—Iw_. COCO-b AN ANALYSIS OF THE DESIGN OF A WELDED PLATE GIRDER BRIDGE Thesis for the Degree of B. S. MICHIGAN STATE COLLEGE G. G. Sangster I949 An Analysis of the Design of a Welded Plate Girder Bridge A Thesis Submitted to The Faculty of MICHIGAN STATE COLLEGE of AGRICULTURE RED APPLIED SCIENCE by G. G. Sangster Candidate for the Degree of Bachelor of Science June 1949 THESIS I 61/ i An Analysis of the Design of a Telded Plate Girder Bridge 4. Sometime in the future a building to be used as a hotel is to be constructed for Iichigan State College and will be located north of the Red Cedar River on Harrison Road in East Lansing, Iichigan. A bridge is to be built to carry pedestrian traffic across the river at the building site. The analysis of the design of this bridge is the subject of this thesis. The bridge will be of welded plate girder design spanning 94' - 0" c/c of bearings. The web is to consist of five sections welded together at panel points two, five, eight, and eleven for a total of thirteen panels. In order to make the structure attractive, the edges of the plate girder are to be cut on two separate parabolic curves with canbers of 12 inches and 30 inches upper and lower, respectively. his allows a mininum clearance of l' — 0" above high water. hese sections will be cut by the fabricator. Preliminary soil investigations allow a maximum soil pressure of 5000 pounds per square foot for the structure. The makimum value for this design is 3920 pounds per square foot and well within the allowable limit. The abutments are to be of reinforced concrete. The design is such hat the fill need not be placed inside - l - 217921 and outside simultaneously. It will be noted that th area of reinforcing steel used generally exceeds the area required. This is due to the fact that the convenience of field layout was considered in determining the spacing and the size of the bars used. The saving in labor costs will exceed the cost of the extra steel. However, if the job were larger with long repeating sections this might not be an economical procedure. The live load on the bridge is extremely light, 100 pounds per square foot, and is purely a pedestrian loading. Specifications for bridges carrying only pedestrian traffic are not available. Except as noted, this design follows the recommendations of the American Association of State Highway Officials. The preliminary sketches of three bridges, each of different design, have been submitted by a consulting engineer to the Board of Agriculture. At the time of this writing a definite acceptance has not been received. Due to the fact that the structure shown as "Preliminary 2“ is the most likely to receive approval, its design was chosen as the subject for analysis. .....-Mw.,.....,...._.,_....,,.WWW”.“... -- . “I"... 3)..“ ~.Vm‘u‘ ‘3‘)07‘1 U- M. .v“ U ......._.- ...,h _ 4 ......— Y1 . I'lt‘u . .Ltl I. . II.. "II.- 0 (I .... ......F . L .1 ..... .. ...v“ .1 e. cwmwwawmnuw. . ...“...akw»... ma..- 1 . 0.4.4.1.. . ... if L.\:A!u e0 c“.(n. 18V: ... e . v e. .0, . . I . .... . . . .nlvu‘..l.i . 4 . . ... ... .. . . .1 .. . . . ,. y e V \ . , r .. . y. 1(\!. nlii$!,l\..e.’.. . ... A . . .. "awn. M 4 _ a . . . w a u . ...rJI 11111 ‘1..- .. a w .. n I I I _' .r--. -.,'.,.. -..-o uv. ,4 O _. ......;:.-....' ..., ......IIIEI Q 0Q.I.ep-.-4'..l.‘ .((..9 I91. n». 3.:- '1?! I. El). T! ul.lflII’.l. e ...... «to... . ' ‘ - I T%“ " 298$ PA '/’) f! .M l e a I . III! (A! (v. .. . ... . . ., _~. atru- ('10-. : ""r. , t ...... ”a-.." .-.-,-.,...-,......,...,,.1 .-.,- . -. .> . . ~ ‘ ....SID‘O.I v’ I 1.1: (PIT- I..Io.n.\) l‘q'v 3TH)?! carame- M. (f . Ialndul‘filfi.1(vnel e: ‘ ..e _ v I . ~ U .m ,.. r.” ._ 0.. . .x .g .. .a . . 4. ... ~ e awed-Ann In . I bazaar: . -. came-2 I IvIl)! .‘.‘l‘00I\E-llt'l3flll'x_1_ :2.65 L L Width : 03500 (1 f 6 x 2.65) x 1 20 20 18.5 3 401 (1 5 .80) l . 2 - o PToe _ 3,20 lbs/ft _ 2 PHeel _ 436 1bS/ft Footi Soil Pressures I: 2010“ 2:4 A36 3620 Pressures: The weight of the concrete is Slope subtracted from the pressures : 3620 - 136 as found above due to the fact 20 that the downward force is : 3484 20 opposed by the upward forces : 174.2 #/ft of the soil. Wc = 150 x 2 = 300 lbs - 11 _ Footing Slab Toe Assume that 3620 #/ft2 acts uniformly over the 2 foot width of the toe. This is logical because of the small dimension. Moment M 3620 x 2 x 1 7240 lb - ft Effective Depth d ;\ M Assume: AT.’ ‘ $8 3/4" 8 steel :{2240 x 12 l O x 12 3 1/2" of cover : 6.75 inches Use 20 1/2 inches ' Area of Steel A - M s - -——r- fS 3 d .7240 x 12 20000 x .885 x 20.5 .240 in2 Use 3/4 inch ¢ 6’21 inches in the direction of the Width. A .25 in2 S Zn 1.35 inches Check fC 2m k j b d2 Ft 0 H 2 x 2240 x 12 .344 x .885 x 12 x 20.52 113 #/in2 Allowable fc = 1050.0 #/in2 - l2 _ Footing Slab (Continued) Toe Minimum Steel A : .002 bd .002 x 12 x 20.5 .493 in2 Minimum Allowable 3/4" a @ 10.5 inches. This is the minimum allowable reinforcing steel which can safely be used running lengthwise in the top section of the floor slab. Floor It is desired that the stress be distributed with uniform intensity between the steel running lengthwise and that running across the width of the floor. First it will be necessary to find the maximum loading which can be carried by a steel area of 0.493 square inches. Loading M = 1/8 WI2 1 z 16 1/2 ft c/c of walls A : ___AL_. 5 re 3 d w = rs 3 d AS 1/8 x 12 20000 x .885 x 20.5 X 1491 .125 x 12 x 16.52 : 438 #/ft _ 13 _ Footing Slab Floor (Continued) Second, find the length of cantilever from the front wall over which this loading acts. Length of Cantilever Referring to soil pressure diagram 10 = 20 - (2 / 2 f X) X : distance of action from heel Slope = 174.2 #/ft w = 438 - 136 : 302 lbs/ft X.: .392_ 174.2 : 133 ft lc : 20 - (2 / 2 / 1.73) : 2O - 5.73 14.27 ft from front wall Homent Required to Carry 433 #zrt M : Slope X 1C X.l£ x.l£ 2 3 174.2 x 14,223 85000 1b - ft _ l4 _ Footing Slab Floor (Continued) Check fC f = 211:1 c k j b d2 '2 x 85000 x 12 .344 x .885 x 12 x 20.52 1370 #/in2 Allowable rc : 1050 #/in2. This will necessitate an increase in the area of steel. The new loading will be recomputed in the same manner as above. Try 7/8 in. ¢ @36 inches c/c _ 2 AS _ 1.2 in Z0 : 505 in Loading w : f8 3 d AS 1/8 x 12 3 20000 x .885 x 20.5 x 1.2 .125 x 12 x 16.52 : 1070 lbs/ft - 15 _ Footing Slab Floor (Continued) Length of Cantilever 1:20-(2712/X) C X =.lQZQ_ 174.2 ; 6.14 ft 1c : 20 - (2 / 2 / 6.14) : 20 - 10.14 = 9.86 ft quent M : Slope x 10 X.12 X.l§ ; 174.2 x 21562 28000 1b - ft Check fc ° 2 k j b d 2 X 28000 X 12 .344 x .885 x 12 x 20.52 452 #/in2 Allowable re _ 16 - = 1050 #/in2 Footing Slab Floor (Continued) Check AS A - M s - -———r-—- fS J d : 28000 X 12 20000 x .88 x 20. : .926 in? Use 7/8 in 0 6‘6 inches c/c for widthway steel in the top section of the floor slab 1.2 in.2 5.5 in. A S It - 17 _ Sidewalls The loading against the sidewalls is partially transferred through the wall base and into the footing floor slab. Due to this, it will be assumed that the design load is equivalent to that caused by a fluid depth of 4 feet and weighing 33 1/3 #/ft3. As an added safety measure the entire load at this point will be considered to act on each wall, individually. Load Acting Over 4 Ft Depth : 633 - 2 X 100 3 567 lbs/ft2 MOment M 567 X 4 X 2 4536 lb - ft Effective Depth If possible, a depth of 12 1/2 inches with a 2 1/2 inch cover will be used. d = I k b {4536 x 12 1 O x 12 5.4 inches Use d z 12 1/2 inches - 18 - Sidewalls (Continued) Check fC k 3 b d2 2 x 4536 x 12 .344 x .885 x 12 x 12.52 : 197 #/in2 Allowable fc a 1050 #/in2 Area of Steel AS = M f5 j d 4536 x 12 20000 x .885 X 12.5 .246 in2 ' 5/8" ¢ @’15 inches is allowable. However use 5/8" d @’12 inches for convenience in field layout. - 2 AS = 31 ln .2 2 inches 0 "U" Barg It is chosen to place "U" bars of 5/8" ¢ 0’12 inches c/c in addition to the above steel. The length of protrusion into the vertical wall is 48 bar diameters. The depth of placement into the floor slab is governed by the effective depth of the section. The horizontal portion of the "U" will be placed at the depth of the reinforcing steel of the floor slab. - 19 _ Sidewalls (Continued) Temperature Steel .002 b d At Shear V : 5: II .002 X 12 X 12.5 .30 in2 This area is divided between the front and back surface leaving an effective _ 2 At - .15 in Use 5/8" S 6’24 c/c inches on each face Shear Intensity Va At : .155 in2 26 z 1 inch wl' 567 x 4 ‘ ‘b 2268 lbs t I i __1___. 