•iSF ^ 1 1 ii GRAPHICAL CONSTRUCTION OF THE BALCOM STEPPING By "The Isacoustio curve" The slope of the Balcory and height of the stoppings shall first be considered, so that the steel used in the construction of the balcony may be designed accordingly. TOien "setting up" the sections of the Parquette circle or balcony in the theater, it is desirable to sight from the eye level of the spec tator, which will be considered as 4 feet 2 inches from the floor when the spectator is seated, (.and 4 feet 10 inches to 5 feet when standing). The theoretical principles used when fixing the heights of the steppings upon which the seats are placed are as follows: A point is fixed on the curtain line 4 feet below the stage level, and from this point, after the distance from the stage, the stepping, and the floor level is placed, set up the spectator's eyes 4 feet 2 inches above the floor, vertical with the back rail of the seat. Now from the 4 feet point on the curtain line, a line should be drawn cutting through the eye of the spectator in the first row, and produced until it cuts a vertical line set ip at the back of the second row. Then from the point where the vertical and radial lines intersect 5 inches is measured up and that point gives the eye level of the second row. From the point below the stage, a line is drawn through the eye level of the second row, and produced until it intersects the vertical line set up at the back of the third row, and from that point again measured up 3 inches for each row, and from each eye level, measured down 4 feet 2 inches will give the floor level for each stepping. After getting the heights of the stepping in this way, the nosings are not tangent to a straight line, but to a concave curve, and the steppings are n<^ equal in height but become steeper as they recede fmm the stage. Tliis curve has been named "the isacoustic" or equal hearing curve, and is a refinement seldom practiced. Thus each row of spectators the stage. This curve has been named "the isacoustic" or equal hearing curve, and is a refinement seldom practiced. Thus each row of spectators in this balcony have a sight line 3 inches above the sight line df the row just ahead. See next sheet for graphical representation. V' " Mathematical Computations of Balcony Seating To find a mathematical equation by which the height and difference in height of the balcony steps l,i,e. seating of the balcony) may be e^ipressed exactly to conform to the sight lines as drawn. That is, the line of sight of each person is S" above that of the person occupying the seat directly ahead. 3ICLTCH Of 5lCjWr L1NL5 Not DHAwr; to Dcml €■ oy Z'^ifotf - L = 43' 2'\« 518" h = 15' from a - 34" drawing e » 3' Each step (platform) is measured down a constant distance (50") from the line of sight (or eye) of the person seated. Therefore the height of each step; step 2 ■ e -f- a tan 1; step 3 ■ e -f a tan 2; etc. hi = 15' -J- 7" -/• 4« 2" hi - 19' 9" s 237" h2 ® hi + e +• a tan 1 hg = h2 + e 4" a tan 2 tan 1 ■ hi ® 19.75 - .4575 "i; 43.1666 tan 2 = hp. = h, 4- e + a tan 1 = 257 +3 + 54C .4575) = 2S7 +3 4-15.5jS L + a 43.1666' + 34' 518" +- 34' " 255.55 = .4625 552 tan 3 = _h^ = hp +■ e 4- a tan 2 = 255.55 +• 3 +-34^,4625) = 258.55 + 15.71 ir+2a L + 2a 5l8 +• 68 586 « 274.26 = .468 586 tan 4 = h^ = h.