. Practical structural design; a text and reference work for engineers, architects, builders, draftsmen and technical schools;. rs, the panel length being10 ft. and the height 10 ft. The length of a diagonal = ft., +V. • • 7, X 25,000 „_„^^ ,, _so the compression m gh = ~— = 35,950 lbs. The . . X 15,000 ^^ _.„ ,, ^,compression m ef = i^j— = 21,750 lbs. The com- . , . , X 5000 _,^^ „pression m dc = -^ = 7190 lbs. With the load considered as applied on the upper chord thetension in gf = 15,000 lbs.; the tension in ec = 5000 lbs.; andthe tension in hd = 0. With the load co
. Practical structural design; a text and reference work for engineers, architects, builders, draftsmen and technical schools;. rs, the panel length being10 ft. and the height 10 ft. The length of a diagonal = ft., +V. • • 7, X 25,000 „_„^^ ,, _so the compression m gh = ~— = 35,950 lbs. The . . X 15,000 ^^ _.„ ,, ^,compression m ef = i^j— = 21,750 lbs. The com- . , . , X 5000 _,^^ „pression m dc = -^ = 7190 lbs. With the load considered as applied on the upper chord thetension in gf = 15,000 lbs.; the tension in ec = 5000 lbs.; andthe tension in hd = 0. With the load considered as applied onthe lower chord the tension in gf = 25,000 lbs.; the tension in ec= 15,000 lbs.; and the tension in bd = 5000 lbs. 126 PRACTICAL STRUCTURAL DESIGN Wl The compression and tension per panel in the chords = —r> a therefore compression in eg = tension in fh = 25,000 lbs., for theratio J = Trj = 1- The compression in de = tension in cf = 25,000 + 15,000 = 40,000 lbs. The tension in he = 25,000 + 15,000+ 5000 = 45,000 object of the computations being to obtain the stresses so. Fig. 80—Shear Diagram for Truss with Even Number of Panels the members may be proportioned, the method above given offollowing the loads from joint to joint and obtaining the coefficientsfor uniformly and symmetrically loaded beams, or of obtainingthe shear at panel joints for unsymmetrically loaded beams, isadequate- and simple. It can be proven that a truss is merely a skeleton beam byfinding the shear and bending moment at each joint and then divid-ing the bending moment by the depth obtain the stresses in thechords, the web members carrying the shear. In Fig. 80 the en^reactions each equal 25,000 lbs. Then M, at / = 10 X 25,000 = 250,000 ft. lbs. M, at gf = 0, for the top chord rests on gh. M, at c = (20 X 25,000) - (10 x 10,000) = 40,000 ft. lbs. M, at e = 10 X 25,000 = 250,000 ft. lbs. GIRDERS AND TRUSSES 127 M,at& = (30x25,000)-(10x10,000+20x10,000) =4
Size: 1796px × 1391px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1920, bookidpracticalstr, bookyear1921
