, for lengths of go radii or under....12,000 “ “ “ « Compression, for lengths of over go radii........ 17,100—57)—r Combined stress due to tension or compression und travisverse logdimg 4 ok ck a ee 16,000. cae Beet OI) WEN PICS So Siscasecy cw csess cee RES it <8 ee 6 Ee Oh 0 | See EAS A ol rr ¥T.G00 {0S Severe RN OM TIVON IO ease rs diese ieucs FO; ere TERE MPO OING See Cah ien es th eee cece wise s 22 O00 o> eae eR IR TEPC ee Sangeet wes oes BOIS: SEU Fig ene Bending, extreme fibre of pitis.............5.. BEANS | POIs cae Bending, extreme fibre of beams Lateral connections have 25 per cent greater unit strains than the above. In Case II of trusses, the above unit strains were increased one- third. The main and radial arch trusses are built as shown in Fig. 170, except that above the haunches the ribs of the radial arches are T-shaped instead of I-shaped, i. e., they have no inside flanges. The purlins are triangular trusses 414 feet deep, made of angles. The brac- ing between main arch trusses terminates at the bottom with heavy portal struts of triangular box section. The lateral rods are not car- ried to the ground on account of the obstruction they would make. The radial trusses are coupled together in pairs with lateral rods down to the ceiling line. The thrust due to wind is transmitted from them into the line of girders around the structure at this point, and into the ad- joining floor systems. The compression ribs of the main and radial arches are stayed laterally by angle iron ties, connecting to the first panel-point in the bottom chord of the purlins. In the planes of the first diagonal braces of the trusses above the haunches, diagonal rods | connect the bottom ribs of the trusses to the upper ribs of the next trusses. No struts were used between the bottom chords, as they would have been directly in the line of vision from the rear gallery seats to the farther end of the arena. The front and rear girders supporting the gallery and main floor beams are tied together with a triangular system of angle iron bracing. To provide for expansion, the radial purlins and all the girders be- tween the arches have slotted hole connections in every alternate bay. The diagonal rods between the two lines of ridge purlins were tightly 366 MISCELLANEOUS STRUCTURES adjusted on a hot day. To prevent secondary strains in the half ring al to which the radial trusses are connected at their tops, there is 1-16-inch clearance in all the pin holes. There is also clearance between the pin plates, so that the trusses and the ring can slide a little sideways on their pins. The lines of the arch trusses were laid out full size and the principal points checked by independent measurements in the template shop, and the work was accurately assembled. In order to avoid the handling of large, heavy pieces before the drill press, the foot of the arch, through which the pin hole was bored, was made separately and afterward riveted on. ‘The total weight of the iron in the entire structure was 1,905,000 pounds, as follows: Main arches, 64,000 pounds, each; radial arches 21,000 pounds, each; purlins between main trusses 1,450 pounds, each; main floor stringers 810 pounds, each; balcony floor stringers, 280 pounds, each; cast shoes 3,000 pounds, each. There were 4,188 days labor spent on the work in the shop and 3,550 days labor during erec- tion, the average number of men in the erecting force being about 50. The stress diagrams and detail plans of the steel frame were made un- der the supervision of Mr. Stern, in the office of the Koken Iron Works, who were contractors for the ironwork, and were submitted for approv- ~_ al to the consulting engineer, Mr. Julius Baier, Assoc. M. Am. Soc. C. E. Mr. C. K. Ramsey was the architect of the Coliseum, and Mr. L. H. Sullivan was the consulting architect. Mr. A. H. Zeller was consulting engineer for the Board of Public Improvements; Mr. J. D. McKee, C. E., was shop inspector, and the Hill-O’Meara Construction Company was the general contractor. a a ee oe em ee ie ee ‘ ‘ Bi Ae tet | go BP iS ee ete ee eae Bs THE LocoMoTivE SHoPs oF THE ATCHISON, TOPEKA AND SANTA FE R. R., Topeka, Kas.* This building is intended for all the locomotive work, including boilers and tenders. It is of particular note for its great size and the peculiar features of its design. In general plan it is 852 ft. long and 153 ft. 10 ins. wide, the width being divided into a center span of 74 ft. 3 ins. and two side spans of 39 ft. 9 ins. ‘It is of self-supporting steel frame construction, with concrete foundations and floor, 13-in. brick walls, and Ludowici tile roof. There is no sheathing under the tiles, which thus constitute the sole covering. The tiles are laid on 2 x 2-in. timber strips to which every fourth tile is fastened by copper wire. The most striking feature of the design is that the saw tooth or weaving shed type of roof is adopted for the side spans, the glazed vertical sides of the ridges facing northward. ‘This feature was intro- duced with the view of making the shop as light as possible. The ar- rangement could not well be used where heavy snows are frequently experienced, as the snow would pack between the ridges, but there are comparatively few heavy snow storms in the vicinity of Topeka. In addition to this arrangement, the greater proportion of the area of the side walls is composed of windows, while the exposed parts of the sides of the central span (between the ridges of the side spans) are also glazed. ‘There are also several windows in the end walls. The roof of the central span has on each side of the ridge a skylight 12 ft. wide, extending the full length of the building. These skylights are fitted with translucent fabric instead of glass. By these various means an exceptionally good lighting effect and diffusion of light are obtained and the shop is in fact remarkably light even on a gloomy day. There is no monitor roof, but ventilation is provided for by Star ventilators 25 ft. apart along the ridge of the main roof. The columns are built up of pairs of 15-in. channels, and independ- ent columns of similar construction carry the double-web box girder runways for the electric traveling cranes which run the entire length of the central span. Fig. 173 shows the elevations, sections and plans *Engineering News, Jan. 3, 1903: and Railway Gazette, Nov. 7, 1902. o>) ON oa) MISCELLANEOUS STRUCTURES Fic. 172. CROSS-SECTION LOCOMOTIVE SHOP. of the steel structural framework, and Fig. 174 is a partial elevation on the east side. Fig. 175 shows the design of the central roof trusses and the lattice girders which form longitudinal bracing between the trusses. This longitudinal bracing is not continuous but is fitted only between ———— — i Se a Locomotive Suors, A. T.& S. F. R.R. 369 alternate pairs of trusses. End trusses are built into the walls, as these walls are pierced by numerous windows and doors and are not relied upon in any way to support the roof. Portal bracing is fitted between the side or wall columns at intervals. No metal less than 3g-in. thick is used in the structural work. LUT Cross Section of Machine Shop. North Lit 5 _ Floor Line|f Ke -5B9 "ne ~~ ~= 741 B --- e- 3B'9 "> Cross Section of Boiler Shop. Fic. 173. PART ELEVATIONS AND PLANS OF STEEL STRUCTURAL WORK OF NEW LOCOMOTIVE SHOPS. Elevation O-t:. The roof trusses are proportioned for a load of 15 Ibs. per sq. ft. for the weight of the roofing, 10 lbs. per sq. ft. for snow, and 25 lbs. per sq. ft. for wind pressure, or 50 lbs. per sq. ft. in all.) The members Bsa > 2 ft se-—Ventil 9 ee SOVSUCSULSUOUOOOMLUUOUOGEUOUGOVOSUUOUUOGOOEOULAWOOUOgnTUOOOASROUUOSUUGGUASREOLUAOORGOLUONOLEOUOUUTNGAVCAUHUTTNEVUETTTTUTTTAA ke aa ‘Riveting Tower PT aii > VYwwYy Ye rw ren 1 ‘mm LO. mmm! mmm: il ute om: Dom mm Mem: WE ES Se ee ee ee ee a ee 1 MP . TMC RPA nT, nF LOM {Hmm MULTI FS ZLENG.NEAS-Z oI Fic. 174. HALF EAST ELEVATION OF NEW LOCOMOTIVE SHOP, (SHOWING RIVETING TOWER AND WEAVING SHED ROOF.) 370 MISCELLANEOUS STRUCTURES. were calculated on a basis of 16,000 lbs. per sq. in. for tension and 14,000 lbs. per sq. in. for compression. Provision for expansion and contraction is made at intervals of 100 ft. The structural work for this shop was built at the Toledo Works of the American Bridge Co. The steel is painted a light grey, and the brick is whitewashed, a pneumatic machine being used for the latter work. B00

: ‘ ; LAN Ss A xa a z ' 5 Sys y y F S Ry Py 3 | ei: Hae He a de ee Se : JHE } i sth iS e 4’ i eke ' =i Y= + A zoo’ 200 o tly oe e a ° Pe r) a a o & o ¥ Retina ee eee aie Ss Mt Urnteen Shima DG. Wee a vie) em ont weet pS, ae th ean nn ean et & =x Latah ts Lantern Stops here-—~~-> s' 8 ik 200 ik 200">i K- 200-2009: Lantern Ss i Pee Kya ates a ° a oe a @ a Kp, o ¥ ® against rst | Li, £ & S Gatie i F * ni 3 in” ue is ik | ] ; 4 ry ef “| S S b> 9 en . a wad « SATS Fre Ma. TP Wo WY a S roa ® tll WE a imicine te. } : rr H ; His PTs be We BL OSE ED sh ee Ei mT 5 > oy }, S > 5 o + c op o a | y va ; y te TI ) 1 Mee hh AOE 2s ties -749'10" ee. Se Tat im i ao Fan ° 10 acelin BSinmntosne haasavonr'y as Sebel eareiee ss Duct Drain 9 (SA House : | Fic. 176. PArtT PLAN, SHOWING PIT AND COLUMN ARRANGEMENT, eh a *Engineering News, May 24, 1900 - 2 Ne AST ) ee OE ey ee a a ee ee a 5) ia LOCOMOTIVE ERECTING SHOP, PHILADELPHIA & READING R. R. 373 tween each pair of columns, 35 pits on each side spaced 20 feet apart c. to c. Over each side bay there are two traveling crane tracks, one above the other. The top track will be for a crane of 120 tons capacity, capable of lifting a locomotive and carrying it the entire length of the building at a height great enough to clear the locomotives standing on the repair pits beneath. On the lower track there will be placed a number of 35-ton cranes for handling parts of locomotives and material. The middle bay composes the machine shop, and it is also covered by a track for a number of traveling bridge cranes of 10 tons capacity each. A material track runs lengthwise of the middle bay at the center, and for bringing the locomotives into the shop there is a track extending transversely through the shop at each end. These transverse tracks connect by means of turntables with a yard track running parallel with the east side of the shop. 3 Taking up the construction in detail, the steelwork is the first part perhaps to demand attention. As already stated, the nature and gen- eral arrangement of the steel frame is shown by Figs. 176 and 177. It may be divided into two parts for description. The first part comiprises Slat Slag Sla Gass Sla l20-+ Grane Slag « !20-t. Grane 55-t Crane 35-t Grane a! 10-t. Grane a is rm S$ Erecting Shop ty Machine Shop Erecting Shop Fan » i Chamber y a sobs coe a ree eb Z Z Fic. 177. ‘TRANSVERSE SECTION OF FRAMEWORK. the main columns and the traveling crane track system which they sup- port directly, and the second part comprises the roof framing. Con- sidering the construction of the main columns first, it will be seen that it had to be devised to withstand an unusually severe combination of loads, including not only the dead and wind loads usual to all building frames, but also a very heavy concentrated moving load from the trav- eling cranes. Careful attention was also required to provide means for longitudinal expansion of the columns and crane track girder sys- tem and still preserve the utmost stiffness possible in the heavily loaded crane tracks. 374 _ MIscELLANEOUS STRUCTURES The interior and wall columns were substantially similar in con- struction, both being composed of two channel sections made up of _ plates and angles, held together by a double lacing of 2% x Y%-in. bars. Fig. 180 shows the column construction in detail. Each column is founded on a concrete pier with a granite capstone. The concrete in these piers and for all other foundation work was composed of I part Portland cement, 2 parts clean sharp sand and 5 parts stone broken to pass a 2-in. ring. In making the concrete the sand and cement will first be mixed dry and then wet, and the wet mortar will then be mixed with the broken stone as it comes moist from washing. The shape of the column piers is indicated in Fig. 177, and the columns are fastened to them by four anchor bolts for each column. The columns are fin- ished by milling machine at both top and bottom, and are capped and stiffened at the tops to provide a bearing for the 4-ft. 13¢-in. plate girders carrying the 120-ton crane tracks. The track construction for the bridge cranes is shown in detail in Figs. 177 and 180. As already stated, the tracks for the 120-ton cranes are carried on plate girders resting directly on the tops of the columns, and running lengthwise of the building. ‘These track girders are 4 ft. 134 ins. deep back to back of flange angles and have spans of 20 ft., c. toc. of columns. They are milled square at the ends and the rivets of the end angles have flat heads. To allow for expansion the ends are not butted -close together, but are separated by a clear space of ™%-in.,. by the construction shown in Fig. 180. From the drawing it will be seen that each girder span has one fixed and one expansion end, the expansion being provided for by the slotted rivet holes in the bearing plate and by the space between the ends of the girders. An exactly simi- lar expansion end construction is provided for the girders carrying the tracks for the 35-ton and 10-ton cranes. These three tracks are on the same level and the girders supporting them are carried by brackets on the main columns. This bracket and girder construction is shown in Fig. 177. The crane tracks proper consist of ordinary railway rails laid di- rectly on the special cover plate forming part of the top flange of each track girder, to which they are attached by stamped steel clips riveted to the cover plate. For the 120-ton cranes the rails weigh 150 lbs. per yard, for the 35-ton cranes they weigh 85 Ibs. per yard, and for the Io- ton cranes they weigh 70 lbs. per yard. The joints are located over the expansion joints at each column and are spaced %-in. open, the angle LocoMOTIVE ERECTING SHOP, PHILADELPHIA & READING R. R. 375 © splices for the rails being arranged to permit expansion and contrac- tion. The rails were rolled to the recommended Am. So. C. E. sections for their respective weights. The specifications require that the crane track rails must be in perfect alinement horizontally and vertically and that the gage must not vary more than %-in. at the maximum. ‘To ebtain vertical alinement it is specified that not more than %-in. thick- ness of shims, each the full size of the bearings, shall be placed between the column top and the girder. The idea, it will be seen, has been to provide for the heavy rolling loads by substantial construction and ac- curate workmanship, and to keep the track smooth and rigid by dis- tributing the expansion over a number of joints instead of having all the allowance made for it at one or two points. Turning now to the roof framing, it will be seen from Fig. 177 that there is a separate roof over each longitudinal bay of the building. The roof trusses for the middle bay rest directly on the tops of the in- termediate rows of columns, but those for the two side bays are car- . ried by special roof columns rising from the main columns, as shown by Fig. 180. The trusses in each bay are connected by purlins and lateral bracing and carry lantern roofs with glazing. Details of the roof trusses, bracing and lantern roof construction are shown in Fig. 178 so fully as to make any further description unnecessary. ‘The roof covering and the glazing in the lanterns, however, deserve brief special notice. The roof covering consists first of a 1 x 8-in. hemlock sheathing, having its upper surface planed. This sheathing is to be covered by four thicknesses of roofing felt spread with granulated slag. The con- _ struction is specified to be as follows: Make the outside course next the edge of five thicknesses of felt, then lay each succeeding layer at least three-fourths of its width over the preceding layer, firmly securing it in place, and thoroughly mop the surface underneath each succeeding layer as far back as the edge of the next lap with a thin coating of roof- ing cement. ‘This cement is in no case to be applied hot enough to in- jure the wooly fibre of the felt. At least 70 Ibs. of felt must be used per 100 sq. ft. of roof. Over the entire surface of the felt laid as de- scribed there is to be spread'a good coating of cement, not less than Io gallons (including what is used between the layers of felt) of cement being employed per 100 sq. ft. of roof. ‘This cement coating is to be covered with a coating of slag, granulated and bolted for the purpose, using no slag larger than will pass through a 5£-in. mesh and none smaller than will be caught by a %4-in. mesh. This slag must be free Re a he eee ee 376 MISCELLANEOUS STRUCTURES from sand, dust and dirt and must be applied perfectly dry while the cement is hot. ? The structural features of the glazing in the windows and lights of the monitor roofs are fully shown by the drawing of Figs. 177 and 178. It will be noted that the sash in the side walls of the monitor roofs are hung on pivots at the middle of each side. There is a similar ar- rangement of sash in the upper side wall windows of the side roofs overlooking the low middle roof, except that the sash in alternate bays only are pivoted. These latter sash are to be provided with “Brand’s” sash openers, with chains operated from the crane runways and ar- ranged to open all the sash of one bay at one time. The feature most worthy of note, however, is that the sash in the side walls of the moni- tor roofs are to be operated by compressed air power from a central point on one of the side walls of the building near the floor. ‘The sash will be operated in sections about 75 feet long and the piping and valves will be so arranged that any one or all of the sections can be operated at one time. Next to the framework of the building, the structural features of most interest are perhaps the pit and floor construction. This is shown in detail by Fig. 180. In constructing the floor the ground inside — the building will be carefully leveled and well tamped. and puddled if necessary. The entire area of the main building, except under the pits, where cement concrete will be used, and under the railway tracks, where stone ballast will be used, will then be covered with a layer of bituminous concrete. This concrete will be composed of well screened cinders and No. 4 coke oven composition mixed in the proportion of at least one gallon of composition for each cubic foot of cinders. ‘This composi- tion is to be laid hot and well rammed. Yellow’ pine floor sleepers, 6 x 6 ins. square, will be bedded in the bituminous concrete at intervals varying from 4 ft. to 5 ft. transversely of the building. To these string- ers there will be spiked an under floor, or lining, of hemlock plank planed on both edges and on the upper side. On top of this lining there will be laid a flooring of 11% x 4-in. maple boards of a uniform and regular width, planed on top and both edges, worked on the back to a depth of 1-16-in. and a width of 234 ins., laid across hemlock floor, bored on a slant for nailing and face nailed with 12d. nails, the two lines of nails in each board being staggered. The nails will be placed in lines on each side of the board and not over 16 ins. apart, and the nails in one line will be opposite the middle point between the nails in the opposite line. eS — = ell CU —~s - Locomotive Erectinc SHop, PHILADELPHIA & Reapvinc R. R. 377 The sidewalls and bottoms of the pits are made of cement concrete of the composition already described as being employed in the founda- tions for the main columns, Fig. 180, and have a facing of I in. of cem- ent. It will be noticed that the bottoms of the pits are crowned trans- versely and have longitudinal side drains of “catch basin” type, run- ning into gutters which connect with the tile drain pipe system of the building. The pit rails are carried on 10 x 12-in. yellow pine side tim- bers resting on the concrete pit walls and connected by anchor rods extending into the bituminous concrete floor foundation. ‘The railway tracks in the building consist of 80-lb. rails spiked to cross-ties laid in stone ballast. Before taking up the exterior construction of the building, a brief reference may be made to the method adopted for heating it. By re- ferring to Figs. 176 and 177 it will be noticed that an underground duct extends along each side of the buildings. ‘This duct will contain all the steam, air and other piping éntering the building and will also serve as a hot-air duct for heating the building. For the latter purpose the duct connects with fan houses placed at intervals along the side of the build- ing, there being four fan houses on the west side and three on the east side. It is expected that the exhaust steam will be ample to heat the building. The general principle of the operation of the heating appara- tus is to use the air over and over again, depending upon the natural ven- tilation to keep it fresh. Structually the ducts are simple, their bottoms and side walls. being of concréte with a cement plaster, and the roof be- ing expanded metal and concrete supported by transverse roof beams. The roofs of the fan houses will consist of I beams carrying T-iron purlins holding book tile, which in turn carry a felt and slag roofing similar to that already described for the main building. From the structural point of view, the wall construction of the main building presents nothing that is particularly notable. These walls are of brick masonry resting on a concrete footing and are anchored to the wall columns, as shown by Fig. 178, but they are independent struc- tures, receiving no support from these columns. The entire wall area is pretty well taken up with doors and windows, there being one in each panel or between each pair of wall columns. This introduces a con- siderable amount of arch work, but nothing that is of unusual character. To illustrate the simple methods which have been adopted to secure a pleasing exterior appearance for the building, the typical details of the wall construction are shown in Fig. 170. . 