^ + e 4- a tan 3 " ^4.26 4- 3 +■ 34(.458) = 277.26 4- 15.9 L 4- 3a L +■ 3a 620 620 = 293.16 - .473 620 The height of the first step being taken as zero, because the first row of seats are to be placed on the floor level of the Balcony directly back of the rail, the height of the following steps up to the first aisle are gxven. Height of steps up to first aisle: Difference in Height e 4- a tan 1 = 34- 34(..4575) « 3 4-15.55 ■ 18.55" of steps to first aisle, .160" e 4 a tan 2 = 3 4- 34C.4625) s 3 4- 15.71 = 18.71" ,190" e 4 a tan 3 = 3 4 34(.468) = 34 15.9 ■ 18.9" .200" e 4 a tan 4 = 3 4 34(.473) = 3 4 16.10 = 19.10" These figures give us the correct difference in height to 3 places of the first 4 rows of seats - which brings us to the first aisle. That is, each step increases by a small amount above the height of thes proceeding step. The increment of difference in height changes so slowly for the first few steps that for the first 3 places it remains almost the same. The total difference of the first 4 steps to be taken as .55" (counting the balcony floor level as step l)« How starting from the first aisle, to find the height of the steps up to the second aisle. Using for the steps above the first aisle (as shown on the graphic diagram sheet ). a* - aisle width - 3' 10" - 46" a^tan 4 +( e+a tan 4) ■ 293.16 +4 6 (.473)+ 3 t34(.47S) = 293.164 21.78^19.1 - 334.04" L« - 12 X 58.333 ■ 700" tan X = h,' - 334.04 = .477 l7 700 tan 2 - hg' ■ h^*f ®4 a tan 1 a S^4.04-^3+34(.477) L'-f a 58.333 + 34 700 + 34 - 337.04 + 16.22 - 353.26 - .482 734 734 tan 3 - hg' ■ hj>« + e+a tan 2 • 352.81+3 +S4(,482 , 2a - 372.81 " .4855 76S Lt+4a L*+4a 700 136 = 395.114-16,63 - 411.74 - .493 5^S ""55^ ® - hg*+ e+a tan 5 „ 411,74 4 3 +34(.493 L«f5a —^i^tsa 700+170 a 414.74-416.75 - 431.49 - .497 870 870 tan 7 - hy« ^ h-M e \-a. tan 6 a 431.49 43+ 34(.497 L*4 6a L*+6a 700+204 a 434.49f 16.9 • 451.39 • .5006 904 904 tan 8 - hgt • h„« -f e-4a tan 7 - 451.49+3-f 34(.5006) Again, considering the first step above the aisle as zero height. Height of steps starting with the Difference in height second step above the first aisle, of steps. e -f- a tan 1 - 3 + 16.2 - 19.2 inches, e + a tan 2^ 3 + 16.35 - 19.35 inches, e + a tan 3 - 3 "f" 16.50 = 19.50 inches, e -t a tan 4=3+ 16.64 = 19.64"inches. 0 + a tan 5 = 3+" 16.77 = 19.77 inches, e + a tan 6 = 3+ 16.9 » 19.9 inches. 0 a tan 7 = 8 + 17.02 = 20.04 inches, eta tan 8 = 3+ 17.1 " 20.13 inches. This gives the correct difference, to 3 places, of the. heights of the stoppings frran the first to the second aisle. A total of approxim ately .93 inches in 8 steps. Starting again above the second aisle: The first step taken as zero heitht. Use h]!" for the steps above the second aisle. h^" = hg' 4- a tan 8 +- (e 4- a tan 8) = 472 + 46(.504)+ ( 8 + 34(. 504) = 472 + 23.15 4 20.1 = 515.25" L" = 74g* = 894" (see graphic diagram sheet ) tan 1 = hi" = 515.25 = .578 894 tan 2 = hp" = h" + e + a tan 1 = 515.25 +- 3 +34(.578) = 518.25 + 19.65 L" + a 894 +- 34 §28 928 = 537.9 = .580 928 tan 3 = h?;" = h2 + e +-a tan 2 = 537.9 + 3 +34(.580) = 540.9 4- 19.7 +Ta 894 -4- 68 962 "962 tan 4 - h4 = h^" 4. e + a tan 5 = 560«6 + 5 + 544.. 584) = 563.6 -h 19.84 L"- f-3a 894 -f-102 996 993 = 583.44 = .586 996 tan 5 = 1^" = h4."-4- e +a tan 4 = 583.44+ 3 + 34U586) = 586,44 + 19,9 L +4a 894 -f- 136 1030 l030 = 606.34 = ,589 1030 tan 6 = hp" - h^" 4- e -t- a tan 5 = 606,34 + 3 +■ 34U589) = 609,34 + 20 5a 1^*"+ 5a 88§ + 170 1064 = 629,34 = .59 1064 tan 7 = hp" = hfi" -f e + a tan 6 = 629,34 + 3 + 34U59) = 632,34 i 20,10 L^4- 6a L" + 6a 1098 1098 = 652,44 = .595 1098 Starting with the second step above the secohd aisle. Height of steps above the second Difference in heists Aisle of steps. e + a tan 1 - 3 419,65 - 22,65 inches, e + a tan 2=34 19,75 " 22,75 inches, e 4- a tan 3 = 3 4 19.84 - 22,84 inches, e + a tan 4 ® 3 4- 19.93 - 22,93 inches, e 4 a tan 5=34 20.1 = 23,1 inches, e + a tan 6=34 20,17 = 23.17 inches, e + a tan 7 ® 3 4 20,221 - 23.221 inches. ,481" The height of the last step at the top of the balcony stepping = 23,22" The first step after each aisle has been omitted in finding the difference in the heights of stoppings in each case because of the difference in the value a (.or width of aisle) for this stepping: a' = 3* 10" = 46" Using the same angle as for the steps; preceding the aisle, the height of the stepping following the first aisle Height