Question

6. A cantilever is loaded as shown in figure. Using a factor of safety of 2, determine whether failure occurs according to the maximum energy of distortion theory. Use Cold-drawn AISI 1020 steel. Given: n = 2, k = 13.4 • 106 mm4 - 0. 4m 0 0kN 10.3 10.3 mm 10 80 mm 6.6 mm

Answer 1

In htis question we have to determine the failure of beam according to maximum distortion energy theory

GORN Cat Cantilever I section beam made of A1S11020 (D Given: - - Sy = 390MPa , n=2, I2 = 13.48106 mm" Bending stres on beam is given as 86= M 4 y = dist from N.A. M= Begading moment = 60x0.4 = 24 kW-m = 24 x106N-mm C N.A. At point A=(06) = 24810 , * 80 = 143,283 N, 2 / M 143.28 mp4 mm 13.4 X106 X 80 = 124.835 MPC At point B = (66)B= 24x106 *(80-10.3) = 124.835 MPa 13.4X106 at point c = (6) e=0 1b Now trong verse shear strees at A, B, C shear fonce v=GOKN=6081000 N (T)n=0 t = va las 60x103x (102x10.3) (80 - 5.15) - 3.45MPa on flange 13.47106 x 102 (T)8 on web = 60x103x102x10.3x180 - 5.15) 13.47106x 6.6 - = 53.35 MPa (T) c = 60 x103 x (1022 10.3 (80-5.15) + 69.7x6.6 X 34.85] 13.4 x 106 x 6.6 - (T)e = 64.225 MPa Now failee stees aconding maximum distovion emendy theory : r's (VB)+31792)n < Sy

cats a V (183.283)2+0 = 286.566 MPa 6526 At point A => At point B3) Aut point ca og'= 21 124.835313(53.35)2 = 310.629 MPa pe=av 107243 164.92572 - 22 nampa alt Strees at all point is the t eam will not of distortion theory lessen than sy=390 m pa so fail acconding to maximum energy SHEAR STRESS 102 mm -> - TI . 10.3 * BENDING STRESS 143.283 MPa 124.835MPa 10. 3.45 MPa 54.35 MPa somm - 6.6 alk 164.225 MPa - Lompa - - 154.35 MPa 13.45