378 MISCELLANEOUS STRUCTURES. ‘uD[g “[2uU01492g *dous eulLDW, 4° $11D49q snap jooy 9492 10 SF. ey aX X Bu1y2223 na aque) vag 109 / fa ——— (22 abel ert era i k fi Hehe rds : fa EM Gy 303 : ei UU) 7 Id” BBR S x 4 ep itt he Oooe oo a. (RT, 205 lbs. eee ae SemNet aa Ca eB x a Sectional Plan _A-B. nm rere: fee : : So ag ees , f ; ae - Rods. i | [ ‘ % it Fae elk 2 on ee Kesh bee Se ae + {Son TE sects SAE amar aE sea a L TTT as Fic. 181. HALF TRANSVERSE SECTION SHOWING STEEL FRAME DETAILS, 2 STEAM ENGINEERING BUILDINGS, BROOKLYN NAVY YARD. a ‘a \ ee STEAM ENGINEERING BuripIncs, Brooktyn Navy YARD 383 on in panels or sections ; each is capable of being swung out to provide ventilation. The side walls of the clerestory consist of corrugated iron covering for the lower part, and glazing above. ‘The bulk of the area of the shed roofs is skylight, and wide skylights are also placed in the clerestory roof. Fully 60 per cent of the area of the external walls and roof is glazed. The large proportion of glazing to the total wall and roof area makes the interior of the shops extremely well-lighted. The entire glazing is on the Paradigm skylight and side light system and was carried out by Arthur FE. Rendle, of New York City, who controls the system. The entire construction is of incombustible materials. There is no wood used in the building, except for the body of the doors, and here it is covered with tin plate. In this respect, and in respect to the amount of glazing used, the buildings are somewhat remarkable, even among modern shop buildings. Turning now to some of the special details, it will be noticed from Fig. 181 that a somewhat unusual construction has been adopted for the main columns. Each column consists of two main members, each com- posed of two channels riveted back to back, the two main members be- ing connected by double latticing. The resulting section is rather re- markable for its length, as compared with its width, but its purpose was evidently to give ample flexural strength to withstand eccentric load of the bridge crane girders:._Details of the crane girders and crane tracks are shown by Fig. 181; there being one 40-ton crane and two ro-ton cranes. It will also be observed from Fig. 181 that the lower portion of the clerestory roof has a concrete and expanded metal cov- ering with roofing slate nailed direct to the concrete. The concrete is composed of Portland cement and cinder, and is 3% ins. thick. The floor construction throughout will be 10 ins. of concrete covered with 1 in. of granolithic or Kosmocrete. In the tool and testing rooms off, the machine shop, and between the boiler and erecting shops, an effort has been made to secure a dust-proof construction, and a roof which will be free from drippings due to condensed moisture. Over these rooms the skylight roofs consist of double-glazed Paradigm skylights, with a I-in. air space betwen them. 3 asenig, Ae = Wood louvres in Wood Frames = a ied 4 Lacing go gods ugated Siding, 212°CS Wood Frames and Doors. “f ~ 4°: Coal Bin, |i Movable Planks» foment lt (ene nee Scale. pM EL Sb AN. 8 THe ENGINEERING RECORD, Part Sipe ELEVATION. Cross-SecTIon oF BoILER Hovsz, 3 SY £ i x Cast Block. ors Seis Fim. a YP) cast Block. Section A-A, 6 Detail at S (Enlarged) ty AR ets G® (2h 23h 1StS' ES rere budiery, J 3 All Gusset Pls yn = ——————————O Ss} | 8'C. a B | be H i-— % xe x SN S S x 386% melpencrv tem 1 Columns. ~ Scale. 48 | ” 4’ 8’ 12” 6’ 20” Hes Lett 4 1 1 1 { e777 hy a [ Finished Floor SO, nana 7 Z | = THE ENGINEERING RECORD. GrEvnERAL Cross-SECTION OF ROLLING MILL. Fic. 182% STEEL ROLLING MILL BUILDING FOR THE AMERICAN ROLLING MILL, COMPANY, MIDDLETOWN, OHIO. *Engineering Record, July 26, (901. GOVERNMENT BuILpINnG, St. Louis ExpositTion.* The Government Building at the St. Louis Exposition had a steel framework with steel arch trusses of the three-hinged type. The span of the arches is 172 ft. c. to c. of pins, and their rise from heel pins to center pins is 66 ft. 9% in. The trusses are spaced 35 ft. apart, and are connected laterally by six lines of lattice girders, carrying the posts of the main roof and monitor roof, and by eight other intermedi- ate transverse struts. A horizontal wind strut truss is located as shown in Fig. 184. The assumed loading was as follows: Dead Load: Roof, 1o Ibs. per sq. ft. on slope; dome, 12 lbs. per sq. ft. of roof surface; side walls of dome and building, 15 lbs. per sq. ft. of surface; trusses, 10 Ibs. per sq. ft. of floor surface. Wind Load: Side walls, 20 lbs. per sq. ft.; dome, curved, 15 lbs. per sq. ft. of projection on a vertical surface. Snow Load: On roof of main building, 20 Ibs. per sq. ft. hori- zontal; reduced to 10 lbs. per sq. ft. on ventilator over center. The revised estimate of loads for a 35-ft. bay. figured out as fol- lows, per sq. ft. of horizontal projection: Lbs. per sq. ft. TERS GE RSE Fog RR a ie RPO Sr foe OA 13.1 . RR eet uke Chak e ca cesie ev eoniabers A eet 6.6 2 SPR ACNE MEN ar, vin CUS v Ava see SET R oe kee eeoee 0.5 RMN S FENE cr gece Whig | TR SK cs hE FRR © cde awie’hs clgtwis'oaale 20.2 SUIS AU EU. f De cae Sal vic yu Sig ale Cio eh¥ bao oe. uis abinw d syereicig’ 21.5 The loading on one truss for the 35-ft. bay was: Actual Weightin lbs, Estimated Weight in lbs GL EEC LOM Ce cya cescd aves tecdes 40,500 70,000 URN SEGET 08 3), Si Giear swans \e ae srp a e's alas 80,000 64,000 RANE ECHL? 6 Sats wabis cass wade ewes 120,500 134,000 The arches are built up of channels, plates and angles, and have 4%4-in. shoe pins and 3-in. center pins. The shoe pins of each truss are connected by a tie bar consisting of a line of g-in. I-beams. The stresses in the arches are given in Fig. 183, while the details are shown in Fig. 184. | * Engineering News. “cham ; oe 2 : . ee | "NOILISOUXY ~ | a | : SINOT “LS ‘ONIGIING LINAWNUAAOD AO NOILIAS ssoug “EgI ‘DIY < a 4 K pL! Ng g j a ; poe i 1 f : - Ds ESL LALLA LL LAL LALLA LLL LLAMA LL SLLMMMMALALLLALLMULL A Leb Z 4 : % 5 on « : a — ; op) i n =) ; fe) Ri w Zz a mai $ é ay ey nN Lom] se s : AOL UO 356 GOVERNMENT BUILDING Sea: | | i ' | . * ' . a ae Cee Pe - yg een ae Set + # Plate: ; FA Filler? ” : 44 Pin a foaper] Eno.News. » 3 “AT Bates Fic. 184. DETAILS oF ARCH, [cena 50" sroreteint Details of Tie Rod, End and Splice. 387 REINFORCED CONCRETE ROUND-HOUSE FOR THE CANADIAN PACcIFIC Ratitway AT Moose JAw, CANADA.* | The round-house is annular in plan and occupies a half circle, the diameter of which is 350’ 5”. The radial width of the house is 80 feet, and it has.22 stalls divided into two groups of 11 stalls each, by a radial fire wall. Each stall subtends an outside wall space of 25’. No 20 Gat. tron Flashing. / CLR fe be ut in When = ww 3°Concrete_and Cohcrete te laid a oe | eS ih Nad0 61. Flashing we fom z [al ig mong laa | : “ill is 10" e ty kel "" ot et Lath encased. b : /% y r Bars in Upper Row *: HF in Concrete GS a Nes » Lower « pts fi + . RES BAe 80’ ----~-——— >I i f & u ‘ | i ‘ 3 : $§ H z od, § +g* cate Sa Fe eae $2... Selipgh ie res LSE q 3 om mt S Si #4 ~ ~~ > “ = ~~ a ee! S ~ ENG. News Fic. 185. Cross SECTION OF REINFORCED CONCRETE ROUND-HOUSE. The end walls, fire wall and outside wall are of plain concrete, the latter being enlarged by buttresses or pilasters between stall spaces to carry the outer ends of the radial roof girders, which are further supported by an outside wall column, and two intermediate columns. The combined roof, wall and column construction is shown in Fig. 185. . Each radial girder consists. of an 18” @ 55 lbs. I-beam encased in concrete as shown in Fig. 186, and is supported by an outside wall pier and three I-beam columns enclosed in concrete as shown in Fig. 187. The space between the main roof girders is spanned by rein- forced concrete beams, carrying a reinforced roof slab and a suspended Section A-B, Fic. 186. Cross SEcTION SHOWING CONSTRUCTION OF Roor SLass, es * Engineering News, February 9, 1905. REINFORCED CONCRETE ROUND-HOUSE 389 B fo | t¥ Bars) . UU eke Beton C=: ’ miedidd EF. fies eet K--/2"--> ACE 7 X aa Sectional Plan A-B. Fig. 187. |-Beam __.. Fic. 188. Hoves 1n Roor ror SMOKE JAcK, CoLUMN. metal lath and plaster ceiling as shown in Fig. 186. The main girders and the angles which knee-brace them to the columns are like the columns, encased in concrete. There is, therefore, no unprotected structural metal in the round-house. The reinforced concrete beams shown in Fig. 186 are of uniform depth, but vary in width and amount of reinforcement with the span and are composed of 1-3-5 Portland cement, gravel concrete. The beam reinforcement consists of plain rods attached to the beams as shown in Fig. 186. The roof slabs are composed of cement and washed cinders, and the ceiling slab consists simply of expanded metal wired to the beams and given two coats of cement plaster, the first coat containing enough lime to make it work well under the trowel, and the second coat being a I cement to I sand mortar. PARANA 49D ce Es > % Sheed Part Side Elevation. Cross Section. 1G. 189. ForMs FOR REINFORCED CONCRETE Roor. to accommodate expansion and contradicts No Venkan ap rE ing the first year, and none are expected to appear, The roof is reinforced where the metal smoke jacks « are as shown in Fig. 185 and Fig. 188. eee APPENDIX I. GENERAL SPECIFICATIONS FOR STEEL FRAME MILL BUILDINGS MILO S. KETCHUM, Assoc. M. AM. Soc. C, EB. 1903 GENERAL DESCRIPTION I. The height of the building shall be the distance from eight of Building the top of the masonry to the under side of the bottom chord ot the truss. 2. The width and length of the building shall be the pimensions of extreme distance out to out of framing or sheathing. Building: 3. The length of trusses» and girders in calculating Length of span. stresses shall be considered as the distance from center to cen- ter of end bearings when supported, and from end to end when fastened between columns by connection angles. 4. The pitch of roof for corrugated steel shall preferably pitch of Roof. be not less than 1%4 (6” in 12’), and in no case less than %. For a pitch less than % some other covering than corrugated steel shall be used. 5. Trusses shall be spaced so that simple shapes may gpacing of Trusses. be used for purlins. The spacing should be about 16 feet for spans of, say, 50 feet and about 20 to 22 feet for spans of, say, too feet. For longer spans than 100 feet the purlins may be trussed and the spacing may be increased. ay Spacing of Purlins. Form of Trusses. Bracing. Proposals, Detail Plans. Approval of Plans. Dead Loads, SPECIFICATIONS 6. Purlins shall be spaced not to exceed 4’ 9” where cor- FE rugated steel is used, and shall be placed at panel points of the — : trusses. 7. ‘The trusses shall ashoubly be of the Fink type with — panels so subdivided that panel points will come under the — purlins. If it is not practicable to place the purlins at panel — points, the upper chords of the trusses shall be designed to take — both the flexural and direct stresses. Trusses shall preferably be riveted trusses. 8. Bracing in the plane of the lower chord shall be stiff; bracing in the planes of the top chords, the sides and the ends may be made adjustable. g. Contractors in submitting proposals shall furnish com- — plete stress sheets, general plans of the proposed structures giving sizes of material, and such detail plans as will clearly show the dimensions of the parts, modes of construction and sectional areas. 10. The successful contractor shall furnish all work- ing drawings required by the engineer free of cost. Working drawings will, as far as possible, be made on standard size sheets 24” x 36” out to out, 22” x 34” inside the inner border lines. 3 11. No work shall be commenced or materials ordered until the working drawings are approved in writing by the en- gineer. The contractor shall be responsible for dimensions and details on the working plans, and the approval of the de- tail plans by the engineer will.not relieve the contractor of this responsibility. LOADS. loads: 13. DEAD LOADS.— Weight of Trusses — The weight of trusses per square foot of horizontal projection, Hp, to 150 feet span shall be calculated by the formula W = % (1+355) 12. ‘The trusses shall be designed to carry the following where W = weight of trusses per square foot of horizontal projection ; SPECIFICATIONS 393 P = capacity of truss in pounds per square foot of hor- izontal projection ; I = span of the truss in feet ; A = distance between trusses in feet. 14. Weight of Covering—The weight of corrugated steel shall be taken from ‘Table I. TABLE I. WEIGHT OF FLAT, AND CORRUGATED STEEL SHEETS WITH 234-INCH CORRUGATIONS Corrugated Steel. _ Thickness|_ Weight per Square (100 sa-ft) | _ [Gage No. in |___Flat Sheets __| Corrugated Sheets _inches Black |Galvanized|Black Painted [Galvanized 16 0625 250 266 275 eo /8 0500 200 2/6 220 2356 i= 2O O3575 130 166 1635 182 ; 4 ; 22 O3/3 125 /4/ 438 154 ; 24 0250 100 116 tts 127 26 0/88 735 39/ 84 99 28 0/56 65 7? 69 66 When two corrugations side lap and six inches end lap are used add 25 per cent to the above weights; when one cor- - _ rugation side lap and four inches end lap are used add 15 per cent to the above weights to obtain weight of corrugated steel laid. For paint add 2 pounds per square. The weight of cov- ering shall be reduced to weight per square foot of horizontal. projection before combining with weight of trusses. 15. Slate laid with 3-inch lap shall be taken at a weight gate of 7%4 pounds per square foot of inclined roof surface for 3-16” slate 6” x 12”, and 6% pounds per square foot of inclined roof surface for 3-16” slate 12” x 24”, and proportionately for other sizes. 16. Terra-cotta tile roofing weighs about 6 pounds per tite. _ square foot for tile I inch thick; the actual weight of tile and 4 other roof coverings not named shall be used. 17. Sheathing of dry pine lumber shall be assumed to pan scare 2m weigh 3 pounds per foot and dry oak purlins 4 pounds per foot board measure. 18. The exact weight of sheathing, purlins, bracing, ven- Miscellaneous tilators, cranes, etc., shall be calculated. oe ; a 394 Snow Loads. Wind Loads. Mine Buildings, Concentrated » Laads. Purlins. Roof Covering. Minimum Loads. SPECIFICATIONS 19. SNOW LOADS.—Snow loads shall be taken from the diagram in Fig. 1. 3: ESR SOE EY EES See We Be, | RS EAM FO SOR, a o 7 ; — 4orFecite Coastand 4 fe a; |_| Aria Freegrons use orie- AO & 3: | Aalt the tabular valves ~ 5 ; 59 Z 80 i . Le bert q - 1 30 ; re she oe : | r fl 2 5 f ait p +h be oy | Pett 20 ; Z a S34 set pch Ss 8 ng 128 io ~ ce ard leer ee. QM slopes x ro te a eR a se 5 30 35 AO 45 50 Latitude in Degrees Fic. I. SNOW LOAD ON ROOFS FOR DIFFERENT LATITUDES, IN LBS. PER SQUARE FOOT. 20. WIND LOADS.—The normal wind pressure on trusses shall be computed by Duchemin’s formula, Fig. 2, with P = 30 pounds per square foot, except for buildings in ex- posed locations, where P = 40 pounds per square foot shall be used. 21. ‘The sides and ends of buildings shall be computed for a normal wind load of 20 pounds per square foot of ex- posed surface for buildings 30 feet and less to the eaves; 30 pounds per square foot of exposed surface for buildings 60 feet to the eaves, and in proportion for intermediate heights. 22. Mine, smelter and other buildings exposed to the action of corrosive gases shall have their dead loads increased 25 per cent. 23. Concentrated loads and crane girders shall be con- sidered in determining dead loads. | 24. Purlins shall be designed for a normal load of not less than 30 Ibs. per square foot. 2s. Roof covering shall be designed for a normal load of not less than 30 lbs. per square foot. 26. No roof shall, however, be designed for an equiva- lent load of less than 30 pounds per square foot of horizontal proje ection. | , Led 2 aS ee ee oe SPECIFICATIONS Bee oo yA L— [7 F, 4 ‘a ? | |Hutton a oe oe EWAS Straight Line ---=— soa as \) : 10° Gay VA il a — / a B giv? ve 4 \b2. ~ Aa 2m ae Sih Ny | 7 | 5. : . Aq Ue = FORMULAS | 7 IAT aye ase 2 Sir f4 } C4 an Duchemin BR=FY + Sh A 2 a Hutton B=PSin. A C4203 AHI pS YN NSS bis SG NN S Straight Line P=5A,(A 2459) « ~ ~ ww ~ ‘5 . #2=Normal Pressure, |bs-per sq.ft. P= Horizontal a“ oe « “ A=Angqle of inclination of surface Normal Wind Pressure. F,.in-Ibs-per square foot - to BN \ ~\ AS Pitch-+, ms Pitch-r Pitch Bs 3 Pitch 4-, . 6 1 5 ne 4 ~| O50. 15 20. 5 3 3 40 45 50556065 7075 GOS Ss Angle Exposed Roof rnakes with Horizontal in Degrees ,A. _ Fic. 2. NorMAL WIND LOAD ON ROOF ACCORDING TO DIFFERENT FORMULAS. PROPORTION OF PARTS. 27. In proportioning the different parts of the structure _ the maximum stresses due to the combinations of the dead and ¥ wind load; dead and snow load; or dead, minimum snow and wind load are to be provided for. Concentrated loads where they occur must be provided for. 28. Allowable Unit Tensile Stresses for Medium Steel. | Pounds per square inch Shapes, main members, net section.......... 16,000 * oN NA Risener Ba 80 et Sale, Ae a eae a 16,000 Bottom flanges of rolled beams....... Dera ters 16,000 Shapes, laterals, net section........... Ds 6 BO,000 PGi FOCUS SOC tate AIS oie oe oe cette y eee: 20,000 Plate girder webs, shearing on net section. ...10,000 Te eae ee ye a eee oe a ae ex > : ny ee 9 sy A 7 ie Drees ts 90 Stresses, Tensile Stress, 395 390 Compressive Stress. Plate Girders, Alternate Stress, t'omb‘ned Stress. SPECIFICATIONS Shapes liable to sudden loading as when used for crane: girders... ci ota ee eee 10,000 Expansion rollers per lineal inch........... .600 X D where D = diameter of roller in inches. a Laterals shall be designed for the maximum stresses due — to 5,000 pounds initial tension and the maximum stress dues to wind. | 29. Allowable Unit Compressive Stress for Medium Steel. For direct dead, snow and wind loads 4 Ss 16,000 — 70 where ¥ S = allowable unit stress in pounds per square inch; 1 = length of member in inches c. to c. of end con- ~ nections ; a r = least radius of gyration of the member in inches. | 30. ‘Top flanges of plate girders shall have the same — gross area as the tension flanges. 3 31. Shear in webs of plate girders shall not exceed 10,000 — pounds per square inch. "3 32. Members and connections subject to alternate stresses shall be designed to take each kind of stress. 