How is E found from the elastic data?

Stress (18000/A0) / strain (0.0367/50) = about 200 GPa

Why is true fracture stress larger than engineering?

It uses final neck area 56.75 mm^2 (722.5 MPa)

Practice · 8 marks

Tensile test data of a steel bar

Short answer

Using A0 = π(12.5)²/4 = 122.72 mm², E = (18000/122.72)/(0.0367/50) = 200 GPa; yield strength = 36000/122.72 = 293.4 MPa; UTS = 52000/122.72 = 423.7 MPa; engineering fracture stress = 41000/122.72 = 334.1 MPa; true fracture stress = 41000/56.75 = 722.5 MPa; % elongation = (62.5-50)/50×100 = 25%; % RA = (122.72-56.75)/122.72×100 = 53.8%; resilience = (293.4)²/(2×200e3) = 0.215 MJ/m³.

A tensile test on a steel bar of diameter 12.5 mm and gauge length 50 mm gave the following data: load of 18 kN produced an elongation of 0.0367 mm (within the elastic range); yield load 36 kN; maximum load 52 kN; fracture load 41 kN. After fracture the gauge length was 62.5 mm and the minimum diameter at the neck was 8.5 mm. Determine (a) Young's modulus, (b) yield strength, (c) ultimate tensile strength, (d) engineering and true fracture stress, (e) percentage elongation and percentage reduction in area, (f) modulus of resilience.

Given

$d_0 = 12.5$ mm, $L_0 = 50$ mm, $d_f = 8.5$ mm, $L_f = 62.5$ mm.

Original area:

A_0 = \frac{\pi}{4}(12.5)^2 = 122.72\ \text{mm}^2

Final (neck) area:

A_f = \frac{\pi}{4}(8.5)^2 = 56.75\ \text{mm}^2

(a) Young's modulus

\sigma = \frac{18000}{122.72} = 146.7\ \text{MPa}, \qquad e = \frac{0.0367}{50} = 7.34\times10^{-4}

E = \frac{\sigma}{e} = \frac{146.7}{7.34\times10^{-4}} = 1.998\times10^{5}\ \text{MPa} \approx 200\ \text{GPa}

(b) Yield strength

\sigma_y = \frac{36000}{122.72} = 293.4\ \text{MPa}

(c) Ultimate tensile strength

\sigma_{u} = \frac{52000}{122.72} = 423.7\ \text{MPa}

(d) Fracture stress

\sigma_f = \frac{41000}{122.72} = 334.1\ \text{MPa (engineering)}

\sigma_{Tf} = \frac{41000}{56.75} = 722.5\ \text{MPa (true)}

(e) Ductility

\%\,EL = \frac{62.5 - 50}{50}\times100 = 25\%

\%\,RA = \frac{122.72 - 56.75}{122.72}\times 100 = 53.8\%

(f) Modulus of resilience

U_r = \frac{\sigma_y^2}{2E} = \frac{(293.4)^2}{2 \times 1.998\times10^{5}} = 0.215\ \text{MJ/m}^3

(Since 1 MPa = 1 MN/m $^2$ = 1 MJ/m $^3$ , the result is in MJ/m $^3$ .)

Quantity	Value
E	200 GPa
Yield strength	293.4 MPa
UTS	423.7 MPa
Fracture stress (eng. / true)	334.1 / 722.5 MPa
% elongation / % RA	25% / 53.8%
Modulus of resilience	0.215 MJ/m³

Answer: E = 200 GPa, σy = 293.4 MPa, σu = 423.7 MPa, σf = 334.1 MPa (true 722.5 MPa), elongation 25%, reduction in area 53.8%, Ur = 0.215 MJ/m³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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