Question

A bar of aluminum is tested under tension, the specimen has a rectangular n with dimensions of 4 mm by 20 mm. The initial length of the sample is 60 mm, the load (force) versus elongation (Aに!finalㅢinitial) graph of this specimen is shown below. a) Determine the ultimate tensile strength. b) Young's modulus c) Total strain after failure 100 63 Fracture ー30 2 3 4 5 Elongation, mm .2 0

Answer 1

(a) The maximum load that can be applied on the specimen, P = 63 x 10^3 N

Cross-sectional area of the specimen, A = 4 x 20 mm^2 = 80 x 10^-6 m^2

So, ultimate tensile strength = Maximum load that can be applied / cross-sectional area

= P / A = (63 x 10^3) / (80 x 10^-6) = 7.875 x 10^8 N/m^2

(b) Look at the elastic region of the curve. This region is for which the graph is a straight line.

Take one point on this region.

Say, load 30 x 10^3 N.

Elongation of the specimen at this point , delta L = 0.2 mm

Initial length of the specimen, L = 60 mm

So, strain = delta L / L = 0.2 / 60 = 0.0033

Stress = load / A = (30 x 10^3) / (80 x 10^-6) = 3.75 x 10^8 N/m^2

Therefore, Young's modulus of the specimen, Y = stress / strain = (3.75 x 10^8) / 0.0033

= 1.136 x 10^11 N/m^2

(c) Total elongation at the point of fracture = 4 mm

So, total strain after failure = 4 mm / L = 4mm / 60 mm = 0.067 (Answer)