4 33. Members subject to combined direct and hoadine 4 stresses shall be proportioned according to the following — formula: j I ———————= ONE x My, ee ks Pee ks: * ~ tO : where =q S = stress per square inch in extreme fibre; 4 P. = direct load; 4 A = area of member; mE M = bending moment in inch-pounds; eo. y, = distance from neutral axis to extreme fibre; _ I = moment of inertia of member ; 4 1 = length member, or distance from point of zero. _ moment to end of member in inches; — : E = modulus of elasticity — 28,000,000. SPECIFICATIONS 397 g When combined direct and flexural stress due to wind is considered add 25 per cent to the above allowable ten- -sile and compressive stresses. 34. Soft steel may be used in mill buildings with unit Soft Steel. _ stresses ten per cent less than those allowed for medium steel. “ q 35. Where the stress due to the weight of the member or _ due to an eccentric load exceeds the allowable stress for direct — giress due to loads by more than ten per cent, the section shall be increased Weight of- Member. " until the total stress does not exceed the above allowable stress . for direct loads by more than ten per cent. The eccentric stress caused by connecting angles by one leg when used as ties or struts shall be calculated, or only one leg will be considered effective. , 36. Rivets shall be so spaced that the shearing stress shall - not exceed 11,000 pounds per square inch; nor the pressure on the bearing surface (diameter x thickness of piece) of the 7 rivet hole exceed 22,000 pounds per square inch. a Rivets in lateral connections may have stresses 25 per cent in excess of the above. Field rivets shall be spaced for stresses two-thirds those _ allowed for shop rivets. Field bolts, when allowed, shall be spaced for stresses - two-thirds those allowed for field rivets. Rivets and bolts must not be tsed in direct tension. 37. Pins shall be proportioned so that the shearing stress _ shall not exceed 11,000 pounds per square inch; nor the pres-. Pins. sure on the bearing surface (diameter x thickness of piece) of the pin hole exceed 22,000 pounds per square inch; nor the ex- treme fibre stress due to cross bending exceed 24,000 pounds per square inch when the applied forces are assumed as act- _ ing at the center of the members. 3 38. Rolled beams shall be proportioned by their moments of inertia. 39. Plate girders shall be proportioned on the assump- _ tion that the flanges take all the bending moment, and that the 3 shear is resisted by the web. | The distance between centers of gravity of the flange areas shall be considered as the effective depth of all girders. q 40. ‘The webs of plate girders shall be stiffened at bear- ings and at all points of concentrated loading, and at inter- Rivets, “e we “a ies Ba a — nee ea aoe ci q Rolled Beams. = Plate Girders. Stiffeners, 398 Timber. ALLOWABLE WorkING UNIT STRESSES, IN PouNnDs, PER SQUARE INCH. — SPECIFICATIONS mediate points, wherever the shearing per square inch exe ¢ the stress allowed by the following formula: i Allowed shearing stress = 12,500 — 90 d + ft where d = depth, and ¢ = thickness of web plate. timbers shall be taken from Table II. TABLE II. Tension. Compression. Transverse. Shearing. © ‘- With Grain. Kind of Timber. ; - ; s is . With | Across} 4 5 Extreme] Modulus With Fibre of End |Columns} Across Grain. Grain. Bear- Under 18 Stress. |Elasticity.| Grain. inz. | Diams. Grain. ‘ Factor of Safetv. Ten. | Ten, | Five. | Five, Four. Six. |, Two. | Four. HILO OO sted esr iis 2 tame ae ae 1000 | 200 | 1400 9°0 500 1000 | 550000 200 INSYRGG BOAO Sess cccic-o sm Si slie go cg cc oe on ee les ees 700 1100 700 200 7ov | 600000 100 South. Long-leaf or Georgia Yellow Pine} 1200 60 | 1600 | 1000 850 120) |} 850000 150 Douglas. Oregon and Yellow Fir. ........ Oy) eer 1600 | 1200 300 1100 | 7ov000 150 co Washington Fir or Pine (Red Fir)....... WOO ow wae aiteataet os beacellcwceseas BOO ATS ee aloanc POPP i re Northern or sarki “teak eevew re a aes 900 50 | 1200 800 250 1000 | 600000 100. | My Red Pine.. : ice oes Jeet C900 50 | 1200 800 200 800 | 600000 | ..........- Norway Wi eons. ashi ss oc re ereeee Se eee 1200 800 200 700 | 60U000 | ........ oe Canadian (Ottawa) White Pine.......... WOO 1S tee 1000: dawn eat tbe eel encom 109 -Canadian (Ontario) Red Pine............ paar Ce eee TODO sata b-aue's 700000 100 Spruce and Eastern Fir...............+4+ 800 50 | 1200 200 700 | 600000 100 PIORIIOOK ion ck op ivecRrwcr ee ant tahun GOO te ieee ee eee 800 150 600 450000 100 Cypress Cesuaxed cst OOF besa 1200 200 800 | 450000 | .........- DS ce Seek ack Sek he ekwals thee mee + (re 1200 800 200 800 850000 | ..... ere piestaut ROR GEich ceue sak eae ney a eae SOO Tacs esah estes 1000 250 8u0 | 500000 150 California Redwood.. ROOM cas sy Fh Somers 800 200 790 | 30000 100 RLOTTIA SPTNOGs 2 i sks ac ss Wadas so dawy eek eee geeks eee REO sores 800.'! 6u0000 fo. oS ee tS = Corrugated Steel. diameters shall be obtained ia means of the formula eo gs ao Sa where C = value in column 4 above; 1 = ag ss of column an & d least width of column, both in feet. COVERING. a 43. Corrugated steel shall generally have 2% inch cor- rugations when used for roof and sides of buildings, and 14 inch corrugations when used for lining buildings. The mini | SPECIFICATIONS : I mum gage of corrugated steel shall be No, 22 for roofs, No. i | 24 for sides and No. 26 for lining. The gage of corrugated steel in U. S. Standard Gane. and weight per square foot shall be shown on the general plan. 44. ‘The span, or center to center distance of purlins, shall not exceed the distance given in Fig. 3 for a safe load of 30 pounds per square foot. Corrugated steel sheets shall prefer- ably span two perue spaces. a 2 = W= 4 Sh2! (Rankine) W = safe load ; $100 S = working Stress =|5000 Ibs, ad h = depth of corrugation-ins. g he -b = width of sheet in inches § 60 * ‘t = thickness of sheet in inches 8 | = clear span in inches ‘S Gg ao} 6 = ‘< ay 5 8 = NS *) £ = -——- 23" 2 ead = & o =. : = VY) 2 40 = ae = 5 2 roa 3 /o = a = eee et ae 3 4 5° 6 S pan,Z, in Feet. Fic. 3. SAFE UNIFORM LOAD IN POUNDS FOR CORRUGATED STEEL . FOR DIFFERENT SPANS IN FEET. 45. Corrugated steel shall be laid with two corrugations side lap and six inches end lap when used for roofing, and one corrugation side lap and four inches end lap when used for siding. 46. Corrugated steel shall be fastened to the purlins by _ means of galvanized iron straps 34 inches wide by No. 18 gage, ‘spaced 8 to 12 inches apart; by clinch nails spaced 8 to 12 inches apart; or by nailing directly to spiking strips with 8d barbed nails, spaced 8 inches apart. Spiking strips shall preferably _ be used with anti-condensation lining. Bolts, nails and rivets shall always pass through the top of corrugations. Side laps shall be riveted with copper or galvanized iron rivets 8 to 12 inches apart on the roof and 1% to 2 feet apart on the sides. 47. Corrugated steel lining on the sides shall be laid with one corrugation side lap and four inches end lap. Girts _ for corrugated steel lining shall be spaced for a safe load of 25 pounds per square foot as given in Fig. 3. 399 Spacing Purlins. End and Side waps. Fastening. Corrugated Steal Lining. 400 Anti-Condexusation Lining. Flashing. Ridge Roll. Corner Finish. Cornice. Gutters, Ventilators. SPECIFICATIONS 48. Anti-condensation roof lining shall preferably be used 4 to prevent dripping in engine houses and similar buildings — and shall be constructed as follows: Galvanized wire poultry netting is fastened to one eave purlin and is passed over the ridge, stretched tight and fastened to the other eave purlin. The edges of the wire are woven together, and the netting is fastened to the spiking strips, where used, by means of small staples. On the netting are laid two layers of asbestos paper 1-16-inch thick, and two layers of tar paper. The corrugated steel is then fastened to the purlins in the usual way. Three- sixteenth-inch stove bolts with 1” x 4%” x 4” ‘plate washers on the lower side are used for fastening the side laps together and for supporting the lining; or the purlins may be spaced one- half the usual distance where anti-condensation lining is used, and the stove bolts omitted. 49. Valleys or corners around stacks shall have flashing extending at least 12 inches above where water will stand, and shall be riveted or soldered if necessary, to prevent leakage. Flashing shall be provided above doors and windows. so. All ridges shall have a ridge roll securely fastened to the corrugated steel. 51. All corners shall be covered with standard corner finish securely fastened to the corrugated steel. 52. At the gable ends the corrugated steel on the roof shall be securely fastened to a finish angle or channel connected to the end of the purlins or, where molded cornices are used, to a piece of timber fastened to the ends of the purlins. CLT. len tant et aes 53. Gutters and conductors shall be furnished at least — S equal to the requirements of the following table: Span of Roof Gutter Conductor Up to 50’ 6” 4” every 40 50’ to 70° rbd ig “ec ee 70'- to 100’ 8” 5” BE 66 Gutters shall have a slope of at least 1” in 15’. Gutters and conductors shall be made of galvanized steel not lighter than No. 24. 54. Ventilators shall be provided and located so as to properly ventilate the building. ‘They shall have a net open- ing for each 100 square feet of floor space of not less than one- SPECIFICATIONS ings; of not less than one square foot for dirty machine shops; _ of not less than four square feet for mills; and not less than _ six square feet for forge shops, foundries and smelters. E- 55. Openings in ventilators shall be provided with shut- ters, sash, or louvres, or may be left open as specified. 3 Shutters must be provided with a satisfactory device for opening and closing. Louvres must be designed to prevent the blowing in of _ tain and snow, and must be made stiff so that no appreciable sagging will occur. They shall be made of not less than No. _ 20 gage galvanized steel for flat louvres and No. 24 gage gal- _ vanized steel for corrugated louvres. a 56. Circular ventilators when used must be designed so as to prevent down drafts. Net opening only shall be used in calculations. 57. Windows shall be provided in the exterior walls equal to not less than 10 per cent of the entire exterior surface in _ mill buildings, and of not less than 25 per cent in machine _ shops, factories and similar buildings. g Window glass up to 12” x 14” may be single strength, over _ 12” x 14” the glass shall be double strength. Window glass _ shall be A grade except in smelters, foundries, forge shops, and _ similar structures where it may be B grade. The sash and _ frames shall be constructed of white pine. 4 58. At least half of the lighting shall preferably be by - means of skylights, or sash in the sides of ventilators. * Skylights shall be glazed with wire glass, or wire netting a shall be stretched beneath the skylights to prevent th: broken 4 glass from falling into the building.Where there is danger of the sf skylight glass being broken by objects falling on it, a wire 4 | netting guard shall be provided on the outside. | Skylight glass. shall be carefully set, special care being a used to prevent leakage. Leakage and condensatfon on the in- ner surface of the glass shall be carried to the down-spouts, or outside the building by condensation gutters. 59. Windows in sides of buildings shall be made with counterbalanced sash, and in ventilators shall be made with sliding or swing sash. % fourth square foot for clean machine shops and similar build- . 4ul Shutters and Louvres. Circular Ventilators, Windows. Skylights, Details. Pitch of Rivets, Diameter of Punch. Net Sections. Minimum Sece tions oh ae ae ty SPECIFICATIONS 60. Doors are to be furnished as specified and are to be provided with hinges, tracks, locks and bolts. Single doors up to 4 feet and double doors up to 8 feet shall preferably 1 be swung on hinges; large doors, double and single, shall be ar- ranged to slide on overhead tracks, or may be counterbale _ to lift up between vertical guides. Steel doors shall be firmly braced and shall be cove with No. 24 corrugated steel with 1%4-inch corrugations. _ a The frames of sandwich doors shall be made of two layers of 7%-in. matched white pine, placed diagonally, and fi aly nailed with clinch nails. ‘The frame shall be covered on each side with a layer of No. 26 corrugated steel with 14-inch cc a rugations. a “a DETAILS OF CONSTRUCTION. 61. All connections and details shall be of sufcient | strength to develop the full strength of the member. 62. ‘The pitch of rivets shall not exceed 6 inches, or si teen times the thickness of the thinnest outside plate in the line ~ of stress, nor forty times the thickness of the thinnest outside — plate at right angles to the line of stress. The pitch shall never be less than three diameters of rivet. At the ends of compression — members the pitch shall not exceed four diameters of the rivet for a length equal to twice the width of the member. = 63. ‘The distance between the edge of any piece and thes center of a rivet hole must never be less than 1/4 inches, except in angles having a 2” leg, where it shall be 1% inches. = 4 64. The diameter of the punch shall not exceed the — diameter of the rivet; nor the diameter of the die exceed the diameter of the punch by more than 1-16 inch. : 65. The effective diameter of a driven rivet will be a sumed the same as its diameter before driving. In deducti the rivet holes to obtain net sections in tension members, the diameter of the rivet-holes will be assumed as % in lar than the undriven rivet. 66. No metal of less thickness than %4 inch shall f used except for fillers; and no angles less than 2” x 2” x 4 No upset rod shall be less than 5% inch in diameter. Sag may be as small as 34-inch. SPECIFICATIONS 67. All rods with screw ends except sag rods must be upset at the ends so that the diameter at the base. of the threads shall be 1-16 inch larger than any part of the body of the bar. . 68. Upper chords shall have symmetrical cross-sections, and shall preferably consist of two angles back to back. 69. All other compression members except sub-struts shall be composed of sections symmetrically placed. Sub-struts may consist of a single section. 70. Side posts which take flexure shall preferably be com- posed of 4 angles laced, or 4 angles and a plate. Where side posts do not take flexure and carry heavy loads they shall pref- erably be composed of two channels laced. Posts in end framing shall preferably be composed of I beams or 4 angles laced. Corner columns shall preferably be composed of one angle. The cross-bending stress due to eccentric oaeiiee’ in col- umns carrying cranes shall be calculated. 71. Laced compression members shall be stayed at the ends by batten plates placed as near the end of the member as practicable and having a length not less than the greatest width of the member. The thickness of batten plates must not be less than 1-40 of the distance between rivet lines at right angles to axis of member. 72. Single lattice bars shall have a thickness of not less than 1-40, and double bars connected by a rivet at the intersec- tion of not less than I-60 of the distance between the rivets connecting them to the member ; they shall make an angle not less than 45° with the axis of the member; their width shall be in accordance with the following standards, generally: Size oF MEMBER. WiptH oF Lacinc Bars. For 15-inch channels, or built | sections with 3% and 4- $ 2% inches (%-inch rivets). inch angles. For 12, 10 and g-inch chan- nels, or built sections with $ 214 inches (34-inch rivets). 3-inch angles. 403 Upset Rods. Upper Chords, Compression Members, Columns. Crane Posts. Lacing. 404 Pin Plates. Maximum Length. Splices. Splices, - Tension Members, Eye-Bars. Pins, SPECIFICATIONS For 8 and 7-inch channels, or - built sections with 24-inch } 2 inches (54-inch rivets), angles. “ot For 6 and 5-inch channels, or oe built sections with 2-inch $ 134 inches (%4-inch rivets). angles. x Where laced members are subjected to bending, the size of lacing bars or angles shall be calculated or a solid web pated shall be used. ‘ 73. All pin,holes shall be reenforced by additicnal mater- é ial when necessary, so as not to exceed the allowable pressure on the pins. These reenforcing plates must contain ence rivets to transfer the proportion of pressure which comes upon — them, and at least one plate on each side shall extend not less than 6 inches beyond the edge of the tie plate. 74. No compression member shall have a length exceed-— ing 125 times its least radius of gyration for main members, nor 150 times its least radius of gyration for laterals and sub- — members. : q 75. In compression members joints with abutting faces planed shall be placed as near the panel points as possible, and _ must be spliced on all sides with at least two rows of rivets on each side of the joint. Joints with abutting faces not planed must be fully spliced. : 76. Joints in tension members shall be fully spliced. 77. ‘Tension members shall preferably be composed of angles or shapes capable of taking compression as well as tension. Flats riveted at the ends shall not be used. sis 78. Main tension members shall preferably be made of 2 angles, 2 angles and a plate, or 2 channels laced. Secondary — tension members may be made of a single shape. 79. Heads of eye-bars shall be so proportioned | as to dee velop the full strength of the bar. The heads shall be forged — and not welded. 80. Pins must be turned true to size and straigat and must be driven to place by means of pilot nuts. The diameter of pin shall not be less than 34 of the depth of the widest bar attached to its — ' SPECIFICATIONS The several members attached to a pin shall be packed so as to produce the least bending moment on the pin, and all vacant spaces must be filled with steel fillers. 81. Long laterals may be made of rods with clevis or sleeve nut adjustment. Bent loops shall not be used. 82. Trusses shall preferably be spaced so as to allow the use of single pieces of rolled sections for purlins. Trussed purlins shall be avoided if possible. 83. Purlins and girts shall preferably be composed of single sections—channels, angles or Z-bars placed with web at right angles to the trusses and posts and legs turned down. 84. Purlins and girts shall be attached to the top chord of trusses and to columns by means of angle clips with two rivets in each leg. 85. Purlins shall be spaced at distances apart not to exceed the span as given for a safe load of 30 pounds, and girts for a safe load of 25 pounds in Fig. 3. &6. ‘Timber purlins shall be attached and spaced the same as steel purlins. 87. Base plates shall never be less than 1% inch in thick- ness, and shall be of sufficient thickness and size so that the pressure on the masonry shall not exceed 250 pounds per square inch. . 88. Columns shall be anchored to the foundations by means of two anchor bolts not less than 1” in diameter upset, placed as wide apart as practicable in the plane of the wind. The anchorage shall be calculated to resist the bending moment at the base of the columns. 80. Lateral bracing shall be provided in the plane of the top and bottom chords, side and ends; knee braces in the transverse bents ; and sway bracing wherever necessary. Later- al bracing shall be designed for an initial stress of 5,000 pounds in each member, and provision must be made for putting this initial stress into the members in erecting. 90. Variations in temperature to the extent of 150 degrees F. shall be provided for. 495 ~ Rods. Spacing Trusses. Purlins and Girts, Fastening. Spacing. Timber Purlins, Base Plates. Anchors, Lateral Bracing. Temperature, 406 Workmanship.. Riveted Work Punching. Holes for Field Rivets, Planing and Reaming. Rivets, Riveters. Bolts. Neat Finish, Contact Surfaces, SPECIFICATIONS WoRKMANSHIP. gi. All workmanship shall be first-class in every partic- ular. Due regard must be had for the neat and attractive ap- pearance of the finished structure, and details of an unsightly character will not be allowed. 92. All riveted work shall be punched accurately with holes 1-16 of an inch larger than the size of the rivet, and when the pieces forming one built member are put together, the holes must be truly opposite; no drifting to distort the metal will be allowed; if the hole must be enlarged to admit the rivet, it must be reamed. 