19 1 £5 3 d ‘ ? a 2268 L 2 x .885 x 12.5 103 #/in2 Allowable u a 15O#/in2 V b j d ______£E¥81_____. 12 x .885 x 12.5 17.1 #/1n2 Allowable v . 60 #/in2 - 20 - Front Wall The loading against the front wall is partially trans- ferred through the base of the wall and into the floor slab of the footing. Due to this, it will be assumed that the design load is equivalent to that caused by a 33 1/3 #/ft3 fluid at the bottom 4 ft of the pressure triangle. As an added safety measure the entire load is considered to act only on the front wall. Load Acting Over 4 ft. Depth ‘W 633 - 2 x.lQQ 3 567 lbs/ft2 Moment M 567 x 4 x 2 4536 lb - ft Effective Depth The wall thickness is 24 inches. We desire a cover of 2 1/2 inches so d = 21 1/2 inches if allowable. 4.2.11. k b {4236 x 12 1 O x 12 128.4 5.4 inches Use d = 21 1/2 inches - 21 - Front Wall (Continued) Check fC fc = 211 k j b d2 : 2x45362gl2 .344 x .885 x 12 x 21.52 = 66 #/in2 Allowable fc = 1050 #/in2 Area of;§tee1 A=_ii__ s fS j d 4536 x 12 20000 X .885 X 21.5 .142 in2 Use 5/8" ¢ bars @’24 inches c/c AS .155 in2 25 z 1.0 "U" bar of 5/8" 6 é‘l2 inches c/c will also be used as was done in the sidewalls. The above steel at 24 inch spacing will be placed on alternate "U" bars. Temperature Steel At : .002 b d This area is divided : .002 x 12 x 21.5 between the two wall : .52 in2 faces leaving an effective At : .26 in2 Use 5/8 in d bars @’12 inches c/c At = .31 in2 lb : 2.0 -22- Front Tall (Continued) Shear V = wl' : 567 x 4 : 2268 lbs Bond u : V 25 j d : 2268 2 x .885 x l2.5 : 103 #/in2 Shear Intensity V V b j d _____2268 12 x .885 x 12.5 17.1 #/in2 A256” . w- 4377/}. “—410——' Allowable u = 150 #/in2 Allowable v - 60 #/in2 Horizontal Reinforcement Sidewalls Loading The load acting on the wall varies with the height of the wall and is greatest at the bottom. Therefore it is not necessary to run the same area of steel the entire height of the wall. Loadings will be assumed as those caused by a 33 1/3 #/ft3 fluid at top 8 feet and 12 feet of the pressure triangle. Results for the "8" foot computations will be 2/3 of those found for the 12 foot loading. W : 1h : 33 1/3 X 12 3 400 #/ft - 24 - Horizontal Reinforcement Sidewalls Letitia Each sidewall is restrained at one end and is considered to be simply supported at the other. Therefore the maximum moment would not exceed 1/10 wl2 x l = 16 ft M 1/10 wl2 1/10 x 400 x 162 10,240 ft lbs Cl/le C11 fc k i b d2 . 2 x 10240 x 12 .344 x .885 x 12 x 12.52 - 430 #/in2 Allowable . 1050 #/in2 Area of Steel fs 3 d 10240 x 12 20000 x .885x 12 .557 in2 - 25 _ Horizontal Reinforcement Sidewalls Area of Sggel (Continued) Each sidewall offers a certain restraint to the front wall requiring an area of steel to withstand this. Loading 15 x 400 = 6000 lbs Reaction _ 6000 X 1/2 = 3000 lbs AS = 3000 a .15 in2 r 20000 . Add .08 in2 of steel to each face of wall to withstand the stress of restraint. :1:- II .557 % .08 .637 in2 Use 3/4 in ¢ bars @38 inches c/c on each face of wall. These will extend 9 ft up from the base of the wall. For the 8 ft Loading Area Steel : .425 11312 Use 5/8 in ¢ 6’8 inches c/c - 26 - Horizontal Reinforcing Front Wall Moment H 1/10 wl2 400 #/ft 17.5 - 1.25 16.25 ft W 1/10 x 400 x 16.252 10560 ft lbs Check fc 2M k j b d2 2 X 10560 X 12 .344 x .885 x 12 x 21.52 150#/in2 fc Allowable f0 = 1050 #/in2 - 27 _ Horizontal Reinforcing Front Wall (Continued) Area of Steel A = M 5 f5 3 d 10560 x 12 20000 X . 