THESII aUPPLEMEWRY MATERE’ fNBACKOF‘ Design of a Timber Bridge at Delta kills, iichigan A Thesis fubmitted to The Faculty of MLCHIGAH fTATE COLLEGE of AQRICULTURE AYD A PLIEU ”CIRTCF by Eugene L. Dexter h.— Cnndidate for the Degree of hachelor of Science June 1942 Table of Contents Introduction Data and Specifications De Sign of Floor System Design of Stringer: Design of Floor Beams Iyesign of 60-ft. fony Truss (Jamber \ Section and Dotaila IDeaign of Split rings in joints and bolts IDesign of abutment: IDesign of Wing Wall Design of piers Design of End Bearings Bibliography Page 11 14 17 17 22 24 29 33 35 3'7 MATH i5 ENE 143078 A Pu iii “.1 K 0F BOO: I Introduction This thesis covers the complete drawings and desim for a timber bridge on 0-533 at Delta Kills, Hichigan. At the present time there is at the site 11 steel through truss bridge which was built by the R. D. Wheaten and Company of Chicago, Illinois in 1391s Present Structure The bridge was one of the best in this section of the state when it was built, but at the present time it has a load limit which inconvicnces the use of the road as a class A county roedehicb it is. Also the bridge is(@narrow fa' the present And future volume of traffic as it carries only one lens or traffic. It is my object in this thesis'to de- sign a bridge wide enough to carry the present and future 1 volume of trait is. 3. .I' . . .1 “.0 J" "' 3; 1" 81’ "". '. -. , . ’1" ., #1 or; {$4st ’ ._ - ‘ i-rxf-7t"‘.' - 4* 15"”.‘fi‘4' -‘ :‘ View Looking North Several types of bridges were investigated and at the time this was dangled there was a war priority on steel end thererore I decided to design one or timber. Concrete was “so considered but due to the mount of steel reinforcing me the cost or concrete as ccunpered to timber it was de- Q Oidcd to be constructed or timber. Timber bridges at the present time are being recognized as one or the foremost typesznm use in.the west and have been experimented eé§h_a great extent. The Forest Products Treating Company at Portland,0regon.has a method or treating the timbers agnthey will last as long as steel if painted and cared for properly. Specifications which.were followed are General Specifi- cations for Timber Bridges and Treaties by Milo S. Ketchum unless otherwise noted. I wish.to take this Opportunity to thank Mr. G. J. Bogus, in.charge of Technical Service or the host Coast Lumberman's Association for his valuable aid and advice in the prepara- tion of this thesis. Computations for 20'x240' 4 span timber Bridge Data. Live load 13-15 log-d (Michigan State Highway Specification) Impact: 59 used throughout design, Dead load: 1?; Floor- 1}" Sheet Asphalt, granite chip rolled in F 14 1bs/sq.ft. Douglas Fir r 12% moisture air dried - lSlbs/sq.ft. Specifications: ’General Specifications for Timber Fridges by Kilo S. Ketchum, except as noted. Stresses: /n\ ) :fs: 16,030 1bs/sq.in. j horizontal shear 120 lbs/ f quins fc 2:. 2,000 IbS/SQQin. ‘3' g. t”? C .L c 525 (W \ \ C I, g l ’ 4 k- 6 ‘1‘ ‘\ \‘. I. a u‘ fun H . 