Question

can someone answer this with explanation because i posted this before but they arent adding up with text book equations

Problem 1 The steel specimen was subjected to a tensile test and fractured shown below. Failure (fracture) occurred when the axial load was 19.8 kips. Determine: a) The average normal stress along the included face at failure b) The average shear stress along the included face at failure

Answer 1

FBD: The angle is: A = 90° - 52° = 38°

The normal stress in x direction: 19.8 (0.5) = 100.841 ksi The normal stress in y direction: 0 = 0 Since, no force is acting in vertical plane. The shear stress is also zero. In = 0

(a) Hence, the average normal stress is along the failure plane is: Ox+ Oy Ox-O Oy= 03+* cos (20)+7q, sin (20) 2 2 100.841+0 100.841-0 !1° cos (2x389)+(0)sin (2x38°) Ox' = 62.6 ksi (b) Hence, the average shear stress is along the failure plane is: 1xy = _ 0972sin (20) +7%, cos(20) 100.841-0 sin (2x38°)+(0) cos(2x380) Txx = -48.92 ksi