of step following the second aisle. Using the same angle as for the step preceding the aisle. = a' tan 8 = 46t.504) = 23.15" To find the distance of the curve, drawn tangent to the nosing of the steppings, above a straight line drawn from a point 2 ft. in front of the stage at the orchestra floor level and tangent to the n nosing of the first step ^see graphical diagram sheet ). All the increases in heights of steppings must be summed up. Thus^ starting at the toe of the balcony and the first step, if iiie additional height of the first step be denoted by ̂ a), the additional height of the second above the first by (b), the third above the second "by Co), etc. Then the total height of the last step above the previous ly mentioned tangent straight line will be a f (a + b) -4- (.a 4 b c) -f -I- (a + b+ 4 n). Numerically; from my calculations. inches 1st step rise = ,16 2nd step rise = ,16 + ,16 3rd step rise = .32 + ,19 4th step rise = ,51 + ,20 35th step rise ,71 +• .20 (the same angle being used for the aisle step as for the preceding step), 6th step rise - ,91 + ,10 (change in height of step preceding and 2nd follow ing the aisle). step rise = 1.01 -f ,15 step rise = 1,16 + .15 step rise = 1.31 + ,14 step rise - 1.44 + il3 step rise - 1.58 + ,13 step rise = 1,71 -4- ,12 step rise = 1,83 ,11 step rise = 1,94 ,11 (step following aisle). step rise = 2,05 + 2,25 (change in height of step preceding and 2nd step following aisle). 16th step rise = 4,30 + ,10 17th step rise = 4,40 4- »09 18th step rise = 4.49 4- ,09 19th step rise « 4^58 4- ,08 20th step rise = 4,66 + ,07 21st step rise = 4,73 4- ,05 48,59" Thus^the rear of the balcony has increased in height 48,59 = 4' 59" due to the isacoustic curvature. In this design 3" has been used for the rise of sight lines of one row of spectators above the other. This is less than the amount generally used in working a balcony problem on this basis^ 6 inches being the usual difference used in heights of sight lines for large theaters. But in order that the rise of the rear of the balcony would not be so great as to make the height of this building out of proportion with its width, an amount of 3" has ) been used merely to whow the calculations and procedure for a refine ment of this type. In this case I have found that at the Girder supported by the columns, ̂ which comes under the 12th step), the isacoustic curve is 12.66 inches above the straight line drawn parallel to the first step nosing. This will be used later in the design of the step frame work. All computations have been made with a slide rule, therefore the figures to several places of decimals may not be entirely accurate. A plan of the balcony framing will now be laid out and the necess ary beams, girders, and columns designed. The balcony floor slab will be assumed in ttsv depth but will not bo actually figured. The process of floor slab design will be taken up later in calculations for the Mezzanine floor. In designing the balcony stringer beams, their lengths have been scaled from the Balcoiy Plan. That is, the lengths iised in the design are the projections of the actual lengths on a horizontal plane. This has been compensated for by using a live load of 125^/ sq, ft. of horizontal area. Otherwise a load of 100/,y sq, would have been used. CALCUIATIONS FOR LOADS OK STRINGER BEAMS Loading: Live Load of Balcony 125^/ sq, ft. for small beams or stringers. lOOH^/ sq, ft. for large girders. Balcony floor is to be of concrete slab construction. Assume the weight of the floor to be 50^/ sq. ft. Weight per sq. ft. for stringer beams = 125 4- 50 - 175^/ sq. ft. The framing of the balcony is symmetrical, therefore only half the stringer beams need to be figured. beam la.