Question

An I-section girder 12 cm deep has the following cross-sectional dimensions: Top flange 6 cm wide by 1 cm thick, bottom flange 12 cm wide by 1 cm thick, web 1 cm thick, web 1 cm thick. The girder is 5 m long simply supported over a span of 3 m, overhanging both supports by the same amount and it carries a concentrated load of 2 kN each end. Find the maximum stress in the material due to bending.

Answer 1

Answer:

Since diagram is symmetrical about y axis i.e. X = 0, x1 = x2 = 0 A1 = 8 × 2 = 16 cm2 A2 = 12 × 2 = 24 cm2 A3 = 12 × 2 = 24 cm2 y1 = (2 + 12 + 2/2) = 15 cm y2 = (2 + 12/2) = 8 cm y2 = 2/2 = 1 cm Y = (A1 y1 + A2y2 + A3y3)/(A1 + A2 + A3) = (16 × 15 + 24 × 8 + 24 × 1)/(16 + 24 + 24) = 7.125cm C.G. = (0, 7.125) Moment of inertia (M.I.) about x-x axis = IXX = IXX1 + IXX2 IXX1 = IGXX1 + A1h12 = (bd3/12)1 + A1(Y – y1)2 = 8 × 23/12 + 16 (7.125 – 15)2 = 997.58 cm4 ...(i) IXX2 = IGXX2 + A2h22 = (bd3/12)2 + A2(Y – y2)2 = 2 × 123/12 + 24 (7.125 – 8)2 = 306.375 cm4 ...(ii) IXX3 = IGXX3 + A3h32 = (bd3/12)3 + A2(Y – y3)2 = 12 × 23/12 + 24 (7.125 – 1)2 = 908.375 cm4 ...(iii) IXX = IXX1 + IXX2+ IXX3 = 997.58 + 306.375 + 908.375 = 2212.33 cm4 ...(iv) Moment of inertia (M.I.) about y-y axis = Iyy = Iyy1 + Iyy2; h = 0 Since X = 0 i.e.; X1 = X2 = 0 Iyy = Iyy1 + Iyy2+ Iyy3 = (db3/12)1 + (db3/12)2 + (db3/12)3 = 2 × 83/12 + 12 × 23/12 + 2 × 123/12 = 381.33 cm4 ...(iv) IXX = 2212.33 cm4; Iyy = 381.33 cm4Read more on Sarthaks.com - https://www.sarthaks.com/512690/an-i-section-has-the-following-dimensions-top-flange-8-cm-2-cm-bottom-flange-12-cm-2-cm?show=512733#a512733