93. All holes for field rivets in splices in tension mem- bers shall be accurately drilled to an iron templet or reamed — 4 while the connecting parts are temporarily put together. 94. In medium steel over 5 of an inch thick, all sheared q edges shall be planed, and all holes shall be drilled or reamed to a diameter of of an inch larger than the punched holes, so as to remove all the sheared surface of the metal. Steel which does not satisfy the drifting test must have holes drilled. 95. The rivet heads must be of approved hemispherical _ shape, and of a uniform size for the same size rivets through- __ out the work. They must be full and neatly finished through- out the work and concentric with the rivet hole.. 96. All rivets when driven must completely fill the holes, the heads be in full contact with the surface, or countersunk when so required. 97. Rivets shall be machine driven wherever possible. Power riveters shall be direct-acting machines, worked by steam, hydraulic pressure, or compressed air. 98. When members are connected by bolts which trans- — mit shearing strains, such bolts must have a driving fit. 99. ‘The several pieces forming one built member must fit closely together, and when riveted shall be free from twists, — bends, or open joints. 100. All portions of the work exposed to view shall be neatly finished. 1o1. All surfaces in contact shall be painted before they are put together. — w.8 ond = my a ca a ae o>) ie: eee ee \ oe ” it WEF ae eee ine Rt . Oe SS ee Pe el ee ave Lo SPECIFICATIONS 102. The heads of eye-bars shall be made by upsetting, rolling, or forging into shape. Welds in the body of the bar 3 will not be allowed. 103. The bars must be perfectly straight before boring. 104. The holes shall be in the center of the head and on the center line of the bar. 105. All eye-bars shall be annealed. 106. All abutting surfaces in compression members shall be truly faced to even bearings, so that they shall be in such contact throughout as may be obtained by such means. 107. Pin holes shall be bored truly parallel with one an- _ other and at right angles to the axis of the member unless otherwise shown in drawings; and in pieces not adjustable for length, no. variation of more than 1-64 of an inch for every 20 feet will be allowed in the length between centers of PS holes. 108. Bars which are to be placed side by side in the structure shall be bored at the same temperature, and shall be of such equal length that, upon being piled on each other, the pins shall pass through the holes at both ends at the same time without driving. 109. All pins shall be accurately turned to a gage, and shall be straight and smooth. 110. The clearance between pin_and pin hole shall ae I-50 of an inch for pins up to 3% inches in diameter, and 1-32 for larger pins. 111. All pins shall be supplied with steel pilot nuts, for use during erection. QuALIty oF MATERIAL. STEEL. 112. All steel must be made by the open hearth process, and if by acid process, shall contain not more than 0.08 per cent of phosphorus, and if by basic process, not more than 0.06 per cent of phosphorus, nor more than 0.05 per cent of sul- phur, and must be uniform in character for each specified kind. 113. The finished bars, plates and shapes must be free from injurious seams, flaws, or cracks, and have a clean smooth finish. 407 Forged Work Eye-Bars. Machine Work Facing. Pin Holes, Pins, Play in Pin Holes, Pilot Nuts, Process of Manufacture. Finish, 408 Test Pieces. Annealed Test Pieces, Marking. Physical Properties. Rivet Steel. Soft Steel. Medium Steel. Full Size Test of Steel Eye-Bars. _ Meditim. required to break in the body, but should a bar break in the SPECIFICATIONS No work shall be done on any steel between the tempera- ture of boiling water and of ignition of hard wood saw dust. — 114. The tensile strength, limit of elasticity and ductility, — shall be determined from a standard test-piece, cut from the — finished material, of at least 14 square inch section. All brok- a en samples must show a silky fracture of uniform color. = 115. Material which is to be used without annealing or a further treatment is to be tested in the condition in which it — comes from the rolls. When material is to be annealed or otherwise treated before use, the specimen representing such ; material is to be similarly treated before testing. 4 116. Every finished piece of steel shall be stamped witha a the blow number identifying the melt. BY 117. Steel shall be of three grades: Rivet, Soft and ~ 118. Rivet steel shall have: Ultimate strength, 50,000 to 58,000 pounds per square inch. Elastic limit, not less than i one-half the ultimate strength. Elongation, 26 per cent in 8 inches. Bending test, after or before heating to a light cherry red and cooling in water, 180 degrees flat on itself, without fracture on outside of bent portion. "II9. Soft steel shall have: Ultimate strength, 54,000 to 62,000 pounds per square inch. Elastic limit, not less than one-half the ultimate strength. Elongation, 25 per cent in 8 inches. Bending test, after or before heating to a light cherry red and cooling in water, 180 degrees flat on. itself, without fracture on outside of bent portion. 3 : 120. Medium steel shall have: Ultimate strength, 60,000 to 68,000 pounds per square inch. Elastic limit, not less than one-half the ultimate strength. Elongation, 22 per cent in 8 — inches. Bending test, 180 degrees to a diameter equal to thickness of piece tested, without fracture on outside of bent portion. 121. Full size test of steel eye-bars shall be required to show not less than 10 per cent elongation in the body of the bar, and tensile strength not more than 5,000 pounds below the minimum tensile strength required in specimen tests of the grade of steel from which they are rolled. The bars will be t a (eS fe s¥ ae Tan eee an? ok ‘5 a ek ie ia NS i ce A MN «7 fo ie SE ae ee le hie aoe he eee ee Soe a ee eS eS re : ‘ i ‘ee PET 7 > ae a ee ee oe ee ee PELL a eR oe a ae > oo Pi ALY oy ETS NI HR REE I EIR I AI I AES I Ey Oe SPECIFICATIONS head, but develop 10 per cent elongation and the ultimate strength specified, it shall not be cause for rejection, provided not more than one-third of the total number of bars tested break in the head; otherwise the entire lot will be rejected. 122. Pins made of either of the above mentioned grades of steel shall, on specimen test pieces cut from finished mate- rial, fill the requirements of the grade of steel from which they are rolled, excepting the elongation, which shall be decreased 5 per cent from that specified. 123. In steel 54 inch or less in thickness punched rivet holes, pitched two diameters from a sheared edge, must stand drifting until the diameter is one-third larger than the origindl hole, without cracking the metal. | 124. The slabs for rolling plates shall be rolled from ' ingots of at least twice their cross-section. 125. Pins up to 7 inches diameter shall be rolled. 126. A variation in cross-section or weight of rolled material of more than 2% per cent from that specified, may be cause for rejection, except in the case of plates which will be covered by the manufacturer’s standard specifications (Cambria Steel, page 345). STEEL CASTINGS. 127. Steel castings shall be made of open hearth steel containing from 0.25 to 0.40 per cent carbon, and not over 0.08 per cent of phosphorus nor 0.05 per cent sulphur, and shall be practically free from blow holes. Cast Iron. 128. Except where chilled iron is specified, all castings shall be of tough, gray iron, free from injurious cold shuts or blow holes, true to pattern, and of workmanlike finish. Test bars one inch square, loaded in middle between supports 12 inches apart, shall bear 2,500 pounds or over, and deflect 0.15 of an inch before rupture. Wroucut [Ron. 129. All wrought iron must be tough, ductile, fibrous and of uniform quality. Finished bars must be thoroughly welded Pin Steel. Drifting. Slabs for Plates. Pins. Variation in Weight. Steel Castings. Cast Iron. Character and -Finish. 409 410 Manufacture. Standard Test Piece Elastic Limit. ’ Tension. Bending Test. Timber, Painting SPECIFICATIONS during the rolling, and be straight, smooth and free from i ‘ jurious seams, blisters, buckles, cracks or imperfect edges. ~ “ai 130. No specific process or provision of manufacture will be demanded, provided the material fulfills the requirements of these specifications. >a 131. The tensile strength, limit of elasticity and sii, shall be determined from a standard test piece of as near | square inch sectional area as possible. The eae s alt a be measured on an original length of 8 inches. 132. Iron of all grades shall have an elastic limit of not less than 26,000 pounds per square inch. is 133. When tested in specimens of uniform sectional area of at least 12 square inch the iron shall show a minimum ulti mate strength of 50,000 pounds per square inch, and a 2 mum elongation of 18 per cent in 8 inches. 3 134. All iron for tension members must bend cold | through 90 degrees to a curve whose diameter is not over twice the thickness of the piece, without cracking. When — nicked on one side and bent by a blow from a sledge, the | fracture must be mostly fibrous. » a TIMBER. 135. ‘The timber shall be strictly first-class spruce, white — pine, Douglas fir, Southern yellow pine, or white oak timber; sawed true and out of wind, full size, free from wind shakes, ; large or loose knots, decayed or sapwood, wormholes or other defects impairing its strength or durability. PAINTING. 136. All iron work before leaving the shop shall be thoroughly cleaned from all loose scale and rust, and be given one good coating of pure boiled linseed oil or paint as speci fied, well worked into all joints and open spaces. : 139; 0 riveted work, the surfaces coming in contact a shall each be painted before being riveted togethe. a 138. Pieces and parts which are not accessible for patina ing after erection shall have two coats of paint. a SPECIFICATIONS 139. The paint shall be a good quality of red lead or graphite paint, ground with pure linseed oil, or such paint as may be specified in the contract. 140. After the structure is erected, the iron work shall be thoroughly and evenly painted with two additional coats of a. paint, mixed with pure linseed oil, of such quality and color as may be selected. Painting shall be done only when the surface of the metal is perfectly dry. No painting shali be done in wet or freezing weather unless special precautions are > taken. 141. Pins, pin holes, screw threads and other finished Phirfaces shall be coated with white lead and tallow before being shipped from the shop. INSPECTION. 142. All facilities for inspection of material and work- manship shall be furnished by the contractor to competent in- spectors, and the engineer and his inspectors shall be allowed free access to any part of the works in which any portion of the material is made. 143. ‘The contractor shall furnish, without charge, such “specimens (prepared) of the several kinds of material to be used as may be required to determine their character. 144. Full sized parts of the structure may be tested at the option of the purchaser; but, if tested to destruction, such material shall be paid for at cost, less its scrap value, if it proves satisfactory. 145. If it does not stand the specified tests, it will be con- sidered rejected material, and be solely at the cost of the con- tractor. ERECTION. 146. ‘The contractor shall furnish at his own expense all necessary tools, staging and material of every description re- quired for the erection of the work, and remove the same when the work is completed. 147. The contractor shall assume all risks from storms or accidents, unless caused by the negligence of the owner, Inspection Testing. Tools. Rigka. 4tt work is ehpapteed’ and oe The contractor shall comply with all ore ulations appertaining to the work. 3 The erection must be carried forward with must be completed promptly, APPENDIX II. PROBLEMS IN GRAPHIC STATICS AND THE CALCULATION OF STRESSES. Introduction.—It is impossible for the student to gain a working knowledge of graphic statics and the calculation of stresses without solving numerous problems. In order to save the time of the student and the instructor the problems must be selected with care, and the data put in working form. The following problems have been given by the author, in connection with a course preliminary to bridge analysis, and are presented here with the hope that they may prove of value to both students and instructors. By slightly changing the quantities and dimensions the data for new problems may be easily obtained. Instructions.—(1) Plate-——The standard plate is to be 9” &K 1014”, with a 1” border on the left-hand side, and a 1%” border on the top, bottom, and right-hand side of the plate. The plate inside the border is to be 7144" & 9%”. (2) Co-ordinates.—Unless stated to the contrary, co-ordinates given in the data will refer to the lower left-hand corner 1500 of border as the origin of co-ordinates. In defining the force, P : 150° (5.0, 3.0”), the force is 1500 fbs., makes 150° with the X-axis (lies in the second quadrant), and passes through a point 5.0” to the right, and 3.0” above the lower left-hand corner of border. (3) Data— Complete data shall be placed on each problem so that the solution will be self explanatory. (4) Scales——The scales of forces, and of frames or trusses shall be given as 1” ( ___) lIbs., or ft.; and by a graphic scale as well. (5) Name—The name of the student is be placed outside the border in the lower right-hand corner. (6) Equa- tions.—All equations shall be given, but details of the solution may be indicated. (7) References.—References are to “The Design of Steel Mill Buildings.” Note.—It should be noted that all the problems have been re- diced so that all dimensions are one-half the original dimensions given in the statements of the problems. 414 PROBLEMS PROBLEM I. RESULTANT OF CONCURRENT FORCES. (a) Problem.—Given the following concurrent forces: O 2200 750. p, 2200 1qerere) 950 1700 eee LEN P ° P : 150° ; a eee Poe peat 120 30 | 315 240 Forces are given in pounds. Required to find the resultant, R, by means of a force diagram. Check by calculating R by the algebraic method. Give all equations. Also construct check force polygon. Give amount and direction of the resultant. Scale for force polygon, . 1” = oo lbs. (b) Methods.—Start P, at point 4”, 314” (4, y, using lower left-hand corner of the border as the origin of co-ordinates) and take the forces in the order, P,, P., P;, P,, P;. Draw check polygon starting at the same point and drawing the forces in the order 7 SOS hea gee notte wit (c) Results.—The resultant is a force R acting through the point of intersection of the given forces, and is parallel to the closing line in the force polygon. It will be seen that it is immaterial in what order the forces are taken in calculating the resultant, R. In the alge-_ braic solution the summation of the horizontal components of the forces, including the resultant, R, are placed equal to zero, and the summation of the vertical components of the forces, including the resultant, R, are placed equal to zero. Solving these equations we have the value of R, and the angle 6, which R makes with the X-axis. PROBLEM Ia. RESULTANT OF CONCURRENT FORCES. (a) Problem.—Given the following concurrent forces: paca beg es pen. a t520°? 2 30° ? ks i ae * 240°’ ae fs Forces are given in pounds. Required to find the resultant, R, by means of a force diagram. Check by calculating R by the algebraic method. Give all equations. Also construct check force polygon. Give amount and direction of the resultant. Scale for force polygon, 1’ = 400 lbs. PROBLEMS Graphic Statics. 1900 cos |20°+ 950 cos 30° +1700c05 315° : # \ Scale |'=400 \ Algebraic Solution. +750 COS 240°+2200c05 150= Rcos8 1900 sin |20°+ 950 $in30°* 1700 Sin 315° +750 Sin 240°+2200sin 150°=Rsin © Roos @=-1206 tan@=-1.135 6=131°23' RsinO=+1369 R=aee 7823" —-—-— Check Force Polygon Problem l. Force Polygon _ Results By Graphics. R= 1820* 6=151°O By Algebra. R=1823" @=131°23' 416 PROBLEMS _ PROBLEM 2. RESULTANT OF NON-CONCURRENT FORCES. (a) Problem.—Given the following non-concurrent forces: 2 7 20 ” ” 7 Cee ” ” 35 “ ” ” P,-—~ (7.0", 6.3"); Pa—, (05°; 9.3"); "2, = 72 oe 150 205 O 210 Pi (BERS). 15 Forces are given in pounds. Find resultant, R, by means of force and equilibrium polygons. Check by calculating R by means of a new force polygon and a new equilibrium polygon. Also check as described below. Scale for force polygon, 1” = 100 lbs. (b) Methods.—Start force polygon at (7.0”’, 1.0’). Take pole | at (3.8”, 0.0”). Start equilibrium polygon at (7.0”, 6.3”). Take new — pole at (2.2”, 0.0”), and draw new polygon starting at (7.0”, 6.3”). (c) Results.—The resultant is a force, R, and acts through the intersection of the strings d, e and d’, e’, and is parallel to the closing line in the force polygon. If corresponding strings in the equilibrium are produced to an intersection, the points of intersection, I, 2, 3, 4, 5, will lie in a straight line which will be parallel to the line O-O’ joining the poles of the force polygons. This relation is due to the reciprocal nature of the force and equilibrium polygons, and may be proved as follows: In the force polygon the force P, may be resolved into the rays c and d, it may likewise be resolved into the rays c’ and d’. In like manner it may be seen that the force O—O’ can be resolved into the d and d’, or into c and c’. Now if the strings d and d’ are drawn through the point 4, and the strings c and c’ are drawn, they must in- tersect in the point 3, and 4-3 must be parallel to O-O’. For the re- sultant of d and d’ is equal to O—O’ and must act in a line parallel to O-O’; likewise the resultant of c and c’ is equal to O—O’ and must act parallel to O-—O’; and in order to have equilibrium 3-4 must be parallel to O-O’. In like manner it may be proved that 1, 2, 3, 4, 5, are in a straight line parallel to O-O’. From the above it will be seen that to have equilibrium in a sys- tem of non-concurrent forces it is necessary that the force polygon and its corresponding equilibrium polygons must close, or that two equi- librium polygons must close. q + # : - j a PROBLEMS i\ Problem 2. seers -O 100 1 720 Scale \"=100* Given F, 7558 (70,63) “- So Pe ree (0.8, 8.5) To find R. ’ 8 / ‘ : TR R= 536 6= 150°10' Force Polygon ————— Equilibrium Polygon ----- Check Equil. Polygon —-— Aas | Opa b : mos iS rh ae ‘ ed iN ‘ \ 1 = Woes Gat it thay ‘\ \ J ; 4 ‘ : , \ [xf i \ XK \ r / ee o © : er ‘ | ! Mi NN e' A see \\ el ee ee Ae, = Ts yet eee 2 iat Ag ee oe en: PROBLEM 2a. RESULTANT OF NON-CONCURRENT FORCES (a) Problem.—Given the following non-concurrent forces: 350 ” ” 2 3527 (70",6.6"), Pi ‘G; Or. 6. yh © i. i = (0. 5”; 9. au). tat 300 ’? ”r Pattee (0.8”, 8.5’). The rest of the statement is the same as for Problem 2 417 418 PROBLEMS ProBLEM 3. TEST OF EQUILIBRIUM OF FORCES. (a) Problem.—tTest the following forces for equilibrium by means of force and equilibrium polygons: 7 8 ” ” 10.8 ” , 12.2 ” ” P,— (4.757, 0.0"); P2— (4751 7-7 Js oie (0.0", 7.0”) ; O 45 Oo 19.35 13-2 ” ” ”" ” Pi ” ” P,—— 4.75", 6.4") Ps——(2.75", 0.4") 3. Po (2.75 gee 300 21 150 The forces are given in tons. Check by using a second pole and equilibrium polygon ; also draw a line through the intersection of the corresponding rays, and check as in Problem 2. Give amount and direction of the equilibrant. Scale of forces, 1” = 5 tons. (b) Methods.—Start force polygon at (0.8”, 1.5’). Take first pole at (3.2”, 2.7”). Start equilibrium polygon at (4.75”, 9.1”). Take second pole at (4.0”, 3.2”). Start second equilibrium polygon at (2.75%, 8.2); (c) Results.—If the system of forces was in equilibrium the equi- librium polygons would close, and the first and last strings f and f, and f’ and f’ would coincide, respectively. The equilibrant will be | equal to a couple with a moment represented by the rays f or f’ multi- plied by the distance h or h’. In general in any system of non-current forces if the force polygon closes the equilibrant of the system is a couple. If the system is in equilibrium the arm of the couple is zero. It is evident that in order that any system of non-concurrent forces be in equilibrium it is necessary that both the force polygon and an equilibrium polygon must close. The check line must be parallel to the line O—O’ joining the poles, and also pass through the intersections of corresponding rays as in Problem 2. ‘ PY att i ee a — ia 5 , si ~ jee ie ee al F ? < = ays le a and ad 2 ‘ act ge ie aS ne, m - — : ay, ion usc ., ‘ a a ao ~~ <- ey ea. 7 Pi " o ihe ‘4 | ont ; x a ee ee ait tat: . oe a Fy _ = a es - v ee _ - wn r PROBLEM 3a. TEST OF EQUILIBRIUM OF FORCES. (a) Problem.