Evaluate young modulus of elasticity - ultimate stress, Mechanical Engineering

Evaluate Young modulus of elasticity - Ultimate stress:

Q: The following observations were made during tensile test on mild steel specimen 40 mm in diameter and 200 mm long. Elongation with 40 kN load (within the limit of proportionality), L = 0.0304 mm

Yield load =161 KN Maximum load = 242 KN

Length of specimen at fracture = 249 mm


(i) Young's modulus of elasticity

(ii) Yield point stress

(iii) Ultimate stress

(iv) Percentage elongation.

Sol.: (i) Young's modulus of elasticity E: Stress,     σ= P/A

= 40/[ ? /4(0.04)2]   

= 3.18 × 104  kN/m2

Strain, e =   δL/L =   0.0304/200   = 0.000152

E = stress/ strain = 3.18 × 104/0.000152

= 2.09 × 10kN/m2                                                                                              .......ANS

(ii) Yield point stress:

Yield point stress = yield point load/ Cross sectional area

= 161/[ ? /4(0.04)2]

= 12.8 × 10kN/m2                                                                                              .......ANS

(iii) Ultimate stress:

Ultimate stress = maximum load/ Cross sectional area

= 242/[? /4(0.04)2]

= 19.2 × 10kN/m2                                                                                             .......ANS

(iv) Percentage elongation:

Percentage elongation = (length of specimen at fracture - original length)/ Original length

= (249-200)/200

= 0.245 = 24.5%                                                                   .......ANS


Posted Date: 10/20/2012 8:35:54 AM | Location : United States

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