~ 33^/'' SIZES FOR STRINGERS Beam la *1 777 ^ R.,= l4.,ZoS^ Load = 15* 10" X 20 X 175#/ sq. ft. ' 2 = 158.333 X 175 - 27,75(^ « load on beam. y/eight / ft. = 27750 = 1387.5^',/' ft. 20 From Carnegie tables - use 14" x 6§" - ZSjf C.B. Allowable load / ft. = 1430#/ fb. From direct computations Max. M s WL a 28410 x 20 = 71025*# x 12 = 853,000"# 8 8 M a I = 853000 a 47.4"^ "S 0 18000 Use 14" X 6f" - 33# C.B. IA = 47.8"'^ Assuming the wt. on the Projection Room floor to be not greater than the live load on the balcony floor. BEAM 5A Uniform load - 175 x 9^ - I660j{=/* Concentrated load at center a 14,205# Max. M = 1660 x 15 5/6 x 15,5/6 14.205 x 15,5/6 = 52100 56,400 8 = 108,500*# Use 16" - 45}# C.B. lA = 73.8"3 BEAM 2A If9 ,00" Uniform load « 175 x (8.5 ^7.93) 14-10*/' « 175 X 8.21 = l,440#/» Concentrated load at center I = 14205 J.1660 X 15^ = 7100 +12.860 = 19.90Ol^ aoorox. 2 2 S.MR2 = 0 Rl ■ 1440 X 29g X 14f 4-19900 x 9^' 62 7500 -+-189000 = 816500 = 27.650# ' I I I ! o/\X' 1 I I ' ' ^ yv mm eeww**** Rg = 1440 X 29|- X 14f + 19903X 20 = 627500 4- 398000 = 1025500 = 3475# 29.5 29,5 29.5 Max. M. occurs at X ■ 27650 7 19.2* from left reaction. 1440 M. = 27650 X 19.2 - 1440 x 19.2^ = 531,000 - 265000 = 266,000 •# T" 12 X 266,000 s i/r = 177,5"3 l8^ Use 24" - 81# C.B. I/t = 190.l"^ Checking- by; taking the wt. of the beam into account. Uniform load ■ 1440 +- 81 - 1521#/' Rt = 27650 + 81 X 29.5 « 27650 -t 1195 - 28.845# _ Max. M. occurs at X ■ 28845 " 18.95' 1521 M. = 28,845 X 18.95 - 1521 x 18.95^ = 547500 - 274000 = 273,500'# 273,500 X 12 = lA " 182.0"3 18000 .*. 24" - 81# C.B. is OK BEAM 8A Assuming 8A to support 2' of floor on each side of its center. W = 4 X 175 X 6.5 = 455# w a 700#/* M ■ TO = 4550 X 6.5 x 12 = 44,200"# 8 8 44200 = l/r a 2,46 18000 Use 4" - 7.7# I IA = 3^0 Assume 7A to support approx. a 3 ft. width of floor slab. Uniform load W s 3 x 175 x 15* a 785C# w a 525#/* Concentrated load a leaction of 8A a 2275# = P H4-- « 2275 X 114-7850 x 7.5 = 250004- 58900 ^ 15 15 R, = 83,900# a 558# R2 a 4,57# 15 Maximum M. occurs at X - 5580 - 2275 - 525 x 4 - 1215 - 2.32 5^5 525 M. a 4920 X 6.32 - 2275 x 2.32- 525 x 6732^ = 15,475 x 12 = 185,500"# 185500 = lA ® 10.3"3 18000 - U se 7" - 17.5# I - beam lA = Assume a uniform load of 3 ft. of floor slab & floor load, = 3 X 175 a 52#/* Supports 2 concentrated loads " P-i a 4,57# Po = 195# (Assumed) 4^70*-p, R-Xi - 4570 X 4.75 4- 1950 x 1.5 4- (525 x 8.5) x 8.5 ̂.7'^ - 43575 = 5,14# R„ = 584# |-r 8.5 r' Max. M. occurs at Pi. Ma 5140 x 3,75 - 525 x 3.75 = 19280 - 3620 a 15,660'# x 12 = 188,000"# 188000 = IA = 10,44""^ 18000 Use 7" - 17.5# I - beam lA = 11.l"® BEAM p. 3A will be designed to take the unifom load - the entire length of the beam, also the 2 concentrated loads from 8A and 6A, Uniform load - 175 x 8.5 = 1490///' 4-80^/^ (assumed wt. of beam) = 1570#- Concentrated loads Pg^ - 4550 - 2275# Pgj^ - 5,840# 2 Z T.'IS^ Rl - 2275x25•4-5840x14+1570 x29.62 1 ISlo^ — ^ -^9:5 7 = 822650 s 28,00C#,^ ' 29.5 . R^a 1570 X 29.5^-'-5840 x 15.5 4 2275 x 4^ s 684000 -+90500 4- 10230 29.5 29.5 a 26,550# Max. M. occurs at X = 28,000 - 2275 - 1570 x 4.5 = 18665 = 11.9 1570 1570 Ifeix. M. occurs underload Psa ■ 584C# M = 26,550 X 14 - 1570 x ̂ = 372000 - 154000 = 218000'# 2 218,000'# X 12 = 2,620,000"# M z I 2620000 = 145.5" S Y 18000 „ Use 24" - 70# C. B. l/f = 161.6'"^ BEAM 4A Uniform load taken as 175 x 8.5 = 149C#/' over entire beam. 1 concentrated load Pqa " 5140# Assumed wt. of beam - 70jf/* Total w - 1560#/' R = 1560 X 5140 X 14 ^ 2 SI4,0^ 29.5 JS'/^ =751900 s 25,500# 29.5 l£r(,o*/i Rg = 25,640# ^9/4 Max. M. occurs under the contentrated load. M. s25,500 X 15| - 1560 x z 395,500 - 187500 z 208,000'# -r~ - 2,500,000"# I/y =2,500,000 = 139' Use 21" - 67# C.B. l/l = 1^2" BEMI 3B Carries a uniform load = 175 x 8«5 a 149(^/* Assume the nvi:. of the beam s 65^/' Total load W » 1555#/» x 25^ S 39600# Use 14" - 58# C. B. Allowable load = 40.1 kips, BEAM EB Use same size as 3B. BEAM 7B s SB Assime a