—Test the following forces for equilibrium by means of force and equilibrium polygons: 8 Ad ”r 8 Ad ”r I bs vr ” P, a (4.75",0.0"); P, ae (4.75, 7.7"); Ps are (0.0",7.0") ; as 4"); Py 333 (278",6.4"); Paves (2.75", 7:7 Js PROBLEMS 419 Graphic Statics. ua, 150° \ \b 10.8 » ‘ we ze NS ‘Ee 2 f) ' x \ Elon) | Ef Gas NS ay y t a Shi ° ' ns , agar ant gas eambetoate f----------- i 3 ‘1 e/ : ‘ ‘ * i Ey | i an. ; i ~ oy RA v7; 19.38 (e 7556.4 i iis | ee the “x 132 (4:75,6.4) 5 10 J j Scale \"= 5 Tons. _—— ee _~ _ Equilibrant= A Couple E=fxh=13.4x.25 = 3.35 inetons =Pexh, =3.26y E by check equil. polygon =f'xh= 16.1%, 19=3.42 inctons } =Pxh, =&3.36 u u >: The forces are given in tons. Check by using a second pole and equilibrium polygon; also draw a line through the intersection of the corresponding rays, and check as in Eee 2. direction of the equilibrant. Give amount and Scale of forces, 1” = 5 tons. 420 PROBLEMS PROBLEM 4. RESOLUTION OF FORCES. (a) Problem.—Given the following forces: 13.2 5.0 3.2 (1.0",—); P, 5 (2.8",—); P, ——(68", —). 270° 270 270 Forces are given in tons. (1) Find the resultant, R, by means of force and equilibrium polygons. (2) Resolve F into two parallel forces P, ? ? sé (3.2”,—) and Sore: Bs (5.4”,—). (3) Find the moment, M, of 270° R about a point Z at (4.0”, —). Check by the algebraic method giv- ing all equations. Scale of forces, 1” = 5 tons. (b) Methods.—Start force polygon at (5.7, 0.5”), and take pole at (1.7”, 3.0’). In the algebraic method take moments about the left border in finding the point of application of R, and take moments in the line of action of P” in resolving R into P’ and P”. (c) Results.—(1) The position of R is at the intersection of strings a and d. (2) Prolong string a until it intersects P’, take the intersection of string d with P” ; then string e is the closing line of the polygon. (3) The moment of R is found graphically by multiplying the intercept, y, by the pole distance, H. (4) To find position of Rk algebraically take moment of P,, P,, and P;, about the left border, and divide by R, which is the sum of P,, P,, P;. (5) The moment of R about Z is equal to R & (4.0 — 2.28)”, = 36.8 in.-tons. _ PROBLEM 4a. RESOLUTION OF FORCES. (a) Problem.—Given the following forces: i T3. 2 ig Je eae 5- O id mB SN AR po ” pen ae ); P, SS. (28. Pea Coe Forces are given in tons. (1) Find the ieee R, by means of sg and equilibrium se igh Fs (2) Resolve R into two parallel forces Pa 2", —) and Ft ao? ~(5.4”,—). (3) Find the moment, VM, of © R about a point Z at (4. on —). Sine by the algebraic method giv- ing all equations. Scale of forces, 1” —5 tons. eS woe lt il a PROBLEMS Graphic Statics. Problem 4. ih cea BP ccs eee mayee ee er fg te ii ee 4” a Oe ar ee a bs | Hee Sd es eg eee Yp | a Fg ON ‘ 5.0 rhe Sach v4 4 1P.-22 5 ee ceed Pyte -T" -" in ' 9 Ra eitiog of aces % 1 O pa | P"s9.0 The, . 4 R= | ae - Be ‘ > - i A wn 21.4] -- a a aS a JP:30.4 } d | yw | e +—-———- - 3.2)~——-3 4-4 rs ; ry [AY Pp. oe or , be ! 7 Zo t— ee wee ee - “4 - — — — y Sie “ | : Algebraic Solution R= 13.2 +5.0+3.2 =21.4 Tons. Rx«m=Px1"+ Pox2.8+ P3x6.8" Center of moments in PR" M=Rxize"=21.4x17e's 36.8 | . 4) eo = = == 5.4 4 --- 2 -o -- 4 A Z | M=H*y=20x1.85=37.0in-Tons. ~>._ ~ aus V \ a Rx3i=P'xez, P3027. P"=30.2-21.4=8.6T. rae coer all © 7 a 2] 5 10 a a a i ) - -—"R Scale 1"s5Tons. ,7 soe - ae aan Pr ah Pa ot Sadi eet -_ p V2 hata ee oe P, ooo _b ORE eer i a} ateker wane he eeoiaaty girtecvaess 1 dR Graphic Solution >. Ree.atons. me=e23" ~s_ i P'=30.4T. P"=o.0T: “>s. RY . . . . » - " M=2.2e8 in-Tons. 421 422 PROBLEMS PROBLEM 5. CENTER OF GRAVITY OF AN AREA. (a) Problem.—Find the center of gravity of the given figure about the X- and Y-axes by graphics. Give the co-ordinates of the C. G. referred to O as the origin. Show all force and equilibrium poly- gons. spat e the algebraic cainare stating all equations. Scale of figure, 1”==1”. Scale of forces, 1” —=1 sq. in. (b) Methods.—Start force polygon (b) at point (2.9”, 8.8” ' and take pole at (5.6”, 5.4”). Start force polygon (c) at (6.9, 0.6”), and take pole at (3.25’, 2.8”). In the algebraic check take moments about the left-hand edge and the lower edge of the figure. (c) Results.—The center of gravity of the figure will come at the intersection of the resultants R and R’, which is at the center of — a area. The areas P,, P,, and P,, may be taken as acting at any angle, but maximum accuracy is attained when the forces are assumed — as acting at right angles. If the figure has an axis of symmetry (an axis such that every point on one side of the axis has a corresponding point on the other side at the same distance from the axis) but one force and equilibrium polygon is required. PROBLEM 5a. CENTER OF GRAVITY OF AN AREA. (a) Problem.—Find the center of gravity of a 6” 4” X 1” angle with the long leg vertical and short leg to the right about the X- and Y-axes by graphics. Give the co-ordinates of the C. G. re- ferred to O as the origin. Show all force ‘and equilibrium polygons. Check by the algebraic method stating all equations. Scale of figure, "= 2". Scale of forces, 1-e=2 64.18. 423 n a oot i = ) 4 oa (rn. [ “sy A *. » . "eet “oO a = —o os -—-~- > -~W 3 c A Wee \— os ar a Z a 77 4 are \ H 4 “y 5 Los Ss ; ( mS 2 , S Ro ee Sake GE ibn ~ / p. Lielanan hes he wee er, aN pas ' Fi / eS aS y. | UAC ee of sf } Y é 7 a= L 18) ' 1 ; Ora Te RS < ae ; ! ; = nw oe SPF eee =; \ a rn) a ' - Fe ens HT I cee tiknay \ ' Oo Tec { pe : ori R ' ! 3 oe ie | 44 al ii ra Le : ms ‘6 a 1 * | = : : ‘ : | Pd I | oH \ = = ” * : - 7 9 wu \ a rt sae 0} \. ! x x ARs le ae ot ies 8 ea a o oO! 7 1 Ace ee xk vy) 5 ee ° a rg i> ! ' vulf a a 3 on! K- yj < wus + ¢ cot | I | te Se S 8 rio ln eet eS Rvaie tsar oO. Se ge . \ Ly + | 3 / =i. aly als The det ti tai [Ot am | ab a ol PVE? Odie 4 fas eek cone ee eer ae ee a re) : Or.ls I ry\ t 1% = ©) oy . ure: aS Ne se ee eG «Set os ° ie \ I-_- © tu « S <1) be Pa : rl uate “ \ \ o a iy ue 'S) = 2 “6 ied ~ - i - & Fee 5 tee Pega Io OR age 0) A o 1 U0 — t = S as a i eSariags. | Mot Po es SX |< a - — \ 7 Te - . 5 Re ee es Se ES SS > gs g . oN o Wooa J Ne | ° a) pe (Ca ea > o = ID . ee ee i 424 PROBLEMS PROBLEM 6. MOMENT OF INERTIA OF AN AREA. (a) Problem. Calculate the moment of ineltia, J, of a standard 9” [@ 13.25 lbs., about an axis through its center at right angles to the web: (1) By Culmann’s approximate method; (2) by Mohr’s approximate method; (3) by the algebraic method. Omit the fillets. Scale of channel, 1”==1”. Scale of forces, 1’’=1 sq. in. (b) Methods.—Divide the channel into convenient sections and consider the areas as forces acting through their centers of gravity. (1) Culmann’s method (Fig. 22). Start force polygon (a) at (3.5”, 9.1”), and take pole at (5.45”, 4.6”). Draw equilibrium polygon (b). Now with intercepts a—b, b-c, c-d, d—e, e-f, f-g, g—-h, h-i, as forces, and a new pole at (4.5”, 0.1”) construct equilibrium polygon (d). The moment of inertia is (approximately) J—=H X H’Xy. (2) Mohr’s method (Fig. 23). Calculate the area of the equilibrium polygon (b) by means of the planimeter or by dividing it into triangles and (ap- proximately) J area equilibrium polygon (b) X 2H. If the area is divided into an infinite number of sections, or if the true curve of equilibrium be drawn through the points determined, this method gives the true value of J. (3) Algebraic method. The moment of inertia about the center line is J=—=J'+ Ad*?+2I]” where J’=moment of inertia of the main rectangle ; A = area of the two flanges ; d = distance of the center of gravity of the flanges from the center line; and J” = moment of inertia of each flange about an axis through its center of gravity parallel to the center line. (c) Results——The algebraic method gives the true value of J; d Mohr’s method gives a value more nearly correct than Culmann’s | method, as would have been expected. The values of J given in the various hand-books are calculated by the algebraic method. PropLeM 6a. Moment or INERTIA OF AN AREA. (a) Problem.—Calculate the moment of inertia, J, of a standard 9” [@ 15 lbs., about an axis through its center at right angles to the web: (1) By Culmann’s approximate method; (2) by Mohr’s approximate method; (3) by the algebraic method. Omit the fillets. Scale of channel, 1’ 1’. Scale of forces, 1” —=1 sq. in. PROBLEMS ~ ep ~- a | BS He i y =e i oo ~~. = —o S ~ Co ae Fa 6 " | - ~ S ‘Say v s ¢ - + aw ——— —- bate Se 7 — Vv) — | Vv a ca a Di ad age fe) n»~ ge 0 1 ~7-= OF aoe GS —< -Or*- ! hone t= "J ~~. iw eka | | 1 i] ~—- —_~_-O*~ ~ ft eRe ay trad o cee Sat es, t . : = - . nr eee Set Sie ee En miele Cpe cla wee ee rH : ; Pe ah 1 ae eo. ee re ee a ee ae 2s 7 (e) 1 wens Mean to as ae em om ae e 1 l 1 wo dn te sro c= —! 1 eer ann Bn om pom ee Re ! geen ee cree hg ne k- > ar eo bea | PR he ~-"-t >4<- ' baht} -~ ' a . ieee fe carel ate soa ere —. he pete -- - ad hin. VW — e oa ae“ ! oO | ! ae ! iy Oe nein eae OD: Jee Te -" 1 Ft od — 4 Te atl at pe Be ee Sap sacaih ls tae on - — —— = — - Se | ied pe pee ast eye) “Shu - - > i ' Sree Fed Se re ee BS eg aay Ol a a Z o- — ~ 6A Ea geert i as Me reser Pa cosh A eg RO ee ie ian L ! a a+ eau eee Eos eabaciaipcth” (ite a hed fe ee _ = _ Saas Ts et we oS ! Vv 1 oO =e A eas 1 P3=.345 as -H a -5* 2,32> 46/98 I= 45% 4 Py=.345 7.345 Ps T= 5.234*9 =47.1 345 eee e Pag vi fl cy = yoo + here S ¢t S ws pes O 3 2) al'6 5 YQ cloud ‘= | wD) a | © oO -4 Or > - -- - Pee BARE ope EEE +O+-——> - -- - -- P) I=Avea Lquil.Poly.(b) x28 f 426 PROBLEMS PROBLEM 7. CONSTRUCTION OF AN INERTIA ELLIPSE AND AN INERTIA CIRCLE. (a) Problem.—Given the following data for an angle 7” 314” xX 1"; A=9.50 sq. in., 1, = 7.53'", [p= 45.37'*3 1 = 0.80"; 1 = 2.19” 3 75 == 2:19” ; tan-a == 0.241; CoG, (2:71", 0.00"); see Canamees pp. 176, 177. (1) Construct the inertia ellipse. (2) Construct the inertia circle. Omit the fillets. Scale of the angle, 1” = 1”. (b) Methods. (1) Jnertia Ellipse—Construct angle a, tan a= 0.241; and draw axes 3-3 and 4-4, which are the principal axes of the inertia ellipse. Calculate r, froni the relation /,-+/,—IJ,-+J1,, from which 72+ 7,2=r7,? +72, and r,==2.25”. Construct the enclosing rectangle of the ellipse on the axes 3-3 and 4-4, and inscribe an ellipse in this rectangle; this ellipse is the central inertia ellipse. Calculate Z,, from the relation Z,,—=A, Xh, Xk, +4.Xh, x k,. Also calculate c, and c, from the relation Z,,—= Ac,r, = Ac,r;. Compare the calculated values of c, and c, with the scaled values on the ellipse. Note that c, and c, are zero for the principal axes. (2) Inertia Circle.—Calculate the product of inertia, Z,_,—=—9.67. From any given point, a, lay off J, = 7.53 to the left extending to B, lay off J, = 45.37 to the right from b, and extending to c. At a erect a perpendicular a-d = Z,,—=— 9.67. Then with center O, midway between a and c, and with a radius O-d describe a circle, which will be the inertia circle. A line drawn through d and e will be parallel to the principal axis 4-4, and the diameter of the inertia circle will be the maximum value of J,. (c) Results.—(1) The inertia ellipse drawn is the central ellipse of inertia, and is the smallest ellipse that can be drawn. The radii of gyration about any axis can be found directly from the inertia ellipse. (2) The moments of inertia about any axis can be found directly from the circle of inertia. PROBLEM 7a. CONSTRUCTION OF AN INERTIA ELLIPSE AND AN INERTIA CIRCLE. | (a) Problem.—Given the data for an angle 7” X 3144" & %"; see Cambria, pp. 176, 177. (1) Construct the inertia ellipse. (2) Construct the inertia circle. Omit the fillets. Scale of the angle, had —y", PROBLEMS Graphic Statics. \! Problem 7: yu “ L7"x36 xl. A=9.505" I,=7.53 r, 70.869. Te=45.37 222.19. =< ! t 1 1 ! ' I ! 1 | 1 1 ! l l | Inertia Ellipse tan &=0.241.. ¥3=0.74, Product of Inertia, Zi-2. Zi-2=Ayxhyxk +Aaxhaxke ! = 35 X.79X-2.21+ 6X-.46 x 1.20 +.19% | =-9.67. C.6.=(.96,2.71) Inertia Circle. — om oe oe ey is | ie o phe @ Sona \ . ir su EY % ‘ on ie: — ee agape ge Ah 70 o@ Z-2=Ac,re=Acan. ae eee : |! 3\ “ Scale of Angle Natural Size. 427 428 PROBLEMS se PropLEM 8. STRESSES IN A Roor Truss By GRAPHIC AND ALGEBRAIC RESOLUTION. (a) Problem.—Given a Fink truss, span 40’-0”, pitch 30° ; trusses spaced 12’-0”; load 40 lbs. per sq. ft. of horizontal projection. Calcu- late the reactions by means of force and equilibrium polygons. Cal- culate the stresses by graphic resolution, and check by algebraic resolu- tion. Scale of truss, 1’ —8’-o”. Scale of loads, 1 = 6000 Ibs. (b) Methods.—Start force polygon at (6.25”, 5.6”). Lay off the loads in the order P,, P,, P’,;, from the top downward. Construct a force polygon, and draw an equilibrium polygon as in Fig. 14, and calculate the reactions R, and R, by means of the closing line as in Fig. 15. Construct stress diagram beginning at L, and analyzing the joints in the order, L,, U,, L,, U2, etc., checking at L’,. Arrows acting toward joints in the truss and toward the ends of the lines in the stress diagram indicate compression, while arrows acting away from the joints and ends of lines respectively, indicate tension. Use one arrow in the stress diagram the first time a force is used, and two arrows the second time. In algebraic resolution the sum of the hori- zontal components at any joint are placed equal to zero, and the sum of the vertical components are placed equal to zero, and the solution — of these two sets of equations gives the required stresses. (c) Results.—The top chord is in compression, while the bottom chord is in tension. In the Fink truss it will be seen that the long web members are in tension, while the short web members are in com- pression. This makes the truss a very economical one. PROBLEM 8a. STRESSES IN A Roor Truss By GRAPHIC AND ALGEBRAIC RESOLUTION. (a) Problem.—Given a Fink truss, span 40’-0”, pitch 14 ; trusses spaced 14’—0”; load 40 Ibs. per sq. ft. of horizontal projection. Calcu- late the reactions by means of force and equilibrium polygons. Cal- culate the stresses by graphic resolution, and check by algebraic resolu- tion. Scale of truss, 1’ 8’-o”. Scale of loads, 1” = 5000 lbs. worn PROBLEMS Problem 8. Sie of Truss 1I"=8*o" Scale of Stress Diagram® I"=6000# Span 40-0’. Trusses \2'o'opart. _ Load=40*per sq.ft. cat horizontal proj. _ -— _ _ —_- ae —— Algebraic Resolution. f she =|-Y¥+\-X Cos 30°%0, ZV= R,-1-X Sin 30%o, ret OX- ¥--t- KY=-12470*% X= +14400* . - ZH=\-Xcos 30°-2-Xcos 30°- I-2-c05 60°= 0, ZV= 4800+2-Xsin30~-lX sin 30°-}-2 singo°=o. > U, Abe =4160% “2-X% =12000% ZH=-I-Y+3-Y+l-2c0s 60% 2-3 cos60°=0. ZV=-l2 sineo °4+2-3 sin 60°%0. “-2-3=-4160" -- 3-Y=-8310* Summar Algebraic.lX=+14400, 52 Ket 12006, FY=-12470. 3-Y:-e316" re: 2-3: 41608 Graphic. FX=+14400 3 .2-X=+ 12000, HY:-12466, 3-Y=-6306, 2=2-3=4200" Li 430 PROBLEMS ProBLEM 9. DEAD LoapD STRESSES IN A TRIANGULAR TRUSS BY GRAPHIC RESOLUTION. (a) Problem.—Given a triangular truss, span 60’-0”; pitch 1%}; camber of the bottom chord 3’-0”; trusses spaced 14’—0”; load 40 Ibs. per sq. ft. of horizontal projection. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1’ == 10'-0”. Scale of loads, 1” = 5000 lbs. (b) Methods.—Start the truss at (0.75”, 7.0"). Start the stress diagram at (6.75”, 6.25’). Calculate the stresses beginning at R,, as described in Fig. 27, using care to analyze each joint beta pro- ceeding to the next. Check at R,. (c) Results.—The upper chord is in compression while the lower chord is in tension. The vertical web members are in compression while the inclined web members are in tension for dead loads. As a check the points 2, 4, 6, should be in a straight line. The partial loads coming on the reactions (not shown) are not considered as they have no effect on the stresses in the truss. This truss is a triangular Pratt and is quite economical, but is somewhat more expensive in material and labor than the Fink truss. This type of truss is much used for com- bination trusses, in which the tension members are made of iron’ or steel, while the compression members are made of timber. The dead joint load will be equal to the horizontal projection of the area sup- ported by a panel point, multiplied by the dead load per square foot, is equal to 12 X 10 X 40 = 4800 lbs. PROBLEM Qa. DEAD Loap Srresses IN A T RIANGULAR Truss” BY GRAPHIC RESOLUTION. (a) Problem.—Given a triangular truss, span 60’-0”; pitch 4%; camber of the bottom chord 2’-0”; trusses spaced 16’-0”; load 40 lbs. per sq. ft. of horizontal projection. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1” 10-0”. Scale of loads, 1” = 5000 lbs. 431 PROBLEMS S * eas 8 2 :" «x? Se gre 5 5™ 3 f ® 3 ong A fy ss D fe) * ° --+-g-- 9 1 oO '/! x 5 aes | ie ae ae a igs ; + oy ats oe sf [" J aa * a Qa; 0 \ ie oc a ea 9 £ * ‘8 ae “3 6! \ s * 7 te i: Be -! en. SS ! # b $: , % + gh 8 | Ooaprayw mg ye 3 oly 9 ' e792" [4 it a vy} Sin § YQ, 4 S. - lon O we > .2) i) ri] x! fu © Wid ea | pos - Vv) i @} Ou! vw xy Pinca a Hee v Bt S! + ike re =| 9/2 e ry) Ss eee eas 3 § AF. 5 432 PROBLEMS PROBLEM 10. WIND LoApD STRESSES IN A TRIANGULAR TRUSS—No ROLLERS—BY GRAPHIC RESOLUTION. (a) Problem.—Given a triangular truss, span 60’-o”, pitch ¥%, camber of lower chord 3’—-0”, trusses spaced 14’-0”, wind load normal component of a horizontal wind load of 30 lbs. per sq. ft., no rollers. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Take P as 3300 lbs. Scale of truss, 1” ==10'-0”. Scale of loads, 1’ = 3000 lbs. (b) Methods.—Start stress diagram at (4.65”, 6.6”). The reac- tions will be parallel to each other and to the resultant of the external loads. The equilibrium polygon may be started at any convenient point in one reaction, closing up on the other one. The closing line of the equilibrium polygon will always have its end in the reactions. The calculation of stresses is begun at R,, and is checked up at R,. (c) Results.—The stresses are of the same kind in the chords as for dead loads as given in Problem 9, while the webs on the leeward side are not stressed. The load P, has no effect on the stresses in the - truss. Calculate the vertical component of the wind load by means of Duchemin’s formula (5), as plotted in Fig. 6 (page 15). The normal wind joint load will be equal to 14 &X 9 X 26== 3276, which is taken as 33v0 lbs. For a discussion on the different conditions of the ends of trusses, see Chapter VII. PROBLEM 10a. WIND LoaAD STRESSES IN A TRIANGULAR TRuss—No ROLLERS—BY GRAPHIC RESOLUTION. (a) Problem.—Given a triangular truss, span 60’-0”, pitch ¥%, camber of lower chord 2’-o0”, trusses spaced 16’-0’, wind load normal component of a horizontal wind load of 30 lbs. per sq. ft., no rollers. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Take P as 3800 Ibs. Scale of truss, 1’ ==10'-0". Scale of loads, 1’ = 3000 lbs. PROBLEMS A Pratt Truss. 6 Wind Load. No Rollers. Trusses spaced 14‘0'c-toc. Problem |0. L Scale 1"=3000* Horizontal Wind Load, 30 per. sq. ft. Normal Comp. =26"p. sq.ft. (from Diagram, Fig.6, for Duchemin's Formula). 433 434 PROBLEMS PROBLEM II. DEAD LOAD STRESSES IN A FINK Truss BY GRAPHIC RESOLUTION. (a) Problem.—Given a Fink truss, span 60’-0”, pitch 14, camber of lower truss 3’—-0”, trusses spaced 14’—-0”, load 40 lbs. per sq. ft. of horizontal projection. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1’ == 10'-0”. Scale of loads, 1’ = 5000 lbs. (b) Methods.—Start the truss at (0.75”, 7.0”). Start the stress diagram at (6.75’, 6.25”). Calculate the stresses as described in Figs. 30 and 34, replacing the members 4—5 and 5-6, temporarily by the dotted member shown. The stress diagram is then carried through to the point 7, and then the stresses in members 5-6 and 4-5 are easily obtained. Carry the stress diagram through and check at R,. (c) Results——The upper chord is in compression and the lower chord is in tension, the stresses being practically the same as in the triangular truss in Problem 9g. In the webs it will be seen that the long members are in tension, while the short members are in com- pression. The loads coming on the reaction are not considered, as they have no effect on the stresses in the truss, as can be seen by comparing with the truss in Fig. 30. As a check the points 1, 2, 5, 6, in the stress diagram should be in a straight line. The dead joint load will be equal to the horizontal projection of the area supported by a panel point, multiplied by the dead load per square foot, is equal to 14 K7% X 40 = 4200 lbs. PROBLEM Ila. DEAD LOAD STRESSES IN A FINK Truss By GRAPHIC RESOLUTION. (a) Problem.