Question

Learning Goal: To use the superposition principle to find the state of stress on a beam under multiple loadings. The beam shown below is subjected to a horizontal force P via the rope wound around the pulley. The state of stress at point A is to be determined. P 1d4 di dz d2 20 mm T 100 mm 200 mm 15 mm 20 mm 150 mm The dimensions are di = 1.85 m, d2 = 0.5 m, dz = 0.9 m, d4 = 355 mm, and r = 220 mm. The applied force has magnitude P = 7.5 kN.

Answer 1

solution: - D ida Cox A Icy k di *x dax do di= 1.85m da=0.5m d3=0.9m du= 355 mm &= 2gomm=0.gam 00355m p=715 kn part A 5 M@D = o 7 Cyx (du+de +d3) + Px (dute) Cyx (-85+65+04 +7.5x (0-355+0.490) = co Cy=-1633 KN EFx=0 7 Cx-P=0 (xc= 7.5kN N N=7.5kN = Cx v=-cy 7 Clu= 1.33KN M = Cyxd3= -1.23x0.9 > M = -1.194 knom

To 10o 4 * 200 K 15min toot + 200x15 x 150mm X B) A= 2x150x20) + 9000 mm2 9x103 m2 I= 150X2403_ (150–15] x 2003 = 82.8x106mmy 12 I= 89.8x100 my o = AJ = 20X150x(100+20) + loox 15 x 100 = 405000mm3 = (4.05x10 m3 6A = N + MCA all А I 75 KN 10194XO KN=m 9X103 m2 82.8x10omy 8.333 xl0² kN/m2 8.33 Mpal ZA= It va 1.33x4.058104 82.8x100x (15x103) 7 ZA = 432.69 kN/m2 = 10.433 Mpa

In N , Cy= -133 N, V= 133N