120 “w" Et = 1,60,000 Temperature: No account has been taken to the temperature which has very little effect on timber. Concrete as noted. Design of laminated wood floor covered with 1%" of sheet asphalt with granitechips rolled in. ;} .us '.”461 ;f .23V1[~KF”?144‘ Aana/f N . F . 2x6 ‘V~ / 2x4 Data/as Fm 7 7 7 ' / Hakka/7‘ 57‘rmjer 45/ Aver-a]: Tllckhcsc /22f gaffe ~2‘ ”cf ('enfer’ H3. / ZaMInafca/ F/aoh Dead Load Pound/sq.ft. Asphalt 14 Laminated wood base 15 Total 29 Then, assume stringers spaced 2' - 6" 0-0 I a: 1/14 “ 9 : 1/14 (29 x 2.52) 12: 155.235 in.-lbs. for maximum” positive moment due to dead load. This moment occurs in the panel. lfiext, M: -1/9.5 $1.2 a -1/9.5 (29 x 2.52) 12-: -229 in.-lbs. :for:maximum.negative moment due to dead load. This moment occurs at the first intermediate support from.the end of the span e D"! Live Load: H-lS M = 1/5 ‘rL =1/5 x 12,080 x 2.5 x 12==500 in.-lbs. for maximum positive moment due to wheel loads. This moment occurs in the ganel. Next, M = ~1/7.7 PL = ~1/7.7 x 12,0oo x 2.5 x 12 £325 in.-lbs. for maximum negative moment due to wheel loads. This moment occurs at the first intermediate support from the end of the span. For the coefficient of impact 0: so _ 50' = r... L 25' “22.5"1'25 "”2 Then, for the maximum positive live-load moment and impact, L: 500 in.-lbs. I= 196 inn-lbs. (50;;- x .392) Total = 696 in.-lbs which occurs at the center of the end panel, and for the maximum negative moment and impact due to wheel loads. L‘-'— 4525 in.-lbs. = ~128 in.-1bg._(325 x .592) Total = ~453 in-lbs. which occurs at the first intermediate floor beam from the end of the span. IDIBtribution of Lheel Loads A wheel load may be assumed distributed by the asphalt cover a retangle with the sides b+2 parallel to the wheel taxis and 2h at right angles to the wheel axle where b=width <3f wheel and h=thickness of asphalt. __ -hb-— ..L h i A 1—A5Fha/r b |-——l TL ‘4 +2}! 2’) "_+— W000, base Fig. 2 theel Distribution Since the wheel load is distributed as shown above the load will be carried by, h= l" the minimum thickness of asphalt, therefore, worse case 2 = ”h: 2" this load is cazried by one plank (2") but maybe assumed that it carried by two planks due the arrangement and fastening of the laminated floor. The flooring is built in the field, the strips being spiked to each other with 20d. spikes spaced about 1ft. centers and then toe~nailed into the string er with one 206. spike for each plank crossing a stringer. For the moment of inertia of two planks take h 4+6+2=5" average h of two planks I 1/12 13113 II 1/12 (4) (5)3 = 530341.? say 42 in. units ”is V Tflhen, for the fiber stress on two planks, when supporting the ehtire woijlt of one wheel, :f:_£y = 15OC)(2.5 : 29.9 say 30 lbs/sq.in. finders K==masiium.moment in inch-pounds plus impact, y==half iflne thickness of the plank, and I==moment of inertia of the b . s .1 mm sun 4-: “1'7 1 S / ‘1 “‘ Slightly high but 0.14.. because this formula indicates greater stresses in a wood team than are usually present, particul rly Lit’n moving or concentrated loads near a support . For moment of'inertia of the beam ;£:1/12 bh3' . =1/12 (14)(:4)3 = 16,128 inches4 