Vierendeel girder and frame - Engineering Class Home Pages

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Vierendeel steel girder Assume: 10” tubing, allowable bending stress F b = 0.6x46 ksi F b= 27.6 ksi Girder depth d = 6’, span 10 e = 10x10’ L = 100’ DL= 18 psf LL = 12 psf � = 30 psf Uniform load w = 30 psf x 20’ / 1000 w = 0.6 klf Joint load P = 0.6 x 10’ P= 6 k Max shear V = 9 P/2 = 9 x 6/2 V = 27 k CHORD BARS Shear (2 chords) V c = V/2 = 27/2 V c = 13.5 k Chord bending M c = V c e/2 = 13.5 x (10’x12”)/ 2 M c = 810 k” Moment of Inertia I = M c c/F b = 810 k” x 5”/27.6 ksi I = 147 in 4 2nd bay chord shear V c = (V–P)/2 = (27-6)/2 V c = 10.5 k 2nd chord bending M c = V c e/2 = 10.5 x 120”/2 M c = 630 k” WEB BAR (2nd web resists bending of 2 chords) Web bar bending M w = M c end bay + M c 2nd bay M w = 810 + 630 M w=1,440 k” Moment of Inertia I = M w c/F b = 1440 k” x 5”/27.6 ksi I = 261 in 4 Vierendeel girder and frame Copyright Prof Schierle 2012 12
Chord bars Moment of Inertia required I= 147 in4 Use ST10x10x5/16 I= 183>147 Web bars Moment of Inertia required I= 261 in4 Use ST10x10x1/2 I= 271>261 Vierendeel girder and frame Copyright Prof Schierle 2012 13
Page 1 and 2: Vierendeel girder and frame Vierend
Page 3 and 4: Vierendeel girder and frame Copyrig
Page 5 and 6: Vierendeel bridge HBF Berlin: Viere
Page 7 and 8: Salk Institute, La Jolla Architect:
Page 9 and 10: Commerzbank, Frankfurt Architect: N
Page 11: Hong Kong Shanghai Bank Architect:
Page 15 and 16: Design end chords Joint load P = w
Page 17: Z U G S P T Z E Vierendeel girder a

Vierendeel girder and frame - Engineering Class Home Pages

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