Question

Part II: Problem 1 (15 points) A tension test was performed on a steel specimen having an original diameter of 12.5 mm and a gauge length of 50 mm. The data is listed in the table below. Plot the stress strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Load (kN) Elongation Load (kN) Elongation Load (kN) Elongation 0 40.9 0.1650 64.5 6.3500 0.0175 43.6 0.2490 62.3 8.8900 31.9 0.0600 53.4 1.0160 58.8 11.9380 37.8 0.1020 62.3 3.0480 0 11.1

Plz show in graph paper if possible. Thanks :)

find stress, strain and draw the graph. if it is blurry. plz zoom

Answer 1

Solution Given → Original dicemeter do = 12.5 mm Area Ao (12.5)2 Gauge ungth Ao=122.71 mm lo =50mm Elongation strain lo load stress= 2 Ao So we will get two more coloumn Stress and strains.

strain load (KM) Elongation stress (MPa) O o O 0 11.1 0.0175 go.u5 31.9 0.0600 259.9 37.8 0.1020 308.04 4o.g 0:1650 330.3 43.6 0.2490 355.3 3.5x154 1. 2x103 2.04x103 3.3X103 4.98x103 20x103 60.96 *103 127 X103 177.8 x103 238.74103 53.4 1. 0160 435.17 62.3 3.0480 507.7 64.5 6.3500 526.6 62.3 8.8900 507.7 58.8 Il. 9380 479.1
stress-strain diagram -
530 с 520 510 500 D 8 450 you А Stress 350 q 360 280 200 150 100 50 40 So 60 90 70 110 120 100 to 130 140 150 20 160 170 180 190 200 210 220 230 240 25 @ stoain (x103)

following are the points mentimed on stress-strain diagramor A → proportionality limit C Hookes law valid till this point) B→ yield point ( yield stress) C- Ultimate stress point. D - fracture point (stress at which fracture occurred ) Now modulus of Elasticity (E)- applying Hooke's law which is stress a strain Stress = Ex Strain Stress slope of stress-strain curve strain till A. (stress, strain vakee at A] 4.98X103 E=71.345 GP a Argwel E: 355.3

yield stress Oy = 435.17 MPa Answers * Ultimate Stress Out = 526.6 MPa * fracture stress Ofail O fail = 479.1 MPa.