Cable Supported Structures George.Hearn@colorado.edu 25 Suspended Beam Roof with Pylons A roof structure is a suspended beam. The roof span is 200 ft. Main cable sag is 20 ft. Suspender length varies. The deck is metal pan plus weather closure that weights 12 PSF. Beams have a transverse spacing equal to 40 feet. The roof carries a transient snow load equal to 20 PSF. Main cables and suspenders are bridge strand furnished by Bethlehem. Beams are compact W shapes in A992 steel. Roof deck provides continuous lateral bracing for beams. AISC specifications are followed for beam selection. Preliminary Design Work with external loads to get a first estimate if forces in members. Then make a first selection of members. Then update for self weights. Then get all values needed for a permanent load reference state. Preliminary design uses a guess about load sharing of beam and suspension cable. We update this in later checks and updates to the design. External Loads Deck self-weight and snow loads are distributed loads along beams. w deck = w snow = 480 plf 800 plf Internal forces for main cable, suspender and beam are estimated as
Cable Supported Structures George.Hearn@colorado.edu 26 Figure 8 wl / 4 deck = wl / 4 snow = 24 k 40 k Suspension Cable R = (3 / 8) wl H f = (wl / 4) L / 2 Since f equals 20 ft and L equals 200 ft, we can write H = 5 (wl / 4) H deck = H snow = Hu = 120 k 200 k 352 tn z-points along parabola, for first estimate of cable slope and cable tension Z: (40, 25, 20 ) Slope = 15 / 50 = 0.3 Tu = Hu * sqrt(1+(0.3)^2) = 1.044 Hu Tu = 367 k Select 2-1/2 strand, Tbreak = 376 tn
Cable Supported Structures George.Hearn@colorado.edu 27 Suspender T = wl / 4 T deck = T snow = Tu = 24 k 40 k 70.4 tn Select 1 1/8 strand, Tbreak = 78 tn Stay Stays are diagonal cables running from the top of the pylon supporting the suspension cable to a foundation at the ground. This design will use stays with rise equal to run (Figure 9). Figure 9 The horizontal component of tension in the stay equals the horizontal component of tension in the suspension cable Figure 10 H deck = H snow = Hu = 120 k 200 k 352 tn The slope of the stay is known. Tu = 1.414 Hu Tu = 498 tn, Select 3 diameter strand, Tmax = 538 tn
Cable Supported Structures George.Hearn@colorado.edu 28 Pylon Pylons carry vertical components of tension in cables. From the first/last element of the suspension cable P = H (15 / 50 ) From the stay P = H (80 /80 ) These add up P = H (0.3 + 1) P deck = P snow = P total = 156 k 260 k 416 k We use an allowable design basis for the pylon. From Chapter E in the AISC specification:
Cable Supported Structures George.Hearn@colorado.edu 29 Pylons are 80 ft tall. Bracing is installed at midheight. L for AISC relations is 40 ft. Select HSS 14 x 14 x ½, b/t = 27, A = 24.6 in2, ry = 5.49, Fe = 37.4 ksi, Fcr = 28.6 ksi, Pallow = 421 k Beam Assume that moment in beam is zero at midspan, the connection to the suspender. This is a good assumption for permanent load and a poor assumption for snow load. M = w L^2 / 128 M deck = M snow = M total = 150 kft 250 kft 400 kft Mallow = Fy Zx / 1.67, Fy = 50 ksi
Cable Supported Structures George.Hearn@colorado.edu 30 Find Zx > 160 in3 Select W24x68, Zx = 177 in3, Mallow = 442 kft First Model Node Coordinates X ft Y ft Z ft Left Cable 0 0 40 a 0 50 24.99822 mid cable 0 100 20 b 0 150 24.99822 Right Cable 0 200 40 Left Beam 0 0 0 True True c 0 50 0 mid beam 0 100 0 d 0 150 0 Right Beam 0 200 0 True True Right Stay 0 280-40 True True True Left Stay 0-80 -40 True True True Left Pylon 0 0-40 True True True Right Pylon 0 200-40 True True True Components Suspension Cable, Suspender, Beam, Stay, Pylon Loads Self weight of components Computed automatically. Self weight of roof deck Distributed Load Wx k/ft Wy k/ft Wx k/ft Beam 0 0-0.48 Snow load This load is used in transient load analysis. This goes in as a distributed load on the beam. Wx k/ft Wy k/ft Wx k/ft Beam 0 0-0.8
Cable Supported Structures George.Hearn@colorado.edu 31 Reference state Using self weight of components + deck weight, after adjustments H k Main Cable 140.55 Suspender 27.43 Beam 0 Stays 140.55 Pylons -187.71 Node X k Y k Z k Left Cable 0 0 0 a 0 0 0 mid cable 0 0 0 b 0 0 0 Right Cable 0 0 0 Left Beam 0 0-13.7 c 0 0 0 mid beam 0 0 0 d 0 0 0 Right Beam 0 0-13.7 Right Stay 0-140.55 139.48 Left Stay 0 140.55 139.48 Left Pylon 0 0-191.3 Right Pylon 0 0-191.3 Check Reference State Structure is at equilibrium
Cable Supported Structures George.Hearn@colorado.edu 32 Analyze for snow load Using load factor equal to 1.0 Deflections Node Dx rft Dy ft Dz ft Left Cable 0 0.33-0.03 a 0-0.04-1.54 mid cable 0 0-2.08 b 0 0.04-1.54 Right Cable 0-0.33-0.03 Left Beam 0 0.03 0 c 0 0-1.59 mid beam 0 0-2.12 d 0 0-1.59 Right Beam 0-0.03 0 Right Stay 0 0 0 Left Stay 0 0 0 Left Pylon 0 0 0 Right Pylon 0 0 0 Member Forces W = 0.48 k/ft + 0.068 k/ft + 0.8 k/ft =1.348 k/ft P k Viy k Mix kft Mjx kft X ft M+ kft Beam 0 0 33.47 0-11.46 24.83 416 Beam 1 0.11 33.71 11.46-11.14 25.01 410 Beam 2 0.11 33.69 11.14-11.46 24.99 410 Beam 3 0 33.93 11.46 0 25.17 416 Pylons 0-415.49 3.59 0 0 0 0 Pylons 1-415.49 3.59 0 0 0 0 Update Beam Zx required = (416kft)(12 / )(1.67)/50ksi = 167 in3, W24x68 is OK Update Pylon P = 415 k, Pallow = 421 k, HSS 14x14x1/2 is OK
Cable Supported Structures George.Hearn@colorado.edu 33 Analyze with 2.2 factor for cables Member P k Main Cable 0 627.21 Main Cable 1 589.83 Main Cable 2 589.83 Main Cable 3 627.21 Suspender 0 148.1 Suspender 1 147.81 Suspender 2 148.1 Suspension cable, Tu = 314 tn, 2-1/2 ok, could go to 2-5/16 Suspender, Tu = 74.1 tn, 1 1/8 ok. Try Lighter Suspension cable