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, b = 285 mm, d = 410 mm, and = 14 mm. Determine: (a) the shear stress at point A, which is located at y = 75 mm below the centroid of the wide-flange shape. (b) the maximum horizontal shear stress in the wide-flange shape. YA by Below, enter the magnitudes of the shear stresses, so enter positive values for both answers. (a) 2.2548 MPa 46.90 MPa

Answer 1

o In inm 21 = 194.5 71 ايا 40 (2x21) - 368 yo A A = 285X21 = .Az = 5958 mm² 214 205mm A = 368x14 = 205-31 21 5152 nm² 194.5 285 Since the section is symmetrip about it axis, Centroid, Neutral axis at d/ from bottom g = 410 Now Moment of Inertia Surface = 205mm I= I + I + Iz 2 - BP 45 -> 285x213 (5988 B8 X 194.5 72 225612578. 3 mont I = Iz = 2256 12 578.3 mont (symmetrical B₂ Dzel + A 2 - 14x368 to 12 12 Iz = 58142037 38 mont I - It Iz + 7 = 509367193.8 mont

&5mm below N.A Now shea stress at A 368 15mm А Az=109814 mm² 109 2 109mn b=14mm F=2058103 N Z= F (zag E (A, J + A 2 J2) (75+ 101 Ixb I b 109 T= 205x108 ((5958 x 194.5) + (109x14x 509367193.8 x 14 39 MPa IT. = 37.703 703 MPa Max. Sheat stress is At Neatral Axis, Imax N.A T = F (A. 5, + A₂5) may л хь 205x106788x174-8) +(2576 256 X 92 A2=184814 = 2576mm J = 184 = 92mm 50367193.8 X 14 Emax = 40.126 MPa Thank you All the Best.
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