5 X 21.5 .333 in2 Additional steel is required to withstand the stress of the restraint which must be offered to the sidewalls 1 Loading = 400 x 16 = 6400 lbs Reaction = 6400 x 1/2 = 3200 lbs AS = 3200 = .16 in2 r 20000 Add .08 in2 of steel to each face of the wall. .333 / .08 .413 in2 Use 5/8 in 8 bars at 8 inches c/c (AS : .47 in2) This is greater than required but the spacing is desirable for convenience in layout. -28... Topwalls Front Steel (Horizontal) Use 5/8 d 6’8 inches c/c for convenience in field layout. As = .47 in.2 Steel (Vertical) A : .002 b d .002 x 12 x 6.5 g .156 Use 1/2" ¢ 6’12" c/c for convenience in field layout. _ 2 As _ .25 in. - 29 - Vertical "Acting" Beam A tie beam is to be used at the upper rear corner of the side walls to render support against the outward thrust acting on each side wall. If possible, reinforc- ing steel of sufficient area will be used vertically at the outer end of each side wall to act as a vertical beam. The horizontal tie beam (12" x 18" x 15') is connected between these acting beams and provides the required support. The wall is considered fixed on one end and supported at the other. Loadings on Side Walls P = w h x 1/2 the distance from the front wall to the center of the horizontal tie beam. T0p of Side Wall Pt 100 x 4.5 x 15.5 3 2 1165 lbs Base of Side Wall Pb : 100 x 17 x 15.5 3 2 : 4400 lbs //6J" 44- 00 - 3o - Vertical Acting Beam (Continued) Moment The moment will be the summation of the moment of a uniform loading and that of a triangular loading acting over a span length of 11.75 feet. 5 H 3235 x 11.75 x 1/2 19000 lbs \\\ //7////////7 WUMM‘ \\\\\ “’ «it-‘aéé-AA\\\\ 1/8 w 12 / .128 w 1 .125 x 1165 x 11.752 A .128 x 19000 x 11.75 20100 7 28600 M 48700 ft - lbs Minimum Value of "b" b : M k d5 48200 X 12 160 x 12.52 23.8 inches - 31 - Vertical Acting Beam (Continued) Area of Stee; A = M S fS 3 d 20000 X .885 X 12.5 Use 5 - 7/8" d bars @F6 inch c/c 2 AS : 3 in Z ,-_- 13.75 inches 0 - 32 - Horizontal Tie Beam. The total reaction on the beam will be an axial load in tension. It is the summation of reactions caused by a uniform loading and a triangular loading from the base of the wall to the center of the horizontal beam. The wall will be considered as fixed at one end and supported at the other. 7 -__.. Reaction RzRu/Rt R : W'; R : 1/3 x P 3 u 2 t 2 1165 x 11,25 2 1/3 x 3235 x 11525 6850 lbs 6350 lbs 6850 x 6350 13200 lbs Horizontal Tie Beam (Continued) Moment Due to the beam being axially loaded the moment is that which is caused by the dead load of the beam. Dead Load w : 18/12 x l x 150 225 lbs/ft As a safety measure in case of any settling of the fill this value is doubled Use W = 450 lbs/ft It is assumed that the maximum moment will be less than that for a simply supported beam with uniform loading, H = 1/8 w 12, and greater than that for a fixed end beam with uniform loading, M . 1/12 w 12. Use M - 1/10 w 12 1 = 16.25 ft 450_x‘16.252 c/c of walls 10 11900 ft — lbs _ 34 -' Horizontal Tie Beam (Continued) Effective Depth d : M k b 11 00 x 12 1 0_x 12 8.64 inches Use d = 9.5 inches This allows the use of a 12" x 12" beam instead of a 12" x 18". Area of Steel AS: M 1 fsjd 11 00 x 12 20000 x .885 x 9.5 .85 in2 Additional steel will be added to withstand the axial loading on the beam. A - R s -— 2 S 13200 2000 [4) .68 in2 Add .34 in2 to each section -35.. Horizontal Tie Beam (Continued) Area of Steel (Continued) AzA 76A S 81 S2 .85 ,1 .34 1.19 in2 Use 2 - 7/8" ¢ bars top and bottom with right angle bends extending downward into each wall. L z 36 inches As = 1.2 in2/ft/section .20 z 5.5 in2/ft/section Floor Beam ...! -...l I L -—---—-'-q l¢————¢#m=——-*4 P. ‘n—I6‘01——h+ I l I l I I I V | ‘T |~ NH Loading Per Foot of Beam Assume a 10 - wf 21 beam Slab Load . 4 ,1 5.5 x 1/12 x 150 x 7.25 = 435 #/ft 2 Live Load 100 x 7.25 : 725 #/ft Beam Weight Assume 21 #/ft ; 21 #/ft Total Load w = 1181 #/ft -37.. Floor Beam (Continued) Moment LII -' 1/8 w 12 l ; 14 ft (face to face) 1/8 x 1181 x (14)2 x 12 347000 in - lbs Section Modulus S : N I um: - 3:;1—140880 19.3 1n3 Allowable Z 3 21.5 in3 Use a 10 WP 21 beam -38- .Plate Girder Panel Loads Each girder is going to carry 1/2 of the total loading. Slab Load : 4 f 5.5 x 1/2 x 150 x 7 = 415 #/ft 2 . Live Load = 100 x 7 x l : 700 #/ft GiI’dGI‘ : [158111516 400 #/ft : 400 #/ft Total W : 1515 #/ft 7' - 3" Panel _ 1515 X 7.25 11 kips *U I The loading for the two end panels is going to be, computed from the average span of 7' - 3" and 7' - 1 1/2". P z 1515 x 7.19 ; 10.9 Kips In computing moments and shears, loadings will be considered as carried to the girders through the floor beams thus rendering concentrated loads at each panel point. i at: M! II II // II I/ J .