Then, for the fiber stress of the beam r: 31 202,700 (12)(s = 526 lbs./sq.in. I - 1U,128. O. K. Alloxable f: 2, 0,0 lbs./so. in. 13 Use 2-7" x 24" beam Design of a 00—ft. tony-Truss The roadway is to be 20ft. wide. The truss rill be 112' - 0" deep at mid-span and 8' - O" at the hip, and tiere rfill.te a 5 panels of 12' - 0" each. The live load will be 14-15 loading. Ileaeroad Stresses it. of floor per ft.::750 lbs. leer foot of truss for the dead load on each truss due to the floor system. Assume wt. of timber in one truss: 6,0q0 lbs. and per foot of truss we have 6.0:? : 100 lbs. 60 ltn additional 10 lbs./ft. is added for wt. of fastenings frbcn, the total wt. per foot of truss is lOO+lO = 110 lbs. tlmwltotal dead wt. per foot for % the bridge 110 + 750 3 860 lbs. 'The total dead wt. per panel is 800 at 12 = 10,3230 lbs. 'Live -load Stresses The equivalent H-15 loading rhovn in (Fig.l0) will be used. 14 /3, 6'00 fir- Momenf d aJ Con?!» frafe lo {/7, 500 for 3604* Uniform /oad 45’a‘t/1’I? of lane Fig. 10 The maximum load on the truss will occur when the load- :iru; is in the position shown in Fig. 11. $ frvsé, ¢ 77,083 I ‘20 A ‘ 27,000 .‘8 2214" 39'0” Taking msnents about B (Fig.11) (4&0 3,2) 12.25 = 520 lbs. .0- 25.5 Ifior the maxinum uniform live load per foot of truss. Then, for the panel load, P = 520 x 12 =6,240 lbs. Augjain, taking moments about B P1: (2 x 13,500) 12.25_: l4,6€8 lbs. say 14,700 lbs. Of) LILo. fHor the panel load due to the concentration to be used in <3eterminin; moment on truss. In the same refiner. r11 =(2 x 1,950) 12.25 2 21,231 lbs. ‘ay 21,23 lbs. 22.5 ifor the panel load due to the concentration to be used in (Esterminin; shear on the truss. Intact is added to the live and bead load stress and the total ganel load is Efi,E“O lbs. by means of "netted of joints" 15 the following, stress table mas ret up. Stress Table U_iL5 : LOU]. 9C", 390 If comp . L41.-5 . LoLl 75,9..0 ‘ ten. L411; = U’J'Ll {58, 55-30 5-: ten. L5L4 -. L1L2 75,900 r. ten. L3U4 :- 1}le 19,0t0 comp. 10,910 ; ten. U 3U4 : U1U2 77,1770 {:5 comp. LsLs 04,710 5 ton. URLs ‘_’. L2U 5 2’ 140 X. comp. 4'. U. l 3 "1270’ .L , 4 Lo 1;: *4, Ir. ‘1‘ er 3 7e4sw' 3Qa&p| aasmr' 36680” Fig. 12 The truss is small and a short sgan tierefore no wind- load stresses were taken into account. 16 «1;: mt er The camber is calculated on the basis of 5 inches rise ef’ loottom chord at center-line the camber curve to be rare- b01100 Sections and ietails The end post ias tie largest force acting upon it, there- fwore design for L0 U1 (Fig.12) and use this design for Whole Of top chord. C 3 -.-. .641 1.6 1: 10° 0 :. 21.2 L.sume the ten chord to Le built up of 3 timers so the Ifiesistanne to bending is increased which is necessary due to tflae fact the top chord is in compression. It is also built pr for construction purposes. Assume it to be made of a rnain.beam.6" x b" with a 5;" x 14" timber fastened.to the two Sides of the main beam (£13.13). Fig. 15 Ratio l/d where l= Span in inches and d= less dimension in Width in inches. d_6+14=1@’ ' 2 l7 1/61 174 _. 