CE Steel Design Class

Transcription

1 Shear Desig of Beams CE Steel Desig Class By: Amit H. Varma Shear Stregth Beam shear stregth is covered i Chapter G of the AISC specificatios. Both rolled shapes ad welded built-up shapes are covered. Rolled shapes is the focus here. Built-up shapes, commoly referred to as plate-girders are beyod the scope of our course. Cosider the behavior of beams i shear

2 Shear Stregth The web will completely yield log before the flages begi to yield. Because of this, yieldig of the web represets oe of the shear limit states. Takig the shear yield stress as 60% of the tesile yield stress. f v = V = 0.6 F y A w \V = 0.6 F y A w This will be the omial stregth i shear provided there is o shear bucklig of the web. Shear bucklig of the web depeds o its h/t w ratio. If the ratio is too large, the the web ca buckle i shear elastically or ielastically. Shear Bucklig Shear yieldig Elastic shear bucklig

3 Shear Bucklig for ustiffeed kv = 5 for stiffeed webs k v 5 = 5 + ( a / h) 2 webs with h / t < 260 é ù ê 260 ú = 5 whe a/h > 3.0 or a/h > ê ú êæ ö ç h ú ëè tw ø û where a = clear distace betwee trasverse stiffers, i. w 2 Shear Desig Equatios

4 Shear Desig Equatios Shear Desig Equatios

5 Shear Desig Equatios! " = area of web, the overall depth times the web thickess,#$ ", %&. ( h= for rolled shapes, the clear distace betwee flages less the fillet or corer radii, i. = for built-up welded sectios, the clear distace betwee the flages, i. = for built-up bolted sectios, the distace betwee fasteer lies, i. = for tees, the overall depth, i. $ " = thickess of web, i. Shear Desig Equatios The web plate shear bucklig co-efficiet,) *, is determied as follows: For webs without trasverse stiffeers ad with +., - < 260: ) * = 5 where except for the stem of tee shapes where ) * = 1.2. For webs with trasverse stiffeers: ) * = : ; (G2-6) =5 whe a/h >3.0 or a/h > (<= :. > - a= clear distace betwee trasverse stiffeers, i. (

6 Shear Desig Equatios 2. Trasverse Stiffeers: Trasverse stiffeers are ot required where +., B. C D, or where the available shear stregth provided i accordace with Sectio G2.1 for ) * = 5 is greater tha the required shear stregth. The momet of iertia, F G,, of the trasverse stiffeers used to develop the available web shear stregth, as provided i Sectio G2.1, about a axis i the web ceter for stiffeer pairs or about the face i cotact with the web plate for sigle stiffeers, shall meet the followig requiremet where F G, I$ " J K (G2-7) K = (.8 9 ; (G2-8) : ad b is the smaller of the dimesios a ad h. Shear Desig Equatios

7 Block Shear Failure of Beam Block shear failure was cosidered earlier i tesio member coectios. To facilitate the coectios of beams to other beams so that the top flages are at the same elevatio, a short legth of the top flage of oe of the beams may be cut away, or coped. If a coped beam is coected with bolts as show, segmet ABC will ted to tear out. Block shear failure of beam eds The applied load i the case of the beam will be the vertical reactio, so shear will occur alog lie AB ad there will be tesio alog BC. Thus, the block shear stregth will be a limitig value of the reactio. Failure is assumed to occurred by rupture (fracture) o the tesio area ad rupture or yieldig o the shear area. R = 0.6 F u A v + U bs F u A t 0.6 F y A gv + U bs F u A t Where, phi =0.75

8 Example Determie the maximum reactio, based o block shear, that ca be resisted by the beam show below. Treat the bolt ed distace of 1.25 i. as stadard. The effective hole diameter is 3/4 + 1/8 = 7/8 i. The shear areas are: A gv = t w ( ) = 0.30 (11) = 3.3 i 2 A v = [ (7/8)] = i 2 Block shear example The et tesio area is: A t = [1.25-1/2 (7/8) ] = i 2 Sice the block shear will occur i a coped beam with stadard bolt ed distace U bs = 1.0. R = 0.6 F u A v + F u A t = kips With a upper limit of R = 0.6 F y A gv + F u A t = kips Therefore, omial block shear stregth = kips Factored block shear stregth for desig = 0.75 x = 81.5 kips.