Question

solve with explanasion please

4. Calculate the axial stress in the x direction at each corner due to the applied moment and the orientation of the neutralaxis (with respect to the z axis). (15 pts) 4 in 50 lb ft 6 in.

Answer 1

sony) D f solb-ft 600 Atify lbin -Y -300 600 x sin 30 My ebrin Mz = 600 x lus 30' = 519.61 lbrin IB 300 tin А c Gin 7zint, Tyy ² G |2 - 32 in 4 63x4 . Izz = 12 a ľ JA M₂ x 3 at My x2 stress comer + Tyy I22 Compressie 2.9 holink 300 x 3 32 519.61X3 १२ + TA - Compressive 300 X 2 32 = 40.4 lblink 519-61 X3 १२ + Tensile. 300 x ² 519.61 X3 १२ - 40.4 lblink C = 32 & Tensile - 300 x 2 32 519.61 X3 72 = -2.9 lblina 2 coo for M N. A & normal stress will be 19 rero My. 2 M2.9 J =0 = I22 Igy 519.61 + => 1.32+ y co 3000 2 32 ג ד 1.3 tanda 무 Ø = 52.430 Anticlock wise from tve z axis.