17.4 ”T6 For built up columns with a seanbetween eleven times the least dimension and K tires the least dimension ere 03. ' :::-‘:- rs intermediate columns. They depend for strength on a combination of crushing strength and resistance to late- ral buckling. Use the following condition in design for the col £1131. ilhen connectors in middle timber are placed at a distance of 1/20 from the ends of this timber. 21.2 x 1,732 =36.8 P/A =maximum load per u-.. it of cross-sectional area. P/A= 0 {3-1/6 ( 31;; d) 4] P/A = 1,466 (1-1/5? 174 _l 4;) 36.8 x 10) P/A= 1,466 (.983) P/A: 1,440 1bs./sq.in. allowable Cross-sectional Area of end post 6 x 6 = 36 134 again. P/A = 90.590 =6'74: lstsq.1n. 0.x. allowable 1,440 134 leQ/SQQino The maximum load which t1 6 column can carry without buckling. 11-; Lafiggfi .,. 108 1n.4 X X 12.; :35 x (1413: 301 1n.4 12 18 I--- 100 +7301 = 303 1n.4 7— -. C. Per-:- 77 11.1.... ._ 9.976 x 1 6 x 10” 0x 909 4 L“ 4 " 2 x 4 Perr- 1231,01. .0 lbs. O.K. since the load is 90,390 lbs. Deflection 6‘ .. £12 ' 251 9Qi390 x (14.51? x 144 2(1.6 x 10°)(909Y‘ .939 inch Di acgonals The diagonals shall be rectangular solid columns all of the same dimensions, therefore design for the largest stress- es in the members which would be U1L2 (Fig.12) for compress- ion and U112 (Fig. 12) for tension. The size assumed will be 6" wide for c-nstruction purposes and 10" deep for strength. 1/6z 198 .1 3:5 6 K = ssme : $11.2 Long: (l/d radios equal to or greater t1'1a.n"K") P/A._._ .274 E g .274 x 1.6 x 106 (17d52 1089 P/A = 403 lbs./sq.in. allowable Cross~sectional Area 6 J: 10:60 sq. in. 2/4 19.000 = 617 lbs./sq.in. O.K. allowable 406 lbs./sq.in. 19 The maximum load which the column can carry without 6” buckling. _‘ [0 x - x I: 6 M109... 5 0 111.4 /Z Per: 77"}: g 9.86 x 1.6 x 10‘5 x 5‘0 4 '1" '4’" 4 (16.5)? (147:) 50,000 lbs. O.K. since the load is 19,0f0 lbs. Deflection For direct tension the same values as for extreme fiber stress in bending is used. Cross-sectional area 6 x 10?- 60 sq. inches P/A = 14,310 g 238 lbs./sq.in. 60 0.x. ollowable 2,0 0 1bs./sq.1n. \ierticals The verticsls shall be a built up colunm for constr- Ilction purposes. The design will be for the vertical "1L1 Which has the larQest load of any vertical. rThe cross- section assumed will be a 4" x 12;" main timber with a 4" x 14" timber fastened to the two 4" sides of the main column. (Fig. 14) l" 1" avg ” ’4Il f) {4 Since the verticals are rlways in direct tension the sauna values as for extreme fiber stress in bending will be used. Cross-sectional area is 2x4xl4zll2 4 x 12g= 50 Total = 162 sq. inches B/A“ERISSH : 258 lbs./sq.in. O.K. allowable 2,0t0 lbs./ 102 sq.in. Lower Chord The liner chord shall be assumed as made of 2 - 4" x 10" tmsams 6" apart to allow for construction at joints. Since 131a 1 wer chord is always in direct tension it will be de- signed re the verticals were. The design will be for LoLl twitch has the largest stress. Cross sectional area is 2 x 4 x 10 r 80 sq. inches P/A= 75.960 4 950 1bs./sq.1n. ed“' 0.x. since allowable is 2.0 0 lbr./sq.in. A tongue made of l - 6" x 12" x 5' timber ta: used at joint Lo for