— rilz’l-u-o— 7'-3"—d-— 71:“+ 713”+ 713"+ 7131—.L3172’“ r 7 W 3 1J l 1 71.4- -39.. Plate Girder (Continued) Depth of Girder d 1/25 span 1/25 x 94 Use 48 inches at center and 66 3.76 ft inches at supports. Thickness of Web (AASHO Spec. 3.6.75) At At Center tc : 1/20 D D = Depth of Web : 1/20 46 Assume 46 3/8 in : .34 inches Use 3/8 inch web (t = .375 inches) at center. Support ts _ 1/20 66 .403 inches Referring to AISC Spec. section 26-b t 1/170 X 66 e .388 inches Use 3/8 inch web at supports. This is under specifications but is considered allowable due to the shape of the girder, its light loading, and the fact that the bridge is subject only to pedestrian traffic. No specifications are established for purely pedestrian loadings. - 40 - Plate Girder (Continued) Moments and Sheagg Mmax : 65.9 x 47 - 11 x 3.625 - 11 x 10.875 - 11 x 18.125 — 11 x 25.375 - 11 x 32.625 - 10.9 x 39.875 3100 - 39.8 - 119.8 - 199.2 - 278.2 - 358.0 - 438.0 : 3100 - 1433 ; 1667 kip - ft at the center of the span Vmax = 11 / 11 / 11 x 11 x 11 x 10.9 % 5.5 71.4 kips at support MO : O kip - ft V0 : 71.4- Kips m1 - 470 kip - ft o-l : 65.9 kips H2 : 868 Kip - ft Vl-2 : 55 kips m 3 . _ V : 44 k1 s -3 1187 kip ft 2_3 p m4 = 1429 kip - ft V3_4 : 33 kips m5 ; 1575 kip - ft V4-5 : 22 kips M6 ; 1656 kip 7 ft V5-6 : ll kips - 41 - Plate Girder (Continued) Check Thickness Against Shear Maximum shear will occur at the end reaction with half going into each girder. V 1/2 x 143 71.5 kips 31>I'U 21500 550x .375 2890 #/in2 Allowable: A.I.S.C. Specs : 13000 #/in2 Stiffeners (A.I.S.C. 26-e) Intermediate stiffeners are required if : 7O 3 176 at: at: : 66 .375 Stiffeners are required. Stiffener Spacing d 11000 t 8s 11000_;,.375 2890 77 inches maximum allowable _ 42 - Plate Girder (Continued) Intermediate Stiffeners The maximum allowable Spacing is found to be 77 inches. Stiffeners will be placed on both sides of web at each welded joint of web; and singly on the inside of web at each panel point and at the midpoint of each panel. This is allowable because of the light loading. w D approximately t 1/12 Width Use 5" x 7/16" stiffener plates for a total of 29 per girder. flélié 3/4" rivets @’5" c/c to 3/8" web = 11200 lbs 4 - 1/4" fillet welds will withstand 9600 lbs If spacing is in 6" lengths L x 5/6 3 11200 9800 Use 2 inch length of 1/4" fillets in each 6 inches of length. - 43 _ Plate Girder (Continued) Data (At Center of Span) Trial Web : 46 3/8" x 3/8" Assumed distance c/c gravity : 47 inches Assumed overall depth : 49 inches Maximum Moment : 1667 kip ft Reduction of Flange Stress s : 18000 x.42 49 17250 #/in2 Required Gross Flange Area A Net flange area minus 1/8 gross web area 1667‘x 12 - 46.375 x .375 47 x 17.25 8 - 24.7 - 2.2 2 22.5 in required in each flange TOp Flange Use a 15" channel @ 50#/ft = 14.64 Add a 14" x 5/8" cover plate =8.75 Total Area A : 23.39 in2 Bottom Flange The flange plate will be formed by splitting a 30" x 1" plate Use a 15" x 1" plate : 14.64 Add a 14" x 5/8" cover = 8.25 Total Area A ; 23.39 in2 - 44 - Plate Girder (Continued) Check Stress maximum moment occurs at center S :.22 I Top Flange s ; 1667000 x 12 x 23.45 29340 16000 #/in2 Bottom Flange 1667000 x 12 x 24.2 29340 16850 #/in2 Allowable - 18000 #/in2 S Point of Cutoff M'; g; I ; 1W / IC / 115 : 19040 3 18000 x 19040 23.11 1230 hip - ft Moment @3panel point No. 3 is 1187 Kip - ft. Cut 14" plate at points No. 3 and No. 10. (25' - 4 1/2" from center line) - 45 - Plate Girder (Continued) Center of Gravity of Girder y AWFW / AC§C / A14§14 / Alsiis / A145’14 AW / AC / A14 / 815 / £14 A 17.4 % 14.64 / 8.75 / 14.64 7 8.75 64.2 (17.4 x 24.811; (14.62 x 47.92)%(8.75 x 49.03) C. £i14.64 x 1.13)£i8.75 x .31) 64.2 : 432 ;_702 z 429 / 16.6 i 217 64.2 = 158213 4.2 24.70 in.from bottom of lower flange or 23.08 in. from bottom of web Toment of Inertia IW : 1/12 bh3 / Ad2 : .375_x 42.3753 : 3100 IC : Ad2 : 14.64 x 23.032 = 7790 114 : Ad2 : 8.75 x 24.132 : 5090 115 3 Ad2 : 14.64 x 23.582 : 8150 114 = Ad2 3 8.75 x 24.392 3 210 '1 29340 in.4 - 46 - Plate Girder (Continued) Flange Welds The maximum shear will occur at the end of reaction. 