Question

Bern 2) Using Mohr Circle, determine the product of inertia (1) of the rectangular cross-section with respect to the inclined u and v axes, shown in the Figure. Centroid of the cross-section is denoted with C and B-20cm and H-30cm and 0 -20°. Answer only what is asked! Hem 40 KN 20 KN 30 KN 3) Determine internal normal force (N), shear force (V), and bending moment (M) at point E and determine the horizontal and vertical components of reaction at the supports A and B. 12 kN/m 7m B 2m 2m2m2m 2m

Answer 1

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The answer is given below.

21 A Ay! 2 ) N sal H=30 cm Ixly' - Product of Inertia want to centroid, 3 B= 20 Ixlyl=0 B= 20CM Moment of Inertia w.rit x-axis, Ix= 43x8 + BxHx () = 18X10%Cm4 Moment of Inertia wort y-axis, Iy= B x H + BxHx(²/2)² = 8x10 cm" 12 Product of Inertia wort a& y axes, Ixy = Ixlyl + Bx HX B, ₂ xH/2 O + 20x301510 • Ixy=9x104 cm4 Product of Inertia wirst u&raxes Turl= (Sy - Tx) cosesino + Try (cos?o-sin20) → Ture = (8x109–18x104) cos 20°Sin20 + 9x1oºx (Cos2209-sin220°) → uv= 36804.6194cm"

30KN 2OkN 3) 40kn F V G H I VD T! 4m BizkN/m Tm 1 TIT 2m' , 2m', 2. 2m², 2m), 2m 2m 2 EY=0+ Ay + By = 90KN EX=0 → Bu = 12X7 = 846N EMB=0> 20x14+ 30x 8+12x7x3,5 = Ayx8 Ay=101.75 kN By=-11.75kN Horizontal & Vertical reaction @ A+ Axe = 0 Ay = 101.75 kN (Upwares) Horizontal &vertical reaction @ B = Bx = 846N (Rightwards) By = 11,756N ( Downwards) Internal Normal Force @ E NFE=0 Shear Force @E+ SFE= - 20 kN , Bending Moment @ E → BME-- 20x2= -40kNm //