construction. Also a l - 6" x 10" x 4' timber was used as s Splice block in the bottom chord. Four foot was to allow for connectors. Due to the truss being low and of short snan no vind stress will be calculated but a knee brace will be place at joints L1 L2 and L3 running from the Joint to the floor beam at a 45 9 angle. Design of Split rings in joints and bolts. Bolt hole shall to of a disaster 'eruittin; bolts to be: driven easily. hinimum s-ac’ng is four times the bolt disameter. 5;? in” l.tte n POTS :Fsuld be at least five times thus bolt diameter. The end margin should be five tires the bcfilt diameter in tension and f-ur tires in conhression. The etigxarmrQin in tension or c.3pression should be, at least one arid.one half times the bolt diameter. iargin nearest the edlge toward which the load is acting is to be at least four tinnes the bolt diameter. Joint L2 Cliords to diagonals Angle of load to brain of diagonal is 35°. Allowable loadiin pounds per connector and belt at angle of 35° 5512 S. R. load is 37,540 lbs. 8 x 5,512 44,096 value of 8 - 4" SR's. Results: use 4 - 4" S.Ii.'s with 5/4" x 14-71;" belts with 2 on each bolt. All bolts are spaced according to Specifications Stated. Vertical to chords Angle of 1 ad to brain of chords is 000 S.R. load is 13,670 4 x 4,675 18,700 value of four 4" S.H.'s Results: use 4 - 4" 53.12. with 3/4" x as" bolt. The 3/4" x 28" bolt carries 5 - 4" S.R.'s; 2 between vertical and chord, 2 between dia50n~l and chord, and one between the kneebrsce and vertical. The rest of the joints were desi¢ned by tie sa"0 method with the following; results: J 01111: Le End post to bottom chord tonQue 12 - 4" S.R.'s with 5/4" x 14;" bolts 2 - 2;" S.R.'s with 5/8" x 14;" bolts Bottom chord tongue to bottom chord 12 - 4" S.R.'s with s 4" x 14;" bolts 2 - 2;" S.R.'s with 5/3" x 14%" bolts Joint L1 Bottom chord to vertical 4 - 4" S.R.'s vith s/4" x s?" bolts 0 h - 4" S.R. .3 cans bolt between kneebrace and vertical. Joint U1 Between chord and vertical 8 - 4" S.R.'s tith 3/4" x 2f" bolts Between chord and diagonal 4 - 4" S.R.'s on 3/4" x 2:" bolts above 2 - sg" S.R.'s with 5/8" x 22" bolts Joint U2 Between chord and vertical 8 - 4" S.R.'s tith 3/4" x 2:" bolts Between chord and diagonal 4 - 4" S.R.'s on 3/4" x as" bolts above 2 - 2;" S.R.‘s with 5/8" x 2w" belts Center line bottom chord 831106 24 - 4” s.R.'s tith 5/4" x 14%" bolts ori the to: chord and the verticals that are made up of three tianers, connectors will be used as a means of fattening. Irl all verticals and chords five connectors is sufficient, except the vertical between joint L1 U1 and L4U4 in \trhich three is sufficient. The fl.nr teams will also be fastened by means of conn- ecrbors. Using the metlod above it was determined that 10-5/8" x :39" bolts with 2 - 2-3;" S.R.'s to each bolt. Due to the fact trlet the floor beam fastens to the vertical 2%" rings had to bee used. The standard WQSJBP made by the Timber Engineering chnpany is used under all bolts thr'ubhout the whole structure. At joints there are end bearings of wood on wood. When 1K>od.members are squared and batted