uniform load s to the weight of Z* of floor load on each side of the center. w s 4 X 175 = 700f^/* of beam Yf = 700 x 11» = 7700# M, a WL = 7700 x 11 a 10,587 '# 8 8 Use 6" - 12,5# I - Beam (American S,) BEAM 4B Assume 4B to cariy only the concentrated loads imposed by 355.01* 7B and 8B. Assume wt, of beam to be 20#/* 1.0^/' _2 M, r 4008 X 8 - 3850 x 3^ - 20 x 8 ■ is'I '2 - s 32064 - 13500 - 640 a 17,924'# ^.4.006' Use 8" - 18,4^ I - Beam (A. S.) BEAM 5B Assume 5B to support 4' of floor space plus the concentrated loads of 7B and 8B, Uniform load = 4 x 175 s 700#/' = w W - 700 x isi? = 11083# = 385C#= P8B Assume v/t, of beam = 30#/*< M, = 9690 X 8 -3850 x 3.5 - 730x8^ - 40,600'# Use 8" - 31# C.B, BEAM IB Assume uniform load 1st 10' from the right end of the beam s 175 X f8.5 , 15U0'\ = 175 x 12.25 = 2140#/* 2 ̂ Ass une for the remaining 15-|-* uniform load of 175 X f8.54 4|n = 175 X 6.5 = 1140^/« Concentrated loads P„ s s 4,008^ 40|Oft* f— % //4.0*// 'LI 40*^/' 25# El = 9690 X 21+ 4008 x 10 + 1140 x 15# x 17f+ 2140 x 10 x 5 + 80 X T" Ri r 203500 4 40080 +~ 324000+ 107000+ 26000 R^ = 700580 = 27,45(Vf. R?. s 54840 - 27,450 - 27,390# 25.5 Max. M. occurs at X = 27450 - 9690 - 1140 x 4# - 80 x 44- Z 27450 - 19770 - 7680 s 6.3' 6.3 4# s 10.8' from left end of beam 15^15 Max. M, - 27450 x 10,8 - 9690 x 6.3 - 1140 x ToTS^- 80 x lO^ " • 2 2 = 296500 - 132.160 r 164,340'# Use 18" - 64# C. B. . H ■ j ■ ' v t ̂ * r.j f " •*'^ -5" ^ - \ 1l r^ » ;'i -l&r' . '■ ;;fe ''- ". 'i Asti^i r'^\:p^z^y S, ■■V-'":"- '-.. ■■ 1. =■- ■ ■ '■ ' _ ' -V-.. - ,.•! J>^- - : V •^'V ^ \ t t'rl ' .'-UNifoe.M LOiD-ii : 'xm If t i V: -/frI n " '• ' ' 4! » M' f M' '' .I I '' I I ' iL . 4i=Ui <.LJ I"--. >"- TT -TT^—•^rn-.-Tfr"—;■»-.• . '■ ■*->~r—~- h I i I >f: -ji ■t-t \ V ;?\ DUGSLA% /HOVJHa folLCL/ j^CTIMQ I-t? >■'> ■-J-. fi Ak-^ V . .. .- '.'^'f, . 't* •' ' rt- r f' . -\ - . s'+ -• 'if IWV . j '- . 2 ■ ft; IlAS X:^iX J . '' 'A • '-'f.' ' 4- / ft . *f' ijyv", - + + „»t, ^ ■ . - - • ' -X-X IV% St.s; r':f^4 4-4- +.!L :^t~rr.rtvft-:--.-; v ..s»3;.r*EA.Si??.c:ii-tsfe +«4 sif . 1 I! ' s.x|piPK» - f -c ft A ̂ r . V, / ' A ^ Dtf^QlLAW THOVmCi 5ALC0NY C At^TILLVLlL. CONNLCTIOk' " X/V^ ' % 1 i ^ I L:- ;f-A3A5 ■, .■ B i ft ViV ' "-v < i- iklH?) - f A f-f A 5r X i I AlNf! ■ . ■■ .- -'^ A '■ A-. ' , ^~ ' -- V . » # .►- i'- y-A- . ; f ■ CC/A. CAHTILEVER BEAMS CAtlTILEVERS IC and 20 Length of beam A - 5.5/ccs 30* - 5.5 " 6g' .8^ Uniform load on A ■ 175 x 6,5 x 15*10" s 9100§= "2 Uniform load on B - 175 x 4-4- x ~ SP p na ic ̂ e D/A(^R.kM 15*10"- = 6300# s Rl = 6300 X 7.75 = 890(^ 5.5 Rp B 6300 X 2.25 s 142C^ c C ' '' ^alc-ontf noic:! The horizontal component of th force of Beam A a H « R^ x cot 30• 6 300'* a 8900 X 1^,732 a 15,40C^ ZloCid^= R-2 Max, M. in B at Rn a 6300 X 4.5 - 14,200 x 12 = 170,000"# 2 f1 Og.CE /f/AajJCAM foe. 5EkfA 5 Try 2 - 7" - 9.8# channels Moment caused by the weight of the channels = 19.6 X 4.5 = 198 »# r 2380"# "T" Total M. a 172,380"# Allowable working stress S. ■ 3.8,00(^/ sq.in. S = MY , P T'*' A In using 2 channels each channel is considered to take half the stress. Therefore in finding the stress S, only the elements of one channel will be used with half the maximum bending moment and half the thrust H, S = 86,190 X 3.5 i_ 7700 a 14350 -f- 2700 = 17,050#/ sq. in, 2T7I 2.85 Which is under the allowable stress smd the 2 - 7" - 9,8# channels will be considered as satisfactory. For Beam A in the diagram Direct stress in A = Ri/sin 30* a 8900 a 17,800# Max. M. of A a 9100 x 6.5 s 7400 x 12 a 88,900"# 8 BEAM 3C , CANTIIoEVER Length of beam A - 4.5/oos 30* - 4.5 - 5,2* .866 Uniform load on A = 5.2 x 8.5 x 175 a 7750§= Uniform load on B ® 175 x 4i- x 8.5 = 6800jl= 3paci Diac,r.am ^ Rt = 6800 X 6.75 r 10,20C^ U .4.5 Rp, s 6800 X 2.25 340C^ 4.5 Thrust H = Rj X cot 30* s 10,200 x 1,732 l7tn:f = 17,700# Max. M. of B a 6800 x 2,25 = 15,300*# 7476*7/ i^n.joo* = 184,000"# Try 2 - 7" - 9.8# channels h A-h s = My , p ,, , fOR.CL DuCjILKW A Allowable stress - 18000#/ sq, in. Each channel takes half the total stress. Therefore using half the moment and thrust for one channel S = 92000 X 3.5 , 8850 a 15,250 +- 3,100 = 18,300 #/ sq, in. 2l7l 2.85 This value comes very near to the allowaT:)le stress, and is therefore considered aatlsfactory. CALCULATIOHS FOR A Direct pull on A = Ri/sin 30* .