Question

HotLW Problem: Giving the engineering stress - engineering strain carve for an aluminum alloy, Calalate: 400FTTTTTTTTTT 10 psi 300 MPa 50 $ Stress (MPa) Stress (10 psi) 8 100- 10/ bd = 110m 1oof lo=50 mm 0 0 0.005 0.010 o 0.05 0. 100 . 150 .20 6 percent elongation at fracture (% elongation = ?) 2 yield strength (Tyield = ?) 3 Tensile strength (TS = ?) ♡ True stress at an engineering strain of 0.04. (I=?) 5 True strain at an engineering strain of 0.04.(8=?) o Elongation De when a load of 25 kN is applied (Al-?) 17 Diameter (d=? at an engineering stress of 350 MPa.

Answer 1

Using Stress strain curve

Please go through photograph for

Required Answers with explanation,

Given. - d do = 11 mm le = 50 mm from stress strain cune 0 Parcant elongation at fracture ( % elongation) - change in length Perrent Clongation -- original length - shroin strain at frochur = 0.16 Percent Etergation = 0.16 x 100 = 16% 2 Yield strength by = 0.2 % E parallel line = 0.002 E = 0.002 x 0.16 Cune & Toy = 300 MPa (Approx) | Tensile strength 16 = 350 MPa (Ollimak Tengile strongly ! ③ Troe stress cat 6 = E=0.09 y = 350 Cito.04) =8C1T & T6 = 364MP ③ True strain Ex a 'lo (it & = ln (1+0.04) = 0.0392 T & = 0.0392

Using Curre ES P= 25 KN 800 25x10w x Toxio E = 30x100 N IX (11810 32 x Boxior Al = 0.0438 m = 43.8 mm d = ? at 6=350 MPa at 6=350 E=0.8 E- change in ora olginal dia 0.8 0.8 = 11-d 11