end to end, the end tends 1x3 bed themselves into each other and the maximum.stren5th will.tb less than the co pressive strength of clear wood. The anwunt of embednent will vary with the percentage of summer wood.and for practical purpose it is not safe to count on more tron 75% of the compressive stress for clear wood. there such and are butted use a yiece of 16 geuve galvanized sheet iron. These bearing plates between tOp chord segments to be placed in the field in a soweut thru the joint made vith a finishing hand.saw. Abutment 8 Both abutments will be of tie same desiQn due to the fact they are the same height and same loading. The load on the abutment due to dead nd live load and impact per truss is 193,900 . This lead will be considered 24 tc> be distributed for 10 ft. along the wall. 192 900 = 19,290 l0 /fto -~'/55'-:— ‘r—ir De Sign of cw._.-tilever wall assume base = .7 h I b \ toe = .726 b BP/{Ie 3P1? I726» 27’ 13:07120: 14' toe dirt. =.36 x 14 = 5' \ 1 fl. £20?- The surcharge due to the live ' load on the fill will be taken as 6" terth tresrure Ce: wh2 2 “27 LQEKELQIE . 5,4<.o lbs./lin.ft. act 1/5 up from bottom or 6.5 ft. wt. of earth (7.? x 18) 100 = 17,860 lbs. wt. of base slab (2.6t x 5)l50-+(1.6b x 7.7) 150 = 5,910 lbs. wt. of stem x (1.3 x 19.53) 150 = 3,764 lbs. 95 homent of forces of base end carth 4? 3'"? toe =0 (13,:':GO+E,'704+I".,910+19,5..LSEO)5E = 13, to (10.15) +3,GlO (7)+ 19,2 .0 (5.t4) +1 ,7o4 (5.:34) 4:.._,z,<_v:24 i=2i>e,17b x = 7.30 ft. from too to 43,8fi0 lbs. resultant. Reasultant of Earth pressure and weight hit base at t.)- 'l:‘r‘ gs. A <1 .3 o “3'. (O (n -q C5 (“j H“ ft. 7.50 "' .97: 6.33ft. ' eccentricity = 7 - 6.33 = .67 ft. Soil Pressure SP- (1: 6e) b b 40 224 (11' 6x.67) 4 520 lbs ./so.ft. ...._L........ ._..____.,_....._ ._ D 1 12 12 ‘ 2.,2o0 lb: ./sq.ft. Sliding; f: .4 factor of safety= 4-f...;,FiQ/'i-X.4 g 2.55 O.K. 6,400 Stem Barth rressure z .27 199.3! (.1432: 4,570 lbs. 2 act 1/3 up or 6 ft. u £3700 ' Bending homent ‘ 4,370 x 6 = 0 ft. lbs. 2U,2‘. use 3,0 0 lbs concrete fc = 1, OLD fs: 20,;o0 n = 12 = 164 K U 100 - 125 lbs./sq.in. T = .0094 V = 50 ' 6‘.) leo/Stloin. d=r 17% #9 x 12 12 x lb4 d: 12.6" use 2.9" of protection over bars therefore D: 15.5" Say 15.5" As = 1" 13d .O'.’_..94 x 12 x 12.5 at 1.51 sq.in. Use 7/8 in round rodfl 4% inches A3 = 1.60 {SC-loin. BOTH}. ._._ V _ 4:570 __~ “ET ‘jci ‘ Toto x1377/8" 2. C- . 5 = 49.5 IbSo/ino 00K. Unitilmar : .31.... g 4,579 “__ __._ 53. 2 lbs./s .in. 0.1:. bjd 12::7/8 x12. 'Temperature changes Temperature steel to keep the wall from cracking on znqrface to crre for stresses due to temperature changes. Steel ratio of temperature .'05 or .52 of area of concrete. Place 2.5 on front face and 1/3 on back. 12 x 12.5 = 150 150 x .005 = .45 sq.in. or temperature steel/ft. 27 .fIHDnt = 1/2 square rod F 6 inches back = 1/2 square rod 12 inches Toe Design .6" ( L I2’ I gm = 31570 x 52 + 9:0 (5)(_§) -2.5 (5)(150)2. 