66 inches .0. - 32.9 inches from bottom of the web _ 71,500 lbs. F4 -< C) t: I 3 41,390 in.4 s = X9 lb 71590 x 14.64 x 33.82 41390 X 1 3 8 #/inch Use 2 ft continuous fron ends then use 1/4" fillets (2400 #/in) stitched 2” in 6". Slippage will occur between the plates making up the flange. The maximum value will occur at the point of cutoff. D 3 52 inches C.G. : 26.08 inches from bottom of web V 3 44,000 lbs. I = 36,490 in.4 _ 44000 x 8.75 X 26.64 36490 0‘) I 280 #/inch Use 1/4" fillets (2400 #/in) stitched 2" in 6". - 47 - Plate Girder (Continued) End Stiffeners Area Reqd. n n H "<3 <3 Use two pair of intermediate stiffeners at each end of each girder. Weldin Required Each pair of end stiffeners will transfer 2/3 of the end shear. P : 2/3 X 71500 : 47700 Length of 1/4" fillet weld : 47700 2400 - 20 inches Use 2” continuous at each end then stitch 2" in 6". -48- Plate Girder (Continued) Check Weight Steel Weight : 490 #/ft3 Web , (46 3/8" x 47' / 19 1/2" x 32 ) x 3/8" x 490 = 3340 Plates 2 x 14" x 5/8" x 50' - 9" x 490 : 3420 1 x 15" x 1" x 94' x 490 3 4300 Channel 1 x 94' x 50 #/ft : 4700 Stiffeners 29 x 5” x 7/16" x 56" x 490 : 1210 ; 170 4 x 5" x 7/16" x 66" x 490 : 17640 lbs lléflg : 138 #/ft of girder 94 Reaction Pin A 16" roller segment with a 24” diameter will be used. Through this, wo 1 1/2" holes will be drilled to hold 1 1/4" anchor bolts, thus allowing for deflection. Length : 16 - 2 x 1 1/2 1 : 13 in. Stress : E l 3 71500 13 5500 #/lineal inch 600D 14,400 #/lineal inch Allowable - 49 - Plate Girder (Continued) Masonry Bearing Plate r'c : 600 #/in2 S : E A A z 1500 2'60?)— 120 in.2 2 ma 1 he Use a 16” x 11" plate, A z 176 in. thickness will be 1 1/2" consisting of 1" of steel acting as a fixed plate faced with a 1/2" bronze expansion plate. 9 f waif-“PL -/13" U 'c‘ : Ibofcl +~z~ ' i /0"- 1539/. -/'-J I at ~ 0 u ; I/ ‘2‘ PL ' l-‘ 1 around - Enfonal'on l-Ijl'n' w " ' " . // «I Pl. -/ -4 gig; chcl- 1'7ch 4 II n ‘ .' . t ' . T . ."'. '1‘ . - ' ." . ~ ‘ 'p ' I'. ‘ - ‘ .' ' ‘. ' ~. . - . . . _ , ‘ _ ..: - 7 a « . -"~- —'~-"“.'-*"' I ' _ : 1 ' - .7 ~ .....-1 ‘ , I i < ' I I , / ~ _ ' ‘ ’ ' . 1' . i ' "- - " ‘ V I I ~ ‘ . _ _. . x , t. .‘ g _ , . . «, . , ‘n ‘. y, , . x . ‘ .- . v, ' v ' . . 4 l . - . . v,_. - I 1 — I . K '. ‘0 ~- . ' \ b V y vv Vlv I ... t . . ". k '_ L ‘ ‘ I ' _. I N '. . . . ' - ' ‘ ' 'r .» ‘ .' - . '1 ' z ‘ . 2., ' . ‘ I ‘ ‘ 8 - ' . ' - 7 — ’ ' 3.... 7 . ‘ ' , . . . ‘ . , — . 1 . l ' - ’ I ’ . ,I .‘ "‘ ‘.‘ ‘ _ . .g _ - ‘ , \ . ' . ' 4 v ’ ,' r ’ . ' A ~ 3 5 :~ 1 ~ 1 . .. . . . - ‘ — . - ‘ . - - - . v, - _ . I . n , ‘ . _ . i , I / . § , . Vs, . , —; - , . - .1 , . . ‘ f . .1' - -' ‘ . _ . _ . .- a ' :\ ' \ 3 I} ' . f ' .- . - ’ ' ,_ - - p . ‘ ‘ I ' ‘ 1'; "'1. 7‘ ' ‘ . . ‘ ' ‘ ' ' I - t ”" " ' ’4 ’ " ’3' '2: , ,r . > ‘ . - . ’ . . J 1 , " d ‘ ' r ' . . ' ' , ‘ . . .. . .‘ 1‘ I , _ ' \ 3. . _. _ . y. a .‘ .‘ ‘ , . _ . r. - . . _ . , ‘ . i' _ . _ . ' ,' ' . ,1, . ' , r ._ - ' ' ' ' ' ' - . A . , - , . . , - , . . _ . . . n - . ‘ ‘ . r . ‘ ‘ _ ‘ - » , . - v 1 . V. . - ' . ‘ ' , . . . ; .. .. > ‘ r ' - ' n - . ‘ I ' ' , - . ' . - .4 _ ' n . .' - i, \ 4 I ' 4 ‘ . - . ' ' ‘ ' N>~_ .. 4 ~ _ . ' . I‘ ' . _ 1 - ' . " ' . q - - . I ’. ‘ . . g." ‘. . _- . - ., I. ) . )1 ..‘7 , 7 ‘ , . I , _ ’. ' f - .. . ‘ ‘_ \v . ‘ ' - “ _ . -. ..., ‘-. .{w ...V»~.V',, , . ‘ 'fl - “- . " . . ‘ _ I \-~('~;-Q-.’-t}--.‘$.~.;-$2p...-vi.m ‘uafi-an-'§'.,i~a‘ue'cl~‘l . ' . . ' ‘ I ' ‘.' ’ . ., - , 1'. -. ~ .-_v . -. ~ ,. ‘. '_ ‘ . ' . - ’ ' " ‘ , ' . . _ . r”... .‘ _ ‘ . _ I e i D‘ ‘ ‘- ' .q ..~{~~—Pi. »--~ —-—«.- v. ‘o‘~<'-' .- - ,- - e - - .. .. . ... ._ . . . . V- r ' 8. .. . - ., - . ._.‘ . - 1 ' _ - ‘ _ \ . , ' .. 1. '. u . / ..A. . 1 __ . ‘ . 1 . y . ' , k , . « , , - ... . . . ~ . - . _ ) , _ ., . . . - . _ ' 1’ v $3 ; .13 _ I. ‘ . , , ‘ .. , 8 , 7‘ _ 7 ‘ .' ; ,r ' . . , ‘ _ :8 - ' . < . ' ' I; ' ' , - . . . Q- , ... ..,. .. _ . .. . , , . \ ~ g . . . "-‘ - -' .4 '3 - -’ _ ; g ,- , _~‘ ,. -' -w' 1 . ' ‘ I , - '~-u-.~'_.'._.{v...u‘\3_..x .4 .fi , ‘- _,\,‘ 1r ‘ - l \‘v ‘ . I . fl... ‘1‘ _.‘¥ . , 1- . _ /4I.l , . .' ' 2‘ . SHIN - I ‘ ’ . ' ' ' ' . ~ . ”~l . . ' 'v . ~ ' . .. J . h -' \. Y w . , . hm... . , ., 1 . . . ‘i- . . . . . ~ ». g " ’1.“ n ' \ l ' . . .; I‘ ’ ' I V S. - . _ .’ ' ‘ u . ~ _. . _ , « . 'i '. 1. ‘ " ‘ x O. , a .. ' ' ' 1.3 " ~ ' ‘ ' - -' ~ - - i. . x 1 I‘ ~ I I ' . . I ‘7’ _ ‘ Q ' ' r I 1 _ t - - . ' ‘ Q I . . I a V ‘-‘_ , '3 : :7 - i - V ‘ - ~. ‘ ,5 . . . - } ~_.. . . I 1'. ' c v . , r 'r' .__ . . . . I . VI. l . '- l . . , . ' ’.,_ I n O I 1 I r I . \\ , . ‘ ‘ , o \ - q t "I . g . > w ' . i ,'. lf-f£.L:crl"-qua . /} n _ . . . .. -b '4 . ‘ I . ' $.14"; .43.----‘4... ...-4....-. ..-. ...v................" ~‘ ,1 ‘ . , . . _ ‘ ,‘\ ‘ 1 ‘ - . . . » W m.-o.~—--—c.—-.... _ - - . - .... ‘ -. ' ‘ - . _ I _ \ r ~ - _ . E . . , - - ~.. ~—.-.e-e~ ..-.» .- . -. . 1 .. . L. . ». ...... . -m g' z ’ '- . ~, -. ‘A. ‘ - : ' '~ <4 .' ‘ l . . . ' . - r J’) _ .. . . -A‘. , g g _.3 ‘ . . . l .. I \ o . .. . - . ~- .- . .- T . —.x ‘ -~--\-.~-V bi“ " /' A ’ . .1!) / b- , A’ f ”a?” W - r ’ 1 ’y' . .-- ~ 1 ‘ ~ - - .. . 4 , ; 370.2 7 94/4713 , 1...; ..zu. / —- ‘ ' ~ . - ' | . '- . .- -- .- - ... .. . .-. . . _.,. ... - ,g,,,... _.3 ,,_ ,_ _,_ . . ... . .- - .. . . . , - - o -_ . .~ ., - -. 7. .. . .- . .4 4"42- 447.18?!» .593::.'- , . . ' ‘ f. . . _. :).-_ _ “eh... -- . "2’; ‘ '_ .‘ V _ . . . . _ _ , ‘ I _ , ’_ ~ . .v , 3 _ 1 ’1. . 7 . -..- . .. _ ‘.' . —‘-« ..._ ..x ... .324 a. ..-w.'--'¢--<..... ‘ - I .7_ . ..- . ',_rl' . ‘ . _-~.——-—-.~.. ..-...— M / 4,. , ‘ . Y".',:—:' "' '.- “ ~ - - ‘ --‘-’ ____"_“_-.....e-~— ,4- - -.-. - -- .- - -..,1-.......~~.«w~ . l . . . l * *‘._. -—1 - s -..... ...,.-. ..r-u- .. --—,~.. .-.. ..-..... ... .--.. "f- ,4 v , AZ; 7" /;;/f4~. “......“ If 23’ {'54, I a .1 -, ..- b” ‘I':': ~n /‘/ ... I. .. .-.—x ~ \ .- ... Mm. m.- m rel/145.2,;-aerroza _ ; ‘ ,' 4 -V--- a... .-.... ...- vim. .. i . r . - -..“-..Jmiwil— A. 8 ' : l . . - l ' 1:. C - 4- 1. ..-.- —_ W“. , . \ I 1 v . u . -——-~o “.-._v~..~\ . ; . . r . - .1 ...- 1 w, C's-pus— -w 'M‘ ; - . , , ‘ , ’ 4 j a I ‘ " ”2 .. v.“ I " w . ’ - 5 ' -' . . ‘ p 4 - - f n' s - -. t- ‘ t J 1 ~ F '- u.’ ' 7"" ' -: w—O 2. Nun F-ro .....o ‘ 4 ~ ..- I, I 6 ~ \ ... . , \ ~*~v- .. ...—.....wz . n 1. t- H ()5 a I. ’ ' _ ,' 8 . 8 . . _ ‘ - - - - . .u . , ‘ . . f . ‘ . ‘ I 1 . r , k- ,. ’ ”d It ‘.,__ k I f ,. . ._ p ., . , . FEB-38,7949 mater .-v‘OMv-v £78.»."v u v 0.4 . - ...» ..,~. .- ..-“. -~.-.-..r-‘~-,a...-4--‘4»-r..‘ . w 1m~n.ou'1 ammo... w! ”evomns —.-. 4— v . u ,--l . x 4. ... ... .... .. . ...U . . - .... ,.__...-.‘ ..-...u..-uu- .-~~-- -.»- or -- ..--.. . .- -...-........-..4 .~..—u..~._—..... vggopw‘- nu-.- .a-n- -\ w. u P.-‘~--Nv~ ...... .wp A“'- ...—.-.- .-. a. .It: I .1bf‘t‘l1’lv. -...{llt .1111 I" 1'... .11, «0:... I: .. .... ‘a<-~‘-: _ _m~wvc-I"- .'~ aw. ' - ..- ~"" ' ......n H... -- ‘ .—----‘ .... .n— .... M v... .- ...... V». 4. , M I l ‘ u. u . u ‘ . 1L7 . . . .11 l‘ ‘1 nil.lv.'"- v‘offhfi?' I II I] h .' F Y. n. n: .9. .. . .. .. . J. v. .Lv . A . O i ' I a. .. u .. . ... . . . v . . _ . . . . . ‘ I. . $1... -vl..:£isll .- ..{bw . .ngwaga a . . . h .lu.A....«u.¢l..§t v I. .9 ...4 - -..-a...» Abo-vtv‘ ~—- v 7 b"- . ' . ' ‘ ' \ avg/J }. Ti. . i‘ MI- -1 .{I . . , L . Wfa‘. -... ...... ..L. “....-.A-.‘A>- . \ .. . 1.2 . . ... . . I .41 it) . . . . . u . ....x;. . . . . . . .1. . ..4 . . T110", o ‘.\\ .Iv...:“tl.:~.l.'Illlotluf . I; n Sly-Ly . . .v . . . . 7 .... - . . u. m .. . .. . ...; . .. .. ... ... . .. v \. . . .. . . . Q - . . H . . . ‘ . . u - W . n \ . .. . ..v A : ~ I .a . .1 . . w. . I .. . . . . , .. . . . . v .. v N . . ... . .. . . .. . . . “T .. 4 O . . v . 1 .. . . . . ~ ... . . .. . . . I... . .... . . v . . . I ll’t‘nllz vfl‘n..\.1.svwnvt.l~“|€|1i.. . .m 1i: ... .. . . . . . . . . . . u 4 . . . . . .. . .. 1. . .. . . . . . . .. . v . . . .. . . ... . . . . . r . ; . . . . . .. x. . . . .. .. v n p m . b . .L . - ... . . . . c s .. . ... . . 2 ~ . . . n I l ’. ~ .. n .. I ... . . . . .. .. . . c I . . ., p . ... . r. n . . .. . A . . . . I . . T . . .. 1 ... _. . . . u. g .. . . . . . ,L . .. . . . .. . . . . . 1 ., . . x . . . . . . .. . . . . . .. .. u ... . . .. .. . . .. . . . . ... . . . . . ... , ...4 X _ . . . x .. x . . . . . r .. .. . . _ . ... . .. .- .. ... .. , . . . . . . . . . .