Question

a-Calculate conventional yield effort at 0.2% (0.002 in./in.)

b- Tensile strength.
c- The modulus of elasticity.
d- Elongation
e- The reduction in area.
f- The engineering effort to fracture
g- The real stress to fracture
h- The modulus of resilience

Magnesium

Load (N)	Calibrated length (mm)
0	30.00000
5000	30.02960
10000	30.0592
15000	30.0888
20000	30.1500
25000	30.51
26500	30.9
27000	31.5
26500	32.1

25000

32.79

- Initial diameter of the test tube: 12 mm
- Initial calibrated length: 30 mm
- Calibrated length after fracture: 32.61 mm
- Diameter after fracture. 11.74 mm

Answer 1

Solution:

a) Yield strength at 0.002

From the following stress-strain curve, at 0.002 strain the yield strength is approx 188 MPa

b) Tensile strength

From the stress-strain curve, the tensile strength is defined as the maximum stress. The approx tensile strength is 238.85 MPa

c) The modulus of elasticity

Therefore, the modulus of elasticity is 45132.65 MPa

d) % Elongation

Therefore, the approx % of elongation is 8.7%

e) The reduction in area

Therefore, the apporx reduction in area is 4.28%

f) The engineering effort to fracture

The stress at fracture is approx 221.16 MPa

g) The real stress to fracture

The relationship between true stress and engg.stress is

Therefore, the real stress to fracture is 241.72 MPa

h) The modulus of resilience

The formula used to find the modulus of resilience is

Therefore, the modulus of resilience is 0.392