—Given a Fink truss, span 60’-0”, pitch 14, camber of lower truss 2’-0”, trusses spaced 16’-0”, load 4o Ibs. per sq. ft. of horizontal projection. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1’’"==10'-0". Scale of loads, 1’ = 5000 lbs. } . re ee ee ’ ee ween tae tet ee = 4 Ss pes a ae i Lh eet P - eal th aed Se ee b. : ba he , PROBLEMS #Ol : # 1) Py Graphic Statics. offs oy. Problem ll. >, A ae OF a ag # Pp Te fe} 3 5 ve XK O| Fs r e/a \! d SESS St Sy g oc e (| 6 e P. ® lp t| 22-57, £43500 ZrN, e Scale Ye) 2, : Lh a" “9 ons x -41PP Fk 1 ; ,\ © F \ =10 QO. #350044 peri : Sa BOO - 24 ie SRE t R,= 14700" | 56 ; \' LK o* 5000 Oe aes Ga Trusses spaced 14'c.t0 c. Fink Truss. “6 Load=40* per sq.ft. hor. proj. Dead Loads. P=40x14 x= a200% v 436 : PROBLEMS PROBLEM I2. WIND LOAD STRESSES IN A FINK TRusS—ROLLERS LEE- WARD—BY GRAPHIC RESOLUTION. (a) Problem.—Given the same truss as in Problem 11. Wind load to be the normal component of a horizontal wind load of 30 lbs. per sq. ft. The truss is assumed to have frictionless rollers under the lee- ward side. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1” =10'-0”". Scale of loads, 1’ = 3000 lbs. (b) Methods.—Start stress diagram at (4.75”, 6.55”). The reaction R, will be vertical, while the direction of R, will be unknown. Use the method of calculating the reactions described on page 51, and in Fig. 33; noting that the vertical components of the reactions are independent of the conditions of the ends of the truss. In calculating the stresses the ambiguity of stresses at point 3-4-7-4-y is removed by substituting the dotted member shown, for members “4-5 and 5-6. The calculation of the stresses is begun at R,, and is checked up at R,. (c) Results.—The load P, has no effect on the stresses in the truss. The stresses in the members are of the same kind as for dead loads as given in Problem 11, except that there are no stresses in the web members on the leeward side. PROBLEM 12a. Wunp LoaAp STRESSES IN A FINK TRUSS—ROLLERS LEEWARD—BY GRAPHIC RESOLUTION. (a) Problem.—Given the same truss as in Problem 11a. Wind load to be the normal component of a horizontal wind load of 30 Ibs. per sq. ft. The truss is assumed to have frictionless rollers under the leeward side. Calculate the reactions by means of force and equili- brium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1” == 10-0". Scale of loads, 1’ = 3000 lbs. PROBLEMS Graphic Statics. Xp 5 Problem (2. oS ad! * 7 por eee pet ti Scale \"=10-0. Same truss as in Problem l\\. S, one ane én om ee an owed oie ee Horizontal Wind Load = 30* per sq.ft. “Normal_Comp.=26"p. sa. ft. P= ia BOa208 =x 2623309" 6 (approx.) Rollers Leeward. pes ly ee a ae et ee. ee, : ; A be ae ne ie aa i Wl a Si | 8 “seni, ta, «rte ©. ~ - <<. - e - true for Fink 1 trusses simply supported. 438 PROBLEMS PROBLEM 13. WIND LOAD STRESSES IN A Fink TRruss—ROLLERS 0 THE WINDWARD SIDE—BY GRAPHIC RESOLUTION. (a) Problem.—Given the same truss and wind load as in P: lem 12, and with rollers under the windward side. Calculate the r tions by means of force and equilibrium polygons. Calculate stresses by graphic resolution. Scale of truss, 1” 10-0”. Scale loads, 1’ = 3000 lbs. | | a (b) Methods.—Start stress diagram at (4.6”, 6.7”). Reactic FR, will be vertical, while the direction of R, will be unknown, the onl ie known point in its line of action being at the right end of the truss. — Use the method for finding the reactions described on page 52 and it Fig. 34. In this solution the equilibrium polygon is started at right reaction, the only known point in R,, and the polygon is drawn. The intersection of FR, in the force polygon, and a line through” O parallel to the closing line is at Y, and R, is then determined in magni tude and direction. The stress diagram i is carried through and check at R,. (c) Results. —The ica P, must be considered as it produces — stresses in the truss. If R, coincides with the top chord there will — be no stresses in the other members of the truss on the leeward side; if line of action of R, passes outside and above the truss the lower — chord will be in tension; while.if the line of action is below the upper chord the lower chord will be in compression. These statements may — : be checked by taking moments about the upper peak of the truss. It — will be seen in Problems 11, 12, and 13 that there will be no reversal a of stress when the dead load and wind load stresses are combined. — This is commonly true for simple Fink trusses resting on walls; but am is not true for Fink trusses supported on columns, nor is it always PROBLEM 13a. Wrnp Loap Stresses IN A FINK Truss—ROLLERS oN THE WINDWARD SIDE—BY GRAPHIC RESOLUTION. = (a) Problem.—Given the same truss and wind load as in Probeat lem 12a, and with rollers under the windward side. Calculate the reac- — 3 tions by means of force and equilibrium polygons. Calculate i Fr stresses by graphic resolution. Scale of truss, 1”==10'-0". Scale of loads, 1 = 3000 lbs. PROBLEMS Problem \3. Same truss as in Problem Il. eo) 1600 ° F —aH—— = = = Aware is Ry a ay ARPS a a ae —— ee _ ae ae i i er \ \ | 1 ' 1 1 ‘ “eg isd i ger { 1 ' + Fa) 3000" 6000 l . ] J Horizontal Wind Load =30* ft. = per sq.ft. 6 Normal Component + =26"per sq, ft. 2 ZOT+20" x 26 =3300" P= 14x 439 440 PROBLEMS PROBLEM 14. WIND AND CEILING LoapD STRESSES IN AN UNSsyM- METRICAL TRUSS, By GRAPHIC RESOLUTION. (a) Problem.—Given the unsymmetrical triangular truss, span 50’-0”, height 16’-0”, rollers leeward, wind joint load 4000 Ibs., ceiling joint load 3000 Ibs. Calculate the stresses due to both systems of loading by graphic resolution. Scale of truss, 1’==8/-o’. Scale of loads, 1” = 4000 lbs. (b) Methods.—Calculate the reactions due to the wind loads, using the method of Fig. 33. Calculate the reactions due to ceiling loads. Then place the y point of the wind load line on the point separat- ing the ceiling load reactions (the x point). The wind loads are laid off in order downwards, while the ceiling loads are laid off in order upwards. The left reaction, F,, is the resultant of R,, and R,, while the right reac- tion, R,, is the resultant of R,y and R,»). The calculation of the stresses is begun at R,, as in Problem 10. The stresses are then calculated by passing to joint C,, then to joint P,, then to C,, etc., until the stress diagram is checked up at R,. (c) Results——The members 1-2 and 7-8 are simply hangers to carry the ceiling loads. The triangular truss in this problem is of the Howe type, the verticals being in tension, while the diagonal web mem- bers are in compression. This truss is expensive to build of iron or steel but is quite a satisfactory type where iron is expensive and wood is cheap, and is used for the struts. PROBLEM 14a. WIND AND CEILING Loap STRESSES IN AN UNSYM- METRICAL TRUSS, BY GRAPHIC RESOLUTION. (a) Problem.—Given the unsymmetrical triangular truss span 50-0”, height, 16’-0”, rollers windward, wind joint load 4000 Ibs., ceiling joint load 3000 Ibs. Calculate the stresses due to both systems of loading by graphic resolution. Scale of truss, 1”’==8’-0”. Scale of loads, 1’ = 4000 Ibs. i ie : poe Th ee eA eee ane = y , lhc es aS ee ol ae! | baie a i or ae ial a r Lore ial shy a — oe es " ¥ y a wi 6 i. ba eve , s ~ 3 a 5 a a a a a ee ee ye . ; os ' : ae aS ee ee a PROBLEMS Rollers Leeward. 44! 442 . PROBLEMS ProptEM 15. Deap Loap STRESSES IN A TRIANGULAR TRUSS BY GRAPHIC AND ALGEBRAIC MOMENTS. (a) Problem.—Given a triangular truss, span 60-0”, pitch a % trusses spaced 15’-0”, dead load 4o lbs. per sq. ft. of horizontal pro-— jection. Calculate the stresses by graphic moments. Check by calcu-— lating the stresses by algebraic moments, giving all equations. Scale of truss, 1’ 10-0”. Scale of loads, 1” = 10,000 Ibs. Use pole dis- tance H = 3000 lbs. | (b) Methods.—Use the methods for algebraic and eraphic mo- ; ments described in Fig. 28 and Fig. 29, respectively. Calculate all moment arms and check by scaling from the diagram. The pole dis- a tance is measured in pounds, while the intercepts are measured to the same scale as the truss. Take the section and choose the center of mo- ments so that but one unknown force will produce moments. Take é # the unknown external force as acting from the outside toward the cut section, the sign of the result if plus will indicate compression, if minus tension. Be careful to take forces on one side of the cut section only. (c) Results.—The kinds of stress in the members are the same as in Problem 9. The center of moments used in calculating each stress can be easily determined from the equations. The method of alge- braic moments is much used for calculating stresses in bridges, and other frameworks which carry moving loads. The method of a moments is used principally as an explanatory method, PRoBLEM 15a. Deap Loap STRESSES IN A TRIANGULAR TRUSS BY 7 -- GraAPHIC AND ALGEBRAIC MOMENTS. (a) Problem.—Given a triangular truss, span 60-0”, pitch 1%, trusses spaced 16’-0’, dead lead 4o Ibs. per sq. ft. of horizontal pro- — jection. Calculate the stresses by graphic moments. Check by calcu- lating the stresses by algebraic moments, giving all equations. Scale of truss, 1” = 10'-0”. Scale of loads, 1/’==10,000"lbs. ~Use pole dis- tance H = 3000 lbs. PROBLEMS Problem \5. o' 10! 20° the l J Scale t"=10-0o" Re= 15000 F, Bs nae nS petal ce Sis elegant?» Fe Span=60-0. Pitch=4 yang oe Ont asad ie cae Spacing =15* ee ¢. ob. =i5l----------- a4 P; Loading =40 “per sq.ft. hor. proj. Ones Basi cre P= SO*1bx40 . =6000* H=30000, 0 s._ PS ee ee = eRe as a vant e 7.08. Foutat 71 =16.64', tg pt C=4.48. Nie 6.94. Lisi caps ‘ Graphic. by: at = ~ 3430000: s56000*% |-K==2— = 5 x 38000 = 2%3Q000 _ # eos yi =a: 566 ~ =~ 8470. aie cea : F2=+6000" 3-4= a SRBC - 4 9000" 3-Y=-EEH -- 84309 OO --54000 Sd p * 4-5-2420 .- Sore =-10800. A-X = Yezt = 422000 - 26 800* 5-Y= im = ee =-1B8000' Algebraic. IHX = HOOS*IO 4 53200. |-Y=-W2EODAS = -30000% ~y_ 15000x20-6000%10 -2= 46000. 2-X=4+33200* 2-3 = 9000%9 2_-ga70. 3 7. Ye} Re vA. 6000 10+ 6000x20_ SE ROOR. A-X _15000%20-6000xi0 be zO 8.94 * A-X =+ 26820. 475 = CLOOMOS S000 X20 _ Og 00% 5-Y= 13000XK 30 - 6000x110 - 6000 x20 =-1g000° 15 444 PROBLEMS PROBLEM 16. DeEap Loap STRESSES IN A CANTILEVER TRUSS BY GRAPHIC RESOLUTION. (a) Problem.—Given the Fink cantilever truss, span 40’-0”, depth 12’-0”, joint load 2000 Ibs. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1” == 8’-0”. Scale of loads, 1” = 1500 lbs. (b) Methods.—Calculate the reactions as in Fig. 16, page 30, noting that the loads in the two cases are laid off in different order. Note that the two reactions and the resultant of the external loads meet in a point, and that the reactions can be determined by means of this principle. The stress diagram is started with the load P, and is closed at R, and R,. (c) Results.—The upper chord is in tension, while the bottom chord is in compression, which is the reverse of conditions in simple trusses. In calculating the stresses due to wind load in this truss, the reaction R, will be in the line of 7-4, and R, will pass through A, as in the case of dead loads. The resultant of the wind loads and the two reactions will meet in a point, and the solution is essentially the same as for dead loads. It should be noted that the closing line of the equilibrium polygon, A-a, has its ends on the line of action of the resultants R, and R,. PropL—EM 16a. Derap Loap STRESSES IN A CANTILEVER TRUSS BY GRAPHIC RESOLUTION. ; (a) Problem.—Given the Fink cantilever truss, span 50’-0”, depth 15’-0”, joint load 2500 lbs. Calculate the reactions by means of force and equilibrium polygons. Calculate the stresses by graphic resolution. Scale of truss, 1’ —=8’-0”. Scale of loads, 1’ = 2000 Ibs. * PROBLEMS Graphic Statics. o* { se toad Scate s"=1500* 6 300" 3000* Truss Scale "= 8-o: L i 445 446 PROBLEMS | PROBLEM 17. CALCULATION OF THE DEFLECTION OF A STEEL BEAM BY GRAPHICS. (a) Problem.—Given a 12” 1 @ 31% lbs. per foot, span 40’-0”, load 5000 lbs. applied 16-0” from the left support. J = 215.8 in.*. E = 28,000,000. Calculate the maximum deflection due to the load, and the maximum deflection under the load by the graphic method. Scale of beam, 1””=6'-0”. Scale of loads, 1’’==2000 lbs. Pole distance, H = 4000 lbs. Scale of areas, 1” = 60 sq. ft. Pole distance, H’ = 240 sq. ft. = (b) Methods.—Construct force polygon (a) and draw bending- moment polygon (b). Divide polygon (b) into segments, and assume that each area acts as a load through its center of gravity. Construct force polygon (c), and draw equilibrium polygon (d). Polygon (d) is a curve which has ordinates proportional to the true deflections. (c) Results.—The maximum deflection comes between the load and the center of the beam. If the area of the polygon (b) was meas- ured in square inches and the ordinates in (d) measured in inches the deflection would be A= y & H & H’+E TJ. In the problem this result must be multiplied by 1728. The closing lines of polygons (b) and (d) need not be horizontal. The solution given above may be very simply stated as follows: Construct the bending-moment polygon for the given loading on the beam. Load the beam with this bending- moment polygon, and with a force polygon having a pole distance equal to EI, construct an equilibrium polygon ; this polygon will be the elastic curve of the beam. It is not commonly convenient to use a pole dis- . tance equal to E J, and a pole distance H is used, where u H equals EJ. For a discussion of this subject see Chapter X Va. ’ PROBLEM 17a. CALCULATION OF THE DEFLECTION OF A STEEL BEAM BY GRAPHICS. (a) Problem.—Given a 12” I @ 31% lbs. per foot span 40’-0”, load 3000 Ibs. applied 16’-o” from the left support, and 3000 Ibs. applied 12’-o” from the right support. J == 215.8 in.*. E = 28,000,- 000. Calculate the maximum deflection due to the load, and the maximum deflection under the load by the graphic method. Scale of PROBLEMS 447 Graphic Statics. Problem \7. ‘ ‘ ‘ + " ° 6 pe Ps 5000" 12"I-Beam@231.5,40-0 span. ey are T=215.8 in* cae tle't-SsI5e © 2 re) 2000 4000 pe ' ’ v Poe Yi re | ge ree Eire ce Scale i" eae Te ! ‘ i | >it a 7 | --ha---ngt-fat- [adkot Sate tzeo0e7 fi 1 \ 4 P,! a 5! i Gos fame H Pee Se (a { ee ot Loe TS 1 [Por Qe lip Ta Wes Se a Wk DL OS eo “d _ 7 . ~~l " ae g wae t POL BS. Po | I 9F=-L 4 ATES Ran art eat aa oa pe Nile eS 5 8, ~_ PR Re | pos e ty tas ee Pan Get ie oe Be, oe 1 ! ' ! i eos a ! t | ! 1 | 1 | ! ’ ! 1 ! , 1 ! pee 1 1 \ " ) 1 ! ! 1 1 | | i ! ! t ! 1 ! 1 ; ; | i Peasy cee gy e H LA = — i oO?! 60" 20" 180"! Epi ET * L l | J SS een eee at 2 ie a aR eee Scale i"=60 sq.ft. pr ---- R ! na FORTE on Toe Se ea eed el i nae » i oF PRP iors, Coad 143 Oo ee lee ae SSR SPR ae = es Max A= ys Tree Saco ieee es Fe Ag 6.6% 4000xK240x1728 4 eee ar A= 2B,000,000* 215.8 =1.61. pe ie ek P, \ ek ie re A under load = vex ett xi728 (c) Sa >~ JP ~3< TP = 0:45 % 4000 X240x1728 _ 77" ~ p- 28,000,000 x 215.8 eh 10 beam, 1”—06'-o0”. Scale of loads, 1 Ud ’— 2000 lbs. Pole distance, H = 4000 lbs. Scale of areas, 1” = 60 sq. ft. Pole distance, H’ = 240 sq. ft. 448 PROBLEMS ProsBLEM 18. DEAD Loap STRESSES IN A WARREN TRUSS BY GRAPHIC — RESOLUTION. (a) Problem.—Given a Warren truss, span 120’-0”, panel length 20’-0”, depth 20’-o”, dead load 7oo lbs. per ft. per truss. Cl the dead load stresses by graphic resolution. Scale of truss, 1”—16’-0 Scale of loads, 1’ = 12,000 lbs. (b) Methods.—The loads beginning with the first load on the left are laid off from the bottom upwards. The calculation of the stresses is started at the left reaction, and the stress diagram is closed at the right reaction. For additional information on the solution see page 70. | (c) Results.—The top chord is in compression, the bottom chord is in tension; all web members leaning toward the center of the truss are in compression, while the web members leaning toward the abut- ments are in tension. All web members meeting on the unloaded chord (top chord) have stresses equal in amount but opposite in sign. The stresses in the lower chord are the arithmetical means of the stresses in the top chord. The Warren truss is commonly made of iron or steel, the most common section for the members being two angles placed back to back. PROBLEM I8a. DEAD LOAD SRESSES IN A WARREN TRUSS BY GRAPHIC RESOLUTION. (a) Problem.—Given a Warren truss, span 126’-0”, panel length 18’-0”, depth 20’-0”, dead load 700 Ibs. per ft. per truss. Calculate the dead load stresses by graphic resolution. Scale of truss, 1” = 15’-0”. Scale of loads, 1’ = 12,000 Ibs. | PROBLEMS Graphic Statics. Warren Truss. Graphic Resolution i] R,\=35000*% Dead Load 700* per ei foot per truss oO 16 Equal Joint. Loads snes, Truss Scal Problem (8. Span, L= 120'-o! Panel length, |,=20-0 Depth,d,=20'-o" 5 ae ax 4 re --- Be *K -_—_S Wigs e < o* \2000" Se ‘ | “e ocate OF Lane \"=12000* Dead Load Stress Diagram. Sell maw tenes ser tts 449 450 PROBLEMS PROBLEM 19. Derap Loap STRESSES IN A HowE Truss By GRAPHIC RESOLUTION. (a) Problem.—Given a Howe truss, span 160’-0”, panel length 20’-0”’, depth 24’—0’’, dead load 600 Ibs. per lineal foot of truss. Calcu- late the dead load stresses by graphic resolution. Scale of truss, 1’ = 25-0". Scale of loads, 1’ = 15,000 lbs. ; (b) Methods.—The loads beginning with the first load on the left are laid off from the bottom upwards. Calculate the stresses by graphic resolution, beginning at RK, and checking at R,, following the order shown in the stress diagram. (c) Results.—The top chord is in compression and the bottom chord is in tension as in the Warren truss. All inclined web members are in compression, while all vertical web members are in tension. The stresses in the verticals are equal to the vertical components of the diagonal members meeting them on the unloaded chord. Stresses in certain panels in top and bottom chords are equal. The Howe truss is commonly built with timber upper and lower chords and diagonal struts, the only iron being the vertical ties and cast iron angle blocks to take the bearing of the timber struts. This makes a very satisfactory truss and is quite economical where timber is cheap. PROBLEM 19a. Deap Loap Stresses IN A Howe Truss By GRAPHIC RESOLUTION. (a) Problem.—Given a Howe truss, span 162’-0”, panel length 18’-0”, depth 24’-0”, dead load 600 lbs. per lineal foot of truss. Calcu- late the dead load stresses by graphic resolution. Scale of truss, 1’ = 25’-0”. Scale of loads, 1’ == 15,000 lbs. PROBLEMS 451 Graphic Statics. Problem 1!9. Howe Truss. Graphic Resolution. | __,+35000 460000 +75000 +75000 +60000 en e i ° ' 4 Ay) 2 4 & 4 ry, 6.9 468 A S52 9 Xe, wt 2) ° tT oY 9 Q| 2 \ oo Noe © 8 39/5 8977 Hl 7 5G) 3 Ne NS — £55000]-50000 |-75008 |-B00001- BOOOOl=7 -75000 “69000-35000 Fy P, F Py Bm 1% $F $F Re «20'0"2 an per lineal ft. y Pil per truss. \ Joint Load, R= 12000% a ~~ Y! Span,L, =160'o" } Rr 2 Fanel\l, l,=BOtO.. 5 ay Depth,d,=24-o'y m af N=8. 3; | 3 4 4 site iy R, o! Lf rs yP2 i \ | v <~ | Dead Load Stress Diagram. 454 PROBLEMS PROBLEM 21. DEAD Loap STRESSES IN A CAMEL-BACK TRUSS BY GRAPHIC RESOLUTION. (a) Problem.—Given a camel-back (inclined Pratt) truss, span 160’-0”, panel length 20’-0”, depth at the hip 25’-0”, depth at the center 32’-0”, dead load 400 lbs. per lineal foot per truss. Calculate dead load stresses by graphic resolution. Scale of truss, 1” == 25’—0”". Scale of loads, 1’ = 10,000 lbs. | (b) Methods.—The loads beginning with the first load on the left are laid off from the bottom upwards. Calculate the stresses by graphic resolution, beginning at R, and checking at R,. Follow the order given in the stress diagram. (c) Results.