66 2 ISSUE Bord. = 44’660 ft. lbs...) Eecesrsry to satisfy sheer tie following .As'=.9094 x 25 x 12 = 2.82 sq.inches Use 7/8" round bars 5 2.5 inches 2.89 sq.1nches Bond fl- 4/:- L600 :12 O.K. (2.7? ) 12 (7/s)25 2.5 Heel Design 1Z7’ ”50 . . "to 4520 W 22“ 28 18 x 100+1 x 150=l,950 lbs. X: 2,260 +( 4,520 -2.2ec) 7.7 = 3,723 1 ( 2 ) {MA=2,260 x 7.7 x 7.7 + Lgfiix 7.7x7.'7 - 1,950 x 7.7 x '2" 2 3 r 7.7 = 25,200 ft. lbs. ‘ 2 d==17” necessary for skear I) '1 17 1- 5 :20 inches As: .0194 x 12 x '17 = 1.92 Use 7/8" C 53;": 2.06 sq.1n. 723.00 (7.7) - 1,950 (7.7) v = 2,100 Unit Shear v = 2!ng : 50 OK 12 x 7/8 x17 Bond 2 __.. £12. .3109, 58 0 K 2.73 (12):: 778 3:17 _L1ng Walls Sfi l3" /5' l Surcharge angle 16° 40' f’v Ce = 31 “I I P . 4% 1):.31 100 x 10‘ = 3,968? _' 2 {I *"' Eerizcntsl force of P==3800 # y lo” ’0" _ M6" get 1/5 h = 5‘75 ft.up base 7- .7 h 70 X 16 211.2. Say 11.5 it. Toe distance =1 b 11.5 g 5.8 ft. —‘= 3 3 PV ‘ f {.3 5?? wt. of earth including stem (7.7 x 14.67) 100 = 11,500 lbs. It. of base slab (1.15 x 1.5) 150: .3, :70 lbs. infitoe = O (11,510+2,50;..-W=(11,5(.7) 7.55 +2.500 (.5 .75) 45:. .54 ft. from tow to 16,6 0 Its. resultvnt _ _j§ = 5 5.0 B ’%600 51/3 3:550 53’. f: 1.49 pt. where R hits base 3590 7.64 - 1.49: 5.85' Eacentricity=:5.75 - 5.85 =.1 ft. Soil pressure p- 15 600 (1 t<6 x .1) = 1,100 lbs. /sq.ft. ‘ 11.5 11.5 1,120 .. .. Sliding; f 2 .4 factor of safety : 1516V0 x .4 =1.45 6,800 Stem [114 . n 100(142/5)2 1 n.P.= .01 2 $7 955 i 3 33° - u, SUO l‘h=3,550 X .958=5,190 IbSo A? .. acts 4.1.9 ft. up 50 BM : 5,190 (4.51:9): 15,600 9075 55:; 10" A8: pbd = .0094 x 12 x 10 =l.15 sq.in. use 5/4” 2 4%"= 1.18 sq.in. Bond 5,201 1__ 2.56 x 12.x7/8 x10 4.5 3' 69 00K. Unit Shear: 3 82:1,... 12x7 8x10 =.36.5 O.K. Temperature Steel 10 x 12: 12 120 X .005 '= .56 front %" f 6%" back %" Q 15" Tee design Zoo Zoo 75 , J L 4/20 ‘¢ I 4’30 1190 - (1190 - 1120) 5.8 11.5 H I) = 1167 {Mai 11672:: 5.8 +23 1: 5:38 3: 2§5é_§_)_ - 73.11.: 7,110 0."? 1'10 - 6 6" 17:10" “154 ' . ,Sey 7" 31 As 2 .0594 (7)(l2) = .79 sQ.in. use ~35" C: 3" == .79 Shear = “9178 x 5.8 - 3.8(206) =.,~,710 Unit {Lear _ ___;_:;719____._51 0.x. 2x778x7 Bond = 3127.110“ : lC' , 00K. 1.57x1gx'7/ex7 5 Heel design [6’2— ‘ $1.21? 1""- I/9o ’9 [4 I II 20 X=1120+ (1190 - 1120) x 4.88 = 1150 - 11.5 15.17 I 100 ‘l' 083:) x 150 = 1642 g Ma 2 O , 1642 (4.38)(2.44)-¥112O (4.8?)(2.44)+u§9 (4.82)(2.4¥) = 6,050 tension in tOp of footing 2 d:= 6050 : 6.05 Say 7" “1'65 D*10" As 2 .0694 x 7 x 12 -- .79 819111. Use 1, <1", 5" = .79 sq.1n. Shear = 1155 x 4.08 - 1642 x 4.89 x 2,460 Unit shear : 173778317 1' >2 PC) Bond = 24-60 gens 0.14.. j“ 1.573lgx7/8x7 5 Piers Design the piers to be 21' - 6" high. Assume that the allowable yressure on the footings can be 9,0 0 lbs./sq.ft. The type shown below will be used. 7’ 7" j; Truss a? a: 26 ’0 fry.” no I”, "a, no ’4'“ J (on. f!) l 1 Fir—L 311/” 20‘6" ‘ r V I 1 1% ” " . \I Ate N [7' 7a [‘0’ 000' II Av - ‘ ' 2‘ ’-‘ n if. I; ’ C 210' a , I, J c [ 'é . ’1 IL: Zfl ',_, FL?