Problem #4 (35 points) (104) [0 - (0.35) (200+ (-300)] Ez = 0.4375 (10 The circular rod has radius r = 0.75 in, moment of inertia I = 0.2485 in and cross-sectional area A = 1.767 in². Determine the absolute value of the normal stress at points A and B. (Note: Provide your answers in units of psi and indicate tension or compression stress at each point.) омч 601b Mya 2015 My2=80 (4) 8016 "Mx = 60(10) Mz=-60 (4) B abs = 60 +0 120 +080 Ому +0Myz + M372008) (Gabs) A = + (160 +16/20 (4-0,75) * P80) + — (My, + My = \1/Mx +Mz)* 10 in. in 80 lb = (1.767) (60 4120+80) + 602435 (120 (19) + (-80 (21)) + 60 (10) + (-00 (8)) (abs) A = 4206/498 psi (X✗() (os) B = (P60 + P₁to + P) + (My, +My+Mx +M₂) (0.75) 4 in. (4-075 120 lb 60 lb )-60 (10.796)) (1.767) (60x120+80) + (0.2485) (120 (10.1996) + (-80 (10.796) + 60 (10.796)-60 (abs)=14560. 482 psi