= 10200 a 20,40C# .5 Max. M. of A = 7750 x 5.2 = 5040*# s 60500"# 8 Try 2-4" x 3" x 5/l6" angles Using one angle and half the Max. Moment and load to figure the stress. S = MY 4. P = 30250 x 1.26 4. 10200 11200 4880 I A 3,4 2.09 " S 16,080#/ sq. in. which is under the allcnrfable and therefore satisfactory. CANTILEVER 40 The uniform load on A & B = 149C^/* Length of A - 3.5 f oos 30* - 3.5 s 4.04* .866 R, of B = 6800 X 5.75 = 11,17# 3.5 Rp of B = 6800 X 2.25 4,37# 3.5 The Thrust H r Hi/ tan 30* = 11,150 r 19,30# .5773 , , =—=== z Direct stress on B. t==—=f Max. M. of B = 6800 x 2.25 ' J'j)ACL D/AC[LA^/I . =15,300«# = 184,000"# = Try 2 - 7" - 12.# chanels M. for wt. of the channels at R = 25 X 253«# ^ -T - 3040"# DjacjZam- Total M. = 187,440"# iAllowable stress - 18,00#/ sq. in. - S - MT 1 P Taking half the Max. Moment and half the load P and 1" X using the values of 1 channel S = 93720 X 3.5 • 96500 ;1 3600 + 2700 = 16,30#/ sq. in. 24.1 "• 3.58 ■Which is below the allowable stress & is therefore (Satisfactory, For Beam A in the diagram. Direct pull on A - R./sin 30* z 11,150 r 22,30# .5 Max. M. of A = 1490 x 3040 «# x 12 = 36,500"# 8 Try 2 - 4" X 3" X #" angles. Using the elements of 1 angle and half the calculated Max. Moment and load. S = MY , P r 18,250 x 1.26_l 11,150 r 8,240-+ 6,600 = 14,84#/ sq. I X 2.8 1.69 This stress is much below the allowable but the angles will be considered OK in order to have the same size member as in 30. Try 2- 4" x 3 x s/s" angles Using the elements of one angle and half the Max, M. and direct stress to figure the stress / sq, in. S = MYj P T""^ X S = 44450 X 1,26 ̂ 8900 « 140004- 3580 = 17,58C#/ sq, in. 4 2 .48 This is below the allowable stress and a more economic sec tion could be picked, but because of the fact that I am trying to pick angles with the same length legs fca* each cantilever con- Btruction. ^ome of the sections chosen may be in excess of the required size. Thus the 2 - 4" x 3"x 3/8" as chosen are preferable. These angles for each cantilever are marked A in the figures. Their weight has not been considered in computing the bending moment as it would be practically negligible, BEAMS ID b = 3* = 36" Load = 175 X 3 X 8.5 = 446# Use a 6" - 12, A, S, I - beam BEMfS 2D L = 7* Load = 175 X 7 x 8.5 = 10,40# Use 6" - 12.# A. S. I - beam. J ̂ BEAMS C & C carry 2 concentratea loads only. The loads will be figured from the plan',- lengths of beams and loads tairen to cover an area which is the horizontal projection of the balcony slope. Pi = 175(8'-#. 7/2') X 8.5» = 17,10C^ = I75(y.5-#- 3/2) X 8.5 = 16,35C^ 5LAM C ̂ C, Ri = 17,100 X 16 + 16,350 x 6 = 273,500 -h 98,000 = 14,280# 26 26 Rg = (17,100-#. 16,350) - 14,280 = 19,170# Assutie the wt. of the beam to be 60#/ sq. ft. Then Ri = 14,280 60 x 26 = 15,060# 2 H2 = 19,170 4 780 = 19,950# Max. M occurs under M = 15,060 X 10 - 60 X " 150,600 - 3000 = 153,600'# 2 For Beam C use an 18" - 57# CB Allowable M = 156,450'# lillcDER A - COFlFUTATlONS FOR DESiGW I ̂ ^ Ei^CT METHOD Hfeoo^ lf\&o6* Z-]ASo^ ■ l^&oo* ]9e^o-^ 1^00 ^ z&ooo* znioSo*- ^ znLS(ff' 7.;;, OOP* Z-^o ° 4-SZoo^ 4-1^00^ S-5;/oom- J4.,ZO-^ 47&00* Max. M = 164,987 x 33 5/12 - 45,300 x 24 ll/l2 - 47,800 x 16 5/l2 - 55,100 X 7 11/12 - 290 x 33 5/12^ 2 = 5,510,000 - 1,131,000 - 789,000 - 440,000 - 97,300 = = 3,052,000 «# = 36,640,000 A Mslx. ohear - 164,9875^ Allowable shearing stress = 12,000 H=/sq, ft 164,987 = 13»y0 sq, in. necessaVy to resist shear. 