Question

Problem #5: A C8x13.75 channels section is subjected to a shear force of 18 kips. Compute the shear stress distribution in the web (vertical) and flanges (horizontal). Also, find the distance of shear center to the centerline of the channels section web.

please show all work.

Answer 1

Given data- 2.343 in A Channel section Es C8% 13.75 channel. From American Standard Steel Channel section data. Depth Degin, width W = 2.343 in thicisness S= 0.303in, 10.303in Area A= 4.04 in Txx = 36.1 in 4 Tyy B 2 K-0.303in. 2 e = 1:53 in 4 Shear stress Distribution: 6 F= shear force 18 kips FAY I b shear stress at point A € O PAD (' A = 0 above that point.) i, shear stress at point B. 18 x CÀxu ) TE {; B= 2.343-6) 36.1 X 2.343 Axū= 2.343x0.303 ( 8 - 030?] AXY = 2.732 in3 NB: 18% 2.732 36.1X2.343 PBS 0.581 KSI ini, shear stress at Point ci 18X (Axü) [szb= 0.303in] 36-1% 0.303 AxÝ 2 2.343X0.303(2-0-303] No (Axy ) = 2.732 in? Te = 18% 2.732 36.1 x 0.303 PC- 4.495 ksi iv) sheall Stress at point o: 18x CAXT) 36.1 x 0.30 3 . 8 0.303) * - - Axū= care vom 0.303.)x0.303 x 0-303 x 2 -343)* (- 0303) (Axų ) = 4.803 in 3 18 x 4.083 To= 36.1 x 0.303 To = 6.72 ksi since Given channel section is Symmetrical can draw following diagram of shear stress Variation. we 0 0.581 ksi 4.495 ksi E 6.72ksi 0.581ksi 4.495 ksi b) shear centre of channel section : e= 62 n²t 4 Ix 2.343x8 X 0.303 44 36.1 -E 2 2 = e=0.737 in from the Centre of web