Question

The internal shear force at a certain section of a steel beam is V = 205 kN. The beam cross section shown in the figure has dimensions of ty = 21 mm, by = 285 mm, d = 410 mm, and is = 14 mm. Determine: (a) the shear stress at point. A, which is located at yA = 75 mm below the centroid of the wide-lange shape. (b) the macimum horizontal shear stress in the wide-flange shape. ЈА by Below, enter the magnitudes of the shear stresses, so enter positive values for both answers. (a) = MPa (b) Tmax MPa

Answer 1

given data AD shear force * CV) 205 KN= 205 x103 21mm 184 Irgas tw=14mm To find i) shear force at point A 1) max. horizontal Shear ress in the wide- flange shape 410mm x YA = 75mm • A 285 y i) Lets find Shear force at point A ý 75mm Inn Bp3 12 as Moment of Inertia about Neutral axis given by (285–14 (410-42) 12 (Tip- consider I section rectangle where B = D= 410) and substract the of remaining two two rectangle) Ixn = 285 4107 (285-14) (410-62)3 12 285 area 3 - 12 Ian = 511.41 x 100 mms Shear at stress Let find Now. Let point A NAY TA Itw TA = 205 x 103 A = 75 14 1050 mm 2

205xlos x 1050 x 75 y = 75mm tw = 14 mm 511.41 x 106 x 14 2.2548 N/mm² 2.2548 N/mm² shear stress at point A = 2.2548 Mpa 5) Now find maximum horizontal shear stress in the wide flange shape (T max) Zmax = VLAYI+ A2 Y2) Ib b =tw = 14mm of flange 285x21 5985 mm² Az= area of web =( 14 X (205-21)] = 2576 mm2 below flange area 194.5mm y = 205-(2/2) 205 - 21 Ya 184 mm 5985 . Iman = 205 x103 x 194.5)+(2576x 189)] 511.41 x 106 x 14 man 46.90 N/mm2 i maximum horizontal shear stress 46.90 N/mm² = 46.90 Mpa