12,000 The web shall be designed to take the shear. Uhoosing a thickness of web = .375 uepth of girder ■ 13,7 - 36.6" Depth shall not be less than 1/15 of span = l/l5 x 802 - 53',i5 Thickness shall not be less than i/l60 of unsupported distance between flanges. Use a 54" depth of web.lplate. l/l60 X 38.5 (assuming 8" x 6" angles to be used) - .24* Use 3/8" = .375" thickness For Flange Areas: M - SI y = ̂ lA = 27.25" S 5 18,000/it/sa.in. y 2 I » 36,640^000 X 27.25 = 55,500"^ necessary I 18,000 ,v ■ of 4 - 8" X 6" X SA angles , = of 54" X 3/8 ffeb plate = 1312.2 x 3/8" = ll_l of 4 - 5/8" X 14 5/8" plates & 2 - l/2" x 14 3/8" plates - 1540 x 14 3/8 +-1229.2 x 14 3/8 -Sf/Af/)c.rx Deductions for rivet holes in angles and plates: 3/4" rivets to be used 2 holes - (.05583 x 1 7/8) 2 + 2 (105 x 21.25 )= 1 ,480.21 64 P. 2 holes - (.05583 x 1 l/s) 2+ 2 (105 x 24.25 )= 1 ,930.21 64 ?. 8 holes - (.94922 x 7/8) 8+ 8 (126 x 28.125 ) 12,456.65 15,867.07 Net I - 71034.8 - 15867.07 = 55,167.73" Reductions for Rivet Holes in stiffeners: Assume rivets in stiffeners at center of G-irder to be max. spaced. 38.5 = 6 rivets necessary in the stiffener. 6 Il_l 2 holes - 2C .05583 x 1 l/s) -+- 2^^ x 3^) = 18.00 64 Il_l 2 holes - 2(.05583 x 1 l/s) -f. 2(.^ x 9^) = 161.00 64 ^ 2 holes - 2U063 -t* 2(^ x 15 ) = 4 48.00 64 727.00"■ 55,167.73—727.00 - 54,440.73 = I of net section at center of Girder. The extreme fiber stress due to bending: = S = My = 36,640,000 x 28.75 - 19,300/)^/sq. in. which is above I 54,400.73 the 18000f/sq. in. allowable strees, but because of using lOC^/' instead of 125^/' (see page, ) we will consider the girder o.k. because 4/5 of 19300 = 15,500(f/sq. in. which is safe enough. Weight of Girder.'4 - 8 x 6 x 3/4 angles = 135.2^^^/' Web and 6 plates 289.2#/' = Wt. of Girder. CALCULATIONS FOR LENGTH OF FLANGE PLATES Moment of inertia about the neutral axis with the outside plate removed top and bottom. I ^ of 4 - 8" X 6" X 3/4" angles = 26364. of 54^X 3/8" web plate - 4920.8 of 4 - 5/8" X 14 3/8" Plates « (1540.3 f 402.6)14 l/2 = 28200. 59,484.8"^ Reductions for Rivet Holes: ^ for rivet holes of stiffener = 727.00 from top of page for rivet holes xn angle legs = 3410.42 ^1-1 " " " flange plates 8 U6667 X 7/8) + (1 3/4 x 27.5 ) = 10604.67 14742.1"^ 59484.8 - 14742.1 = 44,742.7"^ = I for net section Urn N^ZUrr-a/ Si'/z The bending moment that the girder will resist with these two plates off = M - 18,000 X 44,742.7 = 31,100,000"# The bending moment will be figured half w^ between each concen trated load starting at the left end. = 164,987 X - 150 x 47^5 = 700,650'# = 8,410,000"# 2 M? = 164,987 X 12.75 - 45300 x 4.25 - 150 x 12.75 = 1,894,800'# 2 = 22,700,000"# 2 Ml - 164,987 X 21.25 - 45300 x 12.75 - 47800 x 4.25 - 150 x 21.25 = 2,795,500'# = 33,550,000"# " , M. = 164,987 X 29.5 - 45300 x 21 - 47800 x 12.5 - 55100 x 4 - 150 x 29,5 2 = 3,035,500'# = 36,400,000"# , Moment of Inertia about the neutral axis - 2nd plate top and bottom removed. Il_l of 4 angles = 26364. l^.]^ of Web plate = 4920.8 of 2 remaining flange plates = 1375. 32,659.8' Deductions for rivet holes: 1^ ̂ for rivet holes of stiffeners 727.00 II II II of angle legs = 3410.42 " * * " of flange plates ■8 (.21663 X 7/8) +(77 x 27.19 ) = 7130.00 11,267.42"# 32659.8 - 11,267.42 = 21,392.38' 1 of net section. The bending moment which the girder will resist with the two outer plates off of top and bottom of girder = M = 18,000 X 21,392.38 = 13,800,000"# 27.875 The outer plates shall be cut off 14' from the ends of the girder. The second plates shall be cut off 5' from the ends. The third plates shall extend the entire length. SPACi?IG OF STIFFEI-IERS: = 85 X .375V18,000 x 20.25 - 164,987 s 31.85 X 1.205 = 38.4" Stiffeners shall be placed under each concentrated load., V at first concentrated load - 164,987 - 