—The top chord is in compression and the bottom chord is in tension. All inclined web members are in tension; while — part of the posts are in tension and part are in compression. Member I—2 is simply a hanger and is always in tension. This type of truss is quite generally used for steel and combination bridges for spans from ‘150 feet to 200 feet, and also for roof trusses for long span, where it is loaded on the top chord and bottom chord, or on the top chord alone. PROBLEM 21a. DerAp LoaAp STRESSES IN A CAMEL-BACK TRUSS BY GRAPHIC RESOLUTION. (a) Problem.—Given a camel-back (inclined Pratt) truss, span 180-0”, panel length 20’-o” (three panels with parallel chords), depth at the hip 25’-0”, depth at the center 32’-0”, dead load 400 Ibs. per lineal foot per truss. Calculate dead load stresses by graphic resolu- tion. Scale of truss, 1” == 25’-0”. Scale of loads, 1” = 12,000 Ibs. PROBLEMS 455 Sraphic Statics. Problem el. Camel Back Truss. Graphic Resolution. 2090. t40090 +40000 +35 ee ee ee ee ee ee ee ee ee eee Dead Load =400*per lin. ft. per truss,’ << x- Joint Load,R=s8000* ' Span,L,=i6oto" \F Panel,|,=20'-o" <> x k Algebraic Moments. = Center of Moments,A. |O F2x15.9+3B000x 19.3 =0. io -t=-3640*" iB 3-4 X32.0 +3000 X19.34 2000 X34.30, ele -4=-33950* yoo +Po+ Pat Py 7h SS 1 _ AR, 576 %48.5+3000%19.3+ 2000 %34.3 ; +2000%49,.3:0. 5-“G=-4650% ~£°5X34.343000x19.3+2000x34.3 =0, 2-Yx15.6-3000x15 #0 2-3=+369073 (b) 2-Y=+eee55-3-X. -~4-5xX 49,.343000X19.3+2000%34.3 A~YxX22.4-3000X30-2000x15=0 +2000x%49.3=0. 4-5=+4560* (<) 4-Y=+5360%=-5-X. -6-X X64.3+ 3000 X19.342000X 34.3 6-Yx 29.25-3000X45-2000x%30 +2000 X%490.3+R)X15+1000K64.3=0 -2000 X15 =O. 6-X=+62507% x(a) 6-Y=+7600. R ,X30+2000x15+2000 x 304300045 { =o. (h) Ry=+7500=-Re -“--Center of Moments. 457 INDEX. ———_——_———- Algebraic calculation of stresses—see Stresses, Allowable sections ..........ece00. 329 Anchorage of columns ............. 115 Anti-condensation roofing .......... 241 Arch—see Two-hinged and _ three- hinged arches. PM OROR TOOUNE 95. 65s die sk go a wate #4 258 Asphalt paint .......;. RDM TER Aiea 339 PRMIEGED TOOTING. oo Ss caine wid Sotienees 257 A. T. & S. F. Locomotive Shops ...367 Beams, Graphic Methods for calcula- ting the deflection of .......... 158 Bearing power of piles ............ 273 Bearing power of soils ............ 272 Bending moment in beams, 33, 56, 57, 59 OS Ce a ne ene er ae 185 RN SUR ats fei id os wits Cp bo 9s 0b Tee, vie 92 BSTICK( SICN. TOOTS = caidas. cece. mes 290 REPU” 1OOLS © si. oss ie ce bs Ate ae 290 Bridge trusses, stresses in CAIRGl: BSC 2505. co ans li aa Cerice 72 OO OSE La ee Ree te Poy eR ace een 69 MRE oa SWi's ig re ¥ ini acd sp. Behe eters 67 Also, see Stresses. Brown & Sharpe foundry .......... 184 * Buckeye” flooring « ..:. 206s weceues 204 Buckled plates ....... RT eae 296 ROT RI ecb sc vale bias a : - 333 Carey's: TOORNE oo 6 lec ca ees Whe ae ate 259 Center of gravity ..... SOS eee 34 CONSE U Ge BOOTS Fc oi a. .¥-< ess 8's 8 eS adie 282 Choice Of Sections s4.60 ssc vee 328 Circular ventilators ....... Pavan aeRO Cleaning the surface of steel ....... 336 Coai-tar paint 9.0. Carew eens 339 Columns . Design . 68645555 2.3 sa Mie eek Reape re o 217 Details ...205, 206, 207, 208, 209, 380 Pressure on masonry ............ 278 EVDER OF 57 6k Ps eae Ren ak 203 Combined. atresees (sic. osyc pes cats 143 Compression and cross bending ..... 145 Tension and cross bending ....... 148 Stress in bars due to weight ..... 149 Diagram for stress in bars due to WORE G ict Sails eh Saree wad wba earee 149 Concentrated moving loads ......... 61 Concentrated load shear ........... 56 Concentrated load stresses ......... 53 Coneréte. buildings” 22.62.01) 22. aes 268 COonerete G18D8 occ s cck swe hee 266 Conkey printing plant ............. 184 Continuous beans. os5 5 so deldd ead 164 Corrosion-of etee@l i i.:)..60.eGeeek ews k 330 Corrugated GOL ss a nee eke 295 . Corrugated steel Anti-condensation lining ......... 241 Corer: Giaial (5... circa vy ccctencws 234 Complete ya oc. cive oes cea ee 237-241 oe. ae Pare ee RA Shel ES Stay Bem 244 Weta re ees sci aie lat a . 232 Diagram for safe loads .......... 239 DIGS ac sels queens <3 Pawn Onn 233 Fastening ens. vse ac snes COP ery 229 Gutters and Conductors .......... 234 Pyesiea: OF oi. a higtes aie nes 235 PIGDEAN Kotte ic nae RA OP ee Fin rs 244 Plans for transformer ainneen 243 ROORRE Cuiia< ian edi deena a eee 246 Rotary shear for cutting .........227. Midge Toll ois ees hohe anaes Standard sheets .......seese0- «+5286 460 Steel lists for transformer build- Ea = a MS ee FEED Sd FN] OF APES SPE 243 Strength OF \.iieies ae ekes ee ee 230 TeReOf ee eae Pee eee kee ne 231 Weight 08 ies nates prawns 8, 225 Cost—see estimate of cost, and cost under different items. Pe SGimate WE Nee Ae lee laos 346, 349 Of miscellaneous materials ....... 355 Standard hardware lists ......... 357 Of matetias: 2s) dear. eceeee tte 351 Of mill details and shop work ....353 Crane: wivders’: ii ii.vel de ew vat Suen 224 DOAN GE lees a icici oa 380, 382 Deflection of beams ............... 158 Deformation diagram .............. 134 Design—see Article for which design is wanted. Details—see Article for which details are wanted. Diagram for stress in bars due to their own weight ............. 149 Diffusion of Tight: 6 asc s aie -299 Doors PARCIOG | cin ss 5.637% Ste ee ae ee 322 Wioden 555675 J soeaecc aeration 322 SANWICR | 25.2. Scere 323 POOL ean 3 'S5os cee cake SE 323 RIOCALIS: LOK oo Fa Sees Gein ee -325 MOORE OE 55 35 Eb Sark ea ee Roe 325 Draw bridges, reactions of ....... .169 Fastening corrugated steel ......... 227 POPTOINCAVve. 4). ca wee eee 260, 295 Finishing coat of paint ..........:. 338 BROOTE 6 ics OG a bX alate eae ee 281 BTC oe ies ch ee ae eee 284 PPI CR II Se ee Wee eee 290 Buckled plates (..0 525.2. 4 capieues 2096 “Buckeye” fireproof. ..........; -294 ARMOIE es Uh inthe eae bk gc ka ee 282 RAETUBAted sociales ee Cea 205 Corrugated iron arch ............ 2091 Expanded ‘metal 20.0202 i a 292 PETTOMCIAVE (2. gc dete ee eae 2905 PO) ee cc's ick ae Bek eeee 293 INDEX Multiplex steel ........... éé-0:si 0s a Steel plates. che wee > ks on ene Tar concrete.c..0 4 sea ee ose «ee Wooden 2. 6 ck.ws eee vane -282 Floor beam reaction, Maximum .... 81 Foundations . Bearing power of piles .......... 273 Bearing power of soils .......... 272 Design. of -footings =. 6225. daca 278 Pressure of walls on foundation ..275 Pressure of pier on foundation ...276 — Pressure of column on masonry ..278 Girders, ‘erane® v50.:264 Ss pean oe Girders, design of ...........00- ‘ear Glass Amount of light required ........310 Cosh GET. im Anivinis cate 304 Details of windows ......... Perit. y: Diffasion of Tight: ¢ ince vaca -299 Double glazing ........ 5 Scar aon teemeie +307 RinGe::O0) <7c.cauenen ee ene ek: 298 Factory sibbed: oii ceases . ++ +299 PRBES 3325 vA Geeks ae aie vee +299 PHO. 6. Ci ree beeen .298 PIG Sli. ocak ean oe 298 POTOINS 3g se eke es he sonata , +299 BODOG ois 6s 5h cin Gk em cecerah eet . 299 WERE oe 5. Sona 3 otal v va ecole Re - 299 WENGOW 25. Se ite 55 a ee 298 Relative value of different kinds ..299 Placing the glass: ss.2 6033 ewicce eee 302 Sizé-OF Ga oe sae cee 304 Window: shades 3); 6... 5... ssa 303 Government Building, St. Louis Ex- POSTION: © 45 5 s)aio tis ne eee ae - 385 Granite roofitig \)ssi3eF ccs eae 259 Graphic equation of elastic curve ...158 Gravel: foofing.° cs... Aiea ea cn eee .254 Ground oote <6 f5i5 cen ees cee amen Hardware lists ....... iS scene eee Influence diagram. ....0..e0+8000085 ks Iron oxide: 2.56 Geta ee ava laias teehale +333 Iron, classification .......... Sep ee 352 Iron, corrugated—see Corrugated | steel. INDEX 461 BOBHG hrs heh ee Aik ore Oe CE 331 Reinforced concrete round-house BADBCEU) GIs Sines wes Sos eee ee 331 for Canadian Pacific R. R. ..... 388 Lists, standard hardware ...... 357, 358 st.-Lows Coliseum. i232 fs hes 362 Loads Steam Engineering Buildings BGR BadE Foe Sau etek see cla be 4 Brooklyn Navy Yard .......... 381 Weight of covering ........... 8 Steel Dome West Baden Hotel ...359 Weight of cranes .....6..0.... TG" AMISINE DAME 5.) en OS ce ar aes 334 Weise Of -sirte eo Funes ees 8 Modified saw-tooth roof ........... 186 Miscellaneous material ......... 19 Moment of inertia of areas ........ 38 Weight. of purlins: ..%....005.0. 8 Moment of inertia of forces ........ 35 Weight of structure ........... 9 Culmann’s:'method. 2.40 s.6es tiacs 35 Loads on simple roof trusses ... 47 Mohr's: method ..¢4;:. S32 58h cake 36 Concentrated live loads .......... 21 Moments RHIIWE DROME. oy cio eckn a wares 6 whe bes 10 AIQOVBIC. 6. Vedalss wares 44, 56, 57, 72 WT SODUM YS is shales pircka duces: 12 GYADBIC. So sik Bok 32; 46, 56,<57; 73 Miscellaneous loads ............. 17. Moments in beams Live: loads: on: floors... .....<...5. 17 Concentrated loads .............. 56 Locomotive shops Concentrated moving loads ...... 59 PAS? he Bis Css hae ues 197, 367 Maximum moment in a truss ..... 77 Oregon Short’ Line .......5..5.4; 195 Uniform: foarte i i krih ae yee 57 Philadelphia & Reading .......... 372 Uniform moving loads .......... 59 sk Ae DE MENS eke aelen be: 196 Monitor ventilators ................ 317 RIIOT | P AOING ei op aleur rea cwes 196 Moose Jaw Round-house ........... 388 Multiplex steel floor ............... 295 Methods of calculations Algebraic moments ........... Wap 8 eA, SNOOE lo as Wie es ween oe 331 Algebraic resolution .......... BO) GSe Ot Paths ce od hic Wiis zeae eek ne 330 Graphic moments ...........4. 46, 73 Graphic resolution ............ 42, 70 + Paint Also—see Stresses. Applying tho? soi soko ant ce ba 0X 335 Mill buildings Prep bos BG's artic vies os anew 339 General design of ............... 175 Carbon: Voce) < aticewetaieakiemien ss aan spectneations LOT 96.60 sakes ss 3901 Cement paint) sols ciee ve eee uses 339 POEM iy ed aha So esate We ka MO aioe ae I Cleaning the surface ............ 336 Masonry walls ........... rey: 3 Cial-tar pain’) oso ele oe ie ans 339 Masonry filled walls ........... 2 Come OE Sern soar uhoan oe sola, 336 Pts PORES Fig be a veewloky es I Covering: Capacity )).0062 0965 1c 334 Miscellaneous loads RTERING « CORD o 5h. cee alsiad Sa wk we 338 Live loads on floors ............. 17 AFORE VORIGE), Foe oa cere bere ie 333 Weight of electric cranes ........ 18 BT 5 RPE rs AR BO a Ber Pree i) ye 332 Weight of hand cranes .......... 18 LIRECO OLE oS en oda eee 331 Weight of miscellaneous material.. 19 Miting the ese eek hk wee ewes 334 Miscellaneous structures Oil: Gaia. hss Pe ie said eee 330 Fo Poe Ss GRODE” hs. ss, 197, 367 Peiming Get! so 56-05 6b. ese ine 337 Government Building, St. Louis PPOHOIVIONSS Fis oso ie oes oe ee kes 334 MORO OSON Sho oiclcp ers eo Rly ok 385 References on paint ............. 340 Philadelphia & Reading shops ....372 MRE iE Shs g dsl ae Reve ie gabe CRIES oe 333 462 INDEX Painting Roebling floor ....... ova Pattee Applying the paint .......... iva 98! Root, pitch Ors. eae Sa Cost 68 iG ek etn RS 337. Roof trusses—see Trusses. * Cleaning the surface ...... Pe upars 336 Stress in—see Stresses. Paneled “dG6t6 03566 62cb a Sts Cae 322 Roof coverings for railway buildings. 261 Philadelphia & Reading shops ...... 372 ~=Roofing, Pins, stresses in ........ vk eaae es 154 Asbestos ........ eRe tae ...185, 258 Pitch: of “20082 oi 6450). Nees te ee Re 191 Asphalt ~ 22 sscces eee ee -257 Pitelt Of 20NGO6 6 hha ys sla oe eee 192 Carey’S © ip sacecuiuss oe eee 259 Plate: gitderg i. fp wis tics hk Oe 221 Corrugated . steel” . 0... 646 0s ere. 246 Polygon Cost of (io) coeqaakta ees 246, 262 PQUs Orit 5:a 4 Sk a a aeons aw ag 26 Examples 62.50 /050)5 5 Va role 53d se 261 Foeeee ii saute eee eae 24 Ferroinclave: 6.5.5 «cis fanawe eee 260 Mometit 5c. ss eecsaccs 56, 57,.59, 61, 73 GrBnite 6s. be 3 0 bce eee 259, 262 GRR SEC ule sree nem 56, 57, 59, 61, 74 Gravel ig AEC hee a eee 254, 262 Portals Patoid os 3as3 os 6s dae lee <0‘) ne Anchorage of columns ........... II5 Ruaberaide ics a eee .» «250,. 262 Stresses in Slag fia a ee ee ee 256, 375 Continuous portals ........... -117 Sheet steel ...... MON iyo 253, 262 Double: portal: 7. o..n EC ae wate 307 DE hth 3 2S i a cts btn QI, 101 ROE: PORE a Wai as Sale Pik oe aut elbewewe 247 SOO) ROME eas sas hae ae eee 83 SURMES, - WAMGNEIAEN "<2? 5S ie wit av oew ob eiwaats, oe 256, 375 Wind load PONS SR eee as della be woke snes 10 Algebraic calculation .......... 84 APA POOH a 54s 5.5 any woes 2 261 Columns fixed si. Sn ake 87 Specifications for Steel Frame Mill Columns hinged ......... 84, 168 Buildings—Appendix I. ........ 3901 Graphic calculation Standard hardware lists ....... 357, 358 Caner els eee ae 96 Steam engineering buildings for CBS6 Bg carey sale wetaie eels 98 Brooklyn Navy Yard .......... 381 Cae 4c Re las OA aoe 99 Stress in bars due to weight ........ 149 TO ey ghee ee a eta ie 104 Stresses, allowable ........... 178, 214 Graphic calculation of reactions, 103 Also—see Appendix I. ........... 391 ‘Transverse bent with side sheds ....105 Stresses in Transverse bent with ventilator ....103 PRM OSS cs WAP d KAS Role eo 8 92 Trusses Bridge trusses—see Trusses, Portals—see Portals. Roof trusses—see Roof trusses. Transverse bent—see Transverse bent. Three-hinged hinged arch. Two-hinged arch—see Two-hinged arch—see Three- arch, Stresses Calculation of ....... a haeiels Wows. 22 Combined ....... Se ees BE re ere 144 Eeccentrie: 45506 26 0s: ae Petar se Spirits e00 Dracind: .6i6< 665. Pin heyy! 213 Tar concrete floors ....... eat ee bat 284 Three-hinged arch Calculation of stresses ..........- 120 Dea: 16065 collie eee nae ees sel 22 Reactions Algebraic method ...... éaaeae Graphic method .........6%.. 121 Wind load stresses ............125 Timber floors ....0:cessesc002205y 290 Te. FOGUUE sci cca ds ade bese ¥ 8040454250 Economic spacing of ............ 192 Pea OE eae pds we eee oe ed ee 214 PICtAUB OL esac cee’ 197, 199, 378 PHO OR Gan cer abien Me dae aes 190 Saw-tooth—see Saw-tooth roofs, LODE FE ot ve mee aw a cea 180, 190 Trusses, stresses Bridge trusses Algebraic moments ............ 72 Algebraic resolution ........ Pe Oe Graphic moments ....ccseseses 73 Graphic resolution ..........,. 70 Roof trusses Algebraic moments ............ 44 Algebraic resolution ........... 40 Graphic moments .......eese+ss 46 Graphic resolution ............ 42 Concentrated load ............. 53 Dead load ....... Siam gist < cares 47 Dead and ceiling load ......... 48 Sricw loait i... oo esas Sotho, 0 Wind load ....... Rvicnen cS 605° S2 Two-hinged arch Desig 0f -oecixs ven eke keckovae eee Two-hinged arch, stresses Calculation of reactions Algebraic solution .... Graphic solution ... Dead load issn 3s Dead and wind load ... Temperature stresses ... With horizontal tie..... Ventilators .. . . . . . . . . . . . . MONIOF i. ours oes Cost of SSAA EY Mee Circular i... sen oe Cost. OF sb eee oe Walls, masonry ..2.020354. Walls Bide 2s ai. ac ewss - ..128 fake . - 263 Weight of talking oat Weight of Bors Oe West Baden dome ...... Window shades . voeeteeee Windows — os Ses, Amount of Tight PRE RE Cost 08 ies ha 64 o a Details ee EE ets | Double Magee Glass—see Glass. : ie Wooden doors eames Wooden floors Paes Zinc paint | ae TABLES OF STRUCTURAL DETAILS. . PAGE. PMENSIONS CARNEGIE 3-BRAMS® ... 505 6a sc. oDeics Gece bd a he oes 2,3 SIIMENSIONS CARNEGIE CHANNELS % 54 05) scic ds se olds c gues ais 4 DIMENSIONS CARNEGIE Z-BARS .........0ce eee eeees seep ewes 5 Meets On CARNEGIE. ANGLES 40 oss ces vhs Biba skates PROS 6 AREAS OF CARNEGIE ANGLES ............- RMR eR Et fa Sree 7 UPSETS FOR ROUND AND SQUARE BARS ...........0cecceeecee 8 Me a ra eat a MGS Osa AK oa s4iec a o.0% Cada meen 9 Ee VE NITS'AND LURNBUCKLES 66.006 656563 60h bos dey eb wee 10 MRM SENNA 6 cRNA Sk GR > tae Sie 9 Aw Sh E RS 6B a REED ET REE II eed Beis er ee te es eink ig ara Ra PUCK oie a GLa 12 EGMOING INOMENTS IN- PINS 20). sic c css cs So's as ee cutee oe beens 13 PEE PAGE PAGE yogi tos 5 od va ve eee Vad ome awe meeed 14-15 AREAS TO BE DEDUCTED FOR Rivet HOLES ...... BR : 16 CARNEGIE I-BEAMS. 7 “1 bo Set Ie H a9 45 Geer Senfuine « 1 : H ; "ke é e WEIGH GAUGE ITANG'T} DIST. | GRip | MAX. o PER {FLANGE|WES River] 42 ost. | cist. I i | giti ki oj, 35 _ STANDARD FRAMING ale PER ; tNCHES | FOUNDS| INCHES [NCHES INCHES [INCHTS | INCHES | INCHES [INCHES | INCHES | me al - INCHES POUNDS 1000) 7] S14 j20e/ 15) % * 24” 521 3 ee as 96.0] 7-2, | 2 w fo] La < cite 24} 00.0] 7344 " vs lars pas iis ° esis ad < . > wy ak 1 |16)." eT PP e- 5] + | 0.01/04 85.0) Tis} ie] | ” = ye 5 &| & | ss.0 80.0} 7 | 4 " - | a a aS : od 7 : 3 15 ‘ i} Ab. © 5% i _{ 89.0 100.0) 7asj oj “ jiGsj1s] . & ik sd, 521 + loo.0 95.0 a% 3 : > Lp A + 4 # + 95.0} 90.0] 74428) » | « ‘ : ‘ i 96.01 7121 - | ~ x i a id fe ie 20 has] 2 | Btha'x ax Bx 6” wee s7® 5 ib] % | 85.0 80.0}7 Js} ~ | » Oe had fie 20 3 x ye ” + = | 80.0) 98.0) 6234 234 re bees : 18 and 20 xd " 0] Bax 3 27. 1.25 ‘4 © 1) 2 alo as ‘ 5 is] ws | 26.0 70.06%)” | - |” - — 5a} & 170.0 65.0] 62 | + Sal Be ;: 4 70.0} 6+ | 2% )/321i5+112]) + 5 i 3 +6 plea 40.! elarf{Sejl5ai;le}] oc Yh > pe OE Be Cy je | 70.0 1 65.0] 6is| yg = 16 : 2 8 SR Peed : 2 16 as oi i nee tis 5.01 18 } : a) heey” 2 2 60.0 55.016 |% a Tr as Wen Q] awa'xa’ 2, 1-3 wesr® fs! 2 | aso 100.0] 633 [1S] ° [it |2 |4 “s ata we] ie (100.0 95.0| els] ° | » =| 4 i 6ie| wo | 96.0 90.0) 6/5] ” ” £ % (90.0) | 85.0) 6 + " ” - ” 521+ | 85.0 80,0) 625 | ’ ‘ 15 ie : 80. — > = oe 0) 75.0 6s aa Fle 14 = ~e i Al ra at thd “ Sar TF OCR 5%] > | 75.0 185 | 2.0] 63| 3 ” » hs *Te fooly : 65.0] BS a ” * * hs. a a ro nu} 2 se * oy a : 5 is} ie | 65.0 a hig ee is es ernst 52] $ |eo0 b ” 85.0) 52/5) 3eli9t/13| &| * |], 280" a’ gy x 0-10" we. 27 Is | + | 55.0 60.0 5¢ zg . . oj] a ie 5is| = | 50.0 45.0] 522] is “ } 53] i | 450 42.0) 52/22] 2 | - ete i 5 a| + | 420 | si 12” 85.0) 5] i|35| o¢]13] 2 : == ws}, 4 % 2 . 5 te ‘3 55.0 50.0] 5 +| 3 : “ ws ptt bd Besides so u| 2 12 ae Pe ee B&B iio 1_* [st ae 2% wt 5 1s] i | 60.0 1 0" S| is “ > = mel ey ee eal 59] e138 a ate sr = 5 &| = | 45.0 2 40.0] 6242 Cee ae en : 6 4 | 56 ca rreteaee 54| % | 400 35.0) 5 3% Z ot+1%] 3 8 PE ee, a| 5 lag bee Pa Sea Nig 2 2-26"x a's Fx o-7k" we. 20% [5 | i | 85.0 . 32 . * ™ 3 5 = = $1.5 Wee we CARNEGIE I-BEAMS. STANDARD FRAMING . if. 2 2 : a} ey itam ef & ee | - 36. 4 ” staal td 97859; & 10° 21.0 4B ts ° ” Wess { « thf 55] — CARRE GRY ‘Siig ee on re HES ERS ; 2 os . 17.25 3%|%/2 |4$}$}° tp G 14.75) Siz] a > $ 6 12.25] Six| a} 14.75, 3%| 4/13/34] 21 2 5°26” 544 & (14.76 We sare » 2 ee nid bg i fm | a ekak ex 0-3"for 6” Wwe, 8 St] + asy & 9.73.3 | a 6 | Be » 0:22 5 We, y* [5d we 9.75 RES, 2 e) ape a 10.6) 23) ar] 1s] 27 ee — — Sis} We | 105 4 9.6] 233) 2] ° eo © | (if) fee) St} + | 96 4 8.5] 233) 2 ° de Bes we t1 | 36 Sie) a a | eo a Sis] ie | 2.6 2.5) 23) &|1+)1¢ . + 2-8 6x4 Jpxoc2"we. 6 5et i} 2.6} B65) a8) 5) - | z| 6 $] %| 05] 9 6.5) 2 is] a Sis] ie | 5.6 "AM rivets in standard framing angles are} diam, Weights of sé ** include weight of shop rivets only. ‘Whea beams frame ccs each other into another beam with web thickness less or where. beams of short span lengths are loaded to their full capacity, i ‘may be necessary to use framing angles of greater strength.fhan the standards. See table below for minimum span lengths, : : I weIany[Sran ON] & WeronT]eP ayy LT fwercnr)SPan int 7 WEIGHTSPAR Mi] T WEIGHT) a I MEIGHT]SPAR I weionteray 24|80.0/2207 } | 15|80.0}20.0 8 {180| 5.5 | & |9.75| 40 20|80.0|22.0] 28|55.0|/140} ~ 160.0] 15.5|22/40.0] 115.| 20/25.0] 9.0 | 7 115.0) 40 | 4 | 75 | 30. ed 65.0 /18.0 | : _ " [4207110 ft $15} 9.0 | 2 [e10] 7.0 | é 12.25} 60 |} 3 | 55] 20 CARNEGIE CHANNELS. ike _@ ih ey is WEIGHT j cauce | Tano’r. | cist. orp wa o1sT. WeOHT oy anid FLANGE [wee g t k d 7M c woot 55.00] Si |] 23 | 128 |] te] ¢ 3 |55.00 50.00] 8%} #] « “ “ “ & 150.00 w [45.00] 8& |] 2 e a 4 | 45.00 46 40.00] Sir [33] « ie (ie “ ¢ $ 140.00 16 35.00] 8% }%}] « “ “ ‘ $ {35.00 33.00} 83 }a3) « “ “ “ + $3.00 40.00} 38%]/%{ « | 10 + # |40.00 35.00} 8&{e)] « “ “ “ f {85.00 12 130.00] shi] wf « | « « | 2 & {80:00] 22 25.00} Sir }8} « “ “ “ & {25.00 20.50| 2% }% “ Se = + 20,50. 35.00] 38 j4s} « 8] $$] « $ 35.00 80.00] Sis jis} “ “ “ 4 |80.00 10 {25.00} 26 |S] « J « | w | we] 4 25.00] 70 20.00] 2343) 1] « “ ee % |20.00 16,00}; Met] “ foe | wy #% | 15.00 25.00] 3816] “« | wi] i #« 4 |25.00 20.00] 23/8) « “ “ ae $ [20.00 9 15.00] 2¢ [31 @m ] « “ “ $ $ 15.00 9 13.25} Qi feat“ 33 “ a & 13,25 2125} S23 i%F 1 | 64 « + $ 21.25 18.75} Q23h7} “ a “ “e & 18.75 8 fteas;- as }Bp os pe Pw fw te & }1625] 8 13.76] 291%] &} « iy “ § 13.76 11.25] 22 ig} « “ “ “ % 11.25 ‘|i0-75] 23 [¢] & F se] i $19.75 417.261 231%) « “ “ oo 2] 17.25 y 11475) Ah {e] ve tw fw fw f & $ |v5) 12.25] 2218) 2% | « “ “ 4 12.25 9.75] 2eidt « “a “ a + 9.75 19.50] 2% {8) « | ay] “ € [15.50 ~ 113.00} 2% |] @ “s “ “ a + 13.00 6 | i050 ah lal « f« fia do fb ® 4 110.50 6 8,00} 1244 e “ “ + 8.00 11.50] 2% {8} « 3+ i 4 | & $ 11.50 & | 9.00] 7B) « fe fw fa fg 4 1900] ¢ 6.50] 13 [ay * B ae z = 6.50. 