~ . . lJlilJTkI 9.1.4.11... ...vnw.l.!.4ububsl.l.lotl.ill. o .IstGp .. .a . . . p A» . . . U r .. . . w .. . . ., s . . . . p A ... . . . . h . v .. . h l. . .r. .H . .. . .. .9 . . .. . c. . ”.... a u .. .. , W .. ... . . .0 . . ... I. . . . . . . . .I . . . . ... r: . .. . ...H... .. .. .... .0 ., .... . . . .._.. ... . . . . . . .. . . ... . 1 J .. . .. . . .p. i. ... .. . . . . . . . .. . . . . .. . , . . . C . .. .. . . v . . A. . - I ; .. . .. ’ .. . . . H 4 x. u .. . . . ~ 5 w 0 .. . q . I .. . .3. .. .. ., . . ‘ . . Q ‘ .. \ 0 ~— .. n . ... I n l 1 . r . _ . . . .. . .r .. . I. . 1 M. .. . . P . ... .. . . ... ...vAJI.lII“I¢I a ...-54.3.»,144 . .1 .M! n . . .1 . 1 .. D ... .. . H. . . .t , . , ., w . . ... . .. . . .. , H. . .. . . . H . I V . . A . .. . . u . . * . .. . . 3. . . .. , .. . . . . .. . . . 1. 4..- 1.934.. .... . ... .i-‘ . . . . . . . ... .. .. ... . .. . .. . . . .xkhnnfimvfl..N. . .. . .. . O. . . . .. . . V . . . . . . . . .. . . . u . .n . . . . .. . .. . 1 . . H . . . _. . . a . ..... . . .. w v)? 1 :0 .5i.|..fn..l)l . I . . .|,. .w. .. ¢ .l < 9.1)-- u . . . ... r. . . x . . a . . . n. ... A . . . .. n . , 4 . .. . . , . . w \ . . .. .. . . . . . u. . g. . .. .. . x . . . . . _ .. . . \ . . . w m ,. . . . ...Iv)‘ .3. ‘r. .. . ..4. ; 1...... s........o’~....‘«.lccu . \ . n1.‘\ . v .. .\ v . I; .1 o n 9.. . . . . .v. .. . fix. r .§ ‘ . . . . . .\ \ . .I . h~ . . . . .. . .. . « v . I O \. - ..V I \ . A .. u . n 17! la, y I ’....... (I . 4 .o.. ...; \o’i ‘ 'IJIEI‘JT . , v . .I .0 n . . \ .. n... \ . . . u \..|.l..li..\o\.....9o5; .a 1.. u‘I; . j ...!"III‘" . . W ~'.- ~‘..-..‘.—4‘ 'lnthl. ..‘ .... ..f..- . -.., 4... ...w- .a;-.« -— 4 ...»An nil}... It‘lill. ‘n. ttf. 11.1.” 1.... III. 0 ‘0'! . . ... N§§Qfi ... . .. . . . . 4 . a . . .. r _ . . .53? .4. unt‘u‘o‘mh . . . ..L...A.’ .- ‘1 v “A 1.. -, ,3. .v. .4. .... . 4.. ..4‘..¢- . » .... "-...-. .. “...... 44.-.... .-....--4.. -.. . . A ,. .-.».'..-4 f' .f---,.».o ...}... , - n a- “van... .-.‘» - . , ......ru Vulc .- ...4» ......n -. - ‘_ . ......“ ... -..-own. . 'v >A-n ...-...fl.... ....N. 4..."... ... ’wv“. I /¢’0 0% / I) ,. /. . ...!)olfi-ll...) (I . .1! x-Iwa‘i' f .- ~'. V, . ’ t C -11, , . vr‘ f} lu-Ml‘;l~ vf - . ,3“ . .. _ 4. .....r---£.tr.. .... t“. ’7 A AI «... r. f - ..-—no ..p .- / 5' 11,3. .,,,.~..~ ,, .. . - ~. "7 .,n’g (. I. n" m L my“... ..—-..-—. .fiff‘QC'W-vé ...... ,1 i .._.l'-7‘ p- ..‘I¢ />’ 37143314 A1; 3 ..‘I: a v -,‘€ 5.. ‘9” ; I .1: ~rr fr . w. w. . .M ...; . u. ”w h. a .. ... .. . \ .. M n. . . m w a. M .... fl. .. v a . w .... w . H , ... . M . M w w M . N n ’w W . . M . M .loA ... .. .t- . ..Jlrlllillllllii ..Inllllll ll! .... - ......i..:......... w M m T “1 .115: (0...... Incl-... I'll... .... .4 . w. ... I I I O 1 o w 1 II.J v r... n r ....r.» . 4 ......IIIIIL. _ w . /... .. ,. . . / .... . . .....1.. A .. . . . . . 4 . .. J L ... . .2 . M . 5 w . W ......H. ...: . .. . .. ., 4 ... . ...... . . .szt .m.. . . ,. .[ru . .M w _ m-.. I I 7’3. i I § 1 3 1 “...—r... » ;\.. .v.4( .... ‘ ~ . . . . . . . .... . 1’ OD... ..I. ... . 5... .. . . lav .... . ..m. . .....u: z ... , . .11.... , ..k... . “......i, . . . ... . .M r .|§ .... . \ ... y u. . u .....a .I. ... ..c. ... . ...... n. . ... ~.- . ... .. u w . . ’1‘. . g .1. ......Hxx .r.. “.... . . .....1 . . ..I. w . .. ...A .. .. fl ... . .Z} 4 . L .I . xi”; ,. .W . :1. _. . ..~ A . .. 4.. . : . *4 . . v .m; .. . a Y. x . .4. I x . M G I ,w. ... .T. l v . .V .. I. I. 4 . [Wit ..1 I... .. n c I. \n:lw(.l .1. . .n . . . .. . .\\l1Yl . . . ., .....r . . -..—a «. .1- . rm . .u . . .... ..3 . . a w. .. .... .. . . n. . . .. ”hr. .....v . . . . .. . ... . .~.t . . . . . - . . . u r. . ~ I .1 .. . .. . .. .a . ‘V. ' . WNW ... - r\, ...” J . y - . . ~ ‘ M.- ”-.. ova-...... --~¢OM -m~- MW “fl Mn‘. "‘ . .. . .. q. . .. . J _ ... My a u. .... - J .1.-. ._~ III-OIIIIIV I O "I"? h. ..-, u-.. a u v. . f t' - I. u I --._............--. ”...-......" ".1, . ........'..'.- .I‘ ‘ v‘r'r». . \ > a... ......”L,‘ A . o ‘ '2/‘1/54: 8.97722,” .3!» ..wi. b” I - ‘ £4: 5/5 /’ ”49.3 7"fi P . . .u". ‘1 «4' v5- ’15 . r“: 8 w L ‘d M w w . ,' )- x). 3 (1: £3 ..2‘ , f o” J .4 fl "‘5 k. ’90s, . ,1 “an I F} r. I la. 2’ EDGE A: /A?. 4 / O’," I I /. ./"T'£X"T £3 4 {CALE .0!“ .A I i JNA1C3L43