A tension test was carried out on an aluminum alloy using a

A tension test was carried out on an aluminum alloy using a round test specimen with diameter 12.2 mm in the test section

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5. A tension test was carried out on an aluminum alloy using a round test specimen with diameter 12.2 mm in the test section (see Figs. 1.25 and 1.4-3). The elongation was measured with a clip-on extensometer having a gauge length of 50 mm. (it measures the elongation between fixed points on the specimen initially 50 mm apart). The applied load and elongation are given in the table below. Necking began to develop above 37 kN load, and after fracture inspection of the two halves showed that the distance between the extensometer attach points was 53.5 mm and the fracture cross section had a diameter of 9.8 mm 5.1. Construct a plot of stress versus strain (use Excel or Matlab or Mathematica) 5.2. Determine the following material properties: E, proportional limit (on), yield stress (Oy), 0.2% offset yield stress, ultimate stress (auLT) and elongation (EULT), and percent reduction in cross sectional area at fracture 5.3. Convert your results from (5.2) into USCS units. \"Load [kN]\" \"Elongation [mm]\" 12.9 21.9 28.3 32.1 34.4 36. 37.5 39.3 41.1 0.04 0.078 0.157 0.227 0.304 0.421 0.543 0.699 0.897 1.221

Solution

Given Data,Dorig.=12.2mm Lorig.=50mm , Load at Necking=37kN, Lf=53.5mm, Df=9.8mm

5.2==>

A)Yield Stress=(Yield load)/(Orignal load)

                       =(37x10³)/(3.14*12.2²)

   Yield Stress  =316.67N/mm²

B) Fracture Stress= Max. load/Ori. Area

                             =41.1x10³ /(3.14*12.2²)

   Fracture Stress =351.76 N/mm²

C) % of Elongation=(L_final-L_Orignal)/L_Orignal x100

                            =(53.3-50)/50 x100

     % of Elongation=7        

D) % of reduction in cross section=(Area_orignal - Area_final)/Area_orignal

                                                                            ₌₍12.2²-9.8² /12.2² )x100

                                                    =35

E)Young Modulus=(E)=(6_^E (1+e)/(ln(1+e)

                        At load, 37.5kN, 6_^E =37.5x10³/(0.785*12.2² )

                                                  6_^E==320.95N/mm²

           e at load of 37.5 kN

                                  e=0.699/50

                                   e=0.01398

Young Modulus E=320.95(1+0.01398)/(ln(1+0.01398))

          E=23.44x10³ N/mm²

     5.4==>

Residual stain in bar=modulus of resilience(starin energy per unit volume) at 300Mpa

          5.1==>