# Each part of full curve 1 mark total 3 marks each

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Each part of full curve = 1 mark(total = 3 marks)Each point and region = 1 mark(total = 4 marks)Part 1C: Calculation Questions (8 marks)Provide the answers and show complete workings for the following question in youranswer booklet.i)A 10.00 mm round mild steel bar is tested in uniaxial tension and is found to havethe following mechanical properties:Engineering elastic limit stress = 315.00 MPaEngineering elastic limit strain = 1.50 x 10-3Engineering upper yield strength = 345.00 MPaEngineering lower yield strength = 305.00 MPaDistributing prohibited | Downloaded by Bob Cat ([email protected])lOMoARcPSD|1531217244
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Chapter 14 / Exercise 27
An Introduction to Physical Science
Shipman/Wilson
Expert Verified
Page3of23CONTINUED OVER PAGEa)Determine the engineering strain when the load applied is 20 kNAnswer (3 marks):Solving for engineering modulus of elasticity (EE) using engineering elasticlimit stress (!Eel) and engineering elastic limit strain ("Eel),EE=!Eel/"Eel(0.5 mark)EE= 315.00 MPa / 1.50 x 10-3EE= 210,000 MPa = 210 GPa(0.5 mark)Solving for engineering stress (!E) using engineering force (fE),!E= fE/ area = fE/ (#x r2)(0.5 mark)!E= 20,000 N / (#x 5.00 mm x 5.00 mm)!E= 254.65 MPa\$255 MPa(0.5 mark)Solving for engineering strain ("E) using!Eand EE,EE=!E/"E"E=!E/ EE(0.5 mark)"E= 254.65 MPa / 210,000 MPa"E= 0.00121 = 1.21 x 10-3(0.5 mark)Distributing prohibited | Downloaded by Bob Cat ([email protected])lOMoARcPSD|1531217245
Page4of23CONTINUED OVER PAGEb)Determine the true stress and true strain at 50% of the engineering yield stressAnswer (3 marks):Solving for!Eat 50% of engineering yield stress using engineering loweryield strength (!Ely),!E= (50 / 100) x!Ely!E= (50 / 100) x 305.00 MPa!E= 152.50 MPa(0.5 mark)Solving for"Eusing!Eand EE,EE=!E/"E"E=!E/ EE= 152.50 MPa / 210,000 MPa"E= 0.000726 = 7.26 x 10-4(0.5 mark)Solving for true stress (!T) using!Eand"E,!T=!Ex (1 +"E)(0.5 mark)!T= 152.50 MPa x (1 + 7.26 x 10-4)!T= 152.61 MPa(0.5 mark)Solving for true strain ("T) using"E,"T= loge(1 +"E)(0.5 mark)"T= loge(1 + 7.26 x 10-4)"T= 7.26 x 10-4(0.5 mark)c)Determine the ductility if the gauge length after fracture is 70.00 mmAnswer (2 marks):Solving for original gauge length (lo),lo= 5.65 x (area)0.5(0.5 mark)lo= 5.65 x (#x 5.00 mm x 5.00 mm)0.5lo= 50.07 mm(0.5 mark)Solving for ductility by percent elongation (% EL) using and maximum gaugelength (lm) and lo,% EL = ((lm- lo) / lo) x 100(0.5 mark)% EL = ((70.00 mm - 50.07 mm) / 50.07 mm) x 100% EL = 39.80%(0.5 mark)Distributing prohibited | Downloaded by Bob Cat ([email protected])lOMoARcPSD|1531217246
Page5of23CONTINUED OVER PAGEQuestion Two (20 marks)Part 2A: Multiple Choice Questions (5 marks)For each question below, select one correct answer and write down the letter in youranswer booklet.i)Which compound of Portland cement after hydration contributes the most to thedevelopment of strength of concrete after 7 days ageing?a)Tricalcium silicate (C3S)(1 mark)b)Dicalcium silicate (C2S)c)Tricalcium aluminate (C3A)ii)Partial replacement of Portland cement with silica fume in concrete will alter thefresh and hardened properties of concrete by:a)Decreasing extent of bleeding and retarding strength developmentb)Decreasing extent of bleeding and accelerating strength development(1 mark)c)

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Chapter 14 / Exercise 27
An Introduction to Physical Science
Shipman/Wilson
Expert Verified