“ The cap is assumed to be 5' - 0" wide and 1' - 0" thick. Consider the main shaft to be 2' - 6" thick. The base will be assumed to be 6' - 0" wide, 2' - 0" trick and 28' - 6" long. The portion that sum orts tie stringers wiil be assumaito be 20' - 6" long and 1' - 6" wide with the a cap 20' - 6" long, 2' - 0" wide and 1' - 0" thick. Total wt. of pier Base 28.5 x 6 x 2 x 150 = 51,300 lbs. Shaft 26.5 x 2.5 x 17.6 x 150 = 175,0LO lbs. Cap 5 x 1 x 27 x 150 : 12,150 lbs. Ufiper shaft 20.5 x 1.5 x 2.92 x 150 = 15,450 lbs. Cap 20.5 x 2 x 1 x 150 = 6,150 lbs. Total = 250,050 lbs. For ice pressure (assuming the ice to be 20 inches thick) 200 x 201x 20 : 120,000 lbs. 4 For maxisum (irect load on the bott m of the base 252,050-r192,900 x 2 = 645,850 lbs. Dividing tiis by the area of the base 643,850 ::5,760 lbr. 20.5 x 6 for uniform pressure over the base. Next, take moments about point e on base of ice pressure. 120,000 x.12 = 1,44C,OOO ft.lbs. for the moment due to this lead tending to overturn.the pier. T16 moment of inertia of the bottom surface of the base. 1/12 x 6 x (28.6)3=11,7co ft. units Then, f:_l,4flQ,OOO x 14.25 =_ 1,755 lbs. 11,700 for the positive pressure per squere foot at e and the negative pressure per square feet at a on the base due to ice fresrure. Adding this to the uniform load. e :1 5760 + 1755 -= 5515 =-576O ‘ 1755 = 2005 54 Next, consider traction (Fig. 15). The traction force will be 2 x 40,000 x 0.l==5,000 lbs. Taking moments about the bottom of the base 8,050 x 20.92 = 215,300 ft./lbs. for the maximum.moment due to traction. For moments of interia of the bottom surface of the footing about axis K . = 1/12 x 20.5 x 65-. 513 ft. units Then = 215,500 x 3- = 1260 lbs./sq.ft. 515 for the maximum pressure on the footing due to traction. Add- ing this to the 5515 lbs. obtained due to dead wt. and ice pressure 5515-rl260 : 5,775 lbs. for the total maximum pressure per square foot on the footings. 0.K. since allowable assumed was 9,000 lbs. For the moment on the footing along section CC (Fig.15) [@3775 x 7/4) 7/8 - (500 x 7/4)3§]12 = 119,0 -. in.-lbs. Taking d= 2l", j: 0.87 F =_ 119,000 = 6,520 lbs. 21 x .87 Then, for the steel required in the bottom of the base 6520 + 16,000 = .407 Use 5/8" ¢ Inu'ti9" = .41 End Bearings The exyansion of each span viil be approximately .018 of a foot which will be taken by the timber stan itself. C. 1 0‘1 therefore, both ends of each span will be securely fastened by means of a lingo arranuemcnt. Vertical load on each ylate=z96,450 lbs. assume 2w" pin homent on each pin 1: x 48,225 '- .. 60,280 in.-lbs. O.K. allowable 65,000 in./lbs. Shear on yin 4.91 x 44,010 : 210,00 lbs. allowablesactual 96,450 O.K. Bearing area on masonary 100 sq. inches Use 10" x 16"- i H C‘: 0 sq. inches Biblicar aphy Timber Design and Construction~flenry S. Jacoby and Roland P. Davin Theory of Modern Steel Structures-Linton E. Grinter Wood Structural Design Data-Vol. 1~Nationa1 Lumber Manufactu- rers Association Highway Structures of Douglas Firnwoat Coast Lumberman'l Asl- eciation Douglas Fir Use Book-West Coast Lumberman'a Association Highway Bridgeanohn Edward Kirkhmn Design of Highway Bridges-silo 3. 1(0th .I'fll‘i ‘0.’Ir .Ow..’l . . 1 \u 1"!” , I 0 .l 0.. . 'u- '0 L..- . c ’ ‘H*1':~‘oi . ‘C' “V. M] In ,_‘ LP 3 ’ ‘JNIVE‘QS'T " -JBHIARit . $1 .1 T ' .ifl '1 1 v ‘1 ' W I“ 1 ‘ .‘ :7 H ‘ 1 1293 030711026