45,300 - 290 x 8,5 = 117,227^ S = 85tV18>000 X 20.25 - 1 = 31.85 x 2.1 = 67" 177,227 Betvj-een support and 1st concentrated load 2 stiffeners will be used. Between loads 1 stiffener is sufficient. Sizes of stiffeners under loads - 55,100 (largest load) = 3.7 sq. in. 15,000 (considered as short column) Use 2-3 l/2" x 3 l/2" x 3/8 angles on each side of web. Number of rivets necessary in stiffener angles under loads- Safe stress of 3/4" rivet«6750# (bearing of s/s" web governs) 45300 = 7 rivets 47800 = 8 rivets 55100 » 9 rivets 6750 6750 6750 14205 = 3 rivets 6750 END STIFFEmS Rivets in End Stiffeners must resist 164,987^ 164,987 = 25 - 3/4" rivets necessary. 8750 Rivets - Number and spacing to fasten flange angles to web. The longitudinal shear at supports must be figured Plates and angles only considered. Mg figured at K, with 2 outer plates removed (top and bottom) H of 2 angles = 19.88 x 25.69" - 511.0 of plate = 14.5 x 5/8 x 27.562 ~ 249.5 TOTAL STATICAL M 760.5 Longitudinal shear - "7M3 It Zr% ZfXf = 164,987 X 760 = 1560C^/sq. in. x 3/8 21392.38 X .375 » 5850^/" Rivet spacing between support and first stiffener: = 5850 X 4 1/4 X 12 = 45 - 3/4" rivets 6750 necessary, placed in 2 rows 4.25 X 12 B 2.22" spacing. 2.3 Mg with 1st outer plate removed: Mg of 1st plate = 14.5 x 5/8 x 28,187 = 249.5 Mg of other plates and angles = 760.5 1010.0 Statical M. Longitudinal shear at first stiffener, 4.25' from support = 164,987 - 290 x 4.25 x 1010 = 163,757 x 1010 = 3690^/' 44742.7 44742.7 Number of Rivets to center of distance between next 2 concentrated loads = 3690 X 8.5 X 12 - 56 rivets in two rows. 6750 Spacing = 8.5 x 12 = 3.6" Use 29 rivets / row, spaced 3.5" o.c. 28 Mg at a point midway between 1st and 2nd concentrated loads: = 12,75' from end of the beam, all plates on. Mg of first plate = 14.5 x l/2 x 28.747 = 208.5 Is of 2 plates and angles =1010.0 1218.5 Total Mg Longitudinal Shear = 68,187 x 1218.5 - 1500|/' of length. 55167.73 1500 X 8.5 X 12 = 23 rivets necessary in the next 8 l/2' 6750 Spacing = 102 = 8 l/2" Use rivets in tv/^o rows max. spacing of 6" 12 to center of beam. to Longitudinal Shear between the plate and angles at the support: Mg of Plate - 249.5 from page s' = 164,987 X 249.5 = 2,1392.38 for the first 4.25' 1920 X 4.25 X 12 =9.3 rivets necessary in the plates 1st 4 l/4' 10603 Spacing can be adjusted for convenience of riveting and punching spaced not greater than 6" m This Girder will not be used in the Balcony Design as the two column supports under Girder A have been omitted in the above cal culations. I have gone through the calculations for the purpose of showing the method used and as a comparison with the same Girder when the two column supports are figured in the later design. In this Girder, t/s" rivets should be used in the flange angles so that the proper spacing can be obtained. 14 m: I Si454Sf -y GIRDER A DESIGN WITH COLDIOJ 3JPP0RT3 METHOD -- THEOREM OF THREE MOMENTS Girder is considered continuous over four supports. Assumed weight of Girder = 29C^/' This is the correct design for Girder A to be used in the construction of the balcony. ZtLsd* 4-S3oo^ 4.7g,od^ Ss-/oj ■■ • 4": i :' '"•• ^ Hm3i!I! l(J— l~l%Soo^ R. = mti jm V ^ t «^«i %fii»% [ih m* ■+- * / f & O o J f & O O S . Q O G ^ G ' E - D t f i . h . 1 9 9 S 0 o l o c o 1 1 T O = / ^i v t T - a i f J c i - . 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Loads on all stringers were considered vertical and without thrust. To accomodate the steps from the main stair landing to the mezzanine floor level, short columns have been placed above the girder to support the stringer beams of the mezzanine floor. Two thousand pound per square inch concrete was used for floor slabs and footings. The allowable working stresses of structural steel and rivets as given by the Carnegie Steel handbook was used in the design.