7.25] 13 |i) 4 2 { 6] « 2 725 4 | 625) Wispo« f « “ o | + 2 625| 4 §:25] W{&] « “e “ “ Ss 5.25 600} 18 13) @4 &] & ] + & 6.00 2} 500) ts] « fw fw Pw | & 5.00} 3 400] weial « f« Jw fo & | 4200 CARNEGIE Z-BARS. aces uy ..... =o , 7 All dimension’ iu inches NOMINAL FriicKs on J ‘een | wea. aie : c ears — fruicxwess) NOMINAL 4 | x8 xo] 67 | 197 | & 3 2 ras re BR | 2¢x8hx22 | 84 | aas] « ” ‘s 5 8 2 | ates xo8] 07 | ase] « ot # Be Fy 2 w | 2x Qhx 23 | 14 | ase] « ‘2 % a Pa : 4 | gtx3 129%] 125 | 360] « bs i ‘ 4 f& | 2%*38%x22 | 142 | 418] « “ “ 8 4+ | shx4@ x8e | 82 | 241] 2 3 ZL 2 + & | skx4%x3¢ | 103 | 3034 « - ‘i , £ 3 8&x 43x 8% | 124 |. 3.66 “ “ a Fi 3 % | 8&x4 xs} 1a8 | 405] ™ “ ‘ Z 3 8$x4h43h | 158 | 466] « ef i 4 4 % | Sh 4hx8% | 179 | 527] « " “ " % 4 & | shx4 x8h{189 | 555] « ‘ ‘ ry & | Shx4ex33 | 209 | 614] »& “ “ w rt & | 3%. 44x38 | 229 | 675] « “ # ed a & 8i%5 x3t!/ 116 | 340 | 2¢ z % 2h 3 =a) § | S&x5hx3h | 189 | 410} « “ wi] 4 & | 38x5%x88] 164 | 481] « ‘a on ‘ em % | 83x65 282 41784 525] « “ te “ 4 & | 38x 5x35 | 202 | 604) « “ es Ne & & &§ | s825¢x98 } 226 | 66d] « = 7 ‘i rt 5 | Shx5 x8t] 237 |] G96] « " " “ 2, S | hx Shx Se | 260 | 7.644) « “ “ “ 3 % | 325t%3% | 289 | 833] » “ " “ 13 2 | 6x6 13%] w6 | 459] 2 g g 3 3 & | 826%. 3% | 193 | 530] « ” ‘ “ ‘A % | S8e6en 3% | a0 | B19] * " “ " 4 & | 326 «38 | aa7 | ess] « “ “ “ % & | g%.6be 881 254 | 746] « ss x - & 6G | B | s¢sots98 | 280 | 825] + " ‘ ai Be 4 | 93x06 233 ] 203 | sea] « “ w P a iy 3h « 6Hx 83 © 820 9A0 “ ‘ 1 ‘ 33 4 33» 6x 33 | 340 | 1017] « FA s 4 z CARNEGIE ANGLES. Weights in Pounds’ Per Lineal Foot. WEIGHTS OF ANGLES All dimenstons.in inches xe |elele|elelele|2|4|el+elslal 2 | alee] oe 8 «8 26.4 |29.5|32.7135.8|38.9| 42.0| 45.0] 48.0151.0154.0|56.9|8 x 8 6 «6: 14.8 |17.2 |19.6 |21.9|24.2126.5|28.7| 30.9] 33.1| 35.3/37.4 6':6 5 6 12.3 1143 |16.2 |18.1|20.0121.8|23.6| 25.4| 27.2/ 28.9] 30.6 Ss x 5 4x4 8.2 | 9.8 |11.3 12.8 |143} 15.7|17.1| 18.5| 19.9 4x4 8} x 3} 7.1 | 85 | 9.8 |11.1 |12.3] 13.6|148] 16.0] 17.1 Bix 3t 8x3 49|61| 72/83] 94/104/11.4 13.3 28x 23 45|55|6617.6| 85 32. 23 Ox 2 31140] 5.0159] 68| 7.7 2h x 23 a. 2a} {2813.71 45| 5.3] 61| 6s hx 2 2x2 25132140147] 53 22 tx 8 21/28/34|40| 46 fx 1 4+ 13] 12 |18]24] 29/34 tbs 13 tts 1) 10}15}]10]24 le If 14 2 1}o8}12}15 2 14 se |S lis|a |i) s|ie| > | i]s ldo] S| 8] F | | 2 | 24) 2d] 8 7 + Sb 15.0] 17.0| 19.0|21.0123.0|249]2a.8] 28.7}30.5|32.3] 7 «33 6.4 12.3 |14.3}16.3 | 18-3] 20.0]21.8|23.6| 25.4127.2/28.9/30.6 6x4 6 x 8} 11.7]13.5 |15.3 |17.1] 18.9|20.6] 22.3124.0/25.7/27.3| 28.9 6 x 3t eri 11.0}12.8/145 |16.2} 17.8] 19.5] 21.1|22.6 ae Sas 5 x 3} 8.7 |10.4]12.0| 13.6 |15.3 es 19.8|21.3|22:7 5 x BE 5 «3 8.2 | 9.8 |11.3 119.8 1142 | 15.7]17.1118.5] 19.9 5 x3 4» 3} 7.7 | 9.1|10.5 |11.9119.3 | m6} 15.9] 17.2} 18.5 14 x 33 4x8 ! 7.1185} 9.8 }11.1]12.3] 13.6] 148) 16.0} 17.1 {4 «3 3x 3 6.6 | 7.8} 9.1]10.2 /11.4 | 12.5] 13.6/14.7) 15.7 3 Bt « 2 49| 61] 72] 83] 9.4/10.4| 11.4] 12.4 St x 2H 3 x 2h 45|5.5|66| 76] 85] 95 3 x 2 3 «2 40| 5.0] 5.9} 68] 7.7 $32 ky 2 28/3.7|45|5.3| 61| 68 Ob. 2 Ps = wou | & |r| | fo || se | Fe |& aes ae] & | ae] 2 | 2 a8] oe Angles marked * are special CARNEGIE ANGLES. Areas in Square Inches. ANGLES Avea In equare inches, see | lig) F || Fhae|s jae] s 45) 5 1481 Ss S| 2 | ah 2d] sce 8.8 7:75| 8.68] 9.6110.5911.44'12.9413.2 sea pA ge wy «8 626 4.36| 5.06] 5.75| 6,43] 7.11|7.78| 8.441 9.09] 9.74110.37111.00 6 «6 825 3.61|4.18] 4.75|5.31|5.86| 6.43] 6.94| 7.46] 7.99] 8.50] 9.00 Ss 25 424 2,40] 2.86| 3.31| 3.75] 4.18] 4.61|5.03] 5.44] 5,84 4a4 Shs 3H 2.09|2.48|2.87|3.25| 3.62) 3.08|4:34| 460] 9.09} She $33 1.44] 1.78|2.1112.43]2.75|3.06| 3.36 323 ‘OP x 294 1.81] 1.62] 1.92] 2.22] 2.50 On. 2 2, 2: — |0,9011.19]1.47| 1.73] 2.00] 2.25 2k 23 2323] — |o.81}1.06] 1.31] 1.55] 1.78/2.00 Bt 252 0.72] 0,04] 1.15|1.36| 1.56 222 1 13] [0.62] 0.81/1.00]1.17] 1.30 ifs 14 « 13/0.3610,53]0.60| 0.8410.99 tks 1 1} x 13] 0.80] 0.43/0.56/0.69 i¢x i 1» 1]0.24/0.3410.44 ae ee welelelelslelals|elela| sla] e || a | onl ct] ] 4 3h 4.40|5.00| 5.5916.17| 6.75| 7.81] 7.87|8.42|8.97| 9.60 " T7 « 34 614 3.61 4.18] 4.75] 5.31|5.86| 6.41|6.94| 7,47] 7.99|8.50| 9.00 6x4 6 « 3h 3.42] 3.97| 4.50] 5.08|5.55| 6.06] 6.56] 7.06] 7.55]8.03] 8.50 6: 3} 6 54 3.23] 3.75|4.25|4.75|5.23] 5.72|6.19| 6.65] 7.11 6.4 5 «3 2.56| 3.05] 3.63] 4.00] 4.47| 4.9215.87| 5.81]6.25|6.67 5 + 3} 5 x8 2.40| 2.86] 3.31| 3.75|4.18|4.61|5.03|5.44| 5.84 5.3 % 2 3} 2.25] 2.67] 3.09] 3.50] 3.90] 4.30] 4.68] 5.06] 5.43 4 « 3} 4 +3 2,09] 2.48] 2.87] 3.25|3.62/ 3.98] 4.34] 4.69]5.03 43 33x 8 1.93] 2.80] 2.65] 3.00] 3.34/3.67/400]431]462 3i« 3 Shs 2 1.44] 1.78] 2.11| 2.43] 2.75|3.06|3.36] 3.65 Skx 24 3 x 2 1.31] 162]1.92)2.22/2.50/ 2.78 43 x 23 8223 1.19) 1.47) 1.73| 2.00} 2.25 3x2 hx 2 0.81) 1.06] 131] 1.55] 1,78] 2.00 ix 2 Sze | 3 ig | 4 Ey 3 pe = 16 7 a r % % || 2 | 1 15] sze Angles marked * are special. Upsets FOR RouND AND SQUARE Bars. ROUND O BARS SQUARE [2] BARS ROUND UPSET UPSET SQUARE : | re Pa ee Pr eel Giprenge Med ee oe eS ey INCHES | 8a.INS. | IncHEs | INCHES | tncHEs | sa.ins. | 0% ©7, | sais. | incwes | incnes | incHES | squns. | INCHES £ 10307; 4 | 4 4; | 0.420] 368 $ % 10.442! 1 4 33 | 0.550] 244] 206 | 0.694) 3; | 4 13 [0563] 2 2 joen| 4 | 4 5 | 0801] 483] 163 |o8e1| 4 4 14 |o.766} 1 |o7e5] 12 | 4 42 | 1057| 347 | 20.5 | 1205] 4 4 1: |1000] 7 14 | 0.994) 15 4 3; | 1295] 30.3] 10.7 | 1515| 43 43 1, | 1266) 72 22 |1.227] 1¢ | 43 3; | 15156] 23.5 | 311 | 2.0490] 44 | 4: | 12 1.563 1i HZ | 1485) 13 | 45 | 33 | 1744) 174] 217 | 2302] 43 | 5 2 | 1891] 72 12 | 1767) 2 5 4% | 2302] 30.3 | 340 | 3.023} 4$ | 5 2i | 2260] 72 12 | 2.074) 25 | 5 4: | 2.651] 278] 20.6 | 3.410} 4¢ 5% 2; | 2641) 7% WS | 2.405) 25 5 4 3.023] 25.7 | 213 | 3.716) 4: 53 2; |3063) 7% 2z {2.761} 23 | 53 | 43 | 3410] 239] 314°] 4610/ 53 | 6 2 |3.516 | 22 2 3142} 2: | 53 32 | 3.716] 183 | 27.7 |5107| 42 | 6 2 |4000] 2 22 [3547] 23 | 53 | 33 | 4155] 171 | 202 | 5.430] 43 | 6 3 |4516 | 22 22 |3.076] 23 | 6 4% | 5107] 285 | 286 |6510} 5+ | 63 | 3+ | 5.063] 22 23 | 4430] 3 6 43> | 5.430) 226] 338 | 7.548) 6: | 7 33 | 5.641 2S 22 |4909] 33 | e+ | 42 | 5.057] 213] 30.7 | 8170] 6+ | 8 | 3} creo, 2: s /5.412| 3: | 6+ | 4} | 6510| 20.3] 35.0 | 9.905] 63 | 8 | 3 |ese1] 22 ¢ |5.940) 33 | 7 | 4% | 7.088] 193] 821 |9004) 6 |S | 4 {7.563) QF 22 |6492) 3: | 8 5+ | 8170| 25.9] 87.0 |11329] 8 9 4; |8.266| 22 3 | 7.069) 3} | 8 5+ | 8641] 222] 417 |12.753] 7s | 9 4; |9.000| 3 32 | 7.670] 33 | 8 | ‘si | 9305] 213 3 2 [8206] 4 | 8 | 4% | 9.904! 20.7 3t 1 |o6a1] 4 | 9 | 5: |n329] 17.7 3F & {i045} 4: | 9 | 4% |12753] 15.5 35 = a ee Se TO a CLEvIsES. AMERICAN BripGE ComMpANY STANDARDS. All Dimensions in Inches. a (@ fe—---A Vy o_o ety a i Grip G: Seni bea enact to sult connections. DIAM, OF CLEVIS DIAM. OF MAX, PIR CLEVIS FORK NUT WIDTH THICKNESS cCLEvis D FP F N Ww rT A B D 3 1} PS 13 rt 3 6 6 3 “a at i} 1? 13 + 9 8 4 5 3 2% at 23 4 9 8 5 6 8 22 2+ > . 9 8 6 7 4 3t 3t 3t z 9 8 7 Table giving diameter of Clevis for given rod and pin. ROD PINS ROD ROUND | SQUARE | UPSET | 1 14 14 +12 32 2t 2) 2? 3 3: 3} SE 4] vPSET | SQUARE | ROUND g $ 1 3132 6 1 + $ 8 1. ie e1'3 2 4] 4 oe 3 $ 3 1} oN pe Mie ee Sey 1 t g 1 eet 414 4 4 4 13 1 14 1 14 404) -@- 4:55 -8. 8 1+ 1 1% 14 1} 18 A 2 CH h AS 6 8 1g 1} 14 18 i}: 6 515 5 5 & 5 1% 1 1z 13 6S 6 1S. °8. B-56. "8 1} 13 i$ 13 2 6 515 5 5/6 6 6 2 13 14 1 2h OSB 6:16 "6 6)-6 2h 14 13 ly 2r 6-616 6.8 (642? 9 -¢g 2k bt] 13? 1} 14. 22 6 616 617 7.979 97 7 23 4 14 2 13 2h Cee OR ere. F 23 Ey 2 2t 2% Re FET |, Dye af 23 is 23 > ety SN If se SAS 23 13 2} 2 Qt hs HY 2% Ye ee ROUND | SQUARE | UPSET | 1 13 Wo 1f 2 2 2b 23 3 3% 33 3} 4 | uPseT | SQuaRE | ROUND ROD PINS ROD Clevises above and to right of heavy zigzag line, may be used with forks straight. Clevises below and to left of same line, should have forks closcd in until pin is not overstrained, “ SLEEVE Nuts AND TURNBUCKLES. AMERICAN BRIDGE CoMPANY STANDARDS. All Dimensions in Inches. Cisveland Gay Serge : b icah Coane Shier sateh 7 a a Ohio. - 5 REE IES , he >. Xoe== | Satebeaae f-=3 a a eS ee Ss... rao) oe ae Bes me L “4 Estra Lemgchor 9: ist is: 24, 86; 44°-& 72° (Special Prices). woe [OE | ornur | ‘cum. | omm. | ‘oa | "ness [wernt | werowr STANDARD DIMENSIONS ame pes U 7 ee Gee ie ee ete eet PE a ee 2/4] 7 $4) RB pF £98 1 Oh baer at 18) Sit tae Lyn]? $35] Pe PS 1 8 Be 4 ae ae Ot eee Hi uviAnsts3 © | 13) 8 | 33] 4 4 {| «= | ag] | 08 | 1] zie ti “ ‘ ~ |» Ta | 6] 7 ae | as | 1%] OF | 15 7 ae wu) 2 |] 8 | 2 | 2] 1] ¢ | a] @ | « 7 2 | 8a | 18 | 108 | ob | 22 ail j es |e to 1 «bh ) eta | & | | ob) 8] 10k) oti ae | 2 | of | of | 38] 13] & | 8 | ss] = | 2 | 38] 2 | 1% | of] 2 73] “ » fo . w+ 8i 1.10 | * fF 2 33 | 25 | 113 | 2& | 2% az} 2] 9 | 3] af} at | £ | 10 | in] & | = | 83] |g | 18 | 28) 22 2 ok fae wf oe pos pe Rg Pa Pe ae) ae ae ae ee 23 | 2 | OF | S$ | 45 | 28 | & |] 4] 1] SG 25 | 45 | 23 | 128 | 3% | 2% ai, « |-« «}]o« |= | « | 6] 18] 8 | « | 42] 28 | 123 | 3 | 2 22| 38 |10 2} 4: | of | ¢ | 18 | 20 | « | a2 | af | 28 | 133 | 3g | 28 wi “ “ “ “ «| 19} 24] £] 8 53 | 3% | 138 | 33 | 22 2g | st |10: | at | ae | a | 8 | v2 |] 28 | B | | 5B] 3 | a9t | a8 | 2 Co Ea Whee vs ee eae eee (eee me ee ee od ge Se 2z | 3; | 1. | 42 | 53 | sf | 2 | 27 | sa] 12 | * | 62] Bz | 4b | 4% | 22 ‘Rae ee ee ee i ~ | a8 | 3a] + | 3$ | 68 | 3 | 15 | 413 33] st | uy] 5 | se] ot | # | 34 at Bi] sf" |“ | = Js | = | 35 | 60 | 1 | 4 | 6F | 85 | 15 | 45 | SF 33] 4 {12 | 53 | 6 | 3$ | 2 | 30 33 aes We of “ | 40 | 65 | 15] 4 | 7h | 43 | 165 | 6¢ | SF 8% | 45 | 12: 2 | 62 | St] & | 45 e 33 x * “ “ « | 47 s4 5 8: | 4, /18 | 6 | 3% 35 | 42 [13 | 6 | 7 | 45] 1 | 52 32 Z * SE AE 55 ig | 5 | 8$ | 4¢]/18 | 6 | 4 42 | 42/135 | 6: | 7% | 43 | 15 | 65 4t 43) 5 |14 6 | & | 4 | | 75 43 IO AMERICAN BripGE CompANY STANDARDS. 7p) e4 a 4 Ry © 2) —) "19953 OG PIa0xV Jou prnoys yisuo] Suyddrys unuarxeyy :-qQLON 9 | e8f | Tle] Ze] LE| foe] ge] tse] sel sre} tee] fee | tee] tee] tee] tte| ste] te] toe] of | 9 $2 | 42¢] Le} foe] of] tee] ce] gre] sve] tee] tee] tee | tee ste | 4te | te! foe] oe] fee] ee | £9 #° | foe | ge] fee] ge] ¢ the | gee | Fee | #2e | FE] tte | ete] Te | foe] oe] sez} ez| see] 882 | Fo Fo | #98 | SE} Fe] tee] fee | tee} eee} Fee | tte] ete] Te|] Foe] of | fee] ez] sez] s9e¢| f22| tue Eg 2] fhe | fhe | eee | Fes | ee] Te} fle} ete] Te] Foe hk oe | fez] et] see] sec] sic] 2ze] toe] t9z ¢ £F | fee | tee | tze | tee] tte | ste] te} foe] oe] tec} ec] gec| 2 $4t | $42] t0¢] +9¢] 29%] tee | fF fF | tee | ee] ste | tte] te] £ O€ | 762 | 62} 488) 782 | 442 | FL | F9% | y9z | Sez | Zee | fez] toe | ZF fF | t1€ | Ie} Te | foe] of | tee} 62} 49a | tee] #42] tue] f9e | toe | oz | #ec]| she] eee] wel! tee SF Fi] te | 408} of | fee | 62 | F8e | 49%] $42 | Fue | toe | toe | £92 | soe | & 4he] be) fez | e2]| fee F #@ | of | ez | 62 | #e2 | tee | $2 | 222 | toe | 292 | 2o2 | sez £ ve | ve] tec] ec] tee] see| tte | Fe r& | 68 | 8S | 782 | 442 | FLB | 49S | 79S] £92 | Fee | fhe] Boe | we | tee | ee | fee} see] S12] ste] 202] ze fe | 38% | #42 | +42 | toe | tow | fee | Soz-] see | 2 be | FES | CS | ges | fee | fs] He] soz] For] ter | Fe £ | 7Le | 9% | 79% | Fee | foe | he | foe | He | vet | ee |‘ fee | tee | gta | stz #08 | ¢ $61| #61 | 28T £ id 490 | 106 | tho | be | fee | e2 | Fee | fee | ete | Ft2 | toe | foe | tet] get] set] fet] er te Prd f€S | €S | $e | ToS | ETS | FTS | FOS | 40S | FET | fet | eet} ger] stl] fzt| uztl Fe te eI | 418 | OS | FOS} F6t | e6T | eet | eet} st| fut] Lt} got] tor] Fe é e6T |} £61] $8 | $8t} ST] F4t] LI] got | fot] $et] ser é rr : St} £41] Lt | sot] for] tet] tor| ter] tet] Sr 41 cot |4or| tet} ter] ter] fet} ter | #2 . Min, Length C, toend 6° 6,’ preferably 7’ 0” WIDTH MIN. HEAD SCREW END | wots oF THICKNESS + THICKNESS oF Sead Utell Winehouse) ee i Ins. Ins, INS. INS. FT.. & NS, FT. & INS. ine. INS, ins. INS. 4 4: 13 O- 77 4 53 37 1-93 abs ° . 8 a 1 = 53 2t 0- @3 S ot t 2 Pa a 6 34 1-14 1-1 23 5 w ols 25 3 + 7 3 1- 3 1- 5 3 5+ 1 to ly; 3 “a 8 4 1- 6 1- 5 a5 6 1} to 14 4 + 93 4+ 1-8 1- 8 3 6 lto 1}, d “ 10% 5+ 1-10 i- 8 3% 63 = divs tol} - 11+ 5 1- 9 1-9 3% 6s lw ly 5 7 3 T 5 1 123 6 2-1 1-9 33 7 1} to 13 6 2 133 53 1-11 1-11 oe 8 1} to 14, 1 147 67 2-2 1-11 4 8 1} to 1} 6 y + 16 6t 2-3 2-3 4% cs) 1} to 1y, y 4p 17 iis 2-8 2-3 4r 9 1} to 1} 1 17 6+ | a-3 8 i% | 18 73 2-6 8 at 18+ 8 2-10 1+ 19% 72 2-6 9 as 21+ o% 3-1 9 1% 22 9 2-11 10 « 23 10 3-3 10 12 12 < q Note: Eye bara are hydraulic forged, and are guaranteed to develop the fall strength of the bar, under conditions given m the-above table, when tested to destruction, I2 Maximum ALLOWABLE BENDING MOMENTS IN PINS FOR VARIOUS FIBRE STRESSES. | PIN MOMENTS IN INCH POUNDS FOR FIBRE STRESSES PER SQ. IN. OF [lam] area | 15,000 | 18,000 | 20,000 | 22,000 | — 25,000 1 0.785 1470 1770 1960 2160 2450 1} 1.227 2880 3450 3830 4220 4790 13 1.767 4970 5960 6630 7290 8280 1? 2.405 7890 9470 10500 11570 13200 2 3.142 11800 14100 15700 17280 19600 2} 3.976 16800 20100 22400 24600 28000 23 4.909 23000 27600 30700 33700 38400 23 5.940 30600 36800 40800 44900 51000 3 7.069 39800 47700 53000 58300 66300 3t 8.296 50600 60700 67400 74100 84300 33 9.621 63100 75800 84200 92600 105260 3% 11.045 77700 93200 103500 113900 129400 4 12.566 94200 113100 125700 138200 157100 44 14.186 113000 135700 150700 165800 188400 44 15.904 134200 161000 178900 196800 223700 43 17.721 157800 189400 210400 231500 263000 5 19.635 184100 220900 245400 270000 306800 5+ 21.648 213100 255700 284100 312500 355200 54 23.758 245000 294000 326700 359300 408300 BY 25.967 280000 335900 373300 410600 466600 6 28.274 318100 381700 424100 466500 530200 6} 30.680 359500 431400 479400 527300 599200 63 33.183 404400 485300 539200 593100 674000 63 35.785 452900 543500 603900 664200 754800 7 38.485 505100 606100 673500 740800 841900 (es 41.282 561200 673400 748200 823000 935300 re; 44.179 621300 745500 828400 911200 1035400 (¢ 47.173 685500 822600 914000 1005300 1142500 8 50.265 754000 904800 1005300 1105800 1256600 8} 53.456 826900 992300 1102500 1212800 1378200 84 56.745 904400 1085200 1205800 1326400 1507300 8} 60.132 986500 1183800 1315400 1446900 1644200 23" TABLE FOR RIVET SPACING. ey ee ee ey oe dS cae 8 PITCH IN INCHES g 5 [42] a2 | 2] 22] 22] 221 23] 2 23)arls 1 1 2 | -23 -23 -223 -s] -3) +3) -39 -4 - 53] - 54 2 3 | -33 -33 -4H - 4)! - 43 -53] - 53] -6 -8} -89 3 ¢ dsl 8) /se pth i ee TABLE FOR RIveT SPACING. % PITCH IN INCHES % g y o | 3 | 3$| 34 | 33] 32 | 33) « | 42) 42] 49 | 5 | of | oF | oF] 6 1S Pf 1 2] 6] -6t] -64] -63]/-7 |-74] 2 Re -9] -93/ -10] -103 -11] -114] 10] 2 3 | -9 | -98| -932] -108] -103|- 113] 1-0 | 1- 03] 1- 13] 1- 24 1-3] 1- 33] 1- 43] 1- 53] 16] 3 4 | 1-0 |1-08]1- 1 ]1-1$]1-2 |1-3 | 1-4] 1-5] 1-6] 1-7] 1-8| 1-9] 1-10] 1-11] 20] 4 & | 1-3 |1- 3$]1- 43] 1- 43]1- 54] 1- 62] 1-8 | 1- O4f 1-103] 1-113] 2- 1 | 2- 23 2- 34] 2- a 2615 G | 1-6 |1- 63] 1- '73/1- 83]1-9 |1-103| 2-0 | 2-14] 2- 3| 2- 43] a- 6| 2-74) 2-9] 2-10) 3-0] 6 7 | 1-9 | 1- 92] 1-103] 1-118 |2- 0} |2- 2] 2-4 | 2- 5a] 2- 74] 2- 93] 2-11 | 3- 02 3- 2)| 3-4) 3-6] 7 8 | 2-0 /2-1 |2-2 |2-3 |2-4 |2-6 | 2-8] 2-10] 3-0] 3-2] 3-4] 3-6] 3-8] 3:10| 40] 8 9 | 2-3 |2- 43/2- 54/2- 63 |2- 74/2- 93] 3-0 | 3- 2H 3- 4)! 3- 6H 3-9] 3-113] 4- 13] 4- al 46] 9 10 |} 2-6 |2- 74}2- 83]2- 93 /2-11 |3- 14] 3-4 | 3- 6) 3-9] 3-114] 4-2] 4- 4) 4-7] 4- 93) 5-0 | 20 11} 2-9 |2-103|2-113|3- 13|3- 23 |/3- 52] 3-8 | 3-109 4- 14] 4- 4) 4-7] 4- oi 5- 03] 5- 3:] 5-6 | 22 12] 3-0 |3-14]/3- 3 |3- 44|3-6 |3-9 | 40 | 4-3] 4-6| 4 9] 5-0] 5-3] 5-6] 5-9] 6-0] 12 13 | 3-3 |3- 43|3- 63|3- 73 |3- 93 |4- 02| 4-4 | 4- 7] 4-103 5- 13] 5-5 | 5- 83] 5-114] 6- 23 6-6 | 23 14] 3-6 |3- 73]3- 93/3-113 14-1 |4-44] 4-8 | 4-113] 5- 3| 5- 6}! 5-10] 6- 13] 6-5 | 6-84 7-0 | 74 15 | 3-9 |3-103]4- 02|4- 28 |4- 44]4- 81] 5-0 | 5- 33] 5- 73| 5-113] 6- 3] 6- 63] 6-10:| 7- 24 7-6 | 15 16 | 4-0 |4-2 |4-4 |4-6 |4-8 |5-0 | 5-4 | 5-8] 6-0/ 6-4] 6-8| 7-0] 7-4] 7-8] 8-0 | 26 17 | 4-3 |4-5i|4- 73|4- 92 |4-115 |5- 3%] 5-8 4p 6- 4) 6- 83 7- 1] 7- Si 7- 93] 8- 13] 8-6 | 17 18 | 4-6 |4- 83]4-101|5- 02|5- 3 |5- 74| 6-0 | 6-41} 6- 9| 7- 14] 7- 6| 7-10] 8-3] 8-73] 9-0 | 28 19 | 4-0 4-112|5- 13|5- 4: |5- 64/5-113] 6-4 | 6- 83 7- 13] 7- 63] 7-11] 8- 33 8- i] 9- 19 9-6 | 79 20 | 5°0 |5- 24/5-5 |5- 71]5-10 |6-3 | 6-8] 7-1] 7-6| 7-11] 8-4] 8-9] 9-2] 9-7/10-0 | 20 21] 5-3 |5- 58/5- 83|5-102 |6- 14 |6- 62| 7-0 | 7- 53) 7-10} 8- 33 8-9 | 9- 23 9- 73]10- 0% 10-6 | 27 22] 5-6 |5- 83/5-115 |6- 23 |6- 5 |6-103| 7-4 | 7- 9H 8-3] 8- Bi] 9- 2] 9- 73/10- 1]10- 6111-0 | 22 23 | 5-9 |5-113|6- 23|6- 53 |6- 84|7- 23] 7-8 | B- 13 8- 73] O- 14] 9- 7 |10- OF10- By 11- 011-6 | 23 241 6-0 |6-3 |6-6 |6-9:|7-0 |7-6 | 8-0 | 8-6] 9-0] 9- 6 /10- 0/10- 6/11- 0} 11- 6/12-0 | 24 25 | 6-3 |6- 63/6- 9:|7- 08 |7- 34 |7- O3| 8-4 | 8-103] 9- 43) 9-103]10- 5 |10-11)/11- 5y 11-113]12-6 | 25 26 | 6-6 |6- 9:/7- 03/7- 327-7 |8- 14] 8-8 | O- 2H 9- 9 |10- 33]10-10 |11- 4111-11 |12- 53] 13-0 | 26 27 | 6-9 |7- 03|7- 3217- 73 |7-10} |8- 54] 9-0 | 9- ae. 14]10- 83/11- 3 |11- 93]12- 4)/12-113| 13-6 | 27 28 | 7-0 |7- 33|7- 7 17-103 |8- 2:|8-9 | 9-4 | 9-11 |10- 6 [11- 1 [11- 8 |12- 3 |12-10 ]13- 5 |14-0 | 28 29 | 7-3 |7- 6$|7-103|g- 12 |8- 54 |9- OF] 9-8 |10- 3410-10} 11- 53]12- 1 |12- 83)13- 3413-10314-6 | 29 30 | 7-6 |7- 93|8- 13|\8- 5: 18-9 |9- 43] 10-0 |10- H11- 3 |11-101112- 6 |13- 14113- 9 |14- 4315-0 | 30 2 | 3 | 3¢| 33 | 33 | 32] 3 | 4 | 43 | 4¢| 43 | 5 | 53 | of] oF] 6 | 8 s PITCH IN INCHES $ 15 TABLE OF AREAS IN SQUARE INCHES, TO BE DEDUCTED FROM RIVETED PLATES OR SHAPES TO OBTAIN NET AREAS. Sg SIZE OF HOLE. INCHES. 8 “1; l%)8 | wtat& |] & |B] Ea EO OB OD ee pe pas) PERSE cae +} .06 | .08| .09| .11) .13 14; .16 17 | ..19 | ..20:) .22) 235 — 250 zs] 08 | .10| .12/ .14| .16 18} .20 21) .23)..25:) .27 |) 29: 32 $) .09 | .12| .14| .16} .19 y AS Mal Cire 26| .28|..30| .33'| ..45:|. .ae eee “1111 | 114] .16|.19| .22| .25| .27| .30| .33|-.36| .38| .41| 44] .46 4] .13| .16!].19| .22| .25 y >. ane. 9 34| .38| .41| .44] .47 50! .53 zs] -14 | .18} .21 25| .28 32 | 335 39| .42!| .46] .491 .53] .56| .60 2] .16| .20| .23 Pp 7 eo eae, $i ¢ 35| .39 43 | .47) .51) .55| .59} .63 }9 ee fi] 17 | .21 | .26 | .30) .34 39 .43 47| .52| .56| .60| .64] .69]| .73 $1.19 | .23} .28| .33| .38 42 .47 52| .56| .61! .66| .70) <75) 60 43) .20 | .25| .30| .36] .41 46; .51 56| .61| .66| .71| .76| .81]| .86 %|.22| .27| .33| .38| .44 49; .55 60| .66; .71| .77| .82| .88)} .93 13] -23 | 129] .35|.41| .47| .53| .59| .64| .70| .76| .82| .88| .94| 1.00 1 .25 | .31| .38 | .44] .50 56! .63 69} .75| .81)| .88| .94| 1.00} 1.06 1] -27 | .33 | .40| .46| .53 60! .66 73 .80| .86| .93/ 1.00} 1.06} 1.13 1 .28 | .35 | .42| .49] .56 63 .70 77| .841 .91| .98/ 1.05 | 1.13 | 1.20 1,3, .30 | .37| .45| 52) .59 67 | .74 82; .89| .96| 1.04/1.11/ 1.19] 1.26 14] .31| .39| .47| .55| .63 70| .78 86| .94| 1.02) 1.09 | 1.17 | 1.25} 1.33 1,5,| .33 | .41] .49 | .57| .66 74| .82 90} .98| 1.07 | 1.15 | 1.23 | 1.311 1.39 12] .34/| .43| .52| .60| .69 77 | .86 95 | 1.03 | 1.12 | 1.20 | 1.29 | 1.38} 1.46 174) .36 | 145 | .54).63| .72| .81| .90| .99| 1.08 | 1.17 | 1.26 | 1.35 | 1.44] 1.53 13] .38| .47 | .56| .66| .75 84] .941 1.03 | 1.13 | 1.22 | 1.31 | 1.41] 1.50} 1.59 15%] .39 | .49 | .59 | .68| .78 88] .98 | 1.07 | 1.17 | 1.27 | 1.37 | 1.46 | 1.56 | 1.66 128] .41)| .51|.61/.71| .81 91] 1.02 | 1.12 | 1.22 | 1.32 | 1.42 | 1.52 | 1.634 1:73 141) .42| .53| .63| .74|] .84 95 | 1.05 | 1.16 | 1.27 | 1.37 | 1.47 | 1.58 | 1.69 | 1.79 1#] .44) .55 | .66|.77| .88 98 | 1.09 | 1.20} 1.31 | 1.42 | 1.53 | 1.64 | 1.75 | 1.86 148] .45 | .57 | .68| .79| .91/ 1.02] 1.13) 1 25 | 1.36 | 1.47 | 1.59 | 1.70 | 1.81 | 1.93 14] .47| .59|.70| .82| .94/| 1.05 | 1.17 | 1.29 | 1.41 | 1.52 | 1.64 | 1.76 | 1.88 | 1.99 148] .48| .61| .73 | .85| .97 | 1.09] 1.21 | 1.33 | 1.45 | 1.57 | 1.70 | 1.82 | 1.94 2.06 2 .50! .63 | .75 | .88 | 1.00 | 1.13 | 1.25 | 1.38 | 1.50 | 1.63 | 1.75 | 1.88 | 2.00 | 2.13 In calculating the net area add % inch to diameter of rivet before entering Zhe ‘table. 16 Pe oe PLEASE DO NOT REMOVE CARDS OR SLIPS FROM THIS POCKET UNIVERSITY OF TORONTO LIBRARY < oa es, aN By) ven qt tie Se